问:
Let $A$ be a totally bounded subset of a complete metric space $X$ and $f:X\to Y$ a continuous map.
Let $A$ be a totally bounded subset of a complete metric space $X$ and $f:X\to Y$ a continuous map. Then $f|_A:A\to Y$ restriction of $f$ to $A$ is uniform continuous.

My intuition: Since it is totally bounded $\forall \delta>0$ and by finite balls say centered at $$x_1,x_2,...,x_n$$

and inside each ball the $\delta$ is the maximum distance and since $f$ is continuous we can relate our $\delta$ only for $\epsilon$ to prove uniform continuity.

I cannot construct to proper way to prove it.

Can we do it in more deep way not just using epsilon delta? How to do it?

答:
The closure of $A$ is compact  because it is totally bounded and complete. $f$ restricted to the closure is therfore uniformly continuous and hence its restriction to $A$ is also uniformly continuous.

I am using the well known fact that the closure of a totally bouded set is totally bounded. (This is not hard to prove).


问:
Prove this formula $\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$
I am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula.
$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$$

I have been given hints and instructions from this thread

Write using Euler’s identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Factor the denominator

Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again).

This is how I have done:

RHS: $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\dfrac{1-r(\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$

Then from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $1=e^0$.

$$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$

The third step is to decompose the fraction:

$$\dfrac{A}{r-e^{ix}}+\dfrac{B}{r-e^{-ix}}=\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$

$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I notice that let $x=0$, we will have

$$A(r-1)+B(r-1)=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I don't know to find $A$ and $B$

Also, could you provide in details how to finish the part "expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again)".

My symbolic manipulation skill is not very good, so a detailed answer is great!

Thanks!

答:
$\cos(nx)=2\cos(x)\cos((n-1)x)-\cos((n-2)x)$, then $$\sum_{n=1}^\infty r^n\cos(nx)=2\cos(x)\sum_{n=1}^\infty r^n\cos((n-1)x)-\sum_{n=1}^\infty r^n \cos((n-2)x) $$
$$
=2r\cos(x)\sum_{n=0}^\infty r^n\cos(nx)-r^2\sum_{n=0}^\infty r^n \cos(nx)
$$
Then, after some algebra, $$\sum_{n=1}^\infty r^n\cos(nx)=\frac{r\cos(x)-r^2}{1-2r\cos(x)+r^2}$$

ContentLicense: CC BY-SA 4.0


问:
Probability of two variables of having the same value
Let $X$ and $Y$ be two random variables, whose PDFs $f_X$ and $f_Y$ are uniform. $f_X$ and $f_Y$ may overlap. For instance, they could represent two score distributions for two tuples $x$ and $y$ in a database.

Which is the probability for $X$ and $Y$ of having the same value $v$ (e.g., for the tuples $x$ and $y$ of having the same score)?

I have tried with an integral, but this returns me back $0$ as a result, since it is evaluated on an interval that is a point (the value $v$), and an integral that is evaluated on a point returns $0$ as a result.

答: Since $X, Y$ are continuous it is indeed the case that $\text{Prob}(X=Y)=0$.</p>

ContentLicense: CC BY-SA 3.0


问:
Submersions between the same manifolds having the same kernels on their derivative are locally the same
I came across the following lemma in a lecture notes set that I was reading.

Let $U$ be an open set in $\mathbb{R}^m$, $f,g : U \to \mathbb{R}^{m-d}$ be two submersions such that $Ker(Df_p) = Ker(Dg_p)$ for all $p \in U$. Then for all $p \in U$, there is an open neighborhood $W$ of $f(p)$ and a diffeomorphism $h : W \to h(W) \subset g(p) $ such that $h \circ f = g$ in a neighborhood of $p$.

The paper says that this is obvious from the implicit function theorem, but I am not able to see how.

Here is the link to the article.

答:
HINTS: Note that the condition that the kernels agree everywhere tells you that every level set of $f$ is a level set of $g$ and vice versa. So if we set $X_c = \{x: f(x)=c\}$ and $Y_c = \{x: g(x)=c\}$, then clearly $X_c = Y_{h(c)}$ for some function $h$. (Here's where you use a variant of the implicit function theorem.) Choose a smooth local section $\sigma$ of the map $f$ (so that for all $c\in W\subset\Bbb R^{n-d}$ you have $f(\sigma(c)) = c$). Note that $h(c) = g(\sigma(c))$, so $h$ is smooth.

Remark: The local submersion theorem, e.g., in Guillemin & Pollack, tells you that in appropriate local coordinates $f(x_1,\dots,x_n) = (x_1,\dots,x_{n-d})$. Then you can define $\sigma(y_1,\dots,y_{n-d}) = (y_1,\dots,y_{n-d},0,\dots,0)$.


问:
The formulas of prostapheresis: memorization technique
This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,

\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\
\cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\
\cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2}
\end{cases}

and while I am able to find them, I am not able to find a technique to memorize them.

Is there a technique to be able to memorize them?

答:
This answer is not apt to the general high school public, but it can be useful for particularly curious students.

I like very much how these formulas are derived by Feynman in his “Beats” lecture (Lectures on Physics, volume 1, https://www.feynmanlectures.caltech.edu/I_48.html, section 48-1). He uses complex exponential, something that has already been mentioned in comments.

I have always loved his explanation. These apparently obscure formulas actually express the adding and the subtracting of two waves. Since there is a real and an imaginary part, this amounts to four real formulas. The physical phenomenon behind them is the “Beats” one, and it can be heard easily by picking two strings of a guitar. It can actually be used to tune it.


问:
Show Sp($n$)$\subset$SU($2n$)
On the Wikipedia page on symplectic groups, it is stated that Sp($n$)$\subset$SU($2n$). How can this be shown?

答:
There is another way to do it which is perhaps more concrete. Recall that you can write any quaternion $q$ in the form $q=a+bj$ where $a,b \in \mathbb{C}$. Similarly, one can write any matrix in $Sp(n)$ as $A+Bj$ for some matrices $n \times n$ comlex matrices $A$ and $B$. Then you can check that the embedding

Sp$(n) \rightarrow $ SU$(2n)$ which maps the element $A+Bj \in$ Sp$(n)$ to the element

\begin{pmatrix}
A & B \\
-\overline{B} & \overline{A}
\end{pmatrix}

is an injective Lie group homomorphism. Thus you can identify Sp$(n)$ as a subgroup of SU$(2n)$ via the image of this homomorphism.

Title: Show Sp($n$)$\subset$SU($2n$)


问:
How to prove that the characteristic polynomial of this specific matrix is not the power of a linear polynomial?
I was reading an algebra paper, and the problem that appeared to me is the following:

The authors defined a group $G = A \rtimes \left<x\right>$, where $A$ is a finitely generated free abelian group and $\left<x\right>$ is the infinite cyclic group and such that $x$ does not normalize any non-trivial cyclic subgroup of $A$. Then the element $x$ acts on $A$ in the following way for suitable integers $t_{i,j}$, given that $\{a_{1}, ..., a_{r}\}$ is the independent set that generates $A$:

$$x^{-1}a_{i}x = \prod\limits_{j}a_{j}^{t_{i,j}}.$$

Now, the authors defined the matrix $T = [t_{i,j}] \in GL_{r}(\mathbb{Z})$.

The characteristic polynomial $f(y) = \det(yI - T)$ is such that $f(0) = \det(T) = \pm 1$ (because $[t_{i,j}] \in GL_{r}(\mathbb{Z})$) and $f(1) \neq 0 \neq f(-1)$ - this part is ok, given that $x$ does not normalize non-trivial cyclic subgroups of $A$. Now the authors also say that $f(y)$ is not the power of a linear polynomial, "since the composition factors of $x$ on $\mathbb{Q} \otimes A$ cannot all be linear".

I simply could not understand this last part, and I don't know how to justify the fact that $f(y)$ is not the power of a linear polynomial. What does this last sentence mean after all?
If anyone could provide some sort of insight I would be really grateful.

Thanks in advance.

Edit:
By definition, an element $x \in G$ normalizes a subgroup $H \leq G$ iff $xHx^{1} = H$.

答: Suppose that $f(y)=(X-a)^n$. It implies that The map $L_x:A\rightarrow A$ defined by $L_x(u)=x^{-1}ux$ verifies $(L_x-aId_A)^n=0$. This implies that $Ker(L_x-aId_A)\neq 0$, we deduce that there exists $u\in A, u\neq 0$ with $xux^{-1}=au$. This implies that $L_x$ normalize the non trivial subgroup generated by $x$. Contradiction.


问:
What is wrong in this calculation of pdf?
Assume I have a continuous random variable $X$ with pdf $f_X$. Then I have another random variable $Y = X/m$ for some $m \neq 0$. Consider $g:x \mapsto x/m$. Then $g^{-1}(x) = mx$. Hence $f_Y(y) = \left | \left ( g^{-1} \right)' (y) \right | f_X(g^{-1}(y))  = |m| f_X(my)$.

For $m < 0$, I have $$\int_\mathbb R f_Y (y) \, \mathrm{d}y = \int_\mathbb R |m| f_X(my) \, \mathrm{d}y = - \int_\mathbb R m f_X(my) \, \mathrm{d}y = - \int_\mathbb R f_X(x) \, \mathrm{d}x = -1.$$

Could you explain what's wrong with my computation?

答:
This equality : $- \int_\mathbb R m f_X(my) \, \mathrm{d}y = - \int_\mathbb R f_X(x) $ is not true (obviously). When you do the change of variables $x = my$, the fact that $m < 0$ implies that the bounds of the integral should be reversed, which then implies multiplying the integral by $-1$.

If you want to do it rigorously just integrate over $[-t,t]$ :

$$ - \int_{-t}^t m f_X(my)dy = - \int_{-t}^t f_X(my)d(my) = - \int_{-t/m}^{t/m} f_X(x)dx $$

And finally, since $m<0$, the bounds need to be inverted and the integral has to be multiplied by $-1$, so :

$$ - \int_{-t}^t m f_X(my)dy = \int_{t/m}^{-t/m} f_X(x)dx $$

and then let $t \to \infty$ to get the correct result.


问:
I have almost no idea how to factor numbers this big.
$14425638854646469646839767613420413647898432138735230192512819$ is the product of two prime numbers. Each factor is an answer.

I have to give both factors as answers. How would I get to them and which numbers are them?

My knowledge of factors is extremely limited.

答:
It's $3658265762777653112540162274181 \cdot 3943299855747317419837769926999$ :)

edit:
To find the factorization, I used the function ifactor in Maple, which itself relies on the quadratic sieve method, as explained in its documentation.


问:
Finding if estimator is correct using method of moments
b) Is $\hat\theta$ an absolutely correct estimator? Explain.

For the point a) I got $\nu_1=E(X)=\int_0^\theta x^2\frac{2}{\theta^2}dx$ which is $\frac{2}{3}\theta$. And $\nu_1=\bar X$(sample mean). So $\hat \theta=\frac{3}{2}\bar X$.

And for they point b) we have to check if $E(\hat \theta)=\theta$ and $lim_{n\to \infty} V(\hat \theta)=0$.

答:
Hint:

$E(\hat \theta)=\frac{3}{2}E(\bar X)=\frac{3}{2}\mu$ is not a problem. It in fact shows $\hat \theta$ is an unbiased estimator of $\theta$; it is also a consistent estimator as $\mathrm{Var}(\bar X) \to 0$ in probability as the sample size increases.

To illustrate the actual problem here, suppose you happen to see the data $7,1,5,3,6,2$.

What would your estimate of  $\theta$ be?

Would your all your data satisfy $0 \lt x \lt \hat \theta$?


问:
mathematical maturity
So, I finished my undergrad with a degree in applied mathematics, but when reading some graduate level texts and/or papers, I often find myself struggling.  I eventually get there, but I often feel like I lack the intuition necessary to be able to come up with concepts on my own.  I feel like I'm just missing some pivotal step in the journey to mathematical maturity.

Does anyone have any books/references/advice?

PS:
If this means anything at all, as I was not a mathematics major, I did not take analysis or abstract algebra.

答:
I attended engineering school after I three years active military service. Although I received good grades I felt that something was missing. Over the years since I graduated and then retired I have read many algebra, geometry and precalculus texts to gain insight but my goal still eluded me.

Since I retired I had the time needed to better focus on my lack of mathematical insight. At present I have identified three issues impeding my developing mathematical maturity.

Language. It should be obvious that just writing and simplifying formulas is not sufficient to understand the math involved. Mathematical proofs are important for those folks who are sufficiently mathematically mature but are premature for those of us who have not developed that maturity. Developing the skill to write clearly and concisely about mathematical solutions is essential. The best place to start is to tell the story of how a particular type of problem was solved. Describe how the result was obtained in terms that an intelligent individual who graduated 8th grade could understand.

