问:
Modify the rules of Gomoku (Five-in-a-row) or Connect Four type games to enforce the fairness among players
One colleague and me were discussing this problem during lunch today, and I did a little bit digging for several hours after returning to my office.

Fact: For an $(m,n,k)$-game, there does not exist a strategy that assures that the second player will win. For example, in Five-in-a-row, first player wins with perfect play.

Now our question is: How to modify the rules to make the game fair for both players?

I googled a bit more, found an MSE question that addresses the fairness of this kind of games.

I think the fairness in that question can be strenghthen by the following definition: A game is fair if and only if that two players play with perfect strategy, the game will always be a draw. This is fair in that this game favors player with less mistakes. For example, Tic-tac-toe is fair.

Several proposals to modify the rules to enforce the fairness:

Impose more restrictions on the player who plays first. For example, Five-in-a-row bans black to play "three and three", "four and four", and "overlines".

Change the $(m,n,k)$-game to an $(m,n,k,p,q)$-game: $k$-in-a-row on an $(m\times n)$-board, first player put $p$ stones on board, in subsequent moves, players put $q$ stones on board. For example, [Connect 6].

Go to higher dimension. For example, Five-in-a-row played in a $19\times 19\times 19$ "board".

I am no expert in combinatorial game theory or computational complexity in board games, but it is always nice to learn some new things in summer time. My question is: Any papers or treatises dealing the fairness of this kind of games? or more specifically, any proof of the fairness using above three modifications of rules? especially, for Five-in-a-row, do those additional rules make the game fair?

Any analysis of fairness on simpler cases like Connect 4 is welcome too.

Lastly just out of curiousity, as an avid Go player myself, I wonder is there any mathematical analysis on addressing the fairness using komi in the Go game? Or Go is just too complex to analyze...

答:
Going to higher dimension will not help; it makes the problem worse.  The Hales-Jewett theorem implies that for any fixed $p$ and $n$, the first player can win the $p$-in-a-row game on the $n^d$ board for all sufficiently large $d$.  So adding dimensions makes these games less fair, not more fair.

For example:

The first player obviously wins 2-in-a-row on a $2^n$ board for $n>1$, but when $n=1$ he cannot do better than a tie.

Standard $3\times 3$ tic-tac-toe is an easy tie.  The first player easily wins tic-tac-toe on a $3^n$ board for $n \gt 2$.

The 4-in-a-row game on a 4×4 board is tied, by an easy strategy.  The 4-in-a-row game on a 4×4×4 board is a first-player win, but the strategy is complicated enough to be a good game anyway.  The 4-in-a-row game on a $4^n$ board for $n\gt 3$ is an easy first-player win with straightforward play.


问:
Proof that if: a^2=b^2 then a=b or a=-b
I have to prove that if a^2=b^2 then a=b or a=-b, and I came up with the following and I am wondering if it is valid use of absolute values?:
a^2=b^2 |square root of both sides

|a|=|b| |particularly interested in whether or not this next step is valid:

a=b or a=-b

Thank you in advance.

Edit: I do understand that perhaps easier way to prove is the following:
a^2=b^2

a^2-b^2=0

(a+b)=0 or (a-b)=0

a=-b or a=b

But I am interested in whether or not my prior proof was valid.

答:
Overkill

Consider the quadratic equation

$x^2-b^2=0;$

Two real solutions

$x_{1,2}=\sqrt{b^2}=\pm|b|;$

$x_1=|b|;$ $x_2=-|b|;$ which for

1)$b\ge 0:$ $x_1=b;$ $x_2 =-b;$

2)$b< 0:$ $x_1=-b;$ $x_2=b;$ or

with the original $a:$ $a=\pm b;$

ContentLicense: CC BY-SA 4.0

Title: Proof that if: a^2=b^2 then a=b or a=-b


问:
Relation between continuity and differentiability
I have noticed that authors of introductory analysis textbooks always show that a function f is continuous at a point before trying to prove that it is differentiable there. Ahlfors, for instance, does this in "Complex Analysis: An Introduction to The Theory of Analytic Functions of One Complex Variable" when he proves the extended Cauchy's integral formula by induction:
Check here

Bartle does the same when proving the second form of the Fundamental theorem of calculus in "Introduction to Analysis". Is it always necessary to prove continuity at a point before showing that a function is differentiable there? Why? (Please provide a technical explanation)

答:
Being differentiable in a point is stronger than being continuous in that point (this means that being differentiable in a point implies being continuous in that point)
Hence it is just make sense to start from the simpler attribute.
Moreover, (maybe someone will correct me here),
as being differentiable means (at least in the common cases) that the following exists:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
then, the logical requirement is that when $h$ goes to $0$ you want $f(x+h)$ to become close to $f(x)$, and that is continuity.


问:
Boundaries in Spaces where Quasicomponents and Components Coincide
Let's call a space $X$ geometric if its components and quasi-components coincide.  Let's also define a property called the boundary bumping property:

$X$ has the boundary bumping property ("bbp") if for  every proper, closed subset $A$ and a component $C$ of $A$, we have $C \cap \partial(A) \neq \varnothing$.

It is clear that $X$ must be geometric to have the boundary bumping property by the definitions.  It is known that if $X$ is a compact, connected Hausdorff space then it has the bbp; if $X$ is a connected and locally connected $T_1$ space then it, too, has the bbp.  Both are elementary but a bit long; I think their proofs are floating around on MSE if you want to see them.  It's also clear that the space needs to be connected, just by definition of separation.

QUESTION: 1) What might be some sufficient conditions for a connected metric space to have the bbp, besides locally connected or compact?  2) What are necessary conditions?  3) Classify connected, geometric metric spaces with the bbp.

DISCUSSION: Taking a comb with the bottom half of the 'limit tooth' lopped off gives a different sort of counterexample, though not compact.  By renormalizing the metric this shows that topological completeness doesn't get you anywhere in the non-compact case.

When $X$ is not locally connected, it may still have the bpp without being compact: Stick a two-sided topologist's sine curve in the middle of $\mathbb{R}$ and we'll get a space that has the bbp but not be compact or locally connected.  Removing a point from the critical limit arc will still have the bbp (check it), and this space won't have an exhaustion by continua.

If you take a 'necklace' of pseudo-arcs (it's a continuum that's not locally connected anywhere) attaching one to the next at a single point, then it will have the bbp but be locally connected nowhere; for this one just trust me, if you're not already familiar with the pseudo-arc.  Also note that $\mathbb{R}^\omega$ is connected, locally connected and $T_1$ so has the bbp, but is not locally compact.

Summarizing the above examples:

A connected metric space with the bbp might be nowhere-locally connected, it need not be compact, need not be locally compact, need not be topologically complete, and need not have an exhaustion by continua.

Similar questions can be asked for geometric $T_1$ or $T_2$ spaces rather than metric spaces, ofc.

答:
A subset $A$ of $X$ is called a connector if for every component $C$ of $X \setminus A$ we have $\overline{C} \cap \overline{A} \neq \varnothing$.  A space is closely connected if every proper, non-empty subset of $X$ is a connector.

As proved here, a connected space $X$ has the bbp if and only if it is closely connected.

This condition is like asking for the bbp on $X \setminus A$ for arbitrary subsets $A$.  It shows that having the bbp implies this seemingly-stronger condition.  So in a sense, this characterization is a bit of a cheat since one direction implies the other immediately.  I'm still interested in finding properties that every bbp space must have, or sufficient conditions other than the ones mentioned above.

So I'm not gonna give this the check mark.


答:
No, you prove a function differentiable by evaluating the limit

$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ and differentiability implies continuity (and not conversely, as the function $|x|$ at $x=0$ shows).

By contraposition, a discontinuous function is not differentiable, and proving the first may be simpler.

Title: Relation between continuity and differentiability

Body:
I have noticed that authors of introductory analysis textbooks always show that a function f is continuous at a point before trying to prove that it is differentiable there. Ahlfors, for instance, does this in "Complex Analysis: An Introduction to The Theory of Analytic Functions of One Complex Variable" when he proves the extended Cauchy's integral formula by induction:
Check here

Bartle does the same when proving the second form of the Fundamental theorem of calculus in "Introduction to Analysis". Is it always necessary to prove continuity at a point before showing that a function is differentiable there? Why? (Please provide a technical explanation)


答:
Differentiability implies continuity: since
$$
\lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}
$$
exists, $f(x) \to f(x_0)$ when $x\to x_0$. If not, then the limit does not exist, because
$$
\lim_{x\to x_0} x-x_0 = 0.
$$
So it’s not really necessary to check continuity first, but technically it’s often inevitable.


问:
Generation of random sample from a given density function
How to generate random samples associated with a pdf is given by

\begin{equation}
 f_V(v) = \frac{1}{\pi\sqrt{v(2-v)}}
\end{equation}
where $v \in (0,2)$

in MATLAB. As we know in MATLAB 'randn()' is used to generate random numbers which follow a normal distribution and 'rand()' is used to generate random number associated with uniform distribution. But how to generate random samples whose associated pdf is $f_V(v)$.

答:
You can calculate analytically in this case the cdf of your random variable (for lazy people https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x*%282-x%29%29 ). Than:

$$F(x)=1-\frac{2}{\pi} \arcsin \sqrt{1-\frac{x}{2}}$$

for $0\le x\le2$

And $F^{-1}(z)=2-2 \sin^2(\frac{\pi-\pi z}{2})$ (please check my calculations)

So now take $Z\sim U(0,1)$, i.e. use your random uniform enerator. According to theory:

$$G(Z)=2-2 \sin^2\left(\frac{\pi-\pi Z}{2}\right)$$

will be distributed according to the distribution you want. See also here:

https://en.wikipedia.org/wiki/Inverse_transform_sampling


问:
In whiche case do we have $\left(\sum_{i=1}^n x_iy_{i}\right)^2 = \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (y_{i})^2$?
The exercice asked me first to prove that $( \sum_i x_i y_i )^2 \leq (\sum_i x_i^2)(\sum_i y_i^2)$ ( Cauchy inequality) and i managed to prove but then it asked me in which case we have the equality ($\left(\sum_{i=1}^n x_iy_{i}\right)^2 = \sum_{i=1}^n (x_i)^2 \sum_{i=1}^n (y_{i})^2$) but i don't Know how ????

Body: Suppose $x,y\neq 0$. $$|\left<x,y\right>|=\|x\|\|y\|\iff |\left<\frac{x}{\|x\|},\frac{y}{\|y\|}\right>|=1\iff \frac{x}{\|x\|}=\pm \frac{y}{\|y\|}.$$


问:
$V=\{A\in M_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$ is isomorphic to $\text{span}\{AB-BA:A,B\in V\}$
Background: Let
$$V:=\{A\in M_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$
be the vector space of $3\times 3$ real matrices with vanishing trace, and let $[\cdot,\cdot]:V\times V\to V$ be defined by
$$[A,B]=AB-BA$$
Note that this is well defined since $\text{trace}(AB-BA)=\text{trace}(AB)-\text{trace}(AB)=0$ for any matrices $A,B$. Finally, let
$$\hat{V}:=\text{span}\{[A,B]:A,B\in V\}$$

Question: I have to give an isomorphism
$$\varphi:(V,[\cdot,\cdot])\to(\hat{V},[\cdot,\cdot]),$$
i.e. a bijective map such that $\varphi([A,B])=[\varphi(A),\varphi(B)]$, $\forall A,B\in V$.

I am aware that this is related to Lie algebra; namely $V=\mathfrak{sl}(3,\mathbb{R})$. However, I did not study Lie algebra yet. I am suppose to be able to prove it without referring to any well-known theorem of Lie algebra (unless I prove it first).

I used a very computational argument. I am wondering if

My argument is correct.

There is a more fundamental way to prove it, i.e. without having to tediously compute all possible $[X_i,X_j]$ from a given basis.

Atempt: Let

$$
X_1 = \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_2 = \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & -1
\end{array}
\right),\quad
X_3 = \left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),$$
$$\quad
X_4 = \left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_5 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 1 \\
 0 & 0 & 0
\end{array}
\right),\quad
X_6 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 1 & 0 & 0 \\
 0 & 0 & 0
\end{array}
\right),$$
$$\quad
X_7 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 1 & 0 & 0
\end{array}
\right),\quad
X_8 = \left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 1 & 0
\end{array}
\right)
$$
Then, $\{X_1,\ldots,X_8\}$ forms a basis for $V$, and we find
$$[X_1,X_2]=0,\quad[X_1,X_3]=2X_3,\quad[X_1,X_4]=X_4,\quad[X_1,X_5]=-X_5$$
$$[X_1,X_6]=-2X_6,\quad[X_1,X_7]=-X_7,\quad[X_1,X_8]=X_8,\quad[X_2,X_3]=X_3$$
$$[X_2,X_4]=2X_4,\quad[X_2,X_5]=X_5,\quad[X_2,X_6]=-X_6,\quad[X_2,X_7]=-2X_7$$
$$[X_2,X_8]=-X_8,\quad[X_3,X_4]=0,\quad[X_3,X_5]=X_4,\quad[X_3,X_6]=X_1$$
$$[X_3,X_7]=-X_8,\quad[X_3,X_8]=0,\quad[X_4,X_5]=0,\quad[X_4,X_6]=-X_5$$
$$[X_4,X_7]=X_2,\quad[X_4,X_8]=X_3,\quad[X_5,X_6]=0,\quad[X_5,X_7]=X_6$$
$$[X_5,X_8]=X_2-X_1,\quad[X_6,X_7]=0,\quad[X_6,X_8]=-X_7,\quad[X_7,X_8]=0$$
Thus, we have
$$
\begin{array}{cccc}
X_1=[X_3,X_6],& X_2=[X_4,X_7],& X_3=[X_2,X_3],& X_4=[X_1,X_4] \\
X_5=[X_2,X_5],& X_6=[X_5,X_7],& X_7=[X_7,X_1],& X_8=[X_7,X_3]
\end{array}
$$
which shows that $V$ is a subspace of $\hat{V}$, and since $\hat{V}$ is clearly a subspace of $V$, then $V$ and $\hat{V}$ are in fact the same vector space. Thus, we can take $\varphi$ to be simply the inclusion map. Indeed, $\varphi$ is then clearly bijective and we have
$$\varphi([A,B])=[A,B]=[\varphi(A),\varphi(B)]$$

答:
I think your argument is correct, but you can simplify it a bit. Let $E_{ij}$ denotes the matrix whose $(i,j)$-th entry is $1$ and all other entries are zero. Then for any three distinct indices $i,j,k$, we have
\begin{align*}
E_{ii}-E_{jj} &= [E_{ij},E_{ji}],\\
E_{ij} &= [E_{ik},E_{kj}].
\end{align*}
This shows that the subspace traceless matrices $V$ lies inside the span $\widehat{V}$ of all commutators of pairs of traceless matrices. Note that this proof works for every $n\ge3$.

