问:
How to find the nth pair from a generated list of pairs?
I have a mathematical question applied on informatic (python). I would like to find the fastest way to get the pair from a given index.

As example, we have all pair combinations of values from 0 to 10:

```<code>impor itertools
combinations = list(itertools.combinations((i for i in range(0,10)),2)
</code>```

like this the variable combinations is equals to

```<code>[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9), (7, 8), (7, 9), (8, 9)]
</code>```

Thus, I have this function to get the nth pair

```<code>def row_col_from_index(index: int, no_discrete_values: int):
    row_num = 0
    col_num = 0
    seq_value = 0
    level = 1
    no_columns = 10
    while seq_value <= index + 1 - no_columns:
        no_columns -= 1
        seq_value += (no_discrete_values - level)
        level += 1
    row_num = level - 1
    col_num = level + (index - seq_value)
    return row_num, col_num
</code>```

Here we get the 24nth pair (indeed array is 0 based)

```<code>row_col_from_index(23,10)
-> (2, 9)
combinations[23]
-> (2, 9)
</code>```

This work but that will be slow to find the corresponding pair with a huge number of values.
So is there a mathematical trick to speedup the computation like a dichotomy approach or other…

And the reverse question for a given pair how to get its corresponding index ?

thanks

答:
Let $n$ play the role of the number $10$: the pairs are: $(0,1),(0,2),\ldots,(0,n-1),(1,2),(1,3),\ldots,(1,n-1),\ldots,(n-2, n-1)$ (all $n\choose 2$ of them).

Let's find a (zero-based) $k$th pair: say it is $(x,y)$. We have $n-1$ pairs starting with $0$, $n-2$ pairs starting with $1$, etc., ending with one pair starting with $n-2$. Thus, the number of pairs starting with something smaller than $x$ is $(n-1)+(n-2)+\ldots+(n-x)=xn-\frac{x(x+1)}{2}$. Obviously, this is $\le k$. Therefore, we have an inequality:

$$xn-\frac{x(x+1)}{2}\le k$$

and we are looking at the largest such $x$. The inequality above is quadratic:

$$x^2+(1-2n)x+2k\ge 0$$

and the solutions are $x_{1,2}=\frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}$. In addition, we are looking for the solutions that are smaller than the "smaller" of the two solutions, as the "larger" one is obviously greater than $\frac{2n-1}{2}$, which is outside the range of $x$ that we are after.

Thus, $x=\left\lfloor \frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}\right\rfloor$. From there, we find the number of pairs preceding the pair $(x, x+1)$ as $xn-\frac{x(x+1)}{2}$ and so $y$ can be found as: $y=k-\left(xn-\frac{x(x+1)}{2}\right)+x+1$.