Consider the story of the child prodigy Carl Gauss whose class was told to solve problem of adding all the integers from 1 to 100. Gauss immediately answered that the sum of these integers is 5050. When asked how he was able to answer correctly so quickly he explained that he added to each element of the set of consecutive integers of 1 through 100 the elements of this set but in decreasing order. The young Gauss noted that each of the 100 elements of the resulting set of integers had a value of 101. So the sum of a 100 integers with the same value being 101 is twice the sum of the consecutive integers of 1 through 100. Gauss then answered that 5050 was equal to the sum of 50 integers which had the same value of 101. Note that Carl did not provide his answer in the form of the proof.

Learn basic mathematical problem solving skills that are assumed by many mathematical textbook authors to be so basic that they too often state that "It can be easily seen that Blah Blah  Blah is Blah Blah Blah". Oh yah! Those who math can't teach math because they assume someone else has taught the basic math learning skills such as focus on detail, which in turn requires motivation, and understanding that periods of intense concentration should be short and followed by other activities such as walking or even a short nap.

Take a look at No Bullshit Guide to Mathematics by Ivan Solov.
ISBN 9780992001032 and

Again look at a mathematical
dicription that you found difficult to comprehend. Ask yourself what is information do I need? For example, as I have tried to improve my understanding  of mathematical proofs I realized that the proof that the square root of 2 is irrational was the wrong place to start.

So then reviewed my Geometry texts are considered the proof for the Pythagorean theorem. They classic proof based upon Euclid's axioms is not sufficient algebra had not existed. Then I found a tiling proof that resonated for me at Wikipedia.

As I scrolled down to the rearrangement proof I instinctively understood. To improve my understanding of how this proof was developed I sketched congruent right triangles with sides of lengths a, b and c.  To the triangle on the left I sketched a square which had one of its sides coincident with the hypotenuse of the right triangle. On the right I sketched the congruent shape for the sum of the squares of sides a and b of the right triangle.

I noted that on both figures the smallest congruent squares bounding these shapes, that sides on both bounding  measured a plus b. Since both areas would measure the square of the sum of, a plus b. On the right  shape  this is equal to area of the small square a plus the combined areas of the four congruent right triangles plus the area of the large b.

On the left the combined areas is equal to the area of the largest square whose side equal the hypotenuse of the right triangle, plus the combined areas of the four congruent right triangles.
Now when the combined areas of the four congruent right triangle are removed from the shapes on left and  right, what remains is that square of the hypotenuse is equal to the sum of the squares of the sides that are adjacent to the right angle of the right triangle.

This description may seem a bit wordy but it provided to me a better understanding of the process by which this proof was developed.

Title: mathematical maturity


问:
Convergence of integral of a positive function implies the series converges as well
If $f(x)$ is a function having the following properties:

Positive

Continuous

$\int f(x)dx$ converges

Then I think the series $\sum_{n=1}^{+\infty}f(n)$ converges. I would reason that if $f(x)$ wasn't positive then this statement would be false. Is my reasoning correct?

答:
Not true. Draw a triangle with base $(n-\frac  1{n^{3}}, n+\frac  1{n^{3}})$ and height $n$. Using these traingles you get a  continuous non-negaive function $f$ whose intergal is finite but $f(n)=n$ for each $n$.  If you want the function to be strictly positive just add $e^{-|x|}$ to this function.

Explicit definition of $f$: $f(x)=n^{4}(x-n+\frac1  {n^{3}})$ for $n-\frac  1 {n^{3}} \leq x \leq n$, $f(x)=n^{4}(n+\frac 1  {n^{3}}-x)$ for $n \leq x \leq n+\frac  1 {n^{3}}$, $n=2,3,4...$ and $f(x)=0$ if $x$ is not in any of these intervals.


问:
For what $a$ and $b$ is  $9x^4-12x^3+28x^2+ax+b$ a perfect square?
If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)</p>

答: You might want to try writing the polynomial as $(3x^2 + \alpha x + \beta)^2$ and solving for $\alpha$ and $\beta$.</p>


问:
Eratosthenes-like sieve - infinitely many left unstruck?
Given any infinite sequence $c_1,c_2...$ of natural numbers, if all of the natural numbers $x$ such that there exists $n$ such that $x\equiv c_n (\mod p_n)$ and $x \geq 2p$, where $p_n$ is the nth prime, are deleted from a list, must there be infinitely many natural numbers left on the list afterwards?

e.g. $c_i=1 \ \forall i\\
1,2,3,4,\not5,6,\not7,8,\not9,\not10,\not11,12,\not13,14,\not15,\not16,\not17,18,\not19,20...$

So far I only know about the trivial case where $c_i=0$, and I may have missed obvious counterexamples.

答:
Yes, there are infinitely many natural numbers left afterwards.

When performing Eratosthenes' unmodified sieving operation, the striking out of multiples of $2$ eliminates $\frac12$ of the values greater than $2$; the striking out of multiples of $3$ eliminates $\frac13 (1 - \frac12) = \frac16$ of the values greater than $3$; the striking out of multiples of $5$ eliminates $\frac15 (1 - \frac12 - \frac16) = \frac1{15}$ of the values greater than $5$; etc.

With your offset sieve, the numbers struck out are different, but the proportions are exactly the same. The reason is simple: each prime number is (by definition) coprime with all smaller prime numbers. Therefore there is no correlation between the numbers which are struck out before reaching that prime and the numbers which are struck out by that prime, and regardless of the value of $c_i$ we strike out $\frac1{p_i}$ of the unstruck numbers $\ge 2p_i$.

Slightly more rigorously: since the $p_i$ are distinct primes, the LCM of a set of them is simply their product, and in particular $p_i$ is coprime with $\prod_{j<i} p_j$. Then consider the equivalent classes of $\mathbb{Z} / \prod_{j\le i} p_j \mathbb{Z}$: the Chinese remainder theorem gives us a natural bijection with the Cartesian product $(\mathbb{Z} / \prod_{j\lt i} p_j \mathbb{Z}) \times (\mathbb{Z} / p_i \mathbb{Z})$. Then all elements who share a projection onto $\mathbb{Z} / \prod_{j\lt i} p_j \mathbb{Z}$ share a current status (struck out or not); and when we strike out the elements in which the projection onto $\mathbb{Z} / p_i \mathbb{Z}$ is $c_i$ we will strike out precisely $\frac1{p_i}$ of those which were not struck out (and re-strike the same proportion of those which were already struck out).

Therefore for any sequence $c_i$, and for any positive integer $C$, after striking out the values corresponding to the $j$th prime the number of unstruck numbers smaller than $C \prod_{k\le j}p_k$ will be at least as many as for the sequence $c_i = 0$ (and will not exceed that number by more than $j$, where the excess is due to numbers in $(p, 2p]$ not being struck out).


问:
Chain rule for multivariable functions confusion. General formula
I'm struggling to work out what is what in the formula for the chain rule in general.
Say we have $f(x,y)=(u(x,y),v(x,y)).$ And then we have $g(x,y)=(p(x,y),q(x,y))$.
So $(f \circ g)(x,y)=u(p(x,y),q(x,y)),v(p(x,y),q(x,y)).$
What is the formula for $\frac{\partial (f \circ g)}{\partial x} $?
I can't seem to get it to make any sense.

答:
You have the chain
$$
(x,y)\to (p(x,y), q(x,y))\to (u(p(x,y),q(x,y)), v(p(x,y),q(x,y)).
$$
and therefore
$$
\frac{\partial (f\circ g)}{\partial x}=\left(\frac{\partial u}{\partial s}\cdot \frac{\partial p}{\partial x}+\frac{\partial u}{\partial t}\cdot\frac{\partial q}{\partial x},\frac{\partial v}{\partial s}\cdot \frac{\partial p}{\partial x}+\frac{\partial v}{\partial t}\cdot\frac{\partial q}{\partial x}\right),
$$
where the derivatives $\frac{\partial u}{\partial s}$, $\frac{\partial u}{\partial t}$ etc. are evaluated at $(p(x,y), q(x,y))$.

It is easier to use the matrix language and multiply the corresponding Jacobian matrices.


答:
Yes you are right. If you set X and Y to be continuous r.v.'s, then the probability of getting the exact same value is 0 (because the integral of the density function from a point to the same point is always 0).

You could try an approximation by stating a range instead of a number. For example, you can ask what is the probability that both X and Y fall in the interval $(v - \epsilon, v + \epsilon)$, where $\epsilon$ is some small number (such as 0.005).

Title: Probability of two variables of having the same value

Body:
Let $X$ and $Y$ be two random variables, whose PDFs $f_X$ and $f_Y$ are uniform. $f_X$ and $f_Y$ may overlap. For instance, they could represent two score distributions for two tuples $x$ and $y$ in a database.

Which is the probability for $X$ and $Y$ of having the same value $v$ (e.g., for the tuples $x$ and $y$ of having the same score)?

I have tried with an integral, but this returns me back $0$ as a result, since it is evaluated on an interval that is a point (the value $v$), and an integral that is evaluated on a point returns $0$ as a result.


问:
Guess the value of the limit
Simplify and guess the value of $\lim_{x\to 2}\frac {x^3-8}{x-2}$.

I'm very confused on how to do this.  Please help!

答:
The above answers are not the solution you are looking for.  When you are asked to "guess" a limit, you are to construct a table and test input values that approach 2.  To be complete you should consider both values greater than 2 and values less than 2.

x         1.8    1.9   1.95   1.99  Und.  2.01   2.05   2.10   2.20
f(x)     10.84  11.41 11.70  11.94  ?    12.06  12.30  12.61  13.24

Now look for the trend.  As x->2, f(x)-> 12.  Since the output values are approaching 12, we can reasonably "guess" that the limit is equal to 12.

(NOTE: in a later section of your course you will be asked to "evaluate" the limit.  That is what others were trying to explain, and unfortunately probably caused you even more confusion.)

Title: Guess the value of the limit


问:
If a sequence $f_{n}\neq0$ converges uniformly and $\frac{1}{f_{n}}$ converges pointwise then it also converges uniformly
I have this question I'm unable to solve (you need to prove this or give a counterexample:

If a sequence $f_{n}\neq0$ ($f_{n}\left(x\right)\neq0,\forall n\geq N\in\mathbb{N}$ for some N) converges uniformly and $\frac{1}{f_{n}}$ converges pointwise then it also converges uniformly.

I tried proving $$\lim_{n\to\infty}\sup\left|\frac{1}{f_{n}\left(x\right)}-\frac{1}{f\left(x\right)}\right|=0$$ but it didn't work. I tried looking at function of the form $\frac{1}{n}$ but $\frac{1}{\frac{1}{n}}=n$ doesn't converge. I tried looking at $$f_{n}\left(x\right)=\begin{cases}
\frac{1}{n} & x\leq\frac{1}{n}\\
0 & x>\frac{1}{n}
\end{cases}$$ but it doesn't fit the demand of $f_{n}\neq0$.

So I wasn't able to prove nor provide a counterexample

答:
A counterexample is $f_n(x) = \frac{1}{x} + \frac{1}{n}$ where $f_n(x) \to f(x) =\frac{1}{x}$ uniformly on $(0,\infty)$ and $\frac{1}{f_n(x)} \to x$ pointwise.

The uniform convergence of $f_n$ is obvious since $\left|f_n(x)  - f(x)\right| = \frac{1}{n}$. Choosing any $N(\epsilon) > \frac{1}{\epsilon}$ we have $|f_n(x) - f(x)| <\frac{1}{n} < \frac{1}{N(\epsilon)}< \epsilon$  for all $n > N(\epsilon)$ and $x \in (0,\infty)$.

However,

$$\left|\frac{1}{f_n(x)}- \frac{1}{f(x)} \right|= \left|\frac{1}{\frac{1}{x} + \frac{1}{n}}- x\right|= \frac{\frac{x}{n}}{\frac{1}{x} + \frac{1}{n}} =\frac{x^2}{n + x},$$

and the convergence of $\frac{1}{f_n}$ is not uniform on $(0,\infty)$ since

$$\sup_{x \in (0,\infty)}\left|\frac{1}{f_n(x)}- \frac{1}{f(x)} \right| = \sup_{x \in (0,\infty)}\frac{x^2}{n + x}\geqslant \frac{n^2}{n+ n}= \frac{n}{2}\underset{n \to \infty}\longrightarrow+\infty \neq 0 $$


问:
Unique Permutations Of 'LYLY' If Each Permutation Must Contain 'LY'
I recently started working on permutations, and came across this problem:

How many possible unique permutations of 'LYLY' contain 'LY'?

I understand the basic method of solving this type of problem:

n! , where n = number of letters

And if there are repeating symbols:

n!/n1! * n2! * n3!  . . .* nk! , where n = number of letters

However, I cannot solve this problem, because I do not know the method of finding the number of permutations if there are multiple copies of the letters that can make up 'LY'.