Remark. Actually, the set $\mathcal{C}_0:=\{AB-BA: \operatorname{trace}(A)=\operatorname{trace}(B)=0,\ A,B\in M_n(\mathbb{R})\}$ is identical to $\mathcal{C}:=\{AB-BA: A,B\in M_n(\mathbb{R})\}$. (That $\mathcal{C}_0\subseteq\mathcal{C}$ should be clear; $\mathcal{C}\subseteq\mathcal{C}_0$ because $AB-BA=\widehat{A}\widehat{B}-\widehat{B}\widehat{A}$, where $\widehat{A}:=A-\frac1n\operatorname{trace}(A)I$ and $\widehat{B}:=B-\frac1n\operatorname{trace}(B)I$ are traceless.) Since the set $\mathcal{C}$ of all commutators is known to identical to the subspace of traceless matrices (I have given an elementary proof here), the "$\operatorname{span}$" in your problem statement can actually be removed. In short, $V=\mathcal{C}=\mathcal{C}_0=\operatorname{span}(\mathcal{C}_0)=\widehat{V}$. But certainly, with the "$\operatorname{span}$", the problem is much easier to solve.

ContentLicense: CC BY-SA 3.0


问:
Limit of average of bounded linear functionals on finite dimensional normed vector space
Suppose I have a finite dimensional normed vector space $V$, and a sequence of $f_i$ in $V^*$ with norm all equal to 1. I want to know when does $\lim_{n->\infty} \frac{1}{n}\sum_{i=1}^{n} f_i$ converges. Now because $||f_i|| = 1$ for all $i$, I know if such limit exists then it is a bounded linear operator.

And if in addition I know for some {$x_1, x_2 ,...$} in $V$, $f_i(x_j)$ is eventually always non-negative (i.e. for each $x_j$, there exists some large enough $N$ such that for all $i>N$, $f_i(x_j) \geq 0$), does that help determine whether the limit exists?

答:
In general, the limit may not exists.
For example, take an increasing sequence of integers $n_{k}$ such that $\lim\frac{n_{k}}{n_{k+1}} = 0$, for $V$ take $R^{2}$ and lastly define $f_{n}(x_{1},x_{2}) = x_{1}$ if n is between $n_{k}$ and $n_{k+1}$ for event k and $f_{n}(x_{1},x_{2}) = -x_{1}$ if n is between $n_{k}$ and $n_{k+1}$ for odd k.

Now it is easy to see that the average for $n = n_{k}$ , where k is even will be close to $x_{1}$ but for odd $k$ it will be close to $-x_{1}$.

As far as the second case is concerned, since $V$ is finite dimensional it is isomorphic to Hilbert space (any two finite dimensional Banach spaces are isomorphic) so we can assume that any functional has form $<x,y>$. In words, $f_{n}(x) = <x,y_{n}>$ for some $y_{n}$. By assumption if $e_{k}$ is part of orthogonal basis on $V$ we have $f_{n}(e_{k})$ is $>=0$ for $n$ large enough. But this is also true for $-e_{k}$ so we must have $<e_{k},y_{n}> = 0$  for $n$ large enough. Since this is true for any $k$ and the space is finite dimensional, for $n$ large enough we must have $y_{n} = 0$. Then of course the average of $f_{n}$ converges to $0$.


问:
Norm $||.||$ on $C(X)$ is equivalent to $||.||_{\infty}$ norm if evaluation linear functionals on $(C(X),||.||)$ is continuous.
Let $X$ be compact Hausdorff space. Let $||.||$ be a norm on $C(X)$ which makes it into a Banach space and also assume that the linear functional $\lambda_x$ given by $\lambda_x(f)=f(x)$ is continuous for each $x$.
Show that there exist positive constants $A, B$ such that $$A||f||_{\infty}\le ||f|| \le B||f||_{\infty}.$$

I was trying to show that $f\to f$ is a bounded linear map from $(C(X),||.||)$ to $C(X)$ equipped with the sup norm. The map $f\to f$ is clearly one to one and onto, therefore open mapping theorem will give me the desired result.

In order to show the continuity of the above map, I observe that
$$|f(x)|=|\lambda_x(f)|\le ||\lambda_x|| ||f||.$$
I would want to supremum over all $x$ in both sides so that I can get the sup norm in LHS. But my issue is that I do not know a uniform bound on Norms of functionals $\lambda_x$. I do not know how to by pass this, I have a feeling that Uniform boundedness principle can be useful to get a uniform bound on the Norms of these functionals but I am unable to carry out the task. Any help in this regard or any other way to attack this problem would be appreciated.

答:
Since OP never posted the solution (as promised in the comments!) I will add the answer:

We consider the identity operator $I:(C(X),\|\cdot\|)\to(C(X),\|\cdot\|_\infty)$ and we want to show that it is bounded. Since both spaces are Banach, we apply the closed graph theorem. Let $(f_n)\subset C(X)$ so that $f_n\to0$ in $\|\cdot\|$. We want to show that $f_n\to0$ in $\|\cdot\|_\infty$ and by the closed graph theorem we can assume that $f_n\to f$ in $\|\cdot\|_\infty$ for some function $f\in C(X)$. All we need to do is show that $f=0$. Let $x\in X$. Then $|f(x)|=|\lambda_x(f)|=\lim_{n\to\infty}|\lambda_x(f_n)|$. But $f_n\to 0$ in $\|\cdot\|$ and $\lambda_x$ is continuous for $\|\cdot\|$, so $\lambda_x(f_n)\to\lambda_x(0)=0$. This shows that $f(x)=0$ and since $x$ was arbitrary we have that $f=0$.

We also want to show that the identity operator $I:(C(X),\|\cdot\|_\infty)\to(C(X),\|\cdot\|)$ is bounded. Again, we apply the closed graph theorem. Let $f_n\to 0$ in $\|\cdot\|_\infty$ and we want to show that $f_n\to0$ in $\|\cdot\|$. By the closed graph theorem we can assume that $f_n\to f$ in $\|\cdot\|$ for some $f\in C(X)$. All we need to do is show that $f=0$. By the previous result we also have that $f_n\to f$ in $\|\cdot\|_\infty$, so since both $0$ and $f$ are uniform limits of $(f_n)$ we have that $f=0$. This concludes the proof.


问:
Poisson Distribution - How does it come to whatever it is right now?

The Question

Two cafes are side by side and are open 7 days a week.

(a) The first cafe sells cappuccinos at 5 dollars each and has 30 customers a
day. Half the customers buy a cappuccino, the rest don't. What is the
distribution of dollar sales made by the first cafe in a week?

(b) The second cafe sells cafe lattes at 6 dollars each. Customers arrive at
a rate of 60 a day and all the customers buy a cafe latte. What is the
distribution of dollar sales made by the second cafe in a week?

(c) Now the Poisson distribution is approximately Gaussian when the
rate is larger, and this applies moderately well when the rate is 15
or more. Give Gaussian approximations to the distributions of weekly
dollar sales for each of the two cafes.

My Understanding

I have the answers for this but I just don't know how it has been calculated. As in the whole process of coming to the conclusion of those answers and would appreciate if someone helps me understand or gives me step by step solution in order to understand.

The solutions given:

(a) The number of sales per week is Poisson with rate 105. The dollar sales is 5 times this, i.e. distribution is Poisson with rate parameter of 525 dollars per week.

(b) The number of sales per week is Poisson with rate 420. The dollar sales is 6 times this, i.e. distribution is Poisson with rate parameter of 2520 dollars per week.

(c) The question isn't very clear about when to apply the Gaussian approximation. One possible answer is to apply the normal approximation to the Poisson distribution of sales per week. In this case the sales rates respectively are, and. Now we scale to get dollars. The first is, and the second is
.Note since we already applied the Gaussian approximation to the sales rates, we have to square the dollars as the units for the variance are sales, so to get variance of total dollars we compute (sales times dollars per sale)=(sales)(dollars per sale)
$N(105, 105) N(420, 420) N(525, 105.5.5) = N(525, 2625) N(2520, 420.6.6) = N(2520, 15120)
σ2 2 2 2$

答:
Though the question is badly worded, assume that the number of coffee bought at Cafe A per week is poisson distributed with $\lambda=105$ and sells for 5 dollars and the number of coffees bought per week at Cafe B is poisson distributed with $\lambda=420$ and sells for $6 each. (This isn't what the problem actually says.)

The solution for a and b isn't necessarily correct per se because a poisson multiplied by a constant doesn't follow a poisson distribution. (Though it follows poisson probabilities) Therefore the answer for a is $P(d)=f(\frac d 5;\lambda=105)$ where d=0,5,10,...,525 is dollars sold and f is the pmf of a poisson distribution.

How they got c is that a poisson distribution approximately follows a gaussian distribution with mean lambda and and variance lambda. So for cafe A, the distribution of purchases per week is about $N(105,105).$ Multiply this by the dollar amount to get $5*N(105,105)=N(105*5,105*5^2)=N(525,2625)$. (The mean is multiplied by 5, but the variance is multiplied by 5 squared.) The distribution of purchases per week for cafe B can be calculated similarly, and is $N(105,105). 5*N(420,420)=N(420*6,420*6^2)=N(2520,15120)$.


答: If a function is not continuous at a point $a$, it not differentiable at that point. Continuity does not guarantee differentiability, but continuity is necessary. If the function is not continuous, it's not differentiable, so checking for continuity can provide a preliminary check.


问:
Fundamental solution of the Laplace operator
We know that in the entire plane $\mathbb{R}^{2}$ the fundamental solution of the Laplace operator is $$
G(x)=-\frac{1}{2 \pi} \log |x|,
$$
so we can find how $K_{\mathbb{R}^{2}}(x, y)=K(x-y)=\nabla^{\perp} G(x-y)$ behaves, where $\nabla^{\perp}=\left(\partial_{2},-\partial_{1}\right)$. But if instead of the entire plane we have a general connected set $\Gamma$ of $\mathbb{R}^{2}$ how does $K_{\Gamma}(x, y)$ behave?
I'm looking for this because I want to find some estimates on $K_{\Gamma}(x, y)$.

答:
Let us consider the Green's function (for the Dirichlet problem) on the unit disc $D=\{z\in \mathbb{C}\mid |z|<1\}$ (I will use complex numbers, but it can easily be rewritten with real numbers) which is given by
\begin{align}
G_D(z, z_0) = \frac{1}{2\pi}\log\left|\frac{1-z\bar z_0}{z-z_0} \right|.
\end{align}

Suppose $\Gamma$ is a simply connected proper region of the complex plane. By Riemann's mapping theorem, there exists a conformal transformation $f:\Gamma\rightarrow D$. In particular, we have that
\begin{align}
G_\Gamma(\zeta, \zeta_0) = G_D(f(\zeta), f(\zeta_0)) = \frac{1}{2\pi}\log\left| \frac{1-f(\zeta)\overline{f(\zeta_0)}}{f(\zeta)-f(\zeta_0)}\right|
\end{align}
since conformal (analytic) mapping preserves "harmonicity" of harmonic functions.

You can choose some concrete $\Gamma$ and find estimates for $G_\Gamma$.


问:
Prove: symmetric positive matrix multiplied by skew symmetric matrix equals 0
My teacher gave me this task as preparation for the exam but I'm stuck and not sure if it's true anymore.</p>

答:
I think your teacher means Frobenius product.
In the context of tensor analysis (e.g. widely used in mechanics, think about $\int \boldsymbol{\sigma}:\boldsymbol{\epsilon}\,\mathrm{d}\Omega$, if you know the weak form of elastostatics), it is a natural inner product for 2nd order tensors, whose coordinates can be represented in matrices.

The result is straightforward
$$A:B = \operatorname{tr}(A^T B) =a_{ij}b_{ij} = -a_{ji}b_{ji} = - A:B \Rightarrow A:B = 0.$$


问:
A problem from Knuth's The Art of Computer Programming about middle square method
This isn't a 'homework' question or anything similar - just for my own personal interest.

