问:
Euler Lagrange equation, example
can anyone help me solve the following minimisation problem ( w.r.t the matrix $A\in \mathbb{R}^{n\times n}$) using the Euler Lagrange equations?

$$ \inf \Big\{ \int_0^1\sqrt{\langle A\dot{\sigma},\dot{\sigma} \rangle},~~ \sigma:[0,1]\to \mathbb{R}^n ~\text{absolutely continuous on $[0,1]$},\sigma_0=x,\sigma_1=y \Big\} $$

答:
Here is an approach not directly using the Euler-Lagrange equations, but supporting Ben Grossman's answer.

We can assume $A=A^T$; otherwise, replace $A$ by $\frac12\left(A+A^T\right)$. That is, since $A\in\mathbb{R}^{n\times n}$,
$$
\begin{align}
\langle Au,u\rangle
&=\left\langle u,A^Tu\right\rangle\tag{1a}\\
&=\left\langle A^Tu,u\right\rangle\tag{1b}\\
&=\left\langle\tfrac12\left(A+A^T\right)u,u\right\rangle\tag{1c}
\end{align}
$$
Minimizing yields
$$
\begin{align}
0
&=\delta\int_0^1\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}\,\mathrm{d}t\tag{2a}\\
&=\int_0^1\frac{\delta\langle A\dot\sigma,\dot\sigma\rangle}{2\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}\,\mathrm{d}t\tag{2b}\\
&=\int_0^1\frac{\langle A\dot\sigma,\delta\dot\sigma\rangle}{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}\,\mathrm{d}t\tag{2c}\\
&=-2\int_0^1\frac{\langle A\ddot\sigma,\delta\sigma\rangle\langle A\dot\sigma,\dot\sigma\rangle-\langle A\dot\sigma,\delta\sigma\rangle\langle A\dot\sigma,\ddot\sigma\rangle}{\sqrt{A\dot\sigma\cdot\dot\sigma}^3}\,\mathrm{d}t\tag{2d}\\
&=-2\int_0^1\langle A(\kappa_1\ddot\sigma-\kappa_2\dot\sigma),\delta\sigma\rangle\,\mathrm{d}t\tag{2e}
\end{align}
$$
where $\kappa_1=\frac1{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}$ and $\kappa_2=\frac{\langle A\dot\sigma,\ddot\sigma\rangle}{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}^3}$

Therefore, if $A$ has full rank and $\dot\sigma$ doesn't vanish, $\ddot\sigma\parallel\dot\sigma$, which means that $\sigma$ follows a straight line.

Title: Euler Lagrange equation, example


问:
Fourth moment of Z - why does this not work
I thought I could find $E(Z^4) = Var(Z^2)+E(Z^2)^2$ from the variance formula using $X$ as $Z^2$ (std normal square).  i got $Var(Z^2) = 2$ because it's a 1 degree of freedom chi square and $E(Z^2)^2 = 1$ because $E(Z^2)=1$. So it should equal 3. but when i was told that $E(Z_i^4) = 6$, where $Z_i \sim N(0,1)$. What am i doing wrong?

Thanks!

答:
The fourth moment of a normal should be 3 not 6.

Here is the rule,

For a standard normal, the odd moments are always 0.

The even moments (i.e. when m is even) are given by $2^{(-m/2)} \frac{m!}{(m/2)!}$

So if you plug in $m = 4$ you will get 3 not 6.

ContentLicense: CC BY-SA 3.0


问:
$30$ red balls and $20$ black balls are being distributed to $5$ kids. Each kid gets at least one red ball. In how many ways can we distribute balls?
$30$ red balls and $20$ black balls are being distributed to $5$ kids, so that each kid gets at least one red ball. In how many ways can we distribute balls?

Circle the correct answers:

a) $\binom{29}{4}$ $\binom{24}{20}$

b) $\binom{29}{5}$ $\binom{24}{5}$

c)$|Sur(N_{30},N_{5})|S(20,5)$ Note: $|Sur(N_{30},N_{5})|$ is the number of surjections, and $S(20,5)$ is a Stirling number of the second kind

d) None of $3$ previous answers are correct

My approach:

First I gave each of $5$ kids one red ball, which leaves me with $25$ red balls. Now I used the stars and bars method to distribute the balls I am left with.

Red balls: $x_1+x_2+x_3+x_4+x_5=25,x_i\geq 0, i=1,..,5$. This equation has $\binom{5+25-1}{25}=\binom{29}{25}=\binom{29}{4}$.

Black balls: $x_1+x_2+x_3+x_4+x_5=20,x_i\geq 0, i=1,..,5$. This equation has $\binom{5+20-1}{20}=\binom{24}{20}=\binom{24}{4}$.

So I would say a) is the correct answer.. am I right?

答:
Number of ways of distributing balls into boxes tells us the following:

Theorem 1: The number of ways in which $n$ indistinguishable balls can be distributed into $k$ distinguishable boxes is $\binom{n+k-1}{n}, $ which also equals $\binom{n+k-1}{k-1}$.

So... give one red ball to each kid because you have to, i.e. you have no choice here because there cannot be a kid without a red ball. Now you have 25 red balls and 20 black balls remaining.

But after this you have total freedom: any ball can go to any kid with no restrictions. But how many ways are there of doing this? Using Theorem 1, there are $\binom{25+5-1}{25} = \binom{29}{25}$ ways of distributing the red balls out, and $\binom{20+5-1}{20} = \binom{24}{20}$ ways to distribute the black balls out. Since "distributing red balls" and "distributing black balls" in this way are independent, there are no duplicate scenarios to worry about, so we may use the product rule of counting, and the answer is:

$\binom{29}{25} \times \binom{24}{20} $, which is answer (a).

ContentLicense: CC BY-SA 4.0


问:
Coherence of topology and finest topology such that inclusion maps are continuous
Consider the following problem from Lee's Introduction to Topological Manifolds:

Let $(X,\mathcal{T})$ be a topological space and $\{X_{\alpha}\}$ be a collection such that $\bigcup_\alpha X_\alpha=X$. Show that $(X,\mathcal{T})$ is coherent with $\{X_{\alpha}\}$ if and only if $\mathcal{T}$ is the finest topology such that $X_\alpha \hookrightarrow X$ is continuous for all $X_\alpha$.

I am having some trouble finishing the proof of the forward direction, in particular with proving that $\mathcal{T}$ is indeed the finest topology. My attempt is as follows:

Let $\mathcal{T}'$ be another topology such that $X_\alpha\hookrightarrow X$ is continuous. My first question here is that, for the inclusion map $\iota_\alpha:X_\alpha\to X$, is the topology on $X_\alpha$ is the subspace topology induced by $\mathcal{T}'$ and, further, is $\mathcal{T}'$ the topology on $X$? I will proceed as if those assumptions are true.

We want to show that $\mathcal{T}'\subseteq \mathcal{T}$. Let $U\in\mathcal{T}'$, since $\{X_{\alpha}\}$ is coherent with respect to $(X,\mathcal{T})$, it suffices to show that $U\cap X_\alpha$ is open in $X_\alpha$, where $X_\alpha$ has the subspace topology induced by $\mathcal{T}$. Notice that
$$\iota_\alpha^{-1}(U) = U\cap X_\alpha,$$
which is open in $X_\alpha$ with respect to the subspace topology on $X_\alpha$ induced by $\mathcal{T}'$. I'm not sure how to connect the above concepts. That is, I am having trouble showing that if $U\cap X_\alpha$ is open with respect to the subspace topology on $X_\alpha$ induced by $\mathcal{T}'$, it is also open with respect to subspace topology on $X_\alpha$ induced by $\mathcal{T}$.

However, if we assume that, for the inclusion map $\iota_\alpha:X_\alpha\to X$, the topology on $X_\alpha$ is the subspace topology induced by $\mathcal{T}$, then the problem is resolved. I'm not quite sure which of these interpretations is correct, thanks!

答:
Let us begin with some notation: By $\mathcal T_\alpha$ we denote the given topology on $X_\alpha$. The set $X = \bigcup_\alpha X_\alpha$ does not yet have a topology, but we define the topology $\mathcal T$ coherent with the family $(\{X_\alpha,\mathcal T_\alpha) \}$  as the set all $U \subset X$ such that $U \cap X_\alpha \in  \mathcal T_\alpha$ or all $\alpha$. This topology by definition has the property that all  $\iota_\alpha : (X_\alpha, \mathcal T_\alpha) \hookrightarrow (X, \mathcal T)$ are continuous (simply because $\iota_\alpha^{-1}(U) = U \cap X_\alpha \in  \mathcal T_\alpha$ for $U \in \mathcal T$).

Let us emphasize that we start with a collection of topological spaces $\{X_\alpha \}$. In your question the topology on the set $X_\alpha$ is not named, here we explicitly give it the name $\mathcal T_\alpha$. Note that without a given topology on the set $X_\alpha$ it does not make any sense to define (as you did in your comment) that $(X, \mathcal T)$ is coherent with the family $\{X_\alpha\}$ if, for all $U \subset X$, the set $U$ is open in $X$ if and only if $U \cap X_\alpha$ is open in $X_\alpha$ for all $\alpha$.

Now you consider any topology $\mathcal{T}'$ on $X$ such that all $\iota_\alpha : (X_\alpha, \mathcal T_\alpha) \hookrightarrow (X, \mathcal T')$ are continuous. So yes, $\mathcal{T}'$ is the topology on $X$. This does not mean that the topology $\mathcal T_\alpha$ on $X_\alpha$ is the subspace topology induced by $\mathcal{T}'$. It only says that the original topology $\mathcal T_\alpha$ on $X_\alpha$ is finer than the subspace topology $\mathcal T'_\alpha = \{ U \cap X_\alpha \mid U \in \mathcal T'\}$ induced by $\mathcal T'$ on the set $X_\alpha$. This will be shown below.
In your proof you want to show that for $U \in \mathcal T'$ we have $U \cap X_\alpha \in  \mathcal T'_\alpha$. But this does not require a proof because $\mathcal T'_\alpha$ was defined as the subspace topology induced on $X_\alpha$ by $\mathcal{T}'$.

So let us give a correct proof of $\mathcal T' \subset \mathcal T$ which is also more or less obvious.

Consider $U \in \mathcal T'$. Then, because the $\iota_\alpha$ are assumed to be continuous, $U \cap X_\alpha = \iota_\alpha^{-1}(U) \in \mathcal T_\alpha$ for all $\alpha$ and thus by definition of the coherent topology, we have $U \in \mathcal T$.

The argument in the preceding sentence also shows that $\mathcal T'_\alpha \subset \mathcal T_\alpha$, i.e. that $\mathcal T_\alpha$ is finer than  $\mathcal T'_\alpha$ (as claimed above).


问:
Image of a coadjoint orbit of SU(3) under a moment map
Let M be a coadjoint orbit of dimension 6 of $SU(3)$, and let T be the maximal torus in $SU(3)$. If we denote $\mu : M \longrightarrow \mathbb{R}^2$ the moment map associated to the action of T on M, then the image of the moment map is a hexagon with vertices are image of $M^T$ by $\mu $.

My questions are:

$1.$ What is  $M^T$? (My attempt was to choose a regular element $ X \in \mathfrak{t} \simeq  \mathfrak{t}^*$, and consider M to be the orbit of X, and then I get  $M^T=\lbrace y=gxg^{-1} \in M, ty=yt ,\forall t\in T \rbrace=   \mathfrak{t} ?).$

$2.$ why is the image of the moment map a hexagon? Well, I know from convexity theorem that the  image of the moment map, $\mu(M)$, is the convex hull of $\lbrace \mu(F)$, F connected component of $M^T\rbrace$, and that $\mu$ is constant on each connected component of $M^T$ and this implies that the set $\mu(M^T)$ is finite, but how can we find the components of $M^T$ and the cardinal number of the set $\mu (M^T)$ without having an explicit formula of $\mu$?.

Any feedback would be greatly appreciated!

答:
The elements in $su(3)$ satisfy $X+X^T=0$ and are unitarily diagonalizable, with purely imaginary eigenvalues. This means that, identifying $su(3)$ with its dual and adjoint and coadjoint actions, the orbits are labeled by the triples of real numbers, the imaginary parts of the eigenvalues. The highest dimensional orbits are the ones corresponding to distinct eigenvalues. A maximal torus consists of those matrices in $SU(3)$ that are diagonal in some basis. We can take the standard basis, and then "the" maximal torus consists of unitary diagonal matrices, $diag\{e^{\theta_1}, e^{i\theta_2}, e^{i\theta_3}\}$ with $\theta_1+\theta_2+\theta_3=0$. Fix an  orbit corresponding to $\lambda_1, \lambda_2, \lambda_3$. The matrices in this orbit invariant under the torus are diagonal matrices, of which, having fixed the triple $\lambda_1, \lambda_2, \lambda_3$, we get 6 -- all the diagonal matrices with all the permutations of the eigenvalues $i\lambda_1, i\lambda_2, i\lambda_3$ on the diagonal. These give the 6 vertices of the hexagon you are after.

