问:
Finiteness of a $C^{*}$-algebra
A unital $C^{*}$-algebra is said to be finite if $x^*x = 1$ implies $xx^* = 1$.

A unital $C^{*}$-subalgebra $B$ of a finite $C^{*}$-algebra $A$ is finite if $1_{B} = 1_{A}$.

Let $\lbrace A_n \rbrace$ be a sequence of finite unital $C^{*}$-algebras. Then both $\prod_{n=1}^{\infty}A_n$ and $\prod_{n=1}^{\infty}A_n / \oplus_{n=1}^{\infty}A_n$ are finte.

Could anybody help me to prove these two theorems?

答:
There is no magic happening here:

The unit of $\prod_nA_n$ is simply $(1_n)_{n\geq1}$ where $1_n$ denotes the unit of $A_n$. Now suppose that $(1_n)_{n\geq1}=x^*x$ for some $x\in\prod_nA_n$, say $x=(x_n)_{n\geq1}$, where $x_n\in A_n$ and $\sup_{n}\|x_n\|=\|x\|<\infty$. Then for each $n\geq1$ we have that $1_n=x_n^*x_n$ since the operations in $\prod_nA_n$ are pointwise. Since each $A_n$ is unital, we have that $x_nx_n^*=1_n$ as well, for all $n\geq1$. But this means that $1_{\prod_nA_n}=xx^*$. This shows that $\prod_nA_n$ is finite.

For the quotient, let $x\in\prod_nA_n$, $x=(x_n)$ so that $1+\sum_nA_n=(x+\sum_nA_n)^*\cdot(x+\sum_nA_n)$, thus $1-x^*x\in\sum_nA_n$. Equivalently, for each $\varepsilon>0$ there exists $n_0\geq1$ so that for all $n\geq n_0$ we have that $\|1_n-x_n^*x_n\|<\varepsilon$.

Lemma: If $p,q$ are projections in a $C^*$-algebra and $\|p-q\|<1$, then they are homotopic in the set of projections.

But homotopic projections are also Murray-von Neumann equivalent (see Rordam's book "An introduction to K-theory of $C^*$-algebras" for a proof of this and the above lemma). Now let $\varepsilon>0$. We have that there exists $n_0\geq1$ so that $\|1_n-x_n^*x_n\|<\min\{\varepsilon,1\}$ for all $n\geq n_0$. Therefore (by the lemma and our observation) we can find $y_n\in A_n$ so that $1_n=y_n^*y_n$ and $x_n^*x_n=y_ny_n^*$, for all $n\geq n_0$. But since $A_n$ is finite we have that $y_ny_n^*=1_n$, so $x_n^*x_n=1_n$ and again, since $A_n$ is finite we have that $x_nx_n^*=1_n$. This shows that $0=\|1_n-x_nx_n^*\|<\varepsilon$ for all $n\geq n_0$. In particular, $1-xx^*\in\sum_nA_n$, so $1+\sum_nA_n=(x+\sum_nA_n)\cdot(x+\sum_nA_n)^*$.

Title: Finiteness of a $C^{*}$-algebra


问:
Tensor product of fixed ring isomorphism.
Question :

Let $A$ be a ring and $\phi : A\to A$ be a ring homomorphism that is an involution. Suppose $A$ is a finitely generated $\mathbb{C}$ algebra under the map $\beta : \mathbb{C} \to A$ and
$\psi: \mathbb{C} \to  \mathbb{C}$ be conjugation map such that $\phi\circ\beta=\beta\circ\psi$.

Define $B=[a\in A : \phi(a)=a]$. The map $\beta$ then restricts to $\mathbb{R}\to B$. Then show that $B\otimes_\mathbb{R}\mathbb{C}$ is isomorphic to $A$.

Attempt :

Define the map $\Delta : B\otimes_\mathbb{R}\mathbb{C}\to A$ as sending an element $b\otimes z$ to $b\beta(z)$.

I am trying to show that this map is injective and surjective but I am not able to prove either of them.

For example, for injectivity, if $b\beta(z)=0$, then how can I conclude that $b\otimes z=0$.

答:
Note that $A$ is a $\Bbb C$ vector space with scalar multiplication defined as $$z \cdot a = \beta(z)a$$
for $z \in \Bbb C$ and $a \in A$. This makes $A$ an $\Bbb R$ vector space as well.