In his discussion of random numbers (vol 2),Knuth poses this question:

Prove that the middle-square method using 2n-digit numbers to the base b has the following disadvantage: if the sequence includes any number whose most significant n digits are zero, the succeeding numbers will get smaller and smaller until zero occurs repeatedly.

(He rates this as '14' which I take to mean it should take somewhat less than 10 minutes - but I'm really struggling with it.)

A quick explanation: the middle-square method is a naive (though actually first suggested by none other than John von Neumann) random number generation method where we take a number n-digits long, square it and take the middle n-digits as our next seed or random number.

But, actually, it looks to me that it's not too difficult to show an example where Knuth’s proposition appears to be false.

Let n=2, b=10 and the seed number be 60

Then every subsequent number in the sequence is also 60 (obviously that’s as useless as repeated zeros for a random number generator.)

But maybe using n=2  is a bit silly, so let’s look at the general case.

Any (base-b ) integer is then of the form $ b^n - x $ where $ 0<x<b^n $, but I think that without any restriction to the generality of we can look at $ b^n-1 $, the largest number we can have (and therefore the one that will produce the maximum number of digits when squared.)

So we get $ b^{2n}-2b^n+1 $ as the square.

Now we need to get the middle digits so we lose the right $\frac{n}{2}$ digits by dividing by $b^{\frac{n}{2}}$:

$\frac{b^{2n}-2b^n+1}{b^{\frac{n}{2}}} = b^{\frac{3n}{2}}-2b^{\frac{n}{2}}+\frac{1}{b^{\frac{n}{2}}}$

In the general case that final term would be $\frac{x^2}{b^{\frac{n}{2}}}$ and the biggest that could be is $x\rightarrow b$ but as the smallest n  can be is 2, this term dwindles very rapidly indeed compared to the others as n  increases.

So we can ignore the final term and are left with $b^{\frac{3n}{2}} - 2b^{\frac{n}{2}}$, but we still have up to $ \frac{n}{2} $ too many significant digits here – so we need to get rid of them too.

What we are now looking for is the remainder when we divide our current result by $b^{\frac{3n}{2}} - 1 - (b^n-1)  = b^{\frac{3n}{2}}-b^n $

This is where I just run out of puff. How do I do that, and given the complexity of this approach and Knuth's view that it's a relatively simple proof - have I taken a wrong turn somewhere here?

答:
Your example doesn't make Knuth's proposition false. In fact, Knuth talks about the mid-square method with $2n$-digit numbers. So we use a $2n$-digit number, square it to a $4n$-digit number, and take the middle $2n$ digits (i.e., drop the first $n$ and the last $n$ digits). I suppose he does so to make the notion of "middle" digits unambiguous.

So with $n=2$ (as you wrote) and $x=60$ (or rather $0060$ as four-digit number) matches the condition of the proposition: The leading two digits of this four-digit number are zero. As $x^2=3600$ ($=00003600$ as an eight-digit number), the next "random" four-digit number is $0036 = 36$, which is indeed $<60$. As $36^2=1296$, the next is $12$, then (as $12^2=144$) comes $1$, and after this only $0$.

Apparently, you wanted to work with $2$-digit numbers (chopped from the middle of $4$-digit squares), which in the context of Knuth's proposition would mean $n=1$! You are right, that the sequence that is constant $60$ is bad - but that does not contradict Knuth's proposition. This is because your seed $60$ does not have leading digit $0$ and so the  proposition makes no claim about it. Instead, you have found a different weakness of the mid-square method.

Now to showing the original proposition:
If the leading $n$ digits of a $2n$-digit number $x$ are $0$, then $x<b^n$. So $x^2<b^{2n}<b^{3n}$. Then the next number is just
$$\left\lfloor \frac{x^2}{b^n}\right\rfloor \le \left\lfloor\frac{x(b^n-1)}{b^n} \right\rfloor
= \left\lfloor x-\frac x{b^n}\right\rfloor=x+\left\lfloor -\frac x{b^n}\right\rfloor$$
which is $<x$ as long as $x>0$. Hence the "random" sequence is a sequence of integers that strictly decrease, until they hit $0$, as claimed.


问:
Got $e^{-x}=\cos(ix)$ from Euler's formula. Where is my mistake?
When I was messing around with Euler's formula, I came across this: $$e^{ix}=\cos(x)+\sin(x)i$$ Then, let $x$ be an imaginary value, $ix$, so then: $$e^{i(ix)}=\cos(ix)+\sin(ix)i$$ which we can simplify to $$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$. so that means that $$e^{-x}=\cos(ix)$$ This doesn't seem right, so could someone please point out where I made a mistake?

答:
$$e^{-x}=\cos(ix)+\sin(ix)i,$$ but since $e^{-x}$ is a real value for all real inputs, then $\sin(ix)i$ must be equal to $0$.

That does not follow unless you know that $\sin(ix)$ is real, and there's no reason to think that.

$$
\sin(ix) = \frac{e^{i(ix)} - e^{-i(ix)}}{2i} = \frac{e^{-x} - e^x}{2i}.
$$


问:
Defining the Product of Ideals

Where does the "naive definition" of the product of two ideals $I$ and $J$, $IJ = \{ ij \mid i \in I, j \in J \}$ fall apart?

(The product of two ideals $I$ and $J$ is defined to be $IJ := \sum_i a_ib_i$, where each $a_i$ is in $I$ and each $b_i$ is in $J$.)

Note: This is a question from a problem set I'm working on. I am not expecting a full solution for the answers, but I'd like some help knowing where to look. It seems to work fine in $\mathbb{Z}$ and $\mathbb{C}[x]$.

答:
If you read Remark 2.7 of my paper "On a property of pre-Schreier domains"
[Comm. Algebra 15 (1987) 1895-1920] you would find that the question of
discrepancy between products of two ideals $A,B$ $(AB=\{\Sigma $ $%
a_{i}b_{i}|a_{i}\in A,b_{i}\in B\}$ in a ring and the product of two ideals $%
I,J$ $(IJ=\{a_{i}b_{i}|a_{i}\in I$ and $b_{i}\in I\}$ in a semigroup was
raised in the above mentioned paper of mine. I was naive and thought
everyone was my friend. So I would send preprints of even half written
papers around. I have a feeling that the Anderson-Dobbs paper, that is
mentioned by Bill Dubuque, was an effort to lift the idea from a pre-print
of the above paper with very little credit (they do mention the pre-print,
but without giving a clue to why they mention a result from Zafrullah.) At
the end of that remark I do mention the Anderson-Dobbs' paper. I think I
handled it beautifully, mentioning the heist and at the same time praising
it. But now I am bitter, as I don't see any light at the end of the tunnel.
For as I look back I see this kind of activity around most of my solo
papers. In any case, the property $\ast $ came right out of this discussion
of the discrepancy of the definitions of products of ideals. Recall that a
domain $D$ is a $\ast $-domain if $((\cap (a_{i}))(\cap (b_{j}))=\cap
_{ij}(a_{i}b_{j})$ for all $a_{1},...,a_{n},b_{1},....b_{m}\in D$
equivalently, $\in K=qf(D).$ I address an effort to lift the $\ast $%
-property in https://lohar.com/mithelpdesk/hd2006.pdf

Anyone interested in products of ideals or pre-Schreier domains may like to
study the $\ast $-property, as a Prufer domain has the $\ast $-property and
a so called Prufer $v$-multiplication domain (PVMD) that is a $\ast $-domain
is a locally GCD domain.

ContentLicense: CC BY-SA 2.5

Title: Defining the Product of Ideals


问:
Limit of fourier series
I want to compute the limit $f(x)$ of

$$
f(n,x)=\sum_{k=1}^n\frac{1}{k^2+1}\sin(kx)
$$

and didn't succeed. I tried with the means used to compute the >>easy<< limits of trigonometric series like those listed in (e.g.) Bronstein, however, the above series is not in Bronstein. I plotted the graph with sagemath (see my post on ask.sagemath.org: https://ask.sagemath.org/question/55596/limit-of-fourier-series/). At least this suggests that f(x) exists and is not a polynomial.

Is it possible to compute the above limit and if so, how?

答:
The answer was already given in terms of the hypergeometric function.

Here I'd like to make some remarks which might be useful to understand the sum better.

Let us first study the convergence of the sum

$$s(x) = \sum_{k=1}^\infty \frac{\sin(k x)}{1+k^2}$$

and then tun to some related sums to get a better feeling of the behaviour $s$.

We have the estimates

$$|s| \lt \sum_{k=1}^\infty \frac{|\sin(k x)|}{1+k^2}\\\lt \sum_{k=1}^\infty \frac{1}{1+k^2}=\frac{1}{2}\sum_{k=-\infty}^\infty \frac{1}{1+k^2}-\frac{1}{2}=\frac{1}{2}\left(\pi \coth(\pi)-1 \right)\simeq 1.07667\\\lt \sum_{k=1}^\infty \frac{1}{k^2}=\zeta(2) \simeq 1.64493$$

While the maximum is given (numerically) by $$s_{max}\simeq 0.573975 $$

Another related sum is

$$s_2(x) = \sum_{k=1}^\infty \frac{\sin(k x)}{k^2}= \frac{1}{2 i} \left( \text{Li}_2\left(e^{i x}\right)-\text{Li}_2\left(e^{-i x}\right)\right)=\Im (\text{Li}_2\left(e^{i x}\right))$$

Here $\text{Li}_2\left(z\right)$ is the polylog function of second order.

Although I could not prove it from the sums it seems that $s(x) \le s_2(x)$ by a broad margin.

Finding the exact expression for the sum

We can reduce $s(x)$ to simpler sums by observing that we can write

$$\sin(x) = \frac{1}{2 i} (e^{i x} - e^{-i x})$$

$$\frac{1}{1+k^2} = \frac{1}{2} \left(\frac{1}{1+ i k}+\frac{1}{1-i k} \right)$$

Hence defining the sums

$$s_2(y,z) = \sum_{k=1}^{\infty} \frac{z^k}{1+ y k}$$

$$s_1(z) = \sum_{k=1}^\infty \frac{z^k}{1+k^2}$$

we have the identities

$$s_1(z) = \frac{1}{2}(s_2(i,z)+s_2(-i,z))$$

and

$$s(x) = \frac{1}{2 i}(s_1\left(e^{i x})-s_1(e^{- i x})\right)$$

Hence we only need to calculate $s_2$ to find $s$ by simple replacements.

There are several ways to deal with $s_2$. For example we could try to directly identify the series or, what we adopt here, replace the denominator by means of the formula

$$\frac{1}{1+ y k} = \int_{0}^{1} t^{k y} \,  dt $$

Then we have

$$s_2(y,z) =\sum_{k=1}^{\infty} \frac{z^k}{1+ y k}=\sum_{k=1}^{\infty} z^k \int_{0}^{1} t^{k y}\, dt
\\ =\int_{0}^{1}\sum_{k=1}^{\infty} z^k t^{k y} \, dt =\int_{0}^{1}\frac{z t^y}{1-z t^y} \, dt
\\ \overset {t^y\to u}=\frac{z}{y} \int_{0}^{1} \frac{u^{\frac{1}{y}}}{1- u z}\, du  $$

The final integral can be identified by the formula of Euler for the hypergeometric function (see https://en.wikipedia.org/wiki/Hypergeometric_function Euler type)

$$\int_0^1 u^{b-1} (1-u)^{c-b-1}(1-u z)^{-a} \, du \\ =\frac{\Gamma (b) \Gamma (c-b)}{\Gamma (c)} \,_2F_1(a,b;c;z)$$

Comparision leads to $$a\to 1, b\to 1+\frac{1}{y}, c\to b+1=2+\frac{1}{y}$$

Hence $$\frac{\Gamma (b) \Gamma (c-b)}{\Gamma (c)}= \frac{\Gamma (b) }{\Gamma (b+1)}=\frac{1}{b} \to \frac{1}{1+\frac{1}{y}} $$

and we have

$$s_2(y,z) =f_2(y,z)= \frac{z}{1+y}\, _2F_1(1,1+\frac{1}{y};2+\frac{1}{y};z)$$

Here and in what follows we denote the closed form of a sum $s_j$ by $f_j$.

Now we get

$$f_1(z) =  \frac{1}{2}(f_2(i,z)+f_2(-i,z))$$

and

$$f(x) = \frac{1}{2 i}(f_1\left(e^{i x})-f_1(e^{- i x})\right)$$

And we arrive finally at the symmetric expression

$$f(x) = \left(\frac{1}{8}+\frac{i}{8}\right) e^{-i x} \, _2F_1\left(1,1-i;2-i;e^{-i x}\right)\\-\left(\frac{1}{8}+\frac{i}{8}\right) e^{i x} \, _2F_1\left(1,1-i;2-i;e^{i x}\right)\\-\left(\frac{1}{8}-\frac{i}{8}\right) e^{-i x} \, _2F_1\left(1,1+i;2+i;e^{-i x}\right)\\+\left(\frac{1}{8}-\frac{i}{8}\right) e^{i x} \, _2F_1\left(1,1+i;2+i;e^{i x}\right)$$

Done.

