问:
Properties of a specific type of random network
Suppose I have a directed network constructed in the following way:  There are $n$ vertices. Each vertex has $e$ out-edges, which point to other vertices chosen uniformly at random.  (If it matters, I'm generally working with smaller $e$.)

First question, does this type of graph have a name?  Keywords for searching would be helpful.

Specific questions:

Suppose I choose a subset $T$ of vertices of size $s$.  What is the probability that there is a path from every node outside of $T$ to at least one node in $T$?

Suppose two non-overlapping subsets of vertices both of size $s$.  What is the probability that a single node has a path to one but not the other?

It's easy enough to solve numerically, which I can do on my own.  But I was wondering if this was already studied and could be determined exactly.

Thanks for your help!

答:
The case $k=1$ is very special: we are looking at a random function $f : \{1, 2, \dots, n\} \to \{1,2,\dots,n\}$, and lots of the dynamics change. It's much likelier that there will be multiple large components.

For $k\ge 2$, and asymptotically as $n \to \infty$, the first probability already goes to $0$ for some $s = O(\log n)$. I've given some bounds that go to $0$ below, not very quickly. We can get better bounds of the form $\frac1{n^C}$ for any constant $C$ by doing things more carefully, and varying the constant in $O(\log n)$. So certainly when $s$ is a constant fraction of $n$, the probability goes to $0$.

This tells us that actually, every vertex reaches almost all others, so for your second question, the probability is essentially the same as the probability a vertex $v$ cannot reach a set of size $s$. There's more to be said on what the probability is for a very small $s$; I can expand on that if you're interested.

I'll begin with a slightly different question: if there are no edges from a set $S$ to $V-S$, how large can $S$ be? Clearly, $|S| \ge k+1$, since each vertex in $S$ has edges to at least $k$ more.

When $|S|=r$, there are $\binom nr \le (\frac{ne}{r})^r$ ways to choose $S$. For a vertex $v \in S$, the first edge we choose out of $v$ has a $\frac{r-1}{n-1} < \frac rn$ chance of landing in $S$; after that, the probability decreases (because we can't pick the same edge twice) so this is an upper bound for all edges. So the probability is at most $(\frac rn)^{kr}$ that all edges out of vertices in $S$ go back to $S$. The expected number of such sets of size $r$ is at most $(\frac{ne}{r})^r (\frac rn)^{kr} = (e(\frac rn)^{k-1})^r$.

Let's look at this for various ranges of $r$:

For $r = o(n)$, say for concreteness $r \le \sqrt n$, we get $e(\frac rn)^{k-1} < \frac{e}{\sqrt n}$. The sum over all such $r$ is bounded by $\sum_{r=1}^\infty (\frac{e}{\sqrt n})^r = \frac{e/\sqrt n}{1 - e/\sqrt n} = (1+o(1))\frac e{\sqrt n}$, which goes to $0$ as $n \to \infty$. We could actually give sharper bounds, since the dominant terms are the first few, and $\frac rn$ is much smaller for those. Not to mention that we've just used $k \ge 2$ here instead of the actual value of $k$.

For $\sqrt n < r < 0.1n$, say, we get $(e(\frac rn)^{k-1})^r < (0.1e)^r < (0.1e)^{\sqrt n}$, so the sum over all such $r$ is at most $0.1n (0.1e)^{\sqrt n}$, which also goes to $0$ as $n \to \infty$. Actually, the constant $0.1$ can be anything less than $e^{-1/(k-1)}$.

So now suppose we're looking at a set $T$ of $s$ vertices in your problem. We know all vertices outside $T$ expand to reach at least $0.1n$ vertices with high probability. The probability such a vertex has no path to $T$ is less than $(1 - \frac sn)^{0.1kn}$. Using the inequality $1+x \le e^x$, this is at most $e^{-0.1ks}$. Just taking $s = 10 \log n$, we already get a bound of $\frac1{n^2}$ for all $k\ge2$, which still is less than $\frac1n$ (and therefore going to $0$) if we take the very weak union bound over all vertices $v$ outside $T$.

On the other hand, for very small $s$, there's a good chance that there are no edges to $T$, which is the easiest way for the event we want to happen. Specifically, that probability is $(1+o(1))(1 - \frac sn)^{kn} \sim e^{-ks}$. That's not the whole story: the set not reachable from anywhere else could be a bit bigger than $T$. But it says something about the size of the set we're looking at, if we want the probability to be non-negligible.


问:
Smallest possible area for triangle
I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:

Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$

Let's find the tangent of the ellipse:

$2x^2 + y^2 = 1$

$4x + 2y \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{-2x}{y}$

Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:

$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$

So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$

Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.

So now area is $1/2 a.b = 1/2 a^2$

Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.

I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.

答:
For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,  the equation of the tangent line at $(a\cos t, b \sin t)$ is given by
$$\frac{\cos t}ax+\frac{\sin t}by=1$$
which intersects the axes at $\frac a{\cos t}$ and $\frac b{\sin t}$, respectively. Then, the area of the triangle is
$$Area = \frac12 \frac a{\cos t}\frac b{\sin t}=\frac{ab }{\sin2t}\ge ab
$$
Here, $a=\frac1{\sqrt2}$ and $b=1$, allowing the smallest area $\frac1{\sqrt2}$.

ContentLicense: CC BY-SA 4.0

Title: Smallest possible area for triangle


问:
Find a circle orthogonal to two other circles
Given two circles

$x^2+(y-4)^2=4^2$

and

$(x-8)^2+(y+2)^2=2^2$

I need to find the circle that's orthogonal to both the above circles, and contains the point $P=(8,4).$

Is there an algebraic way to find this circle? Or is a geometric way easier?

Any advice will be greatly appreciated.

Thanks in advance!

答:
Given $S_1=x^2+(y^2 -8y + 16)-4^2=0$ and S_2=
$(x-8)^2+(y+2)^2-2^2=0$

Alternate to Apurv's method, after getting condition for orthogonality as:
$-8f-c=0$ and $16g-4f+c=-64$

We can find the radical axis of two circles by substracting their two equation:

$$ y =\frac{4}{3}x-  \frac{6}{3}$$

Now it must be that $(-g,-f)$ solves this equation and by this we have three line equations which solves to:

$$ g= \frac{4}{3} f +  6$$

Solving the following linear system:

$$-8f-c=0 $$
$$ 16g-4f+c=-64$$ $$16g+8f+c=-80$$

We get, $ c= \frac{32}{3}$, $ f= \frac{-4}{3}$ , $g=-5$

ContentLicense: CC BY-SA 3.0


问:
Who was responsible for finding sufficient conditions for functional extrema?
In the calculus of variations, there is a well-known sufficient condition for weak functional extrema, involving conjugate points and the strengthened Legendre condition ($f_{y'y'} > 0$). Who was responsible for this discovery? Charles Fox (An Introduction to the Calculus of Variations, p. 42) credits Jacobi (1837), whereas Morris Kline (Mathematical Thought From Ancient to Modern Times, Vol. 2, p. 748) credits Weierstrass (1879). This is no small discrepancy of forty-two years. Can anyone give an authoritative answer on who is right? References would be good.

答:
Jacobi thought that the Euler condition, strengthened Legendre condition, and absence of conjugate points on $(a,b]$ were together sufficient conditions for $y$ to be a minimum of the integral [1,2]
$$
I = \int_a^b F(y,y^\prime,x) dx.
$$
However, Weierstrass proved that this is not true, that the given conditions are sufficient only for weak variations, and that the given three together with a fourth gave sufficient conditions also for strong variations [1,2,3].

So, who you want to credit the theorem to is somewhat up to you. Jacobi came first, but stated the theorem without a complete proof and without the concept of weak VS strong variations. Weierstrass rigorously proved both cases.

The above story is actually laid out pretty well in the book by Kline that you referenced:
"Jacobi also concluded that an extremal (a solution of Euler's equation) taken between A and B for which $f_{y^\prime y^\prime} > 0$ along the curve and for which no conjugate point exists between $A$ and $B$ (or at $B$) furnishes a minimum for the original integral... Actually, these sufficient conditions were not correct... Weierstrass did prove in 1879 that for weak variations the three conditions ... are indeed sufficient conditions" [1] (emphasis on weak variations added by me).

Here are some additional references corroborating the story:

"Jacobi's condition was originally erroneously stated as a sufficient condition, and was shown by Weierstrass to be insufficient for strong minima." [2] (See pp. 18.)

"In his earlier work, prior to the year 1879, he [Weierstrass]... proved for the first time that the condition $\delta^2 I > 0$ is sufficient for the existence of a weak minimum." [3]

Note also that any of the references crediting Jacobi with the theorem for the weak minimum don't provide these historical details [4, 5]. It seems to me that some authors simply credit the correct part of Jacobi's statement to him, whereas other authors give the credit mainly to Weierstrass, who corrected and extended Jacobi's statement, and gave rigorous proof.

[1] M. Kline. Mathematical Thought From Ancient to Modern Times, Vol. 2, p. 747--748.

[2] E. R. Hedrick. On the Sufficient Conditions in the Calculus of Variations. Bull. Amer. Math. Soc. 9, 11-24 (1902).

[3] M. Giaquinta and S. Hildebrandt. Calculus of Variations I: The Lagrangian Formalism. Springer-Verlag Berlin (2004). ISBN 3-540-50625-X. pp. 261.

[4] C. Fox. An Introduction to the Calculus of Variations, p. 42.

[5] Sufficiency and the Jacobi Condition in the Calculus of Variations. F. H. Clarke and V. Zeidan. Can. J. Math. 38 (5), 1199-1209 (1986).


问:
Find $ax^2+3x+b$ that goes through the given points using the least squares
The points are $ (2,1.5),(−1,1.7),(1,1.9)$.
I did:

$A = \left [ \begin{matrix}
    1 & 2 & 4 \\
    1 & -1 & 1 \\
    1 & 1  & 1
   \end{matrix} \right ] \\
x = \left [ \begin{matrix}
    b   \\
    3   \\
    a
   \end{matrix} \right ]\\
b = \left [ \begin{matrix}
    1.5 \\
    1.7 \\
    1.9
   \end{matrix} \right ]$

I solved for $$A^TAx = A^Tb$$

I got the Cholesky decomposition of $A^TA$

Then I solved $Ly = A^Tb$ and got $$
\left [ \begin{matrix}
    51/10 \\
    16/5 \\
    48/5
   \end{matrix} \right ]$$$

The next part is where I am having a problem. Now I have to solve $L^tx = y$. I already have some values for x, so I multiplied $L^tx$ and made it equal to y and got a system of equations

How do I solve this?

答:
$3$ is not an unknown variable, so your solution is incorrect.
We have
$$
\left\{\begin{array}{rrrrr}
2^2\cdot a &+3\cdot 2&+b&= &1.5 \\
 (-1)^2\cdot a&+3\cdot(-1)&+b&= &1.7 \\
 1^2\cdot a&+3\cdot 1&+b&= &1.9 \\
\end{array}\right.
$$
or
$$
\left\{\begin{array}{rrrrr}
4a&+&b&= &-4.5 \\
a&+&b&= &4.7 \\
a&+&b&= &-1.1 \\
\end{array}\right.
$$
or
$$
Ay=d,
$$
where
$$
A=\left(\begin{array}{ll}
4 & 1\\
1 & 1\\
1 & 1\\
\end{array}\right),\quad
d=\left(\begin{array}{r}
-4.5 \\
4.7 \\
1.9 \\
\end{array}\right),\quad
y=\left(\begin{array}{r}
a \\
b \\
\end{array}\right).
$$
This system has no solutions, thus, there is no curve of the form $ax^2+3x+b$ which goes exactly through the given points. It is only possible to find a least squares solution. This solution won't make a lot of practical sense since $a+b$ can't be equal to $4.7$ and $-1.1$ at the same time even approximately.
$$
y_{ls}=A^{+}d=(A^TA)^{-1}A^Td=\left(\begin{array}{r}
-2.1 \\
3.9 \\
\end{array}\right).
$$
Hence, the answer is $-2.1x^2+3x+3.9$.


问:
More understandable explanation for Cat-Rabbit-Dog quiz
This is a quiz from Brilliant.org Algebra Fundamentals courses.

Quiz:

At Step 40, how many cats will there be in the 7th row?

Answer: 14

Explanation given by Brilliant.org is not clear to me so if someone can give a better explanation it will be helpful.

Explanation by Brilliant.org

In rows where dividing by 3 has a remainder of 1, the pattern is
cat-rabbit-dog.

7 when divided by 3 has a remainder of 1, so cats occur at 1, 4, 7,
etc., i.e. at points where dividing by 3 has a remainder of 1.

In step 40, the grid is 40 columns wide. The full cat-rabbit-dog
pattern happens for $39\div 3=13$ times, but since the pattern
at row 7 starts with a cat, column 40 has one extra cat. So there are
$13 + 1 = 14$ cats.

Note: A more compact way of saying "dividing $x$ by 3 has a remainder
of 1" is $x \mod 3 = 1$.

答:
It strikes me that this is a case where talking about division and remainders clouds the issue. I would explain things as follows:

Every third row is identical, so the seventh row is the same as the fourth is the same as the first.