Since $\phi$ is a homomorphism, it is clear that $B$ is closed under addition (and that it contains $0$). The fact that $\beta$ restricts as a map from $\Bbb R$ to $B$ shows that $B$ is actually an $\Bbb R$-subspace of $A$. (Indeed, if $x \in \Bbb R$ and $b \in B$, then $$\phi(x \cdot b) = \phi(\beta(x)b) = \phi(\beta(x))\phi(b) = \beta(\psi(x))b = \beta(x)b = x \cdot b$$ and thus, $x \cdot b \in B$.)

Thus, showing that $B \otimes_{\Bbb R} \Bbb C \cong A$ is showing the isomorphism of two $\Bbb R$ vector spaces. This can be done by showing that the dimensions are equal. Thus, we need to show that $$2\dim_{\Bbb R}B = \dim_{\Bbb R}A.$$
(Since $\dim_{\Bbb R}\Bbb C = 2$.)

Define $$\iota B = \{\iota \cdot b \mid b \in B\}.$$

Clearly, the above is an $\Bbb R$-subspace of $A$. Moreover, $\dim_{\Bbb R}B = \dim_{\Bbb R}(\iota B)$.

Claim 1. $B \cap \iota B = 0$.

Proof. Let $x \in B \cap \iota B$. In particular, $x = \iota \cdot y$ for some $y \in B$. Now, note that

\begin{align}
x = \phi(x) &= \phi(\iota\cdot y)\\
&= \phi(\beta(\iota)y)\\
&= \phi(\beta(\iota))\phi(y)\\
&= \beta(\psi(\iota)) y\\
&= \beta(-\iota)y\\
&= (-\iota)\cdot y\\
&= -(\iota\cdot y) = -x.
\end{align}

Thus, $x = 0$. $\blacksquare$

Claim 2. If $a \in A$ and $\phi(a) = -a$, then $a \in \iota B$.

Proof. It suffices to show that $\iota^{-1}\cdot a = -\iota \cdot a \in B$.

To this end, note that
\begin{align}
\phi(-\iota \cdot a) &= \phi(\beta(-\iota) a)\\
&= \phi(\beta(-\iota))\phi(a)\\
&= \beta(\psi(-\iota))(-a)\\
&= \beta(\iota)(-a)\\
&= -\iota\cdot a.
\end{align}
Thus, $-\iota\cdot a \in B$. $\blacksquare$

Claim 3. $A = B \oplus \iota B$.

Proof. By Claim 1, we only to show that $+$ part. To this end, note that every $a \in A$ can be written as

$$a = \underbrace{\dfrac{1}{2}\cdot(a + \phi(a))}_{\in B} + \underbrace{\dfrac{1}{2}\cdot(a - \phi(a))}_{\in \iota B}.$$

To see the underbraced statements, simply apply $\phi$ to the term and use the fact that $\phi(\phi(a)) = a$ ($\phi$ is an involution). This directly gives that the first term is indeed in $B$. For the second term, note that

$$\phi\left(\dfrac{1}{2}\cdot(a - \phi(a))\right) = -\dfrac{1}{2}\cdot(a - \phi(a))$$

and thus, $\dfrac{1}{2}\cdot(a - \phi(a)) \in \iota B$, by Claim 2. $\blacksquare$

Thus, it follows that

$$\dim_{\Bbb R}A = \dim_{\Bbb R}B + \dim_{\Bbb R}\iota B = 2\dim_{\Bbb R}B,$$

as desired.

ContentLicense: CC BY-SA 4.0


问:
$f$ such that $\frac{an}{n-b} \leq f(a,b)\cdot n $ for all $n \geq 1$
I have been trying to find upper bound of following thing with no success. Is there any inequality s.t following holds .

assume that $a \geq ,b \geq 1$ then find $f(a,b) \rightarrow \mathbb{R}^{+}$ s.t forall $n\geq 1$

$$\frac{an}{n-b} \leq f(a,b)\cdot n $$

答: Assuming that $n$ is a natural number, you should divide the inequality by $n$ as @RossMillikan mentioned in the comments. Then you get $\frac{a}{n-b} \leq f(a,b)$ for all $n \geq 1$. Now it is easy to show that $\frac{a}{n-b}>\frac{a}{n+1-b}$ if $n>b$ and $\frac{a}{n-b}<0$ for all $n<b$. Thus in the case $b \notin \mathbb{N}$ the function $f(a,b)=\frac{a}{\lceil b \rceil -b}$ works, since it holds for the smallest $n>b$ and is positive. In the case $b \in \mathbb{N}$ the same argumentation holds for $f(a,b)=a$.