Remark 1: The $t^*$ can be thought of as the subspace $v_1+v_2+v_3=0$ inside a 3D space "dual to tangential directions to $\theta$s". The images of the 6 fixed point under momentum map are permuted when we permute the coordinates in this 3D space, so they are either one point, vertices of a triangle, or of a hexagon. These should correspond to the cases of all equal eigenvalues, two equal eigenvalues, and all eigenvalues distinct, respectively.

Remark 2: The permutation group taking the 6 fixed points to each other is the Weyl group of $su(3)$. In general, there should be a description of everything in appropriate representation-theoretic language; but I'm actually not an expert on this, so I'll leave it to someone else to supply it.


问:
Combination problem with balls and colors, how many ways?
A bag contains 40 balls from 4 different colors: red, white, black and green. There are 10 balls
numbered (1, 2, 3, ..., 10) from each color. 5 balls are randomly selected from the bag.

In how many ways can 5 balls be chosen so that no more then 2 balls are of the same color?

Can you help me with this? I know that there are 2 cases, 1: 2 balls with same color & 3 balls with different colors 2: 2 balls with same color, another 2 balls with same color and 1 ball with different color. However, I could not understand my instructor's explanation.

答:
Responding to comment:

The reason why your attempt is incorrect is because you are double counting. For instance,

"black 1, black 2, red 4, red 3, and green 7"

and

"red 4, red 3, black 1, black 2, and green 7"

are counted separately in your method, due to the $\binom{4}{1} \binom{3}{1}$ computation. If you divide your answer by $2$ to account for this, you match your instructor's solution.


问:
In graph theory, is there a term for the value of the difference between indegree and outdegree?
I'm working on a Graph Theory research project, and one of the key components is talking about the indegrees and the outdegrees of a few particular vertices. I need to define a term for the value of indegree-outdegree for a particular vertex. Or for vertex $v$ does the value $\deg^{-}(v) - \deg^{+}(v)$ have a name?

Some possible terms have been "netdegree" or "flow value," which would suffice for my needs, but I would like to make sure I use the existing terminology if possible.

答:
I believe this is called the imbalance of a vertex, see for example this paper: https://core.ac.uk/download/pdf/42932978.pdf

Note that this is defined as indegree - outdegree, but clearly this can be easily reversed.


问:
Who was V. Viskovatov?
I'd be interested to learn some biographical detail about Vasilii Viskovatov, whose name is associated with a method for converting (a ratio of) power series to a "corresponding" continued fraction, c.f. Handbook of Continued Fractions for Special Functions, p. 20 (Springer, 2008).

A few Internet hits (prominently Google books links, not wholly accessible) provide the following literature reference to his work, albeit with apparent mis-transliterated initial:

B. Viskovatov: 'De la méthode générale pour réduire toutes sortes des quantités en fractions continues', Memoires de L'Academie Impériale des Sciences de St. Petersburg, 1(1803-1806), pp. 226-247

[Trans.: 'A general method for reducing all types of quantities to continued fractions'.]

The algorithm itself can be described simply.  Given the ratio of two power series $p(z)/q(z)$ with $q(0) = 1$, remove as a constant term $b_0 = p(0)$, and after factoring out the leading term $z$ in the numerator, invert the remaining factor from $p(z) - p(0)q(z)$:

$$ \frac{p(z)}{q(z)} = b_0 + \frac{a_ 1 z \cdot r(z)}{q(z)} = b_0 + \frac{a_1 z}{q(z)/r(z)} $$

where for simplicity I've assumed a single power of $z$ in the numerator.  Continue to apply the process to $q(z)/r(z)$ and a corresponding-type continued fraction emerges.

答:
It might be Vasilii Viskovatov:

http://encyclopedia2.thefreedictionary.com/Viskovatov,+Vasilii+Ivanovich

The time-frame and location is correct.

Don't forget that 'Vasilii' written in Cyrillic starts with a glyph that looks like a roman B:

Title: Who was V. Viskovatov?


问:
$A+B=\prod_{i=1}^kP_i^{\min(a_i,b_i)}$ and $A\cap B=\prod_{i=1}^kP_i^{\max(a_i,b_i)}$
Question: Prove that $A+B=\Pi_{i=1}^kP_i^{\min(a_i,b_i)}$ and $A\cap B=\Pi_{i=1}^kP_i^{\max(a_i,b_i)}$

My Thoughts: In the above question we are assuming that $R$ is a Dedekind domain with field of fractions $F$.  Also, $A=P_1^{a_1}\cdots P_k^{a_k}$ and $B=P_1^{b_1}\cdots P_k^{b_k}$ are nonzero ideals of $R$ with $P_1,\dots,P_k$ distinct primes of $R$. So I suppose that $R$ is a hereditary ring, and so every nonzero ideal is invertible... but I am not quite sure if this is a good strategy to take... any help is greatly appreciated!  Thank you.

答:
There are a lot of different characterizations of a Dedekind domain, so if I'm using some result or property you haven't seen let me know. The main slogan here is that, in a Dedekind domain, "to contain is to divide"; ie, if $I$ and $J$ are nonzero ideals, $I$ divides $J$ if and only if $I$ contains $J$.

Now, the case where either $A=R$ or $B=R$ are straightforward (check the details yourself!), so we may assume that there is some $a_i$ non-zero and some $b_j$ non-zero. Likewise, we may assume the $P_i$ are all nonzero.

Now, certainly $P_i^{\min(a_i,b_i)}\supseteq P_i^{a_i}$ and $P_i^{\min(a_i,b_i)}\supseteq P_i^{b_i}$ for each $i$. Hence $A\subseteq\prod_{i=1}^kP_i^{\min(a_i,b_i)}$ and $B\subseteq\prod_{i=1}^kP_i^{\min(a_i,b_i)}$, and so $A+B\subseteq\prod_{i=1}^kP_i^{\min(a_i,b_i)}$. Thus $\prod_{i=1}^kP_i^{\min(a_i,b_i)}$ divides $A+B$, and so – by unique factorization into prime ideals – to show equality we just need to show (a) that no prime ideal of $R$ other than the $P_i$ divides (contains) $A+B$, and (b) that no power of $P_i$ greater than $\min(a_i,b_i)$ divides (contains) $A+B$.

For (a), let $Q$ be a prime ideal such that $A+B\subseteq Q$. Then (eg) $\prod_{i=1}^kP_i^{a_i}\subseteq Q$, and there is some $a_i>0$, and so by definition of prime ideals we have $P_i\subseteq Q$. But "to contain is to divide", and so $Q$ divides $P_i$, and thus by unique factorization of ideals $P_i=Q$. (This is just the fact that Dedekind domains have Krull dimension $1$, if you prefer that language.) Thus the $P_i$ are the only prime ideals containing $A+B$, and hence are the only prime divisors of $A+B$.

For (b), fix some $i$; without loss of generality we may assume $a_i\leqslant b_i$. (The other case is identical.) Now, $A\subseteq P_i^{a_i}$. But also, by unique factorization into primes, we have $A\nsubseteq P_i^{a_i+1}$, or else $P_i^{a_i+1}$ would divide $A$. Clearly this means that $A+B\nsubseteq P_i^{a_i+1}$, and so $P_i^{\min(a_i, b_i)+1}=P_i^{a_i+1}$ does not divide $A+B$.

So, summing everything up, we have that the $P_i$ are the only prime ideal divisors of $A+B$, and that every $P_i$ divides $A+B$ with degree precisely $\min(a_i,b_i)$. By unique factorization of prime ideals this gives us $A+B=\prod_{i=1}^k P_i^{\min(a_i,b_i)}$, as desired. The second problem is very similar; do you think you can see how to do it now?


问:
Posterior as Proportional to the Product of Likelihood and Prior
In many accounts of Bayesian inference the posterior is written as being proportional to the product of the likelihood and the prior:

$$
P(H \mid D) \propto P(D \mid H) P(H)
$$

Why write it as a proportionality ?  How does that impact actual estimation procedures? Why not work directly with Bayes' formula ?:

$$
P(H \mid D) =  \frac{P(D \mid H) P(H)}{P(D)}
$$

答:
In some situations, computing the denominator $P(D)$ can be computationally prohibitive, since it can be a sum (e.g. $P(D) = \sum_h P(D \mid H=h) P(H=h)$) or an integral (for continuous random variables).

If the information you need from the posterior probability doesn't require this normalizing constant (e.g. all you care about is how the posterior probabilities for different values of $H$ compare relative to each other), then you don't need to compute the normalizing factor $P(D)$ at all.


问:
$P(a < X \leq b) = F(b) - F(a)$?
Let X be a random variable with distribution function F, $a,b \in \Re$ with $a \lt b$ and $F(x-)$ is the limit of $F$ in $x$ from the left.

I tried to proof that the equality is correct, but I'm stuck by not comprehending what to do next, so far I had:

$(x,y] = (-\infty,y]-(-\infty,x]$

$\Rightarrow \forall x \in \Re$ we have

$F_X(x)=P(X \leq x) = P(X \in (-\infty,x])$

Then I assume that I need to prove that the continuity property of distribution functions from the right still satisfied, but I don't know how to do it with this elements. Can someone clear my mind a little bit? Thanks

答:
You have the right idea. Since $(-\infty,a]$ and $(a, b]$ are disjoint, you have
$$P(X \in (-\infty, a]) + P(X \in (a, b]) = P(X \in (-\infty, a] \cup (a, b]) = P(X \in (-\infty, b]).$$

Title: $P(a < X \leq b) = F(b) - F(a)$?


问:
Probability of a random set of binary numbers XORing to $0$
Question - Given a randomly generated set of binary numbers, what is the probability that all of them, when XORed with each other, yield $0$?

Additional information -

All of the binary numbers in the set are different from each other. As an example, $\{0001, 0010, 0011\}$ is a valid set, but not $\{0001, 0010, 0010\}$ or $\{0010, 0010, 0010\}$.

All of the binary numbers have equal length $n$ which is some given. For example, one could have a given length of 4, meaning numbers such as $00001$ and $010$ are not allowed.

The length of the set $m$ is a given, its members are randomly generated.

Each possible valid binary number has an equal likelihood of being generated.

Context - For a school math project, I decided to analytically investigate the average percentage of errors an error correction code can correct given a data set and error rate. This question I am asking was a part of said project that I am unable to solve. If necessary, I can provide additional context.

I already know how to do this problem if restriction 1 is lifted, but did not really know how to proceed. I visualized the randomly generated data set as being one number stacked on top of another, kind of how we write when adding large numbers on paper by hand, and then taking my answer to be the probability that each column of bits had an even number of $1$s.

Updated context - Looking at the answers already provided and doing some of the math myself, it is clear that using this in my project will take me well over the page count in all of the explanations. I have come up with a simpler path for my investigation, but am leaving this question up here as I am still interested in this problem and I see no reason to stop anyone interested in answering or seeing answers to this question from doing so.

1rst edit - I have assigned $m$ as the length of the set and $n$ as the length of the binary sequences

2nd edit - I have added additional context

答:
Computation for the cases $\ n=1\ $ to $\ n=4\ $ strongly suggests that the probability of the elements of a randomly selected non-empty set of binary sequences of length $\ n\ $  XORing together to zero is $\ \frac{2^{2^n-n}-1}{2^{2^n}-1}=\frac{1}{2^n}\left(1-\frac{2^n-1}{2^{2^n}-1}\right)\ $.

The computations showed that for the above-mentioned values of $\ n\ $, $\ 2^{2^n-n}\ $ of the non-empty subsets of binary sequences of length $\ n\ $ have elements which XOR together to each of the $\ 2^n-1\ $ non-zero binary sequences of that length, and $\ 2^{2^n-n}-1\ $ of them have elements which XOR together to the zero sequence.  I haven't yet managed to prove the result for arbitrary $\ n\ $.


问:
Probability of 3x3 cube
A cube with side length as 3, paint all faces red (initially all white), then cut it into 27 small cubes with side length of 1, randomly pick 1 cube, roll it once and see five faces white, what is the probability that cube has the other face as red?

Attempt:

6 cube with 1 red face, 1 cube with 0 red face.

So the answer is 6/7?

答:
Baye's theorem is the way to go here: let $W$ be the event the chosen cube is all white, let $R$ be the event the chosen cube has $5$ white faces, and let $E$ be the event that when the cube is rolled, five white faces appear. By Baye's theorem,
$$
P(R|E)=\frac{P(E|R)P(R)}{P(E)}=\frac{P(E|R)P(R)}{P(E|R)P(R)+P(E|W)P(W)}
$$
The reason for the denominator is the law of total probability, and the fact that the events $R$ and $W$ cover all possible outcomes in $E$. You should be able to fill in these probabilities.

To give this an intuitive angle, given that you see five white faces alone, all possible instances of $5$ white faces are equally likely. Six of these instances come from the cubes with one red side (one per cube), and six of these come from the central white cube. Therefore, the probability is $6/12$, not $6/7$.