Something like:

```<code>def pair(n, k):
    x = math.floor((2*n-1-math.sqrt(4*n*n-4*n+1-8*k))/2)
    y = k - x*n + x*(x+1)//2 + x + 1
    return x, y
</code>```

For the inverse, notice that $k=y-(x+1) + xn-\frac{x(x+1)}{2}$.

Body:
Let $n$ play the role of the number $10$: the pairs are: $(0,1),(0,2),\ldots,(0,n-1),(1,2),(1,3),\ldots,(1,n-1),\ldots,(n-2, n-1)$ (all $n\choose 2$ of them).

Let's find a (zero-based) $k$th pair: say it is $(x,y)$. We have $n-1$ pairs starting with $0$, $n-2$ pairs starting with $1$, etc., ending with one pair starting with $n-2$. Thus, the number of pairs starting with something smaller than $x$ is $(n-1)+(n-2)+\ldots+(n-x)=xn-\frac{x(x+1)}{2}$. Obviously, this is $\le k$. Therefore, we have an inequality:

$$xn-\frac{x(x+1)}{2}\le k$$

and we are looking at the largest such $x$. The inequality above is quadratic:

$$x^2+(1-2n)x+2k\ge 0$$

and the solutions are $x_{1,2}=\frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}$. In addition, we are looking for the solutions that are smaller than the "smaller" of the two solutions, as the "larger" one is obviously greater than $\frac{2n-1}{2}$, which is outside the range of $x$ that we are after.

Thus, $x=\left\lfloor \frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}\right\rfloor$. From there, we find the number of pairs preceding the pair $(x, x+1)$ as $xn-\frac{x(x+1)}{2}$ and so $y$ can be found as: $y=k-\left(xn-\frac{x(x+1)}{2}\right)+x+1$.

Something like:

def pair(n, k):
    x = math.floor((2*n-1-math.sqrt(4*n*n-4*n+1-8*k))/2)
    y = k - x*n + x*(x+1)//2 + x + 1
    return x, y

For the inverse, notice that $k=y-(x+1) + xn-\frac{x(x+1)}{2}$.


问:
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero
According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.

答:
Just to add a little bit to @Igor Rivin's answer - Stirling's formula gives you an asymptotic approximation for ${\Gamma(x)}$. That means that
$$
\lim_{x\to\infty}\frac{\Gamma(x)}{S(x)} = 1
$$
(where ${S(x)}$ is the Stirling formula). We write ${\Gamma(x)\sim S(x)}$ to represent this fact (of course we are assuming ${S(x)\neq 0}$). You need to use the following proposition:

If ${f\sim g}$ and ${h\sim j}$, we have
$$
\lim_{x\to\infty}\frac{f(x)}{h(x)} = \lim_{x\to \infty}\frac{g(x)}{j(x)}
$$
(provided ${h(x)\neq 0}$ and ${j(x)\neq 0}$). It's easy to prove this by using properties of limits.

ContentLicense: CC BY-SA 4.0


问:
Find the matrix projection of a symmetric matrix onto the set of symmetric positive semi definite (PSD) matrices
Consider $\mathbb{S}^n$, the set of all $n\times n$ real symmetric matrices. Let $A\in \mathbb{S}^n$ and $A=U\Lambda U^{\top}$ be its spectral decomposition. I want to know how to prove that $\Pi_{\Bbb S_+^n}(A)=U\Lambda ^+U^{\top}$, where $\Lambda^+$ is the $n\times n$ diagonal matrix given by $\Lambda_{ii}^+=\max\{\Lambda_{ii}, 0\}$, for $i=1,...,n$；$\Pi_{S}(x)=\arg\min_{z\in S}\|x-z\|^2_2$.

This projection means that we can project any symmetric matrix on $\mathbb{S}_+^n$ and get the closest positive semi-definite matrix, but how can we prove that it is the closest one?

答:
There is a subtle point missing in Rajat's answer, so I will repeat the proof:

Let $A = U\Lambda U^\top$ be the eigenvalue decomposition of $A$.
Then we have
$$
\min_{X\succeq 0} \|X-A\|_F^2 = \min_{X\succeq 0} \|X-U\Lambda U^\top\|_F^2=\min_{X\succeq 0} \|U^\top X U-\Lambda\|_F^2.
$$

Now here is the subtle point: The matrix $B\equiv U^\top X U$ must be diagonal. Why?

First, whenever $\lambda_{ii}\geq 0$ we can set $b_{ii}=\lambda_{ii}$ to get zero error for those elements. Regarding the terms where $\lambda_{ii}< 0$, since $B$ is also psd it must have non-negative diagonal, so the best we can do is put $b_{ii}=0$. One might think that it's possible to put there a negative value which will decrease the error, and somehow take care of the psd constraint by carefully distributing the off-diagonal elements, but negative diagonal elements would violate the constraint. As for the off-diagonal elements - anything diffeent than $0$ would increase the error.

The solution is then given by $X = U \left(\max\left(\Lambda,0\right)\right) U^\top$.


问:
Is it possible that a surjective map on a non contractible space be null-homotopic?
Let $X$ be a non contractible space and $f:Y\to X$ a surjective map.

Is it possible that one have $f$ null-homotopic?

If yes,

Is there some result to help to prove that some specific surjective
map is not null-homotopic?

My searches with keywords did not have any effect, I only got inverse results, like "if $ X $ is a sphere and $ f $ is not surjective, then it is null-homotopic".

Thank you so much.

答:
As mentioned in the comments, if $Y$ is contractible, then any map $f:Y\rightarrow X$ is nullhomotopic.  So, in particular, the covering projection $\mathbb{R}\rightarrow S^1$ is an example where $X$ is non-contractible, $f$ is surjective, and $Y$ is contractible.

There are some exceptions, but generally speaking, surjectivity is not well behaved under homotopies:  two maps can be homotopic with one being surjective and the other not being surjective.

Here's a large class of examples.

Proposition:  Suppose $M$ and $N$ are connected compact manifolds.  If $M$ has positive dimension, then there is a surjective null homotopic map $f:M\rightarrow N$.

Proof:  Choose $p \in M$ and let $U$ be a closed neighborhood of $p$ which is homeomorphic to a closed ball.  Abusing notation, I will identity $U$ with a ball of radius $1$ centered about $0$ in $\mathbb{R}^n$.  Now, there is a surjective continuous function $f:U\rightarrow [-1/2,1/2]$ which is constantly $0$ on the boundary of the ball.  For example, you can define identify $[-1/2,1/2]$ as a small part of a coordinate axis in $U$, define $f:[-1/2,1/2]\rightarrow [-1/2,1/2]$ as the identity, $f:\partial U\rightarrow [-1/2,1/2]$ as constantly zero, and then use the Tietze extension theorem to define $f$ on all of $U$.

Now, $f$ extends to a surjective continuous map $F:M\rightarrow [-1/2,1/2]$ by defining it to be $0$ outside of $U$.

Armed with $F$, we now use the Hahn-Mazurkiewicz theorem which says that a Hausdorff space $N$ is the continuous image of $[-1/2,1/2]$ if $N$ is compact, connected, locally connected, and second countable.  All of these hypothesis are true for a connected compact manifold $M$ (and even a much larger class of spaces!).  Then the composition $M\rightarrow [-1/2,1/2]\rightarrow N$ is surjective by construction.  Note that the composition factors through a coontractible space, so is null homotopic.  $\square$

However, this is not to say that surjectivity has nothing to do with whether or not a map is null homotopic.  For example,

Proposition.  Suppose $M^n$ and $N^n$ are compact oreintable connected $n$-manifolds with no boundary.   Suppose $f:M\rightarrow N$ is continuous and that the induced map $H^n(M;\mathbb{Z})\rightarrow H^n(N;\mathbb{Z})$ is non-zero.  Then any map $g$ which is homotopic to $f$ must be surjective.

Proof:  As $H^n(M;\mathbb{Z})\cong \mathbb{Z}$, the induced map is given by multiplication by some integer called the degree of the map.  The degree can be computed by homotoping a map to be smooth, and then doing a signed count of preimages of a regular value.  If $g$ is not surjective, any point which is not in the image of $g$ is a regular value, so $g$ has degree $0$.  Homotopic maps induce the same map on cohomology, so $f$ also has degree $0$. $\square$


问:
Exponential Function Regression
I need to find the exponential function y = c0e^(c1*x) using fit in Mathematica for the data points {1850, 1.3}, {1900, 1.6}, {1950, 3.0}, {1980, 4.4}, {2000, 6.0}}.</p>

Title: Exponential Function Regression

Body:
Try this

data = {{1850, 1.3}, {1900, 1.6}, {1950, 3.0}, {1980, 4.4}, {2000, 6.0}};
f = a Exp[b (x/1850-x0)];
fit = FindFit[data, f, {a, b, x0}, x];
fit0 = f /. fit // Simplify
gr0 = ListPlot[data, PlotStyle -> Red];
gr1 = Plot[fit0, {x, 1850, 2000}];
Show[gr1, gr0]

so the adjust is

$$
y = 0.330376 e^{22.8667 \left(\frac{x}{1850}-0.955521\right)}
$$

or also

$$
y = 1.0711413556895699\times 10^{-10}e^{0.0123604 x}
$$


问:
Why proofs which provide algorithms for construction are valid?
10 days ago I asked question about validity of a proof (Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis?), but as I later realised I didn't explain well what I really wanted to know and this resulted in the answers that didn′t quite answered my real question, so I will try to rephrase it. So in that previous post there was a proof of the fact that any subspace of finite-dimensional vector space has a finite basis. In it I basically constructed this basis by adding vector if the current set of vectors is not maximal linear independent set. Then I said that this "algorithm" will terminate because of the fact that the vector space is finite dimensional. But my concern about this proof is "Why algorithm for constructing/finding an object is a valid proof of its existence ?". Such question may sound strange, but my concern is the fact that you can′t say by reading a proof how many steps execution of such algorithm will take, because it depends on what vector space and what subspace you take. So basically we don't know how many steps the proof has and it seems kind of strange to me. And this concern is not just about proof of this fact about subspaces, but about any proof which proves existence of an object with an algorithm which has number of steps which depends on something. For example, proof of the fact that any tree has a vertice of degree 1, where you take any vertice and then go along the edges of the tree until you find the vertice of degree 1 (which you will, because tree has only finite amount of vertices and tree is acyclic).

So why are proofs, which provide algorithms for constructing an object, valid and can we formalise (maybe using set theory) them in such a way that number of steps of a proof will be specific ?

答:
In most cases where such questions arise, you will prove things like $\forall x \in M: P(x)$, i. e., you have to prove by definition of $\forall$ for every single object $x \in M$ that the statement $P(x)$ holds.
An algorithm shows, given a particular $x$, the validity of $P(x)$ using finitely many steps.

Please note that the "finitely many steps" is essential in proofs. In mathematical logic, we define that we can "deduce" a statement $S$ from an axiom system $Ax$, if we can get $S$  from $Ax$ by applying finitely often inference rules (things line $A \land B \implies A$).

Otherwise, we could prove that $1 + 1 + 1 + \dots$ converges using that $0 + 0 + 0 + \dots$ converges and changing infinitely often a Zero to a One.


问:
What is correct answer for this algebraic expression?
If $(x+2)$ and $(x-2)$ are factors of $ax^2+2x-3x^2+bx-4$, then what would be the value of $(a+b)$ from following given options and how to arrive at it?

a) -7

b)  7

c)  14

d)  -8

答: You could use Polynomial Division. You could divide the polynomial by $(x+2)$ and $(x-2)$ and add the condition that the rest must be zero hence solve for $a$ and $b$.</p>

ContentLicense: CC BY-SA 3.0


问:
Prove that three segments which intersect a circle pass through the same point
In a scalene triangle $\triangle ABC$ with $AB\ne AC$, I state we have $Y$ which is the point of intersection of the bisector of $\angle A$ with $BC$ and $D$ is the point where the perpendicular line from $A$ towards $BC$ intersects $BC$. If $E$ and $Z$ are the intersections of the circumcircle of $\triangle AYD$ with $CA$ and $AB$ respectively, prove that $AD, BE, CZ$ pass through the same point

I don't know how to solve this question. The first thing I thought of was maybe proving that their intersection is the orthocenter of the triangle or something like that, but that did not lead me to a solution. I then thought of using power of a point towards the circle, but that didn't work either. Could you please explain to me how to solve this question?

答:
By intersecting secant theorem,

$BZ \cdot BA = BD \cdot BY$

$CY \cdot CD = CE \cdot CA$

So we have, $BZ \cdot CD \cdot BA \cdot CY = BD \cdot CE \cdot CA \cdot BY$

But by angle bsector theorem we have, $BA \cdot CY = CA \cdot BY$

Also as $AY$ is angle bisector of $\angle EAZ$ and it is also the diameter of the circle, $AE = AZ (\triangle AZY \cong \triangle AEY$)

That leads to $BZ \cdot AE \cdot CD = BD \cdot CE \cdot AZ \ $ and hence by Ceva's theorem, Cevians $AD, BE$ and $CZ$ concur.


问:
Distance between the minimum of two strongly convex functions and the minimum of their sum
Let $f$ and $g$ be two strongly convex functions from a convex set $\mathcal{X}$ to $\mathbb{R}$, with minimum $x$ and $y$ respectively.
Let us denote by $z$ the minimum of $f+g$, and by $\delta$ the distance $\lVert x-y\rVert$.

I would like to know if it is possible to prove that $\lVert x-z\rVert \leq \delta$ and $\lVert y-z\rVert < \delta$.

Rk:

This statement is true in one dimension (see here)

This statement is also true when $f$ and $g$ are two quadratic functions.

答:
This result is true at least for any $p$-norm with $p\geq 1$.
Indeed, we know that the result is true in one dimension (see here).
Thus, the result is still true if we apply it to each of the coordinates of the functions $f$ and $g$.

In particular, we get:
$$ \forall 1 \leq i \leq n,\ \min(x_i, y_i) \leq z_i \leq \max(x_i, y_i) $$

This shows us that
$$ \lVert x-z \rVert_p = \left(\sum_{i=1}^n (x_i-z_i)^p\right)^{1/p} \leq \left(\sum_{i=1}^n (x_i-y_i)^p\right)^{1/p} = \lVert x_i - y_i \rVert_p =\delta $$
By symmetry, the same result holds for $y$: $\lVert y-z \rVert_p \leq \delta$.


问:
Integrating $x \sin(nx)$
It should be simple but I'm overthinking this so hard, that I can't see what I'm supposed to do anymore. Instead of taking a break from it, I'll ask here.

$\int x\sin(nx) \, dx$, $\sin(nx)$ would be $\frac{-\cos(nx)}{n}$ if I'm not very mistaken, but what should I set as $u$ and $v'$ here?

Thing is, this is part of a Fourier series calculation of an odd function, where this is $b_n$ (and also the whole Fourier series), and the limits are $0$ and $\frac{\pi}{2}$.

I'm also getting confused when I get $\frac{\pi}{2}$ for $\sin(nx)$ and $-\cos(nx)$, as these would vary between $-1$, $0$ and $1$. I might be overthinking ahead of what I've already calculated here, so it might be some stuff I can cross out that I can't see just now.

答: You can do it as follows\begin{align}\int\overbrace{x}^{\phantom u=u}\overbrace{\sin(nx)}^{\phantom{v'}=v'}\,\mathrm dx&=-x\frac{\cos(nx)}n+\int\frac{\cos(nx)}n\,\mathrm dx\\&=-x\frac{\cos(nx)}n+\frac{\sin(nx)}{n^2}+C.\end{align}

Title: Integrating $x \sin(nx)$


问:
Submodules and $T$-invariant subspaces.
My professor wrote this on the board:

Claim: there is a bijection between submodules and $T$-invariant subspaces {$T(w) \subset W$}.

Proof:

If $W \subset V$ is $T$-invariant, then $f(X).w = f(T)(w) \in W.$ so $W$ is a submodule.
Conversely, let $W \subset V$ be a submodule, then for $w \in W,$ $X.w = T(w) \in W$ and hence $W$ is $T$-invariant.

Before that she wrote this:

Let $k$ be a field, $V$ a $k-$ vector space, $n = \dim_kV < \infty.$$R = k[x], T \in L(V)=$ linear operators on $V.$

$T$ has minimal polynomial and characteristic polynomial in $R.$

$V$ has the structure of an $R$-module, a structure induced by $T.$

$$\varphi : R \rightarrow \operatorname{End_k(V)} = L(V)$$
$$1 \mapsto I$$
$$X \mapsto T$$

So $\varphi(f(X)) = f(T)$

Scalar multiplication in $V$ is $f(X).v = f(T)(v).$ then she asked what are the submodules of $V$ and she began to prove the above claim.

But I do not understand the following:

1- Why she defined $\varphi$ like that?

2- why the scalar multiplication in $V$ looks like that?

3- I do not understand at all how the forward direction proves that it is a submodule, why we are applying $f(X)$ to $w$ to prove this?

4- why in the backward direction we are applying $X$ to $w$?

could anyone help me understand all these points?

答:
$k[X]$ is a free $k$-algebra on one generator and it has the universal property that $k$-algebra homomorphisms from $k[X]$ into any $k$-algebra is defined by the choice of the image of the indeterminate $X$. Also note that $\operatorname{End}_k(V)$ is a $k$-algebra, in which multiplication is simply the composition of endomorphisms.

Scalar multiplication by a polynomial results from considering a homomorphism  from $k[X]$ into $\operatorname{End}_k(V)$, i.e. from the choice of an endomorphism $T$ associated to $X$. If you're interested in the properties of a given endomorphism, you'll choose for the image of $X$ this endomorphism.

We actually apply $f(T)$ to a vector in $V$ because by definition, scalar multiplication by the polynomial $f(X)$ consists in applying the endomorphism $f(T)$ to this vector.

I don't understand what you call the backwards direction is.

Title: Submodules and $T$-invariant subspaces.


问:
if two computers are playing tic-tac-toe, but they are choosing their squares randomly, what is the chance for X to win?
Tic-tac-toe is a children's board game that's notorious for draws. It's easy to write a program for either player (X or O) that always draws the game. I would like to know how the outlook changes if both players play completely randomly. Ie, X places their first move uniformly among the 9 squares, then O does the same, and so on until someone wins. Intuitively, X should be better because they'll usually get more squares and first player advantage, but I'm not sure.

I really have no Idea, but I would really like to know. I am good at math, but not that good.

答:
There's a relatively nice way to do this by hand: pretend they play until the board is full, then grade it. There are $\binom 9 5$ such ending boards, most of which have a unique winner or no winner, and each multiple winner (ambiguous) case will have the same analysis.

So, first off are the multiple winner (I'll call these ambiguous) cases: These must be $2$ parallel horizontal lines of X's and O's, where the remaining row/column will have $3$ empty spots filled by $2$ X's and one $O$. So we get: $2$ orientations (row vs column), $3$ choices for the X row/column, $2$ remaining choices for the O row/column, and then $3$ ways to fill the remaining row/column. A total of $36$ such boards. In these cases, order of play matters. We'll return to them later.

Now, for the cases with X as the unique winner: Either we win on a diagonal, or a row/column. If a diagonal, then the other entries wont matter: $\binom 6 2$ possible ways to fill in a board with a winning diagonal, and $2$ possible winning diagonals.
If a row/column, we have again $2$ choices for row/column, $3$ choices for which, and then $\binom 6 2$ choices for filling in the remaining squares.

We have double counted somewhat: we could win with both a diagonal and a row/column, both diagonals, we could win with both a row and a column, or we could have an ambiguous board. Fixing a particular row/column, there are $2$ ways to also get a diagonal. Likewise, fixing a particular row, there are $3$ ways to also get a column.

In total, we have:

$$2 \binom 6 2 -1 + 3 ( \binom 6 2 - 2) + 3 ( \binom 6 2 - 2 - 3) - 36 = 62$$

Where these are the diagonal wins, the column wins (remove the diagonal wins), the row wins (remove the diagonal or column wins), subtract the ambiguous boards.

The unambiguous O wins are much simpler - if O wins a row/column, then there must be a row/column that X wins (there aren't enough O's on the board to prevent it!). So we only have the diagonals:

$$2 (\binom 6 1) = 12$$

Finally, there are the remaining $16$ cases, all of which are draws. You can also verify these by hand, there are $8$ with X in the middle and $8$ without.

To finish, let's return to the 36 ambiguous cases. In each of these, there is a unique win for X and a unique win for O. So the question is just "does X make the 3 winning moves before O." Abstracting the moves to winning or not, based on if they've played one of the moves along their 3 in a row, there are only $\binom 4 3 \binom 5 3 = 40$ strategies to consider. Note that each ordering is equally likely, as our bots do not know if a move is good or not. So X wins in:

$$ \binom 4 3 + 3 (\binom 4 3 - 1) + 0 = 13$$

(based on if X wins on their third or fourth move, they cannot win on their fifth as O will have finished their line first.)

So the total probability that X wins is $$\big ( 62 + 36 \cdot (\frac {13}{40}) \big ) / \binom 9 5 = \frac{737}{1260}$$

Exactly as predicted in other answers.


问:
Determine the integer part
Let $(a_n)$ be a sequence given by $a_1=\frac{1}{2}$ and $a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$ , $n=1,2,...$
Let $b_n=a_1+a_2+...+a_n$ for each positive integer $n$

Determine the integer part $[b_n]$

答:
Let's start by showing that $b_n = 2-\frac{1}{1-a_{n+1}}$. Proof by induction:

$n=0$: $0 = 2-\frac{1}{1-\frac{1}{2}}=0$. Check.

Going from $(n-1)$ to $n$:
$$\begin{eqnarray}b_n = b_{n-1}+a_n & = & 2-\frac{1}{1-a_n} + a_n\\
& = & 2 - \frac{1-a_n+a_n^2}{1-a_n}\\
& = & 2 - \frac{1}{1-\frac{a_n^2}{a_n^2-a_n+1}}\\
& = & 2 - \frac{1}{1-a_{n+1}}\end{eqnarray}$$

The other ingredient is the inequality $0<a_n\leq \frac{1}{2}$, provable by induction as well:

$n=1$: $0<\frac{1}{2}\leq \frac{1}{2}$. Check.

Going from $n$ to $(n+1)$:
$$\begin{eqnarray}
0 < a_n \Rightarrow 0 < a_n^2-a_n+1 \Rightarrow 0 < \frac{a_n^2}{a_n^2-a_n+1} = a_{n+1} \\
a_n \leq \frac{1}{2} \Rightarrow a_n^2 \leq \frac{1}{4} \wedge \frac{1}{2} \leq a_n^2 - a_n + 1 \Rightarrow a_{n+1} = \frac{a_n^2}{a_n^2-a_n+1} \leq \frac{1}{2}\end{eqnarray}$$

Combining these two observations tells us that $0\leq b_n<1$, which implies $\lfloor b_n\rfloor = 0$.

Title: Determine the integer part


问:
Algebraic expression for the period of $\cos (\log (x))$?
This question relates only to $x \in \mathbb{R}^+$. The function $f(x) = \cos (\log (x))$ is clearly defined on the positive reals, with a monotonic decreasing period $p(x)$ which is defined at the limits of this range by

$$\underset{x\to 0^+}{\text{lim}}p(f(x)) = \infty$$
$$\underset{x\to \infty}{\text{lim}}p(f(x)) = 0$$

$p$ still needs clear definition for any given $0 < x_0 < \infty$. So, let us define $p(x_ 0)$ as the smallest positive real $a$ for which this pair of equations hold:

$$\cos (\log (x_0)) = \cos (a \log (x_0))$$
$$\text{sgn}\left(\frac{\partial \cos \left(\log \left(x_0\right)\right)}{\partial x_0}\right) = \text{sgn}\left(\frac{\partial \cos \left(a \log \left(x_0\right)\right)}{\partial  x_0}\right)$$

Is the right approach? And how do I proceed?

答:
$\cos x = \cos y\iff y = \pm x + 2k \pi$ for some integer $k$

$\log w = \log u \iff w = u$ so

$\cos (\log x) = \cos (\log y) \iff \log x = \pm \log y + 2k \pi $

$\iff \log x \mp \log y = 2k\pi\iff \log (\begin{cases}xy\\\frac xy\end{cases}) = 2k\pi$

And if there were a period, $p > 0$, so that $\cos (\log (x + p))=\cos (\log x)$ for all $x$ we would have that for every  $x$ that either $\log (x(x+p))$ or $\log \frac x{x+p} = 2k\pi$.

But $\log(x(x+p))$ and $\log\frac x{x+p}$ are both continuous and not constant they can't be restricted to only multiples of $2\pi$, $\cos (\log x)$ is not periodic.


问:
Epsilon delta proof of a simple sequence convergence
Determine whether the following sequences converge or diverge. Prove
your findings. That is, (directly or indirectly, by contradiction) provide
an $\epsilon$ − N proof to establish your findings.

(a) $\{ \frac{n^4-1}{n^4+n-6}\}$. I know it converges to one just by setting the limit but I don't get how to do the proof. My attempt:

For all $\epsilon > 0$, there exist $ N > 0$, such that for all $n \in N$, if $n > N$, then $|\frac{n^4-1}{n^4+n-6} - 1| = \frac{|-n+5|}{|n^4+n-6|} = \frac{|n-5|}{|n^4+n-6|} \leq \frac{|n-5|}{n^4}$. I'm lost.

答:
I will post the scratch work, because I think this is where the biggest challenges come in, once this is done, constructing the proof is straight forward from the definition of convergence:

$$ \left\lvert \dfrac{n^4 - 1}{n^4 + n - 6} -1\right\rvert  < \epsilon$$

$$ \left\lvert \dfrac{n -5}{n^4 +n-6} \right\rvert = \dfrac{\left\lvert n -5 \right\rvert}{\left\lvert n^4 +n-6 \right\rvert} = \dfrac{n+5 }{\left\lvert n^4 +n-6 \right\rvert} \leq \dfrac{X}{Y}< \epsilon$$

Okay, we've simplified this enough just enough to make some moves. Note that we used the triangle inequality ($\lvert a \pm b \rvert \leq \lvert a \rvert + \lvert b \rvert$) for the numerator.

Our goal now is to figure out what $\dfrac{X}{Y}$ is and we will do this based on the information we have. Notice that in the numerator we can write $ n+5 \leq n + n $ for what $n$ is this true? ($ n \geq 5$, keep track of this somewhere). So let's update our expression:

$$ \dfrac{n+5 }{\left\lvert n^4 +n-6 \right\rvert} \leq \dfrac{2n}{Y}< \epsilon $$

Now let's look at the denominator.

If we increased that 6 to some number greater than 6, then the  denominator that we have would get smaller. So let's write this out as

$$ \lvert n^4 + n - 6 \rvert \geq \lvert n^4 + n - n \rvert \quad \text{for} \quad  n\geq6 $$
$$ \dfrac{1}{\lvert n^4 + n - 6 \rvert} \leq \dfrac{1}{\lvert n^4 + n - n \rvert} \quad \text{for} \quad  n\geq6 $$

So now we have

$$ \dfrac{n+5 }{\left\lvert n^4 +n-6 \right\rvert} \leq \dfrac{2n}{ n^4 } < \dfrac{1}{n^3}< \epsilon $$

So what should/could you choose for your $N$ to be? How about $N = \left\lceil \dfrac{1}{\epsilon ^{1/3}} \right\rceil$. And we include the symbols $\lceil X \rceil$ to signify that we always round up in such a way that whatever we chose for $N$.

But wait! We forgot that to get this expression, we had to impose some extra restrictions, remember?

So instead, let $N = max\left\{ \left\lceil \dfrac{1}{\epsilon ^{1/3}} \right\rceil , 6\right\}$ so that our assumptions and assertions are all covered.


问:
Why $\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)}$
Given $A\succ 0$ (positive-definite) and $c>0$, I am trying to show

$$\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)} \tag{1}$$

using information like this but without success so far. Could you please help to prove it? Any help is welcome.

EDIT: This inequality is an abstract form of the inequality

$$\left\| \left[f''(x_k) - r_M(x_k)\frac{M}{2}I \right]^{-1}f'(x_k)\right\| \leq \frac{\|f'(x_k)\|}{\lambda_{min}\left(f''(x_k)\right)}\tag{2}$$ appeared here (Theorem 3). Here is the part of the proof

答:
This is not true. E.g. when $c>0$ is small, $A=\operatorname{diag}(1,c)$ and $x=(0,1)^T$, we have
$$
\|(A+cI)^{-1}x\|=\frac{1}{2c}>1=\frac{\|x\|}{\lambda_\max(A)}.
$$


问:
Cross product, but with angles
I have two unit-length vectors $\vec{v_1}$ and $\vec{v_2}$ and I would like to find a unit-length vector that's perpendicular to them, so basically $\vec{v_3} = \vec{v_1} \times  \vec{v_2}$.

However the vectors are not expressed as numbers, but as rotations around axes. so $\vec{v} = (\alpha_x, \alpha_y, \alpha_z)$ where $\alpha_x$ is rotation of a $(1,0,0)$ vector around x-axis, $\alpha_y$ around y-axis, etc.

I know it's easy to work this out by converting angular representation to more typical coordinates, then computing cross products and then converting back to angles. But is there an easier way to do it that only relies on angles?

答:
But is there an easier way to do it that only relies on angles?

No, not really.  Also, note that the order of the rotations also matters, because it affects the conversion to/from Cartesian coordinates.

Let's look at the expressions you get by converting your angular coordinates to Cartesian coordinates.

Using spherical coordinates ($r$ being distance from origin, $\varphi$ latitude, and $\theta$ longitude with zero at equator, positive north, negative south), we have
$$\vec{v}_1 = (r_1, \theta_1, \varphi_1) = \left[\begin{matrix}
r_1 \cos(\varphi_1) \cos(\theta_1) \\
r_1 \sin(\varphi_1) \cos(\theta_1) \\
r_1 \sin(\theta_1) \\
\end{matrix} \right], \quad \vec{v}_2 = (r_2, \theta_2, \varphi_2) = \left[\begin{matrix}
r_2 \cos(\varphi_2) \cos(\theta_2) \\
r_2 \sin(\varphi_2) \cos(\theta_2) \\
r_2 \sin(\theta_2) \\
\end{matrix} \right]$$
and therefore their vector cross product is
$$\vec{v}_1 \times \vec{v}_2 = \left[\begin{matrix}
r_1 r_2 \sin(\varphi_1) \cos(\theta_1) \sin(\theta_2) - r_1 r_2 \sin(\theta_1) \sin(\varphi_2) \cos(\theta_2) \\
 - r_1 r_2 \cos(\varphi_1) \cos(\theta_1) \sin(\theta_2) + r_1 r_2 \cos(\varphi_2) \sin(\theta_1) \cos(\theta_2) ~ \, ~ \\
r_1 r_2 \cos(\varphi_1) \cos(\theta_1) \sin(\varphi_2) \cos(\theta_2) - r_1 r_2 \sin(\varphi_1) \cos(\theta_1) \cos(\varphi_2) \cos(\theta_2) \\
\end{matrix}\right]$$
Now, because
$$\begin{aligned}
2 \cos(\theta_1) \cos(\theta_2) &= \cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \\
2 \sin(\theta_1) \cos(\theta_2) &= \sin(\theta_1 + \theta_2) + \sin(\theta_1 - \theta_2) \\
2 \cos(\theta_1) \sin(\theta_2) &= \sin(\theta_1 + \theta_2) - \sin(\theta_1 - \theta_2) \\
2 \sin(\theta_1) \sin(\theta_2) &= \cos(\theta_1 - \theta_2) - \cos(\theta_1 + \theta_2) \\
\cos(\varphi_1)\sin(\varphi_2) - \sin(\varphi_1)\cos(\varphi_2) &= -\sin(\varphi_1 - \varphi_2) \\
\end{aligned}$$
we can write the cross product as
$$\vec{v}_1 \times \vec{v}_2 = \displaystyle \frac{r_1 r_2}{2} \left[ \begin{matrix}
~ \, ~ \sin(\varphi_1)\bigl( \sin(\theta_1 + \theta_2) - \sin(\theta_1 - \theta_2) \bigr) - \sin(\varphi_2)\bigl( \sin(\theta_1 + \theta_2) + \sin(\theta_1 - \theta_2) \bigr) \\
-\cos(\varphi_1)\bigl( \sin(\theta_1 + \theta_2) - \sin(\theta_1 - \theta_2) \bigr) + \cos(\varphi2)\bigl( \sin(\theta_1 + \theta_2) + \sin(\theta_1 - \theta_2) \bigr) \\
-\sin(\varphi_1 - \varphi_2)\bigl( \cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \bigr) \\
\end{matrix} \right]$$
which shows that the final rotation angle (here, $\theta$) does simplify – that is, we do get expressions where we use the sum of and the difference in the final angles, $\theta_1+\theta_2$ and $\theta_1-\theta_2$ – but the other rotation angles cannot be separated: the above becomes way more complicated if we try to use only $\varphi_1+\varphi_2$ and $\varphi_1-\varphi_2$ instead of $\varphi_1$ and $\varphi_2$ alone.  In other words, see how we use $\varphi_1$ and $\varphi_2$ alone, but only $(\theta_1+\theta_2)$ and $(\theta_1-\theta2)$, never $\theta_1$ or $\theta_2$ alone?  That indicates that we can only really separate the last rotation in this expression in an useful manner.

Title: Cross product, but with angles


问:
Fraction of probability terms
I repost the following question, since it has been deleted.

Let $A$ and $B$ be independent events in a probability space which are
each less than certain. Determine the value of the following quantity.
\begin{align*}
     \frac{1 - P(A \cup B)}{(1 - P(A))(1 - P(B))} \end{align*}

I hope you find this question interesting as well.

答:
Another, and a slightly faster way would be using the fact that if two events are independent, then their complements are also independent. Also, note that $P((A \cup B)^c)=P(A^c \cap B^c)$ by De Morgan’s. Then the fraction simplifies quite easily:
$$
\frac{1 - P(A \cup B)}{(1 - P(A))(1 - P(B))}=\frac{P(A^c \cap B^c)}{P(A^c)P(B^c)}=\frac{P(A^c) P(B^c)}{P(A^c)P(B^c)}=1,
$$
by independence of $A^c$ and $B^c.$

Title: Fraction of probability terms


问:
Find the line integral for the following vector field
$\vec{G} = \big(\frac{x}{\sqrt{x^2+y^2-1}}, \frac{y}{\sqrt{x^2+y^2-1}}\big)$.

I need to calculate the line integral $\int_{\mathrm{CR}}\vec {G}\, d\vec{r}\, $ where $CR$ is a circle with the center in origin and the radius $1 \lt R.$
I know that $x^2+y^2 \gt 1,$ which means all points outside of the circle with radius $1$ and center in origin. How do you calculate the line integral of such a vector field?

答:
You can establish that $\vec G$ is conservative, meaning there is some scalar function $g(x,y)$ such that $\nabla g(x,y)=\vec G$. In particular, you have

$$g(x,y)=\sqrt{x^2+y^2-1}+c$$

since

$$\begin{cases}
\dfrac{\partial g}{\partial x}=\dfrac x{\sqrt{x^2+y^2-1}}\\[1ex]
\dfrac{\partial g}{\partial y}=\dfrac y{\sqrt{x^2+y^2-1}}
\end{cases}$$

where $c$ is an arbitrary constant. Then by the gradient theorem, the line integral over the given curve is $0$.


答: You can reason this out in words. The numerator is the probability that neither event occurs. So is the denominator, since the events are independent. So the quotient is $1$.

Title: Fraction of probability terms

Body:
I repost the following question, since it has been deleted.

Let $A$ and $B$ be independent events in a probability space which are
each less than certain. Determine the value of the following quantity.
\begin{align*}
     \frac{1 - P(A \cup B)}{(1 - P(A))(1 - P(B))} \end{align*}

I hope you find this question interesting as well.


问:
How to find the nth pair from a generated list of pairs?
I have a mathematical question applied on informatic (python). I would like to find the fastest way to get the pair from a given index.

As example, we have all pair combinations of values from 0 to 10:

```<code>impor itertools
combinations = list(itertools.combinations((i for i in range(0,10)),2)
</code>```

like this the variable combinations is equals to

```<code>[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9), (7, 8), (7, 9), (8, 9)]
</code>```

Thus, I have this function to get the nth pair

```<code>def row_col_from_index(index: int, no_discrete_values: int):
    row_num = 0
    col_num = 0
    seq_value = 0
    level = 1
    no_columns = 10
    while seq_value <= index + 1 - no_columns:
        no_columns -= 1
        seq_value += (no_discrete_values - level)
        level += 1
    row_num = level - 1
    col_num = level + (index - seq_value)
    return row_num, col_num
</code>```

Here we get the 24nth pair (indeed array is 0 based)

```<code>row_col_from_index(23,10)
-> (2, 9)
combinations[23]
-> (2, 9)
</code>```

This work but that will be slow to find the corresponding pair with a huge number of values.
So is there a mathematical trick to speedup the computation like a dichotomy approach or other…

And the reverse question for a given pair how to get its corresponding index ?

thanks

答:
One way to solve this problem is to work backwards. I'll illustrate my method with a small example, the pairs generated by combinations(range(5), 2). We can write them in a triangular table:

$$\begin{array}{|c|c|c|c|}
\hline
0,1 & 0,2 & 0,3 & 0,4\\
\hline 1,2 & 1,3 & 1,4\\
\hline 2,3 & 2,4\\
\hline 3,4\\
\hline
\end{array}$$

There are 10 pairs in total, which we can calculate by $4\times 5 / 2$. This is the well-known formula for triangular numbers
$$T(n) = n(n+1)/2$$
The first few triangular numbers are $1, 3, 6, 10, 15, 21, 28$

The triangular number formula is a quadratic equation, which we can easily invert, so if we're given some $y=T(n)$ we can find $n$:
$$n = \frac{\sqrt{8y+1}-1}{2}$$

So to find a pair, we just need to find the largest triangle that's below it in the triangular table. Eg, counting from the bottom of the table $(1, 4)$ has an index of 4, the largest triangular number $<4$ is $3=T(2)$. So $(1, 4)$ is in the row above those rows containing the $T(2)$ triangle, and it must be at the end of its row.

Here's some Python code that does these calculations.