So, how, and with what method, would you solve this(or this type of/similar) problem? I have tried searching for a solution or other question, but I cannot seem to find a solid, valid solution.

Thanks in advance

答:
This particular case is fairly simple to solve by examination, but for more complicated case you probably want to work with inclusion-exclusion, where you make an initial estimate of cases, then remove cases that are double-counted and (if needed) restore cases that were triple-counted and hence removed twice in the previous step, etc.

So, as an example of this, we can first arrange the elements with an $\fbox{LY}$ composite element and two solo letters of $L$ and $Y$, giving $3!=6$ arrangements. However we can see that double-counting is possible, so we can arrange two composite elements $\fbox{LY}$ and subtract off the result. However there is a danger here - because the elements are identical, there is only one arrangement of these two elements, not two. So subtract off this single double-counted option and you have the result, $6-1=5$.

By the same process, for example, arranging $LAYLA$ to include $LA$ somewhere would have $4!-3!/2$ options.


问:
Numbers whose decimal digits are the coefficients of its continued fraction form
A curious question recently crossed my mind: can we construct decimal numbers of the form $$\text{"a.bcdefghij…"}$$
where each letter represents a digit $0-9$ (where the number may or may not be rational), so that it is equal to a continued fraction of the form
$$a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{d+\cfrac{1}{e+\ddots}}}}$$

For example, I found that

$$0.32062241134\dots=0+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{0+\cfrac{1}{6+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{5}}}}}}}}}}}}$$

Experimenting a bit, we see that we can certainly construct infinitely many numbers that are close solutions, but how can we make them the most efficient? To clarify, we prefer not to go too deeply nested into the continued fraction to make the previous digits work, because then we have to make those new digits work.  So some questions are:

~What algorithm will construct such numbers?

~Does that algorithm construct numbers that are “efficient” as described above? Is there one number that is the "most efficient"?

~Does an exact non-repeating rational example exist?

答:
A "correct" continued fraction does not contain any zeroes after the first digit. For example, [8;6,7,5,3,0,9] is not a valid continued fraction; the number that computes to will actually have a continued fraction of [8;6,7,5,12]. (This is because 3+1/{0+1/9}=3+9=12). The numbers before and after the zero are added together. Similarly, the continued fraction of the number 0.32062241134... actually begins [0;3,8,2,2,4,1,1,3...]

So, you might be able to generate a sequence of [a;b,c,d,e,...] that gives you a+(b/10)+(c/100)+(d/1000)+(e/10000)+... , but you will not be able to successfully reverse it. You can calculate the decimal digits from the continued fraction, but not the other way around, unless there is a solution that contains no zeroes.


问:
Balanced $2019\times 2019$ grids (BMO $2020$ round $2$)

A $2019 \times 2019$ square grid is made up of $2019^2$ unit cells. Each cell is coloured either black or white. A colouring is called balanced if, within every square subgrid made up of $k^2$ cells for $1 \le k\le 2019$, the number of black cells differs from the number of white cells by at most one. How many different balanced colourings are there?
(Two colourings are different if there is at least one cell which is black in exactly one of them.)

Let $f(n)$ be the number of different balanced colorings for an $n\times n$ grid. A few unfruitful ideas:

$\mathbf{1) Inclusion-Exclusion}$

Call the event that the $i^{th}$ subgrid is balanced, $B_i$. Then

$$ f(n) = \left | \bigcap B_i \right | = \left | \left( \bigcup (B_i)’ \right)’ \right | = 2^{n^2}-\left | \bigcup (B_i)’ \right|$$

but the latter term isn’t easily calculated.

$\mathbf{ 2) Recursion}$

$f(n)$ will include the number of ways where the top left $(n-1)\times (n-1)$ grid is balanced, but there doesn’t seem to be a fixed $t(n)$ so that $$f(n) = t(n) f(n-1) \ \text{or} \ f(n-1) + t(n) $$

$ \mathbf{3) Manual \ Counting } $

I found $$f(1)=2 \\ f(2) = 6 \\ f(3) = 10 \\ f(4) = 18 \\ f(5)=26 $$

from where I guess $$f(2n+1)= 2^{n+3} -6 $$ but that’s just a guess.

答:
If two adjacent squares in a row have the same colour.

For $2\times2$ balancing, the two columns, $C_1$ and $C_2$, containing these squares must consist of rows of same coloured squares, the colours alternating from row to row.

If there were also two adjacent squares in a column with the same colour then we could make the equivalent statement about the rows, $R_1$ and $R_2$ containing these two squares. But then we would obtain a contradiction where $C_1$ and $C_2$ cross $R_1$ and $R_2$.

Therefore no column contains adjacent same coloured squares and so the colours strictly alternate as one goes down any column and are completely determined by the colours in the top row.

This alternation means that 'even' blocks are automatically balanced whereas an 'odd' block is balanced if and only if its top row is balanced. The top row of the whole grid must therefore consist of alternating runs of white and black squares, each run of length $1$ or $2$, where the colours of the runs of length $2$ must alternate across the row.

From any given run of length $2$, successively pair up the squares on each side and note that all runs of length $2$ will be in one of these pairs. Note also that there will be a single square left on one side or the other.

We can colour the single square with either colour and then as we go across the row we must change colour when we come to a new pair of squares but we can arbitrarily choose the colour of the second square of the pair.

We have $2$ choices for the end of the single square and $2$ choices for its colour. There are then $2^{1009}-1$ choices of colour for the pairs. (Since we are assuming at least one run of $2$ squares we must discard the one choice of colours without such a run.) The total number of boards is
$$2^{1011}-4.$$

Otherwise.

We have the same number of boards when two adjacent squares in a column have the same colour. Plus two boards where no adjacent squares have the same colour.

Total number of balanced boards.
$$2\times (2^{1011}-4)+2=2^{1012}-6.$$


问:
What is this matrix/operator notation $\otimes$?
Some notes I'm reading say that:

$$x = \sum_{i=1}^m c_i\langle x,u_i\rangle u_i$$ for all $x\in\mathbb{R}^n$ (where $c_i > 0$ are positive numbers and $u_i$'s are unit vectors) can be equivalently written in matrix (or operator) notation as $$\sum_{i=1}^m c_i u_i\otimes u_i = I_n$$ where $I_n$ is the identity map on $\mathbb{R}^n$, and for any unit vector $u$, $u\otimes u$ is the rank-one orthogonal projection onto the span of $u$, i.e. the map $x\mapsto \langle x,u\rangle u$. The trace of this projection is $1$.

So, I haven't seen the above notation $\otimes$ earlier - and I'm a little confused what it means? Could someone help me understand?

What is the meaning of rank-one orthogonal projection here, and what does the operator really do? How is this equivalent to the first condition?

答:
In particular, if u and v are written as "column" matrices, $u= \begin{pmatrix}u_1 \\ u_2 \\ u_3 \end{pmatrix}$ and $v= \begin{pmatrix}v_1 \\ v_2 \\ v_3 \end{pmatrix}$, then $v^T= \begin{pmatrix}v_1 & v_2 & v_3\end{pmatrix}$, a"row" matrix, and
$uv^T= \begin{pmatrix}u_1 \\ u_2 \\ u_3 \end{pmatrix}\begin{pmatrix}v_1 & v_2 & v_3\end{pmatrix}= $$\begin{pmatrix} u_1v_1 & u_1v_2 & u_1v_3 \\ u_2v_1 & u_2v_2 & u_2v_3 \\ u_3v_1 & u_3v_2 & u_3v_3 \end{pmatrix}$, a 3 by 3 matrix.


问:
$\int_K |u_n-u|^{p-\epsilon} \to 0$ implies that $\int_K|u_n-u|^p \to 0$
Let $K$ be compact set on $\mathbb{R}^n$ with non-empty interior. Assume that $p \geq 2 $ and that it is given a sequence of measurable functions $\{u_n : K \to \mathbb{R}\} \subset L^{p-\epsilon}(K)\cap L^p(K)$, for each $\epsilon > 0$, such that there is $u\in L^p(K)$ such that
$$\lim_{n\to\infty}\int_K|u_n-u|^{p-\epsilon} = 0.$$

My question is: is it true that
$$\lim_{n\to\infty}\int_K|u_n-u|^p = 0?$$

I believe so. These are my thoughts: take $\left\{\dfrac{1}{n} : n > 0\right\}$ and consider the function
$$\Phi : \bigcup_{n\geq 1}L^{p-\frac{1}{n}}(K)\cap L^ p(K)\times \bigcup_{n\geq 1}[p-\frac{1}{n},p]\to \mathbb{R}$$
$$(f,q) \mapsto \|f\|_q.$$

I think that since $\mu(K) < \infty$ then $\Phi$ is continuous, so the result should follows.

Is this correct?

答:
This is false. Consider $K = [0,1]$ and let $u_n = n 1_{[0,n^{-p})}$. Then
$$\int_K |u_n|^{p-\varepsilon} = n^{-\varepsilon} \to 0$$ for any $\varepsilon > 0$
but $$\int_K |u_n|^p = 1.$$


问:
How to know when a system of axioms is 'complete'?
Here, I (basically) stated the group axioms as follows.

$(xy)z=x(yz)$

$xe=x, ex=x$

$xx^{-1}=e$

In that post, answerers Martin and Ittay were critical of the above list for not including $x^{-1}x=e$, even though it follows from the above three. Pece's answer also included $x^{-1}x=e$ without comment.

How can I tell whether a system of axioms of 'complete'? Is this even a rigorous concept?

答:
even if we delete ex=x, it is still a group. which follows like this:

$$
x^{-1}x=x^{-1}xe=x^{-1}x(x^{-1}(x^{-1})^{-1})=x^{-1}(xx^{-1})(x^{-1})^{-1}=x^{-1}(e)(x^{-1})^{-1}=x^{-1}(x^{-1})^{-1}=e
$$

and

$$
ex=xx^{-1}x=x(x^{-1}x)=xe
$$

so it is a group.

I am not sure if there is a general way to tell a group, but the basic rule of a group are those four: closure, associativity, identity and inverse.

As long as we can conclude the four rules from what we have given, then it is complete. So I think concentrate on the basic idea of the system is important, all others come from the basics.

As for group, we often consider subgroup, which will simplify the judging process, we just consider $xy^{-1}$ belongs to the subset or not.


问:
Prove that $6|n(n+1)(2n+1)$, $n\in\mathbb{Z}$
I am trying to solve this question without modular arithmetic.

Let $n\in\mathbb{Z}$. Prove that $6$ divides $n(n+1)(2n+1)$.

Using induction:

$\text{For }n=1
\\P(1) = 1(1+1)(2(1)+1)=6
\\\text{For }n=k
\\P(k)=k(k+1)(2k+1)
$

$P(k)$ is divisible by $6$ so $k(k+1)(2k+1)=6h, h\in\mathbb{Z}$. Once by the induction hypothesis, $P(k)$ and $P(k+1)-P(k)$ being divisible by $6$ so $P(k+1)$ will be divisible by $6$. as well.

$P(k+1)= (k+1)(k+1+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)$

So,

\begin{align}
P(k+1)-P(k) & =(k+1)(k+2)(2k+3)-k(k+1)(2k+1) \\
& = 2k^3+9k^2+13k+6 - (2k^3+3k^2+k) \\
& = 6k^2+12k+6 \\
& = 6(k^2+2k+1) \\
& = 6(k+1)^2
\end{align}
Therefore, $P(k+1)$ is true and $6$ divides $n(n+1)(2n+1)$.

But once $n\in\mathbb{Z}$ the proof is not complete. I was wondering about considering $6|(-n)(-n+1)(-2n+1)$, $n\in\mathbb{Z}$ but I am not really sure how to proceed. It doesn't seem to be the right approach. How can I keep doing that or what would be a good proof? Proofs without modular arithmetic are preferable, but if you got a nice proof it would be welcomed. I've seen similar questions, but they considered $n\in\mathbb{N}$.

答:
Hint $ $ You proved $\,6\mid P(k\!+\!1)-P(k),\,$ thus $\ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow\!\color{#c00}\Rightarrow 6\mid P(k).\,$ You used the $\,\color{#0a0}{{\rm direction}}\ \ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow 6\mid P(k)\,$ to inductively ascend its truth from $\,k\,$ to $\,k\!+\!1.\,$ Now use the reverse $\rm\color{#c00}{{\rm direction}\ (\Rightarrow)}$ to inductively descend its truth from $\,k\!+\!1\,$ to $\,k,\,$ so to all negatives too.

Remark $ $ You may find it helpful to first prove the following Lemma for any integer predicate $p$.

Lemma $ $  (Bidrectional Induction) $ $ Suppose that  $\ p(k\!+\!1)\!\iff\! p(k)\ $ is true for all $\,k\in \Bbb Z$.

