答:
The lowest roll a single dice could be and this work would be $4$, since we need $16$ and the maximum of the other two is $12$. Now the "closest" the numbers can be to each other would be $6,5,5$ and so far we have:

$6,6,4$

$6,5,5$
Now these are the only two combinations possible, now its just whether or not you care about permutations or combinations

Title: In how many ways can 3 dice add up to 16?

Body:
I am trying to use the stars and bars technique to calculate the number of ways 3 dice rolls can add up to 16. To my understanding, this can rephrased as:

$$x + y + z = 16 | x,y,z \in \mathbb{Z}^+ \land x,y,z \leq 6$$

I am not anywhere close to being a mathematician, so simple solutions are much appreciated. My main motivation to the question, is that I am trying to see if there is a cooler way to do it rather than count by hand, and stars and bars came to my mind as a potential solution. However, other simple solutions are also welcome.


问:
Parametrization of a circle, why is it $(\sin t, \cos t)$ instead of $(\sin^2t,\cos^2t)$?

The equation of a circle is $x^2 + y^2 = r^2$.

Then why is the parameterization of a circle $(r\cos(t), r\sin(t))$ and not $(r\cos^2(t), r\sin^2(t))$?

I thought what I'm supposed to do here is to eliminate $x$ and $y$ and write them in a way where I can represent them both with $t$ and then put it back into the equation? If I do that, I get

$$x = r\cos(t)$$
$$y = r\sin(t)$$

then if I put it back into the equation of the circle I get them both squared. Shouldn't the result then be $(\sin^2t,\cos^2t)$?

What's wrong here?

答:
If $x = r \cos t$, $y = r \sin t$, then $$x^2+y^2 = r^2 \cos^2 t + r^2 \sin^2 t = r^2 (\cos^2 t + \sin^2 t)$$ which is indeed $r^2$.

If $x = r \cos^2 t$, $y = r \cos^2 t$, then $$x^2 + y^2=(r \cos^2 t)^2 + (r \sin^2 t)^2 = r^2 \cos^4 t + r^2 \sin^4 t$$ This is not equal to $r^2$ in general.

ContentLicense: CC BY-SA 4.0


问:
Finding best players in a tournament with a probabilistic comparison function
I am currently facing the following problem in my research and I have no clue how to tackle this kind of question.

The problem

Imagine you have a tournament with $n$ players $P=\{p_1,...,p_n\}$. My goal is to determine one of the best players in my tournament.

I do have a comparison function $f: P x P\to \{0,1\}$ that can tell me which of two given players is better, i.e. $f(p_1,p_2)=1$ iff player two is better than player one and $f(p_1,p_2)=0$ iff player one is better than player two. You can think of $f$ as the $<$ relation.

The kicker is that my comparison function $f$ has an error, meaning that it will give me the correct result of my comparison with a probability $p>0.5$. Calculating $f$ will take some time and thus I want to find a good player for my tournament with the least amount of queries. My current approach is to compare all players with each other which gives me a total amount of $b \in O(n^2)$ comparison calls. I then chose the player $p_i$, which "won" the most comparisons.

Edit:

Please be aware that my comparison function will give me the same result for a call $f(p_i,p_j)$ no matter how often I call it. So the probability that the result is correct is $p$, but the function itself is deterministic. My example below is a bit misleading. However, each comparison call is only done once so this won't be a problem.

Key questions

What is the probability that the chosen player is the best player?

What is the probability that the chosen player is in the top k percent?

My thoughts

I think that question one might be easier to calculate as my best player will win all comparisons if $p=1$ and I can deduce the probability that $k$ comparisons were correct. However, I am stuck at the point at which I have to calculate the probability that it in fact is the player that "won" the most comparisons as others might be evaluated incorrectly.

My dream is to get a formula that allows me to calculate the desired probabilities for different $p,n$, and budget $b$.

Simulation

I wrote a small simulation in Python which revealed some interesting facts about the influence of $p$. In my example, the tournament players are represented as numbers $0,...,63$. The function $f$ is the standard $<$ relation with a given probability. In the plot below I have plotted the mean position (y-axis) that was selected as the best individual for different $p$ (x-axis). You can find the source code below.

```<code>import random
import numpy as np
from itertools import combinations
from tqdm import tqdm
import matplotlib.pyplot as plt

x, y = [], []

n = 64 # How many players
nums = np.arange(n).tolist() # Player strengths
count = 1000 # The amount of tests (O(n^2)) combinations that should be made

for p in tqdm(np.arange(0, 1, 0.01)):
    x.append(p)

    def compare(a, b):
        r = random.random()
        if r <= p:
            return a < b
        else:
            return a >= b

    def tournament():
        scores = [0] * n
        for a, b in combinations(nums, 2):
            result = compare(a, b)
            if result:
                scores[b] += 1
            else:
                scores[a] += 1

        best = max(nums, key=lambda x: scores[x])
        return best

    vals = []

    for _ in range(count):
        vals.append(tournament())

    y.append(np.mean(vals))

plt.plot(x, y)

plt.show()
</code>```

答:
In the following I suggest a framework where all relevant probabilities can be calculated - in theory. In practice the combinatorial explosion of possibilities will require additional heuristics and short cuts. But all these can be addressed in this framework.

Framework

To ensure that all relevant information can be processed, it is best to work on the set of all permutations. Every possible configuration of player strengths is captured by a permutation of order $n$. E.g. for $n=3$ the permutation $(2,3,1)$ means player 2 is worst player 1 is best and 3 is in-between. The unknown permutation describing the true strengths is called $\pi_*$.

A probability measure on permutations can be interpreted as $P(\pi) = P(\pi=\pi_*)$, i.e. giving the probability that $\pi$ is the true order of player strengths. From these basic probabilities all others can be derived. For example the probability of "Player $i$ is best" is given by the probability $P(A)$ where $A$ is the set of all permutations with player $i$ at the n-th (i.e. best) position or $A=\{\pi\mid \pi(n)=i\}$. It can be calculated as
$$ P(i \text{ is best}) = \sum_{\pi\in A}P(\pi).$$

Probabilities for other subsets of interest, such as "$i$ is among k-best", can be derived in the same fashion by adding up certain atomic probabilities.

Comparisons

Comparisons allow to access information about the true permutation. Due to the uncertainty of the comparisons, they have to be described as probabilities. Let $A(i,j)$ be the set of all permutations where $i$ is before $j$ and $\bar{A}(i,j)$ its complement. A comparison $f(p_i,p_j)$ which evaluates TRUE (with probability $p$), can then be interpreted as a condition on the measure $P$:
$$ P(A(i,j)) = p \text{ and accordingly } P(\bar{A}(i,j)) = 1-p.$$

If $f(p_i,p_j)$ evaluates FALSE the conditions are reversed:
$$ P(A(i,j)) = 1-p \text{ and } P(\bar{A}(i,j)) = p. $$

Performing several comparisons $c_1,\ldots,c_M$ just leads to more, i.e. $M$, conditions.

Example:

Take $n=4$ and the two comparisons between 1 and 2 as well as between 3 and 4.
Denote by $A_{00}, A_{01}, A_{10}, A_{11}$ the partition of all permutations where $A_{00} = A(1,2)\cap A(3,4)$ and $A_{10} = \bar{A}(1,2)\cap A(3,4)$ and so on. If $f(p_1,p_2)=TRUE$ and $f(p_3,p_4)=FALSE$ then $P(A_{00})=P(A_{11})=p(1-p)$, $P(A_{01})=p^2$ and $P(A_{10})=(1 - p)^2.$

Note that all conditions are linear constraints on the atomic probabilities.

Unique compatible measure

Since all constraints are linear, the atomic probabilities $q_1=P(\pi_1), \ldots, q_{n!}=P(\pi_{n!})$ compatible with the comparisons can be found by solving a system of linear equations together with the probability constraints $q_N\geq 0$ and $\sum q_N=1$. Depending on the number of comparisons this system may be underdetermined, and will not have a unique solution. The example above is a case in point. Each group $A_{xx}$ contains 6 permutations. But there is no information about the probabilities within each set. A standard trick which enforces uniqueness and is consistent with our ignorance about probabilities within each group is to choose among all measures compatible with the constraints the measure with maximal entropy.

Numeric example

Entropy maximisation with linear constraints can be solved by convex solvers. The following examples were calculated using CVRX interface and the ECOS solver in R.

Parameters were $n=4$ and all 6 round-robin comparisons "allcomp" and "inccomp" three comparisons $1<2$, $2<3$, $3<4$. The table shows the probabilities of $i=1,2,3,4$ being the best assuming all comparisons evaluate TRUE.

For $p=0.9$:

```<code>|   |   allcomp|   inccomp|
|:--|---------:|---------:|
|1  | 0.0402612| 0.0345995|
|2  | 0.0404158| 0.0535842|
|3  | 0.0983847| 0.1008493|
|4  | 0.8209382| 0.8109670|
</code>```