Title: Probability of 3x3 cube


问:
Eigenvalues for solving linear recurrence equations
After discovering the way to come up with formulas for such sequences as the Fibonacci sequence using matrices it occurred to me this may not necessarily always possible. What might skew one’s endeavor is a case in which one of the eigenvalues is 1, 0 or -1 for then applying the linear transformation repeatedly does not do much to put it mildly. I was wondering if there is a way around this. Is it possible to solve recurrence relations using compositions of transformations even if the above conditions apply?</p>

答:
Consider the recurrence relation $x_{n+1}=x_{n-1}$ for $n\geq1$, $x_0=a, x_1=b$.  For $n\geq1$ we have:
$$
\left( \begin{array}{cc}0&1\\1&0
\end{array}\right)
\left( \begin{array}{cc}x_n\\x_{n-1}
\end{array}\right)=
\left( \begin{array}{cc}x_{n+1}\\x_{n}
\end{array}\right)
$$

The above matrix has eigenvalues $1,-1$ and the general solution to the recurrence relation is as usual $$x_n=u1^n+v(-1)^n=u+v(-1)^n.$$

Solving for $u,v$: $$u+v=a, \qquad u-v=b,$$
so $$u=\frac{a+b}2, v=\frac{a-b}2.$$

As you can see the fact that the eigenvalues were $1,-1$ did not change anything.

Next consider the recurrence relation $y_{n+1}=y_{n}$, $n\geq1$ with $y_0=a,y_1=b$.

Then for $n\geq1$ we have:

$$
\left( \begin{array}{cc}1&0\\1&0
\end{array}\right)
\left( \begin{array}{cc}y_n\\y_{n-1}
\end{array}\right)=
\left( \begin{array}{cc}y_{n+1}\\y_{n}
\end{array}\right)
$$

Now we have eigenvalues $0,1$ and the general solution has the form: $$u1^n+v0^n=u+v0^n.$$
Solving for $u,v$ (remembering $0^0=1$) we get: $$u+v=a,\qquad u=b,$$
so
$$
u=b,\qquad v=a-b.
$$

Again nothing about the calculation was different.


问:
Compute the chromatic polynomial of the $n$-cycle graph, i.e., the only connected graph on $n$ vertices each of which has degree $2$.
Compute the chromatic polynomial of the $n$-cycle graph, i.e., the only connected
graph on $n$ vertices each of which has degree $2$.

I can only think of how to compute this recursively. That is, using the fact that
$$p_G(k)=p_{G-e}(k)-p_{G/e}(k)$$
where $p_G{k}$ is the chromatic polynomial of $G,$ who has some edge $e$, $G-e$ is the graph obtained by the deletion of $e$, and $G/e$ is the graph obtained by the contraction of $e$, we can see that for some $n$-cycle
$$p_{G-e}(k)=\text{some tree on }n \text{ vertices}\\p_{G/e}(k)=\{\text{a }(n-1) \text{-cycle}\}$$
We know that a tree on $n$ vertices has chromatic polynomial
$$p_{G-e}(k)=k(k-1)^{n-1}$$
we can then use this same trick for $p_{G/e}(k)$, and then again and again, eventually recursively building our desired polynomial. I'm not sure this is what was intended for this exercise, as my professor made it seem like we should have some closed formula. Any help would be amazing!

答:
Let $p_n$ be the chromatic polynomial of the $n$-cycle; then $$p_n = k(k-1)^{n-1} - p_{n-1}.$$
If we set $q_n = (-1)^n p_n$, then multiply the recurrence above by $(-1)^n$, we get
$$
   q_n = -k(1-k)^{n-1} + q_{n-1} \implies q_n = q_3 + \sum_{i=4}^n -k(1-k)^{i-1}
$$
which we can solve for $q_n$, because the sum is just a finite geometric series and $q_3 = -p_3 = -k(k-1)(k-2)$. This gives us a closed form for $p_n$.


问:
What's the problem with $-2=(-8)^{\frac{1}{3}}=(-8)^{\frac{2}{6}}=\sqrt[6]{(-8)^{2}}=2$?
$-2 =(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^{2}}=2$
 first glance I want to show for $a^{rs}$ to work, both $a^r$ and $a^s$ need to be valid, but as you can see, both $-8^{\frac{1}{3}}, -8^{\frac{2}{2}}$ is defined. so I wanna know how do I prove the above statement is wrong

答:
The sixth root of a number is considered even and for any even root, the answer can be positive or negative. You pick which one is best fit or both, which in this case is -2 as shown from your starting point.

Another issue is by stating that -2=2, you imply that 0=4 or 0=1, which is a tell-tale sign something is wrong.


问:
Bijection with constraints between sets of polynomial
How can I show the existence (or even better: construct explicitly) of a bijection (if one exists) between the set of all polynomial that have integer coefficients and a integer root and the set of all polynomials that have positive integer coefficients and positive integer roots ?

If I drop the additional "root" constraint, the problem is very simple to solve. But with the root constraint I have no idea how to solve it, since my knowledge of algebra is rather limited. For example if I use the function $h:\mathbb{Z} \rightarrow \mathbb{N}, \ h:=\begin{cases}
2t & ,t\geqslant0\\
-2t-1 & ,t<0\end{cases} $ to map the integer coefficients bijectively to positive integer coefficients, I can find counterexample that have no integer roots whatsoever.

答:
As has been pointed out, with the way you have worded things, there is no bijection, since one of your collections of polynomials is infinite, while the other collection is empty.

Let us look at a generalization of the problem.  Suppose you have an explicit listing of two types of polynomials with integer coefficients, Type $1$ and Type $2$ (in your case, Type $1$ is all polynomials with integer coefficients, Type $2$ is positive integer coefficients).

Now suppose that you want to exhibit a bijection between Type $1$ polynomials that satisfy a certain additional condition $A$, and Type $2$ polynomials that satisfy an additional condition $B$.  This will not be possible if the cardinalities of the two collections are different.  So suppose for example that there are infinitely many Type $1$ polynomials that satisfy condition $A$, and infinitely many Type $2$ polynomials that satisfy condition $B$.

We can list the Type $1$ polynomials that satisfy condition $A$ as $P_1$, $P_2$, $P_3$, and so on, by letting $P_1$ be the first polynomial in the listing of Type $1$ polynomials that satisfies $A$, letting $P_2$ be the second one, and so on.

Similarly, we can list the Type $2$ polynomials that satisfy condition $B$ as $Q_1$, $Q_2$, $Q_3$, and so on by using the same strategy.

Now we get an explicit bijection between Type $1$ polynomials that satisfy $A$ and Type $2$ polynomials that satisfy $B$ by mapping $P_k$ to $Q_k$ for all $k$.

Is this bijection really explicit?  Suppose we have a computer program for listing  Type $1$ polynomials, and a computer program that can recognize whether a polynomial satisfies condition $A$.  Suppose also that we have similar programs for Type $2$ polynomials and for recognizing condition $B$.  Then we can write an explicit computer program that will produce the bijection.

In general, one cannot expect that the bijection will be given by a simple short formula.  But what I have outlined is more than an existence proof, it is, under appropriate conditions, a recipe.


问:
Uniform Vs Pointwise convergence - tricky problem
Consider $$ f_n (x) = \frac{log(x^{2n})} {(1 +x) ^{2n} } $$ for $x \in (0,1]$.
What can I say about the convergence of this sequence of functions?

My attempt:

The sequence converges pointwise to the constant function 0, because $$ lim_{n \to \infty} f_n (x) =0$$ for $x \in (0,1]$.
Now I don't know if the sequence converges uniformly. I suspect it doesn't. However, I get stuck trying to bound the supremum of $\vert f_n \vert$.
I think it would be easier in a  compact interval, e.g. $[a,1], 0\lt a \lt 1$. There, I would say $sup \vert f_n (x) \vert = \frac{2n\vert log (a) \vert} {(1+a)^{2n}} $, which tends to 0.
Am I mistaken?
Many thanks.

答:
To confirm, for all $x \in (0,1]$, we have the pointwise convergence

$$\lim_{n \to \infty}\frac{\log x^{2n}}{(1+x)^{2n}} = \lim_{n \to \infty}\frac{2n\log x}{(1+x)^{2n} } =0$$

since $(1 +x)^n  >  nx $ and $\displaystyle\frac{n}{(1+x)^{2n}} < \frac{n}{n^2x^2} = \frac{1}{nx^2} \underset{n \to \infty} \longrightarrow 0$.

The sequence does not converge uniformly on $(0,1]$ since

$$\sup_{x \in (0,1]}\left|\frac{2n\log x}{(1+x)^{2n} }\right| = +\infty,$$

and, thus, $\lim_{n\to \infty} \sup_{x \in (0,1]} |f_n(x)| \neq 0$.

Alternatively, we can argue since $\frac{1}{n} \in (0,1]$ it follows that

$$\sup_{x \in (0,1]}\left|\frac{2n\log x}{(1+x)^{2n} }\right| = \sup_{x \in (0,1]}\frac{-2n\log x}{(1+x)^{2n} } \geqslant \frac{-2n\log \frac{1}{n}}{(1+\frac{1}{n})^{2n} } = \frac{2n\log n}{(1+\frac{1}{n})^{2n}} \geqslant \frac{2n\log n}{e^2},$$

and the limit of the LHS is $+\infty$ since $n \log n \underset{n \to \infty} \longrightarrow +\infty$.


问:
In a group $G$, if $a \in G$ is the only element such that its order is 2, prove that $G = C(a)$
(i) As $C(a) \leq G $, $x \in C(a) \implies x \in G$. Hence $C(a) \subset G$.

(ii) How to prove $G \subset C(a)$?

I tried to start with $a^2 = e$ and manipulate this to get $ax = xa$, but it doesn't work.

答: Hint: What is the order of $x^{-1}ax$?

Body: Hint: What is the order of $x^{-1}ax$?


问:
If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$
If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$

It is easy to check the equality if $y$ is a multiple of $x$.

I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.

Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so

$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$

That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

答:
If $i$ is an irrational number is a limit of a sequence $r_n$ of rational numbers
$$ i= \lim_{n \rightarrow +\infty} r_n .$$
Thus
$$\forall N \in \mathbb N \quad \exists n_0 \in \mathbb N  \quad \forall n \in \mathbb N \quad  n \ge n_0  \Rightarrow  |i-r_n| \le \frac{1}{N}
 \Rightarrow  \quad |ix-r_nx| \le \frac{|x|}{N} \quad -\frac{|x|}{N} + r_n\cdot x \le ix \le \frac{|x|}{N} + r_n \cdot x.$$
If $x \ge 0$ then
$$f(x)^{r_n-\frac{1}{N}} \le f(ix) \le f(x)^{r_n + \frac{1}{N}}.$$
This implies that
$$f(i\cdot x) = f(x)^{i}.$$
The case $x \le 0$ is similar.


问:
Prove Ramanujan's formula for nested cubic roots $\sqrt[3]{{a}+b\sqrt[3]{r}}$
Ramanujan found that

$$\begin{align*} & \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}\\ =&\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9}-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9}+\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
\end{align*}$$
for arbitrary $m$ and $n$. The problem is that I am not sure how to prove it.

Question: Is there a way to prove it?

Similarly to the Ramanujan's formula for $$\sqrt{\sqrt[3]{a}+\sqrt[3]{b}}$$
I tried starting with a polynomial and deriving above formula through some clever manipulation. Unfortunately, none of that worked. If I were to do it that way, the terms in the polynomial would have to be a square root.

The book provides its way of proof by cubing both sides brute-force and slugging out the difficult algebra. Is there another way to prove it?

Body:
Denote
$$c_1=\sqrt[3]{\tfrac {(m-n)(m+2n)^2}9},\>\>\>
c_2=-\sqrt[3]{\tfrac {(2m+n)(m-n)^2}9},\>\>\>
c_3=\sqrt[3]{\tfrac {(m+2n)(2m+n)^2}9}
$$

and it is straightforward to verify that

\begin{align}
c_1c_2c_3 &=- \frac{1}{9}(m-n)(2m+n)(m+2n)\\
c_1^3+c_2^3+c_3^3 &= \frac{1}{3}(m^3+6m^2n+3mn^2-n^3)\\
c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 &=
\frac{1}{3} c_1c_2c_3(m^3-3m^2n-6mn^2-n^3)\\
\end{align}
Next, let
$$A=c_1+c_2+c_3, \>\>\> B=c_1c_2+c_2c_3+c_3c_1,\>\>\>C= c_1c_2c_3$$
and evaluate
\begin{align}
A^3 &= 3AB+ c_1^3+c_2^3+c_3^3-3c_1c_2c_3\\
&=3AB+ (m^3+3m^2n-n^3)\tag1 \\
B^3 &= 3c_1c_2c_3 AB + c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3-3(c_1c_2c_3)^2\\
&=3(AB)C +(m^3-3mn^2-n^3)C\tag2
\end{align}
and their product
$$
A^3B^3 = 9(AB)^2C-27(AB)C^2+27C^3
-\frac13C(m^2+mn+n^2)^3\tag3
$$
where the followings are recognized in arriving at (3)
\begin{align}
& (m^3+3m^2n-n^3)+(m^3-3mn^2-n^3)=-9C \\
& (m^3+3m^2n-n^3)(m^3-3mn^2-n^3)=27C^3-\frac13(m^2+mn+n^2)^3\\
\end{align}
Rearrange (3)
$$(AB-3C)^3=-\frac13C(m^2+mn+n^2)^3$$
and substitute $AB$ via (1) to obtain the equation for $A$
$$[A^3-(3mn^2+n^3-m^3)]^3 = -9C(m^2+mn+n^2)^3$$
which leads to the Ramanujan formula
$$A= \sqrt[3]{(m^2+mn+n^2)\sqrt[3]{(m-n)(m+2n)(2m+n)}+3mn^2+n^3-m^3}
$$


问:
How to expand the Fourier series for $f(x)=\max \{0, \frac{\pi}{2}-\lvert x\rvert \} $?
My Question: My Goal is to determine the Fourier series for $f(x)=\max \{0, \frac{\pi}{2}-\lvert x\rvert \} \quad$ for $x \in [-\pi, \pi ]$ This function is $2\pi$-periodic.