问:
Conditional Markov Chain
I am interested in a method of how to compute the expected number of steps for absorpion in a Markov Chain with only one absorption node and given the starting and pre-final node (before absorption).

So, if the transition probability matrix is given, for example by:

$P = \begin{bmatrix}
0 & 1/5 & 2/5 & 1/5 & 1/5\\
1/5 & 0 & 1/5 & 1/5 & 2/5\\
2/5 & 0 & 1/5 & 1/5 & 1/5\\
1/5 & 1/5 & 0 & 1/5 & 2/5\\
0 & 0 & 0 &0 & 1
\end{bmatrix}$

what will be the expected number of steps for being absorbed given that we start in node 1 (first line) and the last state before absorption is node 2 (second line) ?

So what I tried so far... I know the very known technique of using the fundamental matrix $N$ and then computing $N = (I-Q)^{-1}$ where $Q$ is the recurrent probabilities matrix (from the canonical form of $P$) and the expected number of steps would be the first entry of the vector $(I-Q)^{-1}1$, where $1$ is the five dimensional vector with all entries equal to $1$, in the case where we dont know the information about where we are absorbed.

Then, as we know that we are absorbed from node two, I tried to replace all the probabilties of the last columns by $0$ (execept for line $2$ and for the aborption node that I kept $2/5$ and $1$, respectively), then I am not pretty sure of how re-normalizing all the others...

Thank you very much!

答:
Let $S := \{1,2,3,4,5\}$ be the state space of this homogeneous Markov chain (MC) and $\xi_n$ be the random state at time $n \in \mathbb{N}_0$. Let $\mathbb{P}_s$ be the joint distribution of $\xi_0,\xi_1,\ldots$ if the MC start in $s \in S$,  $(\Omega,\mathcal{A},\mathbb{P}_s)$ be the underlying probability space and $\tau$ be the entrance time in $s_a = 5$. Let $A := \{\xi_{\tau-1} = 2\}$ be the event that immediately before stopping the state is $2$ (with $\xi_{-1} := 5$). Of interest are $$\mathbb{E}_s[\tau ~|~ A] = \frac{g(s)}{f(s)}$$
with $f_s = \mathbb{P}_s(A)$ and $g(s) = \mathbb{E}_s Y$ with $Y = \tau \cdot 1_A$ ($1_A$ the indicator function of $A$) for $s \not= s_a$.

Let $P = (p_{si})_{s,i \in S}$ be the transition matrix of the MC. By the Markov property (here we don't need the strong Markov property) we get
$$\mathbb{P}_s(A) = 2/5 \cdot \delta_{s2} + \sum_{i=1}^5 p_{si} \cdot \mathbb{P}_i(A)$$
resp. $f = Pf + (0,2/5,0,0,0)'$. Since $f(5) = 0$ we can omit the state $s = 5$ and with $\tilde P := (p_{si})_{s,i \not= 5}$ we have to solve the equation
$$(\mathbb{1} - \tilde P) \tilde f = (0, 2/5, 0, 0)'$$. We get
$$\tilde f = (0.1897, 0.5, 0.1379, 0.1724)'$$

Similarly for $g$ we get
$$\mathbb{E}_s Y = \sum_{i \in S} p_{si} \mathbb{E}_i [Y ~|~ \xi_1 = i, \xi_0 = s]$$
with $\mathbb{E}_i [Y ~|~ xi_1 = i, \xi_0 = s] = 1$, if $s = 2, i = 5$, $ = 0$, if $s \not=2, i = 5$ or $s = 5$ and $= g_i+1$, else. Thus
$$g_s = (\tilde P (g+1))_s + 2/5 \cdot \delta_{s2}$$
resp. with $x = (g_1,\ldots,g_4)'$
$$x = \tilde P (x+1) + (0,2/5,0,0)' = \tilde P x + \tilde P (1,1,1,1)' + (0,2/5,0,0)'$$
with the solution $x = (2.9612, 2.6250, 3.0172, 2.1466)'$. In particular we get the solution $x(1)/f(1) = 15.6136$.