For $p=0.6$

```<code>|   |   allcomp|   inccomp|
|:--|---------:|---------:|
|1  | 0.1752993| 0.1753445|
|2  | 0.2063212| 0.2115260|
|3  | 0.2614596| 0.2640440|
|4  | 0.3569198| 0.3490855|
</code>```

Further considerations

Better than static approaches where comparisons are defined in advance and are independent of the outcomes, seem dynamic approaches to me. With these later comparisons are determined by earlier outcomes. This makes greedy and divide-and-conquer approaches possible.

Body:
In the following I suggest a framework where all relevant probabilities can be calculated - in theory. In practice the combinatorial explosion of possibilities will require additional heuristics and short cuts. But all these can be addressed in this framework.

Framework

To ensure that all relevant information can be processed, it is best to work on the set of all permutations. Every possible configuration of player strengths is captured by a permutation of order $n$. E.g. for $n=3$ the permutation $(2,3,1)$ means player 2 is worst player 1 is best and 3 is in-between. The unknown permutation describing the true strengths is called $\pi_*$.

A probability measure on permutations can be interpreted as $P(\pi) = P(\pi=\pi_*)$, i.e. giving the probability that $\pi$ is the true order of player strengths. From these basic probabilities all others can be derived. For example the probability of "Player $i$ is best" is given by the probability $P(A)$ where $A$ is the set of all permutations with player $i$ at the n-th (i.e. best) position or $A=\{\pi\mid \pi(n)=i\}$. It can be calculated as
$$ P(i \text{ is best}) = \sum_{\pi\in A}P(\pi).$$

Probabilities for other subsets of interest, such as "$i$ is among k-best", can be derived in the same fashion by adding up certain atomic probabilities.

Comparisons

Comparisons allow to access information about the true permutation. Due to the uncertainty of the comparisons, they have to be described as probabilities. Let $A(i,j)$ be the set of all permutations where $i$ is before $j$ and $\bar{A}(i,j)$ its complement. A comparison $f(p_i,p_j)$ which evaluates TRUE (with probability $p$), can then be interpreted as a condition on the measure $P$:
$$ P(A(i,j)) = p \text{ and accordingly } P(\bar{A}(i,j)) = 1-p.$$

If $f(p_i,p_j)$ evaluates FALSE the conditions are reversed:
$$ P(A(i,j)) = 1-p \text{ and } P(\bar{A}(i,j)) = p. $$

Performing several comparisons $c_1,\ldots,c_M$ just leads to more, i.e. $M$, conditions.

Example:

Take $n=4$ and the two comparisons between 1 and 2 as well as between 3 and 4.
Denote by $A_{00}, A_{01}, A_{10}, A_{11}$ the partition of all permutations where $A_{00} = A(1,2)\cap A(3,4)$ and $A_{10} = \bar{A}(1,2)\cap A(3,4)$ and so on. If $f(p_1,p_2)=TRUE$ and $f(p_3,p_4)=FALSE$ then $P(A_{00})=P(A_{11})=p(1-p)$, $P(A_{01})=p^2$ and $P(A_{10})=(1 - p)^2.$

Note that all conditions are linear constraints on the atomic probabilities.

Unique compatible measure

Since all constraints are linear, the atomic probabilities $q_1=P(\pi_1), \ldots, q_{n!}=P(\pi_{n!})$ compatible with the comparisons can be found by solving a system of linear equations together with the probability constraints $q_N\geq 0$ and $\sum q_N=1$. Depending on the number of comparisons this system may be underdetermined, and will not have a unique solution. The example above is a case in point. Each group $A_{xx}$ contains 6 permutations. But there is no information about the probabilities within each set. A standard trick which enforces uniqueness and is consistent with our ignorance about probabilities within each group is to choose among all measures compatible with the constraints the measure with maximal entropy.

Numeric example

Entropy maximisation with linear constraints can be solved by convex solvers. The following examples were calculated using CVRX interface and the ECOS solver in R.

Parameters were $n=4$ and all 6 round-robin comparisons "allcomp" and "inccomp" three comparisons $1<2$, $2<3$, $3<4$. The table shows the probabilities of $i=1,2,3,4$ being the best assuming all comparisons evaluate TRUE.

For $p=0.9$:

|   |   allcomp|   inccomp|
|:--|---------:|---------:|
|1  | 0.0402612| 0.0345995|
|2  | 0.0404158| 0.0535842|
|3  | 0.0983847| 0.1008493|
|4  | 0.8209382| 0.8109670|

For $p=0.6$

|   |   allcomp|   inccomp|
|:--|---------:|---------:|
|1  | 0.1752993| 0.1753445|
|2  | 0.2063212| 0.2115260|
|3  | 0.2614596| 0.2640440|
|4  | 0.3569198| 0.3490855|

Further considerations

Better than static approaches where comparisons are defined in advance and are independent of the outcomes, seem dynamic approaches to me. With these later comparisons are determined by earlier outcomes. This makes greedy and divide-and-conquer approaches possible.


问:
absolutely continuous with respect to a finite measure, then $\Sigma$-finite
This is a question from Erhan Cinlar's Probability and Stochastics book:

If $\mu$ is absolutely continuous with respect to a finite measure $\nu$, then $\mu$ is $\Sigma$-finite.

Can i have a hint on how i should construct the sequence of finite measure that sum to $\mu$?

答: This is false. Let $X=\mathbb N$,  $\mu (\emptyset)=0$ and $\mu (E)=\infty$ for all non-empty subsets of $X$. Let $\nu (E)=\sum_{n \in E} \frac 1 {2^{n}}$. then $\nu$ is a  finite measure and $\mu <<\nu$. But $\mu$ is not sigma finite.


问:
The rigorous definition of continuity in a topological space
Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.

Let $x \in S_1$.

In this page Proof Wiki, it is said that the mapping $f$ is continuous at (the point) $x$ (with respect to the topologies $\tau_1$ and $\tau_2$) if and only if: For every neighborhood $N$ of $f(x)$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f[M]\subseteq N$.

But in my knowledge, the condition should be $\forall y \in S_1 $, $y \in M \Rightarrow f(y) \in N$ instead of $f[M]\subseteq N$. The problem is that the two statements are not equivalent: the former statement is equivalent to $M \subseteq f^{-1}[N]$ because
$$
y\in M \iff \{y\} \subseteq M \implies f[y] \subseteq f[M] \iff f(y) \in f[M]
$$
As a result, we have a condition $f(y) \in f[M] \Rightarrow f(y) \in N$, which is equivalent to $M \subseteq f^{-1}[N]$.

Would you check this out?

答:
I realized that
$$f[M] \subseteq N\iff M \subseteq f^{-1}[N]$$.
I am sorry for the noise.


问:
10 to the power of 3.5: $10^{3.5}$
So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!

答:
$10^{3.5}$ is equal to $10*10*10*10^{0.5}$. So you just need to know what $10^{0.5}$ is.

One of the property of exponent is this.

$(10^x)^y = 10^{xy}$, i.e. the power of a power, is just the exponents multiplied.

So if $(10^{0.5})^2=x^2=10^1$ when $x=10^{0.5}$ then we just solve for $x^2 = 10$.

I'm not sure if you learned about square roots, yet, but $x=\sqrt{10}$.

Since $\sqrt{10}$ is approximately equal to 3.16227, your calculator gives you $10^{3.5} = 3162.27...$.

ContentLicense: CC BY-SA 3.0

Title: 10 to the power of 3.5: $10^{3.5}$


问:
$u_n\to u \implies u_n'(x)\to u'(x)$ if $u_n, u\in C^1(I)$?
Let $I$ be a closed bounded interval (will this matter?). Let $u_n(x)$ be a sequence of $C^1(I)$ functions such that $u_n\to u\in C^1(I)$. It seems we should have in that case (is it true?) that $u_n'(x)\to u'(x)$. But I am not sure if or why I can swap the two limits $\lim\limits_{n\to \infty}u_n'(x)=\lim\limits_{n\to \infty}\lim\limits_{h\to 0}\frac{u_n(x+h)-u_n(x)}{h}=\lim\limits_{h\to 0}\lim\limits_{n\to \infty}\frac{u_n(x+h)-u_n(x)}{h}=u'(x)$.

答:
Whether this is true depends on what you mean by the hypothesis $u_n\to u$. If you mean convergence in the usual norm for $C^1(I)$ then this implies $u_n'\to u'$ more or less just by definition. But assuming just $u_n\to u$ uniformly is not enough.

For example let $$u_n(t)=\sin(n^2t)/n.$$Then $u_n\to0$ uniformly on $\Bbb R$ but $u_n'(0)$ diverges as $n\to\infty$.


问:
About the roots of the derivative of a special polynomial
Let $p$ be an odd prime, and let $n_1,\dots,n_{p-2}, m$ be even integers such that $n_1 < n_2 < \dots < n_{p-2}$ and
\begin{equation}
2m > \sum_{i=1}^{p-2} n_i^2.
\end{equation}
Consider the polynomial
\begin{equation}
g(x)=(x^2 + m)(x - n_1) \dots (x - n_{p-2}).
\end{equation}
From Rolle's Theorem, we know that for each $i=1,2,\dots,p-3$, there exists $x_i \in (n_i,n_{i+1})$ such that $g'(x_i)=0$. So $g'(x)$ has at least $p-3$ distinct real zeroes. My question is: can $g'(x)$ have more than $p-3$ distinct real zeroes?

I do not know the answer, but for sure the constraints on the parameters are relevant here. For example the polynomial $g(x)=(x^2+1)(x-4)(x-2)(x+2)$ has derivative $g'(x)=5x^4-16x^3-9x^2+24x-4=(x-1)(5x^3-11x^2-20x+4)$ which has four distinct real roots, as you can check on WolframAlpha.

NOTE This strange polynomial arises in the construction given by R. Brauer of a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $p$ whose Galois group over $\mathbb{Q}$ is isomorphic to the symmetric group $\mathcal{S}_p$: see Jacobson, Basic Algebra I, $\S 4.10$. The question I asked is clearly irrelevant for the construction, but has intrigued me, since I could not answer it in the negative nor I could find some counterexample.

答:
Expanding $g'(x)$, we have
$$g'(x) = px^{p-1} - (p-1)A x^{p-2} + (p-2)B x^{p-3} + \cdots$$
where $A = \sum\limits_k n_k$ and $B = m + \sum\limits_{i < j}n_in_j$ and indicies $i,j,k$ run over $\{ 1, \ldots, p-2 \}$.

Notice $g'(x)$ is a polynomial with real coefficients and degree $p-1$. If it has more than $p-3$ real roots, then all its roots are real.
By Newton's inequalities, $A$ and $B$ need to satisfy

$$\left(\frac{(p-1)A}{\binom{p-1}{1}}\right)^2 \ge p\frac{(p-2)B}{\binom{p-1}{2}}
\quad\iff\quad A^2 \ge \frac{2p}{p-1}B$$
Substitute above expression of $A, B$ into RHS, the condition can be reexpressed as
$$\begin{align} & (p-1)\left(\sum_k n_k\right)^2 \ge 2p \left(m + \sum_{i<j} n_i n_j\right)\\
\iff & (p-1)\sum_k n_k^2 \ge 2pm + 2\sum_{i<j} n_i n_j\\
\iff & p \sum_k n_k^2 \ge 2pm + \left(\sum_k n_k\right)^2
\end{align}
$$
When $2m > \sum\limits_k n_k^2$, the last condition cannot be satisfied. The coefficients of $g'(x)$ doesn't satisfy Newton's inequalities. As a result, $g'(x)$ has some complex roots. From these, we can conclude

When $2m > \sum\limits_k n_k^2$, $g'(x)$ has at most and hence exactly $p-3$ real roots.


问:
Use the two properties of $S$ to show that $S =\mathbb N$
Assume $S \subseteq \mathbb N$ such that:

$\forall k \in \mathbb N :2^k \in S$

$\forall k \ge 1: k \in S \implies k-1 \in S$

Show that $S =\mathbb N$.

It's left to show that $\mathbb N \subseteq S$.

Let $P(n)$ be the proposition such that $\left\{ 2^k-n, ...,2^k\right\} \subseteq S$.

The base case is true since $P(0)$ means $\left\{ 2^k\right\} \subseteq S$ which is true by the hypothesis.

Assume $P(n)$ does hold and consider $P(n+1)$,from the assumption $\left\{ 2^k-n, ...,2^k\right\} \subseteq S$ which implies $2^k-n \in S$ From the second property of $S$ it's clear that $2^k-(n+1)= 2^k-n-1 \in S$ and so $\left\{ 2^k-(n+1), ...,2^k\right\} \subseteq S$.
Since this is true for all natural $n$ so $S =\mathbb N$.

Edit:

Another way I tried:
Let $P(n)$ be the proposition such that $\left\{ 0, ...,2^n\right\} \subseteq S$.

The base case is true since $2^0 \in S$ which is true by the first property of $S$ and from the other property, we see that $0 \in S$.

Assume $P(n)$ does hold and consider $P(n+1)$,from the assumption $\left\{ 0, ...,2^n\right\} \subseteq S$.The set $\left\{0,...,2^{n+1}\right\}$ can be written as $\left\{0,...,2^{n}\right\} \cup \left\{2^{n}+1,...,2^{n+1}\right\}$,again from the first property of $S$, $2^{n+1} \in S$,the second property implies $2^{n+1}-1 \in S$,continuing this way we see that $\left\{2^{n}+1,...,2^{n+1}\right\} \subseteq S$,and so $$\left\{0,...,2^{n+1}\right\}=\left\{0,...,2^{n}\right\} \cup \left\{2^{n}+1,...,2^{n+1}\right\}\subseteq S$$

It would be highly appreciated if someone checks the validity of the proofs.

答:
You might consider the following:
Let $n \in \mathbb{N}$. If $n$ is a power of two we are done. Otherwise $n=2^ka$ with $(a,2)=1$ for some $k$. Since $2^a > a$ for all $a$ by a simple induction, $2^ka<2^{k+a} \in S$. and by property $(2)$, $n \in S$.


问:
In how many ways can 3 dice add up to 16?
I am trying to use the stars and bars technique to calculate the number of ways 3 dice rolls can add up to 16. To my understanding, this can rephrased as:

$$x + y + z = 16 | x,y,z \in \mathbb{Z}^+ \land x,y,z \leq 6$$

I am not anywhere close to being a mathematician, so simple solutions are much appreciated. My main motivation to the question, is that I am trying to see if there is a cooler way to do it rather than count by hand, and stars and bars came to my mind as a potential solution. However, other simple solutions are also welcome.

答:
We wish to find the number of solutions of the equation
$$x + y + z = 16 \tag{1}$$
in the positive integers subject to the constraints that $x, y, z \leq 6$.  Observe that since $x, y, z \leq 6$,
\begin{align*}
x' & = 7 - x\\
y' & = 7 - y\\
z' & = 7 - z
\end{align*}
are also positive integers which are at most $6$.  Substituting $7 - x'$ for $x$, $7 - y'$ for $y$, and $7 - z'$ for $z$ in equation 1 yields
\begin{align*}
7 - x' + 7 - y' + 7 - z' & = 16\\
-x' - y' - z' & = -5\\
x' + y' + z' & = 5\tag{2}
\end{align*}
which is an equation in the positive integers. Moreover, we cannot violate the constraints that $x', y', z' \leq 6$.

A particular solution in the positive integers corresponds to the placement of $3 - 1 = 2$ addition signs in the $5 - 1 = 4$ spaces between successive ones in a row of five ones.
$$1 \square 1 \square 1 \square 1 \square 1$$
For instance, placing addition signs in the second and third spaces corresponds to the solution $x' = 2$, $y' = 1$, $z' = 2$ of equation 2 and $x = 5$, $y = 6$, and $z = 6$ of equation 1.  The number of such solutions is the number of ways we can select two of those four spaces in which to place an addition sign.

 $$\binom{5 - 1}{3 - 1} = \binom{4}{2} = 6$$


问:
Is Radon-Nikodym derivative with respect to a finite measure real-valued a.e.?
This question comes from this question. The answer therein missed an argument that the Radon-Nikodym derivative is real-valued a.e. Without this, the proof in that answer has flaw because either the sum of $f_n$ is not equal to $f$ (a.e.) or the measure corresponds to $+\infty$ is not finite. The following is a complete formulation of my question.

On an arbitrary measurable space $(E,\mathcal{E})$, $\mu\ll\nu$ and
$\nu$ is a finite measure. Let $p$ denote the Radon-Nikodym derivative
$d\mu/d\nu$. Show that $p$ is real-valued $\nu$-almost everywhere.

I can find no way to exclude the case that $\nu(\{x\in E|p(x)=+\infty\})=0$. Can you please help me show that this measure is zero? Thanks a lot.

答:
Application of Radon Nikodym Theorem requires sigma finiteness of $\mu$. Under this condition we can argue as follows:
Let $\mu(A_n) <\infty$ and $X =\cup_n A_n$. Then $\int_{A_n} fd\nu <\infty$ so $f <\infty$ a.e. $[\nu]$ on on $A_n$. This true for each $n$ and hence $f <\infty$ almost everywhere $[\nu]$.

A counter-example:
Let $X=\mathbb N$,  $\mu (\emptyset)=0$ and $\mu (E)=\infty$ for all non-empty subsets of $X$. Let $\nu (E)=\sum_{n \in E} \frac 1 {2^{n}}$. Then $\nu$ is a  finite measure and $\mu <<\nu$. If there exist $f$ such that $\mu(E)=\int_Efd\nu$ for all $E$ then $\infty=\mu(\{n\})=\int_{\{n\}}fd\nu=f(n) \frac 1 {2^n}$, so $f(n)=\infty$ for all $n$.


问:
Relating condition number of hessian to the rate of convergence
While minimizing a Lipschitz continuous strongly convex functions, the rate of convergence of the gradient descent method depends on the condition number of the hessian of the function, where a high condition number leads to slow convergence. Can someone give an intuitive explanation explaining why the condition number of the hessian and the rate of convergence are related?</p>

答:
To give a more visual answer, would suggest you download and play with this fantastic package (I am not the author): https://github.com/lilipads/gradient_descent_viz

In particular, try out gradient descent (both the continuous and stepwise versions). Here is a visualization I generated using this package (also see the full resolution version):

Let the function we are trying to minimize (which generates this convex paraboliod surface) be $f(x)$. Recall that if the surface is an upward-facing paraboloid, it is given by the vector-valued function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, where $f(x) = \frac{x^TAx}{2} - x^Tb$. Things to keep in mind:

$A \in \mathbb{R}^{n \times n}$ is a particular kind of matrix called Symmetric Positive Definite. An SPD matrix is analogous to a positive scalar ($\alpha > 0$) when the number of dimensions is $n \gt 1$. The important property is that $x^TAx > 0, \forall x\neq 0$, which is what makes the parabolid open upwards rather than downwards. The values in $A$ determine how curved the paraboloid is.

$b$ is some vector which we assume to be non-zero. Varying b changes what the minima of this surface is.

The vector which gives the direction of the (negative) gradient at any point $x \in \mathbb{R}^n$ is: $ - \nabla f(x) = b - Ax$.

In our current discussion, $x \in \mathbb{R}^2$.

Notice that in the screencap below, we are currently at $x = (\chi_0, \chi_1) = (-1.68, 0.97)$. Even though $f(x)$ is convex, the gradient (black arrow) does not point towards the minima (red region). You can conjecture that, given a certain $A$ and $b$, there are many values of $x$ which cause this to happen, particularly when $\chi_0$ and $\chi_1$ have relatively large magnitude.

Now, when we update the ball to the second location in the video, we are at $x = (-0.56, 0.04)$. Notice that $\chi_1$ here is much smaller than before (0.97 to 0.04). As a result, when we compute the gradient at our second location, is mostly along the $\chi_0$ (notice the stubby little blue arrow pointing east).

Whether or not the gradient vector points in the direction of the minima is immaterial of where the minima is located; it is simply the tangent to the loss surface at some point $x$.

One case where the gradient vector definitely points to the minima is when $A$ is the identity matrix and $b=0$, then the gradient vector is $-x$, which will point exactly at the minima (which will be the origin).

Now, the previous surface was nicely convex: the width and height of the parabola were roughly equal, and we could converge quickly.

Let's consider an ill-conditioned problem, i.e. due to some skewed values in $A$, the paraboloid $f(x)$ is highly stretched along one axis ($\chi_0$ as per the image below). In this case, the value $f(x)$ will fluctuate rapidly with small changes in $\chi_1$, i.e. the component of the gradient vector $- \nabla f(x)$ corresponding to $\chi_1$ is large. This means the gradient vector will mostly point along $\chi_1$, which may or may not be in the direction of the minima.

On the other hand, assume even that in this stretched parabola, we start due south of the minima. In that case, the gradient will generally point towards of the minima, so gradient descent will converge quickly. The issue occurs when you start somewhere on the far left or the far right. In particular, if your random initialization starts you at some point far from the minima in $\chi_0$ (the stretched axis) then you start getting the zig-zag pattern.

To avoid this, we often precondition the problem, so that the contour plots looks like circles. In that case, no matter where you start, convergence takes roughly the same amount of (short) time.

(image credit: https://mc.ai/my-machine-learning-diary-day-65/)


问:
Transformation and its adjoint
For $V$ a vector space.

Let $T\colon V\rightarrow V$ be a linear transformation of a finite dimensional inner product space.

How can I prove that $\ker(T^* T + TT^*) = $\ker(T)$\cap$$\ker(T^*)$?

And what is the meaning of $\ker(T^*T + TT^*)$ ?

Is  $\ker(T^*T + TT^*)=\ker(T^*T)+\ker(TT^*)$ ?

答:
$\ker(T^*T+TT^*)$ is the set of all $x\in V$ such that $$(T^*T+TT^*)x=0, $$
that is
$$ T^*Tx+TT^*x=0.$$
This is not the same as the set of all vectors $x$ that can writtten as $x=y+z$ with $T^*Ty=0$ and $TT^*z=0$. I.e., in general $\ker(T^*T+TT^*)\ne\ker(T^*T)+\ker(TT^*)$.</p>

Title: Transformation and its adjoint


问:
Why is the following method incorrect when deciding on the possible choices?
"Six men and three women are standing in a supermarket queue. Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?"

I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.

$$
 {3 \choose 1} \ \times \ {6 \choose 2} \ + \  {3 \choose 2} \ \times \ {6 \choose 1}  \ + \  {3 \choose 3}
$$

Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.

This approach is in fact correct. However, another approach came to my mind as well.

If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.

$$
{3 \choose 1} \ \times \ {8 \choose 2}
$$

This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.

答:
There is a much simpler approach. "At least one woman" is the opposite of "zero women" or "all men" from the total, so subtracting from the total gives us:

$${9 \choose 3} - {6 \choose 3} = 64$$

If you are interested in another subtractive approach, following from N. F. Taussig's answer, you can again subtract from ${3 \choose 1} \cdot {8 \choose 2} = 84$.

The cases where there are two women and one man (Abigail, Beatrice, Charles) are counted $2$ times, when they should only be counted $1$ time. Therefore, you need to subtract the number of combinations of this type $1$ time. This is just ${3 \choose 2} \cdot 6 = 18$.

Similarly, the cases where there are three women are counted $3$ times, when they should only be counted $1$ time, as there is only one combination possible with all $3$ women. Therefore, the number of combinations should be subtracted by $2$.

Hence $84 - 18 - 2 = 64$, which is the correct answer.


问:
Binomial theorem real world use
I understand binomial theorem helps expand and calculate two terms  raised to nth power (a+b)^n easily.
Can someone explain briefly how they are used and applied in a real world application? I see lot of mentions about their use in weather forecasting, IP subnetting, economic forecast etc. But couldn't find anything more than names of applications that uses it.

Would love to see its usage explained at a high level in any one of the real world application.

答:
I don't think this is an answer to your question so feel free to downvote..

I've thought about this question a lot...but the truth is that the binomial theorem really isn't the important part. It's the binomial coefficient that's the more important part. The binomial theorem is a result of the binomial coefficient!

So how is that? We have $(a + b)^n$ which means $(a + b)(a + b)\dots(a+b)$.

Let's look at $(a + b)^2$:

$$
(a + b)(a + b)= a(a + b) + b(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2
$$

Now let's do $(a + b)^3 = (a + b)(a + b)^2$:

$$
a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3= a^3 + 3a^2b + 3ab^2 + b^3
$$

Ok, so why is it that the binomial coefficient keeps showing up in the binomial theorem?

The reason has to do with combinatorics. Think of it this way: if I have $(a + b)^n$, then I can get $a\cdot a\cdot a\  ...\cdot a$, $n$ times, by choosing $a$ each time from the $n$ number of $(a + b)$'s or I can get some combination of $a$'s and $b$'s, e.g. $a^kb^{n - k}$ (where $0 \leq k \leq n$) by choosing $k$ of the $(a + b)$'s to be $a$ and the rest $b$.

So now the question is if I choose $a^kb^{n - k}$, how many ways can I get that by choosing among the $n$ possible $(a + b)$'s? The answer is the binomial coefficient. Think of it like this, there are $n$ distinct $a$'s and $n$ distinct $b$'s. How many different $k$-groups of $a$'s can I get by choosing among the $n$ distinct $a$'s? Easy: $\binom{n}{k}$. Meaning that if I choose $a^k$ there will be $\binom{n}{k}$ different combinations of the different $(a +b)$ groups that can give me $a^k$ with the "rest" being $b^{n - k}$, giving the binomial theorem:

$$
(a + b)^n \rightarrow\ ?\ +\ ?\ +\ ...\ +\ \binom{n}{k}a^kb^{n - k}\ +\ ?\ ...\ +\ ?
$$

Since we can obviously do that for every possible combination of $a$ and $b$ we finally get:

$$
(a + b)^n = \sum\limits_{k = 0}^{n} \binom{n}{k}a^kb^{n - k}
$$

Here's one last concrete example, let's say I have $(a + b)^5$ and I want to find the coefficient for $a^3b^2$. It's going to be $\binom{5}{3} = 10$. We can see this by enumerating all of the possible ways of choosing $a^3b^2$:

$$
aaabb + aabab + aabba + abaab + ababa + abbaa + baaab + baaba + babaa + bbaaa\\ $$

$$
=10a^3b^2
$$

Title: Binomial theorem real world use


问:
Check ALL of the following Cartesian products to find where element belongs