Then $\,p(k)\,$ is true for all $\,k\in \Bbb Z\!\iff\! p(k)\,$ is true for some $\,k_0\in\Bbb Z$.

OP is the special case $\ p(k) := 6\mid P(k),\, $ and $\,k_0 = 0$.


问:
Question form my book about Zorn's lemma
I found that question at the end of a chapter on Zorn's lemma in my math book. There are only questions to check knowledge, but there are no answers. Could anybody tell me if I did it right? If it is correct, I don't need explanation since I understand that. However, if some of them are wrong I would like to know why exactly.

$X$ and $Y$ are arbitrary sets and $\mathcal{X}=\{⟨A,f⟩:f∈Y^A, A⊆X,f $ is injective$ \}$.
$⟨A,f⟩,⟨B,g⟩∈\mathcal{X}$, and $⟨A,f⟩≤⟨B,g⟩$, only if $A⊆B$ and $g↾A=f$. Then:

If $\langle A,f\rangle$ is a maximal element in $\mathcal{X}$, then $A=X$. $\implies$ correct

If $\langle A,f\rangle$ is a maximal element in $\mathcal{X}$, then $A=X$ or $f$ is onto $Y$. $\implies$ correct

If $L\subseteq \mathcal{X}$ is a string, then $\bigcup\{f:\exists A\langle A,f\rangle\in L\}$ is an injective function $\bigcup\{A:\exists f\langle A,f\rangle\in L\}\to Y$. $\implies$ I don't understand that one

If $\langle A,f\rangle$ is a maximal element in $\mathcal{X}$, then $f$ is onto $Y$. $\implies$ wrong

The existence of a maximal element in this order proves that $|X|≤|Y|$ or $|Y|≤|X|$. $\implies$ correct

I didn't split my question into $5$ independent ones since I think some of them are connected $(1, 2, 4)$.

答:
I’m going to use $\mathscr{X}$ for the partially ordered set and $X$ for the set that is a superset of the first components of the ordered pairs in $\mathscr{X}$. Thus, I interpret (1) as saying that if $\langle A,f\rangle$ is a maximal element in $\mathscr{X}$, then $A=X$. This is not necessarily true. For instance, if $X=\{0,1\}$ and $Y=\{0\}$, there is no injection from $X$ to $Y$, and the maximal elements of $\mathscr{X}$ are $\langle\{0\},\{\langle 0,0\rangle\}\rangle$ and $\langle\{1\},\{\langle 1,0\rangle\}\rangle$. You will always run into this problem when $|X|>|Y|$.

(2), on the other hand, is (as you say) true, assuming that the first $X$ is really $\mathscr{X}$: if $A\subsetneqq X$ and $f$ is not onto $Y$, then there are $x\in X\setminus A$ and $y\in Y\setminus f[A]$, and $\langle A\cup\{x\},f\cup\{\langle x,y\rangle\}\rangle$ is a member of $\mathscr{X}$ strictly larger than $\langle A,f\rangle$.

(3) is true. Let $\mathscr{L}$ be a chain in $\mathscr{X}$, let $g=\bigcup\{f:\exists A\subseteq X(\langle A,f\rangle\in\mathscr{L})\}$, and let $B=\bigcup\{A\subseteq X:\exists f\in Y^A(\langle A,f\rangle\in\mathscr{L})\}$; we want to show that $g$ is an injection from $B$ into $Y$. First we must show that $g$ is a function. Suppose that $a\in A$, $y_1,y_2\in Y$, and $\langle a,y_1\rangle,\langle a,y_2\rangle\in g$. Then there are $\langle A_1,f_1\rangle,\langle A_2,f_2\rangle\in\mathscr{L}$ such that $\langle a,y_1\rangle\in f_1$ and $\langle a,y_2\rangle\in f_2$. $\mathscr{L}$ is a chain, so without loss of generality we may assume that $A_1\subseteq A_2$, and $f_2\upharpoonright A_1=f_1$. Then $a\in A_1$, so $f_2(a)=f_1(a)$, and $g$ is therefore a function. Now suppose that $b\in B$. Then there is some $\langle A,f\rangle\in\mathscr{L}$ such that $b\in A$, so $b\in\operatorname{dom}f$, $\langle b,f(b)\rangle\in f\subseteq g$, and $b\in\operatorname{dom}g$. It’s equally easy to check that $\operatorname{dom}g\subseteq B$, so $B=\operatorname{dom}g$. The last step, which I’ll leave to you, is verifying that $g$ is injective; the argument is very much like the proof that $g$ is a function.

Yes, (4) is correct: this time the problem arises when $|X|<|Y|$. And as you say, (5) is also correct: either you get an injection from $X$ into $Y$, or you get the inverse of an injection from $Y$ into $X$.


问:
Confused by "orthogonal representation" of a finite group. How to show the matrix under some basis from the new defined inner product is orthogonal?
Suppose we have a finite group $G$ and a representation $\rho:G\to GL(V)$ on the finite vector space $V$. So $V$ is a $G$-module.

Now I am asked to show that there is a basis of $V$ under which the matrix of $\rho(g)$ is orthogonal for all $g\in G$. Here is what confused me and I still don't know in what sense is this possible. I have done some searching and know typically the steps are the following:(here we assume $\rho(g)$ is the matrix under standard basis)

(1) Define a $G$-invariant inner product $[v,w]=\sum_{g\in G}\langle\rho(g)v,\rho(g)w\rangle$, where $\langle,\rangle$ is the usual inner product under standard basis.

(2) Choose a basis of $V$ that is orthogonal w.r.t this inner product.

(3) Under such basis, the matrix of $\rho(g)$ is orthogonal for all $g\in G$.

Questions: does the (2) mean we find a matrix $B$ s.t. $B^H\sum_{g\in G}\rho(g)^H\rho(g)B=I$? If yes, then the matrix of $\rho(g)$ under the basis $B$ should become $M(g)=B^{-1}\rho(g)B$ right? But how could we see $M(g)^HM(g)=I$?

I guess I must be misunderstanding somewhere. Any help would be appreciated. Thanks.

Update: I think I got this. I was stupid. I think in this way: if let $\sum_{g\in G}\rho(g)^H\rho(g)=Q^HQ$ positive definiteness, then for each $g\in G$ we have $\rho(g)^HQ^HQ\rho(g)=Q^HQ$ from $G$-invariance. Then this shows $Q\rho(g)Q^{-1}$ is unitary(orthogonal). Now let $B=Q^{-1}$ then we are done. This is is the same as the one in (2).

答:
I think I'm misunderstanding your question, but (2) means we find vectors $\{w_i\} \subset V$ such that $[w_i, w_j] = \delta_{ij}$. In other words, $[w_i, w_j] = 0$ for $i \neq j$ and $[w_i, w_i] = 1$. This is done with the Gram-Schmidt process.

Update: I'm having trouble understanding your notation and the matrix $B$. But maybe if I put (3) in my notation then you can translate?

Note that by $G$-invariance, for all $w_1, w_2 \in V$ we have
$$[\rho(g)w_1, w_2] = [w_1, \rho(g^{-1})w_2] = [w_1, \rho(g)^{-1}w_2].$$
Moreover, the Hermitian transpose satisfies
$$[\rho(g)w_1, w_2] = [w_1, \rho(g)^H w_2]$$

In other words
$$[w_1, \rho(g)^{-1}w_2] = [w_1, \rho(g)^H w_2].$$

Now plugging in the $G$-invariant basis from (2), you can get the equality of the matrices $\rho(g)^{-1}$ and $\rho(g)^H$


答:
Let $\gamma\in G$. Then for all $v$, $w\in V$ one has
$$
[\rho(\gamma)v,\rho(\gamma)w]=
\sum_{g\in G}\langle\rho(g\gamma)v,\rho(g\gamma)w\rangle=
\sum_{g^\prime\in G}\langle\rho(g^\prime)v,\rho(g^\prime)w\rangle=
[v,w]
$$
because the map $g\mapsto g\gamma$ is a bijection of $G$ onto itself.

But the above equality tells exactly that $\rho(g)$ leaves $[\cdot,\cdot]$ unchanged, so that its associated matrix with respect to an orthonormal basis will be orthogonal

Title: Confused by "orthogonal representation" of a finite group. How to show the matrix under some basis from the new defined inner product is orthogonal?

Body:
Suppose we have a finite group $G$ and a representation $\rho:G\to GL(V)$ on the finite vector space $V$. So $V$ is a $G$-module.

Now I am asked to show that there is a basis of $V$ under which the matrix of $\rho(g)$ is orthogonal for all $g\in G$. Here is what confused me and I still don't know in what sense is this possible. I have done some searching and know typically the steps are the following:(here we assume $\rho(g)$ is the matrix under standard basis)

(1) Define a $G$-invariant inner product $[v,w]=\sum_{g\in G}\langle\rho(g)v,\rho(g)w\rangle$, where $\langle,\rangle$ is the usual inner product under standard basis.

(2) Choose a basis of $V$ that is orthogonal w.r.t this inner product.

(3) Under such basis, the matrix of $\rho(g)$ is orthogonal for all $g\in G$.

Questions: does the (2) mean we find a matrix $B$ s.t. $B^H\sum_{g\in G}\rho(g)^H\rho(g)B=I$? If yes, then the matrix of $\rho(g)$ under the basis $B$ should become $M(g)=B^{-1}\rho(g)B$ right? But how could we see $M(g)^HM(g)=I$?

I guess I must be misunderstanding somewhere. Any help would be appreciated. Thanks.

Update: I think I got this. I was stupid. I think in this way: if let $\sum_{g\in G}\rho(g)^H\rho(g)=Q^HQ$ positive definiteness, then for each $g\in G$ we have $\rho(g)^HQ^HQ\rho(g)=Q^HQ$ from $G$-invariance. Then this shows $Q\rho(g)Q^{-1}$ is unitary(orthogonal). Now let $B=Q^{-1}$ then we are done. This is is the same as the one in (2).


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
While I agree with Michael Hardy's answer, there is a sense in which Bayes' theorem is more important than any random identity in basic probability. Write Bayes' Theorem as

$$\text{P(Hypothesis|Data)}=\frac{\text{P(Data|Hypothesis)P(Hypothesis)}}{\text{P(Data)}}$$

The left hand side is what we usually want to know: given what we've observed, what should our beliefs about the world be? But the main thing that probability theory gives us is in the numerator on the right side: the frequency with which any given hypothesis will generate particular kinds of data. Probabilistic models in some sense answer the wrong question, and Bayes' theorem tells us how to combine this with our prior knowledge to generate the answer to the right question.

Frequentist methods that try not to use the prior have to reason about the quantity on the left by indirect means or else claim the left side is meaningless in many applications. They work, but frequently confuse even professional scientists. E.g. the common misconceptions about $p$-values come from people assuming that they are a left-side quantity when they are a right-side quantity.


问:
Prove that a map is Injective
How can I prove that $f: (0, \infty) \times (0,\pi) \to \mathbb{R}^2$ where $f(x,y) = (\sinh(x)\sin(y),\cosh(x)\cos(y))$ is injective? </p>

答:
Let $u:=\sinh x\sin y$, $\>v:=\cosh x\cos y$, $\>z:=x+iy$. Then
$$u-iv=-i\sinh x\sinh(iy)-i\cosh x\cosh(iy)=-i\cosh(x+iy)=-i\cosh z\ .$$
We have to prove that $\cosh z_1\ne\cosh z_2$ when $z_1\ne z_2$ and both $z_1$ and $z_2$ lie in the strip $S:\ 0<{\rm Im}(z)<\pi$.

Put $e^{z_i}=:w_i\ne0$. Then $\cosh z_1=\cosh z_2$ is equivalent with
$$(w_1-w_2)\left(1-{1\over w_1 w_2}\right)=0\ .$$
Since $\exp$ is injective on $S$ the assumption $z_1\ne z_2$ implies $w_1\ne w_2$. Furthermore, $w_1w_2=1$ would imply $z_1+z_2=2k\pi i$ for some $k\in{\mathbb Z}$. But there are no $\>z_1$, $z_2\in S$ satisfying this equation. Altogether it follows that $\cosh z_1=\cosh z_2$ is impossible for two different $z_i\in S$.

Title: Prove that a map is Injective


问:
Figure 7.7. and 7.8 in Loring Tu's An Introduction to manifolds
Figure 7.7 and 7.8 are showed as follows in Loring Tu's An Introduction to manifolds:

However, I cannot understand why $\mathbb RP^2$ cannot be embedded as a submanifold of $\mathbb R^3$. The  the orientation of the boundary of Fig7.7 and Fig 7.8 seems not the same, how do they relate to each other? I would appreciate it if someone could explain this for me.