My Approach: i found out, $f(x)$ is an even function, because its graph is symmetric to the y-axis. The Expression: $\frac{\pi}{2}-\lvert x \rvert$ has to "zeros":

$\frac{\pi}{2}-\lvert x \rvert = 0$

$\frac{\pi}{2}= \lvert x \rvert  $

$x_{0}=-\frac{\pi}{2}$

$x_{1}=+\frac{\pi}{2}$

the graph must look somehow like that.. so the function is linear all over the Intervall:

But now Comes the hard stuff. I'm stuck in building the Fourier series for $f(x)$
The integral should be builded maybe this way, but i am not sure if if am right:

$\int (\frac{\pi}{2}-x)\cdot \cos (nx)\ dx = \frac{\pi}{2n}\sin (nx) - \frac{x}{n}\sin (nx) - \frac{1}{n^2}\cos (nx)$

What do you think, is that right? And what would be the next step? My textbook doesn't describes the further calculation.

p.s. edits for improving language and latex

答:
Since the function is even, the coefficients of $\sin nx$ are all zero.  The constant term is the average value of the function on $[-\pi,\pi]$:
$$
\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,dx\;=\; \frac{1}{2\pi}\left(\frac{\pi^2}{4}\right) \;=\; \frac{\pi}{8}
$$
(The integral here was evaluated using the formula for the area of a triangle.)

The coefficient of $\cos nx$ is given by the formula
$$
a_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\,\cos nx\;dx
$$
Since $f(x)$ is nonzero only on $[-\pi/2,\pi/2]$, this is the same as
$$
a_n \;=\; \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} f(x)\,\cos nx\;dx
$$
Moreover, since $f(x) \cos nx$ is an even function, we can restrict to $[0,\pi/2]$ and double the integral:
$$
a_n \;=\; \frac{2}{\pi}\int_0^{\pi/2} f(x)\,\cos nx\;dx \;=\; \frac{2}{\pi}\int_0^{\pi/2} \left(\frac{\pi}{2} - x\right)\,\cos nx\;dx
$$
The integral on the right can be evaluated using integration by parts.  The result is:
$$
a_n \;=\; \begin{cases}\dfrac{2}{\pi n^2} & \text{if }n\equiv 1,3\pmod{4}, \\[6pt] \dfrac{4}{\pi n^2} & \text{if }n\equiv 2\pmod 4, \\[6pt] 0 & \text{if }n\equiv 0 \pmod{4}.\end{cases}
$$
Thus
$$
f(x) \;=\; \frac{\pi}{8} + \frac{2}{\pi}\cos x + \frac{1}{\pi}\cos 2x  + \frac{2}{9\pi}\cos 3x + \frac{2}{25\pi}\cos 5x + \frac{1}{9\pi}\cos 6x + \cdots.
$$
In summation form,
$$
f(x) \;=\; \frac{\pi}{8} \,+\, \sum_{k=0}^\infty \frac{2}{\pi(2k+1)^2} \cos\bigl((2k+1)x\bigr) \,+\, \sum_{k=0}^\infty \frac{1}{\pi(2k+1)^2} \cos\bigl((4k+2)x\bigr).
$$
By the way, the following animation shows thee convergence of this Fourier series for the first twelve terms:


问:
Number of extrema for given function

Given a function $f(x)=x^{2} + \ln(\cos(x))$. Find the number of its extrema.

I tried plotting the graph and found the local minimum at $x=0$ and two local maximum.

But am not able to proceed post it and how to get the stationary points for the given function so that in the extrema calculation we can ignore them.

答:
Starting by looking at the graph is a good approach but the problem with $f(x)$ is that its behaviour is so strange that a graphing program like Desmos might make it difficult to see what is really going on.

Have you tried finding $f'(x)$?

What do you know about the value of $f'(x)$ when $f(x)$ hits a max or min?

If you try this and rearrange the equation you find you will end up with the equation $2x=tan(x)$. Now, solving this equation would be very tricky, but luckily you don't have to! Think about the graphs of $y=2x$ and $y=tan(x)$. How many points of intersection do they have? What does this tell you about the number of extrema (max or min)?

Title: Number of extrema for given function


问:
Abstract Algebra books (for undergraduate)
I'm looking for good books of abstract algebra. In particular, I need some books with a lot of exercises, solved and not (in particular about modular arithmetic, groups and rings) Do you have any hints? Thanks.</p>

答: I remember enjoying Pinter's A Book of Abstract Algebra in undergrad.


问:
Connected components of the zero set
Let $f:[c,d] \rightarrow \mathbb{R}$ be a continuous function which is piecewise analytic, i.e., there exists $N\in \mathbb{N}$ and an increasing sequence $\left(i_{n}\right)_{1 \le n \le N}$, $i_{1}=c, i_{N}=d,$ such that
$$
\forall x \in(c, d) \quad f(x)=\sum_{n=1}^{N-1} \chi_{\left[i_n, i_{n+1}\right)}(x) f_{n}(x),
$$
where $f_n$ are analytic functions defined on $\left[i_{n}, i_{n+1}\right]$ and $\chi_{\left[i_{n}, i_{n+1}\right)}$ are characteristic functions for $n=1, \ldots, N-1$.

Consider the set $Z(f)$ of zeros of $f$ in $(c,d)$.

Does the set $Z(f)$ have a finite number of connected components?

答: If there are infinitely many components  of $Z(f)$, then there must be an $1\le n\le N$ such that $(i_n,i_{n+1})$ contains infinitely many of them, which we can take to be countable, say $(C_j)_{j\in \mathbb N}$ Now, choose $x_j\in C_j$. Then the sequence $(x_j)_{j\in \mathbb N}$ has a limit point $x\in [i_n,i_{n+1}]$ but $x\neq (i_n,i_{n+1})$ for otherwise $f=0$ on $(i_n,i_{n+1})$ (because $f_n$ is analytic), which is a contradiction since $(i_n,i_{n+1})$ is connected. Without loss of generality, $f(i_n)=0.$ But now, note that $f_{n-1}\cup f_n$ is analytic on $(i_{n-1},i_{n+1})$ and $x$ is a limit point of a sequence in $Z(f)$ interior to this interval, so again we have a contradiction because, in this case, $f=0$ on $(i_{n-1},i_{n+1}).$

Title: Connected components of the zero set


问:
Prove the complex polynomial $P(z,\bar{z})$ is zero if and only if all the coefficient is zero.
Let $F(z,\bar{z}) = \sum_i^n a_i z^i + \sum_j^m b_j \bar{z}^j$ which $a_i,b_i \in \mathbb{C}$ .

Prove if $F(z,\bar{z}) = 0$ for all $z\in \mathbb{C}$,then $a_i,b_j = 0$.

I can prove this by taking derivative on $z$ and $\bar{z}$(it's possible since Frechet differentiable).Is there some other alternative proof?

答:
W.lo.g. we may assume that $n \ge m$. Hence we may even assume $n = m$ (otherwise fill with coefficients $b_{m+1} = 0,\ldots, b_n = 0$). Thus for all $z$
$$F(z) = \sum_{j=0}^n (a_j z^j + b_j \bar{z}^j)= 0 .$$
Note that I suppressed $\bar z$ from $F(z, \bar z)$ because $\bar z$ is uniquely determined by $z$.

Write $z = e^{i\theta}x$ with $\theta, x \in \mathbb R$. Then
$$F(z) = \sum_{j=0}^n (e^{i\theta j}a_j + e^{-i\theta j}b_j) x^j= 0 .$$
Separating real and imaginary part of $F(z)$ yields two real ploynomials in the variable $x$ which are both $0$. This shows $\Re(e^{i\theta j}a_j + e^{-i\theta j}b_j) = 0$ and $\Im(e^{i\theta j}a_j + e^{-i\theta j}b_j) = 0$ for all $j$ and all $\theta$, hence
$$e^{i\theta j}a_j + e^{-i\theta j}b_j  = 0 .$$
Taking $\theta = 0$ we get
$$a_j + b_j = 0. \tag{1}$$
For $j > 0$ we take $\theta = \pi/2j$ and get
$$ia_j  - ib_j = 0, \text{ i.e. } a_j = b_j. \tag{2}$$
$(1)$ and $(2)$ show that $a_j =  b_j = 0$ for $j > 0$. Thus
$$F(z) = a_0 + b_0 = 0 .$$

For $j = 0$ we cannot conclude that $a_0 = b_0 = 0$. In fact, we may take any $a_0 \in \mathbb C$ and $b_0 = -a_0$.

Update:

Daniel Schepler suggests in a comment to consider the more general
$$P(z)  = \sum_{j,k} a_{jk}z^j \bar z^k .$$
This has an additional benefit: The constant term is $a_{00}$ and does not split as $a_0 + b_0$.

With $z = e^{i\theta}x$ we get
$$P(z) =  \sum_{j,k} e^{i\theta (j - k)}a_{jk}x^{j+k}$$
and conclude that for all $n$ and $\theta$
$$\sum_{j+k=n} e^{i\theta (j - k)}a_{jk} = \sum_{j=0}^n e^{i\theta (2j - n)}a_{j(n-j)} = 0 .$$
Via multiplying with $e^{i\theta n}$ we see that this is equivalent to
$$\sum_{j=0}^n e^{i\theta 2j}a_{j(n-j)} = \sum_{j=0}^n (e^{2i\theta})^ja_{j(n-j)} = 0 .$$
That is, for each $\theta$ we have a linear equation for the $n+1$ variables $a_{0n},\ldots,a_{n0}$.

Let $\theta_l = \frac{\pi l}{2(n+1)}$ for $l = 0,\ldots,n$. These are $n+1$ distinct points and we get a system of $n+1$ linear equations for the $a_{0n},\ldots,a_{n0}$. The matrix of this system is the Vandermonde matrix $V = V(\theta_0,\ldots,\theta_n)$ whose determinant is $\prod_{0\le r < s \le n} (\theta_s - \theta_r)$. Thus $\det V \ne 0$ which implies that $a_{0n} = \ldots = a_{n0} = 0$.


问:
Correspondence between maximal ideals and solutions of the equations
Assume we have a commutative algebra $A$ over a an algebraiclly closed field F of characteristic zero defined by generators  $x_1, \dots, x_n$ and relations:

$f_1=0$

$f_2=0$

$f_3 =0$

where $f_1$, $f_2$, and $f_3$ are in $x_1, \dots, x_n$.

That means $A=F[x_1, \dots, x_n]/(f_1, f_2, f_3)$.

my question:

Is true that if we want to find all maximal ideals of $A$, we need to solve the system of equations:

$f_1=0$

$f_2=0$

$f_3 =0$

and why?

Can someone help me to understand this correspondence?

答:
Let $I = (f_1, f_2, f_3)$. By the correspondence theorem, the maximal ideals of $A$ are exactly the (reductions of) maximal ideals $\DeclareMathOperator{\m}{\mathfrak{m}} \m \subseteq F[x_1, \ldots, x_n]$ with $\m \supseteq I$. As you say, by the Weak Nullstellensatz the maximal ideals of $F[x_1, \ldots, x_n]$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$, so $\m = (x_1 - a_1, \ldots, x_n - a_n)$ for some $a_1, \ldots, a_n \in F$. Now $I \subseteq \m \iff f_i \in \m$ for all $i = 1,2,3$ and
$$
f_i \in (x_1 - a_1, \ldots, x_n - a_n) \iff f_i(a_1, \ldots, a_n) = 0 \, .
$$
(For details, see this post.) Thus the maximal ideals of $A$ are exactly the reductions of the ideals $(x_1 - a_1, \ldots, x_n - a_n)$ such that
$$
f_1(a_1, \ldots, a_n) = f_2(a_1, \ldots, a_n) = f_3(a_1, \ldots, a_n) = 0 \, .
$$


问:
Product Series Formula through Induction
For the series:
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$ $(1+\frac{1}{n})^n$ I have the formula  $\frac{(n+1)^n}{n!}$ for n$\in$ $\Bbb N$

I used induction to try and solve but I'm stuck at trying to prove it for n+1 since
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$$(1+\frac{1}{n})^n$$(1+\frac{1}{n+1})^{n+1}$ =  $\frac{(n+1)^n}{n!}$$(1+\frac{1}{1})^{n+1}$ = $\frac{(n+1)^n}{n!}$$\sum_{k=0}^n \binom{n}{k}(\frac{1}{n+1})^k$ using the binomial theorem
but I can't tell properly how to get  $\frac{(n+2)^{n+1}}{(n+1)!}$

Any hints? because I swear I'm not seeing something.

Thank you!