Q:
A =  {3,5,7}
B = {4,5}
Check ALL of the following Cartesian products to which the following element belong:
(5,5)

ANS:

I got
{3,4}
{5,4}
{7,4}
{3,5}
{5,5}
{7,5}

But like what element does {5,5} belong to? The options are

BxB
BxA
AxA
AxB

I was thinking either AxA or BxB or ??

答: They both have a $5$ so $(5,5)$ is an element of every one of those Cartesian products.


答:
Do you know what the cartesian product is?

The cartesian product, AxB is the set of all possible ordered pairs (a,b).

What pairs do you get by creating the cartesian product of your options?

Hint: it's in all of them. Do you see why?

Title: Check ALL of the following Cartesian products to find where element belongs

Body:
Q:
A =  {3,5,7}
B = {4,5}
Check ALL of the following Cartesian products to which the following element belong:
(5,5)

ANS:

I got
{3,4}
{5,4}
{7,4}
{3,5}
{5,5}
{7,5}

But like what element does {5,5} belong to? The options are

BxB
BxA
AxA
AxB

I was thinking either AxA or BxB or ??


问:
Prove $AA^+$ projects onto the column space of $A$

Prove that $AA^+$ is the projection operator onto the column space of $A$

If $A$ has independent column vectors $A^TA$ is invertible and the projection operator onto the column space of $A$ is $P=A(A^TA)^{-1}A^T$.

It is stated that, If $A$ has linearly independent columns, then the pseudoinverse of $A$ becomes $A^+=(A^TA)^{-1}A^T$ and $AA^+=A(A^TA)^{-1}A^T$ is the projection operator on to the column space of $A$.

I know where the first statement comes from in the case for $A$ has independent column vectors, but how do I bring in the pseudoinverse into projection ?

Note: Similar question had been asked Show $AA^+$ is projection matrix onto the column space but the post failed to address the mentioned doubts.

$$
(AA^+)^2=AA^+AA^+=I\\
(AA^+)^T=(U\Sigma V^TV\Sigma^+U^T)^T=(U\Sigma \Sigma^+U^T)^T=U(\Sigma\Sigma^+)^TU^T=U(\Sigma\Sigma^+)U^T\\
=U\Sigma V^TV\Sigma^+U^T=AA^+
$$
$\implies AA^+$ is an orthogonal projector, but how do we show that it projects on to the column space of $A$ ?

Is it enough ?

If $y\in C(A)$, ie., $y=Ax$, for some $x$
$$PA=AA^+A=U\Sigma V^TV\Sigma^+ U^TU\Sigma V^T=U\Sigma\Sigma^+\Sigma V^T=U\Sigma V^T=A\\
Py=PAx=Ax=y$$

If $x\in N(A^T)$, ie., $A^Tx=0$ then
$$(I-P)x=(I-AA^+)x=x-AA^+x=x-$$

答:
A matrix $P$ is an orthogonal projection if and only if $P^2=P$ (idempotence) and $P^T=P$ (symmetry).  The subspace that $P$ projects onto is its image/column space.  So if we wish to show that $AA^+$ is projection onto the column space of $A$, we need to show 4 things:

$AA^+AA^+=AA^+$

$(AA^+)^T=AA^+$

$C(A)\subset C(AA^+)$

$C(AA^+) \subset C(A)$

You proved the first 3 of these.  The fourth is a general fact: no matter what $B$ is (assuming the multiplication makes sense), $C(AB)\subset C(A)$.


答:
We start with: $$x+y+z=16$$ where $x,y,z$ are positive integers with $x,y,z\leq 6$.

Then we make it: $$x'+y'+z'=13$$ where $x',y',z'$ are nonnegative integers with $x',y',z'\leq 5$.

(here we take $x'=x-1$, $y'=y-1$ and $z'=z-1$)

Finally we make it: $$x''+y''+z''=15-13=2$$ where $x'',y'',z''$ are nonnegative integers with $x'',y'',z''\leq 5$.

(here we take $x''=5-x'$, $y''=5-y'$ and $z''=5-z$)

Now observe that the extra condition $x'',y'',z''\leq 5$ can be neglected in this setting because that is a consequence of the other conditions.

So we can apply stars and bars and find: $$\binom{2+3-1}{3-1}=6$$ solutions.

This is a trick that works here but not always.


问:
radical of sum of two ideals
$I$ and $J$ are ideals in $k[x_1,\cdots,x_n]$.

Show that $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$.

I have no idea how to prove it. Can someone help?

答:
"$\subset$" is trivial. For the other inclusion, assume $x \in \sqrt{\sqrt{I} + \sqrt{J}}$.
Then there exists $m$ such that $x^m \in \sqrt{I} + \sqrt{J}$. Say $x^m = y + z$ with $y^n \in I$ and $z^k \in J$. What can you say about $(x^m)^{n + k - 1}$?</p>

Title: radical of sum of two ideals


问:
A limit with an arctan
I need to calculate another limit, $$ \underset{x\rightarrow\infty}{\lim}\left(\frac{2}{\pi}\arctan x\right)^{\sin\left(2/x\right)}$$

I have no idea how to start or what to do because I've never dealt with an arctan in a limit before.

What should I do first?

答:
$\lim_{x \to \pi/2} \tan x = \infty$, so $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$, as $\tan x$ and $\arctan x$ are inverse functions of each other for $x \in [0, \pi/2)$.

The rest is just direct substitution. You should get:

 $$\lim_{x \to \infty} \left(\frac{2}{\pi} \cdot \frac{\pi}{2} \right)^{\sin(0)} = \lim_{x \to \infty} 1^0 = \boxed{1}.$$

Title: A limit with an arctan


问:
Equation of a Equilateral Triangle in $\mathbb{R}^{2}$
I need to find a metric where a Sphere of radius $r$ is a equilateral triangle, but I think it is necessary to find a way to describe a equilateral triangle in $\mathbb{R}^{2}$, just like we do with the ellipse and hyperbole.

答:
Let us assume that $ABC$ has direct orientation.

Let us make a detour by complex considerations.

Let $\omega=\tfrac12(-1+i \sqrt{3})$ (a cubic root of unity). [please note that $\overline{\omega}=\omega^2$].

A classical necessary and sufficient condition for a triangle with direct orientation with vertices $a,b,c \in \mathbb{C}$ to be equilateral is

$$a+b\omega+c \overline{\omega}=0\tag{1}$$

(see here).

Now convert (1), after having multiplied it by 2, under a real form:

$$2(x_A+iy_A)+(x_B+iy_B)(-1+i \sqrt{3})+(x_C+iy_C)(-1-i \sqrt{3})=0\tag{2}$$

We can stay there or expand (2) and equate separately its real and imaginary part to $0$, giving two simultaneous (necessary and sufficient) conditions for being an equilateral triangle:

$$\begin{cases}2x_A-x_B-\sqrt{3}y_B-x_C+\sqrt{3}y_C&=&0\\
2y_A-y_B+\sqrt{3}x_B-y_C-\sqrt{3}x_C&=&0
\end{cases}\tag{3}$$


问:
Balls into bins question. Where is the mistake?
I am not able to figure out a loophole in my argument related to popular "Balls into bins" problem.

Suppose there are $m$ balls and $n$ bins. The $m$ balls are thrown independently and uniformly at random into these $n$ bins. What is the probability that the first bin is empty ?

The answer to this question mentioned in : Probability that first ball is empty
is $\Big(1-\frac{1}{n}\Big)^m$ which seems fine.

This answer should match with the probability found using manual counting method.

My argument is:
The $m$ balls can be distributed into $n$ bins in ${m+n-1} \choose {n-1} $ ways. If first bin is empty, then $m$ balls can be distributed into remaining $(n-1)$ bins in ${m+n-2} \choose {n-2} $ ways. This implies,
$$P(\text{first bin is empty})=\frac{{m+n-2} \choose {n-2}}{ {m+n-1} \choose {n-1} }=\frac{n-1}{m+n-1}$$

The values of both probabilities are very close to each other but have different answers. What is wrong with my argument ?

答:
If you want a counting argument then each case must be equally likely.

Your ${m+n-1 \choose n-1}$ ways of distributing $m$ balls into $n$ bins are not equally likely.  There are $n^m$ equally likely ways of distributing the balls but some look similar to each other.

Take a simple example of $m=3,n=2$.  Then ${m+n-1 \choose n-1}= {4 \choose 1}=4$ wheile $n^m = 2^3=8$.  The different ways are:

Three in the first bin, zero in the second bin $(AAA)$ with probability $\frac18$

Two in the first bin, one in the second bin $(AAB, ABA, BAA)$  with probability $\frac38$

One in the first bin, two in the second bin $(ABB, BAB, BBA)$   with probability $\frac38$

Zero in the first bin, three in the second bin $(BBB)$ with probability $\frac18$

This last case is leaves the first bin empty and has probability $\frac18 = \left(1-\frac12\right)^3$.


问:
Calculating acceleration based on a starting and ending distance
I have been trying to figure this problem out and just can't.

The answer is supposed to be (A): 1.0 m/s/s.

Can anyone tell me how they got that answer?  I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer.  What am I doing wrong? Or is the correct answer not listed?  Thanks for any help.

Problem:
  A Pinewood Derby car rolls down a track.   The total distance travel is measured each second.   Use the chart to calculate the acceleration of the car.

Time(s) |  Dist(m)

0       |     0

1       |     1

2       |     3

3       |     6

4       |    10

A) 1.0  meters/sec/sec

B) 1.5  meters/sec/sec

C) 2.0 meters/sec/sec

D) 2.5 meters/sec/sec

答:
If $x(t)$ denotes the total distance travelled at time $t$, then you can check (according to your data) that

$x(t) = \dfrac{t^2+t}2$

where $x$ is in m, $t$ in s. The acceleration is given by $a = \ddot x$, hence $a = 1\,\text m/\text s/\text s$.


问:
Find a linear transformation that $T((1,-1, 1))=(1, 0)$ and $T((1,1, 1))=(0, 1)$
Is it possible to find a linear transformation
$T\colon\mathbb{R}^3\to\mathbb{R}^2$ only knowing this:  $T((1,-1, 1))=(1, 0)$ and  $T((1,1, 1))=(0, 1)$ ?

答:
Such transformation is defined by 6 paramaters $a,b,c,d,e,f$ which obey:
$$
a-b+c=1\\
d-e+f=0\\
a+b+c=0\\
d+e+f=1
$$ There are 4 equations for 6 unknowns, so we can, for example, solve it f0r $a,b,d,e$:
$$
a=\frac12-c\\
b=-\frac12\\
d=\frac12-f\\
e=\frac12
$$
Thus $c,f$ are left as free parameters. The transformation
$$T(x,y,z)=(v,w)$$ is then defined by:
$$
v=(\frac12-c)x -\frac12y+cz\\
w=(\frac12-f)x+\frac12y+fz
$$


问:
Proving $\int \int _Rf\left(\sqrt{\ \frac{x^2}{3}+\frac{y^2}{5}}\right)dA=2\pi \sqrt{\ 15}\int _0^1\ f\left(p\right)pdp$
Where where f is a continous function on $[0,1]$ and The region R is bounded by the ellipse $5x^2 + 3y^2 = 15$

I'm confused how to approach this, I just recently started with double integration and i have only completed the topic Change of order of integration and now i moved onto Change of variable.
So far i am familiar with question in change of variable where the transformation of the plane is given clearly for example $$T:ℝ^2\rightarrow \ ℝ^2,\ T\left(x,y\right)=\left(u+av,v\right)$$

In such cases like this transforming from xy-plane to uv-plane is fairly easy for me as the instructions are given clearly, but the question at hand is a bit confusing since its a little unclear to me. The only progress i made is draw the region R which is attached below:

Any help to walk me through the derivation is greatly appreciated

答:
I think i figured out how to solve this, I'm simply posting the answer in case someone else is also stuck in the same question:

First we can use polar coordinate system to solve this, our region R is enclosed in the ellipse given by $5x^2 +3y^2 =15$ rewriting it as $\frac{x^2}{3}+\frac{y^2}{5}=1$, take the substitution as : $$x=r\sqrt{\ 3}\cos \theta, y= r\sqrt{\ 5}\sin \theta  $$ putting it into the ellipse equation we get: $r=1$ and taking the Jacabian we get
$J\left(r,\theta \right) = \left|\begin{array} {}
\frac {dx}{dr} & \frac {dx}{d\theta}\\
\frac {dy}{dr} & \frac {dy}{d\theta}\end{array}\right|=r\sqrt{\ 15}$

$\int _0^1\ \int _0^{2\pi}f\left(\sqrt{\ r^2}\ \right)r\sqrt{\ 15}d\theta dr$

$2\pi \sqrt{\ 15}\int _0^1\ f\left(\sqrt{\ r^2}\ \right)rd\theta dr$


问:
Example for Riemann integrable function such that $\int^{a+1}_a f(x)dx=0$ but $ f(x)\neq 0$
I'm looking for Riemann integrable function $f:\mathbb{R}\to\mathbb{R}$ with $\int^{a+1}_a f(x)dx=0$ for all $a\in\mathbb{R}$,  but $ f(x)\neq 0$.

I suspect that floor function involves here, if so, then how?

Thank you all!

To clarify: $f$ must not be identically equal to $0$, and it should be integrable over any finite interval.

答: How about trigonometric functions?</p>

Body: How about trigonometric functions?


问:
Are there disjoint open sets separating closed sets satisfying $d(A, B) = 0$?
Let $f(x) = 1/x$. I was wondering if it was possible to find two disjoint open sets of $\mathbb{R}^2$ that contain $A = \{(x, f(x))~|~x \in \mathbb{R}\}$ and $B = \{(x, 0)~|~x \in \mathbb{R}\}$ respectively. Intuitively I would have said it wasn't true because
$$d(A, B) = \inf\{d(x, y)~|~x \in A \land y \in B\} = 0$$
and since $A,B$ are closed sets in $\mathbb{R}^2$, it should not be possible to find disjoint open sets that contain $A$ and $B$ respectively (in my intuition). However, for each point of $A$ and $B$, one can find open balls of radii small enough to be disjoint. By taking $U$ and $V$ the unions of such balls, did we not explicitly construct the desired open sets? This is breaking my mind, does someone have an explication? Thank you very much!

答:
Assuming $A = \{(x, 1/x) : x \neq 0\}$, then we can separate $A$ and $B$ by open sets
\begin{align*}
U &= \{(x, y) : xy < 1/2\} \\
V &= \{(x, y) : xy > 1/2\}.
\end{align*}
Note that $(x, y) \mapsto xy$ is continuous, hence $U$ and $V$ are open. Trivially, they are also disjoint. Further,
$$(x, y) \in A \implies xy = 1 > 1/2 \implies (x, y) \in V,$$
and
$$(x, y) \in B \implies y = 0 \implies xy = 0 < 1/2 \implies (x, y) \in U.$$


问:
How to solve optimization problem $\max_{x,y} \ ax + by^3 \ \text{subject to}: \ 0\leq x \leq 1,\ 0\leq y\leq 1$
The following is the mathematical expression for my model's rate expression. Variables $x,y$ are the controlling parameter, while the rest are positive constants.

$$\max_{x,y} \ ax + by^3 \ $$ $$\text{subject to}: \ 0\leq x  \leq 1,\ 0\leq y\leq1$$

Can I mathematically say that it is a convex problem within the limits of variables $x,y$?

The graph for the equation strictly follows the definition of convexity.
I have learned to solve the problems with KKT method, but I cannot understand how to resolve the inequality constraints $0 \leq (x,y) \leq 1$.

答:
This is not a convex optimization problem. Actually it is a concave optimization (programming) problem.

The objective function, $ax + by^3$, is a convex function over the constraint region $ 0 \le x \le 1, 0 \le y \le 1$, but it is being maximized, which is equivalent to minimizing its negative, which is a concave function.

A concave function being minimized over compact convex constraints, as in this problem, has a global optimum at an extreme of the constraints. In this case, that means at $x = 0$ or $1$, and $y =0$ or $1$. Because $a$ and $b$ are both positive, the optimum occurs at $x = y = 1$, and has optimal objective value equal to  $1$.


问:
$X/\ker A \cong \operatorname{ran} A$ iff $\operatorname{ran}A$ is closed
$X, Y$ are Banach spaces, $A \in B(X, Y)$. To show that $X/\ ker A \cong \operatorname{ran}A$ iff $\operatorname{ran}A$ is closed.

My approach: I can define $A' : X/ \ker A \to Y$ by taking $A'([x])=Ax$ and now I have an injective map from $\operatorname{ran}A$ to $Y$. I think I have to apply open mapping theorem somehow but I'm not sure.

答:
($\implies$): Apply the open mapping theorem on the operator you have defined (since the range of A is closed, it is a Banach space).

($\impliedby$): Completeness is a three space property, so if  X and $\ker A$ are complete (Banach) spaces, so is $X / \ker A$.


问:
Why is $\frac{d}{dt}[\frac{1}{y}] = \frac{d}{dt}[\cos{x}]$ result in an application of the chain rule in related functions?
Why does the below step result in a chain rule?
Intuitively, I would just differentiate it against y and x resulting in $-y^{-2} = -\sin{x}$.

Khan Acad mentioned something about relating y = y(t) and x = x(t), but my brain is struggling to comprehend why it can just be related like this... (Though I could just be getting it completely wrong)

答:
I actually think it is easier if you separate out differentiation from derivatives.  Differentiation is the process of converting an equation of variables to an equation of differentials.  A derivative is a ratio of derivatives.  I think that separating these processes makes the whole of calculus easier to understand.

So, in your case, you have:

$$y^{-1} = \cos(x)$$

If we just differentiate it, we get:

$$ -y^{-2}\,dy = -\sin(x)\, dx$$

Here, no variable has a privileged place.  We aren't differentiating with respect to anything.  Every expression is followed by the differential of the variables.

The question wants $\frac{dy}{dt}$.  To do that, start by solving for $dy$:

$$ dy = \sin(x)\,y^2\,dx $$

Now, just divide both sides by $dt$ (you can do anything you want to an equation, as long as you do it to both sides - even dividing by a brand new differential):

$$\frac{dy}{dt} = \sin(x)\,y^2\,\frac{dx}{dt}$$


问:
Estimate derivatives in terms of derivatives of the Fourier transform.
Let us suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is a smooth function. Furthermore, for every $\alpha$ multi-index, there exists $C_\alpha > 0$ such that
$$
|D^\alpha f(\xi)| \leq \frac{C_\alpha}{(1+|\xi|)^{|\alpha|}}.
$$
Does it follow that, for every $\alpha$, there exists $C'_\alpha > 0$ such that
$$
|D^\alpha (\mathcal{F}^{-1}(f))(x)| \leq \frac{C'_\alpha}{|x|^{n+|\alpha|}}
$$
where $\mathcal{F}^{-1}$ is the inverse Fourier transform (which exists since $f \in \mathcal{S}'$)?
I tried to do it using the definition, but it is really messed up because $\mathcal{F}^{-1}$ is in general in $\mathcal{S}'$. For instance, if $f$ is a constant function, then its inverse transform is a dirac $\delta$, then I should give it a pointwise meaning, and I don't know when this is possible. Any help would be really appreciated.</p>

答: This settles it. See Theorem 9, it also settles regularity issues.</p>


答:
First of all: don't panic! :-)

There is a nice identity for the arctangent:
$$
\arctan x+\arctan\frac{1}{x}=
\begin{cases} \pi/2 & x>0 \\[4px] -\pi/2 & x<0 \end{cases}
$$
Since the limit is for $x\to\infty$, it's not restrictive to assume $x>0$, so
$$
\arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}
$$
and, setting $y=1/x$, the limit becomes
$$
\lim_{y\to0^+}\Bigl(1-\frac{2}{\pi}\arctan y\Bigr)^{\sin2y}
$$
which is not an indeterminate form because it is $1^0=1$.

Title: A limit with an arctan

Body:
I need to calculate another limit, $$ \underset{x\rightarrow\infty}{\lim}\left(\frac{2}{\pi}\arctan x\right)^{\sin\left(2/x\right)}$$

I have no idea how to start or what to do because I've never dealt with an arctan in a limit before.

What should I do first?


问:
How to check the convergence of a complex series?
Consider the series $\sum_{n=1}^{\infty} \frac{e^{nz}}{1+e^{nz}}$ where $z \in \mathbb{C}$. Does this series converge?
I tried to use Cauchy root test but $\lim_{n \to \infty}(\frac{e^{nz}}{1+e^{nz}})^{\frac{1}{n}}=1$. Do you have any hint?

答: Hint: It is not true that the $n$'th root tends to $1$ for every $z\in\mathbb{C}$, and when it does, the general term does not converge to zero so the series diverges in that case.


问:
How to write the summation along an axis of a matrix?
I'm trying to write an example for an equation involving summing a matrix across its columns.

The example has to show the matrix values, so it would ideally be in the form:

$
\begin{bmatrix}
a_1 \\ \vdots \\ a_m \\
\end{bmatrix}
= $ op $
\begin{bmatrix}
b_{1,1}&\dots&b_{1,n}\\ \vdots&\ddots&\vdots \\ b_{m,1}&\dots&b_{m,n}
\end{bmatrix}
$

Where: $a_t = \sum_{i=1}^n b_{t,i}$

What should $op$ be replaced with here to illustrate the summing across columns?

答:
$
\begin{bmatrix}
a_1 \\ \vdots \\ a_m \\
\end{bmatrix}
=
\begin{bmatrix}
b_{1,1}&\dots&b_{1,n}\\ \vdots&\ddots&\vdots \\ b_{m,1}&\dots&b_{m,n}
\end{bmatrix}
\begin{bmatrix}
1_1 \\ \vdots \\ 1_m
\end{bmatrix}
$

$A_m=B_{m,n} \times J_m$

($J$ typically being used when referring to a matrix of ones)


问:
Set of branch points isn't discrete, but branch points are isolated?
I refer to Chapter II.4 of Rick Miranda - Algebraic curves and Riemann surfaces, which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.)

Question 1: Do I understand correctly, and is this indeed true?

What else I understand:

$A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.

This overflow question (Basic question about branch points on Riemann surfaces) seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).

2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a different text): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.

This stackexchange question (Are branch points always isolated?) seems to suggest this is true.

I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the '3.2. Branch points' on p. 7 in text by Armin Rainer in (2.1), which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.

Kinda related: If $F$ has a ramification point, then is $F$ necessarily not injective?

This page (rigtriv: Hurwitz’s Theorem) says

Now, the ramification and branch points must form a discrete set

I don't think the page assumes anything like $F$ is proper.

The Armin Rainer text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.

Kinda related (see Question 2 below): When closed and continuous maps discrete to discrete

S. K. Donaldson - Riemann Surfaces says

If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.

This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.

Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.

(again) Kinda related: When closed and continuous maps discrete to discrete

答: If you want an explicit counter-example to Miranda's claim about discreteness of the set of branch-points, consider the entire function $f(z)=e^z (1+\sin(z))$ as a map ${\mathbb C}\to {\mathbb C}$. The set of branch-points will accumulate to zero and the latter is a branch-point. One can even construct examples where the set of branch-points is dense in the complex plane. I suspect, Miranda had in mind algebraic maps between complex algebraic  curves but forgot to mention this assumption.


问:
Is this a mistake in Volume 1 of Apostol's Calculus?
I am referring to page 112, near the bottom. Apostol claims that

Note that Cavalieri's principle has been stated in the form of inequalities. If $a(S \cap F) = a(T \cap F)$ for every plane $F$ perpendicular to a given line, we may apply Axiom $5$ twice to deduce $v(S) \le v(T)$ and $v(T) \le v(S)$, and hence $v(T) = v(S).$

Specifically, I am wondering about the “apply Axiom $5$ twice” part. Here is Axiom $5$:

Choice of scale. Every box $B$ is in $\mathscr{A}$. If the edges of $B$ have lengths $a,b,$ and $c$ then $v(B) = abc.$

But right above it is Axiom 4, which states Cavalieri's principle:

Cavalieri's principle. If $S$ and $T$ are two Cavalieri solids in $\mathscr{A}$ with $a(S \cap F) \le a(T \cap F)$ for every plane $F$ perpendicular to a given line, then $v(S) \le v(T).$

If it is not a mistake, then I am confused on how to deduce the aforementioned claim using Axiom $5$. Thanks.

答: It's clearly a mistake and supposed to refer to "Cavalieri's principle" instead of "Axiom 5". So many decades of computing already, and people inexplicably still have not learnt to use auto-numbering?</p>


问:
Intersection of a compact subset and a closed subset of a metric space is compact.
I got this question in a test assignment: Let $K$ and $C$ be subsets of a metric space $X$. Show that if $K$ is compact and $C$ is closed, then $K \cap C$ is compact.

The tutor opted for the method starting with an open cover of $K \cap C$. I used the fact that compact subsets of metric spaces are closed, to then show that $K \cap C$ is closed and since $K \cap C$ is a closed subset of compact set $K$,  $K \cap C$ is also compact.

However the tutor barely gave me any marks and left a note: "how do you justify the fact that K is a metric space or subspace, for you to be able to invoke the result that  K n C, a closed subset of a compact metric space or a compact metric subspace is compact? So far, K is just a compact subset of X with no mention of any induced metric."

I don't really understand the comment or the need to define a metric on $K$.

答:
What you've done is almost complete. You showed that since $K$ is a compact subset of $X$(where $X$ is your parent space) it is closed. Therefore $K \cap C$ is a closed subset of $X$. But this does not (depending on what facts you're allowed to use) mean $K \cap C$ is a compact subset of $X$. Instead, you used the fact that $K \cap C$ is a closed subset of $K$(not $X$) to show that $K\cap C$ is a compact subset of $K$. You were required to show that $K \cap C$ is a compact subset of $X$. You're almost done, since compactness is a topological property, $K \cap C$ is automatically a subset of $X$.

So basically, you're required to stipulate the additional(important) fact that if $A \subset B \subset X$, where $X$ is a metric space, then $A$ is compact as a subset of $B$ if and only if $A$ is compact as a subset of $X$.


问:
How to find all solutions of the optic equation $\frac{1}a+\frac{1}b = \frac{1}c$
Optic equation says for

$${\frac  {1}{a}}+{\frac  {1}{b}}={\frac  {1}{c}}$$

All solutions in integers $a, b, c$ are given in terms of positive integer parameters $m, n, k$ by

$$a=km(m+n) , \quad  b=kn(m+n), \quad c=kmn$$

where $m$ and $n$ are coprime.

How is this proved? I went through the reference, which says "A. Palmström, J. Sadier, and C. Moreau" "Ibid, 299-302". No clue what that means though.