答: I think it is easier to see it if, instead of an equator, you remove a square from $S^2$, identifying the opposite sides (which will have opposite orientations). This is what you see in figure 7.8. The double arrows correspond to a pair of opposite sides of the square and the single ones to the other pair. When you identify them, you end up with the last figure in 7.8, which shows that you necessarily have self intersections (all the points on the vertical segments are double)


答:
In the first of the four subfigures of Figure 7.8, the author has taken the liberty of breaking the boundary of the disc into four segments instead of just two, and has labelled how to identify those four segments in pairs. But if you compare that with the identifications shown in Figure 7.7, you'll see that what actually happens in both cases is that each point on the circle is identified with the antipodally opposite point of the circle. So the identifications in Figure 7.7 and the first subfigure of 7.8 are really identical, hence both yield $\mathbb RP^2$.

What happens in the next three subfigures of Figure 7.8, though, is that certain additional identifications are taking place in order to get a map into $\mathbb R^3$: besides the antipodally opposite 2-to-1 A,C identification, and the antipodally opposite 2-to-1 B,D identification, the remaining points on the circle are identified 4-to-1. That means that the induced map $\mathbb R P^2 \to \mathbb R^3$ has certain 2-to-1 identifications and hence is not injective.

But that only raises the question: Why isn't there some other continuous function $\mathbb R P^2 \to \mathbb R^3$ which is injective? Well, there isn't any such function. Proving it takes some substantial work in algebraic topology, and it's not immediately obvious how to prove it. But that's really beyond the scope of your question.

Title: Figure 7.7. and 7.8 in Loring Tu's An Introduction to manifolds

Body:
Figure 7.7 and 7.8 are showed as follows in Loring Tu's An Introduction to manifolds:

However, I cannot understand why $\mathbb RP^2$ cannot be embedded as a submanifold of $\mathbb R^3$. The  the orientation of the boundary of Fig7.7 and Fig 7.8 seems not the same, how do they relate to each other? I would appreciate it if someone could explain this for me.


问:
Prove that \{a^{3^n} is not regular \n>=1 using pumping lemma
I'm practicing for my CS exam and I got stuck on following language $L:= \{ a^{k} b u a^{k} | k \geq 1, u \in \Sigma^{*}\}$

$\Sigma = \{a,b\}$

I've tried to prove this language using a pumping lemma, not sure if it's the best way to prove the irregularity. I don't think I got far enough to have a strong solution.

$w = a^pbu^pa^p$

$|w| = 3p \geq p$

Which then gives me
$xy^2z = a^{n+k}bu^na^n$

But I'm not sure how to proceed from here. This is the I've gone through examples like $0^n1^m$ etc. but this one seems to be a bit more complicated for me. Any help would be highly appreciated

答:
Note that $u$ in the definition of $L$ is in $\Sigma^*$, not $\Sigma$, so your $u^p$ in your definition of $w$ doesn’t actually specify anything in particular. I would simply take $w=a^pba^p$, where $p$ is the pumping length. If $L$ is regular, there is a decomposition $w=xyz$ such that $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for all $k\ge 0$. This means that $xy$ is contained in the first block of $p$ $a$s, so there are integers $r\ge 0$ and $s\ge 1$ such that $x=a^r$, $y=a^s$, and $r+s\le p$. It follows that $z=a^{p-r-s}ba^p$ and hence that

$$xy^kz=a^r(a^s)^ka^{p-r-s}ba^p=a^{p+(k-1)s}ba^p\,.$$

This is in $L$ if and only if $p+(k-1)s=p$, which is true if and only if $k=1$. That is, $k=1$ is the only value of $k$ for which $xy^kz\in L$, whereas if $L$ were regular, $xy^kz$ would be in $L$ for every $k\ge 0$. This shows that $L$ cannot be regular.


问:
Generalized angle sum identity for $\arctan$?
The angle sum  identity for arctan is:

$$\arctan (\alpha)+\arctan(\beta)=\arctan\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)$$

I was wondering if there exists a relationship for any linear combination of $\arctan$ functions:

$$A\arctan (\alpha)+B\arctan(\beta)$$

$A>0, B>0$ and both real numbers.

答:
Edit: Yes, such a formula indeed exists:

https://functions.wolfram.com/ElementaryFunctions/ArcTan/16/03/03/

Old post

I know this question has been asked a long time ago, but I figured this would still be of use. There must be such a formula for a linear combination of $\arctan$s! I don't know what exactly it looks like, but I accidentally came across this which is similar (long, painful way to get there at all, but anyway):

These two expressions (maybe someone else can format more nicely, sorry:) Mathematica code at the end)

$$A:=\frac{1}{2 a^{3/2}}
\\\left(\frac{ \left(2 \sqrt{a} \sqrt{a+b}-2 a+b\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a}-\sqrt{a+b}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a}-\sqrt{a+b}\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}+3 \sqrt{a}\right)}\\
-\frac{ \left(b-2 \left(\sqrt{a} \sqrt{a+b}+a\right)\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a+b}+\sqrt{a}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a+b}+\sqrt{a}\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}-3 \sqrt{a}\right)}\right)$$

$$B:=-\frac{\sqrt{b-a} \arctan\left(\sqrt{\frac{a}{b-a}} (\sin (\phi)+\cos (\phi))\right)}{2 a^{3/2}}$$

don't look very similar. However, deriving them with Mathematica by $\phi$ yields the same expression

$$\frac{(a-b) (\cos (\phi)-\sin (\phi))}{2 a (a \sin (2 \phi)+b)}.$$

Consequently, I think that $A$ and $B$ differ by a constant, if at all.

What strikes me is that both the coefficients and the arguments to the $\arctan$s seem to contain complex conjugate-like terms everywhere, for example $\sqrt{a+b}-3 \sqrt{a}$, $\sqrt{a+b}+3 \sqrt{a}$ in the denominators. Maybe someone else can resolve the mystery and come up with a linear combination formula for $\arctan$s?

These two links are also quite insightful on such a linear combination :