```<code>from itertools import combinations
from math import floor, sqrt

def tri(n):
    return n * (n + 1) // 2

def unrank(idx, n):
    i = tri(n - 1) - idx - 1
    y = (floor(sqrt(8 * i + 1)) - 1) // 2
    j = tri(y) + n - 1 - i
    return n - 2 - y, j

# Test
hi = 5
a = combinations(range(hi), 2)
for i, t in enumerate(a):
    u = unrank(i, hi)
    print(i, t, u, t==u)
</code>```

Output

```<code>0 (0, 1) (0, 1) True
1 (0, 2) (0, 2) True
2 (0, 3) (0, 3) True
3 (0, 4) (0, 4) True
4 (1, 2) (1, 2) True
5 (1, 3) (1, 3) True
6 (1, 4) (1, 4) True
7 (2, 3) (2, 3) True
8 (2, 4) (2, 4) True
9 (3, 4) (3, 4) True
</code>```

Here's an interactive version to play with, running on the SageMathCell server.

Body:
One way to solve this problem is to work backwards. I'll illustrate my method with a small example, the pairs generated by combinations(range(5), 2). We can write them in a triangular table:

$$\begin{array}{|c|c|c|c|}
\hline
0,1 & 0,2 & 0,3 & 0,4\\
\hline 1,2 & 1,3 & 1,4\\
\hline 2,3 & 2,4\\
\hline 3,4\\
\hline
\end{array}$$

There are 10 pairs in total, which we can calculate by $4\times 5 / 2$. This is the well-known formula for triangular numbers
$$T(n) = n(n+1)/2$$
The first few triangular numbers are $1, 3, 6, 10, 15, 21, 28$

The triangular number formula is a quadratic equation, which we can easily invert, so if we're given some $y=T(n)$ we can find $n$:
$$n = \frac{\sqrt{8y+1}-1}{2}$$

So to find a pair, we just need to find the largest triangle that's below it in the triangular table. Eg, counting from the bottom of the table $(1, 4)$ has an index of 4, the largest triangular number $<4$ is $3=T(2)$. So $(1, 4)$ is in the row above those rows containing the $T(2)$ triangle, and it must be at the end of its row.

Here's some Python code that does these calculations.

from itertools import combinations
from math import floor, sqrt

def tri(n):
    return n * (n + 1) // 2

def unrank(idx, n):
    i = tri(n - 1) - idx - 1
    y = (floor(sqrt(8 * i + 1)) - 1) // 2
    j = tri(y) + n - 1 - i
    return n - 2 - y, j

# Test
hi = 5
a = combinations(range(hi), 2)
for i, t in enumerate(a):
    u = unrank(i, hi)
    print(i, t, u, t==u)

Output

0 (0, 1) (0, 1) True
1 (0, 2) (0, 2) True
2 (0, 3) (0, 3) True
3 (0, 4) (0, 4) True
4 (1, 2) (1, 2) True
5 (1, 3) (1, 3) True
6 (1, 4) (1, 4) True
7 (2, 3) (2, 3) True
8 (2, 4) (2, 4) True
9 (3, 4) (3, 4) True

Here's an interactive version to play with, running on the SageMathCell server.


问:
N'th derivative with chain rule
I am trying to find a general form of the chain rule for higher derivatives, using the general Leibniz rule I got to the following formula. However it doesn't seem to work. I suspect it has something to do with the $n-1$ but I am unsure.
$$D^{\left(n\right)}\left(f\left(g\left(x\right)\right)\right)=\sum_{k=0}^{n-1} \binom {n-1}{k}D^{\left(k+1\right)}g\left(x\right)D^{(n-k)}\left(f'\left(g\left(x\right)\right)\right)$$
(Where $D^{\left(n\right)}f(x)$ is the $n$'th derivative of $f(x)$)

The way I derived this is by using $\left(f\left(g\left(x\right)\right)\right)^{\prime}=g^\prime(x)f^\prime(g(x))$ and then applying Leibniz rule on this for higher derivatives.

EDIT- I have managed to proof that the following formula works. Can it be proofed it is equivalent to Faà di Bruno's formula?.
$$\sum_{k=0}^{n-1} \binom {n-1}{k}D^{\left(n-k\right)}g\left(x\right)D^{\left(k\right)}\left(f'\left(g\left(x\right)\right)\right)$$

答:
This is the Faà di Bruno's formula:

\begin{align*}
\frac{\operatorname{d}^n}{\operatorname{d}x^n}f(g(x))&=\sum\frac{n!}{m_1!1!^{m_1}m_22!^{m_2}\cdots m_nn!^{m_n}}f^{(m_1+\ldots+m_n)}(g(x))\prod_{i=1}^ng^{(i)}(x)^{m_i},\\
&=\sum\frac{n!}{m_1!\cdots m_n!}f^{(m_1+\cdots+m_n)}(g(x))\prod_{i=1}^n\frac{g^{(i)}(x)^{m_i}}{i!},
\end{align*}
where the sum is taken over $n$-tuples of nonnegative integers $(m_1,\cdots,m_n)$ such that:
$$1\cdot m_1+2\cdot m_2+\cdots+n\cdot m_n=n.$$

Title: N'th derivative with chain rule


问:
Solving a PDE by constrained minimization
Consider the Hénon problem in the ball:

\begin{equation}
\left\{
\begin{array}{rcll}
- \Delta u & = & |x|^\alpha u^{p - 1} & \quad \text{ in } B \\
u & > & 0 & \quad \text{ in } B \\
u & = & 0 & \quad \text{ on } \partial B
\end{array}
\right.
\end{equation}
where $2 < p < 2^*$.

My book (Morse Index of Solutions of Nonlinear Elliptic Equations, by Damascelli and Pacella) claims that there exists a solution which is the minimizer of
$$
S_\alpha = \inf_{v \in H_0^1(B), v \not\equiv 0} \frac{\int_B |\nabla v|^2 \ \mathrm{d}x}{(\int_B |x|^\alpha v^p \ dx)^{2/p}}. \label{1}\tag{*}
$$

Now, as I understand, we indeed have that there is a solution $u$ that is the minimizer of
$$
J(u) = \inf_{\frac 1 p \int_B |x|^\alpha v^p \, \mathrm{d}x = 1/p} \int_B |\nabla v|^2 \ \mathrm{d}x \label{2}\tag{**}
$$

My question is,

Why are \eqref{1} and \eqref{2} equivalent?

Thanks in advance.

答:
David Stolnicki helped me:

Upon setting $a(v) = \frac{1}{(\int |x|^\alpha v^p)^{1/p}}$, we have
$$
\inf_{v \neq 0} \frac{\int |\nabla v|^2}{(\int |x|^\alpha v^p)^{2/p}} = \frac{a^2}{a^2} \inf_{v \neq 0} \frac{\int |\nabla v|^2}{(\int |x|^\alpha v^p)^{2/p}} =  \inf_{\int|x|^\alpha w^p = 1} \int |\nabla w|^2
$$

As $v$ runs through $H_0^1$, $w$ runs through the restriction and we are done.


问:
Differences of # and the reverse of # problem

Millie wrote a five-digit whole number on a blackboard and she also wrote it in reversed order. She considered the difference of her two numbers, and then told Lucy the last three digits of this difference. Can Lucy figure out the first two digits of the difference from this information alone or there could be more than one possibility?

This is an extremely difficult question, and I really have no idea how to solve this!

I came up with the equation $9999A+990B-990D-9999E=Z$ if $Z$ is the difference of $ABCDE$ and $EDCBA$ if $ABCDE$ is the original number! But I cannot solve for $A, B, C, D$ or $E$ with just a single statement.

I feel you can only solve this with logic (and probably computing brute force) e.g. for ### as the last three digits the beginning can be # or #. . .

答:
Recall the tests for divisibility by $9$ and $11$:

The sum of digits of the number is $\equiv\pmod 9$ to the original number

The sum/difference of the digits of the number (units minus tens plus hundreds minus thousands plus tens of thousands etc.) is $\equiv\pmod{11}$ to the original number.

If the numbers $\overline{ABCDE}$ and $\overline{EDCBA}$ are written using digits $A, B, C, D, E$ in those two orders, then they are both $\equiv\pmod{9}$ to $A+B+C+D+E$ and they are both $\equiv\pmod{11}$ to $A-B+C-D+E$. This means that their difference is divisible by $9$ and by $11$.

Thus, if someone has provided last three digits $\overline{RST}$ of the difference $\overline{PQRST}$, then the whole difference is given by:

$$\begin{array}{rcll}\overline{PQRST}&\equiv&\overline{RST}&\pmod{1000}\\\overline{PQRST}&\equiv&0&\pmod{9}\\\overline{PQRST}&\equiv&0&\pmod{11}\end{array}$$

As per Chinese Remainder Theorem, this determines $\overline{PQRST}$ up to $\pmod{9\cdot 11\cdot 1000}$ (as $9$, $11$ and $1000$ are all coprime). As this difference is a difference of two $5$-digit numbers, it is between $0$ and $99999-10000=89999$, so being uniquely determined up to $\pmod{99000}$ means it is uniquely determined, period.

Answer: yes, knowing the last three digit of the difference, one can recover the first two digits by demanding that the whole $5$-digit difference is divisible by $9$ and $11$.


问:
Font for math paper
I need to write a paper for math and I don't know which font to use. I would like to use this font, but I don't really know what font it is. Can someone help me?</p>

答:
The default font used in $\LaTeX$, a standard piece of software for document typesetting in mathematics and many scientific fields, is Computer Modern. This appears to be the font shown in your screenshot.

If you wish to learn LaTeX, there are many online resources and websites (such as Overleaf) that will allow you to start learning and writing LaTeX code without needing to download anything; there is also a dedicated StackExchange site devoted to the software.

Title: Font for math paper


问:
Placing a rectangle on a 3D triangle

Background

I have a 3D triangle (let's call the points XYZ) and trying to place a rectangle on it. By that I mean the points of the rectangle should stick to the surface of the triangle in 3D space. The position and size of the rectangle should be defined by some offset vectors relative to X.

My Understanding

My approach would be to calculate two normalized vectors, one which points from X to Y and the other one which is rectangular to the first one and points to the direction of Z. By multiplicating some offset's with the two unit vectors it should be possible to calculate ABCD on the surface of the triangle.

Could that work? How can I calculate this unit vectors?

答:
The dot product of $\vec{XZ}$ with the unit vector $\vec{XY} / \left| \vec{XY} \right|$ gives you the projection
of $\vec{XZ}$ onto $\vec{XY} $.

That multiplied by the unit vector $\vec{XY} / \left| \vec{XY} \right|$ itself is the component along it of $\vec{XZ}$.

$\vec{XZ}$ minus the last  is the orthogonal component you are looking for.

Make it unitary, ...

Title: Placing a rectangle on a 3D triangle


问:
If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has..?
I have a question that is as follows:

If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has:

(a) no  solution

(b) has exactly two solutions

(c) is satisfied by any real $\alpha$ and any real $x \in (0,1)$

(d) is satisfied by any real $\alpha$ and any real $x \in (1,\infty )$

The answer is given as (c)

Here's what I have tried, and I cannot proceed further ..... :

The given inequation is equivalent to:

$\log_2x+\frac{1}{\log_2x}+2\cos\alpha \leq0$

Now, if I let $x \in (1,\infty)$ then clearly $\log_2x >0$ and, by AM-GM inequality, I get that

$\log_2x + \frac{1}{\log_2x} \geq 2$

I also know that $2\cos \alpha \in [-2,2]$
This is what I've managed to analyse and cannot proceed further, plz help.

答:
We have

$$\log_2x+\log_x2+2\cos\alpha \leq0, 1≠x>0.$$

If $\log_2(x)>0,$ then you get

$$\log_2x + \frac{1}{\log_2x} =u(x)≥2$$

$$u(x)+2\cos \alpha ≤0$$

$$-1≤\cos \alpha ≤- \frac {u(x)}{2}≤-1$$

$$\cos \alpha =-1, x=2$$

If $\log_2(x) <0 \Longleftrightarrow 0<x<1$, you get

$$\log_2x + \frac{1}{\log_2x} =-v(x), v(x)\geq 2$$

$$-v(x)+2\cos \alpha ≤0, v(x)≥2$$

$$\cos \alpha ≤\frac{v(x)}{2}, v(x)≥2$$

which is correct for any $\alpha \in\mathbb R$.


问:
help with $\lim_{(x,y) \to (0,0)} {\ln(xy + 1) \over x}$
Compute:

$\lim_{(x,y) \to (0,0)} f(x,y)$ if $f(x,\ y)={\ln(xy + 1) \over x}$

According to Wolfram the limit is 0.

I'm trying to verify it by approaching the origin from different paths.

When approaching the origin from $x$:

$\lim_{(x,y=0) \to (0)} f(x,0) = {\ln(1) \over x}={0\over x}= 0$

However, I don't know how to proceed when approaching the origin from $y$. In this case, $x=0$ and the denominator is zero.

$\lim_{(x=0,y) \to (0)} f(0,y) = {\ln(1) \over 0}={0\over 0}$

I guess I should try get rid of that zero in the denominator, but I don't see how as if I set $x=0$ all the variables disappear.

Body:
We have $\ln(1+x)\sim_0 x$ hence we find
$$\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{(x,y) \to (0,0)}{\ln(xy + 1) \over x}=\lim_{(x,y) \to (0,0)}\frac{xy}{x}=\lim_{(x,y) \to (0,0)}y=0$$


答:
That font is Computer Modern, the default font for TeX and LaTeX.  Usually you type in some macro code and have it compiled:

```<code>\documentclass{article}

<span class="math-container">\begin{document}
The point is that we're \emph{given} the beginning and the end, and somehow
we have to fill in the middle. But we can't just fill it in randomly -- we
have to fill it in with something that ``gets us to the end''.
\end{document}</span>
</code>```

gives you

Usually you'd use some LaTeX installation to use the font, but I think there font files for use in other applications.  I've never tried it, but I'm aware of https://sourceforge.net/projects/cm-unicode/

Title: Font for math paper

Body: I need to write a paper for math and I don't know which font to use. I would like to use this font, but I don't really know what font it is. Can someone help me?</p>

Body:
That font is Computer Modern, the default font for TeX and LaTeX.  Usually you type in some macro code and have it compiled:

\documentclass{article}

\begin{document}
The point is that we're \emph{given} the beginning and the end, and somehow
we have to fill in the middle. But we can't just fill it in randomly -- we
have to fill it in with something that ``gets us to the end''.
\end{document}

gives you

Usually you'd use some LaTeX installation to use the font, but I think there font files for use in other applications.  I've never tried it, but I'm aware of https://sourceforge.net/projects/cm-unicode/


问:
Replacement of the tagged partition displacement of the Riemann Integral
In specific, my question is if I can make the following by the definition on Riemann integral...

$$\int_a^bf(x)dx = \lim_{\lambda \rightarrow 0}\sum_{i=1}^n f(c_i)\Delta x_i = \lim_{\lambda \rightarrow 0}\sum_{i = 1}^n f(c_i)\lambda$$

That is, replacing the displacement $\Delta x_i$ with $\lambda$, where $\; \lambda = \max{\Delta x_i} \;\; (i = \overline{1,n})$

Where $f$ is a real-valued function defined on the interval $[a, b]$.

And the Riemann sum of $f$ with respect to the tagged partition $x_0, ..., x_n$ together with $t_0, ..., t_{n − 1}$

I think succeeded to prove it but I am not sure if this is right...

$$\forall_{(i = \overline{1, n})} \quad \lim_{\lambda \rightarrow 0} \Delta x_i = \lim_{\max{\Delta x_i \rightarrow 0}}\Delta x_i \leq \lim_{\max{\Delta x_i \rightarrow 0}}\max{\Delta x_i} = 0$$

$$\therefore \forall_{(i = \overline{1, n})} \quad 0 \leq \lim_{\max{\Delta x_i \rightarrow 0}}\Delta x_i \leq 0 \quad \Rightarrow \lim_{\lambda \rightarrow 0} \Delta x_i = \lim_{\lambda \rightarrow 0} \lambda$$

答:
This is obviously true if the partition is uniform and $\Delta x_i = \lambda$, but not true in general.

For a counterexample, consider the function $f : [1,e] \to \mathbb{R}$ where $f(x) \equiv 1$ and the partition with points $1 < e^{1/n} < e^{2/n} < \ldots <  e^{(n-1)/n}< e$. The partition norm is

$$\lambda= \max_{1 \leqslant i \leqslant n}\Delta x_i = \max_{1 \leqslant i \leqslant n}(e^{i/n} - e^{(i-1)/n}) = \max_{1 \leqslant i \leqslant n}e^{(i-1)/n}(e^{1/n} - 1) =e^{(n-1)/n}(e^{1/n} - 1) $$

Note that $\lim_{n \to \infty} e^{(n-1)/n}(e^{1/n} - 1) = e\cdot 0 = 0$ and $\lambda \to 0$ if and only if $n \to \infty$.

For any choice of intermediate points $c_i$, we have

$$\sum_{i=1}^n f(c_i) \lambda = \sum_{i=1}^n [1 \cdot e^{(n-1)/n}(e^{1/n} - 1)] = ne^{(n-1)/n}(e^{1/n} - 1),$$

and

$$\lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = \lim_{n \to \infty} ne^{(n-1)/n}(e^{1/n} - 1) = \lim_{n \to \infty} e^{(n-1)/n}\cdot \lim_{n \to \infty}\frac{e^{1/n} - 1}{1/n} = e$$

However,

$$\int_1^e f(x) \, dx = e-1\neq \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = e$$

Since $\lambda \geqslant \Delta x_i$, it is true in general for $f \geqslant 0$ that

$$ \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i) \lambda = \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i)\, \Delta x_i \frac{ \lambda}{\Delta x_i}\geqslant \lim_{\lambda \to 0}\sum_{i=1}^n f(c_i)\, \Delta x_i = \int_a^b f(x) \, dx$$

The counterexample clearly satisfies that condition.


问:
How to represent combinations of plus and minus
Just wondering if there is a shortcut for representing the following:
$$x=1+i,1-i,-1+i,-1-i$$
I know that you can do:
$$x = 1\pm i,-1\pm i$$
But can you use something like:
$$x = \pm 1 \pm i$$
NOTE: I know the example above just gives $x = 1+i, -1-i$, but I just wanted to provide an example

答:
Another variation is
\begin{align*}
x^4=-4
\end{align*}


问:
Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero
According to WolframAlpha, the following sequence
\begin{align*}
a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}
\end{align*}
seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.

答:
First approach. By the beta integral
\begin{align*}
\frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi  }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi  }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ &  = \frac{1}{{\sqrt \pi  }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - s} } }}ds}  = \frac{1}{{\sqrt \pi  }}\int_0^{ + \infty } {e^{ - (x - 1/4)s} s^{ - 1/2} \sqrt {\frac{{s/2}}{{\sinh (s/2)}}} ds}
\end{align*}
for all $x>0$. It is well known that $w < \sinh w$ for all $w>0$. Thus, for $x>\frac{1}{4}$,
$$
\!\! \frac{{\Gamma (x)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}} \le \frac{1}{{\sqrt \pi  }}\int_0^{ + \infty }\! {e^{ - (x - 1/4)s} s^{ - 1/2} ds}  = \frac{1}{{\sqrt {\pi \!\left( {x - \frac{1}{4}} \right)} }}\int_0^{ + \infty }\! {e^{ - t} t^{ - 1/2} dt}  = \frac{1}{{\sqrt {x - \frac{1}{4}} }}.
$$
Consequently,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \sqrt {\frac{2}{n}}
$$
for all $n\geq 1$. From the asymptotics I gave in the comments, it is seen that this bound is sharp.

Second approach. By the log-convexity and the functional equation of the gamma function, we have
$$
\log \Gamma \left( {x + \tfrac{1}{2}} \right) \le \frac{1}{2}\log \Gamma (x) + \frac{1}{2}\log \Gamma (x + 1) = \log \Gamma (x + 1) - \frac{1}{2}\log x,
$$
i.e.,
$$
\frac{{\Gamma \left( {x + \frac{1}{2}} \right)}}{{\Gamma (x + 1)}} \le \frac{1}{{\sqrt x }}
$$
for all $x>0$. Consequently,
$$
\frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \frac{2}{{\sqrt {2n - 1} }}
$$
for all $n\geq 1$.


问:
Show that the Scattered Line (also called the Michael Line) is $T_4$
I just finished reading through the section about normal spaces (resp. $T_4$-spaces) from Willard General Topology text, and came across the following problem:

Show that the Scattered Line (also called the Michael Line) is $T_4$.

I'm not sure how to prove this, let alone get started on the problem. Any help would be much appreciated.

Note: The Scattered Line (also called the Michael Line) is given as follows: Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.

$$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$$

The real line with the topology generated by $\mathcal{B}$ is called the Scattered Line (also called the Michael Line).

答:
A LOTS (linearly ordered topological space) is a triple $\langle X,\tau,\le\rangle$, where $\le$ is a linear order on $X$, and $\tau$ is the associated order topology. The desired result follows quite easily from an important result about thise spaces: if $\langle X,\tau,\le\rangle$ is a LOTS, then $X$ is $T_5$ (i.e., hereditarily normal and $T_1$). I posted a proof of this result here.

Let $X=(\Bbb P\times\Bbb Z)\cup(\Bbb Q\times\{0\})$, and give $X$ the lexicographic order topology. Let $M$ be the subspace $\Bbb R\times\{0\}$ of $X$; then $M$ is homeomorphic to the Michael line by the map $\langle x,0\rangle\mapsto x$. Thus, the Michael line is a subspace of a LOTS and as such is normal.


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
Just as an addition:

A deeper reason is that $5$ and $6$ are idempotents in the ring $\mathbb Z/10\mathbb Z$, meaning that they have the property $x^2 = x$.

By the Chinese remainder theorem, $\mathbb Z/10\mathbb Z \cong \mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$.
In the latter representation, it is not hard to see that the idempotents are given by $(0,0)$, $(1,1)$, $(0,1)$ and $(1,0)$. Their preimages in $\mathbb Z/10\mathbb Z$ are $0$, $1$, $6$ and $5$, respectively.

In other words: In $\mathbb Z/10\mathbb Z$, $5$ is the unit element of the subring $\{0,5\}\cong \mathbb Z/2\mathbb Z$ and $6$ is the unit element of the subring $\{0,2,4,6,8\} \cong \mathbb Z/5\mathbb Z$ of $\mathbb Z/10\mathbb Z$.


答:
Writing $$x = \pm 1 \pm i$$
is ambiguous as it could either mean $x \in \{1+i,-1-i\}$ or the intended $x \in \{1+i,1-i,-1+i,-1-i\}$. You can make it unambiguous by specifying (in words) that the $\pm$-signs can be chosen independently. I've also seen $x = \pm 1 \pm' i$ to make the independence of the $\pm$-signs explicit in the notation, but I'm not sure that's very common.

Title: How to represent combinations of plus and minus

Body:
Just wondering if there is a shortcut for representing the following:
$$x=1+i,1-i,-1+i,-1-i$$
I know that you can do:
$$x = 1\pm i,-1\pm i$$
But can you use something like:
$$x = \pm 1 \pm i$$
NOTE: I know the example above just gives $x = 1+i, -1-i$, but I just wanted to provide an example


问:
$\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.
Suppose $X_n \rightarrow X$ a.s. as $n \rightarrow \infty$ and that $X_n \geq0$ for all $n, \omega$. Furthermore, $\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.

I know that a sequence of random variables converges almost surely to $X$ if for every $\epsilon > 0$ we have
\begin{equation}
P(\lim_{n \rightarrow \infty}|X_n - X|< \epsilon) = 1
\end{equation}

But I'm not sure how to proceed. Some hints would be greatly appreciated.