Does anyone know how why all the solutions are covered by these formulas?

答:
If there were no limits on the right side, then it would be

$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$.

Because of the limits on the right side, the denominator has to be divisible by $a+b$, thats why one multiplies this to the denominators. This would be enough, but for saying that $m$ and $n$ are coprime, its necessary to multiply the denominators by k.


问:
Differential forms and behaviour respect product
Let $M$ a smooth $n$-manifold and $E = M \times \mathbf{R}^m$ the product manifold. Let $\pi: E \to M$ the projection onto the first factor. Let $k \leq n$. It's known that
$$
\{dx_{i_1} \wedge \cdots \wedge dx_{i_k} : i_1 < \cdots < i_k\}
$$
is a local frame of $\Omega^k(M)$. Let $\omega \in \Omega^k(M \times \mathbf{R}^n)$. Locally
$$
\omega = f(x_1,\dots,x_n,y_1,\dots,y_m)dx_{i_1} \wedge \cdots \wedge dx_{i_r} \wedge dy_{j_1} \wedge \cdots \wedge dy_{j_s}
$$
where $r+s = k$ and $y$ is coordinates in $\mathbf{R}^m$.

Therefore, can we deduce that
$$
\Omega^{k}(M \times \mathbf{R}^m) = \sum_{p =1}^{k} \Omega^{p}(M) \wedge \Omega^{k-p}(\mathbf{R}^m)
$$
?

I'm very confused because it's true that $\Lambda(V \times W) = \Lambda V \otimes \Lambda W$ and here appears the wedge product. In particular, I want to prove that every $k$-form in $M \times \mathbf{R}^m$ is a linear combination of
$$
\pi^* (\phi) f(x,y) dy_{j_1} \wedge \cdots \wedge dy_{j_s}
$$
where $\phi \in \Omega^{*}(M)$. Notice that the last formula is global I don't need charts

答: Your last equation gives you the answer: while you indeed have the wedge products of the two coframe, the "function" part may mix the coordinates. In fact you write $f(x,y)$ and not every such $f$ can be written as a product of a function on $M$ with a function on $R^n$, for example $\sin(xy)$. If you prefer, $C^\infty (M\times N) \neq C^\infty(M) C^\infty(N)$.


问:
Finding $1998x + 5y + 3z$ given $\frac{1}{x} + \frac{1}{y} = \frac{3}{z}$
Suppose $x,y,z$ are positive integers so that $y$ and $z$ are prime and $x = yz.$ If they satisfy $$\frac{1}{x} + \frac{1}{y} = \frac{3}{z},$$ find $1998x + 5y + 3z.$

I first moved $\frac{1}{y}$ over to the other side and expanded out, but from there I got stuck. Can someone help me?

答:
You are on the right track. After that you should have

$$\frac{1}x=\frac{3}z-\frac{1}y\Longrightarrow \frac{1}{yz}=\frac{3y-z}{yz}\Longrightarrow 1=3y-z\Longrightarrow 3y=z+1$$

Now note that if $y$ is odd, then $z$ has to be even. And since $z$ is a prime, $z$ has to be $2$, but in this case we will have $y=1$ which is not a prime.

Therefore $y$ has to be an even prime. That is, $\color{red}{y=2}$.

Can you end it now?


答:
The equation $$\frac 1a+\frac 1b=\frac 1
c$$ is equivalent to $$(a+b)c=ab\quad \text {or}\quad (a-c)(b-c)=c^2$$

Note that $(ka,kb,kc)$ is a solution iff $(a,b,c)$ is, so we may assume that $\gcd(a,b,c)=1$.

If $p$ is any prime dividing $c$ then $p$ must divide either $a-c$ or $b-c$.  It can't divide both, else $p$ would divide $\gcd(a,b,c)$.  Thus $p^2$ divides one of $(a-c),(b-c)$.   It follows that both $(a-c)$ and $(b-c)$ are perfect squares, so we may write, e.g., $(a-c)=m^2,\,(b-c)=n^2$.  Then, of course, $c=mn$.

Thus every relatively prime solution is given by $$a=m^2+mn=m(m+n)\quad \quad b=n^2+mn=n(m+n)\quad \quad c=mn$$

as desired.

Title: How to find all solutions of the optic equation $\frac{1}a+\frac{1}b = \frac{1}c$

Body:
Optic equation says for

$${\frac  {1}{a}}+{\frac  {1}{b}}={\frac  {1}{c}}$$

All solutions in integers $a, b, c$ are given in terms of positive integer parameters $m, n, k$ by

$$a=km(m+n) , \quad  b=kn(m+n), \quad c=kmn$$

where $m$ and $n$ are coprime.

How is this proved? I went through the reference, which says "A. Palmström, J. Sadier, and C. Moreau" "Ibid, 299-302". No clue what that means though.

Does anyone know how why all the solutions are covered by these formulas?


答:
What is true is what you stated. If $M$ and $N$ are manifolds, then $\Lambda^k\left(M\times N\right) = \bigoplus_{r+s=k} \Lambda^r(M)\otimes\Lambda^s(N)$. The projections $\pi_M : M\times N \to M$ and $\pi_N : M\times N \to N$ give a morphism $\bigoplus_{r+s=k}\Omega^r(M)\otimes \Omega^s(N) \to \Omega^{k}(M\times N)$. There is no reason that it will be surjective.

For example, if $k=0$, you have a morphism
\begin{align}
\Omega^0(M)\otimes \Omega^0(N) &\longrightarrow \Omega^0(M\times N) \\
(f,g) &\longmapsto (p,q)\mapsto f(p)g(q)
\end{align}but it is clearly not surjetive in many examples. The fact is that in the fibers, everything works fine, but when you take sections, that are global notions, you cannot act like you were working pointwisely.

Title: Differential forms and behaviour respect product

Body:
Let $M$ a smooth $n$-manifold and $E = M \times \mathbf{R}^m$ the product manifold. Let $\pi: E \to M$ the projection onto the first factor. Let $k \leq n$. It's known that
$$
\{dx_{i_1} \wedge \cdots \wedge dx_{i_k} : i_1 < \cdots < i_k\}
$$
is a local frame of $\Omega^k(M)$. Let $\omega \in \Omega^k(M \times \mathbf{R}^n)$. Locally
$$
\omega = f(x_1,\dots,x_n,y_1,\dots,y_m)dx_{i_1} \wedge \cdots \wedge dx_{i_r} \wedge dy_{j_1} \wedge \cdots \wedge dy_{j_s}
$$
where $r+s = k$ and $y$ is coordinates in $\mathbf{R}^m$.

Therefore, can we deduce that
$$
\Omega^{k}(M \times \mathbf{R}^m) = \sum_{p =1}^{k} \Omega^{p}(M) \wedge \Omega^{k-p}(\mathbf{R}^m)
$$
?

I'm very confused because it's true that $\Lambda(V \times W) = \Lambda V \otimes \Lambda W$ and here appears the wedge product. In particular, I want to prove that every $k$-form in $M \times \mathbf{R}^m$ is a linear combination of
$$
\pi^* (\phi) f(x,y) dy_{j_1} \wedge \cdots \wedge dy_{j_s}
$$
where $\phi \in \Omega^{*}(M)$. Notice that the last formula is global I don't need charts


问:
Integral Hammerstein-like equation solution
In signal processing theory, I found this integral equation which I suppose to be of Hammerstein type:
$$u(t)-\int_0^1\frac{\cos(\omega t+\phi)}{u(\phi)}d\phi=0$$
I didn't find anything in literature apart this:

http://www-users.mat.uni.torun.pl/~tmna/files/v15n2-02.pdf

Could someone give me some hints how to solve this equation?
Thanks.

答:
Note that we have
$$u''(t) + \omega^2 \int_0^1 \dfrac{\cos(\omega t+ \phi)}{u(\phi)} d\phi =0\implies u''(t)+\omega^2 u(t) = 0 \implies u(t) = ae^{i\omega t}+b e^{-i \omega t}$$
Obtain $a$ and $b$ by plugging it back into the equations.</p>


问:
Non-linear system of two equations (Method of Moments for Generalized Pareto)
In a Method of Moments Estimation problem involving the generalized Pareto Distribution, the following system of 2 non-linear equations arise

\begin{align*}
\bar{X} &= \frac{\alpha \beta}{\alpha - 1}. \\
\frac{1}{n}\sum_{i=1}^{n} X^2_i  &= \frac{\alpha \beta^2}{\alpha - 2}.
\end{align*}

The problem is to solve the system for $\alpha$ and $\beta$ in terms of $\bar{X} $ and $\frac{1}{n}\sum_{i=1}^{n} X^2_i $. The thing is, it's easy to isolate $\alpha$ or $\beta$ from any of the two questions, but after this has been done you cannot isolate the other parameter in the other equation. Using Mathematica you can get two solutions but I cannot just say "According to Mathematica, the solution is ..."

答:
$$
\begin{cases}
\frac{a b}{a-1}=x\\
\frac{a b^2}{a-2}=y\\
\end{cases}
$$
multiply the first equation by $b$
$$
\begin{cases}
a b^2=(a-1)bx\\
a b^2=(a-2)y\\
\end{cases}
$$
thus
$$(a-1)bx=(a-2)y\to b=\frac{(a-2) y}{(a-1) x} $$
Now plug this in the first, the very first equation
$$ab=x(a-1)\to a\cdot \frac{(a-2) y}{(a-1) x}=x(a-1)$$
You get a quadratic equation in $a$ which gives
$$a=\frac{x^2-y\pm\sqrt{y^2-x^2 y}}{x^2-y}$$
plug back to get $b$
$$b_1=\frac{y \left(\frac{x^2-y-\sqrt{y^2-x^2 y}}{x^2-y}-2\right)}{x \left(\frac{x^2-y-\sqrt{y^2-x^2 y}}{x^2-y}-1\right)}=\frac{y}{x}\cdot \frac{x^2-y-\sqrt{y \left(y-x^2\right)}}{\sqrt{y \left(y-x^2\right)}}=\frac{y-\sqrt{y \left(y-x^2\right)}}{x}$$
and
$$b_2=\frac{y+\sqrt{y \left(y-x^2\right)}}{x}$$


问:
Finding out all questions in test bank
Let’s Say that a school has a test bank for an online multiple choice math exam which contains 100 questions. When a student starts his exam the system randomly selects 15 questions for him.
If my friends and I decide to find out all the 100-questions, how many trials do we need if that is possible??

Edit:(as suggested in comments)

How many trials are needed, on average, to see all the questions??

答:
Consider $n$ trials. There are ${100\choose 15}^n$ options on how to choose them. There are ${99\choose 15}^n$ of them which do not cover some question $j$. Using exclusion-inclusion principle,
$$
\text{count(non-covering cases)} = \sum_{i=1}^{85} (-1)^{i-1} {100\choose i} {100-i\choose 15}^n
$$
so
$$P_n(\text{covering}) = 1 - \frac{\sum_{i=1}^{85} (-1)^{i-1} {100\choose i} {100-i\choose 15}^n}{{100\choose 15}^n}
$$
and "in average", you would wait
$$\sum_{n=1}^\infty n P(\text{first covering is on trial } n)
=\sum_{n=1}^\infty n (P_n - P_{n-1})
$$ ($P_0 = 0$).

Not sure how to simplify to something numerically feasible. Practically, I would use computer simulation.

Title: Finding out all questions in test bank


问:
How do I solve a quadratic for $k$ using the discriminant
Picture of the question

Given that the equation $$kx^2-2x+3-2k=0$$ has equal roots, find the possible values of the constant $k$.

So far I have only attempted to factor $-2$ which I'm not sure is even right. I understand I need to use the discriminant later on in the Question. Completely lost, any guidance would be helpful. If you post the solution please show it step by step.

Thanks!

Body:
$$kx^2-2x+3-2k=0$$
Since this equation has equal roots, therefore discriminant is $0$.
$$\Delta=0$$
$$b^2-4ac=0$$
$$2^2-4(3-2k)(k)=0$$
$$1=(3-2k)(k)$$
$$2k^2-3k+1=0$$
$$k=\frac{1}{2},1.$$


问:
Find $\sum_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$
This question comes from "adjusting the order of an infinite series".

We are talking about whether a series' value would change or not by changing the position of some terms.
For example, we have

$$\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum\limits_{n=1}^{\infty}\left({\frac{1}{2n-1}-\frac{1}{2n}}\right)$$

By changing the order, we get

$$H=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots$$

and
$$H=\sum\limits_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$$

And to generalize, we construct a series which has firstly, k positive terms and then 1 negative term, and the same story repeats. We write it as $H_{k}$.

But how to calculate the values of all these $H_{k}$ becomes a problem, in fact, I'm stuck in $H_{2}$.

Thank you for your attention.

Body:
Edited after the solution of @Christian Blatter:

Let $$S=\sum_{n=1}^{\infty} \left( \frac{1}{4n-3}+ \frac{1}{4n-1}-\frac{1}{2n} \right)~~~~(1)$$
Let (1) be re-written as $S=S_1+S_2$
$$S_1=\sum_{n=1}^{\infty} \left( \frac{1}{4n-3}-\frac{1}{4n} \right), ~S_2=\sum_{n=1}^{\infty} \left( \frac{1}{4n-1}-\frac{1}{4n} \right) $$
In both  cases we have difference of TWO divergent series  and the divergence will cancel out appropriately.
$$S=\sum_{n=1}^{\infty}\int_{0}^{1} [t^{4n-4}- t^{4n-1}] dt=\int_{0}^{1}\left( \frac{1}{1-t^4}-\frac{t^3}{1-t^4}\right) dt= \int_{0}^{1} \frac{1-t^3}{1-t^4} dt$$
$$\implies S_1=\int_{0}^{1} \left(\frac{1}{2(1+t)}+\frac{1}{2(1+t^2)}+\frac{t}{2(1+t^2)}\right) dt=\frac{1}{2}\ln 2+\frac{\pi}{8}+\frac{1}{4}\ln 2.$$
Similarly,
$$ S_2=\sum_{n=1}^{\infty} \int_{0}^{\infty} [t^{4n-2}-t^{4n-1}] dt=\int_{0}^{1} \left(\frac{t^2}{1-t^4}-\frac{t^3}{1-t^4}\right) dt= \int_{0}^{1} \frac{t^2-t^3}{1-t^4} dt$$
$$\implies S_2= \int_{0}^{1} \left(\frac{1}{2(1+t)}-\frac{1}{2(1+t^2)}+\frac{t}{1+t^2}\right) =\frac{1}{2}\ln 2-\frac{\pi}{8}+\frac{1}{4}\ln 2.$$
So finally, we get $$S=\frac{3}{2} \ln 2$$


问:
is the set of values where two continuous functions are equal open or closed?
Suppose f, g : R → R are continuous. Determine if the set
$E = \{x\;|\; f(x) = g^2(x) + 2020\}$
is open, closed or not necessarily open or closed in R.

I know that the function need not necessarily be open, as I was able to give a counterexample. However, I am not able to come up with a counterexample to show that the set need not be closed (I have a hunch that the set is always closed). However, I am unable to prove that the set would be definitely closed. Please help.

