问:
How to solve this permutation problem?
John went to the supermarket with a red and a green bag and bought 10 carrots & 6 radishes. On the way back he divided the vegetables between the two bags in such a way that no bag was left empty. What is the maximum number of ways in which he can divide the vegetables between the two bags?
If one of the bags was allowed to be empty then in how many ways he could divide the vegetables?

Update: My guess is the answer will be $2^{16}$.But I am a bit confused.

答:
HINT

That the bags are of different color means they are distinguishable. One direct approach is to

Pick the number of carrots in the red bag. (How many ways are there to do this?)

Pick the number of radishes in the red bag. (How many ways are there to do this?)

Put the rest of the vegetables in the green bag. (There is 1 way to do this.)

Can you compute (1) and (2) and combine everything to get the total number of ways?

Title: How to solve this permutation problem?


问:
Independent and uniformly distributed on $(\frac{1}{2},1]$
I have two random variables $X,Y$ which are independent and uniformly distributed on $(\frac{1}{2},1]$. Then I consider two more random variables, $D=|X-Y|$ and $Z=\log\frac{X}{Y}$. I would like to calculate both, the disitrbution functions $F_D(t), F_Z(t)$ and the the density functions $f_D(t),f_Z(t)$

To do that I think the first thing we need to do is to evaluate the density of the common distribution of $X$ and $Y$, but I do not know how to do that.

The only thing which is clear to me is the density and distribution function of $X$ and $Y$ because we know that they are uniform.

EDIT: Please read my own answer to this question. I need someone who can show me my claculation mistakes.

答:
My goodness ... what a lot of work. The nice thing about using a computer algebra system to solve this is: a) it's basically a one-liner, and b) if it is for a uni course, you will still need to know yourself how to get there, but at least you will know if your workings are correct. Here I am using the mathStatica add-on to Mathematica to do the grunt work.

By independence, the joint pdf of $(X,Y)$ is say $f(x,y)$:

(source: tri.org.au)

Let $Z$ = Abs$[X-Y]$. The cdf of $Z$ is $P(Z<z)$ = $P($Abs$[X-Y] < z)$:

(source: tri.org.au)

Let $Z = Log[X/Y]$. The cdf of $Z$ is $P(Z<z) = P(Log[X/Y] < z)$:

(source: tri.org.au)

All done. Differentiate to get the pdf ...

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Evaluate and explain $\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{\left(1+2\cos x\right)}\right)dx$
$$\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{1+2\cos x}\right)dx$$

I found this integral on a math discord server. I was unable to apply any standard integration technique to solve it (I am just an advanced high school student so my experience is limited). What methods can I apply to solve such an integral?

The expression does not seem to have an analytical solution.

graph of the function

答:
Just a welcoming gift to you and a New Year present.

As you wrote, I do not think that an antiderivative exists. So, you are left with numerical integration or approximations.

The plot of the integrand is quite nice (looking more or less like a parabola). So, why not compose Taylor series and integrate termwise ?

Composing
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^{8}\right)$$
$$\frac{\cos (x)}{2 \cos (x)+1}=\frac{1}{3}-\frac{x^2}{18}-\frac{x^4}{72}-\frac{7 x^6}{2160}+O\left(x^{8}\right)$$
$$\cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right)=\cos ^{-1}\left(\frac{1}{3}\right)+\frac{x^2}{12 \sqrt{2}}+\frac{23
   x^4}{1152 \sqrt{2}}+\frac{3727 x^6}{829440 \sqrt{2}}+O\left(x^{8}\right)$$

If you integrate termwise the above truncated series, you will get something $\sim 2.047$ while numerical integration would give $2.056$. Add as many terms as you can to improve accuracy.

Edit

What is amazing is that, using this $1,400$ years old approximation

$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ there is an antiderivative for
$$\int \cos ^{-1}\left(\frac{\pi ^2-4 x^2}{3 \pi ^2-7 x^2}\right)\,dx$$ The result is given in terms of an elliptic integral  of the first kind and a complete elliptic integral of the third kind.

For the integral, the result is
$$\frac{\pi}{84}   \left(21 \pi +20 \sqrt{3} \left(F\left(\csc
   ^{-1}\left( \sqrt{\frac{8}{3}}\right)|\frac{11}{6}\right)+\Pi
   \left(\frac{14}{9};-\csc ^{-1}\left(
   \sqrt{\frac{8}{3}}\right)|\frac{11}{6}\right)\right)\right)$$ which, numerically is $2.056016$ while numerical integration gives                 $2.056168$


问:
Proving or disproving basic facts about infinite series
I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I'd like to ask, if my proof to the below question on convergence of infinite series is rigorous and sufficient, and checks out.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Give an example of each or explain why the request is impossible referencing the proper theorem(s).

(a) Two series $\sum {x_n}$ and $\sum{y_n}$ that both diverge but where $\sum x_n y_n$ converges.

(b) A convergent series $\sum x_n$ and a bounded sequence $(y_n)$ such that $\sum x_n y_n$ diverges.

(c) Two sequences $(x_n)$ and $(y_n)$ where $\sum x_n$ and $\sum (x_n + y_n)$ both converge but $\sum y_n$ diverges.

(d) A sequence $(x_n)$ satisfying $0 \le x_n \le 1/n$ where $\sum (-1)^n x_n$ diverges.

Proof.

(a) The simplest examples I could come up with are:

(i) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent.

(ii) $\sum x_n = \sum_{n=1}^{\infty}\frac{1}{n}$ and $\sum y_n = \sum_{n=1}^{\infty}\frac{1}{n+1}$ are both divergent sequences, but $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.

To see that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent, we simply note that $\sum_{n=1}^{\infty}\frac{1}{n^p}$ is convergent for $p > 1$, so $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent. And $\frac{1}{n(n+1)} < \frac{1}{n^2}$, so by the Comparison test,  $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ is convergent.

(b) I think that this request is impossible. The series $\sum x_n$ is convergent. By the Cauchy Criterion, given any $\epsilon > 0$, there exists $N \in\mathbf{N}$, such that
\begin{align*}
 \absval{x_{m+1} + x_{m+2} + \ldots + x_{n}} < \frac{\epsilon}{M}
\end{align*}
for $n > m \ge N$.

Also, the sequence $(y_n)$ is bounded, so $\absval{y_n} \le M$ for all $n\in\mathbf{N}$.

Consider the expression $\absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}}$. We can write,
\begin{align*}
 \absval{x_{m+1}y_{m+1} + \ldots + x_{n}y_{n}} &\le \absval{x_{m+1}\absval{y_{m+1}} + \ldots + x_{n}\absval{y_{n}}}\\
 &\le \absval{x_{m+1}M + \ldots + x_{n}M}\\
 &=M \absval{x_{m+1} + \ldots + x_{n}}\\
 &<M \cdot \frac{\epsilon}{M} = \epsilon
\end{align*}

So, by the Cauchy criterion, $\sum x_n y_n$ is a convergent series.

(c) This request is impossible. By the Algebraic Limit Theorem, if $\sum (x_n + y_n)$ converges and $\sum y_n$ converges, then $\sum (x_n + y_n) - \sum (y_n) = \sum x_n$ is also convergent.

(d) I think that this request is impossible as well. We know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges. I have a hunch, that the the alternating series $\sum_{n=1}^{\infty}(-1)^n x_n$ is always bound by $\sum_{n=1}^{\infty}(-1)^n /n$.

By the Algebraic Limit theorem, $\lim_{n \to \infty} 0 \le \lim_{n \to \infty} \le \lim_{n \to \infty} \frac{1}{n}$, so $\lim_{n \to \infty}x_n = 0$. I would like to show that $(x_n)$ is a decreasing sequence.

答:
(a) What you did is fine.

(b) $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n$ converges, but $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}n\times(-1)^n$ diverges.

(c) What you did is fine.

(d) Take$$x_n=\begin{cases}\frac1n&\text{ if $n$ is odd}\\2^{-n}&\text{ otherwise.}\end{cases}$$


问:
Set theory exercise with universal set
Let $A$ and $B$ be two subsets of a universal set $U$. Show that:

$ A\subseteq B ⇔ A' \cup B = U$

Does anyone happen to know how to prove this correctly? I am struggling with this, as I cannot make anything sensible out of this.
Best regards

答:
"$\Rightarrow$" of course $A'\cup B\subseteq U$ by definition so we only need to show the opposite inclusion. Assume $u\in U$. We have two cases

$u\in A$. Then $u\in B$ by "$A\subseteq B$" assumption and thus $u\in A'\cup B$.

$u\not\in A$. Then $u\in A'$ and so $u\in A'\cup B$

Both give us $U\subseteq A'\cup B$.

"$\Leftarrow$" Let $a\in A$. Since $a\in U$ then $a\in A'\cup B$. But $a\not\in A'$ by assumption. Thus the only possibility is that $a\in B$ and so $A\subseteq B$.

Title: Set theory exercise with universal set


问:
What am I doing wrong in these quartic formula calculations?
I was a bit surprised that there is a general formula for the roots of a quartic equation, so I decided to test Wikipedia's version of it myself. To my surprise, I have arrived at a correct answer only once in about five attempts, using only integer coefficients of factorizable polynomials.

I am testing the formula for the equation $(x-1)(x-2)(x-3)(x-4)=0$, which should obviously return 1, 2, 3, and 4. However, my calculations tell a different story:

Expanding gives $x^4 - 10x^3 + 35x^2 - 50x + 24$. Therefore, $a = -10, b = 35, c = -50, d = 24$.

$u$ is therefore equal to $\frac{3(-10)^2 - 8(35)}{12} = \frac{300 - 280}{12} = \frac{20}{12} = \frac{5}{3}$.

$\Delta_0$ is $35^2 - 3(-10)(-50) + 12(24) = 1225 - 1500 + 288 = 13$.

$\Delta_1$ is $2(35^3) - 9(-10)(35)(-50) + 27(-10)^2(24) + 27(-50)^2 - 72(35)(24)$, which works out to 70.

The discriminant is $-\frac{\Delta_1^2 - 4\Delta_0^3}{27} = -\frac{-3888}{27} = 144$. Because it is positive, the equation must have either four real roots or four complex roots.

$Q$ is equal to $\sqrt[3]{\frac{70 + \sqrt{-3888}}{2}} = \sqrt[3]{\frac{70 + 36i\sqrt{3}}{2}} = \sqrt[3]{35 + 18i\sqrt{3}}$.

$v$ is $\frac{\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2 + 13}{3\sqrt[3]{35 + 18i\sqrt{3}}}$. After rationalizing the denominator it becomes $\frac{1 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$.

$w$ might make things even more complicated, but mercifully its numerator (and therefore value) is zero.

Placing these values into one of the final expressions for the formula gives

$$\frac{1}{4}\left(10 + 2\sqrt{\frac{6 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}} \pm 2\sqrt{\frac{9 - \sqrt[3]{35 - 18i\sqrt{3}}}{3}}\right),$$

a number which is neither rational nor even fully real, and the other two roots are the same.

I expected that the error would become self-evident as I typed this question (which took a while, as you can imagine) — but it has not. So what is wrong with these calculations?

答:
My error was made in rationalizing the denominator of $v$. I multiplied both parts of the fraction by $Q^2$, but rendered the fraction as

$$\frac{35 + 18i\sqrt{3} + 13\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{3(35 + 18i\sqrt{3})}$$

(The first $35 + 18i\sqrt{3}$ should be $(35 + 18i\sqrt{3})\sqrt[3]{35 + 18i\sqrt{3}}$.) Following from this, we can multiply both parts of the fraction by $35 - 18i\sqrt{3}$:

$$\frac{2197\sqrt[3]{35 + 18i\sqrt{3}} + 13(35 - 18i\sqrt{3})\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{6561}$$

And, simplifying, we find that the true value of $v$ is

$$\frac{\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$$

Or, to put it more simply, $\frac{7}{3}$. (This is what Wolfram Alpha evaluates as the decimal — I have no idea how to obtain this from the cube roots; any help in that respect will be appreciated.)

The expressions for the roots are then

$$\frac{1}{2}(5 + 2 \pm 1), \frac{1}{2}(5 - 2 \pm 1)$$

Which yield the expected 1, 2, 3, 4.


问:
Finding a system of equations of which solution is a span
The question is how to find a system of equations when I have the solution. The solution is subspace: $\operatorname{span}\{(2,4,-6), (4,8,-10)\}$.

I can see they describe a plane, however, the equation that describes this plane omits the last variable and looks like this: $$8x - 4y = 0.$$

Doesn't this mean that the solution to this is from $\mathbb{R}^2$ and not $\mathbb{R}^3$? So it would look like this $\{(1,2)\}$?
It is just a technical question if it's okay to just add a dimension.

It seems to be a bit odd. Could I add to the original solution a zero for example? $\operatorname{span}\{(2,4,-6,0)(4,8,-10,0)\}.$
Sorry for the English.

答: The two vectors span a plane.  You want to find a vector that is perpendicular to both so it is perpendicular to the plane, then write an equation that says the dot product of a vector with this perpendicular is $0$.  What do you know about finding a vector that is perpendicular to two others in three space?