At the fortieth step, the first row will have $13$ groups of three (cat, rabbit, dog) and one extra cat, for a total of $14$ cats. So the seventh row will also have $14$ cats.


问:
How should one think to get the radius of the resulting curve?
For example, the curve C is given as the intersection between
$$
C: x²+y²+z²=1, x+y+z=0
$$
Radius: 1

Another one:
$$
C: x²+y²+z²=1, x+y=1
$$

Radius:

$$
\frac{1}{\sqrt{2}}
$$

How should I think to get these? I know they're ellipses, but substituting them into each other yields "hard" expressions.

答:
The two examples given above are circles. Unless the coefficients of $x^2,y^2,z^2$ are not identical, the intersection would remain as circles.

The first equation gives a unit ball in $\mathbb{R}^3$, the second equation gives plane.
I would tackle this problem by finding the distance from the origin $(0,0,0)$ to the plane. Then by simple geometry computation you can find the radius of the circle.

Notice the equation of plane is given by
$$\boldsymbol{n\cdot(\boldsymbol{r}-\boldsymbol{r}_0)} = \boldsymbol{0},$$
why is it? The argument is given here.

By picking a convenient $\boldsymbol{r}_0$, you can find $\boldsymbol{n}$, along this direction, you can find the distance between the origin and the plane.

Now you know the radius of the unit ball and you know the distance to the cutting plane, by Pythagorean you can find the radius of the circle.


问:
Calculate Integral $\iiint_{K}\frac{1}{(z+1)}\,dx \,dy \,dz $
I need to calculate this integral

$$\iiint_{K}\frac{1}{(z+1)}\,dx \,dy \,dz $$

with $K=x^{2}\le y\le z\le x$

I have trouble visualizing $K$ so I am unable to find the extremities of the integrals

Any help would be much appreciated

答: You have $x^2\leqslant x$ when $x\in[0,1]$ and only on that interval. For each $x\in[0,1]$, $y$ can take any value from $x^2$ to $x$. And, for each such $y$, $z$ can take any value from $y$ to $x$. Therefore, you are after the integral$$\int_0^1\int_{x^2}^x\int_y^x\frac1{z+1}\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$Can you take it from here?


问:
Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$

If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$

To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?

I tried this way as an alternative ansewr:

Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$

$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.

Body:
I'll expand on my comments above.

You want to find a cubic polynomial satisfied by $y=x+x^{-1}$ whenever $x$ is a root of $P$.  Note that $x^3+3x+5=0$ gives $5x^{-1}=-(x^2+3)$ and $5x^{-3}=-(3x^{-2}+1)$.  Therefore
\begin{align*}
-5x^{-1}&=x^2+3\\
25x^{-2}&=(x^2+3)^2=x^4+6x^2+9\\
&=x(-3x-5)+6x^2+9\\
&=3x^2-5x+9\\
-125x^{-3} & = 25(3x^{-2}+1)\\
&=9x^2-15x+52.
\end{align*}
Thus
\begin{align*}
5y &=5x+5x^{-1}=-x^2+5x-3\\
25 (y^2-2) &=25 x^2 +25 x^{-2} =28x^2-5x+9\\
125 (y^3-3y) &=125 x^3 + 125 x^{-3} =-9x^2-360x-677
\end{align*}
Eliminating $x^2$ gives
\begin{align*}
25 (y^2-2)+140y &=135x-75\\
125 (y^3-3y)-45y &=-405x-650
\end{align*}
or equivalently,
\begin{align*}
5y^2+28y &=27x-5\\
25y^3-84y &=-81x-130
\end{align*}
Now eliminate $x$ and we have
\begin{align*}
-145=-81x-130+3(27x-5)&=(25y^3-84y)+3(5y^2+28y)\\
&=25y^3+15y^2
\end{align*}
i.e.
$$
5y^3+3y^2+29=0
$$
so the product of roots $y$ is $-29/5$.

Remark: This is essentially the "standard method" for finding a polynomial equation for $y_i=q(x_i)$ where $q$ is a polynomial and $x_i$ are the roots of some polynomial $P$.


问:
Deriving integer solutions to quadratic equation without solving quadratic congruence
I want to generate positive integer solutions of $x$ to the equation:

$x^2-x-aT=0$

where a is an integer $>$ 0 and T is a very large positive number.

I noticed that when plugging this into wolfram alpha is looks like it was formulating the equation in the form:

$x^2-x≡0 (mod T)$

And then solving this quadratic congruence to get the equations to generate all integer solutions to my equation.

I'm not keen on using this method to solve as in order for it to work it relies on first factorising T. If T is really large this will cause the solution to run in linear time (with respect to the number of digits of T) which doesn't scale to the size I want.

So what I am asking is:

Is there a way to generate integer solutions to my initial equation in logarithmic time without solving the quadratic congruence equation I showed above and without factorising T?

答:
Erm... sorry, but this is just a quadratic equation, and the positive solution is
$$x=\frac12+\sqrt{\frac14+T}.$$ I think it's fairly easy to check if this is an integer, even for very large $T$


问:
Graph decomposition into cycles + edges consistent with a linear order over the vertices
I'm interested in decomposing a graph $G$ into a set $\{H_1, \dots, H_n\}$ of cycles and edges of $G$, with an added twist. By decomposing $G$ I mean the usual notion that the $H_i$'s form a partition of $G$, i.e., $\bigcup_{i=1}^n H_i = G$ and $H_i\cap H_j = \emptyset$. For clarity, I identify the $H_i$'s and $G$ with the sets of their edges.

The twist is that I want to ensure that there exists a linear order $L$ over the vertices of $G$ that satisfies each subgraph $H_i$ in the decomposition. What I mean by this is the following (note that I write $xy$ for the edge $(x,y)$):

if $H_i$ consists of a single edge $xy$, then $L$ satisfies $H_i$ if $x$ appears before $y$ in $L$, i.e., it holds that $L = \dots x\dots y\dots$;

if $H_i$ is an $x_1\dots x_m$-cycle in $G$, i.e., $H_i = \{x_1x_2, \dots, x_{m-1}x_m, x_mx_1\}$, then $L$ satisfies $H_i$ if $L$ contains exactly $m-1$ of the comparisons in $H_i$.

To give a quick example, for the set of vertices $\{a,b,c,d,e\}$, the order $L = bcdea$ satisfies the subgraph $H = \{bd\}$, corresponding to an edge from $b$ to $d$, because $L$ orders $b$ and $d$ in the order specified by $H$; and $L$ also satisfies the $abc$-cycle $H' = \{ab, bc, ca\}$ on $a$, $b$, $c$, because it contains the comparisons $bc$ and $ca$, i.e., $L$ breaks the cycle in the most 'efficient' way, by sacrificing only one edge of $H'$.

Now, as an example of what I'm interested in, say we have the graph $G$:

One way to decompose $G$ into cycles and edges is to start with the $abc$-cycle $\{ab, bc, ca\}$ on $a$, $b$ and $c$, and add the remaining edge subgraphs $\{bd\}$, $\{da\}$, $\{ae\}$ and $\{ec\}$. However, this decomposition does not satisfy the extra consistency requirement I mentioned above: any linear order consistent with the edge-subgraphs would have to contain comparisons $bd$, $da$, $ae$ and $ec$ and, by virtue of transitivity, also $ba$ and $ac$, the only such linear order being $L = bdaec$. However, $L$ does not satisfy the $abc$-cycle since it flips two of its edges, i.e., $ab$ and $ac$.

On the other hand the decomposition of $G$ into the $abd$- and $aec$-cycle, plus the edge $bc$, works just fine: $L = abdec$. for instance, satisfies all the components of the decomposition.

My main interest at this point is in knowing whether such a decomposition exists for any graph $G$. Some intuitive procedures I've tried for the last couple of days, that I thought should work, have turned out not to. So at this point any hints to understanding the problem a bit better would be appreciated.

答:
In order for this decomposition to exist, the linear order $L$ must be compatible with the direction of at least $\frac 23$ of the edges in the graph. (It's compatible with every edge that is its own $H_i$, and with $\frac{m-1}{m} \ge \frac23$ of the edges in each $m$-cycle.)

Many graphs are not going to have such a linear order. For example, take a random tournament on $n$ vertices: for each pair of vertices $\{x,y\}$, add edge $xy$ or $yx$ with equal probability. We can show that with a very high probability, especially as $n \to \infty$, no such linear order exists.

For a fixed linear order $L$, the number of edges of the random tournament compatible with $L$ is a random variable $\mathbf X_L \sim \text{Binomial}(\binom n2, \frac12)$. By a Chernoff bound, taking $\mu = \frac12 \binom n2$ and $\delta = \frac13$,
$$
   \Pr[\mathbf X_L >\tfrac23 \tbinom n2] < e^{-\delta^2 \mu/(2+\delta)} = e^{-\binom n2/42}.
$$
There are, of course, $n!$ possible linear orders $L$. But the probability that $\mathbf X_L$ exceeds $\frac23 \binom n2$ for any of them is still at most $n! \cdot e^{-\binom n2/42}$, which is still tiny: $n! < n^n = e^{n \log n}$, and $n \log n$ loses to $\binom n2/42$.

In general, we're going to get a similar result for a random directed graph with $m$ edges: $\Pr[\mathbf X_L > \tfrac23 m] < e^{-m/42}$. So almost all graphs with more than $\Omega(n \log n)$ edges will not work.


问:
Example of complete formula on constants.
I have question considering example of complete formula, that was presented in Keisler/Chang - Model Theory:

2.3.1. EXAMPLES

(1). Let $T$ be a complete theory and let $c_0,c_1,c_2,\ldots$ be constant symbols of $\mathscr{L}$. Then any formula of $\mathscr{L}$ of the form
$$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$
is complete in $T$. If $\mathfrak{U}$ is a model of $T$ such that every element of $A$ is a constant, then $\mathfrak{U}$ is an atomic model.

Why so? Am I right, when I think, that if theory is complete than -> every of it's formulas is complete too? So that formula with conjunction too?

Also, I pin basic definitions about it:

Consider a complete theory $T$ in $\mathcal{L}$. A formula $\varphi(x_1\ldots x_n)$ is said to be complete (in $T$) iff for every formula $\psi(x_1,\ldots x_n)$ exactly one of
$$T\vDash\varphi\rightarrow\psi,\quad T\vDash\varphi\rightarrow\neg\psi$$
holds. A formula $\theta(x_1\ldots x_n)$ is said to be completable (in T) iff there is a complete formula $\varphi(x_1\ldots x_n)$ with $T\vDash\varphi\rightarrow\theta$. If $\theta(x_1\ldots x_n)$ is not completable it is said to be incompletable.

A theory $T$ is said to be atomic iff every formula of $\mathcal{L}$ which is consistent with $T$ is completable in $T$. A model $\mathfrak{U}$ if is said to be an atomic model iff every $n$-tuple $a_1,\ldots,a_n\in A$ satisfies a complete formula in $\mathrm{Th}(\mathfrak{U})$.

答:
$T$ is complete iff for every sentence $\phi$ either $T \models \phi$ or $T \models \lnot \phi$.  This requirement only applies to sentences, i.e., closed formulas: if $\phi = \phi(x_1, \ldots, x_n)$ has free variables then $\phi$ might hold under some interpretations of $\phi$ but not under others. E.g., consider $\phi(x, y) = x < y$ in the theory of a dense linear order without endpoints.

The formula $x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ is complete in the sense of the definition you give, because for any $\phi$,
$T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ iff $T \models \phi(c_1, \ldots, c_n)$. Now, $\phi(c_1, \ldots, c_n)$ is a sentence, so if $T$ is complete either $T \models \phi(c_1, \ldots, c_n)$ or $T \models \lnot\phi(c_1, \ldots, c_n)$ and so either $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ or $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \lnot\phi$.

If every element of a model $\mathfrak{A}$ is a constant, then every $n$-tuple, $a_1, \ldots a_n$ satisfies the formula $x_1\equiv a_1\wedge x_1\equiv a_2\wedge\cdots\wedge x_n\equiv a_n$, which we have just shown is complete. Hence such an $\mathfrak{A}$ is an atomic model according to the definition you have quoted.


问:
calculating norm for a vector with complex element on it
I would really appreciate it if someone can help me with complex numbers. I am haven't worked on it since graduation and decided to study again. I know this is pretty basic but I feel stuck on it.

I got the following vector below and I need to find the norm

$v = \begin{pmatrix} a \\ ai \end{pmatrix}$

My work as follows:

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + (ai)^2}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + a^2 * (i)^2}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + a^2 * (-1)}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 - a^2}$

$\lvert\lvert v \lvert \lvert = 0$

But the answer for the solution is $$\lvert\lvert v \lvert \lvert = \sqrt{2}a$$

Can someone please point out the part that I am doing wrong here?