答:
You can always write

$$ \mathbb{E}\big[ \vert X_n-X \vert \big]= \int_F\vert X_n-X\vert dP+ \int_{F^c}\vert X_n-X\vert dP\leq \int_{F} \vert X_n\vert dP+ \int_F \vert X\vert dP+ \int_{F^c}\vert X_n-X\vert dP$$

Since $X$ is  integrable and $\{X_n\}$ is unifromly integrable, for every $\epsilon>0$ there exists $\delta_\epsilon>0$ such that

$$ \int_F\vert X_n\vert dP<\epsilon \quad \text{and} \quad \int_F\vert X\vert dP<\epsilon $$
if $P(F)<\delta_\epsilon$. Using almost sure convergence, we know that

$$ P( \vert X_n-X\vert>\epsilon )\to0. $$

Therefore, for $n$ large enough we have $ P( \vert X_n-X\vert>\epsilon )<\delta_\epsilon$ and therefore

$$ \mathbb{E}\big[ \vert X_n-X \vert \big] \leq  \int_{ \{  \vert X_n-X\vert>\epsilon \}} \vert X_n\vert dP+ \int_{ \{  \vert X_n-X\vert>\epsilon \} } \vert X\vert dP+ \int_{\{  \vert X_n-X\vert\leq\epsilon \}}\vert X_n-X\vert dP < $$

$$ < 2\epsilon+ \int_{\{  \vert X_n-X\vert\leq\epsilon \}} \epsilon\cdot dP\leq 3\epsilon.   $$

This is true for all $\epsilon>0$, so $\mathbb{E}[X_n]\to \mathbb{E}[X]$.


问:
Fourier transform of compact supported function is entire
Suppose $f\in L^1(\mathbb{R})$ has compact support, say $\operatorname{supp}f\subset[-r,r]$.
I want to show that its Fourier transform
$$ \hat{f}(z) = \int_{-a}^{a} f(t) e^{-2\pi itz}dt$$
for $z\in\mathbb{C}$ is entire, so $\hat{f}\in H(\mathbb{C})$. I think this follows from Morera's theorem, but I am not very familiar with contour integrals. Could someone help me out with the first steps?

答:
Here is a proof based on Moreras theorem that $\hat{f}(z) = \int_{-a}^a f(t) e^{-2 \pi i tz} dt$ is holomorphic.

Let $\gamma(t)$ be a simply closed curve in the complex plane.

Now, $\int_\gamma \hat{f}(z) dz=\int_\gamma \int_{-a}^a f(t) e^{-2 \pi i tz} dt dz = \int_{-a}^a \int_\gamma f(t) e^{-2 \pi i tz} dzdt$ by Fubinis theorem.

But the function $g(z)=f(t)e^{-2 \pi i tz}$ is holomorphic so $\int_\gamma g(z) dz = 0 $ by Cauchys theorem.

Hence, we have shown that for any simply closed curve $\gamma$, $\int_\gamma \hat{f}(z) dz =0$.  Moreras theorem, then, implies that $\hat{f}$ is holomorphic.


问:
Let $f(x)= \frac{x^4-\lambda x^3-3x^2+3\lambda x}{x-\lambda}$ Condition for $\lambda$ according to question?
Let $f(x)= \frac{x^4-\lambda x^3-3x^2+3\lambda x}{x-\lambda}$, If range of $f(x)$ is the set of entire real numbers then the true set in which $\lambda$ lies is?

The options are: (a) $[-2,2]$ (b) $[0,4]$ (c) $(1,3)$

The answer is $[-2,2]$

What I did was, factorize it out and then what I got was, $f(x)= \frac{(x^3-3x)(x-\lambda)}{(x-\lambda)}$

I can see that the local maximum and local minimum values of the polynomial $x^3-3x$ is $2,-2$ respectively, but what do I do to find the value of $\lambda$?

答:
With $g(x) =x^3 - 3x$ we have $g(4) = 52$  If your $\lambda = 4$ there is no longer a way to represent $52$ by the original fraction, which is undefined at $4.$ The graph would have one of those little open circles at $(4,52)$

In brief, for target numbers larger than $2,$  there is just one $x$ value that works, and this $x$ value is also bigger than two.


问:
Adding a constant to each weight of a network does not alter its minimum spanning tree
Consider a network $(G,w)$ and a function $m:E(G)\rightarrow \mathbb{R}$, with $m(e)=w(e)+c$, for every $e \in E(G)$. Show that $T$ is a minimum spanning tree of $(G,w)$ if and
only if it is a minimum spanning tree of $(G,m)$. I have an idea, that adding the same constant to each weight will not alter the order of the increasing sequence of weights, so for example running Kruskal's algorithm or the reverse-delete algorithm will add (for the former) or delete (for the latter) exactly the same edges in the same order, resulting in the same tree. Is this correct or is it too simple?

答:
It's correct, but it's even a little complicated for this kind of result. You could just say that, for each solution of cost $t^*$ in one problem, you can give a solution in the other problem that has a cost of $t^* \pm (|G|-1)c$ (which is the same tree, the $\pm$ depends on the direction), so the optimal solution of both are just the same with a cost differing by $\pm (|G|-1)c$.

The advantage of doing this is that you don't have to use the existence of an algorithm to solve the minimum spanning tree problem, and this kind of method translates very well to other (typically NP-complete) problems...

Moreover, in your approach, you might have problems when there are multiple solutions to the problem. Here, every solution of one problem corresponds to the solution of the translated problem, and each solution is optimal iff the translated version is optimal.


问:
Compact subset of a t.d.l.c. group contained in compact-open subgroup?
Let $G$ be a totally disconnected, locally compact group which is the union of its compact-open subgroups (or perhaps only $\sigma$-compact), and let $C \subset G$ be a compact set. Can one conclude that $C$ is contained in some compact-open subgroup of $G$?

答:
The famous Burnside problem asks whether a  finitely generated group, whose  elements all have finite order,  must be
a finite group.

This question was answered in the negative in 1964 by Golod and Shafarevich.

If $G$ is a counter-example to the Burnside problem, once it is equipped with the discrete topology, it is also a counter-example to the question posed by the OP.
The finite generating set is a compact set not contained in any compact group!


问:
Probability of arrangement of letters following specific rules
I have letters ABCDEFG (7 letters). I want to find the probability that if I shuffle the letters and arrange them randomly that B is first and A is last (but I can pick any arbitrary two letters).

I tried to simplify this:

If I have just two letters: A and B, there are $2! = 2$ ways to arrange them. And the probability of B being first and A being last is just $1/2$.

If I have three letters: A, B, C, there are $3! = 6$ ways to arrange them. The probability of B being first and A being last is $1/6$.

But this pattern breaks down somewhat for A, B, C, D. There are $4! = 24$ ways and the probability of B being first and A being last is $2/24$ because there are two ways for that to be true: BCDA, BDCA.

Is there a general rule for this problem?

答:
Yes. If you know the first and last letters, all you need to count is the number of ways to arrange the letters in between. There are $(n-2)!$ ways to do it, where $n$ is the number of letters (including $A$ and $B$).

Since there are a total of $n!$ ways to arrange the $n$ letters (disregarding the restriction on $A$ and $B$), this gives us a probability of

$$\frac{(n-2)!}{n!} = \frac1{n(n-1)}.$$

Notice that this assumes (as in your given examples) that $A$ and $B$ figure in your $n$ letters (otherwise the probability is of course $0$), and also that the $n$ letters are all distinct.


问:
Prove that $^{n+r}C_r \rightarrow \infty$ as $n \rightarrow \infty$

How do I prove that $^{n+r}C_r \rightarrow \infty$ as $n \rightarrow \infty$?

Writing $^{n+r}C_r = \frac{(n+r)!}{n!r!}$ does not help as it is indeterminate.

I think it tends to infinity because the number of ways of choosing $r$ objects from an collection of an increasing number of objects would also increase without bound. It is quoted as part of a STEP question as an obvious statement, but I can't see an obvious proof.

答: We have $$\binom{n+r}{r} \ge \left\lfloor \frac{n+r}{r} \right\rfloor > \frac{n}{r}$$ because $n+r$ objects contain $\lfloor \frac{n+r}{r} \rfloor$ disjoint groups of $r$ objects each, that could each be chosen. When $n \to \infty$, clearly $\frac{n}r \to \infty$.


问:
Why $\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)}$
Given $A\succ 0$ (positive-definite) and $c>0$, I am trying to show

$$\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)} \tag{1}$$

using information like this but without success so far. Could you please help to prove it? Any help is welcome.

EDIT: This inequality is an abstract form of the inequality

$$\left\| \left[f''(x_k) - r_M(x_k)\frac{M}{2}I \right]^{-1}f'(x_k)\right\| \leq \frac{\|f'(x_k)\|}{\lambda_{min}\left(f''(x_k)\right)}\tag{2}$$ appeared here (Theorem 3). Here is the part of the proof

答:
Since $A\succ 0$, so is $A+cI\succ 0$, so $\lVert (A+cI)^{-1}\rVert = \lambda_{\min}(A + cI)^{-1}$, and
$$
\lVert (A+cI)^{-1}x\rVert\leq \lVert (A+cI)^{-1}\rVert\lVert x\rVert = \frac{\lVert x\rVert}{\lambda_{\min}(A+cI)}\leq \frac{\lVert x\rVert}{\lambda_{\min}(A)}.
$$


答:
Write $$(A+cI)y=x$$ What you wish to show is
$$
\| y \| \leq \frac{1}{\lambda_{min}(A)}\| x\|.
$$
Let us show a little bit more, namely;
$$
\| y \| \leq \frac{1}{\lambda_{min}(A)+c}\| x\|.
$$
By definition of positive definiteness, $$ Ay \cdot y \geq \lambda_\min(A) \| y \|^2 $$
And of course, $A+cI_d$ is also positive definite, and
$$ Ay \cdot y \geq \left(\lambda_\min(A)+c \right)\| y \|^2 $$.

By Cauchy-Schwarz
$$
\| y \| \| x \| \geq x \cdot y
$$
So
$$
\| y \| \| x \| \geq x \cdot y = Ay \cdot y \geq \left(\lambda_\min(A)+c \right)\| y \|^2
$$
If $x=0$, then $y=0$, and the result is trivial. If not, then $y\neq0$, and we obtain the result by simplifying by $\|y\|$.

Title: Why $\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)}$

Body:
Given $A\succ 0$ (positive-definite) and $c>0$, I am trying to show

$$\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)} \tag{1}$$

using information like this but without success so far. Could you please help to prove it? Any help is welcome.

EDIT: This inequality is an abstract form of the inequality

$$\left\| \left[f''(x_k) - r_M(x_k)\frac{M}{2}I \right]^{-1}f'(x_k)\right\| \leq \frac{\|f'(x_k)\|}{\lambda_{min}\left(f''(x_k)\right)}\tag{2}$$ appeared here (Theorem 3). Here is the part of the proof


问:
Infinite series raised to a power being a power series
After thinking about how

$$\left(\sum_{k=0}^\infty \frac{1}{k!}\right)^{z} = \sum_{k=0}^\infty \frac{z^k}{k!}$$

I wondered about what kind of sequence $(a_n)_{n=0}^\infty$ satisfies

$$\left(\sum_{k=0}^\infty a_k\right)^{z} = \sum_{k=0}^\infty a_k z^k \tag{1}$$

for all $z \in \mathbb{C}$. For instance, any such sequence would satisfy

$$\left(\sum_{k=0}^\infty a_k z^k\right) \left(\sum_{k=0}^\infty a_k w^k\right) = \left(\sum_{k=0}^\infty a_k\right)^z \left(\sum_{k=0}^\infty a_k\right)^w = \left(\sum_{k=0}^\infty a_k\right)^{z+w} = \sum_{k=0}^\infty a_k (z+w)^k$$

I will use the notation $(a_n)$ to denote $(a_n)_{n=0}^\infty$ and $(a_n)^z$ to denote $(a_n z^n)$. I would like to find a product $(a_n) \cdot (b_n)$ of such sequences which satisfies

$$(a_n)^z \cdot (a_n)^w = (a_n)^{z+w} \tag{2}$$

for all $z,w \in \mathbb{C}$. If we only consider sequences which satisfy $a_n \binom{n}{k} = a_k a_{n-k}$ for all $k$, then a discrete convolution would be such a product since

$$(a_n)^z \cdot (a_n)^w = (c_n)$$

where

$$c_n = \sum_{k=0}^n a_k a_{n-k} z^k w^{n-k} = a_n \sum_{k=0}^n \binom{n}{k} z^k w^{n-k} = a_n (z+w)^n$$

as required.

Are there any other sequences which satisfy (2) where the product is a discrete convolution? Are there any other products which satisfy (2), for some subset of sequences satisfying (1)? And more importantly, is $(\frac{1}{n!})$ the only sequence satisfying (1)?

答:
Suppose $a_k$ is a sequence of such that such that $\sum_{k=0}^\infty a_k$ converges.  Write $c = \sum_{k=0}^\infty a_k$.  Assume $c>0$, let $\log c$ be the natural logarithm of $c$.  Then
$$
\left(\sum_{k=0}^\infty a_k\right)^z = c^z
= \exp(z\log c) = \sum_{k=0}^\infty \frac{(\log c)^k}{k!} z^k
$$

So, in order to get the equation you request, there must be a number $c>0$ such that
$a_k = \frac{(\log c)^k}{k!}$.

Of course the famous example is where $c=e$, and $\log c = 1$, so $(\log c)^k = 1$.

I suppose, in case $c > 0$ fails, then we have to inestigate choosing branches of the logarithm.