问:
Length of a curve (?) tangent to a circle
I'm preparing for my (hopefully) future university's entrance exams, and one example question reads as follows:

In the cargo bed of a truck with width 2d a tube with radius R is placed as seen in the picture below. The cargo bed itself also has a height of R. A belt is attached over the tube, perpendicular to the length direction of the tube and the cargo bed, with its ends at points A and B, on the edges of the cargo bed. Which expression displays the length of the belt?

(A) $~2R \cdot \arcsin\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2-R^2}$

(B) $~2R \cdot \arcsin\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2+R^2}$

(C) $~2R \cdot \arccos\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2-R^2}$

(D) $~2R \cdot \arccos\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2+R^2}$

I quite frankly have no idea how to approach this problem, and it's the first real roadblock I've encountered on the example tests. Any help is much appreciated!

答:
Cheating:

The first terms relate to the curved part and the second to the straight part, by Pythagoras.

If $d$ tends to infinity, the first term must vanish, and that excludes C and D.

Then the belt must be longer than $2d$, so A.


问:
$\sigma^p$ is the product of p cycles of length q
Let $p, q$ be two integers and consider $n = p  q$. Let $\sigma$ be a cycle of length $n$ of the symmetric group $S_n$.

Show that $\sigma^p$ is the product of $p$ cycles of length $q$.

I've tried it out with the cycle $\sigma = (1\ 2\ 3\ .... 10\ 11\ 12)$ in $S_{12}$:

$\sigma^2 = (1\ 3\ 5\ 7\ 9\ 11)(2\ 4\ 6\ 8\ 10\ 12)$

$\sigma^3=(1\ 4\ 7\ 10)(2\ 5\ 8\ 11)(12\ 3\ 6\ 9)$

$\sigma^4=(1\ 5\ 9)(2\ 6\ 10)(3\ 7\ 11)(4\ 8\ 12)$

Now I need to show it in the general case.

答:
Hints: First prove the result for the $n$-cycle $\sigma = (1 2 \dots n)$. Notice that $\sigma^p$ sends 1 to $p + 1$, 2 to $p + 2$, and so on. For instance, in your example for $\sigma^3$ we have $1 \mapsto 4 \mapsto 7 \mapsto 10 \mapsto 1$, where the addition of $+ 3$ is being done mod 12. How does this description relate to its cycle decomposition?

Second, can you reduce to this case? Think about conjugation in $S_n$.


问:
How that R squared is squared correlation coefficient for $X$ given in the matrix form?
How to show that in the linear regrssion with intercept, the coefficient of determination $R^2$ is equal to the square of the sample coefficient of correlation between the variables $X$ and $Y$ given in the matrix form. I have an idea how to show this fact using the deffiniton of R squared $R^2 = \frac{ESS}{RSS}$, and formulas for ESS and RSS. However, my idea works only for one $X$ not a matrix of x's... Can you give an idea?

答:
Note, that $R^2$ is $\frac{ESS}{TSS}$ (not $\frac{ESS}{RSS}$).

But I will suggest another approach:
$$R^2 = \frac{Var(\hat{y})}{Var(y)} = \frac{Var(\hat{a} + \hat{b}X)}{Var(y)} = \frac{Var(\hat{b}X)}{Var(y)} = $$ $$ = \hat{b}^2 \frac{Var(X)}{Var(y)} = \Bigg(\frac{Cov(X, y)}{Var(X)}\Bigg)^2 \cdot \frac{Var(X)}{Var(y)} = $$ $$ = \Bigg(\frac{Cov(X, y)}{\sqrt{Var(X)Var(y)}}\Bigg)^2 = Corr^2(X,y)$$


问:
Two terms that I want to understand: weakest topology and jointly continuous (in the following context).
I was reading an article online, please help me to understand the following lines (in bold letters). -
Topological structure:
If (V, ‖·‖) is a normed vector space, the norm ‖·‖ induces a metric and therefore a topology on V.the distance between two vectors u and v is given by ‖u−v‖.

This topology is precisely the weakest topology which makes ‖·‖ continuous and which is compatible with the linear structure of V in the following sense:

1.The vector addition + : V × V → V is jointly continuous with respect to this topology. This follows directly from the triangle inequality.

2.The scalar multiplication · : K × V → V, where K is the underlying scalar field of V, is jointly continuous. This follows from the triangle inequality and homogeneity of the norm.

Please explain me the Weak topology and how does it makes norm ‖·‖ continuous. what does it mean by "Addition + : V × V → V is jointly continuous with respect to this topology"

Thank you so much in advance.

答:
The norm $\|\cdot\|$ is a function from $V$ to $\Bbb R$. There are many topologies that can be placed on $V$ that make this function continuous. Let $\mathscr{T}$ be the set of all such topologies. Then it turns out that there is a topology $\tau_0\in\mathscr{T}$ such that $\tau_0\subseteq\tau$ for all $\tau\in\mathscr{T}$. That is, every topology on $V$ that makes the norm a continuous function has to include all of the open sets in $\tau_0$. This, by the way, is equivalent to saying that $\tau_0=\bigcap\mathscr{T}$. Weakest topology here means coarsest topology, i.e., the one with the absolute minimum of open sets needed to make the norm function continuous.

Vector addition is a function $+$ from $V\times V$ to $V$. Once we impose the topology $\tau_0$ on $V$, we automatically get a product topology on $V\times V$, and we can ask whether the function $+:V\times V\to V$ is continuous with respect to that product topology on $V\times V$ and the topology $\tau_0$ on $V$. It turns out that it is. In this context jointly continuous is just a synonym for continuous with respect to the product topology, so (1) is just saying that vector addition is a continuous function. Similarly, the field $K$ has a natural topology $\tau_K$, so $K\times V$ has a product topology defined from $\tau_K$ and $\tau_0$, and (2) is just the claim that the scalar multiplication function

$$\cdot:K\times V\to V:\langle\alpha,v\rangle\mapsto \alpha v$$

is continuous with respect to this product topology in $K\times V$ and the topology $\tau_0$ on $V$.

The reason for the term jointly continuous is that it is possible for a function $f:X\times Y\to Z$ that is not continuous nevertheless to be continuous in each variable separately. That is, it’s possible for each function

$$f_x:Y\to Z:y\mapsto f(x,y)$$

with $x\in X$ to be continuous and for each function

$$f^y:X\to Z:x\mapsto f(x,y)$$

with $y\in Y$ to be continuous, without $f:X\times Y\to Z$ being continuous as a function of two variables.

When each function

$$f_x:Y\to Z:y\mapsto f(x,y)$$

with $x\in X$ to be continuous, we say that $f$ is continuous in the second variable, and when each function

$$f^y:X\to Z:x\mapsto f(x,y)$$

with $y\in Y$ to be continuous, we say that $f$ is continuous in the first variable. If both of these are the case, $f$ is separately continuous, but, as I said, this does not guarantee that it is actually continuous as function from the product space $X\times Y$ to the space $Z$.


问:
Example where natural density exists but binomial density does not
I was reading Persi Diaconis's PhD Dissertation, Weak and Strong Averages in Probability and the Theory of Numbers (warning: pdf), and found the following theorem on page 25:

Theorem 3. If a set of integers has binomial density $\ell$, then it has natural density $\ell$ but not conversely.

If needed, I define what is meant by natural and binomial density in the section below (separated out by horizontal bars).

We may define the italicized terms as follows:

Let $A$ be a set of positive integers. We define the natural density of $A$ to be

$$d\left(A\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a\left(k\right),$$

where $a\left(k\right)$ is $1$ when $k\in A$ and $0$ otherwise. In other words, this is the limit of uniform probabilities of selecting an element of $A$ in the intervals $\left\{1,\ldots,n\right\}$ as $n\to\infty$.

In order to define binomial density, let's first rewrite the natural density (if it exists) as

$$d\left(A\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a\left(k\right)=\lim_{n\to\infty}\sum_{k=1}^{n}a\left(k\right)\cdot\frac{1}{n}=\lim_{n\to\infty}\sum_{k=1}^{n}a\left(k\right)u\left(k,n\right),$$

where $u\left(k,n\right)=\frac{1}{n}$ is the (uniform) probability of selecting $k$ in $\left\{1,\ldots,n\right\}$.

Let's replace the uniform distribution with the binomial distribution. Let $b\left(k,n,p\right)={n\choose k}p^{k}\left(1-p\right)^{n-k}$ for $k=0,1,\ldots, n$ and $0$ otherwise (making sure to select $p\neq0,1$). We define the binomial density of $A$ [with parameter $p$] (if it exists) as

$$B\left(A,p\right)=\lim_{n\to\infty}\sum_{k=0}^{n}a\left(k\right)b\left(k,n,p\right).$$

(Note: Our sum also now starts at $0$ instead of $1$.)

Diaconis gives the following combinatorial interpretation of binomial density:

Flip a coin $n$ times and count how many heads show up. For large $n$, the probability that the resulting number of heads flipped is in the set $A$ is approximately $\ell$.

Question: Since Theorem 3 says that binomial density $\implies$ natural density (where implication here means that the existence of the former guarantees the existence of the latter and their values coincide), but not the converse, what does a set with natural density but no binomial density look like?

In the proof of Theorem 3, Diaconis alludes to a counterexample to the converse of the following Theorem 4.

Theorem 4. If $A\subseteq \mathbb{N}$ has Borel density $\ell$, then $A$ has Abelian density $\ell$.

He cites a previous paper of his, Limits of measures of the integers with application to random number generators and the distribution of leading digits (pp. 19-20), which I am unable to find. This counterexample to the converse of Theorem 4 would also answer my question, since binomial density $\implies$ Borel density $\implies$ Abelian density $\iff$ natural density.

I will accept an answer which either: 1) constructs a counterexample to the converse of Theorem 3 directly, or 2) provides a link to where I can find Diaconis's previous paper along with an explanation of the counterexample to the converse of Theorem 4.

答:
Take a very rapidly increasing sequence of natural numbers $N_1 < N_2 < \dots$.

Let $A = \cup_{i=1}^\infty [N_i-\sqrt{N_i}\log(N_i),N_i+\sqrt{N_i}\log(N_i)]$.

First note that $A$ has natural density $0$; indeed, the times it has the highest local density is at points of the form $N_i+\sqrt{N_i}\log(N_i)$, at which the density is roughly $\frac{2\sqrt{N_i}\log(N_i)}{N_i+\sqrt{N_i}\log(N_i)}$ (since the $N_i$'s are rapidly increasing).

However, for any fixed $p$, the $p$-binomial density of $A$ does not exist. I claim it oscillates between $0$ and $1$. Indeed, for $n=p^{-1}N_i$, basic probability (or binomial coefficient estimates) dictate that the number of heads when flipping a coin $n$ times is within $N_i\pm \sqrt{N_i}\log(N_i)$ with probability $1-o_{N_i \to \infty}(1)$, so for these $n$, we have $\sum_{k=0}^n a(k)b(k,n,p) \approx 1$. However, for $n$ not close to any $p^{-1}N_i$, we have $\sum_{k=0}^n a(k)b(k,n,p) \approx 0$ for the same reason.


问:
Is $\frac{\sin x}{x}$ undefined at $x=0$?
I have read that any function (correction: expression) divided by zero is undefined, so is $\frac{\sin x}{x}$ undefined at $x=0$?

Out of curiosity, I entered $\frac{\sin x}{x}$ into Wolfram Alpha, and the plot does not show any sign of being undefined at $x=0$.

答:
A function is defined on a domain. Since we are (implicitly) considering functions from the real line with values in the real line, you cannot divide by zero, hence the domain of the function is $\mathbb{R} - \{0\}$.

Plotting the function, you probably see that the limit of $\frac{sin(x)}{x}$ is well defined for $x \rightarrow 0$, since limits are defined in a neighborhood of a point that does not contain the point itself (as you explain in your nice blog :) )


答:
Wolfram Alpha explicitly says that the domain (in the reals) is

$$\{x\in\mathbb R:x\ne 0\}.$$

Don't trust a plot blindly.

Title: Is $\frac{\sin x}{x}$ undefined at $x=0$?

Body:
I have read that any function (correction: expression) divided by zero is undefined, so is $\frac{\sin x}{x}$ undefined at $x=0$?

Out of curiosity, I entered $\frac{\sin x}{x}$ into Wolfram Alpha, and the plot does not show any sign of being undefined at $x=0$.


问:
How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$?
How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$ ?

MY WORK :-

Well we know that $a+b$ divides $a^3+b^3$ thus $n+7$ divides $n^3+7^3$ thus it divides $n^3+343$. Now if $n+7$ divides $n^3+72$ then it divides their difference. Thus $n+7$ divides $343-72=271$. Now the factor of $271$ are $1$ and $271$ only. So if $n\lt100$ then there are no values possible; hence answer should be $0$. However in answer key it's said $1$. I want to know why it's $1$ ??