1, 2

Mathematica code:

```<code>FullSimplify[D[((2 (-2 a+b+2 Sqrt[a] Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])] ArcTan[(b (Sqrt[a]+(Sqrt[a]-Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]-Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])])])/(b (3 Sqrt[a]+Sqrt[a+b]))-(2 Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])] (b-2 (a+Sqrt[a] Sqrt[a+b])) ArcTan[(b (Sqrt[a]+(Sqrt[a]+Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]+Sqrt[a+b]) Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])])])/(b (-3 Sqrt[a]+Sqrt[a+b])))/(4 a^(3/2)),phi],{a+b>0,b-a>0}]

FullSimplify[D[-((Sqrt[-a+b] ArcTan[Sqrt[a/(-a+b)] (Cos[phi]+Sin[phi])])/(2 a^(3/2))),phi],{a+b>0,b-a>0}]
</code>```

Body:
Edit: Yes, such a formula indeed exists:

https://functions.wolfram.com/ElementaryFunctions/ArcTan/16/03/03/

Old post

I know this question has been asked a long time ago, but I figured this would still be of use. There must be such a formula for a linear combination of $\arctan$s! I don't know what exactly it looks like, but I accidentally came across this which is similar (long, painful way to get there at all, but anyway):

These two expressions (maybe someone else can format more nicely, sorry:) Mathematica code at the end)

$$A:=\frac{1}{2 a^{3/2}}
\\\left(\frac{ \left(2 \sqrt{a} \sqrt{a+b}-2 a+b\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a}-\sqrt{a+b}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a}-\sqrt{a+b}\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}+3 \sqrt{a}\right)}\\
-\frac{ \left(b-2 \left(\sqrt{a} \sqrt{a+b}+a\right)\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a+b}+\sqrt{a}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a+b}+\sqrt{a}\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}-3 \sqrt{a}\right)}\right)$$

$$B:=-\frac{\sqrt{b-a} \arctan\left(\sqrt{\frac{a}{b-a}} (\sin (\phi)+\cos (\phi))\right)}{2 a^{3/2}}$$

don't look very similar. However, deriving them with Mathematica by $\phi$ yields the same expression

$$\frac{(a-b) (\cos (\phi)-\sin (\phi))}{2 a (a \sin (2 \phi)+b)}.$$

Consequently, I think that $A$ and $B$ differ by a constant, if at all.

What strikes me is that both the coefficients and the arguments to the $\arctan$s seem to contain complex conjugate-like terms everywhere, for example $\sqrt{a+b}-3 \sqrt{a}$, $\sqrt{a+b}+3 \sqrt{a}$ in the denominators. Maybe someone else can resolve the mystery and come up with a linear combination formula for $\arctan$s?

These two links are also quite insightful on such a linear combination :

1, 2

Mathematica code:

FullSimplify[D[((2 (-2 a+b+2 Sqrt[a] Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])] ArcTan[(b (Sqrt[a]+(Sqrt[a]-Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]-Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])])])/(b (3 Sqrt[a]+Sqrt[a+b]))-(2 Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])] (b-2 (a+Sqrt[a] Sqrt[a+b])) ArcTan[(b (Sqrt[a]+(Sqrt[a]+Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]+Sqrt[a+b]) Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])])])/(b (-3 Sqrt[a]+Sqrt[a+b])))/(4 a^(3/2)),phi],{a+b>0,b-a>0}]

FullSimplify[D[-((Sqrt[-a+b] ArcTan[Sqrt[a/(-a+b)] (Cos[phi]+Sin[phi])])/(2 a^(3/2))),phi],{a+b>0,b-a>0}]


问:
Significant Figure Rules for Measured Bases and Exponents
The following link How to determine significant figures involving radicals and exponents mentions that if you have 5.1^4 "The 4 is (probably) exact, so we ignore that for deciding precision, so the answer should have two significant digits, just as 5.1 does. That give an answer of 680". In the above quoted example it appears the base is considered a measured value and the exponent an exact value.

What is the rule for when both the base and exponent are considered measured values. Consider the following power where the base and exponent are considered measured quantities: 3.21^2.33

What about when the base is an exact value and the exponent is a measured value?

答:
All in all I think you're trying to push significant figures beyond their normal use. The gist is that significant figures are a very simple method to error propagation. The basic interpretation is to essentially think of the values being used as rounded. So $x= 3.21$ implies that $x$ should be in the range $3.205 < x < 3.215$.

In the case $3.21^{2.33}$ it is assumed that there are errors in both values so:

$$3.215^{2.335} = 15.29$$
$$3.21^{2.33} = 15.14$$
$$3.205^{2.325} = 15.00$$

So the high difference is 0.15 and the low difference is 0.14. Take the bigger value and you get 15.14(15) indicating that the error in the last two digits is 15, so $14.99 \le \mathrm{value} \le 15.29$.

Now there is a bit of a further dilemma. Do you want to express just the value in significant figures, or are you satisfied with the odd 15.14(15) which is common in scientific writings? If you want to stick with just the value in significant figures then I'd use 15.1 for the value. Having 0.05 for the error on 15.1 loses some information, but it keeps you from foolishly proposing that the value should be "approximately" $15.141 075 516 144 591 945 214 121 498 099$ implying an error of $\pm 0.000 000 000 000 000 000 000 000 000 000 5$.


问:
How to find $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$ using known result for $\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx$
I need to find integral of $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$. From
this Riemann_zeta_function , it gives $\zeta\left(  s\right)  =\frac{1}{\Gamma\left(
s\right)  }\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$. Hence for $s=3$ this gives

\begin{align}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx  & =\zeta\left(  3\right)
\Gamma\left(  3\right)  \nonumber\\
& =2\zeta\left(  3\right)  \tag{1}%
\end{align}

My question is, how to use the above result to find integral of $\int%
_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$? Mathematica gives

\begin{equation}
\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx=\frac{3}{2}\zeta\left(  3\right)
\tag{2}
\end{equation}

I tried to see if there is change of variable or other trick, but I do not
see one. I do know about a direct method given in this answer here, but I want
to see if it is possible, with some trick, to use the result from (1) to find (2).

答:
Let $\displaystyle \mathcal I =\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}\,\mathrm{d}x $ and $\displaystyle \mathcal J = \int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}\,\mathrm{d}x. $

You already know $\mathcal I =2\zeta\left(  3\right)$ but you seek $\mathcal J$.

$\displaystyle \mathcal I =\int_{0}^{\infty}\frac{(2x)^{2}}{e^{2x}-1}\,\mathrm{d}(2x) = \int_0^\infty \frac{8x^2}{e^{2x}-1}\, \mathrm{d}x$

But  $\displaystyle \frac{8x^2}{e^{2x}-1} = \frac{4x^2}{e^x-1}-\frac{4x^2}{e^x+1}$

Thus $\displaystyle \mathcal I =  4\int_{0}^{\infty} \frac{x^2}{e^x-1}\,\mathrm{d}x-4\int_{0}^{\infty} \frac{x^2}{e^x+1}\,\mathrm{d}x $

So  $\mathcal I = 4 \mathcal I-4 \mathcal J $ and it follows that $\displaystyle \mathcal J =\frac{3}{2}\zeta(3). $


问:
Showing that formulae which are provably $Σ_n$ and $Π_n$ ($n > 0$) in the arithmetical hierarchy are closed under $∃$ and $∀$ respectively
As the title explains, I'm trying to solve the following problem:

Show that for $n > 0$, formulae provably $Σ_n$ with respect to PA are closed under existential
quantification, and formulae provably $Π_n$ with respect to PA are closed under universal
quantification

(where $Π_n$ and $Σ_n$ are in the arithmetical hierarchy).

I assume this is done using induction, and I've done the base cases using the fact that $$PA ⊢ (∃v_j ∃v_i F(v_i , v_j) ↔ ∃v_k (∃v_j ≤ v_k)(∃v_i ≤ v_k)F(v_i
, v_j )).$$

to show that if $\phi$ is provably $Σ_1$ then it's equivalent to some $\phi' = \exists v_1\psi$ where $\psi$ is $\Pi_0$, and since $(∃v_j ≤ v_k)$, $(∃v_i ≤ v_k)$ are bounded it follows that $(∃v_j ≤ v_k)(∃v_i ≤ v_k)\psi$ is $\Pi_0$ making $$∃v_3 (∃v_2 ≤ v_3)(∃v_1 ≤ v_3)\psi$$ be $Σ_1$ and the result following by equivalence of $\exists v_2\phi$ with $∃v_3 (∃v_2 ≤ v_3)(∃v_1 ≤ v_3)\psi$ (and doing similar for the case where $\phi$ is $\Pi_1$).

I'm stuck when I get to the inductive step though - if we take $\phi$ to be $Σ_{n+1}$ the same trick doesn't work because there's no guarantee (I don't think) that $(∃v_2 ≤ v_3)(∃v_1 ≤ v_3)\psi$ will be $Π_n$ since what made $(∃v_2 ≤ v_3)(∃v_1 ≤ v_3)\psi$ be $Π_0$ in the previous case was that $Π_0$ formulae are defined as those with only bounded quantifiers, so the same trick doesn't work.

I though about maybe proving both results together as a single induction, so that I could apply the inductive hypothesis to some manipulation of $\psi$ (call it $\psi'$) which is in $\Pi_n$ to obtain $\forall v_2 \phi'$ in $\Pi_n$ and try to manipulate that somehow to obtain $\exists v_2\exists v_1\psi$ as being $Σ_{n+1}$, but I don't know how I'd make that work.

Any advice would be greatly appreciated.

答:
You need to be a bit more subtle in using bounded quantifiers. Get greedy! Make your new variable do more work!

It's often easier to first work things out for a more expressive theory, and then "pull down" to the actual theory in question. This is such a case. Consider the following equivalence in the expanded language of arithmetic gotten by adding the "factor-counting function" $count(a,b)=\max\{n: a^n\vert b\}$ (we'll address the change-of-language later): $$\exists a\exists b\varphi(a,b)\quad\equiv\quad \exists u\varphi(count(2,u), count(3,u)).$$

Here $\varphi$ is any formula whatsoever - complexity considerations haven't entered the picture yet.

The idea here is that $u$ codes the pair $(a,b)$ and the "count" function decodes it appropriately: we'll want $u=2^a3^b$ (admittedly, $2^a3^bk$ for some $k$ not divisible by $2$ or $3$ would also do the job). For longer blocks of existential quantifiers we just use more primes, e.g. $$\exists a\exists b\exists c\psi(a,b,c)\quad\equiv\quad \exists u\psi(count(2,u), count(3,u), count(5,u)).$$

And of course universal quantifiers will play out the same way, e.g. $$\forall a\forall b\forall c\forall d\theta(a,b,c,d)\quad\equiv\quad\forall u\theta(count(2,u), count(3,u), count(5,u), count(7,u)).$$

Now let's go back to the complexity issue. Note that all the above worked regardless of the complexity of the formula after the displayed quantifiers ($\varphi,\psi,\theta$ respectively). We can see the real power of this approach when we look at mixed quantifiers: the sentence $$\color{green}{\exists a\exists b}\color{red}{\forall c\forall d\forall k}\color{blue}{\exists l\exists m}\chi(\color{green}{a,b},\color{red}{c,d,k},\color{blue}{l,m})$$ is provably equivalent to $$\color{green}{\exists u}\color{red}{\forall v}\color{blue}{\exists w}\chi(\color{green}{count(2,u), count(3,u)}, \color{red}{count(2,v), count(3,v), count(5,v),}\color{blue}{count(2, w), count(3,w)}).$$

(Colors! Muahahaha! I've gone mad with power!)

Via coding finite tuples as individual numbers - which is a trick we use constantly - we've condensed all the homogeneous quantifier blocks at once.

OK, now there's just one issue left: change of language. The above establishes (once some appropriate verification has been performed) the desired result for an extension of Peano arithmetic. But what about $\mathsf{PA}$ itself?

Here we need a "low-complexity definability" result:

There is a $\Pi_0$ formula $\zeta(x,y,z)$ defining the graph of $code$ in $\mathbb{N}$; that is, such that $$\zeta^\mathbb{N}=\{(x,y,z): code(x,y)=z\}.$$

We then use this formula to "simulate" the function symbol $code$ by appropriate introduction of new variables, e.g. $$\exists a\exists b\varphi(a,b)\quad\equiv\quad\exists u\forall a<u\forall b<u(\zeta(2,u,a)\wedge\zeta(3,u,b)\rightarrow \varphi(a,b))$$ (note that those $\forall$s could be $\exists$ if that would be preferred).


问:
Coincidence of the coordinate ring and the set of regular functions
If $X$ is an affine variety, and if we denote by k[X] its coordinate ring and by $O(X)$ the set of regular functions on $X$, then it is well-known that $k[X]\cong O(X)$.

My question is : Do we have the same for (affine) algebraic groups ? How do we prove that?

答:
Well, affine algebraic groups are in particular affine varieties, so of course the same result holds for them.

For non-affine algebraic groups, the ring of regular functions can be very restrictive, and the coordinate ring of a non-affine variety is not really well defined.  For example, a connected abelian variety $G$ is complete (that is, for any variety $Y$, the projection map $G \times Y \rightarrow Y$ is closed), and therefore has trivial global section.

Definitions: if $X$ is a variety, $\mathcal O(X)$ denotes the ring of regular functions on $X$, i.e. morphisms of varieties from $X$ to the variety $k = \mathbb A^1$.

If $X$ is an affine variety, and $X$ is a Zariski closed subset of some $k^n$, then the coordinate ring $k[X]$ is the ring of polynomial functions from $k^n$ to $k$, restricted to $X$.  Equivalently, $X$ is can be identified with the space of maximal ideals (or prime ideals, depending on the formulation) of a finitely generated $k$-algebra $A$, and the coordinate ring $k[X]$ is just $A$.


问:
Leibniz rule X Direct integral computation
This question arose on a very complicated context, but it is very simple:

I have for $0<a<1$ the integral $\int_0^t (t-x)^{a-1} dx$ and 'd like to derive on $t$. The imediate result is $t^{a-1} $, which is a positive function for $t$ positive.

Suppose that I'd like to use Leibniz rule, so $\dfrac{\partial}{\partial t} \displaystyle\int_0^t (t-x)^{a-1} dx=(t-t)^{a-1}+ \displaystyle\int_0^t \dfrac{\partial}{\partial t}(t-x)^{a-1} dx=\displaystyle\int_0^t (a-1) (t-x)^{a-2} dx$

But this is a positive function? Because $(a-1)<0$ and $(t-x)^{a-2} \geq0$, so the integrandum is everywhere nonpositive...!

Thank you so much

答:
The problem lies in the fact that the function $g(x) = (t - x)^{a - 1}$ is not defined at $x = t$. You are treating it as if $g(t) = (t - t)^{a - 1}  = 0$, but this does not actually make sense; the graph of $g$ has vertical asymptote as $x$ approaches $t$ from the left.

If you treat your problem carefully by instead writing $\int_0^{t - \delta} (t - x)^{a - 1} \, dx$, you can get the proper calculation.


问:
How to integrate the function $\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}$
How to integrate the following family of functions? I really do not have too many ideas. The context of this is to find moment generating functions but the context is not too important here.

$$I_n:=\int_{0}^{\infty}n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$

I wanted to try this inductively/iteratively but I got nowhere (by the way I don't have to do this iteratively).  I should also stress that $\lambda>t$ so that I think $I_n$ converges for each $n$. Could someone please given me  a hint or two? I know how to do $n=1$ btw.

Many thanks!

答:
lets collect the terms first:
$$f(x)=\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}=\lambda e^{-(t-\lambda)x}(1-e^{-\lambda x})^{n-1}$$
$$I_n=n\lambda\int_0^\infty (e^{-x})^{t-\lambda}(1-(e^{-x})^\lambda)^{n-1}dx\tag{1}$$
now if we make the substitution $u=e^{-x}\Rightarrow dx=-\frac{du}{u}$ and our integral becomes:
$$I_n=n\lambda\int_0^1 u^{t-\lambda-1}(1-u^\lambda)^{n-1}$$
now make the substitution $v=u^\lambda\Rightarrow dv=\lambda u^{\lambda-1}du=\lambda v^{(\lambda-1)/\lambda}du\Rightarrow du=\frac{dv}{\lambda}v^{(1-\lambda)/\lambda}$ now sub this in:
$$I_n=n\int_0^1 v^{(t-\lambda-1)/\lambda}\,v^{(1-\lambda)/\lambda}(1-v)^{n-1}dv$$
$$I_n=n\int_0^1 v^{(t-2\lambda)/\lambda}(1-v)^{n-1}dv\tag{2}$$

Now, the incomplete beta function is defined as:
$$B(x;a,b)=\int_0^x y^{a-1}(1-y)^{b-1}dy\tag{3}$$
and the beta function is defined as:
$$B(a,b)=\int_0^1x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\tag{4,5}$$
Your integral fits this form nicely, you could also use binomial expansion.


问:
Minimum of a function by inequality
No answers please, hints only!

I am asked to use the fact that $$\cfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\cdots a_n}, n\in \mathbb N \\ \text{with equality if and only if } a_1=a_2=\cdots=a_n$$
To find the minimum value (and when this value occurs) of the function $$f(x)=\cfrac{9x^2(\sin^2x)+4}{x\sin x}, \hspace{0.5cm} x\in (0, \pi)$$

I've decided to set $f(x)\ge K$ to make it easier to use this inequality. \begin{align*} \cfrac{9x^2(\sin^2x)+4}{x\sin x} &\ge K \\ \cfrac{9x^2(\sin^2x)+4}{x} &\ge K\sin x \\ \cfrac{\biggr(3x\sin x\biggr)^2+2^2}{x} &\ge K\sin x\end{align*}
And then this is the point I didn't know how to continue, and I'm not sure if I even have the right approach to this.

答: Hint: Multiplying top and bottom by $\frac2{x \sin(x)}$ gives $$f(x) = \frac{18x \sin(x) + \frac8{x \sin(x)}}{2} $$

Title: Minimum of a function by inequality


答:
HINT:

Consider the function $$g(x)=\frac{x\sin(x)f(x)}2=\frac{(3x\sin x)^2+2^2}2$$

Note that we can apply AM-GM (your fact) to get a lower bound on $g(x)$. Using your fact, you can also show that when $(3x\sin x)^2=2^2$, then $g(x)$ equals the lower bound.

Now multiply your inequality by $\frac2{x\sin x}$ to get a lower bound on $f(x)$.

Title: Minimum of a function by inequality

Body:
No answers please, hints only!

I am asked to use the fact that $$\cfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\cdots a_n}, n\in \mathbb N \\ \text{with equality if and only if } a_1=a_2=\cdots=a_n$$
To find the minimum value (and when this value occurs) of the function $$f(x)=\cfrac{9x^2(\sin^2x)+4}{x\sin x}, \hspace{0.5cm} x\in (0, \pi)$$