答: If $h=f-g^2$, then $h$ is continuous too. and $E=h^{-1}\bigl(\{2020\}\bigr)$. So, $E$ is a closed set, since is the pre-image of a closed set with respect to a continuous function.


问:
Rouché's theorem for $z^7-5z+1$ if $|z|>1$ and $|z|<3$
I want to count number of zeros of  $z^7-5z+1$ for $\{z | 1 < |z| < 3 \}$. Using Rouché's theorem I found that the function has 1 zero ($0$) of multiplicity 1 in $\{z | 1 < |z|\}$ and 1 zero ($0$, the same!) of multiplicity 7 in $\{z ||z| < 3 \}$. And how many zeros do we finally have in  $\{z | 1 < |z| < 3 \}$?

The main problem is how one can count the difference? Why can we just writy that there are 6 zeros because $7-1=6$ and don't pay attention that $0$ is not in $\{z | 1 < |z| < 3 \}$?

答:
We first consider the disc $|z|<1+\epsilon$ with $\epsilon$ small positive quantity (thanks for @saulplatz's comment for this correction), and the 2 functions $g=z^7$ and $f=-5z+1$ since $|g|< 1+\epsilon<5(1-\epsilon) -1)<|f| $ on $|z|=1+\epsilon$ we have by Rouche's Theorem that $|f|$ and $|f+g|$ have the same number of zeros on the disc $|z|=1+\epsilon$- that is 1.
For $|z|<3$ we take $f=z^7$ and $g=-5z+1$. Obviously we have $|g|<|f|=3^7$ on $|z|=3$ so $f$ and $f+g$ have the same number of zeros inside $|z|<3$ which is 7. We thus conclude that in the annulus $1<|z|<3$ the function $z^7-5z+1$ has 6 zeros.


答:
$u(t)-\int_0^1\dfrac{\cos(\omega t+\phi)}{u(\phi)}d\phi=0$

$u(t)=\int_0^1\dfrac{\cos\omega t\cos\phi-\sin\omega t\sin\phi}{u(\phi)}d\phi$

$u(t)=\cos\omega t\int_0^1\dfrac{\cos\phi}{u(\phi)}d\phi-\sin\omega t\int_0^1\dfrac{\sin\phi}{u(\phi)}d\phi$

$u(t)=C_1\cos\omega t+C_2\sin\omega t$

$\therefore C_1\cos\omega t+C_2\sin\omega t\equiv\cos\omega t\int_0^1\dfrac{\cos\phi}{C_1\cos\omega\phi+C_2\sin\omega\phi}d\phi-\sin\omega t\int_0^1\dfrac{\sin\phi}{C_1\cos\omega\phi+C_2\sin\omega\phi}d\phi$

The only problem is to determine $C_1$ and $C_2$ .

Title: Integral Hammerstein-like equation solution

Body:
In signal processing theory, I found this integral equation which I suppose to be of Hammerstein type:
$$u(t)-\int_0^1\frac{\cos(\omega t+\phi)}{u(\phi)}d\phi=0$$
I didn't find anything in literature apart this:

http://www-users.mat.uni.torun.pl/~tmna/files/v15n2-02.pdf

Could someone give me some hints how to solve this equation?
Thanks.


问:
Is there a way to find the center point of this circle?
I am facing a problem that I have difficulty to solve.

I have a line that originates from the origin (point $(0,0)$) and has a known angle to the $x$ axis (angle $\theta$)

Somewhere between the $x$ axis and this line is a circle. This circle touches both lines, but the dimension and center of this circle are not known.

On the right side of the circle is a point $P$ that lies on the circumference of the circle and of which the coordinates are known.

How can I find the coordinates of the center of the circle?

I can't seem to add an image so I have used an external website to make the drawing.

========================================================

I now realize that I have over-simplified the problem because it is hard to explain in words. However, now I know that I can send a drawing in another way I would like to post the original problem:

You can see the drawing over here

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known.
A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines.
On the arc of the circle is a point P of which the coordinates are known.
The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

答:
For the modified problem, draw a vertical line through $C$. Since $C$ is on the diagonal, $x_C=y_C$. That's one of the unknowns. The second is the radius $r$. So you need two equations. One of them is the distance to point $P$:
$$(x_P-x_C)^2+(y_P-x_C)^2=r^2$$
For the other equation, look at the triangle formed by the $x$ axis, the upper line, and the vertical through $C$. This is a right angle triangle, with the angle at origin $\pi/4+\theta$, the opposite side is $x_C+r$, and the adjacent side $x_C$. Then the other equation is: $$\tan\left(\frac\pi4+\theta\right)=\frac{x_C+r}{x_C}$$
Eliminating $r$ from these two equations you get a quadratic in $x_C$. You need to choose the solution where $x_C<x_P$.


问:
If a null space is spanned by only a single vector, is that vector necessarily the zero vector?
I have a homework problem where I need to find the values of a variable within a matrix, A, that result in Nul(A) being spanned by a single vector. My intuition is telling me that the only single vector that can span the null space must be the zero vector, but I'm not 100% certain and would just like to confirm.</p>

答: Not at all. Take, say, $A=\left[\begin{smallmatrix}1&1\\2&2\end{smallmatrix}\right]$. Then $\operatorname{null}(A)$ is spanned by the vector $(1,-1)$. So, it is spanned by a single vector (which is not the same thing as asserting that there is a single vector that spans it).


问:
How to check the convergence of a complex series?
Consider the series $\sum_{n=1}^{\infty} \frac{e^{nz}}{1+e^{nz}}$ where $z \in \mathbb{C}$. Does this series converge?
I tried to use Cauchy root test but $\lim_{n \to \infty}(\frac{e^{nz}}{1+e^{nz}})^{\frac{1}{n}}=1$. Do you have any hint?

Body:
$f_m(z)
=\sum_{n=1}^{m} \frac{e^{nz}}{1+e^{nz}}\\
=\sum_{n=1}^{m} \frac{1+e^{nz}-1}{1+e^{nz}}\\
=\sum_{n=1}^{m} 1-\sum_{n=1}^{m} \frac{1}{1+e^{nz}}\\
=m-\sum_{n=1}^{m} \frac{1}{1+e^{nz}}\\
$

The sum converges for
$\Re(z) > 0$
so $f_m(z)$ diverges for
these as $m \to \infty$.


问:
Factorize $x^{2n}+1$ to evaluate $\int\tan^{1/n}(x)dx$
I've been attempting to find a formula for $\int\tan(x)^{1/n}dx, n\in\Bbb N$.

I started out by performing the substitution $$u=\tan(x)^{1/n}$$

$$dx=\frac {nu^{n-1}}{1+u^{2n}}du$$ to transform the integral into $$n\int \frac{u^n}{1+u^{2n}}du$$
This is a relatively simple rational function, so it should be possible to solve via a partial fraction expansion. For the simple case where $n=2$, I was able to rewrite the integral as $$\frac{\sqrt2}{2}\left(\int\frac{u}{u^2-\sqrt2u+1}du - \int\frac{u}{u^2+\sqrt2u+1}du\right)$$

Which can then be solved by completing the square to yield $$\frac{\sqrt2}{4}\ln\left(\left(\sqrt{\tan{x}}-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\over\left(\sqrt{\tan{x}}+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\right)+\frac{\sqrt2}{2}\left(\arctan\left({\sqrt{2\tan{x}}-1}\right)+\arctan\left({\sqrt{2\tan{x}}+1}\right)\right)+C$$

However, I can't figure out how to factor the denominator for the partial fraction expansion of the general case. It will always be a product of $n$ irreducible quadratic factors, clearly, but I don't know how to locate them. I have managed to find the complex linear factors, using Euler's identity to find that they're of the form $$x\pm e^{im\pi\over{2n}}$$ where m is allowed to vary from 1 to $2n$. How can I combine these linear factors into real, irreducible quadratics for use in my partial fraction expansion? Alternatively, is there a simpler method of evaluating this integral? Any help would be vastly appreciated.

答:
With $a_k=\frac{(2k-1)\pi}{2n}$

$$x^{2n}+1= \prod_{k=1}^{2n}(x- e^{i a_k})$$
and the corresponding partial fractional decomposition is

\begin{align}
\frac{x^n}{x^{2n}+1} & = -\frac1{2n}\sum_{k=1}^{2n}\frac{e^{i(n+1)a_k}}{x-e^{i a_k}}\\
&=-\frac1{2n}\sum_{k=1}^{n}\left(
 \frac{e^{i (n+1)a_k}}{x-e^{ia_k} }+ \frac{e^{-i (n+1)a_k}}{x-e^{-ia_k}} \right)\\
&= -\frac1{n}\sum_{k=1}^{n}
 \frac{x\cos(n+1)a_k+\cos na_k }{x^2-2x\cos a_k +1} \\
 &= \frac1{n}\sum_{k=1}^{n}
 \frac{(-1)^{k+1} x\sin a_k}{x^2-2x\cos a_k +1}
\end{align}
Then, integrate term-wise to obtain
\begin{align}
& \int\tan^{\frac1n}(t)\> dt= \int \frac{nx^n}{x^{2n}+1}dx\\
=& \sum_{k=0}^{n}\frac{(-1)^{k+1}}2\left( \sin a_k
\ln\left(x^2 - 2x\cos a_k + 1\right)
+ 2\cos a_k
\tan^{-1}\frac{x- \cos a_k}{\sin a_k}
\right)
\end{align}


问:
For which values $x,\alpha$ does $\sum_{n=1}^{\infty}\frac{n^{nx}}{(n!)^ \alpha}$ converge?
I want to know for which values of $x$ this series converges:
$$\sum_{n=1}^\infty \frac{n^{nx}}{(n!)^ \alpha};$$ here $\alpha \in\mathbb R$ is a constant

This series is defined $ \forall x \in \mathbb R$.

$$a_n=\frac{n^{nx}}{(n!)^ \alpha} \sim \frac{n^{nx}}{(\sqrt {2 \pi n}* (\frac{n}{e})^n)^ \alpha}= \frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}n^{nx-n \alpha}e^{n \alpha} \sim 0 \Leftrightarrow  \alpha <0 \land x- \alpha <0.$$

Applying the root test:

$$ \sqrt[n]{\frac{1}{(2 \pi n)^{\frac{\alpha}{2}}}*n^{nx-n \alpha}*e^{n \alpha} }=\frac{1}{(2 \pi n)^{\frac{\alpha}{2n}}}n^{x- \alpha}e^{\alpha } <1 \iff x- \alpha<0   \iff x<\alpha $$

Is it right?

My doubt is regarding the necessary condition for the convergence in which I find
$\alpha <0 \land x- \alpha <0$ and the generality of $\alpha$ that doesn't make me say if $\alpha<$ or $>0$.

答:
Your approach is correct! Here's an other approach using the Ratio Test.

Let's first consider the case $x=\alpha$:
$$
\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)\alpha}}{n^{n\alpha}}\cdot\frac{(n!)^{\alpha}}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)\alpha}}{n^{n\alpha}}\cdot\frac{1}{(n+1)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{n\alpha}}{n^{n\alpha}}\right|
$$By continuity, we can pull the $\alpha$ powers outside the limit, which is then well-known:
$$
=\left(\lim_{n\to\infty}\left|\frac{(n+1)^{n}}{n^{n}}\right|\right)^{\alpha} =\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^{\alpha} =e^{\alpha}
$$So when $x=\alpha$, the series converges only when $\alpha<0$. The other two cases can be deduced the same way:
$$
\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)x}}{n^{nx}}\cdot\frac{(n!)^{\alpha}}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{nx}}{n^{nx}}\cdot\frac{(n!)^{\alpha}(n+1)^x}{((n+1)!)^{\alpha}}\right|
$$
$$
=\lim_{n\to\infty}\left|\frac{(n+1)^{nx}}{n^{nx}}\cdot(n+1)^{x-\alpha}\right|
$$
$$
=e^x\lim_{n\to\infty}\left|(n+1)^{x-\alpha}\right|
$$Clearly this converges only when $x<\alpha$. So in summary, either $x<\alpha$, or $x=\alpha <0$.


答:
We can compte the partial sums
$$S_p=\sum_{n=1}^{p}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$$
$$\sum_{n=1}^{p}\frac{1}{4n-3}=\frac{1}{4} \left(\psi \left(p+\frac{1}{4}\right)-\psi   \left(\frac{1}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4n-1}=\frac{1}{4} \left(\psi \left(p+\frac{3}{4}\right)-\psi
   \left(\frac{3}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{2n}=\frac{1}{2}H_p$$ Summing and going from the polygamma to generalized harmonic numbers
$$S_p=\frac{1}{2} \left(H_{2 p-\frac{1}{2}}-H_p+2\log (2)\right)$$ Using asymptotics
$$S_p=\frac{3}{2}\log (2)-\frac{1}{4 p}+\frac{3}{64
   p^2}+O\left(\frac{1}{p^3}\right)$$

Title: Find $\sum_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$

Body:
This question comes from "adjusting the order of an infinite series".

We are talking about whether a series' value would change or not by changing the position of some terms.
For example, we have

$$\ln{2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum\limits_{n=1}^{\infty}\left({\frac{1}{2n-1}-\frac{1}{2n}}\right)$$

By changing the order, we get

$$H=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots$$

and
$$H=\sum\limits_{n=1}^{\infty}\left({\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}}\right)$$

And to generalize, we construct a series which has firstly, k positive terms and then 1 negative term, and the same story repeats. We write it as $H_{k}$.

But how to calculate the values of all these $H_{k}$ becomes a problem, in fact, I'm stuck in $H_{2}$.

Thank you for your attention.


问:
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$

I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.

Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf

答:
Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.

Let $p(x) = (x-2)(x-16)g(x)$.

Thus we see that

$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$

Thus $$ (x-8)g(2x) = 4(x-2)g(x)$$

We now see that $g(4) = g(8) = 0$

Thus $g(x) = (x-4)(x-8)h(x)$

Thus
$$ (x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$

i.e

$$ h(2x) = h(x) $$

This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)

Thus

$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Parametrized ODEs
Suppose we have a parametrized family of ODEs given by

$$\displaystyle \frac{dy}{dt} = \mu(y(t)) + \alpha f(t);\quad\quad y(0) = y_0$$