答:
Do the polynomial division with remainder:
$${n^3+72\over n+7}=n^2-7n+49-{271\over n+7}\ .$$
Now $n^2-7n+49$ is an integer for all $n$. Therefore it remains to check, for which $n\in[100]$ the number $271$ is divisible by $n+7$.


问:
How to derive the expectation of $\max(X,0)$ given the mgf function of $X$?
I got a moment-generating function for an r.v. $X$,

$$M_X(t)=\frac{i\lambda\mu}{(\lambda +t)(\mu-t)},\forall t\in(-\lambda,\mu)$$
where $i$, $\lambda$, $\mu$ are non-negative constants.

I know the expectation of $X$ can be easily derived by $M_X^{'}(0)$.

However, I am more interested in the expectation of $\max(X,0)$. i.e., I want to derive
$$\int_0^\infty x\cdot f_X(x)dx+\int_{-\infty}^0 0\cdot f_X(x)dx=\int_0^\infty x\cdot f_X(x)dx$$

based on the known $M_X(t)$.

答:
Let $F_1$ be the CDF of Exp$(\lambda)$ with characteristic function $ \frac{\lambda}{ \lambda - it} $ and $F_2$ the CDF of $-$Exp$(\mu)$ with characteristic function $ \frac{\mu}{ \mu + it} $.

Then the characteristic function of $\frac{\lambda}{\lambda + \mu}F_1 + \frac{\mu}{\lambda + \mu}F_2$ would be the given expression (see [1, Lemma 3.3.9]).

Now you can simply calculate the expected value of the positive part.

[1] Probability: Theory and Examples. Rick Durrett. Fifth Edition. https://services.math.duke.edu/~rtd/PTE/pte.html


答:
There are $2^{16}$ ways in which john can divide all the vegetables in one or two bags.

But there are only 2 ways to put everything in one bag only.

So, the number of ways he can divide the vegetables is $=2^{16}-2$

Title: How to solve this permutation problem?

Body:
John went to the supermarket with a red and a green bag and bought 10 carrots & 6 radishes. On the way back he divided the vegetables between the two bags in such a way that no bag was left empty. What is the maximum number of ways in which he can divide the vegetables between the two bags?
If one of the bags was allowed to be empty then in how many ways he could divide the vegetables?

Update: My guess is the answer will be $2^{16}$.But I am a bit confused.