Title: Limit of fourier series


问:
Can we choose the kernel of a Hilbert-Schmidt integral operator $T$ s.t. $0\in\sigma_p(T)$, $0\in\sigma_c(T)$ or $0\in\sigma_c(T)$?
Let $(\Omega_i,\mathcal A_i,\mu_i)$ be a $\sigma$-finite measure space and $k\in L^2(\mu_1\otimes\mu_2)$. We know that $$Tf:=\int k(\;\cdot\;,x_2)f(x_2)\;\;\;\text{for }f\in L^2(\mu_2)$$ is a linear Hilbert-Schmidt operator from $L^2(\mu_2)$ to $L^2(\mu_1)$. In particular, $T$ is compact and hence $\sigma(T)\setminus\{0\}=\sigma_p(T)\setminus\{0\}$.

Can we choose $k$ such that

$0\in\sigma_p(T)$;

$0\in\sigma_c(T)$;

$0\in\sigma_r(T)$?

I'm also interested in the same question, but with a more general (but still compact) integral operator and would be very happy if someone could provide a good reference on that.

答:
To make life simple let us assume that both measure spaces coincide with ${\mathbb N}=\{1, 2, 3, \ldots \}$  under counting measure.  Thus the integral
kernel
$$
  k=(k_{i, j})_{i, j\in {\mathbb N}}
  $$
turns out to be the same thing as  the matrix of the associated integral operator on $\ell^2$,  relative to the standard orthonormal basis.

Choosing
$$
  k(i, j) = \left\{ \matrix{
  \displaystyle\frac1{i}, & \text{if } i=j, \cr
  0, & \text{otherwise},
  }\right.
  $$
gives rise to the  diagonal operator with matrix
$$
  \left[\matrix{
  1 & 0 & 0 & \cdots  \cr
  0 & \frac12 & 0 & \cdots  \cr
  0 & 0 & \frac13 & \cdots  \cr
  \vdots & \vdots & \vdots & \ddots \cr
  }\right]
  $$
which is injective and has  dense range, hence $0$ lies in the continuous spectrum of $T$.
If we instead take
$$
  \left[\matrix{
  0 & 0 & 0 & \cdots  \cr
  1 & 0 & 0 & \cdots  \cr
  0 & \frac12 & 0 & \cdots  \cr
  0 & 0 & \frac13 & \cdots  \cr
  \vdots & \vdots & \vdots & \ddots \cr
  }\right]
  $$
we get what is sometimes called a weighted shift, and which is again injective, but its range is not dense because it
is orthogonal to the first basis vector.  Consequently $0$ lies in the residual  spectrum of $T$.

Finally,  the zero operator has $0$ in its point spectrum.


问:
Inner product on a positive-definite matrix
Let $A$, be a positive-definite matrix, $x=(x_1,\cdots,c_n)$,$y=(y_1,\cdots,cy_n)$. How do we prove that $$\langle x,y\rangle = x^TAy$$ is a inner product?

I know that $\langle x,x\rangle > 0$ it's easy to prove. For the other conditions, my attempt is $$\langle x,y \rangle =\sum_{i=1}^n y_i(\sum_{j=1}^n a_{ij}x_i) $$

if this equality is true, i can easily prove the conditions.

There is an easier way to do this?

Thanks!

答:
You need to show

linearity in the first argument

Conjugate symmetry

positive-definite

(1) follows from the linearity of matrix multiplication: $(x+z)^TAy=x^TAy+z^TAy$

(2) follows from the property of matrix transpose:
$$
x^TAy=(x^TAy)^T=y^TA^Tx=y^TAx
$$

You said you have done (3).


问:
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$

If

$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:

$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.

答:
less beautiful

$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 -  \frac{ 1 }{ 2 }  & 1 & 0 & 0 \\
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  & 1 & 0 \\
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 2 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 &  \frac{ 3 }{ 2 }  & 0 & 0 \\
0 & 0 &  \frac{ 4 }{ 3 }  & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  \\
0 & 1 &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 3 }  \\
0 & 0 & 1 &  -  \frac{ 1 }{ 2 }  \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
2 &  - 1 &  - 1 &  - 1 \\
 - 1 & 2 & 0 & 0 \\
 - 1 & 0 & 2 & 0 \\
 - 1 & 0 & 0 & 2 \\
\end{array}
\right)
$$

Back substitution says $t=0$  then $z=0$ then $y=0$ then $x=0.$   All that is going on is that the matrix I am calling $H$ is positive definite. With all real variables, each row in my $Q$ must indicate a zero vector ( when multiplied by $(x,y,z,t)^T \; \; \; \; $ )


问:
Proving that $\angle AOE = 2 \angle COD$ using complex numbers

Suppose $OABC$ is a square, with points $D$ and $E$ on line segment $BC$ such that $BD = CD$ and $BE = DE.$ Prove that $\angle AOE = 2 \angle COD$ using complex numbers.

I noted that $\triangle COD$ was a right triangle with $CD:CO = 1:2,$ so I knew $\sin(\angle COD) = \frac{1}{\sqrt{5}}.$ However, I got stuck from here on out.

答:
Instead of $D$, the midpoint of $BC$, work with $F$, the midpoint of $AB$. Since $\angle COD=\angle AOF$.

Set $O=0,A=1,B=1+i,C=i,E=\frac{3}{4}+i$

Then $F=1+\frac{1}{2}i$

$$z_F^2=\left( 1+\frac{1}{2}i \right)^2=\frac{3}{4}+i=z_E$$

$$\Rightarrow \arg(z_E)=2\arg(z_F)$$

$$\therefore \angle AOE=2\angle AOF=2\angle COD$$


问:
Does Right continuity at each point implies continuous a.e?
Suppose f is a function from non negative real numbers to extended non negative real numbers which is right continuous at each point,does it imply that f is continuous almost everywhere in its domain?</p>

答:
The result is actually stronger.

Let $f [0, +\infty) \rightarrow [0, +\infty]$ be a right continuous function. Then the set of points where $f$ is discontinuous is a countable set.

Proof:

Let $f [0, +\infty) \rightarrow [0, +\infty]$ be a right continuous function. For all $n \in \Bbb N$, $n\geqslant 1$,  let

$$ \Gamma_n = \left \{ I \subseteq  (0, +\infty) : I \textrm{ is an open interval and, for all } x,y \in I, |f(y)-f(x)|< 1/n    \right \} $$

Let $A_n = \bigcup_{I \in \Gamma_n} I$. We have that $A_n$ is an open set.

Let $F_n = A_n^c$. $F_n$ is closed. Now, given any $x \in  F_n$, since $f$ is right continuous, it is easy to see that there is $\delta_x>0$, such that $(x, x+\delta_x) \subseteq A_n$. It means, $(x, x+\delta_x) \cap F_n =\emptyset$. It follows that $F_n$ is countable (see Remark).

Now, $A= \bigcap_{n\geqslant 1}A_n$ is the set of points where $f$ is continuous and we have that $A^c= \bigcup_{n\geqslant 1}F_n$ is countable.

Remark: If $F_n$ is uncountable, then there is $k \in \Bbb N$ such that $F_n \cap [0, k)$ is uncountable, but this is not possible, since for each $x \in  F_n$, there is  $\delta_x>0$ such that $(x, x+\delta_x) \cap F_n =\emptyset$.

As pointed out by @EvangelopoulosF. another way to prove that $F_n$ is countable is, for each $x \in F_n$, to pick a $q_x \in \Bbb Q \cap (x, x+\delta_x)$. It is easy to see that the map $F_n \rightarrow  \Bbb Q$ defined by $x \mapsto q_x$ is injective.


问:
Size of a linear image of a cube in $\mathbb{Z}^d$
Suppose that we have an element $v = (v_1, \dots, v_d) \in \mathbb{Z}^d$ such that $\gcd(v_1, \dots, v_d) = 1$. Then $v$ is contained in some base of $\mathbb{Z}^d$ (seen as a free-abelian group or a free module over $\mathbb{Z}$). In particular, there exists a regular integer matrix $A \in \mathop{GL}_d(\mathbb{Z})$ such that $A(v) = (1, 0, \dots, 0)$.

My question is the following: assuming that $|v_1| + \dots + |v_d| = n$, what is the smallest $\ell \in \mathbb{N}$ (taken over all possible matrices $A$) such that the image $A\left([-1, 1]^d\right)$ is contained in the cube $[-\ell, \ell]^d$? I care about asymptotics of $\ell(n)$ up to multiplicative constant.

答:
If $d$ is fixed and you care about the asymptotic behavior in $n$, then you cannot do better than linear bound because you can take the vector $(p,p,\dots,p,q)$ where $p\approx n$ is prime and $qk\equiv 1\mod p$ for some $k\approx n/2$, so the first row has to be large. Keeping this in mind, you can easily arrange the other rows of size $n$ by putting $0,\dots,0,v_{j+1},-v_j,0,\dots,0$ there (if all $v_j\ne 0$; otherwise arrange them so that the first few aren't and then do this until you hit the zeroes and add the diagonal of ones in the end).

Thus, the only problem is to come up with the first row of size comparable to $n$, i.e., with an integer vector $a$ such that $(a,v)=1$ and $|a|\le C(d)n$. The integer vectors with the property $(a,v)=1$ form a lattice in the corresponding hyperplane with determinant $\|v\|_2\le n$ and the minimal vector length $\ge 1$. Such a lattice cannot miss a ball of radius $C(d)n$, so the solution exists.

Edit Ah, yes, I should have mentioned the last step. Here goes.

We have a lattice $\Lambda$ given by $\langle a,v\rangle=0$ and you want the last $n-1$ rows to be a basis of that lattice. What I created is a set of short (length $\le n$) linearly independent vectors $a_1,\dots,a_{d-1}$ in $\Lambda$. But now just consider the lattices $\Lambda_k=\Lambda\cap L_k$ where $L_k={\rm span}(a_1,\dots,a_k)$. The basis in $\Lambda_1$ is just the primitive vector $a_1'$ in the direction of $a_1$ (so it is $\le n$). To get the basis in $\Lambda_2$, take the closest (in the Euclidean distance) to $L_1$ vector in $\Lambda_2$ that is not in $\Lambda_1$. Its distance from $L_1$ is at most $n$ (because $a_2$ competes). And its projection to $L_1$ can be brought below $|a_1'|\le n$ by adding $a_1'$ with appropriate integer coefficient. Thus we can have $|a_2'|\le 2n$. Now we have a basis in $\Lambda_2$ with the fundamental parallelogram of diameter $3n$. For $a_3'$ take the closest to $L_2$ vector in $\Lambda_3\setminus\Lambda_2$. The distance to $L_2$ is $\le n$ ($a_3$ competes) and the projection to $L_2$ can be brought into the fundamental parallelogram we constructed, so we get $|a'_3|\le 4n$ and the fundamental parallelepiped for $\Lambda_3$ of diameter $7n$. And so on. The final estimate is $2^dn$. It can, probably, be improved but I have to refresh my rather rusty knowledge of Geometry of Numbers to say how and to what. We all have read the same books anyway (Cassels, Rogers, & Co.), so you may also search the answer there yourself :-)


问:
Second derivative near a minimum positive?
If I have a smooth real valued non-constant function $f$ with isolated minimum at $x_M$ does that mean that close to $x_M$ its second derivative must be positive? I.e. must there exist $a > 0$ so that

$$
\begin{align}
f''(x_M + \varepsilon) > 0 \\
f''(x_M - \varepsilon) > 0
\end{align}
$$

for all $\varepsilon \in (0, a]?$ If not can someone give an example?

答:
No, nothing can be said about the sign of $f''$ in the vicinity of an isolated minimum.

An example is the function $f$ defined by
$$
 f(x) = e^{-1/x^2}\left(3 + 2 \sin(\frac {1}{x^2})\right)
$$
for $x \ne 0$ and $f(0) = 0$. $f$ is infinitely often differentiable and has an absolute minimum at $x=0$. For $x \ne 0$ is
$$
 f''(x) = \frac{2e^{-1/x^2}}{x^6 }
\left( 6 - 8 \cos(\frac {1}{x^2}) - 9x^2+6x^2\cos(\frac {1}{x^2})-6x^2\sin(\frac {1}{x^2})\right) \, .
$$
For sufficiently small $|x|$ is
$$
f''(x) \le \frac{2e^{-1/x^2}}{x^6 }
\left( 7 - 8 \cos(\frac {1}{x^2})\right)
$$
and that is negative if $\cos(1/x^2) = 1$, which happens arbitrarily close to zero.


问:
Generation of random sample from a given density function
How to generate random samples associated with a pdf is given by

\begin{equation}
 f_V(v) = \frac{1}{\pi\sqrt{v(2-v)}}
\end{equation}
where $v \in (0,2)$

in MATLAB. As we know in MATLAB 'randn()' is used to generate random numbers which follow a normal distribution and 'rand()' is used to generate random number associated with uniform distribution. But how to generate random samples whose associated pdf is $f_V(v)$.

答:
\begin{align}
v(2-v) & = {}\overbrace{-(v^2 - 2v) = -(v^2 - 2v+1) + 1}^\text{completing the square} \\[8pt]
& = 1-(v-1)^2
\end{align}

The pdf is for $0<v<2.$ Note that if $v<0$ or $v>2$ then $v(2-v)$ is a negative number under the radical.

\begin{align}
F_V(v) & = \int_0^v \frac{du}{\pi \sqrt{u(2-u)}} \\[8pt] & = \frac 1 \pi \int_0^v \frac{du}{\sqrt{1 - (u-1)^2}} \\[10pt]
& = \frac 1 \pi \int_{-1}^{v-1} \frac{dw}{\sqrt{1 - w^2}} \\[10pt]
& = \frac 1 \pi \big( \arcsin(v-1)-\arcsin(-1) \big) \\[8pt]
& = \frac 1 2 + \frac 1 \pi \arcsin(v-1).
\end{align}

$$ \text{So } F_V^{-1}(y) = \sin\left(\pi\left(y - \frac 1 2 \right)\right) + 1 \text{ for } 0\le y\le1.$$

If (capital) $Y$ is a random variable that is uniformly distribution between $0$ and $1,$ then $F_V^{-1}(Y)$ has the distribution that you gave us.