where $\mu$ is a well-behaved nonlinear function, $f$ is some bounded oscillating function, say a finite sum of sinusoids, and $\alpha \in \mathbb{R}$.
I'm interested in the behaviour of the solution at a fixed time, say $y_{\alpha}(t_1)$, regarded as a function of the scalar $\alpha$.

I must confess I don't know very much about ODEs. I think I took one short course on them as an undergraduate. I'm sure this kind of thing must have been studied quite extensively, but I don't know where to begin reading about it.

If someone could recommend some texts to look at or keywords to search for, I'd be very grateful.

答:
You could try to use sub and supersolutions. Let $M=\sup\{f(t):0\le t\le t_1\}$ and  $m=\inf\{f(t):0\le t\le t_1\}$, and let $y^+$, $y^-$ be the solutions of the differential equations $z'=\mu(z)+\alpha M$, $z'=\mu(z)+\alpha m$, with initial condition $z(0)=y_0$. Then $y^-(t_1)\le y_\alpha(t_1)\le y^+(t_1)$.

You can obtain estimates on $y^+(t_1)$ and $y^-(t_1)$ from the identities
$$
\int_{y_0}^{y^+(t_1)}\frac{dz}{\mu(z)+\alpha M}=\int_{y_0}^{y^-(t_1)}\frac{dz}{\mu(z)+\alpha m}=t_1,
$$
obtained by integration of the equations $z'=\mu(z)+\alpha M$ and $z'=\mu(z)+\alpha m$ from $t=0$ to $t=t_1$.

Title: Parametrized ODEs


问:
Space modelled on ring
I am learning some deformation theory. The dimension of the moduli space for some deformation problem is bound by the dimension of (usually) the first cohomology group of the object that is to be deformed.

If $X$ is a complex manifold with infinite-dimensional first cohomology and no second cohomology (i.e. no obstructions), the space of first-order deformations is infinite-dimensional. For a particular example, this cohomology is infinite-dimensional over $\mathbb C$, but finite-dimensional over some ring of functions.

It would make sense to think of the moduli space as a space modelled on this ring of functions (because then it would be "finite-dimensional", which presumably is a nicer space). My question is: Is there a concept for spaces which are modelled on rings/modules, that generalize manifolds/varieties which are usually modelled on fields/vector spaces?

答:
It seems likely to me that you are looking for the notion of scheme (or its generalisations), so I will briefly explain what they are. I will assume you know at least the definition a sheaf on a topological space.

Definition. A locally ringed space is a pair $(X, \mathscr{O}_X)$ where $X$ is a topological space and $\mathscr{O}_X$ is a sheaf of commutative rings on $X$ such that every stalk of $\mathscr{O}_X$ is a local ring (i.e. has a unique maximal ideal). A morphism of locally ringed spaces $(X, \mathscr{O}_X) \to (Y, \mathscr{O}_Y)$ consists of a continuous map $f : X \to Y$ together with a sheaf-of-rings homomorphism $f^{\sharp} : f^{-1} \mathscr{O}_Y \to \mathscr{O}_X$, where $f^{-1}$ denotes the pullback operation on sheaves, such that for each point $x$ in $X$, the induced homomorphism $\mathscr{O}_{Y, f(x)} \to \mathscr{O}_{X, x}$ sends non-invertible elements in $\mathscr{O}_{Y, f(x)}$ to non-invertible elements in $\mathscr{O}_{X, x}$. An open subspace of a locally ringed space $(X, \mathscr{O}_X)$ is a locally ringed space of the form $(U, \mathscr{O}_U)$, where $U$ is an open subset of $X$ and $\mathscr{O}_U = \mathscr{O}_X |_U$.

Example. If $X$ and $Y$ are smooth manifolds and $\mathscr{O}_X$ and $\mathscr{O}_Y$ are the respective sheaves of smooth functions, then a morphism $(X, \mathscr{O}_X) \to (Y, \mathscr{O}_Y)$ is precisely the same thing as a smooth map $X \to Y$. The situation for complex manifolds is a little more subtle, however. In both cases, the open subspaces are precisely the open subsets with the induced smooth/complex manifold structure.

*Theorem.*$\DeclareMathOperator{\Spec}{Spec}$ For every commutative ring $A$, there exist a locally ringed space $(\Spec A, \mathscr{O}_{\Spec A})$ and a ring isomorphism $A \cong \Gamma (\Spec A, \mathscr{O}_{\Spec A})$ such that, for every locally ringed space $(X, \mathscr{O}_X)$, the map that sends a morphism $(X, \mathscr{O}_X) \to (\Spec A, \mathscr{O}_{\Spec A})$ to the ring homomorphism $A \cong \Gamma (\Spec A, \mathscr{O}_{\Spec A}) \to \Gamma (X, \mathscr{O}_X)$ is bijective onto the set of all ring homomorphisms $A \to \Gamma (X, \mathscr{O}_X)$.

Example. If $k$ is a field, then $\Spec k$ is a point, and $\mathscr{O}_{\Spec k}$ is just $k$ itself. Thus, a morphism $(\Spec k, \mathscr{O}_{\Spec k}) \to (X, \mathscr{O}_X)$ is the same thing as a point $x$ in $X$ and a ring homomorphism $\mathscr{O}_{X, x} \to k$. Such a morphism is said to be a $k$-valued point of $X$.

Definition. An affine scheme is a locally ringed space isomorphic to one of the form $(\Spec A, \mathscr{O}_{\Spec A})$ for some commutative ring $A$. A scheme is a locally ringed space that admits an cover by open subspaces that are affine schemes. A morphism of schemes is a morphism of locally ringed spaces.

It is customary to abuse notation and refer to a scheme $(X, \mathscr{O}_X)$ just by the space $X$, omitting mention of the structure sheaf $\mathscr{O}_X$; similarly, one often omits mention of the sheaf-of-rings homomorphism $f^\sharp$. We henceforth adopt this simplified notation.

The most important contribution of Grothendieck is the so-called ‘relative point of view’, which tells us to regard a morphism of schemes $f : X \to Y$ as a family of spaces “continuously” parametrised by $Y$ (in much the same way as one might think of fibre bundles).

Accordingly, let us fix a base scheme $S$, which will often be an affine scheme or even $\Spec k$ for some field $k$.

Definition. An $S$-scheme, or scheme over $S$, is a morphism of schemes $X \to S$. A morphism of $S$-schemes $(X \to S) \to (Y \to S)$ is a morphism $X \to Y$ making the evident triangle commute. An $S$-point of an $S$-scheme $X \to S$ is a morphism $(S \to S) \to (X \to S)$, i.e. a morphism $S \to X$ such that the composite $S \to X \to S$ is the identity.

For deformation theory, my understanding is that one often studies schemes over bases like $S = \Spec \mathbb{C}[\epsilon] / (\epsilon^2)$, which is an infinitesimal thickening of the point $\Spec \mathbb{C}$. Although $S$ is topologically just a point as well, we can see that $S$ carries some extra structure by noting that a morphism $S \to X$ is essentially the same thing as a $\mathbb{C}$-valued point of $X$ together with a specified tangent vector.

There are also bases like $S = \Spec \mathbb{C}(t_1, \ldots, t_n)$, where $\mathbb{C}(t_1, \ldots, t_n)$ denotes the field of rational functions over $\mathbb{C}$ in $n$ variables. This too is topologically a point, but one usually thinks of a scheme over this $S$ as being the generic fibre of some scheme over $n$-dimensional complex affine or projective space. For example, one can define a generic elliptic curve over the function field $\mathbb{C}(t_1, t_2)$, where $t_1$ and $t_2$ correspond to the two parameters in the Weierstrass equation.

Title: Space modelled on ring


问:
$GL(n,R)\cong O(n,R)\times B(n,R) $ as smooth manifolds, but are they isomorphic as Lie groups?
In the title, $B(n,R)$ is the upper tiangular matrix with positive entries on the diagonal. According to Gram-Schmidt normalization, we know that $\forall G\in GL(n,R),\exists U\in O(n,R),T\in B(n,R)$，s.t.
$G=UT$,the corresponding $U,T$ are unique, which induces diffeomorphism between $GL(n,R)$and $O(n,R)\times B(n,R)$ as smooth manifolds. My question is , does there exists a Lie group homeomriphism $f: GL(n,R)\rightarrow O(n,R)\times B(n,R)$?

The polar decomposition does not give a Lie group homeomorphism naturally because it does not preserves multiplicaton of matrix, i.e. $G_i=U_iT_i,i=1,2$，but $G_1G_2=U_1U_2T_1T_2$ does not hold for all elements in $GL(n,R)$.

答: No, the derived group $[Gl(n,R),Gl(n,R)]=Sl(n,R)$ of $Gl(n,R)$ is semisimple not the derived group of $O(n,R)\times B(n,R)$.


问:
Prove that $\overline{A} - A \subset \partial A$
Prove that $\overline{A}-A \subset \partial A$

I did the following:

Let $x \in \overline{A}$ $\wedge x \not \in  A $

$\forall r_x>0$ $\{B(x,r_x) \cap \overline{A} \not = \emptyset\}$

$\implies \forall r_x>0$ $\{B(x,r_x) - \{x\}\cap \overline{A} \not = \emptyset\}$

$\implies \forall r_x>0$ $\{B(x,r_x) - \{x\}\cap  \partial A \not = \emptyset\}$

$\implies x \in  \partial A$

$\therefore \overline{A}-A \subset  \partial A$

答:
definitions used:
Let A be a subset of a metric space (M,d).

$\overline A:=\{p\in M \mid \exists \delta\gt 0 \;\text{such that} \;B(p,\delta)\cap A\ne \emptyset\}$

$\partial A:=\{p\in M \mid \forall \delta\gt 0, B(p,\delta)\cap A\ne \emptyset, B(p,\delta)\cap (M-A)\ne \emptyset\}$

$\overline A -A$ may or may not contain points in $M-A$, if it doesn't, then it's the empty set $\emptyset \subset \partial A;$ if it does, then $\exists \delta'\gt 0 \;\text{such that} \;B(p,\delta')\cap A\ne \emptyset,B(p,\delta')\cap (M-A)\ne \emptyset$, which makes $\partial A$ clearly a subset of $\overline A -A$.


问:
How do you solve recurrent polynomial relation?
I managed to solve simple linear recursive relations with the help of generating functions, but I have trouble applying it to polynomial recurrent relations like this:

$$a_nx_n^2 - (2\sqrt{x_{n+1}} + a_nx_{n+1})x_n + x_{n+1}^{3/2}= 0$$

Where $x \ge 1 $.

答:
By setting
$x_n=y_n^2$,

$a_nx_n^2 - (2\sqrt{x_{n+1}} + a_nx_{n+1})x_n + x_{n+1}^{3/2}= 0
$
becomes
$a_ny_n^4 - (2y_{n+1} + a_ny^2_{n+1})y^2_n + y_{n+1}^{3}= 0
$
which is polynomial but definitely not linear.

Note that $y_{n+1}$
is a quadratic in $y_n$
so it can explicitly be represented in terms of
$y_n$ using the quadratic formula.


问:
Rate if change of surface area when volume of cylinder is increasing.
I get the same answer as the text book for part a) of the following question but cannot agree with it for part b).

The radius of the base of a right circular cylinder is r cm and its height is 2r cm. Find a) the rate at which its volume is increasing, when the radius is 2 cm and is increasing at 0.25 cm/s. b) the rate at which the total surface area is increasing when the radius is 5 cm and the volume is increasing at $5\pi$ cubic cm per second.

I have said let V = volume and A = area.

a)

$V = 2 \pi r^3$ and $A = 4 \pi r^2$

$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$

$= \frac{6 \pi r^2}{4} = 6 \pi$ when r = 2.

b)

$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$

$5 \pi = 6 \pi r^2.\frac{dr}{dt}$

$\frac{dr}{dt} = \frac{5 \pi}{6 \pi r^2} = \frac{5}{6r^2}$

$dA/dt = \frac{dA}{dr}.\frac{dr}{dt}$

$dA/dt = 8 \pi r.\frac{5 \pi}{6 \pi r^2} = \frac{4 \pi}{3}$

But the book says $2 \pi$

Is the book wrong?

答: $A=6 \pi r^2$.  Four of them come from the cylindrical surface, which is $2 \times 2\pi$ and two from the ends, which are $\pi r^2$ each.


问:
Find the time where their velocity is equivalent
The question is:

The equations of movement of 2 particles are:
$$x_1=3t^2+5t+6 \quad m$$ $$x_2=6t+8 \quad m$$
Find the exact time where both will have the same velocity.

So i've tried this: $$\vec{v}=\frac{dx}{dt}$$ basically differentiated both to get $$\vec{v_1}=6t+5\quad m/s$$ $$\vec{v_2}=6\quad m/s$$ and then i got their magnitude $|\vec{v_1}|$ and $|\vec{v_2}|$ as this $$|\vec{v_1}=\sqrt{36t^2+5^2}$$ $$|\vec{v_2}|=6$$ and then equated both to get the instant time where both velocities are equivalent, $$36t^2+5^2=6^2$$ $$t^2=\frac{11}{36}=\sqrt{\frac{11}{36}}$$ is this correct? i'm afraid i should have equated both velocity vectors and not their magnitude

答: The velocities are in the $x$ direction, so the magnitudes would just be the absolute values of $6t+5$ and $6$.  There is no need to take them.  Just equate the two velocities, giving $6t+5=6$ and solve for $t$.


问:
Parametrization of a circle, why is it $(\sin t, \cos t)$ instead of $(\sin^2t,\cos^2t)$?

The equation of a circle is $x^2 + y^2 = r^2$.

Then why is the parameterization of a circle $(r\cos(t), r\sin(t))$ and not $(r\cos^2(t), r\sin^2(t))$?

I thought what I'm supposed to do here is to eliminate $x$ and $y$ and write them in a way where I can represent them both with $t$ and then put it back into the equation? If I do that, I get

$$x = r\cos(t)$$
$$y = r\sin(t)$$

then if I put it back into the equation of the circle I get them both squared. Shouldn't the result then be $(\sin^2t,\cos^2t)$?

What's wrong here?

答:
If
$(x, y) = (r\cos^2(t), r\sin^2(t))
$
then $x+y=r$
which is the equation of a straight line, not a circle.