问:
Density function of waiting time
```<code>    ◮ A post office has two clerks.
    ◮ Service time for each clerk is
      exponentially distributed by parameters λ1 , λ2.
    ◮ When you arrive,
     both clerks are busy but no one else waiting.
    You will enter service when either clerk becomes free.
    ◮ Calculate the density function
 of the time the customer will have to wait until his turn.
</code>```

The answer is:

Can someone explain me why?
I also have the full explanation of the answer but I can't really understand it.
They set 2 expansional variables that indicate about the time of each clerk (T1 and T2), and found the minimum between them and few more things and got to the answer.

I did read about density functions and did see here:
https://en.wikipedia.org/wiki/Exponential_distribution

that the PDF is : ${λe}^{λx}$
and

but I can't figure it out, how should I use it it order to solve the question.
Should I take it as granted? that every PDF is in that form, without the need to prove anything?

Thanks.

答:
The answer is very easy: the time you have to wait before being served is the mininum between the two waiting times.

The only problem is to understand which is the CDF of the minimum: here is an easy proof

Set $Z=min(X,Y)$

$$P(Z>z)=P(X>z;Y>z)=P(X>z)P(Y>z)=e^{-\lambda_1 z}e^{-\lambda_2 z}=e^{-(\lambda_1+\lambda_2)z}$$

Thus the CDF of Z is

$$F_Z(z)=P(Z\leq z)=1-e^{-(\lambda_1+\lambda_2)z}$$

take the derivative of F to get the desired density finding (setting $\lambda=\lambda_1+\lambda_2$) exactly the solution you posted.

$$f_Z(z)=\frac{d}{dz}=(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)z}\cdot\mathbb{1}_{[0;+\infty)}(z)$$

Title: Density function of waiting time


问:
Finding an expression for a multi variate joint CDF.
Let $X,Y$ and $Z$ be random variables with $X$ and $Y$ dependent, and $Z$ independent of both $X$ and $Y$. Let  $f_{X},f_{Y},f_{Z}$  denote the density function's of $X,Y$ and $Z$ respectively and $f_{X,Y}$ denote the joint density function of $X$ and $Y$.

Assuming the support of all three random variables is $[0,\infty]$, we can give an expression for $\mathbb{P}[Z>f(Y),Yf(Y),Yf(X,Y),X+Y<b]\;?$

答:
To get the expression you want, you will need the support for the joint p.d.f. of $X$ and $Y$ (i.e. the set of $(x,y) \subseteq \mathbb{R}^2$ for which $f_{X,Y} \neq 0$). For convenience I will let the support be given by $[u,v] \times [c,d]$ for some real numbers $u < v, c < d$.

Since $Z$ is independent of $X$ and $Y$, we have that the joint p.d.f. of $X, Y,$ and $Z$ is given by $f_{X,Y,Z}(x,y,z) = f_{X,Y}(x,y)f_Z(z)$ and so $$\mathbb{P}[Z>f(X,Y) \cap X+Y f(x,y)$, and then you integrate over the region for which $x + y < b$. That is, you integrate first over all $y$ for which $c < b - x$, and then over all possible values of $x$.

If something about this is unclear or straight-up wrong, let me know and I'll try to correct :)

Body:
To get the expression you want, you will need the support for the joint p.d.f. of $X$ and $Y$ (i.e. the set of $(x,y) \subseteq \mathbb{R}^2$ for which $f_{X,Y} \neq 0$). For convenience I will let the support be given by $[u,v] \times [c,d]$ for some real numbers $u < v, c < d$.

Since $Z$ is independent of $X$ and $Y$, we have that the joint p.d.f. of $X, Y,$ and $Z$ is given by $f_{X,Y,Z}(x,y,z) = f_{X,Y}(x,y)f_Z(z)$ and so $$\mathbb{P}[Z>f(X,Y) \cap X+Y<b] = \int_u^v\int_c^{b-x}\int_{f(x,y)}^{\infty} f_{X,Y}(x,y)f_Z(z) \; dz \; dy \; dx$$

So it's a similar procedure to the one you already have. You start by integrating $f_{X,Y,Z}$ over all values of $z$ for which $z > f(x,y)$, and then you integrate over the region for which $x + y < b$. That is, you integrate first over all $y$ for which $c < b - x$, and then over all possible values of $x$.

If something about this is unclear or straight-up wrong, let me know and I'll try to correct :)


问:
Prove $\overline{MN} \parallel \overline{PQ}$
Line $\ell$ intersects sides $\overline{AB},\overline{AC}$ of $\triangle ABC$ at $D,E$. $P,Q$ are the midpoints of $\overline{CD}$ and $\overline{BE}$ respectively. The lines through $P,Q$ perpendicular to $\ell$ meet the perpendicular bisectors of $\overline{AC}$ and $\overline{AB}$ at $M,N$ respectively. Prove that $\overline{MN} \parallel \overline{PQ}$.

The way I approached it was that $P, Q$ lie on medial triangle of $ABC$ and circumcenter $O$ of (ABC) is orhocenter of medial triangle
So by taking homothety at $G$(centroid of $ABC$) with factor of $2$ maps medial triangle to $ABC$ and $O$ to orthocenter of ABC .
Let ,$CX$ and $BY$ be perps on $AB$ and $AC$ and $PM$ intersect $BY$ at $R$ and $QN$ intersect $CX$ at $S$ then $RS$ is parallel to $MN?$

I checked it with geogebra but its not true could you point out the mistake please

答:
Since $QN\parallel PM\perp \ell$, it is enough to show that $QN=PM$.

$P$ and $Q$ lie on the sides of the medial triangle of $ABC$.

Let $M_2$ be the symmetric of $C$ wrt $M$, let $N_2$ be the symmetric of $B$ wrt $N$ and let us focus on the pentagon $AM_2 DE N_2$: we know that $M_2 D\parallel N_2 E\perp DE$ and $AN_2\perp AD, AM_2\perp AE$. Both $AEDM_2$ and $AN_2 ED$ are cyclic quadrilaterals since they have opposite right angles. This means that both $M_2$ and $N_2$ lie on the circumcircle of $ADE$. Since $\angle M_2 AD = \angle EAN_2$ it follows that $M_2 D = N_2 E$ and $NQ=MP$ as wanted.


答:
It has a limit but not a value.  That is, we can all see where the graph is going.  The value of the function will be infinitely close to 1 as $x$ gets infinitely close to 0.  This means that it is generally okay to use 1 as the value for $\frac{\sin(0)}{0}$ but it is technically non-existent precisely at the number 0.  In nearly all applications, it will work just fine.

As @Peter noted in the comments, this discontinuity is so small that it won't show up on graphs unless the graphing software is written to explicitly make it obvious to you.


答:
A function and an expression are two different things. A function has a domain (and codomain), and for each element of its domain assigns a value from its codomain.

For some functions, this assignment may be denoted by an expression, for all or some of its domain.

As such, "$\frac{\sin x}{x}$" isn't a function. It's an expression. On the other hand,

The function with the real numbers as domain and codomain, given by $x\mapsto\frac{\sin x}x$.

is almost a function. Phrased this way, it is undefined at $x=0$, even though that's an element of the domain, and this makes it not a function. There are two common ways to mend this and make the above into a bona fide function. One is

The function with the real numbers except $0$ as domain and the real numbers as codomain, given by $x\mapsto\frac{\sin x}x$.

The other is

The function with the real numbers as domain and codomain, given by $x\mapsto\frac{\sin x}x$ for $x\neq 0$, and $0\mapsto 1$.

There is no way to just look at "$\frac{\sin x}x$" and know which one of these (or something different) is intended.


问:
Is the formula for standard error for the slope of a linear regression with intercept the same as without?
If we are given sets $X$ and $Y$. The standart error formula for $\alpha$ coefficient of the regrssion $\hat{y} = \alpha x + \beta$ is $$ \frac{\sum{(y_i -\hat{y})^2}/(n-2)}{\sqrt{\sum(x_i-\bar{x})^2}}$$ How the formula for $S.E(\alpha)$ would change if we considered $\hat{y} = \alpha x$

答:
I advise using a matrix form for independent variable $X$.
Then for both cases $$s.e.(\hat{\alpha}) = \sqrt{Var(\hat{\alpha})} = \sqrt{\sigma_u^2(X^TX)^{-1}}$$
where $\sigma_u^2$ is the variance of the disturbance term.

The only difference is $X$:

$$ \text{With intercept: } \quad X =
 \begin{bmatrix}
   1 & x_1\\
   1 & x_2  \\
   \vdots & \vdots  \\
   1 & x_n   \\
   \end{bmatrix}  $$
and
$$ \text{Without intercept: } \quad X =
 \begin{bmatrix}
   x_1\\
   x_2  \\
   \vdots  \\
  x_n   \\
   \end{bmatrix}  $$


问:
Show that matrix is positive definite
Let
$A=\begin{pmatrix} d_1 & e_2 & \ldots & 0 & 0 & 0\\ e_2 & d_2 & e_3 & \ldots & 0 &0\\ 0 & e_3 &d_3 & e_4 &\ldots &0 \\ \ldots & \ldots & \ldots & \ldots & \ldots &e_n\\ 0 & \ldots &\ldots & 0 & e_n & d_n\end{pmatrix}$.

The following are true:

$d_i > 0 $ for $1 \leq i \leq n$

$d_1 > \left|e_2\right|$

$d_i > \left|e_i\right| + \left|e_{i+1}\right|$ for $2 \leq i \leq n-1$

$d_n > \left|e_n\right|$

Show that the matrix A is positive definite.

Firstly, I can show that A has strict diagonal dominance and therefore is invertible. I have found some theorems that use invertibility to prove positive definite but none suited my case I think.

I have also found the Sylvester's criterion which states that if all principal minors of A are nonnegative then A is positive definite. Well, the first two principal minors are indeed nonnegative, but after that I can't prove it.

Which should my approach be?

答:
Consider $x=(x_1,\cdots, x_n).$ We get $$xAx^T=\sum_{i=1}^n d_ix_i^2+2\sum_{i=2}^ne_ix_{i-1}x_i.$$

Now we use the inequality $2 ab\ge -a^2-b^2 $ to arrive at

$$xAx^T\ge\sum_{i=1}^n d_ix_i^2-\sum_{i=2}^ne_ix_{i-1}^2-\sum_{i=2}^n e_ix_i^2.$$ In other words

$$xAx^T\ge (d_1-e_1)x_1^2+\sum_{i=2}^{n-1}(d_i-e_i-e_{i+1})x_{i}^2+(d_n-e_n)x_n^2.$$ Finally, we conclude using the hypothesis that $x\ne 0\implies xAx^T>0,$ which shows that $A$ is positive definite.

Title: Show that matrix is positive definite


问:
If $\lim\limits_{x \to \pm\infty}f(x)=0$, does it imply that $\lim\limits_{x \to \pm\infty}f'(x)=0$?

Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is everywhere differentiable and

$\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)=0$,

there exists $c \in \mathbb{R}$ such that $f(c) \gt 0$.

Can we say anything about  $\lim_{x \to \infty}f'(x)$ and $\lim_{x \to -\infty}f'(x)$?

I am tempted to say that $\lim_{x \to \infty}f'(x)$ = $\lim_{x \to -\infty}f'(x)=0$.

I started with the following, but I'm not sure this is the correct approach, $$\lim_{x \to \infty}f'(x)= \lim_{x \to \infty}\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.$$

答:
Key fact: on small scales, you can change the derivative of a function hugely without changing its values very much. I can introduce a tiny wiggle in a curve that is barely noticeable in size, but very sharp in derivative.

Throw in a few tiny wiggles arbitrarily far along a decaying function of your choice, and you'll get a fucntion which decays but whose derivative keeps spiking.

As a more concrete hint, consider a differentiable function $f$ such that $|f(x)| < 1$ for all $x$, and take $g(x) = \frac{1}{2}f(2x)$. Then $g(x)$ has $|g(y)| < \frac{1}{2}$ for all $y$, so $g$ takes smaller values than $f$, but $g'(z) = f'(2z)$, so the derivative of $g$ gets just as big as the derivative of $f$.

This shows how very small functions can nonetheless have very large derivatives.


问:
natural mapping into ring of fractions is not always injective?
Let $R$ be a ring, $S$ a multiplicatively closed set in $R$, and $S^{-1}R$ the ring of fractions. Let $f:R \rightarrow S^{-1}R$ send $r \to r/1$. I know this mapping is injective for fields of fractions. But I don't think this is true for when $R$ is not an integral domain. Given $r/1=s/1$, by definition we have $t(r-s)=0$ for some $t \in S$, but no guarantee that $r-s=0$.

First, I'm looking for confirmation on whether this reasoning is correct (or not). Also, I'm having trouble thinking of a concrete counterexample, as I don't have much experience with rings of fractions. Any help would be appreciated, thanks!

EDIT: fixed my definition of equivalency in the ring of fractions

答:
If $R$ is an integral domain, then $(r-s)t=0$ in $R$ implies $r=s$. (Since $t=0$ implies $0\in S$, so $S^{-1}R=0$).

For a counterexample, let $R=\mathbf{Z}/6\mathbf{Z}$ and $S$ be the complement of the prime ideal $2\mathbf{Z}/6\mathbf{Z}$. (i.e. $S=\{1,3,5\}$). Then $(5-3)\cdot 3=0$, so $\frac{3}{1}=\frac{5}{1}$ in $S^{-1}R$.


问:
A simple counting and restricting the case from Vietnam National Olympiad 2021
Abstract

This problem is apparently not difficult and I managed to solve it, but I think it is just nice to share it on MSE. This might be considered one of the most "give away" problem (if not the most) in the entire contest, but the idea to solve it is worth knowing, though.

The contest's over: for another question on it, see here

Problem

Dan is given $30$ marbles to divide in to $5$ boxes, each box consist of a non-negative number of marbles. He then color the marbles such that

No marbles of the same box have the same color and

For each two boxes, we cannot choose $8$ marbles (in total) from them so that the number of colors chosen is less than or equal to $4$.

Prove that, in order to color the marbles, Dan has to use at least $10$ colors.

答:
Note, that numbers of marbles in boxes ordered in descending order can have the following patterns (* means, that this number is irrevelant):

10+, *, *, *, *

9, 6...9, *, *, *

8, 6...8, *, *, *

7, 6...7, *, *, *

6,6,6,6,6

Of course in the first case the proof is trivial.

Note, that considering all marbles within box have different color, the only situation when we have a chance of picking only 4 colors is when we pick 4 marbles from each box.

To make sure, that we pick at least 5 colors while picking 4 marbles from 2 boxes each, it's obvious, that second box can't contain more, than 3 colors from the first one. Knowing this it's pretty obvious, that in cases, where the biggest box contains 7-9 marbles, and second biggest contains at least 6 marbles, we have to use at least 10 colors (first box is coloured in at least 7 colors, second one gives us at least 6-3=3 more colors)

Now comes the fun part. Assume, that the balls are divided into 5 boxes, 6 marbles each. Assume, that in 2 boxes we've used only 9 colors (according to the previous step, it's impossible to use less colors). Let's take the third box - it's possible to use already existing colors ( 3 new from the second box, a 3 from the first one, that weren't used in the second one). We still have 9 colors, each one used in two boxes.

Now comes the fourth box. Assume, that we've used only the existing colors - we've picked 6 out of them. Each one was previously used in 2 boxes, which gives us 12 matches. According to the pigeonhole principle, there's at least one box with at least 4 matches, which contradicts with the assumption, that there are at most 3 matches between boxes. Therefore it's impossible to color fourth box using only already used 9 colors and we have to introduce one more, tenth color.

QED.


问:
Why is this identity involving vector valued functions true?
I am reading about stationary harmonic maps and I came across the following calculation.

Let $\mathcal{Q}_t(x) = x+ t\zeta(x),$ where $\zeta\in C^{\infty}_c(B_{\rho}(x_0), \mathbb{R}^n)$ where $x_0\in \mathbb{R}^m$ and $\rho>0.$ We choose $|t|$ small enough such that $\mathcal{Q}_t$ is a diffeomorphism of $B_{\rho}(x_0)$ onto itself. Then I don't understand why the following identities hold.

\begin{align}
D \mathcal{Q}_{t}^{-1}\left(\mathcal{Q}_{t}(x)\right) &= I-t D \zeta(x)+O\left(t^{2}\right) \\
\operatorname{det} D \mathcal{Q}_{t}(x) &=1+t \operatorname{div} \zeta(x)+O\left(t^{2}\right) \end{align}

Any hints/references suggesting how to prove the above two identities will be much appreciated.

答:
Since $DQ_t=I+tD\xi$, the two formulas follow from the following two facts:

If $Inv:GL(\mathbf{R}^n)\to GL(\mathbf{R}^n)$ is the inversion, then
$$D(Inv)_I[A]=-A;$$

The differential of the determinant at the identity is
$$D(\det)_{I}[A]=\operatorname{tr}(A).$$


问:
Why can a system of linear equations be represented as a linear combination of vectors?
I was watching Gilbert Strang's first Linear Algebra lecture, and the very first thing he does is relating the standard view of a system of linear equations as lines -in $\mathbb{R}^2$ of course- (what he calls the row picture) with the notion of taking a linear combination of the vectors given by the columns of the matrix (the column picture).

Now, I can see it works in practice to reach the same solution, but I don't intuitively understand why this is the case. A priori, they seem like very different things, and it's mysterious that these two views somehow correspond to each other.

Could anyone shed some light on this? I'd greatly appreciate it!

答:
@Matt24, I was thinking upon the same set of questions regarding- how did Strang start by equating a system of linear equations to a column form that you have posed in your thread. Not sure if this thread is still open.  Here is my 2 cent.

After pouring over so many books on linear algebra, finally found the answer in  W.W.Sawyer's book titled -"An Engineering Approach to Linear Algebra". This is available as pdf, just search for it.

The clue to the missing link between the row form (SLE) and the column form is the glaring omission of the below 3 items by Strang and others in the depiction of the vector form.

Strang hasn't provided the exact definition of the problem statement that the 2 linear equations are representing. In real world, the 2 equations could be representing anything or for example price (first equation) and weight (second equation). This is a colossal omission for engineers, whereas acceptable for mathematicians as they breath in the abstract form.