Thanks in advance

答:
The formula for the norm you've  used is only for real vectors. When your vector is complex it's

$$ \left\| { \begin{pmatrix} z_1 \\ z_2 \end{pmatrix}}\right\|= \sqrt{|z_1|^2 + |z_2|^2}$$

where those $|\cdot|$ mean "magnitude of a complex number, i.e.

$$|x + iy| = \sqrt{x^2 + y^2}$$

If $z = x +iy$ this can also be written
$$ |z| = \sqrt{z \bar z}$$

where $\bar z$ is the complex conjugate of $z$.

So the place where you went wrong  is when you wrote

$$ a^2 +(ai)^2$$

It should have been

$$ |a|^2 +|ai|^2$$

This actually makes no difference in the first term, but it does in the second.

Also note that how you write this up depends on whether $a$ is a real number or a complex number. The answer you give from the solution actually only works if $a$ is a positive real number, so I guess you're supposed to assume that.


问:
Find a circle perpendicular to two other circles.
Here are two circles $C_1 = [ x^2 + y^2 = 1]$ and $C_2 = [ (x-2)^2 + y^2 = 3 ]$.  The radii are $1$ and $3$ and the two circles are orthogonal to each other $C_1 \cdot C_2 = 0$.  What is the equation of a third circle cutting the other two at right angles?  I suspect there are only two solutions which are equivalent by a reflection.

Looking for a geometric solution if possible.  Counting the number of solutions, finding the equation of the circles and maybe some insights into the method.

Related: Find a circle orthogonal to two other circles

The Inner Product of Two Circles

If $C = [ a(x^2 + y^2) - 2px - 2qy + r = 0]$ the inner product between two such circles would be:
$$ [C_1 \cdot C_2] = \frac{1}{2}[2p_1p_2 + 2q_1q_2 - r_1a_2 - r_2a_1] $$

So we are trying to find three circles $C_1, C_2, C_3$ the first two given, such that:
$$ [C_1 \cdot C_2] = [C_2 \cdot C_3] = [C_3 \cdot C_1 ]= 0 $$
The inner product can also be written in terms of the radii of the circles and the distance between the centers:
$$ [C_1 \cdot C_2 ] = \frac{1}{2} [ R_1^2 + R_2^2 - D^2]$$
and if $C_1 \perp C_2$ we have that this inner product is zero, $[C_1 \cdot C_2 ] = 0$.

Not too many resources on the "algebra of circles", e.g. Geometry a Comprehensive Coruse by Daniel Pedoe.
 Descartes theorem deals with mutually tangent circles.

答:
Length of tangent from circle $C(x,y) = x^2 +y^2 + 2gx + 2fy+ c=0$ from a point $P=(x_i,y_i)$ is given as: $C(P) = C(x_i,y_i)$

Now, if we have two circles $C_1$ and $C_2$, then the radical axis is given as:

$$C_1(x,y) = C_2(x,y)$$

Or,

$$ 2(g_1 - g_2) x + 2(f_1 - f_2)y + c_1 - c_2 = 0 \tag{1}$$

I.e: points for which length of tangent from each circle is equal.

Now, for any point on this line, we can draw a circle which intersects the two original circles orthogonally (Picture from wikipedia). We can write a point for $(1)$ parametrically as $$(t, \frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)$$, now a circle cantered at a point with radius $r$ as given as:

$$ (x-t)^2 + (y-\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)^2 = r^2$$

But what is the radius from the definition of radical axis? it is the length of tangents!

Hence,

$$ r(t)=\sqrt{C_1(t,\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t) }= \sqrt{C_2(t,\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)}
$$

Which leads to:

$$ (x-t)^2 + (y-\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)^2=  C_1(t,\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t) =C_2(t,\frac{c_2 -c_1}{2(f_1  - f_2)}- \frac{g_1 - g_2}{f_1 - f_2} t)$$


问:
How many Distinct Hamiltonian Maximal Planar Graphs are there (n vertices) and could this representation help?
If we make a regular polygon with n vertices (n edges) and triangulate on the inside with n-3 edges, then triangulate on the outside with (n-3) edges (or draw dotted lines inside again), a Maximal Planar Graph is formed.  Edges shouldn't be repeated and there's no loops or directions.

How many distinct graphs of this type are there?

It's connected to an earlier question where it was asked 'How many Distinct Maximal Planar Graphs are there? How many distinct Maximal Planar Graphs exist with $n$ vertices?

Will Orrick gave the OEIS numbers A000109.  It was then wondered if there was a known formula for those numbers or bounds on them.  It's conjectured that graphs of the type in this question might constitute most Maximal Planar Graphs, so a formula for the 'polygon' types might be an approximate formula or a lower bound for all types of Maximal Planar Graphs.

Dividing the polygon on the inside can be done in C(n-2) ways, where C(n) are the Catalan numbers A000108, so a connection was looked for - and A000109 seem to be close to C(n-2)*2^(n-13) at least for n=9 to 23.

So, the answer to this question would be of interest and any thoughts on connecting the A000108 and A000109 numbers.  Lots of ways have been tried so far, e.g. since the next Catalan number can be formed by adding the product of ones before, e.g. C(4) = C(3)*C(1) + C(2)*C(2) + C(1)*C(3), perhaps something similar happens for A000109, or by incorporating numbers from both sequences.  Lots of coincidences (probably) have been found including the 233 number in A000109 say X(10) is C(9-2) - the sum of the Catalan numbers before it.

Think I'm going to go crazy looking for patterns any longer!  Any suggestions please!

答:
Here is the work so far.  In what follows the number of distinct Hamiltonian Maximal Planar Graphs with n vertices is $X(n)$.  It should follow A000109 https://oeis.org/A000109/list up to $n=11$ when the first non Hamiltonian occurs.

Also used are the number of distinct ways to divide a polygon https://oeis.org/A000207 .   $A(n)$ will be a number from this list corresponding to the number of distinct ways to triangulate a polygon with $n$ sides (offset by $2$ from the list in the link) .  There is a formula for these, not written here.

Since a Maximal Planar Hamiltonian graph can be represented by a regular polygon triangulated by 'inside' edges and also triangulated again with 'outside' edges, without repeating edges, the following is suggested.
$$
\begin{array}{c|cc}
n & A(n) & X(n)\\
\hline
3 & 1 & 1\\
4 & 1 & 1\\
5 & 1 & 1\\
6 & 3 & 2\\
7 & 4 & 5\\
8 & 12 & 14\\
9 & 27 & 50\\
10 & 82 & 233\\
11 & 228 & 1249\\
\end{array}
$$

Dividing the pentagon on the inside can be done in $1$ way with a fan $F$ from one vertex.
there is only one distinct way to do it on the outside, another fan, $1\times1=1$

For the hexagon let's start with the maximum number of edges possible from either set of edges, leaving a vertex, it's $3$, making a fan $F$.  Let the hexagon be $ABCDEF$.  If the $3$ inside (say) edges leave node $B$.  The outside edges can't repeat any of these, one of the outside edges must also join $A$ to $C$, to ensure it's triangulated and not repeat edges.  This leaves a pentagon to triangulate by the outside edges, but using one less edge (one was used for $AC$) - the number of distinct ways for that is $A(5)$

There are two remaining ways to divide the hexagon with inside edges, a triangle $T$ and a $Z$ shape.
Next the case where the maximum number from one set of edges leaving a vertex is $2$.  If we draw diagrams using the $2$ remaining ways to divide the hexagon but not using a fan $F$, we find an extra distinct case from this, so $X(6)=A(5)+1$.

In a similar way using tracing paper, with the $4$ ways to divide a heptagon, it's found that $X(7) = A(6) + 2$.  The $A(6)$ is the number using a Fan of $4$ edges at one corner of the heptagon, similar to above, and the $2$ comes by combining the other $3$ ways, each of the $2$ new cases came when one of the set of edges has $3$ edges from one set leaving a vertex.  The case where the maximum number is $2$ edges leaving a vertex gave no extra distinct cases.

So a pattern is now being looked for of this type
$$
X(n) = A(n-1)+aX(n-1)+bX(n-2)+cX(n-3)\ldots
$$
where $a$, $b$, $c$ etc...are integer coefficients.

The pattern isn't clear and might involve the factors of the number $n$, perhaps the Catalan numbers?  So a definite reason isn't given here for the coefficients, perhaps it's coincidence, but it's possible to build up the $X(n)$ in this way e.g.
$$
\begin{array}{r|rcl|l}
n & X(n) &\!=\!& A(n-1)+aX(n-1)+\ldots &\\
\hline
6 & 2 &\!\!=\!\!& 1 + 1 &\\
7 & 5 &\!\!=\!\!& 3 + 2 &\\
8 & 14 &\!\!=\!\!& 4 + 2\times5 &\\
9 & 50 &\!\!=\!\!& 12 + 2\times14 + 2\times5 &\\
10 & 233 &\!\!=\!\!& 27 + 3\times50 + 4\times14 &\\
11 & 1248 &\!\!=\!\!& 82 + 4\times233 + 3\times50 + 6\times14 & \text{1248 instead of 1249 as there is one}\\
& & & & \text{non Hamiltonian graph for $n=11$.}
\end{array}
$$


问:
$n$ such that $ n^2 \mid \varphi(n)^{\varphi(n)} - 1 $. Can $n$ be a prime number $n>2$ or a Carmichael number?
For all $n$ such that $\large n^2 \mid \varphi(n)^{\varphi(n)} - 1 $ with $\varphi()$ denotes Euler's totient function.

It can be shown that $n$ must be a cyclic number. That means $n$ is coprime to $\varphi(n)$.

Implying further that $n > 2$ is odd and squarefree.

Besides the two trivial solutions $n=1$ and $n=2$ OEIS states five nontrivial solutions below $10^{12}$:

$19043= 137 \cdot 139$,

$289627=13 \cdot 22279$,

$6674419= 571 \cdot 11689$,

$49865347= 7 \cdot 173 \cdot 41177$,

and

$185014655= 5 \cdot 7 \cdot 17  \cdot 310949 $.

As Carmichael numbers are also cyclic numbers, I tested the first $1,401,644$ of them below $10^{18}$ from a database. None of them gives a solution.

If $n$ is a prime number then  $n^2 \mid (n-1)^{n-1} - 1$.

Hence $n$ must be a Wieferich prime to base $n-1$.

Can $n$ be a prime number $n>2$ or a Carmichael number?

答:
$n$ can't be a prime number $> 2$ because in the binomial expansion of

$(p-1)^{(p-1)} ~-~ 1$,

$p^2$ must divide all but the last term, and can not divide the last term, which is $\pm \binom{p-1}{p-2}p^1.$


答:
The standard sesqulinear form on $\mathbb{C}^n$ is defined as
$$\langle x,y\rangle=\sum_{j=1}^nx_j\overline{y_j}$$
for vectors $x=(x_1,...,x_n)^{tr},y=(y_1,...,y_n)^{tr}\in\mathbb{C}^n$, where $x\mapsto\overline{x}$ is the complex conjugate. The norm is the squareroot of the quadratic form of this sesquilinear form:
$$||x||=\sqrt{\langle x,x\rangle}=\sqrt{\sum_{j=1}^nx_j\overline{x_j}}=\sqrt{\sum_{j=1}^n|x_j|^2}=\sqrt{\sum_{j=1}^n\Re(x_j)^2+\Im(x_j)^2}.$$

Title: calculating norm for a vector with complex element on it

Body:
I would really appreciate it if someone can help me with complex numbers. I am haven't worked on it since graduation and decided to study again. I know this is pretty basic but I feel stuck on it.

I got the following vector below and I need to find the norm

$v = \begin{pmatrix} a \\ ai \end{pmatrix}$

My work as follows:

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + (ai)^2}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + a^2 * (i)^2}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 + a^2 * (-1)}$

$\lvert\lvert v \lvert \lvert = \sqrt{a^2 - a^2}$

$\lvert\lvert v \lvert \lvert = 0$

But the answer for the solution is $$\lvert\lvert v \lvert \lvert = \sqrt{2}a$$

Can someone please point out the part that I am doing wrong here?

Thanks in advance


问:
Cardinality of set of all straight lines in $\mathbb R^2$, that pass through at least two different rational coordinates.
Definition: A coordinate $(x,y)\in \mathbb R^2$ is rational if $x,y$ are both rational nos.

Let $S$ be the set of all straight lines in $\mathbb R^2$ that pass through at least two distinct rational points.

Let $C$ be the set of all rational points in $\mathbb R^2$. Since $\mathbb Q \times \mathbb Q$ is countable so are its subsets. Therefore, it is possible to write  $C=\{c_1,c_2,...,c_k,c_{k+1},...\}$. Of course $C=\mathbb Q \times \mathbb
 Q$.

Let $S_i$ be the set of all straight lines that pass through $c_i\in C$. For every $s\in S_i$, $\exists$ a rational point $c_{k(s)}\ne c_i$ thus we can have a bijection. Therefore, $S_i$ is countable.

Hence, $S=\cup_{i=1}^{\infty} S_i$ is countable collection of countable sets and therefore countable.

Is my proof correct? Thanks.