问:
How the false positive value affects accuracy?
I have two binary images, one of them is an output produced by my algorithm and the other one is the ground truth. For this two images I calculated the values true positive, false negative, true negative and false positive. I want to interpret this result. For example I have this values:

```<code>TP (true positive) = 2739
TN (true negative) = 103217
FP (false positive) = 43423
FN (false negative) = 5022
</code>```

After I calculated this parameters I compute the accuracy
$$accuracy = \frac{TP +TN}{TP+TN+FP+FN}$$
In this case the accuracy is 0.68. Can I say that I have low accuracy because the value false positive is high? There is any relathion between false positive and the parameters true positive or true negative?

答:
You might be interested in the false positive rate. This is

```<code>FPR=FP/(TN+FP)=43423/(43423+103217)=.3
FNR=FN/(TP+FN)=5022/(2739+5022)=.65
</code>```

Your false positive rate is actually very low. You are mostly classifying negatives correctly as negative. But your false negative rate is higher at .65. You are not detecting enough images that are actually true images. For a more detailed discussion, read on.

The accuracy is an overall summary of the success of your tests. It is sometimes helpful to look at a breakdown of it into sensitivity and specificity. Sensitivity is the true positive rate, i.e. the chance your program predicts a positive given that it's supposed to be positive. Specificity is the true negative rate, i.e. the proportion of negatives correctly predicted as negative.

$sensitivity=\frac{TP}{TP+FN}=\frac{2739}{2739+5022}=.35$

$specificity=\frac{TN}{TN+FP}=\frac{103217}{103217+ 43423}=.7$

Based on these results, it seems that your test is not as good as identifying positives as it is at correctly identifying negatives. One possible problem could be that there is a difference in the prevalence of images from the truth and produced by your computer algorithm. Here a confusion matrix of the data you have collected:

$\begin{matrix}&actual&\\
predicted&truth&algorithm\\
truth&2739& 43423\\
algorithm & 5022& 103217\end{matrix}$

If you compute the accuracies separately for the true images and the algorithm images, you find that the accuracy for the true images is $\frac {2739}{2739+5022}=.35$ and the accuracy for your algorithmic images is $\frac{103217}{103217+ 43423}=.7$. (Notice these are the definitions of sensitivity and specificity, the same above.) The reason that your overall accuracy is high is because you have a much higher prevalence of algorithm-generated images; your algorithm tends to guess conservatively, and the fact that there are more algorithm generated images pushes your overall accuracy up.

A useful one-number summary combining sensitivity and specificity is the $F_1$ score, defined as $$\frac {1}{\frac{\beta^2}{\beta^2+1}\frac{1}{sensitivity}+\frac{1}{1+\beta^2}\frac{1}{specificity}}$$

When $\beta=1$, a common choice, this is the harmonic average of sensitivity and specificity:

$$\frac {1}{\frac{1}{2}\left(\frac{1}{sensitivity}+\frac{1}{specificity}\right)}$$

The $F_1$ score of your test is .47. The closer to 1, the better. You might try to optimize your algorithm based on the $F_1$ score.