Due to the above (1), the X & Y axes definitions which represent the actual problem statement are not provided for the Vector form. I.e. what does X axis and Y axis in vector form represent about the original problem statement from the real world?

The row form of SLEs are represented as a single graph with 2 lines representing 2 equations in this case. The arrangement/intersection of lines represent say either 'a' solution or 'many' solutions or 'no' solution and if solution(s) exist, what are they?  This is not intuitive or suitable for studying or solving a real world problem which has multiple attributes. By the end of this response, we will understand what this means.

I believe, if only mathematicians have spent some more time in explaining the above 3 items in their text books, this could have been intuitively understood by mortals like us.

To make it simpler and understand better, I have provided the problem statement as below.

The price of milk is 2 dollars per bottle and the price of cheese is 1 dollar per box. The weight of milk is 1 pound per bottle and the weight of cheese is 1 pound per box.
Due to a promotion, when you buy a bottle of milk, a box of cheese is free and the delivery to your home is free if the combined weight is up to 5 pounds.
What quantities of milk and cheese can be bought for 1 dollar and make avail the promotion for free delivery (up to 5 pounds) (if at all possible)?

The above problem statement can be represented by 2 linear equations. 1 for price view and 1 for weight view.

Total Price = (cost of milk per bottle * number of milk bottles) - (cost of cheese per box * number of cheese boxes) ---> Price Equation

Total Weight = (weight of milk per bottle * number of milk bottles) + (Weight of cheese per box * number of cheese boxes) --> Weight equation

There is '-' in the Price equation to accommodate free cheese. You don't have to pay for it due to promotion otherwise you would have paid.
The '+' in the weight equation to accommodate the combined weight milk and cheese. The price of cheese if free whereas the weight still needs to be carried and hence counted.

After applying the price, weight and constraints from the problem statement, the 2 linear equations are

2x - y = 1 (price equation)

x + y = 5 (weight equation)

Note: Have deliberately kept the price, weight as 2 dollars, 1 pound for milk and 1 dollar, 1 pound for cheese just to highlight there may or may not be a solution.

This is the key here. Both equations represent different constraints of the problem statement. i.e. both are not price equations/both are not weight equations.
Can both represent same constraint like 2 price equations? Probably yes, but that would be like price equation for 2 separate occasions etc. and not this problem statement, maybe different problem statement.

As explained above, let's assume the first equation is representing price view and the second equation is representing weight view.  With this explanation, let's try to answer your questions.

Simplified progression from SLEs to Vectors

Vector view with dimensions

What is a vector [2, 1] T?

From the 2 linear equations, there are 3 vectors.

Milk Vector    = (+2x dollars from first equation, +1x pounds from second equation) = (2, 1).

Cheese Vector = (-1y dollars from first equation, +1y pounds from second equation) = (-1, 1).

Combined Milk & cheese vector (Resultant vector) = (+1 dollar from first equation, +5 pounds from second equation) = (1, 5).

So the vector form represents 2 attributes (price and weight in 2 dimensions) of milk or cheese or both!!!  Due to different attributes (dimensions) that we are interested, the price and weight cannot be added by usual arithmetic (say number line addition) as this doesn't make any sense. So this calls for different arithmetic (linear algebra) with its own rules and methods.  This is the crux of linear algebra which heavily uses vectors, matrices, complex numbers and all of these aids addition and multiplication of numbers which are represented by 2 or more quantities (tuples).

What does it mean by multiplying a real number x to the vector [2, 1] T?

By now, this should be evident. Multiplying a real number x to the vector (2, 1) - x in this case represent scalar/real number which represents the 'number of milk bottle'. Each milk bottle has 2 attributes. I.e. each milk bottle represents 1 dollar cost and 1 pound weight.

What does it mean by adding two vectors together?

In this case vector addition means, addition of milk vector (price, weight) and cheese vector (price, weight) to obtain combined milk and cheese vector or resultant vector.

When are two vectors the same?

In this case, when price and weight of both milk (say 2 dollars & 1 pound) and cheese (say 2 dollars and 1 pound) are same, milk and cheese vectors shall be same.  The book definition says when magnitude and directions are same, both vectors are same. When same attributes of milk and cheese vectors are added in the X-Y plane, the magnitude and direction will also be same.

What is the linear function or linear map for the above problem statement?
The linear function or linear map for the above example can be depicted as below.

f(x) is mapped to f(x*)

f(milk bottle) is mapped to (milk price + milk weight)

f(y) is mapped to f(y*)

f(cheese box) is mapped to (cheese price + cheese weight)

f(x,y) is mapped to f(x*,y*) or f(x+y) is mapped to f(x*+y*)

f(milk bottle, cheese box) = ( price of milk and cheese, weight of milk and cheese)

f(x,y) = (2x-y, x+y)
The price/weight does not contain any powers or combinations of each other.

f(1 bottle of milk, 1 box of cheese) ==> f(x,y) ==> f(1,1) = (2.1-1,1+1) = (1,2)
ie f(1,1) is mapped to (1,2)

f(2 bottle of milk, 1 box of cheese) ==> f(2x,y) ==> f(2,1) = (2.2-1,2.1+1) = (3,2). Replace x in second equation with 2x.
i.e. f(2,1) is mapped to (3,2)

This is an example of 2 SLE with 2 unknowns. The same can be extended to any set of SLEs with any unknowns. Hope this makes the intuition clear. If there are any glaring holes in my argument, let me know. I can share some of the worksheets and simulations in excel which show the above clearly.


答:
Confusingingly, people often talk about the behaviour of this function as $x$ approaches $0$ (but is not equal to it). This can be easily be confused with evaluating $\sin(0)/0$, which is non-sensical. However, we can say that
$$
\lim_{x \to 0}\frac{\sin(x)}{x}=1 \, .
$$
Roughly speaking, the above expression means 'as $x$ gets closer and closer to $0$, $\sin(x)/x$ gets closer and closer to $1$'. It does not mean that if $x=0$, then $\sin(x)/x=1$.


问:
Triple integral clarification please
I have this question which states:

Let $T$ be the tetrahedron bounded by the planes $x = 0, y = 0, z =0$ and $x+y+z = 2$

Integrate $$J = \iiint_{T} y^2 \,dx\,dy\,dz$$
Using the above integral, or otherwise, evaluate
$$K = \iiint_{P}(x^2+\sin(xy^2)+ y^2) \,dx\,dy\,dz$$

where $$P = \left \{(x,y,z) : 0 \leq z \leq 2 -|x| -|y|\right \}$$

My thoughts are as follows I think the first part of the question is understandable just use the limits and follow through with the solution $J = 0$ but the second part is tougher.

I'm guessing we realise (if we let) $f(x,y) = x^2y$ this is odd w.r.t. y and domain P is symmetric about y-axis. And the same for $g(x,y) = \sin(xy^2)$ which is odd w.r.t. x and P is symmetric about x-axis. We can say their integral over P is $0$

So the final solution is $$K = J = 0$$

I hope someone could either correct me and tell me how to proceed if this is wrong thank you!

答:
I'm going to give you an hint only for the triple integral $J$ and hope I will not do any stupid error.

The region you are considering is $x+y+z = 2$, $x\ge0$, $y\ge0$ and $z\ge0$.
Then you can get for example the following parameterization:
$$y\in [0,2], \quad z \in [0,2-y], \quad x \in [0,2-y-z]$$

Once you got this you can write the integral as follows:
$$\int_0^2 \int_0^{2-y} \int_0^{2-y-z}y^2 \ \ dx\ dz\ dy$$

Now you just have to perform the integrations correctly and you should get $J = \frac{8}{15}$

Title: Triple integral clarification please


问:
Differentiate an equation
I'm learning about implicit differentiation and i'm stuck at understanding the concept of "differentiating an equation".

If I define $ f(x) = x^2$ and $g_1(y) = 9$ and $g_2(y) = 9 + y^2$

Then I define 2 equations:

$(1)$ $f(x) = g_1(y)$ (i.e. $x^2 = 9$)

Then I can't differentiate both side of this equation with respect to x (ortherwise I will get: $2x = 0$)

$(2)$ $f(x) = g_2(y)$ (i.e. $x^2 = 9 + y^2$)

Then I can differentiate both side of this equation with respect to $x$ to get: $ 2.x = 2.y.\frac{dy}{dx}$

Could you please explain me the difference between the 2 cases ? And in general, in which condition could I differentiate both side of an equation concerning more than 1 variable ?

Thank you very much for your help!

答:
A few notes: you are taking derivatives.  I think the more fundamental operator is the differential (which gets confusing because taking a derivative is referred to as "differentiating").  The differential is like a derivative, but not with respect to any variable.  So, for instance the derivative of $x^2$ is $2x$ but the differential is $2x\,dx$.  The derivative of $xy$ depends on which variable the derivative is being taken with respect to, but the differential is just $x\,dy + y\,dx$.

Derivatives are more likely to lead to contradictions.  For instance, take the equation $x = 1$.  The derivative of this with respect to $x$ is $1 = 0$.  However, the differential is $dx = 0$, which is true: the value of $x$ is not changing.  The reason for the contradiction of the derivative is that the derivative with respect to $x$ is just the differential divided by $dx$.  Since $dx = 0$, dividing both sides by $0$ leads to an invalid $\frac{0}{0}$.

So, going back to your problem, when you had $x^2 = 9$, the differential gives you a better picture. $2x\,dx = 0$.  Why?  Because at $x^2 = 9$, $x$ is not moving, so the differential is zero.  If you try to divide by $dx$, you will wind up with an invalid $\frac{0}{0}$.

So, in your second case, differentiating gives you: $2x\,dx = 2y\,dy$.  You can then solve for the derivative by algebraic manipulations.

So, in both cases, finding the differential works.  In the first case, finding the derivative doesn't work because you wind up with a $\frac{0}{0}$ when you try.

Title: Differentiate an equation


问:
Did anybody use numerical analysis before modern computers were introduced?
I was recently introduced to Numerical analysis to solve PDEs and I was wondering wether this kind of approach was also used barehand to reach approximate conclusions in the same way it is now done by employing calculators.</p>

答:
Gauss calculated the orbit of Ceres

"... for it is now clearly shown that the orbit of a heavenly body may
be determined quite nearly from good observations embracing only a few
days; and this without any hypothetical assumption."

https://sites.math.rutgers.edu/~cherlin/History/Papers1999/weiss.html

Perhaps this answers your question.


问:
The power set of a $\Pi_1^0$ set is a $\Pi_1^0$ class.
I'm working through the syllabus of our Computability Theory course in preparation of the exam, and came across the following exercise for which I want to check whether my construction is correct or not.

The statement is the following: if $A$ is a $\Pi_1^0$-set, then $P(A)$ is a $\Pi_1^0$-class.

Small elaboration; $\Pi_1^0$-sets are sets of the form $A=\{x\mid \forall yR(x,y)\}$ for some computable relation $R$. $\Pi_1^0$ classes are of the form $\mathcal{A}=[T]=\{X\in 2^\omega\mid \forall n(X\restriction n \in T)\}$, for some computable $T\subseteq 2^{<\omega}$. So it consists of all infinite paths in a computable tree.

So, given such a set $A$ we want to construct some computable tree $T$. If $A$ were computable; we would define the tree by starting at node $0$. If $0$ were in $A$, we would start two seperate trees; one with a $0$ in the first bit, and one with a $1$ in the first bit. If $0$ weren't in $A$, we only start a tree with the $0$ in the first bit. Then we proceed to the second bit, and do the same construction; we extend both our existing trees by two branches if $1$ is in $A$, and by only one branch if $1$ is not in $A$. This construction clearly gives us a computable tree for which the paths are models for all subsets of $A$.

Unfortunately, the set $A$ we are considering is not computable and hence we have to do an alternative construction of our tree. My idea now is the following; we start at the bottom, and check whether $R(0,0)$ holds. If it holds; we start two trees, if it doesn't, we only start one with $0$ in the first bit. Then, we move to the second bit. First, we check whether $R(0,1)$ holds. If it holds, we do nothing extra and proceed to check whether $R(1,0)$ and $R(1,1)$ hold (and then add two branches if both hold, and only one if it doesnt). If checken $R(0,1)$ would have shown that it doesn't hold, then we destroy the entire tree that starts with a $1$ in the first bit. We then continue inductively, and at step $n$ we check $R(m,n)$ for every $m\leq n$ and destroy branches 'in hindsight' if this doesn't work.

Sorry, this might be totally unreadable, but I could not really find another way to describe the approach. Do you think this construction gives a correct answer to the exercise, or is there something I am missing in my approach?

Thanks!

答:
Your argument is a bit unclear to me, but it sounds like it's got the right idea. The following I think is a snappier way of describing what you're doing:

Suppose $A$ is $\Pi^0_1$, and fix some computable $R\subseteq\omega^2$ such that $A=\{x:\forall y(R(x,y))\}$. Say that a finite binary string $\sigma$ is a plausible subset of $A$ iff for each $x<\vert\sigma\vert$ such that $\sigma(x)=1$ the relation $R(x,y)$ holds for all $y<\vert\sigma\vert$. The set $T$ of plausible subsets is a computable tree - the point being that the "$<\vert\sigma\vert$" clauses make all the relevant searches finite.

Now suppose $f$ is an infinite path through $T$ and $f(x)=1$. Then for each $n$ we have $f\upharpoonright (n+1)$ is a plausible subset of $A$, and so $R(x,n)$ holds. This means in fact that we have $\forall y(R(x,y))$, and so we get $\{x: f(x)=1\}\subseteq A$ as desired.


问:
What is the image of this Linear Map?
In my Linear Algebra class we have been asked to find the kernel and
the image of the following linear map $T : \mathbb{R}^4 \rightarrow \mathbb{R}^4$, then verify the rank-nullity theorem.

$T(x_1,x_2,x_3,x_4) \mapsto(x_1-x_2, x_2-x_3,x_3-x_4,x_4-x_1)$

$\ker(T)$ was trivial, it's obviously equal to the set $\{(v,v,v,v) \mid v \in \mathbb{R} \}$.

However, I am having difficulty finding $\mathrm{im}(T)$. Some observations:

There is no ordered 4-tuple in the image such that all four elements have the same sign.

$T$ is not surjective, since the vector $(1,0,0,0) \not \in \mathrm{im}(T)$.

Since $\{(1,0,0,0), ..., (0,0,0,1)\}$ spans $\mathbb{R}^4$, their images $\{(1,0,0,-1),...,(0,0,-1,1)\}$ must span the image.

Can anyone provide a hint to the solution?

答:
Hints: Let's look at your third point in more detail. You concluded that the images of the four standard basis vectors $$\langle (1, 0, 0, -1)^T, (-1, 1, 0, 0)^T, (0, -1, 1, 0)^T, (0, 0, -1, 1)^T \rangle$$
spans the image. But is this spanning set a basis for the image? In other words, is this set linearly independent dependent? If it's independent, we're in trouble with rank-nullity because you found the 1-dimensional kernel. But if it is dependent, how do you modify this set to get a basis?

Title: What is the image of this Linear Map?


问:
Is log of ETS(MNM) a ETS(ANA) model?
Given two time series models $ETS(MNM)$ and $ETS(ANA)$. If we replace $y_t $ by $\ln y_t $ in $ETS(MNM)$, is it true the model will become $ETS(ANA)$?

答:
They are not exactly the same, however the log transformation makes somewhat $ANA$ model $MNM$. Let's write the equations for the both models.

$$ ETS(MNM) = \begin{cases} y_t = l_{t-1}\cdot s_{t-12}\cdot (1+u_t) \\
 l_t = l_{t-1}\cdot (1+\alpha u_t)\\  s_t = s_{t-12}\cdot (1+\gamma u_t) \end{cases}  $$

$$ ETS(ANA) = \begin{cases} y_t = l_{t-1} + s_{t-12} + u_t \\
 l_t = l_{t-1}+\alpha u_t\\  s_t = s_{t-12} + \gamma u_t \end{cases}  $$
Remember that $\ln(ab) = \ln a + \ln b$. Hence,
$$ \ln y_t = \ln(l_{t-1}) + \ln(s_{t-12}) + \ln(1 + u_t)$$
Same for other equations.
As we can see, the models are not exactly the same, however it became additive ($ANA$).

Title: Is log of ETS(MNM) a ETS(ANA) model?


问:
Can the Complete Graph on ten vertices be edge covered by three copies of the Petersen Graph?
Put another way ... Colour the edges of the complete graph with 3 colours, so that three subgraphs are each a copy of the Petersen Graph.
I heard somewhere that it can be done (Maybe I should not go on MathOverFlow !) but I have spent all weekend trying and have convinced myself it is impossible. Thanks in advance for your comments.</p>

答:
This can be checked computationally.  Here is my GAP code:

```<code>F:=[[1,2],[2,3],[3,4],[4,5],[1,5],[6,8],[8,10],[7,10],[7,9],[6,9],[1,6],[2,7],[3,8],[4,9],[5,10]];
</code>```

In the above, F is the edge set of the Petersen graph.

```<code>count:=0;
S:=[];
for alpha in SymmetricGroup(10) do
  F2:=OnSetsSets(F,alpha);
  count:=count+1;
  if(Intersection(F,F2)=[]) then
    S:=Concatenation(S,[F2]);
    Print(Size(S)," out of ",count," out of ",Factorial(10),"; expecting: ",1.0*Factorial(10)*Size(S)/count,"\n");
  fi;
od;
</code>```

The above generates a list S of graphs isomorphic to the Petersen graph F which do not share an edge.

```<code>B:=[S[1],S[1]];
best:=Size(Intersection(B[1],B[2]));
for i in [1..Size(S)] do
  for j in [i+1..Size(S)] do
    F1:=S[i];
    F2:=S[j];
    int:=Size(Intersection(F1,F2));
    if(int<best) then
      best:=int;
      B:=[F1,F2];
      Print("Best thus far: ",best,"\n",B[1],"\n",B[2],"\n\n");
    fi;
  od;
od;
</code>```

The above attempts to find two members of S, denoted F1 and F2, which are edge disjoint.  It doesn't find any.  The best it finds is the following:

Here the last two graphs share the 6 edges highlighted in bold.