I've decided to set $f(x)\ge K$ to make it easier to use this inequality. \begin{align*} \cfrac{9x^2(\sin^2x)+4}{x\sin x} &\ge K \\ \cfrac{9x^2(\sin^2x)+4}{x} &\ge K\sin x \\ \cfrac{\biggr(3x\sin x\biggr)^2+2^2}{x} &\ge K\sin x\end{align*}
And then this is the point I didn't know how to continue, and I'm not sure if I even have the right approach to this.


答:
Building off of the comment, it is actually not too terrible to use binomial theorem. However, first observe that

$$ne^{-\lambda x}(1-e^{-\lambda x})^{n-1} = \frac{d}{dx}(1-e^{-\lambda x})^n\text{.} $$

Then you can apply binomial theorem to

$$(1-e^{-\lambda x})^n $$

and take the derivative of the resulting sum. Once substituted into the integrand, evaluating the integral (term by term) is straightforward. The result will be a sum of $n$ terms and will not be pretty.

Title: How to integrate the function $\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}$

Body:
How to integrate the following family of functions? I really do not have too many ideas. The context of this is to find moment generating functions but the context is not too important here.

$$I_n:=\int_{0}^{\infty}n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$

I wanted to try this inductively/iteratively but I got nowhere (by the way I don't have to do this iteratively).  I should also stress that $\lambda>t$ so that I think $I_n$ converges for each $n$. Could someone please given me  a hint or two? I know how to do $n=1$ btw.

Many thanks!


问:
Let $p\in (1,a)$ and $y''+py=0$ with $y(0)=2$ and $y(\pi)$=3 then what is the max value of $a$?
Let $p\in (1,a)$ and $y''+py=0$ with $y(0)=2$ and $y(\pi)=3$ then what is the max value of $a$?

What I tried:

I used the characteristic equation $r^2+p=0$ then $r=±i\sqrt p$

general solution $y(x)=c_1cos(\sqrt px)+c_2sin(\sqrt px)$

with $y(0)=2, c_1=2 $

then with $y(\pi)=3$,

$2cos(\sqrt p\pi)+c_2sin(\sqrt p\pi)=3$

but then I am stuck, how can I find max value of $a$? Anyone can help me please? I would be very glad if you help.

答: Notice that if $\sin(\sqrt p \pi) = 0$ then this equation cannot be satisfied. This occurs when $\sqrt p \pi = n \pi$ or when $p = n^2$. Hence the open interval $(1,a)$ will have solutions provided that $a = 4$.


问:
Evaluation of "concentration of measure function" at zero.
Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). We define the concentration function of the triple $X,d$ and $\mu$ as follows:

$\alpha_{(X,d,\mu)}(r)=\sup\{1-\mu(A_r):A\subset X\text{ is measurable and }\mu(A)\geq 1/2\}$, $r\geq0$,

where we define $A_r=\{x\in X:d(x,A)<r\}$ (EDIT: I adopt the convention $A_0=A$).

I am interested in the value of the following (sparing you the elementary details that leads to the equality below):

$\alpha_{(X,d,\mu)}(0)=\sup\{\mu(B):B\subset X\text{ is measurable and }\mu(B)\leq 1/2\}$

If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$,
then it clearly is the case that $\alpha_{(X,d,\mu)}(0)=1/2$.
If there is no set with measure $1/2$, then the situation is not so clear.
I can easily find examples where there is no set of measure $1/2$ and $\alpha_{(X,d,\mu)}(0)$ is not $1/2$ (Dirac probability measure comes to mind), but I am unable to either prove (by finding an example ideally) or disprove that there exists a probability measure on some metric space such that no set has a measure of $1/2$ but $\alpha_{(X,d,\mu)}(0)=1/2$.

答:
Based on your clarification, I think that the question is equivalent to saying the following:

Is it possible to have a sequence of sets whose measure tends upwards to $\frac{1}{2}$, without there being a set who measure is $\frac{1}{2}$?

Clearly, if the sets are nested, this is not possible, not just in a Borel $\sigma$ algebra generated by a metric, but in any measure space.

If we don't have a probability measure, we can take a sequence of disconnected points whose measure tends upwards to $\frac{1}{2}$. If they all have measure above $\frac{1}{4}$, then there could be no set which has measure exactly $\frac{1}{2}$.

Now, maybe you can consider a similar situation, and try and show that a sequence of disconnected points whose total measure is 1, where there are finite subsets whose measure tends upwards towards $\frac{1}{2}$, must have a subset of measure $\frac{1}{2}$.

This is a fact about sums of a sequence of real numbers, rather than about measure theory. I didn't think it through quite yet, so it might be trivial or non-trivial (or false).

But if you can show that there must be a set of measure $\frac{1}{2}$ for the sequence of disconnected points, this might imply the general case. Post back if you manage to carry any of the above out, otherwise I will let you know anything further I think of.


问:
Is a given integer's binary representation ever shorter than its decimal?
In particular, once you get into large numbers, does any given number's binary representation ever become a shorter string of digits than the decimal representation of the same number?</p>

答: No. Think about the positions as bins which empties into the next bin when it fills, and the base being the size of the first bin. You will always need less bins if each bin can hold more things. This means if $n > m$ then the string representing a number in base $n$ will always be shorter than or equal to the length of the same number in base $m$.


问:
Examples of maps $\phi: \operatorname{Hom}(\prod_s N_s, M) \to \prod_s \operatorname{Hom}(N_s, M)$
Let $R$ be a commutative ring, $\{N_s\}_{s \in S}$ is a family of modules and $M$ is a module over $R$. There is canonical map of modules
$$
\phi: \operatorname{Hom} \bigg(\!\prod_s N_s, M \bigg) \to \prod_s \operatorname{Hom}(N_s, M),
$$
that sends map $f \colon \prod_s N_s \to M$ to the "infinite vector" of maps $(f\circ \iota_k)_s$, where $\iota_k\colon N_k \to \prod_s N_s$ is the canonical injection. It is well-known that this map is isomorphism if $\prod_s N_s$ is replaced by $\bigoplus_s N_s$. But I want to know examples when product version fails to be injective or surjective.

I'm interested in the following examples in order of preferences

Ring $R$ is a field and $\phi$ is neither injective nor surjective.

Ring $R$ is a field and $\phi$ is not injective (surjective).

Map $\phi$ is neither injective nor surjective.

Map $\phi$ is not injective (surjective).

If there are theorems that forbid examples of any of the above types, it is also very interesting.

答:
Equivalently, you are asking about the restriction map  $\operatorname{Hom}(P, M) \to \operatorname{Hom}(Q, M)$ induced by the inclusion map $Q\to P$ where $P=\prod_s N_s$ and $Q=\bigoplus_s N_s$.  This map is part of an exact sequence $$0\to \operatorname{Hom}(P/Q,M)\to\operatorname{Hom}(P,M)\stackrel{\phi}\to\operatorname{Hom}(Q,M)\to \operatorname{Ext}^1(P/Q,M)\to \operatorname{Ext}^1(P,M).$$  So, $\phi$ is injective iff there are no nontrivial homomorphisms $P/Q\to M$.  In particular, for instance, this means that $\phi$ is never injective when $R$ is a field as long as infinitely many of the $N_s$ are nontrivial and $M$ is nontrivial.  Or, for an arbitrary family where infinitely many of the $N_s$ are nontrivial, if $M=P/Q$ then $\phi$ will not be injective.  On the other hand, if $M$ is an injective module (so in particular, if $R$ is a field, or more generally if $R$ is a semisimple ring), then $\phi$ is always surjective.

For an example where $\phi$ is not surjective, you could take $R=\mathbb{Z}$ and consider the direct product $P=\prod_p\mathbb{Z}/(p)$ where $p$ ranges over the primes.  Note that then the quotient $P/Q$ is divisible: given an element $x\in P$ and a nonzero integer $n$, you can divide $x$ by $n$ mod $Q$ by just ignoring the coordinates corresponding to the finitely many primes dividing $n$.  But $P$ has no nontrivial divisible submodules, since if $x\in P$ is nonzero on its $p$th coordinate then $x$ is not divisible by $p$.  So in particular, $P/Q$ is not isomorphic to a submodule of $P$ so the inclusion map $Q\to P$ does not split.  Taking $M=Q$, this says exactly that $\phi$ fails to be surjective since its image does not contain the identity map $Q\to Q$.

To get an example where $\phi$ is neither surjective nor injective, you can just take a direct sum of examples of each type.  For instance, with the same product as in the previous example, you could take $M=Q\oplus P/Q$.

Another well-known example where $\phi$ is not surjective is if you take $R=\mathbb{Z}$ and $P$ to be a product of countably infinitely many copies of $\mathbb{Z}$.  Then it is a nontrivial theorem that every homomorphism $P\to\mathbb{Z}$ is a linear combination of the projection maps.  In particular, there are only countably many such maps, whereas there are uncountably many homomorphisms $Q\to\mathbb{Z}$ (since $Q$ is free on infinitely many generators), so the restriction map $\phi:\operatorname{Hom}(P,\mathbb{Z})\to\operatorname{Hom}(Q,\mathbb{Z})$ is very far from surjective.  (On the other hand, it follows from the theorem that $\phi$ is injective in this case, which is rather surprising.)


问:
Show that $\sum_{k=0}^{n} {n \choose k}2^k=3^n$ using mathematical induction.
Ok so I'm not looking for someone to just straight up tell me the answer but if you could nudge me in some direction approaching a valid proof that'd be great.

$\sum_{k=0}^{n} {n \choose k}2^k=3^n \ \ \ \ \ \ \ \ \ \ \ (eq.1)$

(1) for $n=1$:
$$ \sum_{k=0}^{1}{1 \choose k}2^k = {1 \choose 0}2^0+{1\choose1}2^1 = 1+2 = 3 = 3^1 $$

so eq.1 is true for n=1.

(2) Assume eq.1 holds for $n=p$. That is,
$$\sum_{k=0}^{p} {p \choose k}2^k=3^p $$
is a true statement.

(3)  Show that (2) necessitates eq.1 be true for $n=p+1$. That is, show
$$\sum_{k=0}^{p+1} {p+1 \choose k}2^k=3^{p+1}.$$

So anyhow, I've been attempting to start (3) by splitting the sum from k=0 to p then having the last term separate like so :
$$\sum_{k=0}^{p+1} {p+1 \choose k}2^k=\sum_{k=0}^{p} {p+1 \choose k}2^k+2^{p+1}$$

This is the method the prof suggested but I'm getting nowhere with it.  And yes I must use induction.

答:
Hint: In order to use your inductive hypothesis, it'd be ideal if we had $\binom{p + 1}{k}$ in terms of $\binom{p}{k}$. As it turns out, there's a recurrence relation called Pascal's rule:
$$\binom{p + 1}{k} = \binom{p}{k} + \binom{p}{k - 1}.$$
(If you haven't seen this identity before, it's fun to find a combinatorial proof for it!)


问:
Vector fields tangent to a submanifold
Let $M$ be a smooth manifold and $N$ a submanifold of $M$. Consider the set of vector fields on $M$ that are tangent to $N$. I denote it by $\mathcal{F}$.

I am trying to prove that $\mathcal{F}$ is singular foliation on $M$, i.e. a $C^{\infty}(M)$-submodulde of $\mathfrak{X}(M)$ that is closed under the Lie bracket and locally finitely generated.

It is easy to see that the first condition is satisfied and after working with the local formula of the Lie bracket of vector fields I managed to show that the Lie bracket of two vector fields tangent to $N$ is also tangent to $N$.

I am struggling however to deal with the local generators. If $p$ is a point of $N$ then there exist local coordinates $(U,x^1,...,x^n)$ around $p$ such that $N\cap U =\{x^{k+1}=...=x^n=0\}$, where k is the dimension of $N$.BUT vector fields $\partial{x^1},...,\partial{x^k}$ do not necessarily generate $\mathcal{F}|U$. In addition, I cannot think of way to obtain generators of $\mathcal{F}$ at a neighborhood of a point that does not lie on $N$. I have also tried to work the case where $N$ is a closed submanifold, but I end up to linear combinations of vector fields on $M$ with coefficients smooth functions on a neighborhood of N. Maybe I can extend these to a neighborhood of M but I do not see how.

A submodule $\mathcal{F}\subset \mathfrak{X}(M)$ is locally finitely generated if for each point $x \in M$ there exists a neighborhood $U$ of $x$ and vector fields $X_1,...,X_k \in \mathfrak{X}(M)$ such that $\mathcal{F}|U = C^{\infty}(U)X_1|U +...+ C^{\infty}(U)X_k|U$

答:
Throughout, Greek indices will run from $1$ to $n$, roman indices $i,j,k,\cdots$ will run from $1$ to $k$ and $a,b,c,\cdots$ will run from $k+1$ to $n$, with summation convention in effect. I'll also assume that by "submanifold" you mean "embedded submanifold", since you're using adapted coordinates.

Working in adapted local coordinates $x^1\cdots,x^n$, $\mathcal{F}|_U$ consists of vector fields $V^\alpha(x^1,\cdots,x^n)$ such that $V^i(x^1,\cdots,x^k,0,\cdots,0)=0$. I claim that $\mathcal{F}|_U$ is generated by the vector fields $\partial_a$ and $x^i\partial_j$.

To see this, we may write any vector field $V\in\mathcal{F}$ as $V=V^a\partial_a+V^j_ix^i\partial_j$, and define $V^j_i$ by
$$
V_i^j(x^1\cdots,x^n)=\int_0^1\partial_iV^j(tx^1,\cdots,tx^k,x^{k+1},\cdots,x^n)dt
$$
It's straightforward to show that $V_i^j$ are well defined and smooth (assuming without loss of generality that the coordinate chart is convex) and that the decomposition above holds.

It takes a bit more work to fit this into your definition of locally finitely generated, but it can be done by finding a smooth bump function $\psi$ around the point of interest and a a restriction of neighborhoods $U'\subset U$ such that $\psi|_{U'}=1$.

Title: Vector fields tangent to a submanifold


问:
Seeking a formula for a counting problem
Suppose $cnt(i)$ represents the number of occurrences of $i$ in array $A$ of length $n$ whose elements are between $1$ and $n$. An array is called a $k$-good array if and only if $cnt(k)=k$ and $\forall i\not=k, cnt(i)\not=i$. Let $f(n,k)$ be the number of all $k$-good arrays of length $n$ whose elements are between $1$ and $n$.

For example, $4,2,3,3,3$ is a $k$-good array for $n=5,k=3$, but $2,2,3,3,3$ is not a $k$-good array for $n=5,k=3$, because $cnt(2)$ also equals to $2$ so you can't tell from what $k$ the array is derived.

I am wondering if there is a recursive formula for $f(n,k)$.

答:
(Added Update with an inclusion-exclusion approach, see below. Also, see comment by Mike below for a fast python implementation of this.)

Some preliminary results. First some numerical values I've computed brutally for $f(n,k)$ (rows are $n\ge 1$, columns are $k\ge1$):

$$
\begin{array}
{}f(n,k)& k=1& k=2& k=3& k=4& k=5& k=6& k=7 &k=8\\
n=1 &1 \\
n=2 &2 &  1\\
n=3 &9 &  3  & 1\\
n=4 &80 &  30&    8 &   1\\
n=5 &945  & 360 &   90  &  15    & 1\\
n=6 &13884   & 5340  &  1360  &   240   &   24  &     1\\
n=7 &242410  &  93660 &   24255  &   4480   &   525   &    35  &      1\\
n=8 & 4898872  & 1897056 &  497896 &  95270  &   12264   &   1008   &  48 &  1\\
\end{array}$$
(This triangle does not appear to be on OEIS.)
Next, we can make some observations.

Trivially $f(n,n)=1$.

For $n\ge 2$ we also have $f(n,n-1) = n(n-2)$. (We have to use exactly $n-1$ many $n-1$ in our length $n$ vector, the one extra element cannot be $1$ nor $n-1$.)

For $n\ge 5$ we have $f(n,n-2) = {n \choose 2}(n-2)^2$. (This we have to count carefully, but similar to previous by paying attention to how many 1's and 2's we used. Keep inclusion-exclusion in mind.)

Perhaps you can continue to work out the entries diagonally first, namely $f(n,n-k)$, paying attention to edges cases. Certainly a recursion or combinatorial identification of these will be more desirable.

We can also count the number of "bad" arrays, perhaps denote as $f(n,0)$, where it is not $k$-good for any $k$. Then $f(n,0) = n^n - \sum_k f(n,k)$. Some numerical values for $f(n,0)$, for $n\ge 1$, are:
$$
0 ,1 ,14, 137, 1714, 25807, 458177, 9374801,...
$$
which also does not seem to be on OEIS.

Note.

Perhaps a suggestion: Consdier $g(n,k)$ to be the number of of length $n$ arrays with entries from $\{1,\ldots,n\}$ where there is exactly $k$ many entries with value $k$, and denote $S(n,k)$ to be this set of arrays. Then
$$
f(n,k) = |S(n,k) - \bigcup_{i\neq k} S(n,i)|,
$$
this we see clearly the inclusion-exclusion business.

Update. Computing this with inclusion-exclusion
(also, I am going to call arrays vectors, as that is more pleasing to me.)

Let $T\subset \{1,\ldots,n\}$, denote $s(n,T)$ to be the number of length $n$ vectors such that for each $t\in T$, there is exactly $t$ many $t$'s among the $n$ entries, and no restriction on the other values. Then
$$
s(n,T) = {n \choose T}(n-|T|)^{n-\sum T},
$$
where for $T=\{t_1,\ldots,t_k\}$, we write ${n\choose T} := {n \choose {t_1,\ldots,t_k}}$, multinomial coefficient; $|T|=k$ is the cardinality of $T$; and $\sum T = t_1+\cdots t_k$.

Then
$$\begin{eqnarray}
f(n,k) &=& \sum_{T\subset [n]} (-1)^{|T|+1}s(n,T) - \sum_{T\subset [n]-\{k\}} (-1)^{|T|+1}s(n,T) \\
\implies f(n,k)&=& \color{blue}{\sum_{k\in T\subset [n]} (-1)^{|T|+1}s(n,T)}
\end{eqnarray}
$$
by inclusion-exclusion. Note: The quantity $s(n,T)=0$ whenever $\sum T >n$, so you can prune out a lot of terms. The total number of nonzero $s(n,T)$ terms should be on the order of the number of partitions of $n$ into distinct parts, which I believe is on the order of $c^{\sqrt n}$ for some $c$. I didn't implement this pruning in the Octave computation below, however.

Using this and crudely implement it on Octave/Matlab I found $f(n,k)$ for:
$$
\begin{array}
{}n=9 &  112427973 &   43588944&    11564280    & 2274552    &  312102  &     29064     &   1764      &    63    &       1 \\
n=10& 2887731740 &  1120317120 &   299845680   &  60267900  &    8667288     &  878640     &   61800     &    2880   &       80   &         1\\
n=11&82060868406 &  31847443320 &   8585818065  &  1755965640    & 261829260  &    28248528    &   2198130 &       120780      &    4455 &           99   &          1\\
n=12&2555907405204   & 992122408860   & 269098056700  &   55822922100   &   8566352784 &      969619728  &      81279000     &    5005440     &     220660      &      6600   &          120      &         1
\end{array}
$$
which cost the following to compute the whole row:

```
n =  9   10.4786 seconds.
n = 10   26.563 seconds.
n = 11   60.0332 seconds.
n = 12   175.576 seconds.
```

We see this is roughly exponential $O(2^n)$ as expected. You can test out my (poorly written) Octave code at SageMath Here. Pruning out the zero terms may get us to $O(c^{\sqrt n})$ by this approach.

Body:
(Added Update with an inclusion-exclusion approach, see below. Also, see comment by Mike below for a fast python implementation of this.)

Some preliminary results. First some numerical values I've computed brutally for $f(n,k)$ (rows are $n\ge 1$, columns are $k\ge1$):

$$
\begin{array}
{}f(n,k)& k=1& k=2& k=3& k=4& k=5& k=6& k=7 &k=8\\
n=1 &1 \\
n=2 &2 &  1\\
n=3 &9 &  3  & 1\\
n=4 &80 &  30&    8 &   1\\
n=5 &945  & 360 &   90  &  15    & 1\\
n=6 &13884   & 5340  &  1360  &   240   &   24  &     1\\
n=7 &242410  &  93660 &   24255  &   4480   &   525   &    35  &      1\\
n=8 & 4898872  & 1897056 &  497896 &  95270  &   12264   &   1008   &  48 &  1\\
\end{array}$$
(This triangle does not appear to be on OEIS.)
Next, we can make some observations.

Trivially $f(n,n)=1$.

For $n\ge 2$ we also have $f(n,n-1) = n(n-2)$. (We have to use exactly $n-1$ many $n-1$ in our length $n$ vector, the one extra element cannot be $1$ nor $n-1$.)

For $n\ge 5$ we have $f(n,n-2) = {n \choose 2}(n-2)^2$. (This we have to count carefully, but similar to previous by paying attention to how many 1's and 2's we used. Keep inclusion-exclusion in mind.)

Perhaps you can continue to work out the entries diagonally first, namely $f(n,n-k)$, paying attention to edges cases. Certainly a recursion or combinatorial identification of these will be more desirable.

We can also count the number of "bad" arrays, perhaps denote as $f(n,0)$, where it is not $k$-good for any $k$. Then $f(n,0) = n^n - \sum_k f(n,k)$. Some numerical values for $f(n,0)$, for $n\ge 1$, are:
$$
0 ,1 ,14, 137, 1714, 25807, 458177, 9374801,...
$$
which also does not seem to be on OEIS.

Note.

Perhaps a suggestion: Consdier $g(n,k)$ to be the number of of length $n$ arrays with entries from $\{1,\ldots,n\}$ where there is exactly $k$ many entries with value $k$, and denote $S(n,k)$ to be this set of arrays. Then
$$
f(n,k) = |S(n,k) - \bigcup_{i\neq k} S(n,i)|,
$$
this we see clearly the inclusion-exclusion business.

Update. Computing this with inclusion-exclusion
(also, I am going to call arrays vectors, as that is more pleasing to me.)

Let $T\subset \{1,\ldots,n\}$, denote $s(n,T)$ to be the number of length $n$ vectors such that for each $t\in T$, there is exactly $t$ many $t$'s among the $n$ entries, and no restriction on the other values. Then
$$
s(n,T) = {n \choose T}(n-|T|)^{n-\sum T},
$$
where for $T=\{t_1,\ldots,t_k\}$, we write ${n\choose T} := {n \choose {t_1,\ldots,t_k}}$, multinomial coefficient; $|T|=k$ is the cardinality of $T$; and $\sum T = t_1+\cdots t_k$.

Then
$$\begin{eqnarray}
f(n,k) &=& \sum_{T\subset [n]} (-1)^{|T|+1}s(n,T) - \sum_{T\subset [n]-\{k\}} (-1)^{|T|+1}s(n,T) \\
\implies f(n,k)&=& \color{blue}{\sum_{k\in T\subset [n]} (-1)^{|T|+1}s(n,T)}
\end{eqnarray}
$$
by inclusion-exclusion. Note: The quantity $s(n,T)=0$ whenever $\sum T >n$, so you can prune out a lot of terms. The total number of nonzero $s(n,T)$ terms should be on the order of the number of partitions of $n$ into distinct parts, which I believe is on the order of $c^{\sqrt n}$ for some $c$. I didn't implement this pruning in the Octave computation below, however.

Using this and crudely implement it on Octave/Matlab I found $f(n,k)$ for:
$$
\begin{array}
{}n=9 &  112427973 &   43588944&    11564280    & 2274552    &  312102  &     29064     &   1764      &    63    &       1 \\
n=10& 2887731740 &  1120317120 &   299845680   &  60267900  &    8667288     &  878640     &   61800     &    2880   &       80   &         1\\
n=11&82060868406 &  31847443320 &   8585818065  &  1755965640    & 261829260  &    28248528    &   2198130 &       120780      &    4455 &           99   &          1\\
n=12&2555907405204   & 992122408860   & 269098056700  &   55822922100   &   8566352784 &      969619728  &      81279000     &    5005440     &     220660      &      6600   &          120      &         1
\end{array}
$$
which cost the following to compute the whole row:

n =  9   10.4786 seconds.
n = 10   26.563 seconds.
n = 11   60.0332 seconds.
n = 12   175.576 seconds.

We see this is roughly exponential $O(2^n)$ as expected. You can test out my (poorly written) Octave code at SageMath Here. Pruning out the zero terms may get us to $O(c^{\sqrt n})$ by this approach.


问:
How to decompose this fraction
$$\begin{align}\frac{1-r(e^{ix}+e^{-ix})/2}{(r-e^{ix})(r-e^{-ix})}&=\frac{A}{r-e^{ix}}+\frac{B}{r-e^{-ix}}\\\\\end{align}$$

$$=1-r(\dfrac{e^{ix}+e^{-ix}}{2})=A(r-e^{-ix})+B(r-e^{ix})$$

I have received the instruction here (see J.G.'s reply) that I should consider the "$r^0$ and $r^1$" terms. I don't understand this, so he wanted me to set $r^0=1$ and $r^1=r$?

The end result is two equations: $1=-e^{-ix}A-e^{ix}B$ and $-\dfrac{(e^{ix}+e^{-ix})}{2}=A+B$

How do you obtain this system of equations?

Please don't use concepts in complex analysis, I have only studied partially Calculus 2.

答:
Expanding out, we get
$$1 - r\left(\frac{e^{ix} + e^{-ix}}{2}\right) = (A + B)r + (-Ae^{-ix} - B e^{ix}).$$

J.G.'s suggestion is to compare the LHS constant term with the RHS constant term, as well as the LHS linear (or $r^1$) term with the RHS linear term.

Edit: Here's one way to solve the system (although the algebra might get a bit messy). First, one of the equations can be rewritten as
$$Be^{ix} = -1 - Ae^{-ix} $$
Or, after multiplying both sides by $e^{-ix}$,
$$B = -e^{-ix} - Ae^{-2ix}.$$
After substituting in to the other equation,
$$A + (-e^{-ix} - Ae^{-2ix}) =  \left(\frac{e^{ix} + e^{-ix}}{2}\right).$$
Now get all the $A$s on one side, everything else on the other, factor out the $A$ and divide through. Hopefully this is a good start.

Title: How to decompose this fraction