There is some difference between your $F_1$ score and accuracy:

```<code>F_1: .47
accuracy: .68
</code>```

Body:
You might be interested in the false positive rate. This is

FPR=FP/(TN+FP)=43423/(43423+103217)=.3
FNR=FN/(TP+FN)=5022/(2739+5022)=.65

Your false positive rate is actually very low. You are mostly classifying negatives correctly as negative. But your false negative rate is higher at .65. You are not detecting enough images that are actually true images. For a more detailed discussion, read on.

The accuracy is an overall summary of the success of your tests. It is sometimes helpful to look at a breakdown of it into sensitivity and specificity. Sensitivity is the true positive rate, i.e. the chance your program predicts a positive given that it's supposed to be positive. Specificity is the true negative rate, i.e. the proportion of negatives correctly predicted as negative.

$sensitivity=\frac{TP}{TP+FN}=\frac{2739}{2739+5022}=.35$

$specificity=\frac{TN}{TN+FP}=\frac{103217}{103217+ 43423}=.7$

Based on these results, it seems that your test is not as good as identifying positives as it is at correctly identifying negatives. One possible problem could be that there is a difference in the prevalence of images from the truth and produced by your computer algorithm. Here a confusion matrix of the data you have collected:

$\begin{matrix}&actual&\\
predicted&truth&algorithm\\
truth&2739& 43423\\
algorithm & 5022& 103217\end{matrix}$

If you compute the accuracies separately for the true images and the algorithm images, you find that the accuracy for the true images is $\frac {2739}{2739+5022}=.35$ and the accuracy for your algorithmic images is $\frac{103217}{103217+ 43423}=.7$. (Notice these are the definitions of sensitivity and specificity, the same above.) The reason that your overall accuracy is high is because you have a much higher prevalence of algorithm-generated images; your algorithm tends to guess conservatively, and the fact that there are more algorithm generated images pushes your overall accuracy up.

A useful one-number summary combining sensitivity and specificity is the $F_1$ score, defined as $$\frac {1}{\frac{\beta^2}{\beta^2+1}\frac{1}{sensitivity}+\frac{1}{1+\beta^2}\frac{1}{specificity}}$$

When $\beta=1$, a common choice, this is the harmonic average of sensitivity and specificity:

$$\frac {1}{\frac{1}{2}\left(\frac{1}{sensitivity}+\frac{1}{specificity}\right)}$$

The $F_1$ score of your test is .47. The closer to 1, the better. You might try to optimize your algorithm based on the $F_1$ score.

There is some difference between your $F_1$ score and accuracy:

F_1: .47
accuracy: .68


问:
Why radiuses of the circumcircles are equal?
Why radiuses of the circumcircles of triangles ABC, AHC, BHC and ABH are equal? I know it can be proven using the law of sines, but I can't figure how.

答:
As one of my favorite lecturers would say, "Triangle is a democratic figure, if you apply a theorem for one vertex, you should apply it for the others too."
$$180^\circ-m(\angle ABC)=m(\angle AHC)$$
$$180^\circ-m(\angle BCA)=m(\angle BHA)$$
$$180^\circ-m(\angle BAC)=m(\angle BHC)$$
Therefore, all sine values are equal, implying the equality of circumradii.


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:
This is too long to be a comment, but only provides a modest reduction in the $C_{n-2}^2$ upper bound.

For $n\ge3$ let $a_n$ be the number of unordered pairs of triangulations of the $n$-gon with no common diagonals. Let $g(x)=a_3x^3+a_4x^4+a_5x^5+a_6x^6+\ldots$ be the generating function. The first few coefficients are $a_3=\frac{1}{2}$, $a_4=1$, $a_5=5$, $a_6=34$ (see A257887). The non-integer value of $a_3$ makes sense in light of the property that there is one unordered pair of triangulations for every set of two ordered pairs of triangulations in which the triangulations trade places, except in the case of the triangle, where the only choice is for both triangulations to be empty. The OEIS omits the $a_3$ term. To understand why $a_6=34$ let $T$, $Z$, and $F$ denote the triangle, Z-shaped, and fan triangulations described in one of your comments. Then there is one $TT$ pair, six $TF$ pairs, $12$ $ZF$ pairs, nine $ZZ$ pairs, and six $FF$ pairs. The $TT$ pair and three of the $ZZ$ pairs give the octahedron; all others give the graph $K_6$ minus a path of length $4$ described by Dan Uznanski in a comment here. This example shows that one must account, not only for rotations and reflections of a pair of triangulations, but also for different choice Hamiltonian circuit, which can result in different triangulation pairs for the same graph.

Danièle Huguet and Dov Tamari in La structure polyédrale des complexes de parenthésages, J. Combin. Inform. System Sci., 3(2):69–81, 1978 showed that the generating function satisfies
$$
x=\sum_{j=0}^\infty C_j^2\left(x-\frac{2}{x}g(x)\right)^{j+1}.
$$
In other words, $g(x)$ can be obtained by inverting the formal power series
$$
\phi(x)=\sum_{j=0}^\infty C_j^2x^{j+1},
$$
which means finding the formal power series $u(x)$ such that $\phi(u(x))=x$. Then $g(x)=\frac{x}{2}(x-u(x))$. I have not been able to find a copy of Huguet and Tamari's paper, but a derivation is also given in

Dimbinaina Ralaivaosaona, Jean Bernoulli Ravelomanana, and Stephan Wagner, Counting Planar Tanglegrams, in James Allen Fill and Mark Daniel Ward eds., 29th International Conference on Probabilistic,  Combinatorial and Asymptotic Methods for the Analysis of Algorithms (AofA 2018), Leibniz International Proceedings in Informatics (LIPIcs) 110:32:1–32:18, 2018.

Ralaivaosaona, Ravelomanana, and Wagner point out that $\phi(x)$ is the series expansion of a complete elliptic integral and use this to do an asymptotic analysis. They find that $a_n$ grows like a constant times $n^{-3}\left(\frac{4\pi}{4-\pi}\right)^n$. The upper bound $C_{n-2}^2$ grows like a constant times $n^{-3}\cdot16^n$. The revised upper bound has slower growth since $\frac{4\pi}{4-\pi}\approx14.64$.


问:
Finding whether a relation is transitive, reflexive and symmetric.
Let's say we have set: $A = \{1,2,3,...,10\}$, and there is a relation $M=P(A)-{\phi}=\{X | X\subseteq A, x\neq \phi\}.$

And there is this relation:

$T$ is defined as: $\{X,Y\}\in T$ iff  $\{X,Y\}$ is a partition of $A$.

So to find out whether it's transitive, I first tried to find the result of the relation $M$, and then found it to be almost impossible and impractical, as calculating $2^{10}$ of subsets is not the best idea, and the by using this I could calculate the $R^2$, but that's not the case this time.

Then I tried finding whether it's Symmetric, but because I couldn't calculate $M$, as I think I can't also find $M^{-1}$.

And.. I am pretty sure it's not reflexive as it can't have the Identity because it's the $P(A)$ of this set, meaning there isn't a set like this one: $\{1,1\}$.

I would like to get some assistance \ explanation on finding whether a relation is reflexive, symmetric, and transitive.

答:
$M$ isn't a relation, it's the universe on which $T$ is a relation.

Since $\{X,\,Y\}$ is only a partition of $A$ only if $X\cap Y=\emptyset$, in general $(X,\,X)\notin T$, so $T$ isn't reflexive.

Since $\{X,\,Y\}=\{Y,\,X\}$, $T$ is symmetric.

If $(X,\,Y)\in T$ and $(Y,\,Z)\in T$ then $X=A\setminus Y=Z$, so $(X,\,Z)=(X,\,X)\notin T$ (becaise $\emptyset\cup\emptyset\ne A$), so $T$ isn't transitive.


问:
What can be deduced from $P(A or B) = P(A) + P(B)$?
$P(A \cup B) = P(A) + P(B)$

(1) If $A$ and $B$ are independent, then $P(A) = 0$ or $P(B) = 0$.

$\quad$ True, $P(A \cap B)= 0 =P(A)\cdot P(B)$

(2) If $A$ = S(sample space), then $P(B) = 0$

$\quad$ True, since $P(A \cap B) = 0$

(3)  If $P(A) >P(B) > 0$, then $A$ and $B$ are not independent

$\quad$ True, if both are positive, then $P(A \cap B) > 0$, which contradicts $P(A \cap B) = 0$

(4) If $P(A) \neq P(B)$, then $A$ and $B$ are mutually exclusive

$\quad$ True, this is vacuously true since $P(A \cap B)= 0$, $A$ and $B$ are already mutually exclusive

Is my reasoning correct? I am unsure about (4), is there a counterexample that disproves (4)?

答: Well, we know $\Pr(A\cap B)=0$, but that doesn't necessarily mean $A\cap B=\emptyset$.  Choose a real number $X$ uniformly at random from $[0,1]$. Let $A$ be the event $X\in[0,\frac23]$ and let $B$ be the event $X\in[\frac23,1]$.  $\Pr(A\cap B)=0$, but $A\cap B$ is the event that $X=\frac23$.


答: Hint. Prove (say: with angle chasing) that the reflections of the orthocenter across each side $AB, BC, CA$ respectively lie on the circumcircle of $\triangle ABC$. Since the resulting triangles $H_cAB, \ldots$ are congruent to the original ones, you’re done.

Title: Why radiuses of the circumcircles are equal?

Body: Why radiuses of the circumcircles of triangles ABC, AHC, BHC and ABH are equal? I know it can be proven using the law of sines, but I can't figure how.


问:
What is the name for a vector field that is both divergence-free and curl-free?
Consider a smooth vector field $\mathbf u\colon\Omega\to\mathbb R^3$ defined on an open domain $\Omega\subseteq\mathbb R^3$ such that $\mathbf u$ has zero divergence and zero curl on $\Omega$, that is,
$$\begin{align}\nabla\cdot\mathbf u&=0,\\\nabla\times\mathbf u&=0.\end{align}$$
Is there a specific technical name for such a vector field?

Wikipedia calls it a Laplacian vector field, but

it does not cite any references, and

it asserts that any such vector field is the gradient of a harmonic function, but this is only true if $\Omega$ is simply connected (counterexample: $\big(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\big)$ on the region $x^2+y^2>0$),

so I'm disinclined to trust it. Can anyone provide references supporting this or any other name?

答:
Put musical isomorphism aside, I believe what Rahul Narain refers to is just harmonic $1$-form.

In Hodge decomposition for $k$-forms $\omega$:
$$
\omega =\mathrm{d}\alpha +\delta \beta + \gamma
$$
where $\gamma$ is harmonic in that $(\mathrm{d}\delta + \delta\mathrm{d})\omega = 0$, and $\delta = (-1)^{nk+n+1} \star^{n-k+1}\mathrm{d}^{n-k}\star^k$.

In the 3-dimensional case. We have the cochain complex:
$$
\Lambda^0\ \stackrel{\mathrm{d}^0}{\longrightarrow}\ \Lambda^1 \ \stackrel{\mathrm{d}^1}{\longrightarrow}\ \Lambda^2\ \stackrel{\mathrm{d}^2}{\longrightarrow}\ \Lambda^3.
$$
Define $\delta  = \mathrm{d}^*_{k}: \Lambda^*_k\to\Lambda^*_{k-1}$ as the adjoint of $\mathrm{d}^{k-1}: \Lambda^{k-1}\to\Lambda^{k}$ with respect to the inner product. We can have somewhat a dual complex:
$$
\Lambda^*_3\ \stackrel{\mathrm{d}^*_3}{\longrightarrow}\ \Lambda^*_2 \ \stackrel{\mathrm{d}^*_2}{\longrightarrow}\ \Lambda^*_1\ \stackrel{\mathrm{d}^*_1}{\longrightarrow}\ \Lambda^*_0.
$$
For a harmonic 1-form $\gamma$:
$$
(\mathrm{d}\delta + \delta\mathrm{d}) \gamma = (\mathrm{d}^0\mathrm{d}^*_1 + \mathrm{d}^*_2 \mathrm{d}^1)\gamma = 0. \tag{1}
$$
Note: $$\mathrm{d}^1 = \nabla \times, \quad\mathrm{d}^*_1 = (-1)\star^{0}\mathrm{d}^{2}\star^1 = -\nabla \cdot$$
(1) is:
$$
\nabla \times (\nabla \times \gamma) - \nabla (\nabla \cdot \gamma) = 0.\tag{2}
$$
We can say a curl-free and a divergence-free vector field is harmonic under musical isomorphism for
$$
 \nabla \times \gamma = 0\;\text{ and } \nabla \cdot \gamma = 0\Longrightarrow\nabla \times (\nabla \times \gamma) - \nabla (\nabla \cdot \gamma) = 0 .
$$

I am guessing the wikipedia page was using "Laplacian vector fields" in that (2)'s left side is actually the vector Laplace operator (or Laplace-Beltrami) acting on a vector field.

For references, we use this term a lot in computational geometry, a field which inherits a lot of terminologies from vector calculus, it is like almost a tradition that saying a vector field is harmonic means it is curl-free and divergence-free w/o citing anyone's book.


问:
A men's endurance competition has an unlimited number of rounds

A men's endurance competition has an unlimited number of rounds. In each round, a competitor has, independently, a probability p of making it through the round; otherwise, he fails the round. Once a competitor fails a round, he drops out the competition; before he drops out, he takes part in every round. The grand prize is awarded to any competitor who makes it through a round which all the other remaining competitors fail; if all the remaining competitors fail at the same round the grand prize is not awarded. If the competition begins with three competitors, show that the probability that the grand prize is awarded is $\frac{3p(1+p^2)}{(1+p)(1+p+p^2)}$

I have an attempt at the problem which I am pretty convinced by and gives a similar answer, but not correct. I would like to know where my reasoning goes wrong

Suppose we number each player P1,P2,P3. We will find the probability P3 wins the grand prize and by symmetry that is the same for all of them.

P(P3 wins on the rth round) = $2\choose1$ P(A player fails on the rth round) * P(A player fails before or on the rth round) * P(A player passes r rounds)

The choose permutes P1 and P2. I.e. P1 could fail on the rth and P2 before, or vice versa.

P(A player fails on the rth round) $=p^{r-1}(1-p)$

P(A player passes r rounds) $=p^r$

P(A player fails on or before the rth round $=1-p^r$

So now, we want to sum from $r=1$ to infinity. We multiply by three to perform this for each player.

$$P(\text{Grand prize awarded})=\sum_{r=1}^\infty 6p^{2r-1}(1-p)(1-p^r)=\frac{6p}{(p+1)(p^2+p+1)}$$

答:
The first calculation, where you multiply by $\binom21$ is wrong, I think. It's okay to multiply the probability that A fails on round $r$ and B fails before round $r$ by $2$, but you're also multiplying the probability that both fail on round $r$ by $2$, if I understand you correctly.

As noted in the comments, the answer is only valid for $0\leq p<1$.


问:
Question on notation in Eisenbud's Commutative Algebra With a View Toward Algebraic Geometry Exercise 2.19(b)
I will restate the question here:

Let $R$ be a ring and let $M$ be an $R$-module. Suppose that $\{f_i\}$ is a set of elements of $R$ that generate the unit ideal. Prove:
If $m_i\in M[f_i^{-1}]$ are elements such that $m_i$ and $m_j$ go to the same element of $M[f_i^{-1}f_j^{-1}]$, then there is an element $m\in M$ such that $m$ goes to $m_i$ in $M[f_i^{-1}]$ for each i.

I am confused about the notation $M[f_i^{-1}f_j^{-1}]$. If it is indeed the localization with respect to the product, then if $m_i=n_i/f_i^{k_i}$, am I to view $m_i\in M[f_i^{-1}f_j^{-1}]$ as $n_if_j^{k_i}/(f_if_j)^{k_i}$? Otherwise, does he mean $M[f_i^{-1}f_j^{-1}]=M[\{f_i,f_j\}^{-1}]$?

Does anyone have insight into this notation? From other questions on this site it seems that he means the former, but I am still confused what he means by "
$m_i$ in $M[f_i^{-1}f_j^{-1}]$". Does anyone have insight into this?