Body:
This can be checked computationally.  Here is my GAP code:

F:=[[1,2],[2,3],[3,4],[4,5],[1,5],[6,8],[8,10],[7,10],[7,9],[6,9],[1,6],[2,7],[3,8],[4,9],[5,10]];

In the above, F is the edge set of the Petersen graph.

count:=0;
S:=[];
for alpha in SymmetricGroup(10) do
  F2:=OnSetsSets(F,alpha);
  count:=count+1;
  if(Intersection(F,F2)=[]) then
    S:=Concatenation(S,[F2]);
    Print(Size(S)," out of ",count," out of ",Factorial(10),"; expecting: ",1.0*Factorial(10)*Size(S)/count,"\n");
  fi;
od;

The above generates a list S of graphs isomorphic to the Petersen graph F which do not share an edge.

B:=[S[1],S[1]];
best:=Size(Intersection(B[1],B[2]));
for i in [1..Size(S)] do
  for j in [i+1..Size(S)] do
    F1:=S[i];
    F2:=S[j];
    int:=Size(Intersection(F1,F2));
    if(int<best) then
      best:=int;
      B:=[F1,F2];
      Print("Best thus far: ",best,"\n",B[1],"\n",B[2],"\n\n");
    fi;
  od;
od;

The above attempts to find two members of S, denoted F1 and F2, which are edge disjoint.  It doesn't find any.  The best it finds is the following:

Here the last two graphs share the 6 edges highlighted in bold.


问:
Find $\sin\alpha,\cos\alpha,\tan\alpha$ if $\cot\alpha=-2$
Find $\sin\alpha,\cos\alpha,\tan\alpha$ if $\cot\alpha=-2.$

We have defined trigonometry with a circle, but only for angles between $0^\circ$ and $180^\circ.$

We have $\begin{cases}\cot\alpha=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}\iff\begin{cases}\dfrac{\cos\alpha}{\sin\alpha}=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}.$

So I got that $\sin\alpha=\dfrac{\sqrt5}{5},\cos\alpha=-\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2$ or $\sin\alpha=-\dfrac{\sqrt5}{5},\cos\alpha=\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2.$

An angle with sine equal to $-\dfrac{\sqrt5}{5}$ isn't in the inverval $\left[0^\circ;180^\circ\right],$ right? I suppose that it is possible two different angles to have equal $\tan$ and $\cot.$

答: There is no angle in the range $[0,180^\circ]$ with a negative sine.  The angles in $(180^\circ,360^\circ)$ all have negative sine.  You can use the relation $\sin(180^\circ+x)=-\sin(x)$ but if your definition is restricted as you say, only your first answer is acceptable.  There are two angles with the same $\tan$ and $\cot$ in $[0,360^\circ)$.  They are $180^\circ$ apart.


问:
Seeking for an improvement of: $ P_n \sim \frac{10^n[\ln(n)-\frac{1}{4}]}{2\operatorname{Li}(10^n)}$
$$P_n \sim n\ln(n)\tag1$$

Where $P_n$ is the nth prime number

Equation $(2)$ is an improvement of $(1)$

$(2)$ comes about through trial and error using Wolfram alpha.
$$ P_n \sim \frac{10^n[\ln(n)-\frac{1}{4}]}{2\operatorname{Li}(10^n)}\tag2$$
Where $\operatorname{Li}(n)$ is the Logarithmic integral

Example:

$(1):$

$$P_{10,000}\sim 10,000\ln(10,000)$$

$$104,729 \sim 92,103$$

Got an error of $12$%

$(2):$

$$P_{10,000} \sim \frac{10^{10,000}\left[\ln(10,000)-\frac{1}{4}\right]}{2\operatorname{Li}(10^{10,000})}$$

$$104,729\sim 103,155$$

Got an error of $1.5$%

Can anyone help to improve on $(2)$?

答:
Your approximation is very bad because $\operatorname{Li}(x) \sim \frac{x}{\ln x}$ and so \begin{align*}\frac{10^n (\ln n-0.25)}{2 \operatorname{Li}(10^n)} &\sim  \frac{10^n (\ln n-0.25)}{2 \frac{10^n}{\ln 10^n}} = \frac{\ln 10^ n (\ln n-0.25)}{2} \\ & = \frac{n \ln 10 (\ln n-\frac{1}{4})}{2} \sim \frac{n \ln n }{2}  \ln 10 \approx 1.15129 n \ln n\end{align*} for large enough $n$.

M. Cipolla proved that there are arbitrarily better approximation for $p_n$, for example there are $ p_n \sim n \ln n$ and $ p_n \sim n(\ln n+\ln \ln n-1) $ and $p_n \sim n(\ln n+\ln \ln n-1+\frac{\ln \ln n-2}{\ln n}) $


问:
Show that $\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) = \mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3})$
I have seen this same question answered here, but I would like to know if I could arrive to the solution with any of my approaches. Splitting field of x^3-2 as a simple extension

Clearly $\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) \subseteq \mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3})$, so the only thing left to do is prove it the other way.

1st attemp

To see that $\sqrt[3]{2}, e^{2 \pi i/3} \in \mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3})$ would be enough. For this, I have seen that a common approach is to compute $(\sqrt[3]{2} + e^{2 \pi i/3})^{-1}$ and generaly you obtain an element that summed with $\sqrt[3]{2} + e^{2 \pi i/3}$ results in $\sqrt[3]{2}$ or $e^{2 \pi i/3}$. I am having a really hard time trying to rationalize it.

2nd attemp

I try to reason it by using the degrees of the extension. We have that $[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3})] = \frac{[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) : \mathbb{Q}]}$.

I think the esiest way to get $[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}]$ is by presenting a $\mathbb{Q}$-base. Although I think I am not doing it right, because seeing this question Splitting field of x^3-2 as a simple extension, I think it should be of degree 6, not 7.

$\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) = \mathbb{Q}\langle1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{2}e^{2 \pi i/3}, \sqrt[3]{2}e^{4 \pi i/3}, \sqrt[3]{4}e^{2 \pi i/3}, \sqrt[3]{4}e^{4 \pi i/3}\rangle$

To compute $[\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) : \mathbb{Q}]$ we can try to find the irreducible polynomial.

$$
\alpha = \sqrt[3]{2} + e^{2 \pi i/3} \\
\alpha^3 = (\sqrt[3]{2} + e^{2 \pi i/3})^3 \\
\alpha^3 = 2 + 3\sqrt[3]{4}e^{2 \pi i/3} + 3\sqrt[3]{2}e^{4 \pi i/3} + 1 \\
\frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{4}e^{2 \pi i/3} + \sqrt[3]{2}e^{4 \pi i/3} \\
\frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{2}e^{2 \pi i/3}(\sqrt[3]{2} + e^{2 \pi i/3}) \\
\frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{2}e^{2 \pi i/3}\alpha \\
\frac{1}{27}(\alpha^3 - 3)^3 = 2\alpha^3 \\
\frac{1}{27}(\alpha^9 - 9\alpha^6 + 27\alpha^3-27) = 2\alpha^3 \\
\alpha^9 - 9\alpha^6 - 27\alpha^3-27 = 0
$$

So clearly, $\sqrt[3]{2} + e^{2 \pi i/3}$ is a root of the polynomial $p(x) = x^9 - 9x^6 - 27x^3-27$. But it does not seem irreducible.

I am stuck at this point. Thanks in advance, any help would be appreciated.

答:
We have
$$
x^9-9x^6-27x^3-27=(x^6 + 3x^5 + 6x^4 + 3x^3 + 9x + 9)(x^3 - 3x^2 + 3x - 3).
$$
The factors are irreducible, so we have the desired result for degree $6$, which you have linked: Splitting field of x^3-2 as a simple extension


问:
How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$?
How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$ ?

MY WORK :-

Well we know that $a+b$ divides $a^3+b^3$ thus $n+7$ divides $n^3+7^3$ thus it divides $n^3+343$. Now if $n+7$ divides $n^3+72$ then it divides their difference. Thus $n+7$ divides $343-72=271$. Now the factor of $271$ are $1$ and $271$ only. So if $n\lt100$ then there are no values possible; hence answer should be $0$. However in answer key it's said $1$. I want to know why it's $1$ ??

答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^2$.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1,
   10^2}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]=Nothing
</code>```