答:
Yes, but not completely:

The set $S_i$, as defined, is not countable. In order to be countable, the set $S_i$ should be defined as the set of lines in $S$, tha passes through $c_i$.

also, with this definition, $S_i$ the described map is not a bijection. It is true that there is a map from $C$ to $S_i$ that assigns to $c_j$ the line passing through $c_i$ and $c_j$. But this map is not a bijection (it is not injective): Still, it is surjective. This implies that $S_i$ is countable.

Last step of the proof is now correct.


问:
measurability of supremum of a class of functions
Let $f:X\times Y \mapsto R$ be a measurable function on product space $X\times Y$, where $X$ and $Y$ both are some metric spaces. Define $g(x) := \sup_{y\in Y} f(x,y)$. </p>

<p>[Q.] Is $g$ a measurable function on $X$? If not, what is the sufficient condition to be measurable?</p>

答:
This is a very nice discussion and I have a question related to it, though not exactly the same.

If $(f_{j})_{j \in J} $ is a family of continuous functions over a compact set of $K \subset \mathbb{R}^{n}$, which is uniformly bounded, say $\sup_{j \in J} \sup_{x \in K}|f_{j}(x)| = M < \infty $, and such that the index set $J$ is infinite uncountable, is the function
$$ f(x) = \sup_{j \in J} f_{j}(x) $$
measurable?

There is no a measure-theoretic structure on $J$, it is an arbitrary uncountable set.

Thank you very much.


答:
$$
x^{\,2}  - x = x\left( {x - 1} \right) = x^{\,\underline {\,2\,} }  = 2\left( \matrix{
  x \cr
  2 \cr}  \right) = T
$$

So first of all $T$ must be even.

Then  $$
\sqrt T  <  \approx x
$$ and for large T you can check if $$
x = \left\lceil {\sqrt T } \right\rceil
$$
works.

In any case you can use the Newton-Raphson method to find the root of $x^2-x-T=0$, and see if it is an integer.

Title: Deriving integer solutions to quadratic equation without solving quadratic congruence

Body:
I want to generate positive integer solutions of $x$ to the equation:

$x^2-x-aT=0$

where a is an integer $>$ 0 and T is a very large positive number.

I noticed that when plugging this into wolfram alpha is looks like it was formulating the equation in the form:

$x^2-x≡0 (mod T)$

And then solving this quadratic congruence to get the equations to generate all integer solutions to my equation.

I'm not keen on using this method to solve as in order for it to work it relies on first factorising T. If T is really large this will cause the solution to run in linear time (with respect to the number of digits of T) which doesn't scale to the size I want.

So what I am asking is:

Is there a way to generate integer solutions to my initial equation in logarithmic time without solving the quadratic congruence equation I showed above and without factorising T?


问:
Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$
I have recently learned about pattern that goes like this:

$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$

It is a very cool pattern, but after a bit it stops:

$$1,111,111,111^2 = 1,234,567,900,987,654,321$$

My main question:

Why does this pattern work?

A side question that's less important:

Is there an algebraic equation to describe this pattern?