答:
Hint: We have
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n = \frac{(n+1)^{n}}{n!}
\end{eqnarray*}
So
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n \color{red}{\left( 1+\frac{1}{n+1} \right)^{n+1}}&=& \frac{(n+1)^{n}}{n!} \color{red}{\left(\frac{n+2}{n+1} \right)^{n+1}}\\
&=& \cdots
\end{eqnarray*}


问:
Reciprocal of Laurent series
I understand how to use long division to find the reciprocal of a power series.

Are there any theorems that allow you to divide Laurent series that have a finite number of terms with negative exponents?  That is, if I have a Laurent series around $0$, such as

$f(z) = \frac{1}{z^2} + 3z + 5z^4 + \ldots$,

can I use long division to come up with a series for $\frac{1}{f(z)}$ and know anything about its convergence?

答:
If you want to find out how the function behaves around a  tiny neighbourhood of $\,z=0\,$, for example, then you can do

$$\frac1{f(z)}=\frac1{\frac1{z^2}+3z+5z^4+\ldots}=\frac{z^2}{1+3z^3+5z^6+\ldots}=$$

$$=z^2\cdot\frac1{1+3z^3+\mathcal O(z^6)}=z^2\left(1-3z^3+9z^6-27z^9+\ldots\right)=z^2-3z^5+\ldots$$

And the above is valid whenever

$$|3z^3|<1\implies|z|<\frac1{\sqrt[3]3}$$

Thus, our function, which is undefined at $\,z=0\,$, has a removable singularity there and that point behaves as a double zero.

Title: Reciprocal of Laurent series


问:
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.

Find all solutions to the diophantine  equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.

Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.

Assume $y \ge 3$ and $x \ge 1$.

By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.

By $\mod 7$,we get $y \equiv 1 \mod 6$.

I also tried $\mod 2$ but it didn't work.

Please post hints ( not a solution). Thanks in advance!

答:
examples to study

CW:Catalan Thue Ramanujan Nagell Tijdeman p^x - q^y = C

https://math.stackexchange.com/users/292972/gyumin-roh

Exponential Diophantine equation $7^y + 2 = 3^x$

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.  ME!  41, 31, 241, 17

Finding solutions to the diophantine equation $7^a=3^b+100$   343 - 243 =
100

http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847

The diophantine equation $5\times 2^{x-4}=3^y-1$

Equation in integers $7^x-3^y=4$

Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$

Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$.  128 - 125 = 3

Diophantine equation power of 7 and 2

Find natural numbers a,b such that $|3^a-2^b|=1$  did +-1


答:
It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.

Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).

Therefore, $3^a-1$ is a multiple of $13$.

Therefore, $7^b-1$ is a multiple of $13$.

Therefore, $b$ is a multiple of (what?).

Therefore, $7^b-1$ is a multiple of $9$.

Therefore, $3(3^a-1)$ is a multiple of $9$.

Therefore, $a$ is (what?).

Therefore, $x$ is (what?).

Body:
Find all solutions to the diophantine  equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.

Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.

Assume $y \ge 3$ and $x \ge 1$.

By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.

By $\mod 7$,we get $y \equiv 1 \mod 6$.

I also tried $\mod 2$ but it didn't work.

Please post hints ( not a solution). Thanks in advance!


问:
What are the functions to plot this graph?
I'd really like to re-create this graph in latex but I have no idea how to figure out what the functions are to create it. And I don't quite know how to approximate the functions from simply looking at the graph.

If anyone can provide two functions that produce roughly the same graph it would be much appreciated. Thanks.

They're graphical representation of the Big-O definition, this is mostly used in computer science. The graph shows that $f(n)$ is bounded by $cg(n)$.

答:
Here is a graph of
$$
g(x) = x + \sin x \quad\text{and}\quad f(x) = \bigg(\frac{5/3}{x^2+1}+\frac{2}{3}\bigg) (x+\sin x)
$$
from $x=0$ to $x=8$:

(Thought process: the function $g(x)$ looks like a small oscillation added to a straight line. The function $f(x)$ generally tracks with $g(x)$, but it is around twice as large at first and then maybe about $\frac34$ as large near the end of the graph; that suggests a multiplication factor of the form "constant plus function-decaying-to-zero", and some visual tweaking of numerical constants produced this particular choice.)


问:
For fundamental solutions $y_1,y_2$, there exist one and only one zero of $y_1$ between consecutive zeros of $y_2$.

Suppose that $y_1$ and $y_2$ are a fundamental set of solutions, of the linear second order ODE $y''+p(t)y'+q(t)y=0$, on the
interval $-\infty < t < \infty$. Show that there is one and only one
zero of $y_1$ between consecutive zeros of $y_2$. Hint: differentiate
the quantity $y_2/y_1$ and use Rolle's theorem.

I found an existing answer to this question to be unilluminating.

Here's how far I got.

Say, $a$ and $b$ are consecutive zeros of $y_2$. This would mean $\frac{y_2(a)}{y_1(a)} = \frac{y_2(b)}{y_1(b)}= 0$. Rolle's theorem now states that there exists $c \in (a,b)$ such that $\frac{\mathrm d y_2/y_1}{\mathrm{dt}}|_{t=c} = \frac{W[y_1,y_2](c)}{y_1^2(c)}=0$. The Wronskian being zero is in contradiction with $y_1$ and $y_2$ being a fundamental set of solutions. Hence, $\frac{W[y_1,y_2](t)}{y_1^2(t)}$ is nonzero for all $t\in (-\infty,\infty)$. Furtermore, the assumption we made in using Rolle's theorem is not correct.

Namely, $y_2(t)/y_1(t)$ is not continuous and therefore also not differentiable at some points the interval $(a,b)$. This is because the denumerator $y_1(t)$ has a zero in this interval.

Is my argument correct? And how should I now prove that that this zero of $y_1$ is unique?

Context: this is question 2.1.18 of Braun's Differential Equations and Their Applications, 4th edition.

答:
Let $a$ and $b$ ($a<b$) be two consecutive zeros of $y_2$; namely
$$ y_2(a)=y_2(b)=0. \tag1$$
Since $W(t)\neq0$, we can have $y_1(a)\neq0,y_1(b)\neq0$. Suppose that $y_1$ does not have zero in $(a,b)$.
Since $W[y_1,y_2](t)\neq0$ for $t\in[a,b]$, we can assume $W[y_1,y_2](t)>0$ and hence we have
$$
\frac{\mathrm d (y_2/y_1)}{\mathrm{dt}} = \frac{W[y_1,y_2](t)}{y_1^2}>0.
$$
Namely $y_2/y_1$ is strictly increasing and hence
$$ \frac{y_2(a)}{y_1(a)}<\frac{y_2(b)}{y_1(b)}. \tag2$$
Using (1) in (2), we have $0<0$ which is absurd. Thus $y_1$ has a zero in $(a,b)$. You can prove the uniqueness of zero of $y_1$.


问:
Is this propositional logic inference incorrect?
Trying to figure out whether I am applying the rules of propositional logic correctly in the following inference:

[x≠a ∨ x≠b] → [¬(x=a) ∨ ¬(x=b)] → [¬(x=a ∧ x=b)] →X→ [¬(x=a=b)] → [¬(x=a ∧ x=b ∧ a=b)] → [¬(x=a) ∨ ¬(x=b) ∨ ¬(a=b)] → [x≠a ∨ x≠b ∨ a≠b]

My intuition says that the step →X→ is incorrect, but I can't figure out why. It seems like De Morgan's Laws are applied correctly.

When a=1 and b=2 is substituted, the premise [x≠a ∨ x≠b] would then evaluate to 'true', which feels incorrect. Or is it correct after all?

答:
Your steps are correct. $$[\neg(x=a\land x=b)]\to[\neg(x=a=b)]\tag1$$ is true for any $a,b,x$.

Suppose by way of contradiction it's false; then, $\neg(x=a\land x=b)$ must be true but $\neg(x=a=b)$ must be false. This means that $x=a\land x=b$ is false but $x=a=b$ true. Since $x=a=b$ is true, we know that $x=a=b\leftrightarrow x=a\land x=b\land a=b$ (*), in particular $x=a\land x=b$ is true, but this contradicts our deduction of $x=a\land x=b$ false. So $(1)$ holds for all $a,b,x$.

(*) Is very easy to see: $\to$ is pretty obvious; for $\leftarrow$ it is because of transitivity of $=$.

You could also argue by rule of addition which states $p\to p\lor q\equiv\mathrm{T}$ (do the truth table), so from $x\neq a\lor x\neq b$ you end up with $(x\neq a\lor x\neq b)\lor a\neq b$ and, by associativity of $\lor$, $x\neq a\lor x\neq b\lor a\neq b$. This way saves you a few steps.


问:
What is an "induction principle"?
In section 4 Empirical Risk Minimzation of the paper Principles of Risk Minimization for Learning Theory by V. Vapnik, the author says the following:

In order to solve this problem, the following induction principle is proposed: the
risk functional $R(w)$ is replaced by the empirical risk functional

$$E(w) = \dfrac{1}{\mathscr{l}} \sum_{i = 1}^\mathscr{l} L(y_i, f(x_i, w)) \tag{3}$$
constructed on the basis of the training set (1). The induction principle of empirical risk minimization (ERM) assumes that the function $f(x, w^*_\mathscr{l})$, which minimizes $E(w)$ over the set $w \in W$, results in a risk $R(w^*_\mathscr{l})$ which is close to its minimum.

I am familiar with mathematical induction, such as in proofs, but what is an "induction principle", in general?

EDIT: It seems to me like the author's using "induction principle" here to refer to any principle (such as a mathematical expression) that allows one to induce some value? In this case, using the empirical risk functional $E(w)$ to find the risk $R(w^*_\mathscr{l})$ which is close to its minimum

答:
The word comes from the way human think inductively from empirical experiment in contrast to deduction reasoning.

As stated in Statistical Induction Principle:

Inductive statistics is way for scientists to make evidence-based decisions based on empirical/experimental results.

Methods of inductive statistics (so called statistical induction) can under given assumptions to make general conclusions and to objectively enumerate their degree of confidence. The main aim of inductive statistics is to elaborate procedures how to create general conclusions from empirical data that can substitute subjective inductive thinking by objective inductive thinking based on concepts of probability theory.

Title: What is an "induction principle"?


问:
Covering a polygon with an odd number of sides
I have the following elementary problem/question that I do not know how to tackle. It comes with a "math-olympiad-flavor" but I suspect it may be much more difficult than an high-school olympiads problem.

Let $\mathscr{P}$ be a simple polygon (not necessarily convex) in the plane, and such that the number of sides of $\mathscr{P}$ is an odd number $2k+1$.

For each vertex $v$ of $P$ there is an intuitive notion of opposite side (because the number of sides is odd), just label the sides counterclockwise at $v$ from $1$ to $2k+1$, and it would be the one with the label $k+1$.

Now suppose that for each vertex and its corresponding opposite side I consider the triangle the determine in the plane. Let us call $\Delta_v$ this triangle and call it a central triangle.

Is the following:

$$ \mathscr{P} \subseteq \bigcup_{v\in \mathscr{P}} \Delta_v$$

always true? In other words, if one paints all central triangles is it true that the whole polygon is then painted?

答:
The statement is true. In fact, we can generalize it a little bit beyond the simple polygons; we need only the weaker condition that its winding number is $1$ (intuitively, that if you "walk around" the polygon you will only make a turn of $360^\circ$, not any other multiple thereof - its angles add up to the right amount). We can also bend the edges however we like.

To see that it doesn't necessarily hold for polygons with higher winding number, consider the case of a 5-pointed star (darker regions indicating those which are swept out twice):

To answer the question, we move from the discrete to the continuous and solve a more generalized problem.

Claim: Let $C$ be a closed curve given by a map $f:[0,1]\to \mathbb{R}^2$ such that $f(0)=f(1)$, and for notational convenience let $f(x)$ denote $f(x\pmod{1})$ in the event that $x\not\in[0,1]$. We wish to show that the union of the line segments connecting $f(x)$ and $f(x+0.5)$ for every $x\in[0,1]$ contains every point around which $C$ has winding number $1$. (Intuitively, this just means that if you stood at that point and turned to watch the path of $C$ from $f(0)$ to $f(1)$, you would make one turn counterclockwise.)

First, let's see that the claim implies our proposition. Given a polygon $\mathscr{P}$ as in the problem with $2k+1$ vertices $P_1,\ldots,P_{2k+1}$, we partition $[0,1]$ into $4k+2$ equal intervals $I_1,I_2,\ldots,I_{4k+2}$, and define $f$ so that for $x\in I_{2j}$, $f(x)$ is constant at the point $P_j$. For $x\in I_{2j+1}$, $f(x)$ traverses the edge from $P_j$ to $P_{j+1}$ (let's say linearly, but any continuous parameterization works). Now, observe that $f$ is indeed a continuous function with $f(0)=f(1)=P_{2k+1}$, and that the curve $C$ is our polygon. Note also that if you are standing in the interior of a simple polygon, it will make a net revolution of $2\pi$ radians around you, which we can choose to be counterclockwise by reversing the order of the vertices (or flipping the signs on this proof - the sign of the winding number doesn't matter, I'm just taking it to be positive for convenience).