What implies that there are no solutions. If I extend the search to $1\le\text{n}\le10^8$ I found:

```<code>In[2]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1,
   10^8}] //. {} -> Nothing

Out[2]={264}
</code>```

Which means that $\text{n}=264$ is a solution, because:

$$\frac{264^3+72}{264+7}=67896\tag1$$

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^2$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1,
   10^2}] //. {} -> Nothing

Running the code gives:

Out[1]=Nothing

What implies that there are no solutions. If I extend the search to $1\le\text{n}\le10^8$ I found:

In[2]:=Clear["Global`*"];
ParallelTable[
  If[IntegerQ[(n^3 + 72)/(n + 7)], n, Nothing], {n, 1,
   10^8}] //. {} -> Nothing

Out[2]={264}

Which means that $\text{n}=264$ is a solution, because:

$$\frac{264^3+72}{264+7}=67896\tag1$$


答:
Hint:
$1\cdot(1,0,0,-1)^T + 1 \cdot (-1,1,0,0)^T + 1 \cdot (0,-1,1,0)^T + 1 \cdot (0,0,-1,1)^T = \mathbf{0}$, the zero vector.

What happens, if you remove one of those vectors that lie in the span of the other three?

Title: What is the image of this Linear Map?

Body:
In my Linear Algebra class we have been asked to find the kernel and
the image of the following linear map $T : \mathbb{R}^4 \rightarrow \mathbb{R}^4$, then verify the rank-nullity theorem.

$T(x_1,x_2,x_3,x_4) \mapsto(x_1-x_2, x_2-x_3,x_3-x_4,x_4-x_1)$

$\ker(T)$ was trivial, it's obviously equal to the set $\{(v,v,v,v) \mid v \in \mathbb{R} \}$.

However, I am having difficulty finding $\mathrm{im}(T)$. Some observations:

There is no ordered 4-tuple in the image such that all four elements have the same sign.

$T$ is not surjective, since the vector $(1,0,0,0) \not \in \mathrm{im}(T)$.

Since $\{(1,0,0,0), ..., (0,0,0,1)\}$ spans $\mathbb{R}^4$, their images $\{(1,0,0,-1),...,(0,0,-1,1)\}$ must span the image.

Can anyone provide a hint to the solution?


问:
Evaluate and explain $\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{\left(1+2\cos x\right)}\right)dx$
$$\int_{0}^{\frac{\pi}{2}}\arccos\left(\frac{\cos x}{1+2\cos x}\right)dx$$

I found this integral on a math discord server. I was unable to apply any standard integration technique to solve it (I am just an advanced high school student so my experience is limited). What methods can I apply to solve such an integral?

The expression does not seem to have an analytical solution.

graph of the function

答: As @K.defaoite notes, this is Coxeter's integral, equal to $5\pi^2/24$. The usual way (see Inside Interesting Integrals Secs. 6.2 and 6.3, where this takes Nahin just over 11 pages) to evaluate it (whose details I've half-provided as they're worth working through to appreciate how hard this is!) is to prove it's four times Ahmed's integral $A(1)$ where $A(u):=\int_0^1\tfrac{\arctan\left(u\sqrt{2+x^2}\right)}{(1+x^2)\sqrt{2+x^2}}dx$, so$$A(\infty)=\left[\tfrac{\pi}{2}\arctan\tfrac{x}{\sqrt{2+x^2}}\right]_0^1=\tfrac{\pi^2}{12}$$and$$A^\prime=\tfrac{1}{1+u^2}\left[\arctan x-\tfrac{u}{\sqrt{1+2u^2}}\arctan\tfrac{xu}{\sqrt{1+2u^2}}\right]_0^1$$so$$A(\infty)-A(1)=\tfrac{\pi^2}{16}-\int_1^\infty\tfrac{u}{(1+u^2)\sqrt{1+2u^2}}\arctan\tfrac{u}{\sqrt{1+2u^2}}du.$$With $u\mapsto1/u$, this simplifies to$$A(\infty)-A(1)=\tfrac{\pi^2}{16}-A(\infty)+A(1)\implies A(1)=\tfrac{5\pi^2}{96}.$$Now call your integral $C$ so, with some trigonometric identities,$$C=\int_0^{\pi/2}2\arctan\sqrt{\tfrac{1+\cos x}{1+3\cos x}}dx\stackrel{y=\tfrac{x}{2}}{=}4\int_0^{\pi/4}\arctan\tfrac{\cos y}{\sqrt{2-3\sin^2y}}dy.$$Since $\arctan b=\int_0^1\tfrac{bdt}{1+b^2t^2}$,$$C=\int_0^{\pi/4}\int_0^1\tfrac{4\cos y\sqrt{2-3\sin^2y}}{t^2+2-(t^2+3)\sin^2y}dtdy.$$With $\sin y=\sqrt{\tfrac23}\sin w$ followed by $s=\tan w$ (who'd guess that?),$$C=\int_0^{\sqrt{3}}\int_0^1\tfrac{8\sqrt{3}dtds}{(1+s^2)(t^2s^2+3t^2+6)},$$which simplifies by partial fractions to$$C=\tfrac{2\pi^2}{9}-4\int_0^1\tfrac{t\arctan\tfrac{t}{\sqrt{t^2+2}}}{(t^2+3)\sqrt{t^2+2}}dt.$$With $u=\arctan\tfrac{t}{\sqrt{t^2+2}}$, this becomes $C=4A(1)$, as desired.


问:
Evaluating $\int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx$

Q: Suppose $\alpha>0$ and $|\beta|<\pi/2$, show that
  \begin{align*}
\textbf{(1)} \; \int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx &= \frac 1 2 \sqrt{\pi/\alpha}\cos(\beta/2)\\
\textbf{(2)} \; \int_0^{\infty}e^{-\alpha x^2 \cos \beta} \sin(\alpha x^2 \sin \beta) dx &= \frac 1 2 \sqrt{\pi/\alpha}\sin(\beta/2)
\end{align*}

How can I integrate the above with the method of contour?

The integral can be changed into $\displaystyle \int_0^{\infty}e^{-\alpha x^2 \cos \beta} e^{i(\alpha x^2 \sin \beta) }dx = \int_0^{\infty} e^{x^2\alpha e^{i (\pi - \beta)}}dx$. This is similar to $\displaystyle \int_0^{\infty} e^{-x^2}dx$ which has been discussed here except that it has complex coefficients. How do I modify it?

答:
Use a wedge contour $C$ of angle $-\beta/2$, i.e., below the real axis.  That is, consider

$$\oint_c dz \, e^{-a e^{i \beta} z^2} = \int_O^R dx \, e^{-a e^{i \beta} x^2} + i R \int_0^{-\beta/2} d\theta e^{-a R^2 e^{i (\beta+2 \theta)}} + e^{-i \beta/2} \int_R^0 dx \, e^{-a x^2}$$

Note that by using this contour, we get a pure Gaussian integrand along the sloped line to the origin.

That the second integral vanishes in the limit as $R \to \infty$ may be seen by noting that $\cos{(\beta+2 \theta)} \gt 0$ within the integration interval.  Therefore,

$$\int_0^{\infty} dx \, e^{-a e^{i \beta} x^2} = e^{-i \beta/2} \int_0^{\infty} dx \, e^{-a x^2} = \frac12 e^{-i \beta/2} \sqrt{\frac{\pi}{a}}$$

The stated answers come from taking real and imaginary parts of the above.


答:
Yes, more than two angles can have same $\tan$ (and $\cot$ which solely depends on $\tan$). For example, for $\tan\theta=1$, you have
$$\theta=\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4},\cdots$$
For any interval of length $2\pi$, there are two such angles. For $\tan\theta=1$ in $[0,2\pi]$, we have
$$\theta=\frac{\pi}{4}\textrm{ or }\frac{5\pi}{4}$$
For any interval of length $\pi$, you can find only one unique angle. For example, consider your equation. There is only one solution in $[0,\pi]$,
$$\alpha=\text{arccot}\,(-2)$$

Hope this helps. Ask anything if not clear :)

Title: Find $\sin\alpha,\cos\alpha,\tan\alpha$ if $\cot\alpha=-2$

Body:
Find $\sin\alpha,\cos\alpha,\tan\alpha$ if $\cot\alpha=-2.$

We have defined trigonometry with a circle, but only for angles between $0^\circ$ and $180^\circ.$

We have $\begin{cases}\cot\alpha=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}\iff\begin{cases}\dfrac{\cos\alpha}{\sin\alpha}=-2\\\sin^2\alpha+\cos^2\alpha=1\end{cases}.$

So I got that $\sin\alpha=\dfrac{\sqrt5}{5},\cos\alpha=-\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2$ or $\sin\alpha=-\dfrac{\sqrt5}{5},\cos\alpha=\dfrac{2\sqrt5}{5},\tan\alpha=-\dfrac{1}{2},\cot\alpha=-2.$

An angle with sine equal to $-\dfrac{\sqrt5}{5}$ isn't in the inverval $\left[0^\circ;180^\circ\right],$ right? I suppose that it is possible two different angles to have equal $\tan$ and $\cot.$


问:
How to solve this tetration equation analytically?
$ x^{x^{x^{x}}}=5 $

Here is what tetration is about

Note : numerically method isn't what I want. (Sorry for I didn't emphasize it before.)

Body:
$$f(x)=x^{x^{x^x}}-5$$
we find a zero with Newton's method
$$f'(x)=x^{x^{x^x}} \left(x^{x^x-1}+x^{x^x} \log x \left(x^{x-1}+x^x \log x (\log x+1)\right)\right)$$
We set $x_0=1.5$ and $x_{n+1}=x_n-\frac{f(x)}{f'(x)}$

I got the following results

$$
\begin{array}{r|r|r}
n & x_n & error \\
\hline
 0 & 1.5 &  \\
 1 & 1.83748 & 0.337477 \\
 2 & 1.79603 & 0.0414445 \\
 3 & 1.74805 & 0.0479864 \\
 4 & 1.70007 & 0.0479734 \\
 5 & 1.66681 & 0.0332666 \\
 6 & 1.65618 & 0.0106251 \\
 7 & 1.65539 & 0.000786085 \\
 8 & 1.6553947522 & 3.85\times 10^{-6} \\
 9 & 1.6553909024 & 9.16\times 10^{-11} \\
\end{array}
$$

Thus the solution with $9$ exact decimals is $x\approx 1.655390902$.


问:
Variance of ARMA process?
Problem

Given $ARMA(1,1)$ stationary process $$x_t = 0.7 x_{t-1} + u_t + 0.2 u_{t-1} $$ where $u_t$ is white noise, with standard deviation $\sigma(u_t) = 4$

Note, stationarity of $x_t$ implies that
$$ \Bbb E(x_t) = const = \mu$$
$$ \Bbb Cov(x_t, x_{t+k}) = \gamma_k$$

Find $ \Bbb Var(x_{t+1}|x_t, u_t)$

My ideas

We could substitute the expression of $x_t$ into the variance (I can do that), but I do not really understand what the fact that it is $ARMA(1,1)$ process gives us?

答:
Actually, the fact that it is $ARMA(1,1)$ process does not change the procedure much.
$$ Var( x_{t+1}| x_y, u_t) = Var(0.7x_t + u_{t+1} + 0.2u_t |x_t, u_t) = Var(u_{t+1}) = 16$$ as $ 0.7x_t$ and $ 0.2u_t$ are considered as constants so disappear from the variance and $u_{t+1}$ is independent of $x_t, u_t$ so it is usual non-conditional variance.

Title: Variance of ARMA process?


Title: How to solve this tetration equation analytically?

Body:
$ x^{x^{x^{x}}}=5 $

Here is what tetration is about

Note : numerically method isn't what I want. (Sorry for I didn't emphasize it before.)

Body:
If
$$x=\sin \left(\frac{8 \pi }{35}\right)+\cos \left(\frac{\pi }{44}\right)$$
$$x^{x^{x^x}} =4.9999978$$


问:
Maximum number of triangles with area 1
We have n dots on a plane such that no three of them are in the same line. Prove that the number of triangles with area 1 that are made by connecting three of these dots is less than or equal to $\frac{3}{2}(n^2-n)$

I got this far:

Every rectangle can make at most four triangles. So using induction, if we look for the maximum number of rectangles with area 2 that can be made by adding one dot to the plane, we can multiply that by four and find the maximum number added triangles and hopefully complete our proof. Finding rectangles must be easier. But I don't know how to find the maximum number of rectangles.

答:
Pick two points $A,B$. Then any point $C$ such that the area of $ABC$ (in either orientation) is $1$ must be one one of two parallel lines to $AB$. Hence each pair $(A,B)$ can be completeted to an area-1-triangle in at most four ways. This gives us $4\cdot {n\choose 2}$ candidate triangles - but each triangle is counted thrice (once per side). Thus the total number of such triangles is at most
$$\frac13\cdot 4\cdot {n\choose 2}=\frac23(n^2-n).$$

Title: Maximum number of triangles with area 1


问:
Union and Intersection, Problem with {a}
In some book (Deiser, Mengenlehre, 3. Aufl.) about set theory the author defines the general union and intersection:

$\cap$M = {x|for all a $\in$ M also x $\in$ a}

$\cup$M = {x|there exists some a $\in$ M with x $\in$ a}.

He then brings the example: $\cap${a} = $\cup${a} = a. If you just apply the definition it does not make sense to me. Do you see that by just applying the definition? What'd make sense is this: $\cap${{a}} = $\cup${{a}} = a. Am I right or do I make a mistake?

答:
Let's take a few examples.

Let's say $a=\{1,2,3\}$ and $b=\{4,5,6\}$. Then:

$\cup\{a,b\}=\{1,2,3,4,5,6\}$ (everything in either $a$ or $b$).

Now, let's look at $\cup\{a\}$. This is:

$\cup\{a\}=\{1,2,3\}$ (everything in $a$).

So, as you can see, $\cup\{a\}=a$ because they both have the same elements.


问:
Why is the volume of a sphere $\frac{4}{3}\pi r^3$?
I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?

答:
To give you a larger perspective, the volume of an $n-1$-sphere in $\mathbb{R}^n$ is given by $C_n r^n$, where $r$ is the radius and $C_n=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}$. For even $n$, this reduces to $C_n =\frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!}$, and for odd $n$, it becomes $C_n =2^{(n+1)/2}\frac{\pi^{(n-1)/2}}{n!!}$, where $n!!$ denotes the double factorial.

Thus in the case $n=3$, the constant is $2^2 \frac{\pi^1}{3!!} = \frac{4 \pi}{3}$.

The point of this is to show you that the general formula also involves factors in the denominator, and that the formula for $n=3$ is not 'random' but rather fits into a general pattern.

LastEditorDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Tensor product $\mathbb{Q} / \mathbb{Z} \otimes M$
Is it true that if $M$ is a torsion abelian group, then $\mathbb{Q} / \mathbb{Z}  \otimes_{\mathbb{Z}} M =0$? If it is, why?

答:
An element of $\mathbb Q/\mathbb Z \otimes_{\mathbb Z} M$ can be written as a sum of elements of the form $[r]\otimes m$, where $[r]$ represents the equivilence class containing $r\in \mathbb Q$ in the quotient module.  However, if $m$ is torsion, there exists an $n$ such that $nm=0$.  Then inside the tensor product, $$[r]\otimes m=[nr/n]\otimes m=[r/n]n\otimes m=[r/n]\otimes nm=[r/n]\otimes 0 = 0.$$

If every $m\in M$ is torsion, then every element of $\mathbb Q/\mathbb Z\otimes_{\mathbb Z}M$ is thus a sum of elements that are all equal to $0$.


问:
What does $\Bbb{R} \times \Bbb{R}$ denote in sets
Let set $A$ be : $A = \{(a, b) \in \Bbb{R} \times \Bbb{R} : |a – 5| < 1 \text{ and }|b – 5| < 1\}$

What exactly does $(a, b) \in \Bbb{R} \times \Bbb{R}$
mean here ?

答: $\mathbb{R}^2 = \mathbb{R}\times \mathbb{R} = \{(a,b)\in \mathbb{R}\times \mathbb{R}\} = \{(a,b), a\in \mathbb{R}, b\in \mathbb{R}\}$


问:
Prove $f(x)=x^3$ is not uniformly continuous in $\mathbb{R}$
I wanted to post here my solution and check if I'm right.

Also, I wanted to know how can you prove it with "Formal" way with Epsilon-Delta (I saw some posts but most people jump immediately to a solution).

My try :

I will take the sequences $x_n=n$ and $y_n=n+\frac {1}{n}$

$|x_n-y_n|=|n-n-\frac {1}{n}|=\frac {1}{n} \rightarrow 0$ (when $n \rightarrow \infty$)

but $|f(x_n)-f(y_n)|=|n^3-(n+\frac{1}{n})^3|=\cdots=3n+\frac {3}{n} + \frac {1}{n^3} \rightarrow \infty$ (when $n\rightarrow \infty$).

which shows that the function is not uniformly continuous over $\mathbb{R}$.

答:
Yes, that's fine.

If you want's a $\varepsilon-\delta$ approach, it's almost the same thing. You want to prove that $f$ is not uniformly continuous, which means that$$(\exists\varepsilon>0)(\forall\delta>0)(\exists x,y\in\Bbb R):|x-y|<\delta\wedge\bigl|f(x)-f(y)\bigr|\geqslant1.$$Take $\varepsilon=1$. Now, if $\delta>0$, you take $n\in\Bbb N$ such that $\frac1n<\delta$. Now you take $x=n$ and $y=n+\frac1n$ and your computations show that, indeed,$$\bigl|f(x)-f(y)\bigr|\geqslant1.$$


问:
Show that matrix is positive definite
Let
$A=\begin{pmatrix} d_1 & e_2 & \ldots & 0 & 0 & 0\\ e_2 & d_2 & e_3 & \ldots & 0 &0\\ 0 & e_3 &d_3 & e_4 &\ldots &0 \\ \ldots & \ldots & \ldots & \ldots & \ldots &e_n\\ 0 & \ldots &\ldots & 0 & e_n & d_n\end{pmatrix}$.

The following are true:

$d_i > 0 $ for $1 \leq i \leq n$

$d_1 > \left|e_2\right|$

$d_i > \left|e_i\right| + \left|e_{i+1}\right|$ for $2 \leq i \leq n-1$

$d_n > \left|e_n\right|$

Show that the matrix A is positive definite.

Firstly, I can show that A has strict diagonal dominance and therefore is invertible. I have found some theorems that use invertibility to prove positive definite but none suited my case I think.

I have also found the Sylvester's criterion which states that if all principal minors of A are nonnegative then A is positive definite. Well, the first two principal minors are indeed nonnegative, but after that I can't prove it.

Which should my approach be?

答: Hint: The Gershgorin circle theorem implies all of the eigenvalues of $A$ are positive.

Title: Show that matrix is positive definite


问:
Create a graph network flow model for the Computer Science Department
I need to create a network flow model for building (if possible) the new organizing schema for the following problem:

At the Computer Science Department at the beginning of the first semester there are p freshmen
(study) groups: group i contains n(i) students, for all i = 1, p. For the second semester the Department
wants to reorganize these groups in such a way that:

->the new organizing schema has r groups;

->the new group j contains m(j) students, for any j = 1, r;

->any new group cannot contain more than c students which were classmates in a same old group
from the former organizing schema (c ∈ NN∗ \ {1}).

-

My idea would be to use a network that contains a bipartite graph but don't know how to organise students and grups inside it and how to describe this bipartition.

答: Define a complete bipartite graph with $p$ supply nodes and $r$ demand nodes.  The flow from node $i$ to node $j$ indicates the number of students that move from old group $i$ to new group $j$. Each flow variable has an upper bound of $c$.


问:
Non splitting sequence with free module
I am looking for a non splitting short exact sequence of modules $0 \rightarrow M_{1} \rightarrow M_{2} \rightarrow M_{3} \rightarrow 0$ where only $M_{1}$ (and only $M_{1}$) is free. Is it possible?

Also, the same question for $M_{2}$ (and only $M_{2}$) being free. Is it possible?

答:
It seems your restrictions are here to avoid things like the usual $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}\to 0$, but you can just take the variant
$$0\to \mathbb{Z}\to \mathbb{Z}\oplus A\to \mathbb{Z}/n\mathbb{Z}\oplus A\to 0$$
for any $A$ to answer your first question.

For the second one, you just need to take a non-principal ring. So for instance
$$0\to (n,X)\mathbb{Z}[X]\to \mathbb{Z}[X] \to \mathbb{Z}/n\mathbb{Z}\to 0$$
where the base ring is $\mathbb{Z}[X]$.

Title: Non splitting sequence with free module


问:
Exactness on $\sf{QCoh}_X$ versus on $\mathscr{O}_X$-$\sf{Mod}$.
Let $X$ be a scheme and $\sf{QCoh}_X$, $\sf{Coh}_X$, $\mathscr{O}_X$-$\sf{Mod}$ be the categories of quasi-coherent sheaves, coherent sheaves and $\mathscr{O}_X$-modules over $X$, respectively. All these categories are Abelian, which makes me wonder the following:

If $\varphi:\mathscr{F}\to\mathscr{G}$ is a morphism of coherent sheaves, the kernel and the of $\varphi$ in those 3 categories coincide? In other words, a sequence of coherent sheaves is exact in $\sf{Coh}_X$ iff it is in $\sf{QCoh}_X$ and iff it is in $\mathscr{O}_X$-$\sf{Mod}$?

答:
Let $\sf{B}$ be an Abelian category and $\sf{A} \subset \sf{B}$ be an additive subcategory in the sense that $\hom_\sf{A}(M,N) \cong \hom_\sf{B}(M,N)$ as Abelian groups. Let $\phi:M \to N$ be a morphism in $\sf{B}$ and consider its kernel $K$, that is, we have an exact sequence
$$
0 \to K \to M \to N
$$
Suppose now that $M, N$ and $K$ lie in the (essential) image of $\sf{A}$. Then $K$ is the kernel of $M \to N$ seen as a morphism in $\sf{A}$:

By our assumption the composite $K \to M \to N$ is still $0$, and given any map $T \to M$ in $\sf{A}$ with $T \to M \to N$ equal to the zero map then we can find a unique lift $T \to K$ in $\sf{B}$, but since $\sf{A}$ is full, this is a unique lift in $\sf{A}$ as well.

By using a dual argument we conclude that the same is true for cokernels.

The upshot then is that if we have a full subcategory of an Abelian category that is closed under Kernel and cokernel (at least in the essential image) then the inclusion functor is exact, that is, this is an Abelian subcategory.

Now we just note that this is true for quasi-coherent and for coherent modules.


问:
Why is this identity involving vector valued functions true?
I am reading about stationary harmonic maps and I came across the following calculation.

Let $\mathcal{Q}_t(x) = x+ t\zeta(x),$ where $\zeta\in C^{\infty}_c(B_{\rho}(x_0), \mathbb{R}^n)$ where $x_0\in \mathbb{R}^m$ and $\rho>0.$ We choose $|t|$ small enough such that $\mathcal{Q}_t$ is a diffeomorphism of $B_{\rho}(x_0)$ onto itself. Then I don't understand why the following identities hold.

\begin{align}
D \mathcal{Q}_{t}^{-1}\left(\mathcal{Q}_{t}(x)\right) &= I-t D \zeta(x)+O\left(t^{2}\right) \\
\operatorname{det} D \mathcal{Q}_{t}(x) &=1+t \operatorname{div} \zeta(x)+O\left(t^{2}\right) \end{align}

Any hints/references suggesting how to prove the above two identities will be much appreciated.

答:
For the first identity, by the inverse function theorem we have that, $DQ_t^{-1}(Q_t(x)) = (DQ_t)^{-1}(x)=I -tD\zeta(x) +O(t^2).$

For the second identity, we can use the fact, $\det(I+tB)=I+t\operatorname{Tr}(B)+O(t^2)$ and $DQ_t = I+tD\zeta$ where $I$ is the Identity matrix. Then we simply replace the matrix $B$ with the Jacobian matrix $D\zeta$, which gives us the second expression.


问:
Finding Constrained Subsets of Parameters in Larger Poorly-Constrained System of Linear Equations
I have a system of linear equations.  The system is not well-constrained (I have more parameters than independent equations).  What is the easiest way to identify the subset of parameters that are well-constrained?

For a simple example, if I have the system:

$5a + 6b + 7c + 8d= 100$

$9a - 18b +3c - 4d = -50$

$6d + 5e = -25$

$3d - 4e = 62$

$d$ and $e$ are well-constrained but $a$, $b$ and $c$ are not.  How can I take a system like this of arbitrary size and figure out which parameters in that system are well-constrained?

答:
Honestly, I would suggest that you are taking a somewhat limited definition of "well-constrained" here. So first let's attempt to address that definition specifically, and then suggest a better way of looking at it.

To identify individual well-constrained variables, you can construct the kernel or null space of the coefficient matrix. Any variable that does not appear in that basis is well-constrained, according to your definition. For instance, in the example above, the coefficient matrix is
$$A\triangleq \begin{bmatrix} 5 & 6 & 7 & 8 & 0 \\ 9 & -18 & 3 & -4 & 0 \\ 0&0&0& 6 & 5 \\ 0&0&0& 3 & -4 \end{bmatrix}$$
The rank of this matrix is 4, so the null space is spanned by exactly one vector:
$$Z=\begin{bmatrix}3 \\ 1 \\ -3 \\ 0 \\ 0 \end{bmatrix}$$
Since the fourth and fifth elements of $Z$ are zero, those variables are well-constrained.

The kernel may require more than one vector to describe its basis---or if these basis vectors are assembled into a single matrix $Z$, that matrix may have more than one column. In that case, the well-constrained variables are those corresponding to an entire row of zeros in $Z$.

As I said above, the rank of $A$ is 4. That means the linear system $Ax=b$ effectively constrains four directions of the solution vector $x$. And yet, only $2$ of the variables are well-constrained, according to your definition. Note the distinction: $4$ directions are constrained, but only $2$ variables are. Is there a way to bring these two definitions together; that is, find $4$ variables that are "well-constrained" in some sense?

Yes, there is. To do so, we compute the reduced row echelon form of $A$. The RREF is computed by performing Gaussian elimination on the rows of the matrix, and doing some additional elimination to reach a unique canonical form; see the Wikipedia link I've provided. The RREF of your example is
$$R =\begin{bmatrix}
1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1/3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1
\end{bmatrix}$$
Because this matrix is formed via Gaussian elimination on the rows, we have $A=BR$ for some nonsingular square matrix $B$, but we do not need to know $B$ to proceed.

Looking at this matrix $R$, you will see that columns 1, 2, 4, and 5 are the columns of an identity matrix. We call the variables corresponding to those columns---$a$, $b$, $d$, and $e$---as the bound variables, and they correspond in an important sense to your "well-constrained" variables. What this means is that once you have selected a value for $c$, all four of the other variables are uniquely determined. That's not quite the same as your definition of "well-constrained", but it's a more standard way of looking at the problem.

Note that the choice of "bound" variables depends on the ordering of the columns. So for instance, if we swapped the columns for $c$ and $a$, the RREF would identify $b$, $c$, $d$, and $e$ as bound.

It's not difficult to recover your original definition of "well-constrained" from the RREF. Note that rows 3 and 4 have only one nonzero each in columns 4 and 5, respectively. From this we can see that $d$ and $e$ can be determined even without selecting a value for $c$. They are indeed well-constrained.


问:
Finding a system of equations of which solution is a span
The question is how to find a system of equations when I have the solution. The solution is subspace: $\operatorname{span}\{(2,4,-6), (4,8,-10)\}$.

I can see they describe a plane, however, the equation that describes this plane omits the last variable and looks like this: $$8x - 4y = 0.$$

Doesn't this mean that the solution to this is from $\mathbb{R}^2$ and not $\mathbb{R}^3$? So it would look like this $\{(1,2)\}$?
It is just a technical question if it's okay to just add a dimension.

It seems to be a bit odd. Could I add to the original solution a zero for example? $\operatorname{span}\{(2,4,-6,0)(4,8,-10,0)\}.$
Sorry for the English.

答:
We have a plane. A plane is described by one equation in three unknowns.

$(2,4,-6)\times(2,4,-10)=\begin{vmatrix}i&j&k\\2&4&-6\\2&4&-10\end{vmatrix}=-16i+8j$ to get $-16x+8y=0$, for the equation (up to scalar multiple).


答:
To prove $\bigcup\{a\}=a$ we need to prove (i) $\bigcup\{a\}\subseteq a$; and  (ii) $a\subseteq \bigcup\{a\}$.

For (i) let $x\in\bigcup\{a\}$. Then for some $p\in\{a\}$ we have $x\in p$. But there is only one such $p$, namely $a$. So $x\in a$ and we are done.

For (ii) let $x\in a$. Then $x\in a\in \{a\}$, so by definition $x\in\bigcup\{a\}$.

[I have assumed that we have proved that $\{a\}$ (is the definition in your book $\{a\}=\{a,a\}$, where $\{a,b\}$ is already defined?) has precisely one element $a$. ]

Title: Union and Intersection, Problem with {a}

Body:
In some book (Deiser, Mengenlehre, 3. Aufl.) about set theory the author defines the general union and intersection:

$\cap$M = {x|for all a $\in$ M also x $\in$ a}

$\cup$M = {x|there exists some a $\in$ M with x $\in$ a}.

He then brings the example: $\cap${a} = $\cup${a} = a. If you just apply the definition it does not make sense to me. Do you see that by just applying the definition? What'd make sense is this: $\cap${{a}} = $\cup${{a}} = a. Am I right or do I make a mistake?


问:
Is there a difference in meaning between the notations here?
I was going over notes today on properties and proofs of random variables, in particular the proof that $E(cX_{n}) = \lim_{n \to \infty}E(cX_{n})$.

In the definition, it states that $X$ is a non-negative random-variable and this is denoted as $0 \leq X$. Also that $c$ is an element of the positive reals. This is denoted as $c \in \mathbb{R}^{+}$.

I was curious as to why there is a difference in notation, if both $c$ and $X$ are positive, then why are they written differently? And hence is there a difference in meaning between the two or is just down to personal choice?

答:
The expression "$X \ge 0$" does not mean "$X$ is positive". Instead, it means that every value of $X$ is either positive or zero.

If you really wanted notation for "$X$ is positive", then you could write $X > 0$, and it is indeed your own preference.

Keep in mind that there is a subtle distinction between $c>0$ and $X>0$. First, $c$ denotes a constant, having only one value, and so the expression $c>0$ means that this one value is positive. On the other hand $X$ is a random variable, having many possibly different values each time you sample it, and the expression $X > 0$ means that each time you sample it the value will be positive.

In fact there usually are multiple ways to express the same mathematical thought, just as there are usually multiple ways to express the same thought in any branch of human discourse, and the choice of expression is in general, as you say, a matter of personal preference, tempered by context, or clarity, or a number of other factors.