答:
Just do the maths:

```<code>        1 1 1 1 1
      x 1 1 1 1 1
        ---------
        1 1 1 1 1
      1 1 1 1 1
    1 1 1 1 1
  1 1 1 1 1
1 1 1 1 1
-----------------
1 2 3 4 5 4 3 2 1
</code>```

Of course the pattern only goes up to nine $1$'s, because after that you get eight hundred and ninety-ten in the billions column...

Body:
Just do the maths:

        1 1 1 1 1
      x 1 1 1 1 1
        ---------
        1 1 1 1 1
      1 1 1 1 1
    1 1 1 1 1
  1 1 1 1 1
1 1 1 1 1
-----------------
1 2 3 4 5 4 3 2 1

Of course the pattern only goes up to nine $1$'s, because after that you get eight hundred and ninety-ten in the billions column...


问:
,For $y=x^2+1$ and $x=y^2+1$, if PQ is the shortest distance and R,S are points of contact of common tangent, find area PQRS.
From equation, it is evident that curves are symmetric about $y=x$

So I tried writing the equations of tangents
$$k=mh +\frac{1}{4m}$$ and $$h=mk-\frac{m^2}{4}$$

Where $(h,k)$ is a point on curve 1

But on plugging $m=1$ the lines turn out to be coincident, and not two separate parallel lines. I was hoping to find $(h,k)$ by solving these two equations but that clearly isn’t working. What should I do?

答:
(Without calculus) The line $y=c-x$ meets $y=x^2+1$ where $x^2+x +1-c+0$. Tangency occurs when this quadratic equation has equal roots, namely when $c=\frac34$. (You may check that the same applies for $x=y^2+1$.) Thus the tangent points are $$R\;(-\tfrac12,\tfrac54)\quad\text{and}\quad S\;(\tfrac54, -\tfrac12).$$

To find the shortest connection between the two curves, note that this is twice the length of the shortest connector from either curve to the line of symmetry $y=x$, say from the point $(t,t)$ on the line to the nearest point on $y=x^2+1$ that lies on the line $y=c-x$ which passes through $(t,t)$, namely the line $y= 2t-x$. This point is given by first solving $2t-x=x^2+1$ for $x$ in terms of $t$:
$$x=-\tfrac12+\sqrt{2t-\tfrac34},\qquad y=2t+\tfrac12-\sqrt{2t-\tfrac34}.$$
(The sign of the square root is chosen to get the minimum distance, in the first quadrant.)

We need now to minimize the squared distance of this point from $(t,t)$, which is
$$\left(t+\tfrac12-\sqrt{2t-\tfrac34}\right)^2+\left(-t-\tfrac12+\sqrt{2t-\tfrac34}\right)^2=2\left(t+\tfrac12-\sqrt{2t-\tfrac34}\right)^2.$$
To minimize this, write $t=\tfrac12u^2+\tfrac38,$ so that the quantity to be minimized (after dropping unnecessary features) becomes
$$\tfrac12u^2+\tfrac78-u=\tfrac12(u-1)^2 +\tfrac38.$$ Thus $u=1$ and $t=\frac78$. Hence the required points on $y=x^2+1$ and $x=y^2+1$ respectively are
$$P\;(\tfrac12,\tfrac54)\quad\text{and}\quad Q\;(\tfrac54,\tfrac12).$$ The midpoints of $PQ$ and $RS$ are $(\frac78,\frac78)$ and $(\frac38,\frac38)$, and so the distance between them is $\frac12\surd2$. Also, $|PQ|=\frac34\surd2$ and $|RS|=\frac74\surd2$. Therefore the area of $PQRS$ is $\frac12\surd2\cdot\frac12(\frac34\surd2+\frac74\surd2)$ or
$$\frac54.$$


问:
Lifting of arbitrary map to a covering space

Let $(\tilde{X},p)$ be a covering space of $X$, $\tilde{x}_{0}\in\tilde{X}$, $x_{0}=p(\tilde{x}_{0})$, $y_{0}\in Y$ and $\varphi :(Y,y_{0})\rightarrow (X,x_{0})$. Under what conditions does there exist a map $\tilde{\varphi}:(Y,y_{0})\rightarrow (\tilde{X},\tilde{x}_{0})$ such that the diagram  is commutative? If we assume such a map $\tilde{\varphi}$ exists, we obtain the follow commutative diagram of groups and homomorphisms: beacuse $p_{*}$ is a monomorphism the existence of a homomorphism $\tilde{\varphi}_{*}:\pi(Y,y_{0})\rightarrow\pi(\tilde{X},\tilde{x}_{0})$ which makes the diagram commutative, is exactly equivalent to the dondition that the image $\varphi{*}$ be contained in the image of $p_{*}$

The Massey says this before the theroem of the lifting of an arbitrary map. But I don't see where we use the fact that $p_{*}$ is a monomorphism. In general, if $$(p_{*}\circ\tilde{\varphi})=\varphi$$ the condition on the images is tha same without assumptions on $p_{*}$...
If I know that there is that commutative diagram, my necessary condition is image of $\varphi_{*}$ in image of $p_{*}$. I don't need the fact that $p_{*}$ is a monomorphism. Is it wrong?

答:
Given $\varphi_*$ and $p_*$, there may be no $\tilde{\varphi}_*$ such that $(p_* \circ \tilde{\varphi}_*) = \varphi_*$.
It is the existence of such a $\tilde{\varphi}_*$ that is dependent on the condition that the image of $\varphi_*$ is contained in the image of $p_*$.


问:
a perpendicular "object" to another "object" in a n-dimensional space
Sorry if I'm not too good at explaining and formatting(im new here :p)...

In a 2-dimensional space, if you have a line(1-dimensional) there is another line(1-dimensional) that is perpendicular to it. ax + by = 0 => bx - ay = 0

In a 3-dimensional space, if you have a line(1-dimensional) there is a plane(2-dimensional) that is perpendicular to it(and vice versa). ax = by = cz => bcx + acy + abz = 0

I assume if there is a n-dimensional space, with an m-dimensional object, there would also potentially be an (n - m) dimensional object that is perpendicular to that object.

I want to know if

A. whether the assumption is true

B. and if it is, if its possible to find an general formula for it.

答:
The general concept is called orthogonal complement.

The orthogonal complement of a subspace $W$ of a vector space $V$ equipped with a bilinear form $B$ is the set $W^\perp$ of all vectors in $V$ that are orthogonal to every vector in $W$.

A usual case is an inner product space $V$ where the bilinear form $B$ is the inner product. The complement of the subspace $W$ is written as
$$
W^\perp=\{v\in V:\langle w,v\rangle=0 \ \textrm{for all } w\in W\}
$$

In your first example, $V=\mathbb{R}^2$, $W=\{(x,y):ax+by=0\}$.


问:
In a matrix ring, no zero divisors may have an inverse
In a general ring with 1, a right (left) zero divisor cannot have a right (left) inverse. In a matrix ring over a field, a stronger condition is satisfied: a (right or left) zero divisor cannot have a (right or left) inverse, with the proof I know invoking the rank plus nullity theorem (which makes sense, because infinite dimensional linear transformations can have right-inverses and be right-zero divisors). Is this more or less a peculiarity of matrices, or am I missing a more general property that they (and perhaps other spaces) possess?</p>

答:
It's quite a bit more general than just matrix rings, though obviously some finiteness condition is required. It's a property of (left) Noetherian rings that anything that has a left inverse also has a right inverse.

Proof: if $x \in R$ has a left inverse $x^{-1}$, then right multiplication by $x$ is surjective since $y=(yx^{-1})x$. It is also injective: let $I_n$ be the (left ideal) kernel of right multiplication by $x^n$, and choose $n$ with $I_n=I_{n+1}$ (Noether used here). Then if $y \in I_1$, we can write $y=z x^n$ for some $z$, so we have $$0=y x=zx^{n+1} \quad \implies \quad 0=zx^n=y$$ by choice of $n$. Thus right multiplication by $x$ is also injective. Hence
$$0=(1-x x^{-1})x \implies 1=x x^{-1}$$ so $x^{-1}$ is a right inverse for $x$.

PS: If the ring is right Noetherian, then one applies the opposite argument (interchanging right and left) to the element $x^{-1}$ to show that it also has a left inverse. So being Noetherian on one side is enough.


问:
Measurable functions on trivial sigma-algebra must be constant?
I'm self-studying measure theory through Axler's "Measure, Integration & Real Analysis" and have come across the following definition of measurable functions:

Suppose $(X,S)$ is a measurable space. A function $f:X\rightarrow \mathbf{R}$ is called S-measurable if $f^{-1}(B)\in S$ for every Borel set $B\subseteq \mathbf{R}$.

I'm reading this definition and basically interpreting it in English that every measurable image of the function $f$ must also have a measurable pre-image. That is, the Borel sets are measurable, and so for the function to be measurable, the sets that map to the Borel sets via $f$ must be $S$-measurable. So a measurable function is basically a function that transforms measurable sets into measurable sets in a different space (measurable sets of $X$ to Borel measurable sets of $\mathbf{R}$ in this case).

It then gives an example: If $S=\{\emptyset, X\}$, then the only $S$-measurable functions from $X$ to $\mathbf{R}$ are the constant functions.

And here I'm realizing I must be off. Consider $X=\mathbf{R}, S=\{\emptyset, \mathbf{R}\}$. Then $f:\mathbf{R}\rightarrow\mathbf{R}$ is measurable if $f^{-1}(B)\in S$ for all Borel sets $B\subseteq \mathbf{R}$. Consider $f(x)=x$. This is clearly a map $\mathbf{R}\rightarrow\mathbf{R}$. Consider the Borel set $(0,1)$. We then have that $f^{-1}((0,1))=(0,1)$. But $(0,1)\not\in S$. So wouldn't this mean that it isn't measurable?

Thanks for the help!

答:
Your understanding that

a measurable function is basically a function that transforms measurable sets into measurable sets in a different space

is not correct. It is simply as the definition says: the pre-image of a measurable set has to be measurable. This does not hold for the 'other direction' of the mapping in general. For example take any set $X$ and the power set $\mathcal P(X)$ as $\sigma$-algebra and let $\mathcal F$ be a further $\sigma$-algebra on $X$. Since $\mathcal P(X)$ is 'extremly big' every map $f:X\to X$ is $\mathcal P(X)$-$\mathcal F$-measurable. Now consider $\mathcal F=\{\emptyset, X\}$. There is no function which transforms all $\mathcal P(X)$-measurable sets into $\mathcal F$-measurable set if $X$ has more than one element.

The example you gave with the $\sigma$-algebra $S$ and the function $f=\operatorname{Id}_\mathbb{R}$ is correct. In this case $f$ is not $S$-$\mathcal B$-measurable. This is the reason why you actually always have to say $S$-$\mathcal B$-measurable and not only measurable, because beeing measurable depends on both $\sigma$-algebras.

It is similar to the topology and continuous maps. The definition of a continuous map is that the pre-image of an open set has to be open. The other direction, that the image of an open set has to be open, is called an open map. But a continuous map is not open in general. And if you change the topology (analogous to your example) a previous continuous map might not be continuous any more.


问:
is this theorem true for case quasi-projective?
Definition. If $X, Y$ are two varieties, a morphism $\varphi: X \rightarrow Y$
is a continuous map such that for every open set $V \subseteq Y,$ and for every regular function $f: V \rightarrow k$, the function $\varphi^{*} := f \circ \varphi: \varphi^{-1}(V) \rightarrow k$ is regular.

let $O(Y)$ be the ring of all regular
functions on $Y$.

Theorem .Let $X$ be a variety
and $Y \subset \mathbb{A}^{n}$ be Affine
variete and $\varphi: X \rightarrow Y$
be set function ( or any subset of $X$ send to a subset of $Y$)
then:
$\varphi$ is a morphism iff
$\varphi^{*}$ send $O(Y)$ to $O(X)$ .

Is above theorem true when $Y$ be quasi-projective ?

答: No; the problem is that $O(Y)$ could be very small.  For instance, if $Y$ is an irreducible projective variety, then $O(Y)$ will consist of only constant functions, so $\varphi^*$ will always send $O(Y)$ to $O(X)$ (since $O(X)$ always includes the constant functions), no matter what $\varphi$ is.


问:
Why replacing $\cos(x)$ by $e^{ix}$ in an integral and taking the real part doesn't always work? When does it work?
I wanted to solve the following integral using the residue theorem $\begin{align*}\int_{0}^{2\pi} \frac{\cos(3x)}{5-4\cos(3x)} \,dx\end{align*}$ and for that I thought I could replace it by
$\begin{align*} \operatorname{Re}\biggl(\int_{0}^{2\pi} \frac{e^{3ix}}{5-4 e^{3ix}}\, dx\biggr) = \operatorname{Re}\biggl(\oint_\gamma \frac{z^{3}}{5-4 z^{3}} \frac{dz}{iz}\biggr)\end{align*}$ using $z=e^{ix}$ and $\gamma$ beeing a closed loop along the unit circle. Following this method, I could find that there are no singularities enclosed by the path so this should give me $\begin{align*}\int_{0}^{2\pi} \frac{\cos(3x)}{5-4\cos(3x)}\, dx=0\end{align*}$ which is wrong. I thought the mistake is  in the step $\begin{align*}\int_{0}^{2\pi} \frac{\cos(3x)}{5-4\cos(3x)}\, dx=\operatorname{Re}\biggl(\int_{0}^{2\pi} \frac{e^{3ix}}{5-4 e^{3ix}}\, dx\biggr)\end{align*}$. Indeed I calculated these on Wolfra alpha and didn't get the same result.
I don't understand when I can use the trick of substituting an integral with $\cos(x)$ by the real part of an integral with $e^{ix}$ or substituting an integral with $\sin(x)$ by the imaginary part of an integral with $e^{ix}$ multiplied by $i$, since $e^{ix}=\cos(x) + i\sin(x)$. So, I would like to know in general when I am allowed to use it, and when I should be more careful about it.

答:
Yep the mistake is there
the problem is that the $\operatorname{Re}$ function is linear, but it doesn’t distribute over division. $$\operatorname{Re} \left(\frac{a}{b}\right)$$ is not the same as $$ \frac{\operatorname{Re} (a)}{\operatorname{Re}(b)}.$$

Therefore, you cannot change $$\frac{\cos(3x)}{5-4\cos(3x)} $$ to $$\operatorname{Re}\left(\frac{e^{3ix}}{5-4 e^{3ix}}\right).$$

Basically the only time this will work as is, is if you only have cos on the numerator of the fraction.


问:
Given a set X and the distance function d(x,y), prove that (X,d) is a metric space.
I encountered this question while solving the exercises of the book "Metric Spaces" by Satish Shirali and H.L. Vasudeva.

Let $X$ denote the set of all real sequences and $x = \{x_i\}_{i \leq 1}$ and $y = \{y_i\}_{i \leq 1}$  be
arbitrary elements in X with
$$d(x,y)=\sum_{k=1}^{\infty} \frac{1}{k^2}\min\left\{|x_k - y_k|,1\right\}.$$

Prove that $(X,d)$ is a metric space.

I was easily able to show that.

$d(x,y)\geq0$

$d(x,y)=d(y,x)$

$d(x,y)=0 \iff x=y$