答:
It means either of the following things, which are all canonically isomorphic (here, canonically = uniquely if you require that it respects the map coming from $M$):

$M[(f_if_j)^{-1}]$, $M[f_i^{-1}][f_j^{-1}]$, $M[\{f_i,f_j\}^{-1}]$

In particular, you have a morphism $M[f_i^{-1}]\to M[f_i^{-1}][f_j^{-1}]$ and in particular $m_i$ (an $m_j$) is sent to someone in the latter.

If you want to view it as $M[(f_if_j)^{-1}]$, then indeed, you will have that the image of $m_i = \frac{n_i}{f_i^{k_i}}$ is $\frac{n_if_j^{k_i}}{(f_if_j)^{k_i}}$

But you donw't have to view it that way, you can also simply say it's $\frac{m_i}{1}$ for instance (that's the $M[f_i^{-1}][f_j^{-1}]$ point of view) or just $\frac{n_i}{f_i^{k_i}}$ (that's the $M[\{f_i,f_j\}^{-1}]$ point of view)


问:
find the equation of the circle $x^2 +y^2 +ax +by = c$ passing through points $(6,8), (8,4), (3,9)$

Find the equation of the circle $x^2 +y^2 +ax +by = c$ passing through points $(6,8), (8,4), (3,9)$.

How do I go about solving this? I don't have a textbook assigned so I'm not even sure what this relates to, but its the beginning of linear algebra. so I guess I'm supposed to construct some kind of system of equations...

答:
$x^2 +y^2 +ax +by = c$ passing through points $(6,8), (8,4), (3,9)$.

It is given by the determinant:

$$ \begin{vmatrix} x^2 +y^2&  x& y& 1 \\ 6^2 + 8^2&  6& 8& 1 \\ 8^2 + 4^2&  8& 4& 1 \\ 3^2+9^2&  3& 9& 1\end{vmatrix} =0$$

Observe that equation has no $xy$ term, and coefficent of $x ^2$ and $y^2$ same, also notice that subbing $(x,y)$ as any of the point given solves the determinant. Hence it's the unique circle passing through three points QED.

refer here


答:
We have P(A or B)=P(A)+P(B)-P(A and B)

So, P(A or B)=P(A)+P(B) implies P(A and B)=0

As already answered here, note that it does not mean (A and B) is an impossible event.

Title: What can be deduced from $P(A or B) = P(A) + P(B)$?

Body:
$P(A \cup B) = P(A) + P(B)$

(1) If $A$ and $B$ are independent, then $P(A) = 0$ or $P(B) = 0$.

$\quad$ True, $P(A \cap B)= 0 =P(A)\cdot P(B)$

(2) If $A$ = S(sample space), then $P(B) = 0$

$\quad$ True, since $P(A \cap B) = 0$

(3)  If $P(A) >P(B) > 0$, then $A$ and $B$ are not independent

$\quad$ True, if both are positive, then $P(A \cap B) > 0$, which contradicts $P(A \cap B) = 0$

(4) If $P(A) \neq P(B)$, then $A$ and $B$ are mutually exclusive

$\quad$ True, this is vacuously true since $P(A \cap B)= 0$, $A$ and $B$ are already mutually exclusive

Is my reasoning correct? I am unsure about (4), is there a counterexample that disproves (4)?


问:
If weak topology and weak* topology on $X^*$ agree, must $X$ be reflexive?
Let $X$ be a Banach space and suppose that the weak topology on $X^*$ agrees with the weak* topology on $X^*$. Must $X$ be reflexive?

To prove the contrapositive, it will suffice to assume that $X$ is not reflexive and construct a sequence $\phi_n \in X^*$ such that $\phi_n(x) \rightarrow \phi(x)$ for each $x\in X$, but $\lambda(\phi_n)\not\rightarrow \lambda(\phi)$ for some bounded linear fucntional $\lambda$ on $X^*$. However, I have been unable to do so. Does anyone have any ideas?

答:
Let $J : X \longrightarrow X^{**}$ be the natural embedding of $X$ into $X^{**}.$ Suppose that weak-topology and weak$^{*}$-topology on $X^*$ coincides i.e. $\sigma (X^*, J(X)) = \sigma (X^*, X^{**}).$ Then each $\xi \in X^{**}$ being continuous with respect to the weak topology $\sigma (X^*,X^{**})$ on $X^*$ they will be continuous with respect to the weak$^{*}$-topology $\sigma(X^*,J(X))$ on $X^*.$ That means $\xi \in \text {span}\ (J(X)) = J(X).$ So we have $X^{**} \subseteq J(X),$ proving that $J(X) = X^{**}$ i.e. $X$ is a reflexive normed linear space, as required.

Note that completeness of $X$ is superfluous here. The above argument works fine for any arbitrary normed linear space. Also the above argument doesn't involve Banach-Alaoglu's theorem.


答:
I think the third one is incorrect, if you choose $B=A^{c}$ then they are dependent, and you can build an example where both of probabilities is positive, but of course still $P(A\cap B)=0$

As for the fourth one, I think the counter example comes from not countable sample spaces, where you can consider:
A is distributed uniformly over [0,1] ($P(A)=1$)
and B is the rational numbers set ($\mathbb{Q}$) (*)
Then you get $$P(A\cup B)= P(A)+P(B) =1$$
and $1=P(A)\neq P(B)=0$
but of course $B\subseteq A$

(*) or any other countable set


问:
Analytic solution of the heat equation with a source term
I have the heat equation with Dirichlet boundary conditions
$$u_t(t,x)=u_{xx}(t,x)+\sin(x)$$
$$u(t,0)=u(t,2\pi)=0$$
$$u(0,x)=u_0(x)$$
Now, without the source term I could write the solution as
$$u(t,x) = \sum_{n=1}^\infty B_n \sin(n\pi x)e^{-n^2\pi^2t}$$
where
$$B_n=2\int_0^{2\pi}u_0(x)\sin(n\pi x)dx$$
but I'm not sure what it looks like with a source term. I had a problem set question which assumed knowledge of the solution to show it converges to it's stationary form, so I'm guessing it can be derived using the homogenous equation.

答:
Hint:

the $\sin x$ source term is time independent: what happens if you add $t \sin x$ to your solution?

does it respect the PDE ? and the boundary conditions ?


问:
Probability of 3 friends being part of 5 person committee
There are $18$ students that will be forming a committee of $5$ people. The people in the committee will be picked out of a hat randomly. What is the probability that Joe, Joseph, and John will be $3/5$ members of the committee?

I've tried during various things, I know that the total number of combinations of 5 people that can be formed from $18$ is $8568$  $(18C5)$, but I'm not exactly sure what the numerator should be. I would love an explanation as to why the answer is why it is please

答:
You've got one part of the answer: all of the possible committees which is $_{18}C_5$.

Now you'll need the number of committees that have Joe, Joseph, and John.

As a hint, you know three of the five committee members -- Joe, Joseph, and John -- so you only need to choose two from who's left.

Can you take it from here?

After picking Joe, Joseph, and John, there are $15$ left, and we need to pick two, which is $_{15}C_2$. So the probability is $(_{15}C_2)/(_{18}C_5)$.


问:
Technique for proving four given points to be concyclic?
While making my way through an exercise, I stalled on question 7:

7. Prove that the points $(9, 6)$, $(4, -4)$, $(1, -2)$, $(0, 0)$ are concyclic.

The book does not provide any guidance on how to tackle such a question and I can only assume that the authors are assuming that anybody using their textbook will have covered this in a previous course.

I have looked around online for resources that could help me, but none of the ones I've found seem to offer a thorough explanation, building from the ground up.

Would anybody here happen to know where I can find a good explanation of how to prove such a thing?

Body:
From this answer here, the condition is given as:

$$
   \det\begin{bmatrix}
     9^2+6^2 & 9 & 6 & 1 \\
     4^2+(-4)^2 & 4 & -4 & 1 \\
     1+2^2 & 1 & -2 & 1 \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix} = 0
$$


问:
When does variété mean manifold?
Following advice from this post, I am in the process of translating Ehresmann's 1934 paper "Sur la Topologie de Certains Espaces Homogènes" from French to English.

French-English dictionaries online and Google translate are helping me out quite a bit. However, I'm confused about the standard usage of the word variété in French mathematics writing. It can be translated as either variety or manifold.

Does the phrase  variété algébrique translate to algebraic variety, or algebraic manifold?

The difference is subtle, and maybe not all that important for my purposes in studying the Grassmann manifolds which are simultaneously algebraic varieties and topological manifolds, whence algebraic manifolds. Of course, I could think of Grassmann manifolds all the time as algebraic manifolds, but in translating the paper I want to have the same frame of mind as Ehresmann did when he wrote it.

答:
Nowadays (note the caveat!) the translation into English of "variété algébrique" is definitely "algebraic variety". To give an explicit example, the equation $y^2=x^2+x^3$ describes "une variété algébrique affine" and in English "an affine algebraic variety"  ,  not "an algebraic manifold".
Traditionally, if you wanted to translate "algebraic manifold" into French you would say "une variété algébrique non-singulière": the adjectives "régulière" is more recent and "lisse" even more recent (due to Grothendieck I would guess). I even remember reading the clumsy sentence "une variété algébrique non nécessairement non-singulière" just meaning  "an algebraic variety" !

A standard  pun/joke is "une variété algébrique n'est pas une variété" which translated becomes "an algebraic variety is not a manifold",  a true but not sidesplitting statement.

Finally, in (differential ) topology I have never heard anything but variété suitably qualified: topologique, différentiable (différentielle), $\mathcal C^k$, à bord. The translations would be : topological, differentiable (differential), $\mathcal C^k$, manifold and manifold with boundary .

Title: When does variété mean manifold?


问:
How to prove that End(R) is isomorphic to R?
I tried to prove this:
$R$ as a ring is isomorphic to $End(R)$, where $R$ is considered as a left $R$-module. The map of isomorphism is

$$F:R\to End(R), \quad   F(r)=fr,$$
where $fr(a)=ar$. And I define the multiplication in $End(R)$ by $(.)$, where $h.g=g\circ h$ for $g,h$ in $End(R)$.

Is this true?

答:
With your definition of multiplication in $\def\End{\operatorname{End}}\End(R)$ the map is indeed an isomorphism.

Let's go slowly. We define, for $r\in R$,

$$
F(r)\colon R\to R,\quad x\mapsto xr
$$
It's probably better to use $F_r$ instead of $F(r)$, so
$$
F_r(x)=xr
$$
It's clear that $F_r$ is an $R$-module homomorphism; also
$$
F_r+F_s=F_{r+s}
$$
is a trivial verification.

Let's see what happens with $F_{rs}$:

$$
F_{rs}(x)=x(rs)=(xr)s=F_s(xr)=F_s\circ F_r(x)
$$
for all $x\in R$, therefore
$$
F_{rs}=F_s\circ F_r = F_r\cdot F_s
$$

We have proved that $F\colon R\to\End(R)$ is a ring homomorphism, as $F_1$ is clearly the identity.

Is it an isomorphism? If $F_r$ is the zero map, then $F_r(1)=0$, so $0=r1=r$. Thus $F$ is injective.

If $f\colon R\to R$ is a module homomorphism, set $r=f(1)$. Then

$$
F_r(x)=xr=xf(1)=f(x1)=f(x)
$$
We can push $x$ inside $f$ because $f$ is a homomorphism.

A common convention in module theory is to write morphisms of left modules on the right, but composing them "as written": so, if $f\colon L\to M$ and $g\colon M\to N$ are left module homomorphism, their composition is
$$
f\bullet g\colon L\to N,\quad (x)f\bullet g=((x)f)g
$$

With this convention, your map $F$ becomes an isomorphism between $R$ and $\End(R)$ without changing the operation on $\End(R)$: it's just the same, actually, but computations are more straightforward.


问:
Expressing a projective conic in standard form
I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:

$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$

into the form:

$$X^2 + Y^2 + Z^2 = 0$$

Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:

$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end {array} \right]
$$

Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).

In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:

$$  \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\,
\sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3}
\\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array}
 \right]
$$

I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?

答:
So you want to diagonalize $$A=\begin{bmatrix} 2&1&-3/2\\ 1&0&2
\\ -3/2&2&1\end{bmatrix}$$

First find the eigenvalues

$$ {\rm det}(A-w \mathbf{1}) = 0 $$

which are $$\mathbf{w} = \pmatrix{w_1 \\ w_2 \\ w_3} = \pmatrix{ -2.2599587007 \\ 3.1586587920 \\ 2.1012999087} $$ in arbitrary order.

Now form a matrix $X$ from the eigenvectors as columns. There are many ways to get there, but I will demonstrate a linear algebra method.

Solve the system $$ (A -w  \mathbf{1}) \pmatrix{v_1\\v_2\\v_3} = \pmatrix{ \xi \\ \xi \\ \xi } $$ for $v_1$, $v_2$ and $\xi$ and defer the $\xi \rightarrow 0$ for later.

This gives me the eigenvector as a function of $w$

$$ \mathbf{v}(w)= \pmatrix{v_1 \\ v_2 \\ v_3} = \pmatrix{ \left( \frac{2 w^2-3 w-12}{2 w^2-3 w-9} \right) v_3 \\ \left(1+ \frac{ 3(2 w-7)}{2 ( 2 w^2-3 w-9)} \right) v_3 \\ v_3} $$

and we need to normalize the eigenvectos, such that $\mathbf{v}^\top \mathbf{v} = 1$ which leads to $$v_3^2 =  \frac{1}{3+ \left( \frac{3(16 w^3-84 w^2-48w+483)}{4(2 w^2-3w-9)^2}\right)}$$

Now that we can define the eigenvectors $\mathbf{v}$ for each eigenvalue $w$ we construct the following 3×3 matrix

$$ X = \begin{bmatrix} \vdots & \vdots & \vdots \\ \mathbf{v}(w_1) & \mathbf{v}(w_2) & \mathbf{v}(w_3) \\ \vdots & \vdots & \vdots \end{bmatrix} $$

$$ X = \begin{bmatrix}
0.37748124858 & -0.70137614012 & -0.60463163747 \\ -0.70173922139 & 0.20936797859 & -0.68097512046 \\ 0.60421020524 & 0.68134907323 & -0.41315066050 \end{bmatrix} $$

And now for the determination of the diagonal matrix $D$ which factors $A$ as follows $$A = X D X^{-1}$$ or $$D = X^{-1} A X$$

$$D = \begin{bmatrix}
 -2.2599587007 & & \\
  & 3.1586587920 & \\
 & & 2.1012999087
\end{bmatrix}$$

This means a transformation for the solution vector $p = \pmatrix{x \\ y \\ z}$ into  $$p = X u $$ or $$ u = X^{-1} p$$

The conic

$$ p^\top A p = \mathbf{0} $$

becomes

$$ u ^\top D u = \mathbf{0} $$

since $X^\top = X^{-1}$ and $p^\top A p = (X u)^\top X D X^{-1} (X u) = u^\top (X^\top X ) A ( X^{-1} X) u = u^\top D u$.


答:
Note that since you're on $[0,2\pi]$, the appropriate argument of the eigenfunctions is not $n \pi x$. This is not just a minor technicality since it affects how you must deal with the forcing.

Once you get that straight, consider the evolution equation for the $n$th Fourier coefficient (obtained by multiplying both sides by the $n$th eigenfunction and integrating, and pulling the time derivative out of the first term). You should find that all but one of them are just of the form $\frac{d}{dt} \hat{u}_n + \lambda_n u_n = 0$ (which is what you see in the homogeneous heat equation) while the $n=1$ equation is of the form $\frac{d}{dt} \hat{u}_1 + \lambda_1 u_1 = c$. This latter ODE is simple enough that you can just solve it.

Title: Analytic solution of the heat equation with a source term

Body:
I have the heat equation with Dirichlet boundary conditions
$$u_t(t,x)=u_{xx}(t,x)+\sin(x)$$
$$u(t,0)=u(t,2\pi)=0$$
$$u(0,x)=u_0(x)$$
Now, without the source term I could write the solution as