For example, we could define $(-1)^z = \exp(i \pi z)$ so that
$$
a_k = \frac{i^k \pi^k}{k!} .
$$


问:
If $f(x) = (x^2+2\alpha x + \alpha^2-1)^{\frac{1}{4}}$ has its domain and range so that union is $\mathbb{R}$, what does $\alpha$ satisfy?
My question asks:

If $f(x) = (x^2+2\alpha x + \alpha^2-1)^{\frac{1}{4}}$ has its domain and range such that their union is set of real numbers, what does $\alpha$ satisfy? (Answer is stated to be $\alpha \leq -1 $)

My approach was as follows, I can clearly see that the range has to be positive so that the domain must comprise of all negative real numbers, so that the union of range and domain consists of the full set of real numbers. For domain, the part under the radical sign must be $\geq 0$

$$x^2+2\alpha x + \alpha^2-1 \geq 0$$

$$\implies (x+\alpha-1)(x+\alpha +1)\geq 0$$

But now I'm stuck, how do I get the range for $\alpha$, and what condition do I apply such that the domain is negative? How do I proceed from here?

答:
Opting for a different route than the one suggested by @Gregory in the comments.

Going from what you have
$$g(x) = (x-(-\alpha +1))(x-(-\alpha-1)) \geq 0$$
So, if you plot $1-\alpha$ and $-1-\alpha$ on a number line; notice that the function $g$ is positive everywhere to the right of $(1-\alpha)$ and also positive to the left of $(-1-\alpha)$ and thus satisfies the inequality when
$$x \in D = (-\infty, -1-\alpha] \cup [1-\alpha, \infty)$$
Moreover, notice that the vertex coordinates of the upward open parabola representing $g$ is simply $(-\alpha, -1)$ (thereby making its range $[-1,\infty)$). Therefore, the range of $f$, regardless of the value of $\alpha$ is $R = [0,\infty)$ (because of the fourth root on the outside).

Now, we know $$D \cup R = \mathbb{R} \implies (-\infty, -1-\alpha] \cup [1-\alpha, \infty) \cup [0,\infty) = \mathbb{R}$$
Notice that when $\alpha \leq -1$, we have the first interval in the above equality becomes $(-\infty, c]$ where $c$ is non-negative, the second interval becomes redundant because of the third interval. Therefore, the inequality is held true.  So $\alpha \leq -1$.

Edit: If the last bit of the above argument is a little messy, here's neater a way by contradiction. If $\alpha > 1$, then $-\alpha$ can never be in $D \cup R$, can you see why?

Hint: $-1-\alpha + 1 = -\alpha$, and $-1-\alpha + 2 = 1- \alpha$.


答:
Otherwise it is immediate that $$\hat{f}(z)=\int_{-a}^a \sum_{k\ge 0}z^k\frac{(-2i\pi t)^k}{k!} f(t)dt= \sum_{k\ge 0}z^k\int_{-a}^a \frac{(-2i\pi t)^k}{k!} f(t)dt$$

where inverting $\sum,\int$ is allowed by convergence of $\int_{-a}^a \sum_{k\ge 0}\frac{|-2i\pi z t|^k}{k!} |f(t)|dt$

Title: Fourier transform of compact supported function is entire

Body:
Suppose $f\in L^1(\mathbb{R})$ has compact support, say $\operatorname{supp}f\subset[-r,r]$.
I want to show that its Fourier transform
$$ \hat{f}(z) = \int_{-a}^{a} f(t) e^{-2\pi itz}dt$$
for $z\in\mathbb{C}$ is entire, so $\hat{f}\in H(\mathbb{C})$. I think this follows from Morera's theorem, but I am not very familiar with contour integrals. Could someone help me out with the first steps?


问:
How to pass minimum count of apples so that every one around the table has same amount of apples?
With three ($A_1,A_2 and A_3)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=4 $ apples,to ensure each one has equal amount of 3 apples,  A3 just  gives A1 one apple A3 $\to$ A1 =1. In this case, the minimum count of ample exchanged is 1.

With four ($A_1,A_2,A_3 and A_4)$ people sitting around the table, each having $a_1=2,a_2=3,a_3=6,a_3=9 $ apples, to ensure each one has equal amount of 5 apples,  A1 $\to$  A2 =0, A2 $\to$ A3 =-2,  A3 $\to$ A4 ==-1, A4 $\to$ A1 =3. In this case, the minimum count of apples exchanged is 0+2+1+3=6.

Question : With n people $A_i$(i from 1 to n) having  $a_i$ apples initially, what  the minimum count of apples exchanged needed so that each will have  same amount of apples in the end? (please note the question is about the minimum count of apples exchanged, not the minimum count of apple exchanges: we are counting the apples, not the exchanges.

Clarification:   The exchange can only happen between 2 neighbors. For any person  $A_i$, you can  give away $x_i$ to $A_{i+1}$ and take in $x_{i-1}$ from $ A_{i-1}$. $x_{i-1}$ and $x_{i+1}$ can be any integers (including 0).

答: Think of everyone who has too many piling the excess in the center of the table.  Those who don't have enough take the number they need.  This is the minimum number of apples that need to be exchanged.</p>


问:
help with $\lim_{(x,y) \to (0,0)} {\ln(xy + 1) \over x}$
Compute:

$\lim_{(x,y) \to (0,0)} f(x,y)$ if $f(x,\ y)={\ln(xy + 1) \over x}$

According to Wolfram the limit is 0.

I'm trying to verify it by approaching the origin from different paths.

When approaching the origin from $x$:

$\lim_{(x,y=0) \to (0)} f(x,0) = {\ln(1) \over x}={0\over x}= 0$

However, I don't know how to proceed when approaching the origin from $y$. In this case, $x=0$ and the denominator is zero.

$\lim_{(x=0,y) \to (0)} f(0,y) = {\ln(1) \over 0}={0\over 0}$

I guess I should try get rid of that zero in the denominator, but I don't see how as if I set $x=0$ all the variables disappear.

答:
Without further information, the function $f(x,y)$ is not defined on the line $x = 0$. Of course, we can try to define:

$$f(0,y) := \lim_{x \to 0} f(x,y)$$

which by an application of De l'Hôpital's rule amounts to:

$$\lim_{x\to 0} \frac{\log(xy+1)}x= \lim_{x\to 0} \frac{\frac y{xy+1}}1 = y$$

This shows that $f$ may be continuously extended to the line $x=0$ by setting $f(0,y) = y$. It follows at once that the limit $\lim\limits_{(x,y)\to(0,0)}f(x,y)$, if it is to exist, must equal the iterated limit:

$$\lim_{y\to0}\lim_{x\to 0} f(x,y) = \lim_{y\to0} y = 0$$

One can subsequently set out to prove that this is indeed the sought limit, e.g. using the approach of Sami Ben Romdhane, using suitable theorems about limits, or via $\varepsilon$-$\delta$ methods.


问:
How should we use differential "dx" in deriving Tabular formula
I am learning repeated integration with the  Tabular method. I see textbooks or web sites prove the method, but  most of the times, they  either do not explain each step or just skip with (...). I try to prove it myself  by going through  each step.  I am not sure if my manipulations of differential "$dx$" are legitimate, and in general how to justify them.

I started with  with  (uv)'=u'v+vu'

$\frac {d(uv)} {dx}
= \frac {du} {dx}v+ u\frac {dv} {dx}$  $\to$ (Q1: Can I  simply multiply both sides  by $dx$?)

$d(uv)= vdu+udv$

$\int uv =\int udv+\int vdu$

$\int udv=\int uv -\int vdu$  $    $  (I)

Set $\frac {du} {dx}=u^1$ ,  $\frac {dv^{-1}}{dx}=v$ $\to$  $du= u^1dx$,   $v=\frac {dv^{(-1)}} {dx}$

$\int vdu= \int u^1dx$$\frac {dv^{(-1)}} {dx}$  $\to$ $\int u^1dv^{-1}$  (Q2:  Can I directly cancel $dx$ from  both numerator and denominator?)

$\int udv= uv -\int u^1dv^{-1}  $

Repeating similar steps $\to$

$\int udv=$$\sum_{i=0}^{n-1} (-1)^iu^i v^{-i} +\int (-1)^nu^n dv^{-n}$   $    $

$\int udv=$$\sum_{i=0}^{n} (-1)^iu^i v^{-i} +\int (-1)^{n+1}u^{n+1} dv^{-(n+1)}$   $    $  (II)

which is equivalent to common expression

$\int udv=$$\sum_{i=0}^{n} (-1)^iu^i v^{-i} +\int (-1)^{n+1}u^{n+1} v^{-(n)}dx$   $    $  (II')

答:
Treating dx as a fraction is technically wrong but I don't think you'll make any 'big mistakes' due to it. As in, even though it is an incorrect operation, it is one which produces correct results.

Anyways for repetead integration by parts, there is a faster derivation by a switch of notation, let $f_{i} $ denote the ith derivative of $f$, so for example $f_1$ is first derivative of $f$.

Now, we can write: $$(fg)_{-1}= f_{-1} g - (f_{-1} g_1)_{-1}= f_{-1}g -f_{-2} g_{1} + (f_{-2} g_{2})_{-1} $$

Repeating the above method n times,

$$ (fg)_{-1} = \sum_{i=1}^n f_{-i} g_{i-1}(-1)^{i+1}+ (-1)^n (f_{-n}g_{n})_i$$

refer here page-50 of the Joseph Edward book


问:
Determine if the set $A=\{(x,y) \in \overline{B}\mid x \geqslant0 \}$ is open or closed.

Let $\overline{B}= \{(x,y) \in \Bbb{R}^2 \mid x^2+y^2 \leqslant 1 \}.$ Determine if the set $A=\{(x,y) \in \overline{B}\mid x \geqslant0 \}$ is open or closed.

If I let $(x,y)\in A$ and $r >0$ then in order for $A$ to be open it would have to satisfy that for every $(x,y) \in A$ the open ball $B((x,y),r)$ would be a subset of $A$. I think this set is not open, but I'm not sure how to prove it? For example I have that $(-\frac{\varepsilon}{2},0) \in \overline{B}\setminus A$, but $(0,0) \in A.$

答: Yes, this shows that $A$ is not open in $\bar B$. To show that it is closed, note that $\bar B\setminus A=\{(x,y)\in\Bbb R^2:x<0\}\cap\bar B$ which is open in $\bar B$ since the former set is open in $\Bbb R^2$.


问:
Minimal residue making congruence system solvable (CRT)

A number when divided by $902$, $802$ and $702$ leaves remainder of $602$, $502$ and $402$ respectively. What would be the minimum remainder of when the number is divided by $2005$?

My attempt:

For the given number $n$, $n + 300$ is divisible by $902$, $802$ and $702$. Hence $n + 300$ can be written as $\mathrm{lcm}(902,802,702)\cdot K$ where $K$ is the integer and we can keep varying $K$ from $1$ onwards to manually check the minimum remainder. However this is not a great approach. Any simplistic approach to this problem would be appreciated.

答:
Lots of obfuscation.

$n \equiv 502 \pmod {802}$ so $n\equiv 502 \equiv 101 \pmod {401}$.

So $n =101 + 401K$ for some $K$ so $\pmod {2005 = 401 \times 5}$ we have  $n \equiv 101 + 401k$ for $k = 0,1,2,3,4$

So $n \equiv 101 + 401k \pmod {2005}$.  The least that can be is $101$ if $k=0$ and we'd have $n \equiv 101 \pmod 5$.

Now if we want

$n\equiv 602 \pmod{902}$

$n \equiv 502 \pmod {802}$

$n \equiv 402 \pmod {702}$ and

$n \equiv 101 \pmod 5$ we just need

$n \equiv 0 \pmod 2; n \equiv 602 \pmod 451; n \equiv 502 \pmod{401}; n\equiv 402\pmod{351}; $ and $n\equiv 101 \pmod 5$.

And as $2, 451, 401, 351, 5$ are relatively prime.  So by CRT we can have that.

(And thank god we don't have to find that...... well, it wouldn't be that bad[1] but... we don't have to find it; it's enough to know it exists)

And so $n \equiv 101\pmod 5$ and $n \equiv 101 \pmod {401}$ and so $n\equiv 101 \pmod {2005}$ and  $101$ is the smallest possible remainder.

====

[1] If we actually want to solve:

$n \equiv -300 \pmod {902, 802, 702}$ so $n\equiv -300 \pmod{2;451;401;351}$ so $n \equiv -300 \pmod {126957402}$ and $n \equiv 1\pmod 5$ so $n\equiv -300 + 126957402k \pmod {634,787,010}$ where $k = 0,1,2,3, 4$ but as $n \equiv 1 \pmod 5$ we must have $k=3$ so $n \equiv 380871906 \pmod {634,787,010}$ and indeed $380871906 \pmod {2005} \equiv 101$.

And $380871906 \equiv 602;502;402 \pmod{902;802;702}$.


答:
Using a little Taylor series:

$$\log(1+xy)=xy-\frac{x^2y^2}2+\mathcal O(x^3y^3)\implies$$

$$\frac{\log(1+xy)}x=y-\frac{xy^2}2+\mathcal O(x^2y^3)\xrightarrow[ (x,y,)\to (0,0)]{}0$$

Title: help with $\lim_{(x,y) \to (0,0)} {\ln(xy + 1) \over x}$

Body:
Compute:

$\lim_{(x,y) \to (0,0)} f(x,y)$ if $f(x,\ y)={\ln(xy + 1) \over x}$

According to Wolfram the limit is 0.