Finally, note that as $x$ ranges over the interval $I_{2j}$, the locus of all these line segments is exactly the triangle formed by vertex $P_j$ and its "opposite side" $\overline{P_{j+k}P_{j+k+1}}$ (where the indices are to be interpreted modulo $2k+1$). (When $x$ is in an odd-indexed interval,$x+0.5$ is in an even-indexed one, so these triangles cover every line segment involved.)

So, for the given parameterization, this just says that every point in the interior of the polygon $\mathscr{P}$ is covered by one of these triangles.

Having established that the claim will yield what we want, we move on to the proof.

Proof of claim: Let $q$ be a point not on $C$. Then we can associate each point $f(t)\in C$ with an angle $\theta(t)$ of the vector from $q$ to $f(t)$, parametrized in the usual way relative to the positive $x$-axis (but where we only care about the value modulo $2\pi$).

By some basic covering space theory (I think the Wikipedia article on the winding number sketches this out in more formality), we can uniquely choose the multiple-of-$2\pi$ offsets of $\theta(t)$ so that $\theta(t)$ is continuous and $\theta(0)=0$. $q$ has winding number $1$ iff $\theta(1)=2\pi$. As an example: if $q$ is the origin and $f(t)=(\cos(2\pi t), \sin(2\pi t))$, then $\theta(t)=2\pi t$.

Now, observe that if $\theta(s)-\theta(t)\equiv\pi\equiv-\pi\pmod{2\pi}$, the line segment from $f(s)$ to $f(t)$ contains $q$.

Now, consider $g(t)=\theta(t+0.5)-\theta(t)$. $g$ is a continuous function on $[0,0.5]$, and $g(0)+g(0.5)=\theta(1)-\theta(0)=2\pi$ by assumption on $q$, so by the Intermediate Value Theorem $g$ takes on the value $\pi$ at some point. But this is exactly the condition we need for $q$ to be in our locus of line segments, completing the proof.

(Thanks to Calvin Lin for the "beam" framing of the problem, and thinking about things in terms of a rotating line segment instead of a sum of triangles.)


问:
Gaussian Elimination, Question Check.
I'm going through my practice problems, and just want to know if I am doing this right:

$$ 2x_1 - 3x_2 = -2$$
$$ 2x_1 + x_2 = 1$$
$$ 3x_1 + 2x_2 = 1$$

And this is my solution:
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}
\end{align}
$$

Can anyone tell me if I did this right, or if I did a mistake where? Thanks.

答:
In your second matrix, assuming you meant to divide by 2 in the first row, you incorrectly negated the second entry. If you meant to divide by $-2$, then the first entry should be negated, and the third entry should be positive.

In your third matrix, when you doubled row 1 and added to row 2, your entry in the middle row, middle column should be $-3 + 1 =-2$. I'm afraid then, that the error will "trickle down" in later reductions using that row.

Likewise, in the fourth matrix, when you tripled row 1 and added to row 3, your middle value in the bottom row should be $-\frac 92 + 2 = -\frac 52$.

$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

Subtract row 1 from row 2:
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
0 & -2 & 3\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

1/2 row 1:

$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & -2 & 3\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
Add $-3 R1$ to row 3
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & -2 &  3\\
0 & -5/2 & 4
\end{bmatrix}
\end{align}
$$
-1/2 Row 2, 2 times Row 3:
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & -3/2\\
0 & -5 & 8
\end{bmatrix}
\end{align}
$$
5 Row 2 + row 3 = row 3
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & 1\\
0 & 0 & 1/2
\end{bmatrix}
\end{align}
$$
We can now multiply the last row by 2:
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & 1\\
0 & 0 & 1
\end{bmatrix}
\end{align}
$$

Title: Gaussian Elimination, Question Check.


问:
Calculation of CPU Utilization Problem

A computer, has an operating system occupying 1GB of memory. The
  following are the processes with memory sizes and their I/O wait
  percentages: A(512MB, 50%), B(128MB, 60%), C(64MB, 70%). If the goal
  is at least 75% CPU utilization, what would be the minimum system
  memory size requirement? Give examples for how many process A,B, and
  C's should we run at the same time. (Show your formulas and work
  clearly, list all your assumptions if you have any)

The formula for CPU utilization is $1-p^{n}$, in which $n$ is number of process running in memory and p is the average percentage of time processes are waiting for I/O.

$$64 = x, 128 = 2x, 512 = 8x$$

$$0.25 = p^n$$

$$0.25 = 0.5*4x * 0.6*2x * 0.7*x$$

$$0.5 = 16x^3 * 0.42$$

For a limited time, I can't go further. Even if I have time, I don't go further. Is my approach is wrong? Is there any simpler way to find it? It is an exam question.