$$u(t,x) = \sum_{n=1}^\infty B_n \sin(n\pi x)e^{-n^2\pi^2t}$$
where
$$B_n=2\int_0^{2\pi}u_0(x)\sin(n\pi x)dx$$
but I'm not sure what it looks like with a source term. I had a problem set question which assumed knowledge of the solution to show it converges to it's stationary form, so I'm guessing it can be derived using the homogenous equation.


问:
Show that sequence $a_n$ defined by $a_{n+1} := \frac{(a_n + b_n )}{2}$, $b_{n+1} := \sqrt{a_n b_n}$ is decreasing
Let $a_1 ≥ b_1 > 0$, and define $a_n$ and $b_n$ for all $n>
1$ as

$a_{n+1} := \frac{(a_n + b_n )}{2}$
and
$b_{n+1} := \sqrt{a_n b_n}$.

How can I show that the sequence $a_n$ is decreasing? I thought about using that $a_{n+1} - a_n < 0$, but it becomes rapidly unbearable.

答: The case $a_1=b_1$ is trivial, so we assume $a_1>b_1$. Then by the AM-GM inequality we get $b_{n+1}<a_{n+1}$ and $a_{n+2}=\frac{a_{n+1}+b_{n+1}}{2}<\frac{a_{n+1}+a_{n+1}}{2}=a_{n+1}$.


问:
If $f'(x) = L$, then $\lim_{x \to x_0} \frac{|f(x) - [f(x_0)+L(x-x_0)]|}{|x-x_0|} = 0$
In this exercise we are asked to prove that $f$ is differentiable at $x_0$ with $f'(x_0) = L$ iff

$$\lim_{x \to x_0} \frac{|f(x) - [f(x_0)+L(x-x_0)]|}{|x-x_0|} = 0.$$ In my proof sketches, I did
$$\lim_{x \to x_0} \frac{|f(x) - f(x_0)|}{|x-x_0|} - \lim_{x \to x_0} \frac{L(x-x_0)}{|x-x_0|} = L-L = 0.$$ But looking back, I'm pretty sure this isn't allowed with the absolute values. Any hints on other methods I should proceed, or how to fix the absolute value problem?

Thank you!

Body: HINT: $$\dfrac{|f(x)-[f(x_0)+L(x-x_0)]|}{|x-x_0|} = \left|\dfrac{f(x)-f(x_0)-L(x-x_0)}{x-x_0}\right| = \left|\dfrac{f(x)-f(x_0)}{x-x_0}-L\right|.$$


答:
In general,
$
P(A\cup B)=P(A)+P(B)-P(A\cap B).
$

(1) is correct, as you observed that the assumption $P(A\cup B)=P(A)+P(B)$ is equivalent to $P(A\cap B)=0$, and independence says $P(A\cap B)=P(A)P(B)$.

(2) is correct since $P(B)=P(A\cup B)=0$.

(3) is correct since it is the contrapositive of (1).

(4) No, $P(A\cap B)=0$ does not imply that $A\cap B=\emptyset$. A counterexample: $A=\{0\le X\le 0\}$ and $B=\{0\le X<\infty\}$ where $X$ is a random variable with normal distribution.  Then in this example $P(A)\ne P(B)$ and $P(A\cap B)=0$. But $A\cap B\ne\emptyset$.


问:
Modify the rules of Gomoku (Five-in-a-row) or Connect Four type games to enforce the fairness among players
One colleague and me were discussing this problem during lunch today, and I did a little bit digging for several hours after returning to my office.

Fact: For an $(m,n,k)$-game, there does not exist a strategy that assures that the second player will win. For example, in Five-in-a-row, first player wins with perfect play.

Now our question is: How to modify the rules to make the game fair for both players?

I googled a bit more, found an MSE question that addresses the fairness of this kind of games.

I think the fairness in that question can be strenghthen by the following definition: A game is fair if and only if that two players play with perfect strategy, the game will always be a draw. This is fair in that this game favors player with less mistakes. For example, Tic-tac-toe is fair.

Several proposals to modify the rules to enforce the fairness:

Impose more restrictions on the player who plays first. For example, Five-in-a-row bans black to play "three and three", "four and four", and "overlines".

Change the $(m,n,k)$-game to an $(m,n,k,p,q)$-game: $k$-in-a-row on an $(m\times n)$-board, first player put $p$ stones on board, in subsequent moves, players put $q$ stones on board. For example, [Connect 6].

Go to higher dimension. For example, Five-in-a-row played in a $19\times 19\times 19$ "board".

I am no expert in combinatorial game theory or computational complexity in board games, but it is always nice to learn some new things in summer time. My question is: Any papers or treatises dealing the fairness of this kind of games? or more specifically, any proof of the fairness using above three modifications of rules? especially, for Five-in-a-row, do those additional rules make the game fair?

Any analysis of fairness on simpler cases like Connect 4 is welcome too.

Lastly just out of curiousity, as an avid Go player myself, I wonder is there any mathematical analysis on addressing the fairness using komi in the Go game? Or Go is just too complex to analyze...

答: In response to your last question, Elwyn Berlekamp has introduced the concept of Coupon Go (video lecture - the intro stops around 16:15 and the interesting discussion starts there). In Coupon Go, in addition to the board there is a stack of cards ranging from 20 (on top) to 0.5 (on the bottom). When it is your turn, you are allowed to take and keep the top card (worth that many points) rather than making a move on the board. Among other things, this effectively allows the two players to bid on the komi value instead of having it chosen for them.</p>


问:
Show that pair of straight lines $ax^{2}+2hxy+ay^{2}+2gx+2fy+c=0$ meet coordinate axes in concyclic points. Also find equation of the circle
Show that pair of straight lines $ax^{2}+2hxy+ay^{2}+2gx+2fy+c=0$ meet coordinate axes in concyclic points. Also find equation of the circle through those cyclic points

My Attempt:

Given equation to the pair of straight lines is

$ax^2+2hxy+ay^2+2gx+2fy+c=0$

Let the lines be

$l_1x+m_1y+n_1=0$
and
$l_2x+m_2y+n_2=0$

Now what should I do next?

答:
Begin by finding intercepts:

$$ax^2+2hxy+ay^2+2gx+2fy+c=0$$

For x intercepts,

$$ ay^2 + 2fy +c = 0$$

For y intercepts,

$$ ax^2 + 2gx + c = 0$$

Let the points corresponding to root be: $(a,0) , (b,0) , (0,c) , (0,d)$

From this answer here, the equation of circle through these points:

$$
   \det\begin{bmatrix}
     x^2 +y^2 & x & y & 1 \\
     b^2 & b & 0 & 1 \\
     c^2 & 0 & c& 1 \\
    d^2 & 0 & d& 1 \\
   \end{bmatrix} = 0
$$

For checking if $(a,0)$ is on this circle , simply plug it in place of $(x,y)$


问:
Nonlinear recurrence relation over $\mathbb{C}$
A sequence in $\mathbb{C}$ is given: $(a_n)_{n=1}^{\infty}$ s.t.

$a_1 = i$ and $a_{n+1} = \frac{3}{2 + a_n}$.

Assume $\lim(a_n)$ exists.

Using the fact that $\lim(a_n) = \lim(a_{n+1})$ and solving the resulting quadratic gives that $\lim(a_n)\in \{-3,1\}$.

I was trying to figure out which one the limit is. I tried finding a relation on the norm or the argument (using the initial condition, since clearly that is what the limit is determined by) and I also tried looking at individual terms and by this, the real part seems to stay positive and hover around one while the imaginary part we know goes to zero in any case (i.e., I am thinking the limit is 1). I wanted to ask for verification that I did this last part correctly, since $a_2 = \frac{6-3i}{5}$ if we can show that:

$Sgn[Re(a_m)] = Sgn[Re(a_{m+1})]$ $\forall m\geq 2$

we are done and the limit is 1. Suppose that $a_m = x+yi$ ($m\geq 2$) and $x > 0$. Then

$a_{m + 1}= \frac{3}{2 + x + yi}\frac{(2+x - yi)}{(2+x-yi)} = \frac{3(2+x) - 3yi}{(2+x)^2 + y^2}$

meaning that

$Sgn[Re(a_{m+1})] = Sgn\bigg[\frac{6 + 3x}{(2+x)^2 + y^2}\bigg] = 1$ since $x > 0$ and we are done. So the limit is 1. Does that look about right?

答:
A specific approach to convergence is to examine how $a_{n}-1$ changes in each iteration.

\begin{align}
a_{n+1} - 1
&= \frac{3}{2+a_n} - 1 \\
&= \frac{1-a_n}{3-(1-a_n)}
\end{align}
Then if $|a_n-1|$ is small relative to $3$ it is apparent that $|a_{n+1}-1|$ decreases by  a factor of nearly $1/3$ on each iteration and so $a_{n+1} \to 1$.  In fact, as long as $3-|1-a_n|$ is greater than some bound greater than $1$, convergence is guaranteed.   For example if $|a_{n-1}-1|< 15/8$ we have,
\begin{align}
|3-(1-a_n)| &\geqslant 3 - |1-a_n| \\
&\geqslant \frac{9}{8}
\end{align}
whence $|a_{n+1} -1| \leqslant \frac{8}{9} | a_n - 1 |$.

Starting with $a_1 = i$, $|a_1-1| = \sqrt 2 < \frac{15}{8}$ so the condition is met and the sequence converges to $1$.

Another more general approach is to recognise that recurrence is a particular case of a continued fraction.  Consider the expression,
\begin{align}
f = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{p_3}{\ddots}}}
\end{align}
If we define functions of $w \in \mathbb C$ by,
\begin{align}
t_0(w) = q_0 + w, \quad t_n(w) = \frac{p_n}{q_n +w}
\end{align}
then when we combine $t_0$ through $t_n$ we obtain,
\begin{align}
c_n(w) = t_0\circ t_1 \circ \cdots \circ t_n(w) = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{\ddots}{q_{n-1}+\dfrac{p_n}{q_n+w}}}}
\end{align}
and we interpret $f$ to be the limit of this as $n \to \infty$ when $w=0$ (if it exists).  The number $c_n(0)$ is called the $n$th convergent.

A useful recurrence relation exists for $c_n(w)$, $n \geqslant 0$,
\begin{align}
c_n(w) = \frac{P_n+P_{n-1}w}{Q_n +Q_{n-1} w} \tag{1}\label{eq1}
\end{align}
where we use initial conditions, $P_0 = q_0, Q_0 = 1, P_{-1} = 1, Q_{-1}=0$ and the $P_n, Q_n$ do not depend on $w$ and can be obtained from the recurrence,
\begin{align}
\begin{array}{l}
P_{n+1} = q_{n+1}P_n + p_{n+1} P_{n-1} \\
Q_{n+1} = q_{n+1}Q_n + p_{n+1} Q_{n-1} .
\end{array}\tag{2}\label{eq2}
\end{align}
This is proved by induction.  We observe that the given initial conditions imply equation \eqref{eq1} holds when $n=0$.  Then, if \eqref{eq1} holds for some $n \geqslant 0$, from the definition,
\begin{align}
c_{n+1}(w) &= c_n(p_{n+1}/(q_{n+1}+w)) \\
&= \frac{P_n+P_{n-1}(p_{n+1}/(q_{n+1}+w))}{Q_n+Q_{n-1}(p_{n+1}/(q_{n+1}+w))} \\
&=\frac{P_{n}q_{n+1}+P_n w + P_{n-1}p_{n+1}}{Q_{n}q_{n+1}+Q_n w + Q_{n-1}p_{n+1}} \\
&=\frac{P_{n+1}+P_n w}{Q_{n+1}+Q_n w}.
\end{align}
Thus \eqref{eq1} also holds for $n+1$ with the new coefficients $P_{n+1}, Q_{n+1}$ given by \eqref{eq2}.

Applied to the case in hand, we take $q_0 = 0, p_n = 3, q_n=2$ and $w = i$, so that $c_0(i) = i, c_1(i) = 3/(2+i)$ and in general $c_n(i) = a_{n+1}$.

We can solve the two recurrence relations for $P_n, Q_n$ using standard techniques with the initial conditions provided above, obtaining
\begin{align}
P_{n} &= \frac{3}{4} (3^{n} - (-1)^n) \\
Q_{n} &= \frac{1}{4} (3^{n+1} +(-1)^n )
\end{align}
In this form, eliminating the factor $4$, we obtain, for general $w$,
\begin{align}
c_{n-1}(w) &= \frac{3(3^{n-1}-(-1)^{n-1})+3w(3^{n-2}-(-1)^{n-2})}{3^n+(-1)^{n-1}+w(3^{n-1}+(-1)^{n-2})} \\
&=\frac{(3+w)(-3)^{n-1}-3+3w}{(3+w)(-3)^{n-1}+1-w} \\
&=1+\frac{12(1-w)}{(3+w)(-3)^{n}-3(1-w)}.
\end{align}
From which it is clear the recurrence will converge to $1$ regardless of the starting value $w$ provided not $-3$.  Setting $w=i$ this becomes after multiplying numerator and denominator by $(1+i)/2$
\begin{align}
a_n &= c_{n-1}(i) \\
&=1+\frac{12}{(1+2i)(-3)^{n}-3}.
\end{align}


答:
Notice that $f(x) = \sin x$ solves
\begin{align}
f_{xx}+\sin x = 0,
\end{align}
with the given boundary conditions, that is, $f$ is a stationary solution to the above problem.

Next, consider the function $v=u-f$ where $u$ solves the above heat equation. Then, we see that
\begin{align}
v_t-v_{xx} = u_t-u_{xx}+f''= \sin x+f'' = 0.
\end{align}

Here, we see that $v$ measures how $u$ deviates from the stationary solution $f$. The above calculation demonstrates that the deviation satisfies a heat equation that converges to zero given the boundary conditions. In short, $u$ converges to $f$ as $t\rightarrow \infty$.


问:
Automorphisms of $\mathbb{A}^{n}$
Let $\varphi: \mathbf{A}^{n} \rightarrow \mathbf{A}^{n}$ be a morphism of $\mathbf{A}^{n}$ to $\mathbf{A}^{n}$ given by $n$ polynomials $f_{1}, \ldots, f_{n}$ of $n$ variables $x_{1}, \ldots, x_{n} .$ Let $J=\operatorname{det}\left|\partial f_{i} / \partial x_{j}\right|$ be the Jacobian polynomial of $\varphi .$

show that If $\varphi$ is an isomorphism (in which case we call $\varphi$ an automorphism of $\mathbf{A}^{n}$ ) show that $J$ is a nonzero constant polynomial.

solution :

If $\varphi \in \operatorname{Aut}\left(\mathbb{A}^{n}\right),$ then each $f_{i} \notin k,$ since then $\varphi$ is not surjective. Therefore each $f_{i}$ is a linear non-constant polynomial, so $J \in k^{\times}$

why  $f_{i} \notin k$ ? why each $f_{i}$ is a linear non-constant polynomial ? also why $J \in k^{\times}$ ?

答: Hint Try using the inverse polynomial map and the chain rule. What are the invertible  elements of a polynomial ring?

Title: Automorphisms of $\mathbb{A}^{n}$


问:
challenging digit sum problem
I have an interesting problem having to do with digit sum! but the problem is challenging for a novice like me.  I have copied the interesting problem in the lines under.

Consider a number $N$ which is created by concatenating the numbers $1$ through $\underbrace{99\ldots99}_{k\ 9\text{'s}}$ inclusive. For example, if $k=2$, then $N=1234567891011\dots9899$. Find a formula for the digit sum of $N$ in terms of $k\ge1$.

this is very sophisticated problem for a novice like me but i like the challenge! i will be detailing the strategy i am thinking of in the following lines!

so it is going from $1$ to $99...999$ so the number of 1 and 9 will be the same! same with any pair that can be taken! so hear me out here.... we can count the number of 1's in N! now i will call this x! the answer will be $(1+2+3+4+5+6+7+8+9)x=45x$
now the problem is to find x! this is a complicated issue that i cannot seem to get anywhere with! i am asking for everyone's help on this problem! please enter in your thoughts on how to continue.

答:
consider $k=2$
the sum of the digits will be unchanged if we include leading zeros.
i.e. we are using
$$N=00010203...979899$$
there are 200 digits in all, $20$ copies of each digit from $0$ to $9$.
so $$S_2 = 20(45)=900$$
for $k=3$ we have $3000$ digits, $300$ copies of each one
so $$S_3 = 300(45)= 13500$$
This suggests that the general formula is
$$S_k = 45k\cdot 10^{k-1}$$

Title: challenging digit sum problem


问:
Block identity theorem?
If $C$ and $C'$ represent the matrix of a linear transformation with respect distinct bases then we know $C$ and $C'$ are equivalent; in particular we can find invertible matrices $A$ and $B$ such $C'=BCA$. Can we also conclude that $C'$ can be taken to have a partial identity block form? that is an identity block (consisting of only ones) of some size and $0's$ in the remaining entries? Why?</p>

答: Here's an answer emphasizing the linear transformation aspect rather than the matrix aspect.  Suppose $T:V\to W$ is a linear transformation.  Let $K_0$ be a basis for its kernel; since that's a linearly independent set in $V$, it can be extended to a basis $K$ for $V$.  The images under $T$ of the vectors in $K\setminus K_0$ form a linearly independent set $L_0$ in $W$; extend it to  basis $L$ of $W$.  Then the matrix of $T$ with respect to the the bases $K$ and $L$ consists of an identity matrix in the block indexed by $K\setminus K_0$ and $L_0$, and zeros in all other entries.</p>

Title: Block identity theorem?