But what I can't understand is how to approach $d(x,y) \leq d(x,z) + d(z,y)$ due to the presence of the summation over all of k.

Any help/hint will be highly appreciated.

答:
Justify why for any triple of real numbers $x,y,z$ we have that
$$\min\{|x-y|,1\}\leq\min\{|x-z|,1\}+\min\{|z-y|,1\} $$
Conclude that for any $k$
$$\frac{1}{k^2}\min\{|x_k-y_k|,1\}\leq\frac{1}{k^2}\min\{|x_k-z_k|,1\}+\frac{1}{k^2}\min\{|z_k-y_k|,1\} $$

sum up and conclude that $d(x,y)\leq d(x,z)+d(z,y)$.


答:
It is a peculiarity of “vaguely finite” rings.

A right Artinian ring has an even stronger property shared by matrices: if $a$ is not a left zero-divisor (there is no nonzero $b$ with $ab=0$), then $a$ has both a left and right inverse (and so they are equal). Indeed, it need only have the DCC on principal right ideals, a so-called left perfect ring.

The exact rings you are looking for are the “Dedekind-finite” rings: those rings in which $ab=1$ implies $ba=1$. Every one-sided noetherian ring is Dedekind finite.

One also has in general, that if $ab=1$ then $a(ba-1) = aba-a = 0$ so that if $a$ has a right inverse, then either it is a unit (has a left inverse identical to its right inverse) or it is a left zero divisor.

If $ab=1$, but $ba\neq 1$, then notice that $a \cdot \left((1-ba)a^j +b\right) = 0 + ab = 1$ and that for distinct $i,j$, $(1-ba)a^i \neq (1-ba)a^j$, so that if $a$ has a right inverse, but no left inverse, then it must have infinitely many right inverses.

Proof of equivalence: Suppose no one-sided zero-divisor can have a one-sided inverse. Then every element with a one-sided inverse must have a two-sided inverse, but that is precisely the condition that $ab=1 \implies ba=1$, that is, the ring is Dedekind-finite. Conversely suppose that the ring is Dedekind-finite: then any element with a one-sided inverse has a two-sided inverse, and so cannot be a zero divisor.

One can find this and more in Lam's First Course on Non-commutative Rings, especially chapter 1.

Title: In a matrix ring, no zero divisors may have an inverse

Body: In a general ring with 1, a right (left) zero divisor cannot have a right (left) inverse. In a matrix ring over a field, a stronger condition is satisfied: a (right or left) zero divisor cannot have a (right or left) inverse, with the proof I know invoking the rank plus nullity theorem (which makes sense, because infinite dimensional linear transformations can have right-inverses and be right-zero divisors). Is this more or less a peculiarity of matrices, or am I missing a more general property that they (and perhaps other spaces) possess?</p>


问:
Compute the curve integral $\int_{|z|=3}\dfrac{dz}{\sin^2 z}$ [trouble with limit]
My attempt is as follows;

The region $\Omega$ which is enclosed by the curve $\gamma = \{z \in \mathbb{C}:|z|=3\}$ contains one singularity at $z=0$, a pole of order 2. Therefore, we may use the residue theorem which yields
$$\int_{\gamma}\dfrac{dz}{\sin^2z}= 2\pi i\, \text{Res}_f(0)\,.$$

When computing the residue according to definition for pole of order $k=2$, I arrive at
\begin{align*}
\text{Res}_f(0) &= \dfrac{1}{(2-1)!}\lim_{z \rightarrow 0} \dfrac{d}{dz}\left[\dfrac{z^2}{\sin^2(z)}\right]\\[5pt]
&=\lim_{z \rightarrow 0} \dfrac{2z[\sin z-z\cos z]}{\sin^3z}.\\[4pt]
\end{align*}
According to the solution manual, this limit is 0. I am unable to conclude this myself, so I was wondering if anyone here could provide a hint?

Many thanks in advance.

答:
Hint.

$$
\begin{align}
\frac{1}{\sin^2(z)}
&=\frac{1}{z^2}\left(1-\frac{z^2}{6}+\dots\right)^{-2}\\
&=\frac{1}{z^2}\left(1+\frac{z^2}{3}+\dots\right)\\
&=0\cdot \frac1z+\frac{1}{z^2}+\frac13+\dots
\end{align}
$$

where you use the fact that $\sin z=z-\frac{z^3}{6}+\cdots$.

Alternatively, observe that $\sin^2 (z)=\sin^2 (-z)$. So if you take the opposite direction of the contour $\gamma$, the integral remains the same:
\begin{align}
\int_\gamma f(z)dz
=&-\int_{\gamma}f(-w)dw&\textrm{(change of variable $w=-z$)}\\
=&\int_{-\gamma}f(-w)dw&\textrm{(definition of "negative" contour)}\\
=&\int_{-\gamma}f(w)dw&\textrm{($f(z)=f(-z)$)}\\
=&-\int_{\gamma}f(z)dz
\end{align}


问:
Smallest possible area for triangle
I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:

Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$

Let's find the tangent of the ellipse:

$2x^2 + y^2 = 1$

$4x + 2y \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} = \dfrac{-2x}{y}$

Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:

$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$

So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$

Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.

So now area is $1/2 a.b = 1/2 a^2$

Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.

I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.

答:
If we consider just the part of the ellipse in the first quadrant, it is
the graph of the function $f(x) = \sqrt{1 - 2x^2}$ on the interval
$\left[0,\frac1{\sqrt2}\right]$;
and between $0$ and $\frac1{\sqrt2}$, $f$ is a positive function with a decreasing first derivative.

From my answer to The method to calculate area between f(x)
and coordinate axes
we therefore know that the area between the coordinate axes and the line tangent to the graph of $f$ is minimized when the product $x f(x)$ is maximized,
and that area is $2 x_m f(x_m)$ where $x_m$ maximizes the product.

In this case, $x f(x) = x \sqrt{1 - 2x^2}.$
It is simpler to maximize $x^2 (f(x))^2,$
which is $x^2 (1 - 2x^2),$ which is $u(1-2u)$ where $u = x^2.$
The derivative of $u(1-2u)$ is $1 - 4u,$ which is zero at $u = \frac14,$
positive for $u < \frac14$, and negative for $u > \frac14,$
so the maximum occurs at $u = \frac14$,
therefore at $x = \frac12,$ and the area of the triangle is

$$ 2 \left(\frac12\right) \sqrt{1 - 2\left(\frac12\right)^2}
= \sqrt{\frac12}.
$$

Title: Smallest possible area for triangle


问:
Prove or disprove: Let $V$ be a vector space, and $T:V\to V$ a diagonalizable linear operator...
then $Ker(T)=ker(T^{2018})$.

My Work:

My thoughts are a little bit shattered, but here's what I tried to do, I wanted to prove it by proving $Ker(T)\subset ker(T^{2018})$ and $Ker(T^{2018})\subset ker(T)$.

I started the first direction: $Ker(T)\subset ker(T^{2018})$. Let $v\in Ker(T)$, then $T(v)=0$, and it's known that $T(0)=0$ in any linear transformation, so $T(T(v))=T(0)=0$, so $T^{2018}=T(....T(v))=T(0)=0$ so $v \in ker(T^{2018})$.

I'm struggling to prove the second direction, and I don't feel alright with my proof up as I feel like I could have shown my idea in a better way.

But from the given information that $T$ is diagonalizable, I tried to do this: there exists a basis $B$ such that $[T]_B$ is diagonal. Here I'm trying to link between $T$ and $T^{2018}$ but I'm not really getting how to do it.
 Any help is appreciated, thanks in advance.

答:
Since $T$ is diagonalizable there exists $\{v_1,\dots, v_n\}$ basis of $V$ and $\lambda_1,\cdots, \lambda_n\in\mathbb{R}$ such that $T(v_i)=\lambda_i v_i, i=1,\cdots, n.$

Thus if $v=\sum_i \alpha_i v_i$ then

$$T(v)=T\left(\sum_i \alpha_i v_i\right)=\sum_i \alpha_i \lambda_i v_i$$ and

$$T^{2018}(v)=T^{2018}\left(\sum_i \alpha_i v_i\right)=\sum_i \alpha_i \lambda_i^{2018} v_i.$$

Thus we have that

\begin{align}
v\in Ker (T^{2018}) & \iff \sum_i \alpha_i \lambda_i^{2018} v_i=0 \\ & \iff \alpha_i \lambda_i^{2018}=0, \forall i \\ & \iff \alpha_i \lambda_i=0, \forall i \\ & \iff \sum_i \alpha_i \lambda_i v_i=0 \\ & \iff v\in Ker (T).
\end{align}


问:
Manipulating Vector Identity
I would like to expand then simplify (if possible) the following quantity.

$\nabla (a \cdot (C\, a))$

Where $a = a(x)$ is a vector valued function of $x$ and $C$ is a constant matrix.

答:
I personally find that using indices makes things like this way easier.

Write  the $i$th component as $\sum_{j,k}\partial_i (a_j C_{jk} a_k)$. You can now just use the product rule etc. as normal and reconvert into matrix notation if you wish.

Title: Manipulating Vector Identity


问:
Can we make a better estimate of the variance given that the distribution is known?
When we have a sample of numbers, say

-69, 153, -54, 54, -198, -242, -63, 87, -45, -134, ...

we can calculate an estimation of the variance using the formula

$$\hat \sigma^2=s^2=\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

now this formula applies to any sequence of numbers, regardless of its underlying distribution. But what if we know that these numbers are sampled from a normal distribution, can we provide a better estimate of the variance? And/or if we know the true mean?

答:
If the sample is known to be iid from a normal distribution with unknown mean and variance then $\hat{\sigma^2}$ is the Uniform Minimum Variance Estimator of $\sigma^2$.

If $\mu$ is known, then replace $\bar{x}$ by $\mu$ and delete the $-1$ in the denominator. That will be the UMVUE in that scenario.

If you don't know $\sigma^2$, but have correct knowledge about what it might be (a prior distribution), then the posterior mean of $\sigma^2$ will be biased but have smaller mean-squared error than the UMVUE.


问:
Line along shortest distance between two skew lines
We have points $A(1,-1,1),B(6,-3,1),C(2,-1,5),D(5,1,1)$

We also have a line $p$ that goes through the edges $AD$ and $BC$ perpendicularly.

Find the equation of this line:
My try:

$$AD = (4,2,0)$$
$$BC = (-4,2,4)$$
$$s = (s_1,s_2,s_3)$$

Then we know that(*...scalar product):
$$s*AD = 0$$
$$s*BC = 0$$
From this two equations we get: $$4s_1+2s_2+0s_3 = 0 = -4s_1+2s_2+4s_3 $$
$$ 4s_1+2s_2+0s_3 = 0\\-4s_1+2s_2+4s_3 = 0\\
 \text{We add both equations together }\implies 4s_2 + 4 s_3 = 0 \implies 4s_2 = -4s_3\implies s_2 = -s_3$$
Then we get:
$$4s_1 = 4s_3\implies s_1 = s_3$$
However, when I plug this back into the equation I get zero at the end, which is not correct.

The solution is $p:x - 3 = \frac{y}{-2} = \frac{z-1}{2}$

答:
We have $AD=(2,1,0)$ and $BC=(-2,1,2)$. Direction of $p$ can be found by taking cross-product. This will be found as $(1,-2,2)$.

Now we find a point on $p$. A point on $BC$ is given by $(2,-1,5)+t(-2,1,2)$. A point on $AD$ is given by $(1,-1,1)+s(2,1,0)$.

From a specific point on $BC$, if we travel along $p$, we will arrive at a specific point on $AD$. The second point satisfies
$$(1,-1,1)+s(2,1,0)=(2,-1,5)+t(-2,1,2)+r(1,-2,2)$$

This is a system of linear equations in $r,s,t$ which can be solved easily. The intersection point is indeed found as $(3,0,1)$. Thus $$p : \frac{x-3}{1}=\frac{y-0}{-2}=\frac{z-1}{2}$$


答:
Since OP is asking for an algebraic explanation, I will try a different approach, possibly trickier than the other answer which is elegant but not formal.

We consider the polynomial $(1+x+x^2+\dots +x^n)^2$: by expanding the square we get

$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
1 & x & x^2 &\cdots& x^{n-1}& x^n\\
\hline
x & x^2 & x^3 &\cdots& x^{n}& x^{n+1}\\
\hline
x^2 & x^3 & x^4 &\cdots& x^{n+1}& x^{n+2}\\
\hline
\vdots & \vdots & \vdots &\vdots& \vdots& \vdots\\
\hline
x^{n-1} & x^{n} & x^{n+1} &\cdots& x^{2n-2}& x^{2n-1}\\
\hline
x^n & x^{n+1} & x^{n+2} &\cdots& x^{2n-1}& x^{2n}\\
\hline
\end{array}$$
that is, for $n\geq 0$, and for $0\leq j\leq 2n$,

$$\begin{align}
(1+x+x^2+\dots +x^n)^2&=\sum_{j=0}^{n}\sum_{k=0}^{j} x^k\cdot x^{j-k}
+\sum_{j=n+1}^{2n}\sum_{k=j-n}^{n} x^k\cdot x^{j-k}\\
&=
\sum_{j=0}^{n}x^j\sum_{k=0}^{j} 1
+\sum_{j=n+1}^{2n}x^j\sum_{k=j-n}^{n} 1\\
&=\sum_{j=0}^{n}(j+1)x^j
+\sum_{j=n+1}^{2n}(2n-j+1)x^j.
\end{align}$$
In our case, we take $x=10$. The pattern works as soon as
$j+1\leq 9$ for $j=0,\dots,n$ AND $2n-j+1\leq 9$ for $j=n+1,\dots,2n$ where $9$ is the largest decimal digit, that is for $n\leq 8$.

For $n=8$ and $x=10$ we have that
$$(1+x+x^2+\dots +x^n)^2=111111111^2=12345678987654321$$
whereas for $n=9$ we get
$$(1+x+x^2+\dots +x^n)^2=1111111111^2=1234567900987654321.$$

Title: Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$

Body:
I have recently learned about pattern that goes like this:

$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$

It is a very cool pattern, but after a bit it stops:

$$1,111,111,111^2 = 1,234,567,900,987,654,321$$

My main question:

Why does this pattern work?

A side question that's less important:

Is there an algebraic equation to describe this pattern?


答:
Set $z=3e^{it};\;t\in[0,2\pi]$ and $dz=3 i e^{i t}dt$

The integral becomes
$$\int_0^{2\pi}\frac{3 i e^{i t}\,dt}{\sin^2\left(3e^{it}\right) }=\left[-\cot \left(3 e^{i t}\right)\right]_0^{2\pi}=-\cot 3+\cot 3=0$$

Title: Compute the curve integral $\int_{|z|=3}\dfrac{dz}{\sin^2 z}$ [trouble with limit]

Body:
My attempt is as follows;

The region $\Omega$ which is enclosed by the curve $\gamma = \{z \in \mathbb{C}:|z|=3\}$ contains one singularity at $z=0$, a pole of order 2. Therefore, we may use the residue theorem which yields
$$\int_{\gamma}\dfrac{dz}{\sin^2z}= 2\pi i\, \text{Res}_f(0)\,.$$

When computing the residue according to definition for pole of order $k=2$, I arrive at
\begin{align*}
\text{Res}_f(0) &= \dfrac{1}{(2-1)!}\lim_{z \rightarrow 0} \dfrac{d}{dz}\left[\dfrac{z^2}{\sin^2(z)}\right]\\[5pt]
&=\lim_{z \rightarrow 0} \dfrac{2z[\sin z-z\cos z]}{\sin^3z}.\\[4pt]
\end{align*}
According to the solution manual, this limit is 0. I am unable to conclude this myself, so I was wondering if anyone here could provide a hint?

Many thanks in advance.


Body:
$$\lim_{z \to 0} \dfrac{2z[\sin z-z\cos z]}{\sin^3z}
=\lim_{z \to 0} \dfrac{2z\left[\left(z-\dfrac{z^3}{3!}+O(z^5)\right)-z\left(1-\dfrac{z^2}{2!}+O(z^4)\right)\right]}{z^3-O(z^5)}=\lim_{z \to 0}\frac z3.$$


问:
What are these special functions?
I'm writing the documentation for my esoteric language Whispers, which specialises in mathematical functions. I've come across the following function which I'm not sure exactly which special function it is, and I can't remember where I found the definition. The code is