Title: Conditional Markov Chain


问:
Polynomial expressions in localized ideal
Let $A$ be a commutative ring and $S\subset A$ a multiplicatively closed subset. Fix a polynomial $f\in A[X]$ and define $I$ to be the ideal generated by elements of the form $f(a)$ with $a\in A$.

My question is:

Does the localized ideal $I_S$ contain all elements of the form $f(x)$ with $x\in A_S$?

This was motivated by this question/answer. There I prove the statement for polynomials of the form $f=X^n-X$. For $x=\frac{a}{s}\in A_S$ we have $$f(x)=\frac{1}{s^{n+1}}\left(sf(a)-af(s)\right)$$
And therefore $f(x)\in I_S$.

However I didn't manage to generalize this to arbitrary polynomials $f$. I would guess that one can explicitely write down some general formula for $f(\frac{a}{s})$ in terms of $f(a_i)$ where $a_i$ are suitable elements from $A$. But maybe there is also a 'nicer' proof that avoids these messy calculations (assuming the statement is indeed correct).

答:
Notation: For a ring $A$ and a polynomial $f \in A[X]$, let me define $I_{A,f} = \langle \{f(a)\}_{a \in A} \rangle$ to be the ideal of $A$ generated by the values of $f$. For any ring map $A \to B$, there is a natural $B$-module morphism $I_{A,f} \otimes_{A} B \to I_{B,f}$.

Remark: If $(\deg f)! \in A^{\times}$ (e.g. if $f$ is linear), then $I_{A,f}$ is the ideal generated by the coefficients of $f$, hence $I_{A,f} \otimes_{A} B \to I_{B,f}$ is surjective for any ring map $A \to B$.

Proof: Let $f(X) = a_{d}X^{d} + \dotsb + a_{1}X + a_{0}$; then $I_{A,f} \subseteq \langle a_{d},\dotsc,a_{1},a_{0} \rangle$. We plug in $X=0,1,\dotsc,d$ to get $\begin{bmatrix} f(0) \\ f(1) \\ \vdots \\ f(d) \end{bmatrix} = V \cdot \begin{bmatrix} a_{0} \\ a_{1} \\ \vdots \\ a_{d} \end{bmatrix}$ where $V$ is the Vandermonde matrix corresponding to $0,1,\dotsc,d$. The determinant of $V$ is $\pm \prod_{0 \le i < j \le d} (i-j)$, which is invertible in $A$ by assumption; hence $V \in \mathrm{GL}_{d+1}(A)$ and so we can solve for $a_{0},\dotsc,a_{d}$ as $A$-linear combinations of $f(0),\dotsc,f(d)$. $\square$

Claim: For any multiplicative subset $S \subset A$, we have an equality $S^{-1}(I_{A,f}) = I_{S^{-1}A,f}$ of ideals of $S^{-1}A$.