答:
Given that we have 3 types of processes as follows:

```<code>Process A : 512 MB with 50% Idle time
Process B : 128 MB with 60% Idle time
Process C : 64  MB with 70% Idle time
</code>```

Now we have each of these as follows:

(Number of each of these processes) = x1 of A, x2 of B and x3 of C

We can calculate CPU Utilisation $\eta$ as follows:

$\eta$ = 1 - 0.5^x1 * 0.6^x2 * 0.7^x3

Now we need to play with x1, x2, x3 such as $\eta$ >= 0.75

(x1,x2,x3) = (1,0,2) results in 0.755 which is as close as you can come to 0.75 with x1,x2,x3 integer. -- (1)

with, (x1,x2,x3) = (1,1,1) utilisation comes out to be 0.79, hence result obtained in statement (1) is the closest you can reach atleast 75% CPU utilisation.

Now, with result obtained in (1), let's calculate system memory size M

M = x1* 512MB + x2* 128MB + x3*64MB + 1024MB (OS)

M = 512+2*64+1024 MB = 1664 MB = 1.625GB

Title: Calculation of CPU Utilization Problem

Body:
Given that we have 3 types of processes as follows:

Process A : 512 MB with 50% Idle time
Process B : 128 MB with 60% Idle time
Process C : 64  MB with 70% Idle time

Now we have each of these as follows:

(Number of each of these processes) = x1 of A, x2 of B and x3 of C

We can calculate CPU Utilisation $\eta$ as follows:

$\eta$ = 1 - 0.5^x1 * 0.6^x2 * 0.7^x3

Now we need to play with x1, x2, x3 such as $\eta$ >= 0.75

(x1,x2,x3) = (1,0,2) results in 0.755 which is as close as you can come to 0.75 with x1,x2,x3 integer. -- (1)

with, (x1,x2,x3) = (1,1,1) utilisation comes out to be 0.79, hence result obtained in statement (1) is the closest you can reach atleast 75% CPU utilisation.

Now, with result obtained in (1), let's calculate system memory size M

M = x1* 512MB + x2* 128MB + x3*64MB + 1024MB (OS)

M = 512+2*64+1024 MB = 1664 MB = 1.625GB


问:
Prove that limit of $\sqrt{\frac{n^2+3}{2n+1}}$ is $\infty$ as $n$ tends to infinity.
I need to prove that $\lim_\limits{n\to \infty}$$\sqrt{\frac{n^2+3}{2n+1}} = \infty$ (series) by using the definition:

"A sequence $a_n$ converges to $\infty$ if, for every number $M$, there exists $N∈N$ such that whenever $n≥N$ it follows that $a_n>M$."

I came with a proof which I'm not sure is valid, because I just learned the definition and maybe I didn't grasp it yet.

Please let me know what do you think of the following proof:

I'll find an $N$ such that whenever $n≥N$ it follows that $a_n>M$ for every $M$:

$M < \sqrt{\frac{n^2+3}{2n+1}} \Rightarrow M^2 < \frac{n^2+3}{2n+1} < \frac{n^2+n^2}{2n} = \frac{2n^2}{2n} = n$

Hence, we got $n > M^2$, therefore we can say that when $N = M^2$ then $a_n>M$ for every $n > M$.
End of proof.

Do you think it is valid or am I missing something? Any suggestion for a better proof using the definition?

Thanks!

答:
The proof isn't right. What you proved is that

$\frac{n^2+3}{2n+1}>N \implies n>M^2$ for $N=M^2$

This strategy would have worked if you had double implication type deductions... so then your reasoning would work backwards as well as forwards... but your implications are one way.

What you want to prove is that

$n>N \implies \frac{n^2+3}{2n+1}>M^2$, for some $N$

Also $n^2+3<2n^2$ requires $n \ge 2$ in the integers.

A way to check your proof is wrong is to try to go from $n>M^2$ to $\frac{n^2+3}{2n+1}>M^2$

You should do this at the end anyway. That is really your proof. The rest of the work is preliminary.

What you want to do is find an expression which is some constant times n which is less than $\frac{n^2+3}{2n+1}$ and force that expression to be greater than $M^2$

First I'm going to assume $n \ge 1$,

So then

$\frac{n^2+3}{2n+1}>\frac{n^2}{2n+1} \ge \frac{n^2}{2n+n} =\frac{n}{3}$ call this eqn.(1)

So given an $M$ we want $\frac{n}{3}>M^2$ so we want $n > 3M^2$

So here's the proof,

given an $M>0$, we choose $N = \max \{1,3M^2+1\}$

$\begin{aligned}&n \ge N \\ \\ \implies &n \ge 3M^2+1 > 3M^2 \\ \\ \implies &\frac{n}{3} > M^2 \end{aligned}$

Since we know $n \ge 1$ we can use eqn. (1), so we know that

$\begin{aligned} &\frac{n^2+3}{2n+1}>M^2 \\ \\ \implies &\sqrt{\frac{n^2+3}{2n+1}}>M \\ \\ \implies &a_n>M\end{aligned}$

EDIT: Note that I incorrectly had "min" before and fixed it to "max".


问:
Is my Calculus Proof Correct?
Question:
A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(fg)'=f'g'$. If $f(x)=e^{x^2}$, determine, with proof, whether there exists an open interval $(a,b)$ and a nonzero differentiable function $g$ defined on $(a,b)$ such that this wrong product rule is true for $x$ in $(a,b)$.

Is my solution correct? Is there anything I need to improve?

My solution:

We're trying to find a function $g$ such that $(fg)'=f'g'.$
Using the product rule, we get $$f'g + fg' =  f'g'.$$
Plugging in both $f(x)=e^{x^2}$ and $f'(x)=2xe^{x^2},$ we get
$$2xe^{x^2}g + e^{x^2}g' = 2xe^{x^2}g'.$$
Simplifying and canceling $e^{x^2},$ we get $$\frac{g'}{g} = \frac{2x-1}{2x} = 1 +  \frac{1}{2x-1}.$$
Taking the integral of both sides,
$$\int \frac{dg}g = \int 1+ \frac{1}{2x-1} \, dx.$$
$$\log|g| = \frac{\log|2x-1|}{2} + x + C.$$
$$|g| = e^{\frac{\log|2x-1|}{2}}e^xe^C.$$
Letting $e^C=K,$ we get
$$\boxed{g = Ke^x\sqrt{|2x-1|}},$$ Where K is a constant greater than 0.
Therefore, on any interval $(a,b)$ that does not contain the value of $\frac{1}{2},$ there exists nonzero differentiable function $g$ defined on $(a,b)$ such that $(fg)'=f'g'$ is true for $x$ in $(a,b)$.

答:
I think you have gone far beyond the requirements of the question,
which asked merely to prove that there exists an open interval and
a non-zero differentiable function with the stated properties.

To prove that such an interval and function exist, you can simply
choose an open interval -- for example, $(a,b) = (1,2)$ --
and one function on that interval,
for example $g(x) = e^x \sqrt{2x - 1}.$
Then show that $f(x)$ and $g(x)$ (and therefore also $(fg)(x)$)
are defined on your chosen interval and calculate $f'$, $g'$, and $(fg)'$:

\begin{align}
f'(x) &= \frac{\mathrm d}{\mathrm dx} e^{x^2} = 2x e^{x^2}, \\
g'(x) &= \frac{\mathrm d}{\mathrm dx}\left(e^x \sqrt{2x - 1}\right)
= \frac{2x e^x}{\sqrt{2x - 1}}, \\
(fg)'(x) &= \frac{\mathrm d}{\mathrm dx} \left(e^{x^2}\cdot e^x \sqrt{2x - 1}\right) \\
&= \frac{4 x^2 e^{x^2 + x}}{\sqrt{2x - 1}} \\
&= 2x e^{x^2} \cdot \frac{2x e^x}{\sqrt{2x - 1}} \\
&= f'(x)\cdot g'(x).
\end{align}

Note that it was not necessary to take the absolute value of $2x-1$ in any of these formulas, since $2x - 1 > 0$ everywhere in the domain of $f$ and $g$ as defined in this proof.

And that is essentially the proof of existence of an interval and a function with the desired properties.
You could choose the interval $\left(\frac12,\infty\right)$ rather than $(1,2)$ for the existence proof (as I would) since it is one of the two maximal intervals on which such a function $g$ is defined,
and perhaps you might like to show off your clever method of identifying the entire family of functions that might be the desired function $g$ (as I likely would),
but all of that is bonus material beyond the plain meaning of the question.

Title: Is my Calculus Proof Correct?


问:
Approximately finding self intersection points of closed plane curves?
Let $0 = t_0 < t_1 < t_2 < \cdots < t_N = 2 \pi$.
I compute the points on the closed curve $F(t)$ in the complex plane , (Set $\epsilon = +\infty$) :

$$P_i = F(t_i)$$
$$P_{i+1} = F(t_{i+1})$$
if $|P_i - P_{i+1}| < \epsilon$: then set $\epsilon := |P_i - P_{i+1}|$.

In the next step I consider circles around those points on the curve of radius $\epsilon/2$

Let $L = \{\}$

for $i,j$ in $0,1,2,...,N, i \neq j$:

if $| P_i - P_j | < \epsilon$: then add $Q = (P_i+P_j)/2$ to the list of candidate intersection points $L$

In the next step this candidate list is taken to generate an undirected graph with nodes = $L$ and they are connected if $|p-q| < 2 \epsilon$.
Consider the connected components of this graph and take the mean value of each connected component:

$$q_C = \frac{1}{|C|}\sum_{p \in C} p $$

Then the approximate list of self intersection points will be :

$$\{ q_C | C \text{ is connected component } \}$$

This method seems to approximately work on some curves if $N$ is a large number:

Example (Please wait a few seconds for the plot to show up.)

Q1) Is it possible to prove, that this method will always find the intersection points if $N$ goes to infinity?

Q2) What better way would you suggest in the second step when testing if the circles overlap, instead of iterating $N\cdot (N+1)/2$ times over the points?

Thanks for your help!

Edit:
With the help of @PaulSinclair in his answer, I could find an algorithm which solves this question approximately:

One can use the Bentley-Ottmann algorithm after construction a polygon of the curve.

答:
The answer to Q1 is no. As currently described, this will not always find intersection points. You choose $\epsilon$ to be the smallest gap beween adjacent vertices. But you don't do anything to limit how large the gaps are between other vertices.

For example, when $N = 5$, consider:
$$\begin{align}P_0 &= (1,1)\\P_1 &= (-1, -1)\\P_2 &= (-1,1)\\P_3 &= (1,-1)\\P_4 &= (1,0)\\P_N &= (1,1)\end{align}$$

For this curve $\epsilon = 1$ and there are no vertices whose distance is less than that, so $L$ is empty and the rest falls apart.

So what about increasing $N$? One way you can do it is to repeatedly increase $N$ by $1$, with $P_N$ remaining as $(1,1)$, and filling the now vacated $P_{N-1}$ with the new point $\frac 12(P_N + P_{N-1})$. This halves the value of $\epsilon$ each time, but does not change the fact that none of the $P_i$ are in a distance $< \epsilon$ of each other. So no matter how large $N$ gets by this procedure, $L$ remains empty, and the intersection at $(0,0)$ is never found.

You'll have to control in some way the ratio of the largest gap to the smallest gap to guarantee that a limit point will be found.

Personally, once you polygonalized your curve, you are dealing with line segments. And figuring out exactly (up to computational accuracy) when and where two line segments intersect is not that hard to accomplish, so any additional approximation after the polygonalization does not seem worthwhile.


问:
Describe a specific equivalence class
Relation on $\mathbb N$:

$x \sim y \iff xy$ is a square

Give a description of the equivalence classes $[3]$, $[9]$, and $[99]$.

I'm not sure what this question is really asking, but this is what I have for $[3]$:

$\{y\in \mathbb N\ |\ 3y \text{ is a square }\}$

Any $y$ of the form $3$ times a square. Is this sufficient?

答:
You can be a little more specific. Ideally, you could give a formula that generates the entire equivalence class.

Using the work you've already done, you know that in order for $3y$ to be a square, all of its prime factors need to be present with even multiplicity. Thus, $y$ needs to contain another factor of $3$ and takes on the form
$$ y = 3z$$
for some $z \in \mathbb{N}$.

Next, for $3(3z) = 9z$ to be a perfect square, $z$ itself must be a perfect square. Thus, $y$ takes on the form $y = 3n^2$ for some $n \in \mathbb{N}$. Altogether,
$$ [3] = \{3n^2 \mid n \in \mathbb{N}\}$$

In an analogous fashion, you can compute $[9]$ and $[99]$ by doing a prime-factorization on an equivalence class representative.

Title: Describe a specific equivalence class


问:
Closet point theorem proof
I am reading the book Nonlinear Programming: Theory and algorithms by Bazaraa et al.
However, I am not sure if I am undertanding the proof of the following theorem correctly.

Closet Point Theorem. Let $S$ be a non-empty closed convex set in $R^n$ and let $y∉S$, then $∃$ a point $\hat{x}∈S$ with minimum distance from $y$, i.e.,$‖y−\hat{x}‖≤‖y−x‖∀x∈S$. Furthermore, $\hat{x}$ is a minimizing point if and only if $(y−ˆx)^T(x−ˆx)≤0$

I understand the proof of the first part of the theorem. I am just having troubles understanding the proof of $\hat{x}$ is a minimizing point if and only if $(y−ˆx)^T(x−ˆx)≤0$
when stablishing sufficiency

Proof

For the second part of the proof (to prove sufficiency), assume $\left ( y-\hat{x} \right )^{\tau }\left ( x-\hat{x} \right )\leq 0 \forall x\in S$

Now, $\left \| y-x \right \|^{2}=\left \| y-\hat{x}+ \hat{x}-x\right \|^{2}=\left \| y-\hat{x} \right \|^{2}+\left \|\hat{x}-x \right \|^{2}+2\left (\hat{x}-x \right )^{\tau }\left ( y-\hat{x} \right )$ by the parallelogram law

$\Rightarrow \left \| y-x \right \|^{2}\geq \left \| y-\hat{x} \right \|^{2} $because $\left \| \hat{x}-x \right \|^{2}\geq 0$ and $\left ( \hat{x}- x\right )^{T}\left ( y-\hat{x} \right )\geq 0$

Thus, $\hat{x}$ is minimizing point.

$\blacksquare$

The part I am not sure about is why $\left \| \hat{x}-x \right \|^{2}\geq 0$ and $\left ( \hat{x}- x\right )^{T}\left ( y-\hat{x} \right )\geq 0$. My undertanding is that $\left \| \hat{x}-x \right \|^{2}\geq 0$ because the definition of the Euclidean Norm, am I right?, and
$\left ( \hat{x}- x\right )^{T}\left ( y-\hat{x} \right )\geq 0$ follows from $(y−ˆx)^T(x−ˆx)≤0$

答:
$\|\hat{x}-x\|$ is a real number. When we square a real number, we get a nonnegative number.

As for the other inequality is due to we started from the assumption that

$$(y-\hat{x})^T(x-\hat{x}) \le 0$$

Multiplying by negative, we have

$$(y-\hat{x})^T(\hat{x}-x) \ge 0$$

which is equivalent to

$$(\hat{x}-x)^T(y-\hat{x}) \ge 0$$

Title: Closet point theorem proof


问:
How can I prove that $m <n$ then $m^2 < n^2$
I'm having trouble taking this test, I apologize if the question is duplicated and I appreciate any indication of content. My math base is not that good, so please be clear.  If $0 < m < n$, can I prove:

$$m < n \Rightarrow m^2 < n^2.$$

答:
I will assume $n > m >0$ because it may not be true if $m$ or $n$ are negative. Let $n = m + \epsilon$, where $\epsilon$ is positive. Then, $n^{2} =(m + \epsilon)^{2} = m^{2} + 2\epsilon m + \epsilon^{2} > m^{2}$.

This is a way of proving that $x^{2}$ is increasing on positive $x$, which may or may not be assumable in your context.


答: It would be easier if you can consider $n^2-m^2=(n+m)(n-m)>0$.

Title: How can I prove that $m <n$ then $m^2 < n^2$

Body:
I'm having trouble taking this test, I apologize if the question is duplicated and I appreciate any indication of content. My math base is not that good, so please be clear.  If $0 < m < n$, can I prove:

$$m < n \Rightarrow m^2 < n^2.$$


答:
Under the assumption $0<m<n$, we have $m^2<n^2\Leftrightarrow 0<n^2-m^2$.

$n^2-m^2=(n-m)(n+m)$

Now $n-m>0$ by assumption and $n+m>0$. So the product $n^2-m^2>0$.


问:
What is meant by "class of indicator functions"?
In section 5 VC-Dimension of the Set of Functions of the paper Principles of Risk Minimization for Learning Theory by V. Vapnik, the author says the following:

The theory of uniform convergence of empirical risk to actual risk developed in the 70's and 80's, includes a description of necessary and sufficient conditions as well as bounds for the rate of convergence (Vapnik, 1982). These bounds, which are independent of the distribution function $P(x, y)$, are based on a quantitative measure of the capacity of the set of functions implemented by the learning machine: the VC-dimension of the set.

For simplicity, these bounds will be discussed here only for the case of binary pattern recognition, for which $y \in \{0, 1\}$ and $f(x, w), w \in W$ is the class of indicator functions. The loss function takes only two values $L(y, f(x, w)) = 0$ if $y = f(x, w)$ and $L(y, f(x, w)) = 1$ otherwise. In this case, the risk functional (2) is the probability of error, denoted by $p(w)$). The empirical risk functional (3), denoted by $v(w)$, is the frequency of error in the training set.

It is this part that I am confused about:

For simplicity, these bounds will be discussed here only for the case of binary pattern recognition, for which $y \in \{0, 1\}$ and $f(x, w)$, $w \in W$ is the class of indicator functions.

What exactly is meant by "class of indicator functions", and which one of $y \in \{0, 1\}$ or $f(x, w)$ is this "class of indicator functions"? $f$ is the only function I see here, but I'm not sure how it's a "class". Could this be in reference to "equivalence classes"?

答:
Here $w$ is a parameter that indexes the class of functions. As $w$ ranges over $W$ we obtain the class of functions.

The notation $f(x,w)$ should be read as "the function identified by parameter $w$, evaluated at $x$". A more typical notation would be $f_w$ for the function identified by parameter $w$, and $f_w(x)$ would denote the value at $x$ of the function $f_w$, and the class of functions might be denoted $\mathcal{F}:=\{f_w : w\in W\}$. Perhaps the author wanted to avoid excessive subscripting.

To say "$f(x,w)$, $w\in W$ is the class of indicator functions" means that each function returns either $0$ or $1$ when given any input $x$. This returned value is to be compared with the true value $y\in\{0,1\}$.


问:
If sum of triangle angles is $180$ degrees, how $\sin(270)$ is possible?
I'm not new to trigonometry, but this question always bothers me.

As it is in Wolfram MathWorld-

$$
\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}
$$

We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$  is possible?

Edit

After Joe posted his answer, I went back to MathWorld and read it more carefully. I noticed what Brian Drake mentioned in his answer. MathWorld has already explained both definitions. I kept the question because of the brilliant answers below.

答:
I was very surprised to see this question: MathWorld is not the sort of site that would gloss over this, so why can’t we find the answer there?

In fact, MathWorld acknowledges the problem mentioned in the question (emphasis added):

The common schoolbook definition of the sine of an angle $\theta$ in a right triangle (which is equivalent to the definition just given) is …
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}.\tag{1}$

Here is the “definition just given”:

Let $\theta$ be an angle measured counterclockwise from the $x$-axis along an arc of the unit circle. Then $\sin \theta$ is the vertical coordinate of the arc endpoint, as illustrated in the left figure above.

To be fair, this text and the figure given in MathWorld don’t do a great job of explaining this definition to people not already familiar with it. Hopefully some of the other answers and comments here will explain it in a way you can understand.

But I would suggest that you try to read sites like MathWorld more carefully before turning elsewhere.


问:
Basis of a subspace of a matrix
Given
$$ A =
    \begin{pmatrix}
    -1 & -1 & 0 \\
    0 & 1 & 3 \\
   0 & 0 & -2
    \end{pmatrix}
$$
consider the subspace $W$ of $\Bbb{R}^{3}$ given by $W= \{x \in \Bbb{R}^{3}: Ax = −2x\}$. Find a basis for W

答:
Hint. The space $W$ is precisely the null space of $A+2I$. You only need to solve the linear system: $$(A+2I)x=0.$$

Edit. Suppose $x=(x_1,x_2,x_3)^T$. Then $Ax=-2x$ yields
\begin{align*}
Ax=\begin{bmatrix}
-x_1-x_2 \\ x_2 + x_3\\ -2x_3
\end{bmatrix}=-2x=\begin{bmatrix}
-2x_1 \\ -2x_2 \\ -2x_3
\end{bmatrix}.
\end{align*}
Comparing the entries pairwisely,
\begin{align*}
-x_1-x_2=-2x_1 \\ x_2 + x_3=-2x_2\\ -2x_3=-2x_3.
\end{align*}
I believe you can complete the following steps.

Title: Basis of a subspace of a matrix


问:
Translate $\sf CKKCpqCqrpr$ into classical notation
From polish to classical notation

$\sf CKKCpqCqrpr$

I have $((p\to q)\land(q\to r)\land p)\to r~
$ for "$\sf CpqCqrpr
$" but I don't understand what $\sf CKK$ means in the beginning.

答:
No, the string of $\sf CpqCqrpr$ is not a well formed formulae .

A string of the form $\sf CKKabcd$ translates to $\sf((a\land b)\land c)\to d$, as follows:

$$\sf{CKKabcd\\CK(a{\land}b)cd\\C((a{\land}b){\land}c)d\\((a{\land}b){\land}c){\to}d}$$

So it is the entire statement of $\sf CKKCpqCqrpr$ that translates as $((p\to q)\land(q\to r)\land p)\to r~$ as follows:$$\sf{CKKCpqCqrpr\\CKK(p{\to}q)(q{\to}r)pr\\(((p{\to}q){\land}(q{\to}r)){\land}p){\to}r}$$


问:
General solutions of $ \sin (A) = \sin (B)$
In my text book it is directly stated that the general solution of the equation
 $ \sin (A) = \sin (B)$ , is

$ A = 2n\pi + B$   and     $A = (2n +1)\pi - B$

but where is the second solution coming from? Since period of $\sin$ function is $2\pi$, shouldn't the solutions be, $ A = 2n\pi + B$ and $ A = 2n\pi - B$  ?

答:
Equation

The general solution of the equation sin(A) = sin(B) is given by:

$A = B + (-1)^n nπ$

EDIT: Proof

Here is the proof for the above equation. we are given:

$sin(A) = sin(B)$

=> $sin(A)-sin(B)=0$

=> $2cos(\frac{A+B}{2})sin(\frac{A-B}{2}) = 0$

=> $cos(\frac{A+B}{2})=0$ or $sin(\frac{A-B}{2})=0$

=> $\frac{A+B}{2}=(2n+1)π/2$ or $\frac{A-B}{2}=nπ$

=> $A=(2n+1)π–B$ or $A=2nπ+B$

=> $A = (2n+1)π+(–1)^{2n + 1} B$ or $x = 2nπ +(–1)^{2n}B$

Combining both the equations we get,

$A = nπ + (–1)^n B$


问:
For $p \in (0,\infty)$, is it possible to have a function $f \in L_q$ for $q\neq p$ and $f \notin L_p$?
That is, is it possible for a function to be in all $L_p$ spaces except for a single value of $p \in (0,\infty)$?

For $p=1$ $q \in (1,\infty)$, an example would be
\begin{align*}
    f(x) =
\begin{cases}
\frac1x \text{ if }x\geq 1\\
0 \text{ if }x< 1
\end{cases}.
\end{align*}

This example wouldn't work for values of $p,q\in (0,1)$ though.

答:
No, that's not possible: Suppose $f\ge 0,$ $q_1<p<q_2,$ and $f$ is in both $L^{q_1}, L^{q_2}.$ Then

$$\int_{f\le 1} f^p \le \int_{f\le 1} f^{q_1} <\infty$$

and

$$\int_{f>1 } f^p \le \int_{f\le 1} f^{q_2} <\infty,$$

proving $f\in L^p.$


问:
Trying to define relationship between two numbers
I have 2 numbers 97 (F) and 415 (S) this is the key / legend.

Given any number (S), commonly between 200 - 900, I am looking for "F".

Where if (S) is above 415 "F" will be less than 97 and if (S) is below 415 "F" will be more than 97.

The higher "S" is the lower "F" is.

The lower "S" is the higher "F" is.

Here's some estimations:

415 (S) gives 97 (F)

612 (S) should give ~72 (F)

514 (S) should give ~82 (F)

This is not a standard proportion as (F) of 612 (S) would be higher than 97.

What is this relationship called? How can I find the "F" of any number (S)?

答:
I believe I figured this out by doing this.</p>
```<code>const getF = (S) => (1 / (S / 415)) * 97;
</code>```