```<code>def Lc(z):
    return z*ln(2*sin(z*π)) + S(τ*z, 2) / τ

def S(a, b):
    f = lambda k: math.sin(k * a) / k ** b
    return Σ(f, [1, math.inf])
</code>```

which, in $\LaTeX$ is

$$\text{Lc}(z) = z\ln(2\sin(z\pi)) + \frac{S(\tau z, 2)}\tau \\
S(x, 2) = \sum_{k=1}^\infty \frac{\sin(kx)}{k^2}\\
\text{Lc}(z) = z\ln(2\sin(z\pi)) + \frac{1}\tau \sum_{k=1}^\infty \frac{\sin(\tau kz)}{k^2}$$

Looking through lists such as this or this haven't yielded anything.

$S$ appears to be a related function/extension of the $\text{sinc}$ function, but again, googling around hasn't found anything.

Does anyone recognise either $\text{Lc}(z)$ or $S(a, b)$?

I do also have the derivatives and anti-derivatives of both the functions, and the full source code of them can be found here ($\text{Lc}$) and here ($S$)

答: For $S(z, 2)$ have a look at the dilogarithm. For other $S(z, b)$, the polylogarithm.

Title: What are these special functions?


问:
What should be $W$ in $I_W$ when choosing randomly uniformly a point $(x,y)$ inside triangle?
We choose randomly uniformly a point $(x,y)$ inside triangle that it's points are $(0,0)$, $(0,1)$ and $(1,0)$. So we can say that:
$$
f_{X,Y}(x,y)=\frac{1}{S_B}=\frac{1}{0.5}I_{\{0<y<1,y<1-x,0<x<1\}}=2I_{\{0<y<1,y<1-x,0<x<1\}}=2I_{\{0<x<1,0<y<x-1\}}
$$
where $S_B$ is the area of the triangle.

I'm trying to build myself a formula of the indictor $I$ for any three points $(a_1,b_1)$, $(a_2,b_2)$ and  $(a_3,b_3)$. What should be $W$ in $I_W$ if we replace the three points above general $(a_1,b_1)$, $(a_2,b_2)$ and  $(a_3,b_3)$ points?

答:
By contracting and sliding one edge of a triangle towards its opposing vertex, we can see that this covers the triangle interior completely, and that every point in a triangle can be uniquely referenced by two of its sides as all the triangles are similar.

Define a parameter for each edge, say $\alpha, \beta$, and then:

$$W=\{w=(x,y)+\alpha v_1+\beta v_2 : 0\le \alpha,\beta\le 1\}$$

where $v_1,v_2$ are the vector edges, and $(x,y)$ is their point of intersection.


答: $S(a,\,b)=\sum_{k\ge1}\frac{\sin(ka)}{k^b}$ is a generalized Clausen function, typically denoted $S_b(a)$. In terms of the Clausen function $\operatorname{Cl}_2=S_2$, $\operatorname{Lc}(z)=\tfrac{1}{\tau}\operatorname{Cl}_2(z\tau)-z\operatorname{Cl}_2^\prime(z\tau)$.

Title: What are these special functions?