Proof: Let $f(X) = a_{n}X^{n} + a_{n-1}X^{n-1} + \dotsb + a_{1}X + a_{0}$. We will show that \begin{align*} \textstyle f(x/s) = \frac{1}{s^{n}}(a_{n}x^{n} + a_{n-1}x^{n-1}s + \dotsb + a_{1}xs^{n-1} + a_{0}s^{n}) \end{align*} is an $S^{-1}A$-linear combination of $f(x),f(xs),\dotsc,f(xs^{n})$. Thus we want to solve the vector equation \begin{align*} \begin{array}{@{}c@{}}{\begin{bmatrix} a_{0} & a_{1}x & \dotsb & a_{n}x^{n}\end{bmatrix}}\\ \vphantom{\vdots} \\ \\ \\ \end{array} \begin{bmatrix} s^{n} \\ s^{n-1} \\ \vdots \\ 1 \end{bmatrix} = \begin{array}{@{}c@{}}{\begin{bmatrix} a_{0} & a_{1}x & \dotsb & a_{n}x^{n} \end{bmatrix}}\\ \vphantom{\vdots} \\ \\ \\ \end{array} \begin{bmatrix} 1 & 1 & \dotsb & 1 \\ 1 & s & \dotsb & s^{n} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & s^{n} & \dotsb & s^{n^{2}} \end{bmatrix} \begin{bmatrix} b_{0} \\ b_{1} \\ \vdots \\ b_{n} \end{bmatrix} \end{align*} for $b_{i} \in S^{-1}A$. In fact we solve the equation \begin{align*} \begin{bmatrix} s^{n} \\ s^{n-1} \\ \vdots \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & \dotsb & 1 \\ 1 & s & \dotsb & s^{n} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & s^{n} & \dotsb & s^{n^{2}} \end{bmatrix} \begin{bmatrix} b_{0} \\ b_{1} \\ \vdots \\ b_{n} \end{bmatrix} \end{align*} in $\mathbb{Z}[s,s^{-1}]$ where we view $s$ as an indeterminate. For this, we recall that the inverse of the Vandermonde matrix \begin{align*} \begin{bmatrix} 1 & 1 & \dotsb & 1 \\ x_{0} & x_{1} & \dotsb & x_{n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{0}^{n} & x_{1}^{n} & \dotsb & x_{n}^{n} \end{bmatrix} \end{align*} has entries \begin{align*} v_{i,j} = (-1)^{n-j} \frac{e_{n-j}(\{x_{0},\dotsc,x_{n}\} \setminus \{x_{i}\})}{\prod_{m \ne i} (x_{i}-x_{m})} \end{align*} for $i,j=0,1,\dotsc,n$, where $e_{i}$ denotes the $i$th elementary symmetric polynomial (defined as $\prod_{m=1}^{n}(X-x_{i}) = \sum_{i=0}^{n}(-1)^{i}e_{i}(\{x_{1},\dotsc,x_{n}\})X^{n-i}$). In our case, we have $x_{i} = s^{i}$ for $i=0,1,\dotsc,n$. Thus we get the formula \begin{align*} b_{i} &= \frac{1}{\prod_{m \ne i} (s^{i}-s^{m})}\sum_{j=0}^{n} s^{j}(-1)^{n-j}e_{n-j}(\{x_{0},\dotsc,x_{n}\} \setminus \{x_{i}\}) \\ &= s^{n}\frac{\prod_{m \ne i} (\frac{1}{s}-s^{m})}{\prod_{m \ne i} (s^{i}-s^{m})} \\ &= \pm s^{???} \frac{\prod_{m \in \{1,\dotsc,n+1\}\setminus \{i+1\}} (1-s^{m})}{\prod_{0 \le m < i} (1-s^{i-m})\prod_{i < m \le n} (1-s^{m-i})} \\ &= \pm s^{???} \frac{\prod_{m=1}^{n+1} (1-s^{m})}{\prod_{m=1}^{i+1}(1-s^{m}) \prod_{m=1}^{n-i}(1-s^{m})} \end{align*} so it suffices to show that, for any $n,k$, the polynomial $\prod_{m=1}^{k} (1-s^{m})$ divides $\prod_{m=n-k+1}^{n} (1-s^{m})$ in $\mathbb{Z}[s]$. (This seems related to the fact that binomial coefficients are integers but I don't know of a proof that can be made to work here, for example it is not always true that we can rearrange $k$ consecutive integers $n-k+1,\dotsc,n$ so that $m$ divides the $m$th integer $n-k+\sigma(m)$, see $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}$.) So instead let $\Phi_{d}(s) \in \mathbb{Z}[s]$ be the $d$th cyclotomic polynomial, which is irreducible and satisfies $1-s^{n} = \prod_{d | n} \Phi_{d}$ for any $n$. For any $d,k$, the order of $\Phi_{d}$ dividing $\prod_{m=1}^{k}(1-s^{m})$ is $\lfloor \frac{k}{d} \rfloor$ so we conclude using that $\lfloor \frac{k}{d} \rfloor + \lfloor \frac{n-k}{d} \rfloor \le \lfloor \frac{n}{d} \rfloor$ for any $d,n,k$. $\square$

Remark: The analogous statement is true for multivariate polynomials as well, for example the above shows that $f(\frac{x}{s},\frac{y}{t})$ can be written as an $S^{-1}A$-linear combination of $f(s^{i}x,\frac{y}{t})$ for $i \ge 0$, each of which can be written as an $S^{-1}A$-linear combination of $f(s^{i}x,t^{j}y)$ for $j \ge 0$, etc.