I'm trying to verify it by approaching the origin from different paths.

When approaching the origin from $x$:

$\lim_{(x,y=0) \to (0)} f(x,0) = {\ln(1) \over x}={0\over x}= 0$

However, I don't know how to proceed when approaching the origin from $y$. In this case, $x=0$ and the denominator is zero.

$\lim_{(x=0,y) \to (0)} f(0,y) = {\ln(1) \over 0}={0\over 0}$

I guess I should try get rid of that zero in the denominator, but I don't see how as if I set $x=0$ all the variables disappear.


问:
If $p$ is a prime then all the non trivial subgroups of $G$ with $\lvert G\rvert=p^2$ are cyclic

If $p$ is a prime then all the non trivial subgroups of $G$ with $\lvert G\rvert=p^2$ are cyclic.

I tried looking online where does this result come from, but could not find any direct result. I found that all groups with order $p^2$ are isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z_p} \times \mathbb{Z_p}$; is this a consequence of this result?

答: If $H\le G$ is nontrivial, then $|H|=p$ (Lagrange). Then $1\ne x\in H\Longrightarrow \langle x\rangle=H$ (Lagrange, again), so $H$ is cyclic.


问:
homomorphism of $G/N$-modules
Let $G$ be a group with $N$ a normal subgroup (not necessarily of finite index). Let $Q$ and $A$ be $G$-modules and $P$ be a $G/N$-module.

I want to make sense of the term ${\mathrm{Hom}}_{G/N}(P, {\mathrm{Hom}}_N(Q,A))$.

Here $Q$ and $A$ are $N$-modules by restriction of the $G$-action. Is there any standard $G/N$-module structure on ${\mathrm{Hom}}_N(Q,A)$?

${\mathrm{\bf{Some~background~on~the~question:}}}$ If $N \unlhd G$, and $A$ is a $G$-module, then there is a (Hochschild-Serre-) spectral sequence $H^p(G/N,H^q(N,A)) {\Rightarrow}_p H^{p+q}(G,A)$.

The construction of this begin with taking a pair of projective resolution ${\mathrm{\bf{P}}} \rightarrow {\mathbb{Z}}$ (as $G/N$-modules) and ${\mathrm{\bf{Q}}} \rightarrow {\mathbb{Z}}$ (as $G$-modules) and form the double complex $E^{0}_{p,q} := {\mathrm{Hom}}_{G/N}(P_p, {\mathrm{Hom}}_N(Q_q,A))$ ($0$-th level).

答:
Let $f\colon Q \to A$ be an $N$-module homomorphism, let $g \in G$, and let $q \in Q$. Then we can define
$$(g * f)(q) = g f(g^{-1} q).$$
If $g \in N$, then $g * f = f$ because $f$ is $N$-equivariant. It's a straightforward check of the definitions to verify that $g * f$ defined as above is an $N$-equivariant linear map and that this defines a group action of $G/N$ on $\operatorname{Hom}_N(Q, A)$. (It might be helpful to think of $G/N$ as "acting by conjugation" here.)

Title: homomorphism of $G/N$-modules


问:
Are these two definitions of the Riemann mapping theorem equivalent?
I have two definitions of the Riemann mapping theorem

1.Suppose that $G$ is a simply connected domain in the complex plane, $G\neq \mathbb{C}$, and that $z_0$ is a point of $G$. There exists a unique conformal mapping f of $G$ onto the open unit disk $\mathbb{D}$ satisfying the conditions $f(z_0)=0$ and $f'(z_0)>0$.

2.$G$ is a non-empty simply connected Domain of the complex number plane $\mathbb{C}$ which is not all of $\mathbb{C}$, then there exists a biholomorphic mapping $f$ from $G$ onto the open unit disk $\mathbb{D}$.

Regarding that $conformal \Leftrightarrow biholomorphic$ it follows: $1 \Rightarrow 2$. Is $2 \Rightarrow 1$ also true ?

答:
The answer is yes.

So let $f: G\to \mathbb{D}$ be the biholomorphic map you're given. Let $w=f(z_0)$.

Then, $T_w(z)=\frac{z-w}{1-\overline{w}z}$ is a conformal equivalence of the unit disk onto itself mapping $w$ to $0$.

Hence, $g=\frac{1}{f'(z_0)T_w'(w))}T_w\circ f$ is a conformal equivalence with the desired property. This gives you existence.

Must it be unique?

Well, assume $g$ and $\tilde{g}$ are both biholomoprhic maps onto $\mathbb{D}$ mapping $z_0$ to $0$ and such that their derivatives at $z_0$ are positive. Assume without loss of generality that $g'(z_0)\geq \tilde{g}'(z_0)$. Then,
$$
(g\circ \tilde{g}^{-1})'(0)=\frac{g'(z_0)}{\tilde{g}'(z_0)}\geq 1
$$
but by the Schwarz lemma, $|(g\circ \tilde{g}^{-1})'(0)|\leq 1$ so, in fact it must be equal to 1. Thus, by the second part of the Schwarz lemma, $g\circ \tilde{g}^{-1}$ must be a rotation and, since it's derivative is positive, it must be the identity. Hence, $g=\tilde{g}$.


问:
How to calculate line integral if $c2$ has the range $0\le t \lt \infty$
I have to calculate the line integral for $\int_{\mathrm{C2}}F\, dr\,$ where $C2$ is given by $$\frac{(cost,sint)}{1+e^t}$$ and $0\le t \lt \infty$
Where $\vec{F} = \big(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}\big)$.

I have no idea how to get started, I have never worked with ranges with $\infty$ before. Any tips are appreciated!

答:
Alternatively, as pointed out in the comments, $\vec F$ is conservative, with scalar potential

$$f(x,y)=-\sqrt{1-x^2-y^2}+c$$

since $f_x=\frac x{\sqrt{1-x^2-y^2}}$ and $f_y=\frac y{\sqrt{1-x^2-y^2}}$.

The the integral is

$$\begin{align}
\int_{C_2}\vec F\cdot\mathrm d\vec r&=\int_{C_2} \nabla f(x,y)\cdot\mathrm d\vec r\\[1ex]
&=\lim_{t\to\infty}f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right)-\lim_{t\to0^+}f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right)
\end{align}$$

We have

$$\begin{align}
g(t)=f\left(\frac{\cos t}{1+e^t},\frac{\sin t}{1+e^t}\right)&=-\sqrt{1-\frac{\cos^2t}{(1+e^t)^2}-\frac{\sin^2t}{(1+e^t)^2}}+c\\[1ex]
&=-\frac{\sqrt{2e^t+e^{2t}}}{1+e^t}+c\\[1ex]
&=-\frac{\sqrt{2e^{-t}+1}}{e^{-t}+1}+c
\end{align}$$

As $\lim\limits_{t\to\infty}g(t)=-1+c$ and $\lim\limits_{t\to0^+}g(t)=-\frac{\sqrt3}2+c$, we find that the line integral has a value of $\boxed{\frac{\sqrt3}2-1}$.

Alternatively, you can compute the integral directly, if you feel so bold. Composing $\vec F$ with $x(t)=\frac{\cos t}{1+e^t}$ and $y(t)=\frac{\sin t}{1+e^t}$ gives

$$\begin{align}
\vec F(x(t),y(t))&=\left(\frac{\frac{\cos t}{1+e^t}}{\sqrt{1-\left(\frac{\cos t}{1+e^t}\right)^2-\left(\frac{\sin t}{1+e^t}\right)^2}},\frac{\frac{\sin t}{1+e^t}}{\sqrt{1-\left(\frac{\cos t}{1+e^t}\right)^2-\left(\frac{\sin t}{1+e^t}\right)^2}}\right)\\[1ex]
&=\frac{\sqrt{e^t(2+e^t)}}{(1+e^t)^2}(\cos t,\sin t)
\end{align}$$

and differentiating the parameterization of $C_2$ gives

$$\frac{\mathrm d\vec r}{\mathrm dt}=\frac{\left(-(1+e^t)\sin t-e^t\cos t,(1+e^t)\cos t-e^t\sin t\right)}{(1+e^t)^2}$$

Now, in the integral,

$$\begin{align}
\int_{C_2}\vec F\cdot\mathrm d\vec r&=\int_0^\infty \vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt\\[1ex]
&=-\int_0^\infty\frac{e^{\frac t2}}{(1+e^t)^2\sqrt{2+e^t}}\,\mathrm dt\\[1ex]
&=-2\int_{\frac\pi4}^{\frac\pi2}\frac{\cos^3v}{\sqrt{1+\cos^2v}}\,\mathrm dv&\text{where }\tan v=e^{\frac t2}\\[1ex]
&\vdots
\end{align}$$

(The precise steps for finding the antiderivative elude me, but rest assured this integral has the same value as found earlier.)


问:
Correspondence between divergent integrals and analythic functionals
The Dirac delta function can be defined the following way:

$$\delta(x)
= \frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}\, dt={\displaystyle {\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}}$$

The later formula formally follows from the Fourier transform definition.

We also know that $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$

It seems that if the expression under integral has a singularity at a point, the integral may gain wonderful changes there.
Some other examples:

$$\delta''(x)= -\frac{1}{2\pi}\int_{-\infty}^\infty t^2 e^{itx}\, dt={\begin{cases}-\frac1{2\pi}\int_{-\infty}^\infty t^2 dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And $\int_{-\infty}^\infty f(x)\delta''(x)dx=f''(0)$

Another example $(x\in\mathbb{R})$:
$$\delta(ai+x)= \frac{1}{2\pi}\int_{-\infty}^\infty e^{-at} e^{itx}\, dt={\begin{cases}\frac1{2\pi}\int_{-\infty}^\infty e^{-at} dt,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

And we know that $\int_{-\infty}^{\infty}f(x)\delta(ai+x)dx=f(-ai)$

So, given a divergent integral $S$, an expression $$\phi(x)={\begin{cases}S,&{\text{if }}x=0\\0,&{\text{if }}x\ne 0.\end{cases}}$$

What can we say about the functional $\int_{-\infty}^\infty f(x)\phi(x)dx$? Can we, knowing the functional, find the corresponding divergent integral $S$?

It seems to me, the expressions under the Fourier transform are similar to the algebra of linear operators.

答:
Following up on @reuns' points:

It is possible to reasonably think about distributions as "generalized functions", but then one must give up "point-wise values" as a description of "functions". Yes, there is still a notion of localization of distributions, and a notion of "wavefront set" and/or "smooth set", and so on, so it is sometimes possible to talk about pointwise values on certain sets. But not altogether.

Similarly, Fourier transform maps (and such) are indeed linear, and continuous in various topologies. But the integral definition does not literally apply outside $L^1$ functions. Either extend by continuity, or define by duality.

So, again, I don't think it's about "divergent integrals" per se, but about literal definitions of operators (etc) via integrals which fail to converge even when the operator has a reasonable extension. But those extensions very often require giving up tooo-classical pointwise senses of both functions and integrals.


问:
Analysis of $\delta_{k+1} \leq \frac{3}{2} \left(\frac{\delta_k}{1-\delta_k}\right)^2 \leq \frac{8}{3} \delta_{k}^2$
I am reading this paper (Theorem 3) and in some point it is stated that

$$\delta_{k+1} \leq \frac{3}{2} \left(\frac{\delta_k}{1-\delta_k}\right)^2 \leq \frac{8}{3} \delta_{k}^2$$

with $M \leq 2L$. Based on the above, the author states that $\delta_{k+1} \leq 1/4$ and $\delta_{k+1} \leq \frac{2}{3} \delta_k$. How did he come up with these bounds? Could you please give some intuitive explanations? Any help is highly appreciated, I am not familiar with series and their behavior.

Below I attach the part of the proof in this paper

答:
Suppose $\delta_0\leq \frac14$ let us show that if  $$\delta_{k+1} \leq \frac{3}{2} \left( \frac{\delta_k}{1-\delta_k}\right)^2\tag{1}$$  then for all $k$, $\delta_k\leq \frac14$.
By induction, it is true for k=0, and if it is true for $k$, then $$1-\delta_k \geq \frac34$$, $$ \frac{\delta_k}{1-\delta_k} \leq \frac13$$ and therefore $$ \delta_{k+1} \leq \frac 23 \frac 19 < \frac14.$$
Now that this is true, using again $1-\delta_k \geq \frac34$ we find
$$
\delta_{k+1} \leq \frac83 \delta_k^2 \leq \frac23 \delta_k.
$$

On how $\delta_0$ was probably chosen (to put it in the answer and not in the comments). The relationship is of the form
$$
\delta_{k+1} \leq f\left( \delta_k \right)
$$
and the author wishes to show that the sequence was first bounded, then eventually decreasing following the rate of $f$ near the origin. The function $f$ is increasing near the origin, so if $\delta_k \leq a$, then $\delta_{k+1}=f(\delta_k) \leq f(a)$. This provides a bound provided $f(a)\leq a$. Here we see that $f(a)\leq a$ provided $a\leq a_0$, where $a_0=f(a_0)$ is given by $a_0=\frac{7-\sqrt{33}}{4}\approx 0.31$ a simple choice for $a$ is then $\frac{1}{4}$, as $\frac{1}{3}$ is slightly too large.