Body:
I believe I figured this out by doing this.

const getF = (S) => (1 / (S / 415)) * 97;


问:
Uniform continuous proof.
If $f(x)$ define as: $f(x)=x\cos(\log(x))$, when $x>0$. $f(x)=0$, when $x=0$. we need prove that f is uniform continuous in $[0,+∞]$?

I already prove the following equation
$|f(x)−f(y)|=|x−y||f′(c)| ≤2|x−y|$.
I just feel confuse with the  $δ=ϵ$ conclusion and how to write it.

答:
You only need to to complete the last step:

For every $\epsilon>0$, we may set $\delta=\epsilon/2$. As we can see, if $|x-y|<\delta$, then $$|f(x)-f(y)|\leq 2|x-y|<2\delta=2\cdot\frac{\epsilon}{2}=\epsilon.$$

Title: Uniform continuous proof.


问:
Manually Finding Values of Inverse Trigonometric Functions
I'm trying to solve (for $x$) some problems such as $\arctan(0)=x$, $\arcsin(-\frac{\sqrt{3}}{{2}})=x$, etc.

What is the best way to go about this? So far, I have been trying to solve the problems intuitively (e.g. I ask myself what value of sine will give me $-\frac{\sqrt{3}}{{2}}$?), maybe drawing a triangle to help. Is there a better way to solve these problems?

答:
You need to know the basic values of the trigonometric functions, thus for example:

$$\sin\left(-\frac{\pi}3\right)=\sin\left(-\frac{2\pi}3\right)=\frac{\sqrt3}2\implies\arcsin\left(-\frac{\sqrt3}2\right)\in\left\{\;-\frac{\pi}3\;,\;-\frac{2\pi}3\;\right\}$$

so you must know where your values' range is.


问:
Computing Relative Homology
I'm sort of uncertain about how to compute relative homologies, so I would like some feedback on my work. As I understand it, you entirely use the idea that they form long exact sequences.

Suppose $X = S^1 \times [0, 1]$, the cylinder. And let $A = S^1 \times \{0, 1\}$ the upper and lower boundary circles. We want to compute $H_n(X,A)$ for every $n$.

So we use the idea that the homology groups form a long exact sequence $$...H_n(A) \rightarrow H_n(X) \rightarrow H_n(X, A) \rightarrow H_{n-1}(A) \rightarrow H_{n-1}(X) .....$$

Next since $X$ deformation retracts to a circle, $H_n(X) = \mathbb{Z}$ for $n = 1, 2$ and trivial otherwise. Similarly, as $A$ is a disjoint union of two circles, $H_n(A) = \mathbb{Z} \oplus\mathbb{Z}$ for $n = 1,2$ and trivial otherwise. This means that $H_{n}(X,A) = 0$ if $n \geq 4$. For $n = 3$,

$$.. 0 \rightarrow H_3 (X, A) \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$$

is what we obtain. Which means that $H_3(X, A)$ is a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$. Moreover, the map $i_{*} : \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$ is induced by the inclusion $i : C_2(A) \rightarrow C_2(X)$.

However, I'm not sure which properties of $i_{*}$ are to be used to get what $H_3(X, A)$ is exactly. I'm facing similar problems for the other $H_n(X,A)$ ($n = 0,1,2$).

I think I could get it done after looking at a concrete example but such examples are surprisingly hard to find. Most examples, like the ones in Hatcher, are trivial and quick. I'm sure this one is easy as well and I'm just being an idiot but I would still appreciate the help.

答:
To start: $H_n(X) = \mathbb{Z}$ if $n=0, 1$, zero otherwise; and $H_n(A) = \mathbb{Z} \oplus \mathbb{Z}$ if $n=0, 1$, zero otherwise. Furthermore, you also need to know that the map $H_1(A) \to H_1(X)$ sends both $\mathbb{Z}$ summands to $\mathbb{Z}$ by the identity: $(a,b) \mapsto a+b$. Well, maybe you don't need this precise computation, but at least you need to know that the map $H_1(A) \to H_1(X)$ is onto. After all, as you point out, $X$ deformation retracts onto a circle, and so the inclusion of either end of the cylinder induces the identity map on $\mathbb{Z}$.

From there, you can use exactness to see that the map $H_1(X) \to H_1(X,A)$ is zero, and then you can compute $H_2(X,A)$.

Title: Computing Relative Homology


答:
Let $A=\begin{bmatrix}
-1 & -1 & 0\\
0 & 1 & 3 \\
0 & 0 & -2
\end{bmatrix}$
You need to describe the subespace $W$
consider the product of the matrix by the vector $m=(x,y,z)$ you should get
$$Am=(-x-y,y+3z,-2z)$$
but $-2m=(-2x,-2y,-2z)$
therefore $Am=-2m$ if and only if
$$-2z=-2z \rightarrow z=z$$
$$y+3z=-2y \rightarrow y=-z$$
$$-x-y=-2x \rightarrow x=y=-z$$
With this our objetive is characterize all the vectors of $W$ which are of the form
$$(-z,-z,z)=z(-1,-1,1)$$
therefore a basis for $W$ are
$$\beta= \lbrace (-1,-1,1) \rbrace$$
and
$$Dim_W=1$$

Title: Basis of a subspace of a matrix

Body:
Given
$$ A =
    \begin{pmatrix}
    -1 & -1 & 0 \\
    0 & 1 & 3 \\
   0 & 0 & -2
    \end{pmatrix}
$$
consider the subspace $W$ of $\Bbb{R}^{3}$ given by $W= \{x \in \Bbb{R}^{3}: Ax = −2x\}$. Find a basis for W


答:
Graham Kemp has already explained how to translate the specific example you're looking at; let me try to address what I think is the source of your confusion.

I think the issue is with how you're interpreting concatenation - it sounds like you think concatenation is how we write conjunction in Polish notation, so that e.g. "$Cpqr$" means "$(p\rightarrow q)\wedge r$." This is not the case. Conjunction is denoted by "$K$" which behaves syntactically just like "$C$" does, so "$(p\rightarrow q)\wedge r$" in Polish notation would be written "$KCpqr$."

So what does concatenation mean? The answer is:

Nothing whatsoever!

An expression like "$Cpqr$" is not meaningful: it has two maximal well-formed components, namely "$Cpq$" and "$r$," but the whole thing isn't a complete expression yet. Similarly, the expression "$CpqCqrpr$" you consider in the OP is not meaningful on its own: it's just four well-formed formulas put next to each other, namely "$Cpq$," "$Cqr$," "$p$," and "$r$." Additional connectives are needed to combine these into a single expression, and this is exactly what the "$CKK$" at the front is doing.

Title: Translate $\sf CKKCpqCqrpr$ into classical notation

Body:
From polish to classical notation

$\sf CKKCpqCqrpr$

I have $((p\to q)\land(q\to r)\land p)\to r~
$ for "$\sf CpqCqrpr
$" but I don't understand what $\sf CKK$ means in the beginning.


问:
There is a value less than Global minimum!

Here we see that minimum of the function $x+y+1/xy$ is obtained at $(1,1)$ which is $3$

However , $f(1,-1) = -1$. Then how is $f(1,1)$ actually the minimum?

答:
It's not! The key is local minimum, meaning that $f(1, 1)$ is less than $f(x, y)$ for all $(x, y)$ near $(1, 1)$. More precisely, there is some $r > 0$ such that
$$\sqrt{(x - 1)^2 + (y - 1)^2} < r \implies f(x, y) \ge f(1, 1).$$
But, $f(x, y) \ge f(1, 1)$ doesn't hold globally, i.e. $f$ does not have a global minimum at $(1, 1)$.

While local minimums can usually be determined purely with derivatives (which themselves are purely about the local behaviour of functions), global minimums are more tricky. You usually have to consider the behaviour of $f$ as $(x, y)$ gets very large (Can it go to $-\infty$? If not, how small can it get?), as well as compare the function value at each critical point (or at least, the critical points which might be local minimums).

In this case, if we take $(x, y) = (-t, -t)$ as $t \to \infty$, note that
$$f(x, y) = -2t + \frac{1}{t^2} \to -\infty.$$
That is, $f$ can become arbitrarily large and negative, and hence there will be no global minimum.


问:
Why doesn't the GCSE syllabus allow us to use herons formula?
I had this question in an exam:

In the triangle below, $AB = 12cm$, $BC = 19cm$ and $AC = 14cm$. Calculate the area of the triangle.

The answer to this question finds the angle $A$ using the cosine rule and then uses this formula to find the area:

$$ \frac { 1 }{ 2 } ab \sin { A } $$

Why can't I just use Heron's formula, where the area of $\triangle ABC$ with perimeter $S$ is:

$$ \sqrt { s(s - a) (s - b) (s - c) } $$

This is a GCSE question.

答:
Seconding Karl's answer that the exam may not actually disallow Heron's formula (as claimed), due to the British exam boards practising "positive marking" these days:

From a 2018 GCSE mark scheme:

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark
scheme. However, credit is given for valid answers which go beyond
the scope of the syllabus and mark scheme, referring to your Team
Leader as appropriate


问:
Euler's formula for connected, planar graphs.
So I have to prove the graph in the picture is planar using Euler's formula $v - e + f = 2.$ It obviously is planar, but the formula for it is $9 - 12 + 6 = 3$. What am I doing wrong?

答:
I have to prove the graph in the picture is planar using Euler's formula

You can't prove planarity using Euler's formula. You can sometimes prove the opposite: that graph is not planar, by using Euler's formula and fact that each face in planar graph has at least 3 edges. (which will get your criteria that in planar graph it should be that $E \leq 3V - 6$).

Euler's formula is for connected graphs. If your graph is disconnected, calculate Euler's formula for each connected part separately. Alternatively you can use formula $V - E + F = k + 1$, where k $-$ number of connected components.