Body:
I'm writing the documentation for my esoteric language Whispers, which specialises in mathematical functions. I've come across the following function which I'm not sure exactly which special function it is, and I can't remember where I found the definition. The code is

```<code>def Lc(z):
    return z*ln(2*sin(z*π)) + S(τ*z, 2) / τ

def S(a, b):
    f = lambda k: math.sin(k * a) / k ** b
    return Σ(f, [1, math.inf])
</code>```

which, in $\LaTeX$ is

$$\text{Lc}(z) = z\ln(2\sin(z\pi)) + \frac{S(\tau z, 2)}\tau \\
S(x, 2) = \sum_{k=1}^\infty \frac{\sin(kx)}{k^2}\\
\text{Lc}(z) = z\ln(2\sin(z\pi)) + \frac{1}\tau \sum_{k=1}^\infty \frac{\sin(\tau kz)}{k^2}$$

Looking through lists such as this or this haven't yielded anything.

$S$ appears to be a related function/extension of the $\text{sinc}$ function, but again, googling around hasn't found anything.

Does anyone recognise either $\text{Lc}(z)$ or $S(a, b)$?

I do also have the derivatives and anti-derivatives of both the functions, and the full source code of them can be found here ($\text{Lc}$) and here ($S$)


问:
Conditional probability in Galton-Watson
In discussing Galton-watson chains one considers, for a given generation $n$, iid $\nu$-distributed random variables $x_1^n,\ldots,x_k^n$, where $\nu$ is a probability measure on $\mathbb{N}$. At some point in the discussion the following equality is used for a fixed positive integer $k$
$$E[\mathbb{1}_{x_1^n+\cdots+x_k^n=k}|\sigma(x_2^n,\ldots,x_n^n)]=\nu(k-x_2^n + \ldots +x_n^n)$$
and I am failing to understand why this equality holds. This is not really specific to Galton Watson, it was just to give a reference of where this comes from.

答:
An argument based on conditional distributions goes as follow. Let $(P_x)$ be a regular conditional distribution of $Y$ given $X$. Then
\begin{align*}
\mathbb{E}[h(X,Y) \, | \, X = x]
=
\int h(x,y) \, P_x(\mathrm{d}y)
\end{align*}
cf. Theorem 2.2.3 in the following excellent lecture notes: http://web.math.ku.dk/noter/filer/beting.pdf. (Similar results also appear in standard textbook references on the subject.) If $X$ and $Y$ are independent, then one may take $P_x$ to be $Y(\mathbb{P})$, and
\begin{align*}
\mathbb{E}[h(X,Y) \, | \, X = x]
=
\int h(x,y) Y(\mathbb{P})(\mathrm{d}y)
=
\mathbb{E}[h(x,Y)].
\end{align*}
Thus $\mathbb{E}[h(X,Y) \, | \, X] = \mathbb{E}[h(x,Y)]\rvert_{x=X}$. Alternatively, one may show that the latter expression satisfies the conditions of being the desired conditional expectation.

The result now follows by choosing $X$, $Y$, and $h$ appropriately, as we discussed in the comments.


问:
How can I verify if a linear system has a solution
Given $\vec{a}, \vec{b} \in \mathbb{Z}^d$ and a system of two-equation:
$$
\begin{cases}
\langle \vec{a}, \vec{x} \rangle = 0 \\
 \langle \vec{b}, 1-\vec{x} \rangle = 0
\end{cases}
$$
where $\vec{x} \in \{0, 1\}^d$. Here $\langle,\rangle$ refers to inner product and $1-\vec{x}$ referrs to all 1 vector minus $\vec{x}$.

How can I verify such a system exists a solution for $\vec{x}$?

If there are solutions, how can I enumerate all of them?

答:
For small $d$ you can of course enumerate all $x\in \lbrace 0,1 \rbrace^d$ and try them all in the system of equations.

More generally, to determine if a solution exists you can treat the problem as a binary integer program with a trivial objective function (minimize 0 subject to your two equations) and solve it with any MIP solver. One (time-consuming) way to enumerate all solutions is to solve the IP model, add a constraint prohibiting the solution you got (and only that solution), and repeat. To eliminate a solution $x=\hat{x}$, you add the constraint $$\sum_{i:\hat{x}_i=0} x_i + \sum_{j:\hat{x}_j=1} (1-x_j) \ge 1.$$


问:
Variable transformation in evaluating a double integral
Evaluate

$$\iint_R(x+y)\,dA$$

Where $R$ is the trapezoidal region with vertices at $(0,0),(5,0),(5/2,5/2)$ and $(5/2,−5/2)$, using the transformation $x=2u+3v$ and $y=2u−3v$.

Approach:

The Jacobian comes out to be $12$, and the region in the $u−v$ plane consists of the points $(0,0),(5/2,0),(5/2,5/6),(0,5/6)$ which is a rectangle, and therefore, the double integral $\iint_{R'}(4u)[J\,du\,dv]$ is simply $$48\int_{0}^{5/6}\int_{0}^{5/2}u\,du\,dv= 125.$$
However, the answer given is $125/4$.

To kind of "check" where my mistake is, I came up with a small test:

By putting $f=1$ in the variable transformation formula for double integrals, we get:
$$Area(R)=J\times Area(R')$$

However, in my case, area of the original region is $25/2$, whereas $12$*(area of the transformed region) is $25$.

So have I messed up in drawing the transformed region? Or somewhere else?

答:
Your work is correct except for the points of the new region:

$$\begin{cases}
2u + 3v = 5/2\\
2u - 3v = 5/2
\end{cases}$$
yields the point $(5/4, 0)$ and
$$\begin{cases}
2u + 3v = 5\\
2u - 3v = 0
\end{cases}
$$
gives the point $(5/4, 5/6)$.

With these new points:

$$48\int_{0}^{5/6}\int_{0}^{5/4}u \ dudv=\frac{125}{4}$$


问:
Geometric Sequence Proof
Problem:

(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.

(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.

My Solution for $a)$:

$a)$ First, we can analyze the case when $r \geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.

I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?

答:
Actually, for (b) any nonnegative integer $r$ works. If you want exactly $r$ terms to be integers, take $2^{r-1}, 2^{r-2}, 2^{r-3},..., 2^2, 2, 1, \frac{1}{2}, ...$

example: $r=3$

$\text{ratio}=\frac{1}{3}$

we get $2^{3-1}, 2^{3-2}, 2^{3-3}$, and those are all the integer terms.

I'm only an 8th grader so there may be a flaw in my thinking, if so please correct me!

Title: Geometric Sequence Proof


问:
Can there be a different Eigen vector for a particular Eigen value?
Please see the photo.

Here, my answer came $k$
$\begin{bmatrix}
                   1\\
                 1\\
                   -1
\end{bmatrix}$

But their answer is given :
$k$
$\begin{bmatrix}
                     1\\
                    -1\\
                     1
\end{bmatrix}$
My answer is correct?? Please tell--

THE ACTUAL MATRIX]

答:
Sure.  An eigenvalue can be associated with multiple eigenvectors.

For example:  Let $$A=\pmatrix{1&0&0\\0&1&0\\0&0&1}.$$

The eigenvalue $\lambda=1$ has all non-zero vectors as eigenvectors.


问:
Motivation behind Lambda Calculus?
The lambda calculus provides a formalism broad used in theoreretical cs to write functions without
giving them explicit names, it declares anonymous functions.
That is at first glance it's just an alternative
form of function notation to write $\lambda$.$f(x)$ instead of $f(x)=x$.
But I don't see the reason why this lambda calculus is so important and
provides a better way to deal with functions for functional programming
languages and from viewpoint of theoretical computer science instead of convential "mathematical"
notation $f(x)=x$?

The techniques like curring work fine also with later notation, so what what are
the main reasons why lambda calculus is prefered?

答:
Specifically for the question about functional programming, I'd note that lambda calculus is essentially a direct formalization of computation for higher order functions using syntax that is reasonable/desirable to actually use. First-class functions are highly desirable when programming, and once you have them, a 'function literal' is a natural thing to want, which leads to something like λ.

So, from this perspective, λ-calculus is exactly what you want for thinking about programmatic functions, and most other things are inferior. In lambda calculus, you imagine the function just is syntax you wrote, and the way it computes is just by simple rewriting rules that give you other things you might have actually written. Actual programming languages will do something more optimal, but this mental model is pretty close in a lot of ways. By contrast:

Set theoretic encodings tell you that a function is a set of ordered pairs (of sets, probably), and applying a function is just like a lookup table. The actual thing you wrote has been discarded, and probably some notion of computation has to be added back in to explain what computing actually happens.

Turing machines require you to first translate the syntax you'd like to write into a state machine operating on symbol strings, which is not at all easy. Then, the machines are a mathematical construction, and to model higher-order programs, one must have a further level of encoding of machines as symbol strings, and the details of all this are far removed from reasoning about the original program.

μ-recursive functions operate on natural numbers, so the situation is similar to Turing machines, where higher-order programs require a baroque encoding of syntax, and the way the encoding operates may not be obviously related to the original specification.

SKI combinators check most of the same boxes as λ-calculus, but as mentioned in the comments, writing programs exclusively in combinators is often not nice, and not what people actually want to do.

...

That's not to say these other models of computation are worthless. However, when considering a formal, rigorous basis for a programming language based on higher-order functions, lambda calculus or something like it (some variety of type theory) is a pretty natural choice. It gives you exactly what you want with no unnecessary cruft.

Title: Motivation behind Lambda Calculus?


问:
Really confused with unit vectors

Let $\mathbf{u}$ and $\mathbf{v}$ be linearly independent unit vectors. Find the set of all possible values of $\mathbf{u}\cdot\mathbf{v}$. Give your answer in interval notation.

What is the maximum possible length of the vector resulting from the following linear combination?
$$
\frac{1}{\| \mathbf{v_1} \|} \,\mathbf{v_1} + \frac{1}{\|\mathbf{v_2}\|} \,\mathbf{v_2} + \cdots + \frac{1}{\| \mathbf{v_n} \|} \,\mathbf{v_n}
$$

So far I have gotten nowhere with this problem and I cant even look at a similar problem in the textbook. If someone can help me i would really appreciate it.

答:
For the first problem, this formula is very useful: $u \cdot v = |u||v|\cos(\theta)$, where $\theta$ is the angle between the vectors.

For the second problem, call this combination $V$.  The length is given by
$$|V| = \sqrt{V \cdot V}$$.  Expand the dot product like you would with a polynomial.  The self-similar terms will give $\frac{1}{||v_i||^2}v_i \cdot v_i = 1$, while the cross terms will give $\frac{v_i \cdot v_j}{|v_i||v_j|} = \cos(\theta_{ij})$.  Now, how can we maximize these angles?

Title: Really confused with unit vectors


问:
Solving a system of piecewise linear equations with multiple variables in minimum function

General problem

I am facing a problem from the transportation field. For a given network, the maximum flow $q$ on a link is either the sum of the flows from upstream links, entering at the upstream intersection, or the link capacity $c$. Turning ratios $\alpha$ and the link capacity $c$ are assumed to be known.

I consider an example network as shown in the figure attached (see Example network). Only one turning ratio is shown, but they exist for all possible relations.

I have the following system of equations:
$$q_{11} = c$$
$$q_{12} = c$$
$$q_{13} = c$$
$$q_{21} = \min\left(q_{11}\cdot(1-\alpha_{12}) + q_{12}\cdot\alpha_{21}, c\right)$$
$$q_{22} = \min\left(q_{11}\cdot\alpha_{12} + q_{12}\cdot(1-\alpha_{21}), c\right)$$
$$q_{23} = \min\left(q_{13}\cdot(1-\alpha_{31}) + q_{21}\cdot\alpha_{13}, c\right)$$
$$q_{31} = \min\left(q_{13}\cdot\alpha_{31} + q_{21}\cdot(1-\alpha_{13}), c\right)$$
$$q_{32} = \min\left(q_{22}\cdot(1-\alpha_{23}) + q_{23}\cdot\alpha_{32}, c\right)$$
$$q_{33} = \min\left(q_{22}\cdot\alpha_{23} + q_{23}\cdot(1-\alpha_{32}), c\right)$$

I could solve this by inserting $q_{11}$ and $q_{12}$ into the equation for $q_{21}$, etc. However, for large networks this becomes infeasible. Thus, I am looking for other ways to solve this. So far, I have explored the following potential solution techniques:

Max-plus algebra:

I tried to apply techniques from max-plus algebra by converting the min-operator to a max-oprerator. However, since I have two variables in the minimum operation, this becomes non-linear in the max-plus algebra (if I am not taken wrong). For example:
$$ - q_{21} = (- q_{11}\alpha_{12} \otimes - q_{12}(1-\alpha_{21})) \oplus -c$$
Since these are my first steps into max-plus algebra, I am very much open to any comments on this. So far, I did not find any commonly applied solutions methods for a system of non-linear max plus algebra equations.

Mixed-integer linear program:

I could rewrite each eq. to a set of constraints, for example:
$$q_{21} = \min\left(q_{11}\cdot(1-\alpha_{12}) + q_{12}\cdot\alpha_{21}, c\right)$$
$${q}_{21} \leq q_{11}\cdot(1-\alpha_{12}) + q_{12}\cdot\alpha_{21} $$
$${q}_{21} \leq c $$
$${q}_{21} \geq y c + (1-y) (q_{11}\cdot(1-\alpha_{12}) + q_{12}\cdot\alpha_{21})  $$
$$y \in [0,1] $$

However, I do not know what the objective function then is. Could it be an maximization/minimization of the sum of all flows?

Any thoughts, hints or other help on this is highly appreciated!

答:
Your MIP model is close, but there are two things to fix. First, the domain for $y$ should be $y\in\lbrace 0,1\rbrace$, not $y\in[0,1]$. Second, you want to avoid multiplying $y$ and any expressions containing $q$ variables, as the relaxation is quadratic rather than linear.

To save me some typing, let $f=q_{11}\cdot(1-\alpha_{12}) + q_{12}\cdot\alpha_{21}$. To linearize the product with $y$, you need an a priori upper bound $M$ for $f$. The constraints become:
\begin{align*}
q_{21} & \le f\\
q_{21} & \le c\\
q_{21} & \ge f-My\\
q_{21} & \ge cy.
\end{align*}

As far as the objective function goes, if you are looking for any feasible solution, just minimize (or maximize) 0 subject to your constraints. Minimizing or maximizing the sum of all flows will also work. If you are looking for a particular solution, such as a solution with maximum overall flow, or a solution with maximum flow into (out of) certain nodes, then you will want to customize the objective function to get that solution.


问:
Intuitive explanation for negative binomial expansion
I understand the relationship between Pascal's triangle and binomial expansion for nonnegative integer powers by using combination e.g the binomial coefficient for $x^3$ in $(a + x)^8$ is equivalent to how many ways there are to choose 3 $x$'s from 8 possibilities.

I thought this idea would fail with regards to negative binomial expansion but I find out we can have cases where $r<0$ in $nCr$.

Can someone intuitively explain this? I kind of understand binomial expansion using  Maclaurin series but I can't wrap my head around what negative combination is intuitively and why it works for negative binomial expansion.

Basically I don't understand the reason behind the combination formula still working for negative numbers.

答:
$(1+x)^{-n}$ can also be written as $\frac{1}{(1+x)^n}$.

The denominator can be expanded and then viewed as the usual identity $$\frac1{1-x}=1+x+x^2+\dots$$

For example with $n=2$:

$$1+(-2x-x^2)+(2x+x^2)^2+(-2x-x^2)^3+\dots$$

This can have the like terms collected, which gives:

$$\binom{-2}{1}=-2$$
$$\binom{-2}{2}=3$$

etc...


问:
Understanding Bayes' Theorem
I worked through some examples of Bayes' Theorem and now was reading the proof.

Bayes' Theorem states the following:

Suppose that the sample space S is partitioned into disjoint subsets $B_1, B_2,...,B_n$.  That is, $S = B_1 \cup B_2 \cup \cdots \cup B_n$, $\Pr(B_i) > 0$ $\forall i=1,2,...,n$ and $B_i \cap B_j = \varnothing$ $\forall i\ne j$.  Then for an event A,

$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$

The numerator is just from definition of conditional probability in multiplicative form.

For the denominator, I read the following:

$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$

Now this is what I don't understand:

The sets $A \cup B_i$ are disjoint because the sets $B_1, B_2, ..., B_n$ form a partition.$\tag{$\clubsuit$}$

I don't see how that is inferred or why that is the case.  What does B forming a partition have anything to do with  it being disjoint with A.  Can someone please explain this conceptually or via an example?

I worked one example where you had 3 coolers and in each cooler you had either root beer or soda.  So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda.  But I don't see why these would be disjoint.  If anything, I would say they weren't disjoint because each cooler contains both types of drinks.

Thank you in advance! :)

答: Here is a write that describes Bayes theorem in detail along with a bunch of examples on how to use it (different write ups)</p>

Title: Understanding Bayes' Theorem


问:
Example of Complex Closure, Interior and Boundary
I came up with a problem in Alphors that asks the following:
Given $X$ as the set of complex numbers with real and imaginary parts rational, what is, the Interior, the Bounday and the Closure.

If possible I would like examples to add, like other sets also, to comprehend better.

答:
Here we consider the subset $X=\mathbb{Q}+i\mathbb{Q}\subseteq\mathbb{C}$, where $\mathbb{C}$ is equipped with the standard topology.

We know that $\mathbb{Q}$ is dense in $\mathbb{R}$ with the standard topology on $\mathbb{R}$ which means that the closure $\overline{\mathbb{Q}}=\mathbb{R}$.
Similarly we have here the closure of  $\mathbb{Q}+i\mathbb{Q}$ is $\mathbb{C}$.
\begin{align*}
\color{blue}{\overline{\mathbb{Q}+i\mathbb{Q}}=\mathbb{C}}
\end{align*}

We observe that for any point $z\in\mathbb{Q}+i\mathbb{Q}$ and any positive real number $r>0$ the ball $B(z;r)$ with center $z$ and radius $r$ contains points from $\mathbb{Q}+i\mathbb{Q}$ as well as points from the complement $\mathbb{C}\setminus\left(\mathbb{Q}+i\mathbb{Q}\right)$. This means the set is equal with its boundary.
\begin{align*}
\color{blue}{\mathbb{Q}+i\mathbb{Q}=\partial\left(\mathbb{Q}+i\mathbb{Q}\right)}
\end{align*}

It follows that there are no interior points in $\mathbb{Q}+i\mathbb{Q}$:
\begin{align*}
\color{blue}{\left(\mathbb{Q}+i\mathbb{Q}\right)^{\circ}=\emptyset}
\end{align*}


问:
Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$

Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$.

I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$, and proceed with the simplification. Finally I am getting it as $\cos \left(x - \frac {\pi}{4}\right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!

答:
$$\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}$$ It is as simple as that. This is followed by the substitution $v(x)=\frac{\sin x+\cos x}{\sqrt{2}}$. $$ \begin{align} \frac{d}{dx}\arcsin(v(x)) & =\frac{1}{\sqrt{1-v^2(x)}}v'(x) \\
&= \frac{1}{\sqrt{1-\Biggl(\frac{\sin x+\cos x}{\sqrt{2}}\Biggl)^2}}\Biggl(\frac{\cos x-\sin x}{\sqrt{2}}\Biggl) \\
&= \frac{\sqrt{2}}{\sqrt{(\cos x-\sin x)^2}}\frac{\cos x-\sin x}{\sqrt{2}} \\
&= \frac{\cos x-\sin x}{|\cos x-\sin x|} \\
&= \frac{\cos(x+\frac{\pi}{4})}{|\cos(x+\frac{\pi}{4})|} \\  \end{align}$$


问:
Grouping vertices of a graph
Suppose $G$ is a graph with $4n$ vertices. Prove that one can partition the vertices into two groups with $2n$ vertices, such that that there is an even amount of edges between the two groups.

How can one approach this problem? Induction doesn't seem to work and it is just so strange.

答:
Make a partition $A,B,C,D$ of all vertices, so that $|A|=|B|=|C|=|D|=n$.

Now let the number of edges between

$A$ and $B$ be $a$,

$B$ and $C$ be $b$,

$C$ and $D$ be $c$,

$D$ and $A$ be $d$,

$A$ and $C$ be $e$,

$B$ and $D$ be $f$.

So the number of edges between $A\cup B$ and $C\cup D$ is  $$b+d+e+f =:p_1 $$

the number of edges between $A\cup D$ and $C\cup B$ is  $$a+c+e+f=:p_2$$

and the number of edges between $A\cup C$ and $B\cup D$ is $$a+b+c+d=:p_3$$

Now suppose the statment doesn't hold. Then all those sums $p_1,p_2,p_3$ must be odd and thus their sum $p_1+p_2+p_3$ is odd. But this is equal to $$2a+2b+2c+2d+2e+2f$$ which is even. A contradicton, so the statment does hold.

Title: Grouping vertices of a graph


问:
Is map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$ surjective on a hemisphere?
I have a map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$. It seems intuitve believe that $f(\mathbb{D}^2)=\mathbb{S}^2_+$ or $f(\mathbb{D}^2)=\mathbb{S}^2_-$ (hemispheres north and south), but I could not to prove it.

Could you give me a clue?

Thank you

答:
Suppose there are two points not in the image of $f$, one on the North emisphere and one on the South emisphere. Removing the two points, up to homeomorphism you get a map

$$\hat{f} : D^2 \to S^1 \times \mathbb{R}$$

That maps the boundary to a generator of the fundamental group on the right. This is impossible, because an element of the $\pi_1$ extends to a disk precisely when it is trivial.

On the other side, it is possible to have a surjective map with the constraints you prescribed. Firstly, let me notice there is a surjective function $g: D^2 \to S^2$ that maps the boundary to the South Pole: this is kind of "put a disk on a sphere and then collapse the boundary to the South Pole. A similar one can be done with North Pole, that we call $g_1$.

Now define $f: D^2 \to S^2 $ as:

$$ f(x) = g(2x) ,\quad \|x \| \le 1/2$$
$$ f(x) = g_1 \left ( \frac{x}{\| x \| } \left (\|x\| - \frac{1}{2} \right ) \right ) , \quad 1/2 \le  \|x\| \le 1 $$

You can verify that it satisfy your constraints. Use yous imagination!


问:
Line along shortest distance between two skew lines
We have points $A(1,-1,1),B(6,-3,1),C(2,-1,5),D(5,1,1)$

We also have a line $p$ that goes through the edges $AD$ and $BC$ perpendicularly.

Find the equation of this line:
My try:

$$AD = (4,2,0)$$
$$BC = (-4,2,4)$$
$$s = (s_1,s_2,s_3)$$

Then we know that(*...scalar product):
$$s*AD = 0$$
$$s*BC = 0$$
From this two equations we get: $$4s_1+2s_2+0s_3 = 0 = -4s_1+2s_2+4s_3 $$
$$ 4s_1+2s_2+0s_3 = 0\\-4s_1+2s_2+4s_3 = 0\\
 \text{We add both equations together }\implies 4s_2 + 4 s_3 = 0 \implies 4s_2 = -4s_3\implies s_2 = -s_3$$
Then we get:
$$4s_1 = 4s_3\implies s_1 = s_3$$
However, when I plug this back into the equation I get zero at the end, which is not correct.

The solution is $p:x - 3 = \frac{y}{-2} = \frac{z-1}{2}$

答:
Here is another way you can do it -

Points on line $BC$ is given by $(6-2s, -3 + s, 1 + 2s)$ and points on line $AD$ is given by $(1 + 2t, t - 1, 1)$

For some $s$ and $t$, there is a line $p$ that passes through those points on the lines (say $M$ on AD, $N$ on $BC$) perpendicular to both $AD$ and $BC$.

$\vec {MN} = (5-2s-2t, -2 + s - t, 2s)$

$(5-2s-2t, -2 + s - t, 2s) \cdot (2,1,0) = 0 \implies 3s + 5t = 8$

Similarly, $(5-2s-2t, -2 + s - t, 2s) \cdot (-2,1,2) = 0$

$\implies 3s + t = 4$

So $s = t = 1$

Plugging in $\vec{MN}$, we have vector of line $p$ as $(1, -2, 2)$

Point $M$ on line $AD = (3, 0, 1)$

So equation of line $p = (3,0,1) + (1, -2 ,2) t$

or $\displaystyle \frac{x-3}{1} = \frac{y}{-2} = \frac{z-1}{2}$


答:
In barycentric coordinates
a point internal to the triangle (or any other convex polygon) with vertices ${\bf v}_k$
is given by the weighted average of the verices with positive cooefficients
$$
\eqalign{
  & {\bf p} = \left( {\matrix{   x  \cr   y  \cr  } } \right)
 = {{a{\bf v}_1  + b{\bf v}_2  + c{\bf v}_3 } \over {a + b + c}}\quad \left| {\;0 \le a,b,c} \right. =   \cr
  &  = \alpha {\bf v}_1  + \beta {\bf v}_2  + \left( {1 - \alpha  - \beta } \right){\bf v}_3  =   \cr
  &  = \alpha \left( {{\bf v}_1  - {\bf v}_3 } \right) + \beta \left( {{\bf v}_2  - {\bf v}_3 } \right) + {\bf v}_3
  = \quad \left| \matrix{
  \;0 \le \alpha ,\beta \left( { \le 1} \right) \hfill \cr
  \;\left( {0 \le } \right)\alpha  + \beta  \le 1 \hfill \cr}  \right. \cr}
$$
So $\alpha , \beta$ are the coordinates inside a right triangle with unit sides.

Since
$$
\left( {\matrix{
   {dx}  \cr
   {dy}  \cr
 } } \right) = \left( {\matrix{
   {x_1  - x_3 } & {x_2  - x_3 }  \cr
   {y_1  - y_3 } & {y_2  - y_3 }  \cr
 } } \right)\left( {\matrix{
   {d\alpha }  \cr
   {d\beta }  \cr
 } } \right)
$$
then the Jacobian is constant and non null if the vertices are not aligned.

Therefore a uniform distribution of $(\alpha , \beta)$ will translate in a uniform
distribution of $(x,y)$.

Title: What should be $W$ in $I_W$ when choosing randomly uniformly a point $(x,y)$ inside triangle?

Body:
We choose randomly uniformly a point $(x,y)$ inside triangle that it's points are $(0,0)$, $(0,1)$ and $(1,0)$. So we can say that:
$$
f_{X,Y}(x,y)=\frac{1}{S_B}=\frac{1}{0.5}I_{\{0<y<1,y<1-x,0<x<1\}}=2I_{\{0<y<1,y<1-x,0<x<1\}}=2I_{\{0<x<1,0<y<x-1\}}
$$
where $S_B$ is the area of the triangle.

I'm trying to build myself a formula of the indictor $I$ for any three points $(a_1,b_1)$, $(a_2,b_2)$ and  $(a_3,b_3)$. What should be $W$ in $I_W$ if we replace the three points above general $(a_1,b_1)$, $(a_2,b_2)$ and  $(a_3,b_3)$ points?


问:
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?
Where does the function $f(x) = \frac{2x}{x - 7}$ have an increasing slope?

$a. x \le 0, x > 7$

$b. x<7$

$c. x > 7$

$d. x \in \Bbb R, x \neq 7$

This question is from a test of mine in a pre-calculus course (so no calculus allowed in answering the question).I have no idea as to how to solve this problem. I can tell where the function is negative and where it is positive, but that's about it, and I'm fairly sure that's no use here.

Any ideas?

答:
$$\frac{2x}{x-7} = \frac{2x-14+14}{x-7} =2 + \frac{14}{x-7}.$$
Are you allowed to use basic properties of $\frac{1}{x}$?</p>