问:
Riesz' representative $v\in H^1$ of the functional $H^1\ni u\mapsto\langle u,f\rangle_{L^2}$ for $f\in L^1\setminus H^1$
Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$. If $f\in L^2(\Omega)$, then $$\langle u,\varphi\rangle:=\langle u,f\rangle_{L^2(\Omega)}\;\;\;\text{for }u\in H^1(\Omega)$$ is clearly a bounded linear functional on $H^1(\Omega)$. Now $H^1(\Omega)$ is a Hilbert space and hence, by Riesz' representation theorem, there is a $v\in H^1(\Omega)$ with $$\langle u,\varphi\rangle=\langle u,v\rangle_{H^1(\Omega)}=\langle u,v\rangle_{L^2(\Omega)}+\sum_{i=1}^d\langle\partial_iu,\partial_iv\rangle_{L^2(\Omega)}\tag1$$ for all $u\in H^1(\Omega)$.

Clearly, if $f\in H^1(\Omega)$, then $f=v$, but is there something we can generally say about the relation between $f$ and $v$, even when $f\not\in H^1(\Omega)$?

答:
No, you do not have $f = v$, even in case $f \in H^1(\Omega)$. If this would be true, then
$$
\langle u,f\rangle_{L^2} = \langle u,\varphi\rangle = \langle u, f\rangle_{L^2} + \sum_{i=1}^d \langle \partial_i u , \partial_i f\rangle_{L^2}$$
for all $u \in H^1$,
which is clearly false (unless $\nabla f = 0$).

What is true, that $v \in H^1(\Omega)$ is the (weak) solution of
$$
-\Delta v + v = f \text{ in } \Omega, \qquad\frac{\partial}{\partial n} v = 0 \text{ on } \partial\Omega.$$


问:
Singularity of Green's function and eigenfunction expansion
AFAIK, a Green's function $G(x,\epsilon)$ has a singularity for $x=\epsilon$.
This is clear in many analytical expressions for $G(x,\epsilon)$.

I do not understand, however, what happens to the singularity when $G(x, \epsilon)$ is expanded in terms of the linear operator eigenfunctions:

$G(x, \epsilon) = \sum_i \frac{u_i(x)u_i(\epsilon)}{\lambda_i}$.

For example, for rectangular-like domains and the dirichlet laplacian, eigenfunctions $u(x)$ are sines. How can I show that the series of $G(x, \epsilon)$ is still singular at $x=\epsilon$?

答:
I found an answer in Courant & Hilbert's book pgs. 351-2.

The Green's function itself may be continuous and well-defined at $x=\epsilon$. However, the Green's function is singular in the sense that its first derivative does not exist at $x=\epsilon$.


问:
Optimal strategy for 2 players Lights Out game variation
Consider a turn-based game for 2 players. They're both playing on the same board. The board is 8x8, randomly generated and each cell has 0 or 1 (with equal probabilities), for example:

```<code>01101100
01010101
10111100
00001111
10101010
00000001
11100011
00101000
</code>```

In turn 1 player 1 moves, in turn 2 player 2 moves, in turn 3 player 1 moves and so on.

Move consists of choosing coordinates (x, y) where 1 is. It flips elements on positions (x, y), (x+1, y) and (x, y+1), if they are in  bounds of the board. By flips I mean turns 1 to 0 or 0 to 1.

(0, 0) is top left corner. So, after playing in (1, 1) board looks like this:

```<code>01101100
00110101
11111100
00001111
10101010
00000001
11100011
00101000
</code>```

And after playing in (7, 1) like this:

```<code>01101100
00110100
11111101
00001111
10101010
00000001
11100011
00101000
</code>```

You win when after your move there are only 0 on board.

My question is: what is the optimal strategy for this game?

Research so far:

Lights Out description on MathWorld is worth reading.

Presented game is a variation of Lights Out, but instead of flipping 4 neighbor cells, we flip only 2 and we can move only in cells with 1. It's also similar to Nim.

In the same way like MathWorld describes we can find $S$ - a minimum set of moves to win. For 8x8 board there will be always exactly one solution.

Since matrix addition is commutative, the order in which the moves are performed is irrelevant.

If $|S|$ (minimal number of moves to win) is odd - it's a winning position. For example if it's 1 - we make only one move and we win.

If it's 3 - we make a move from $S$ and opponent gets a board with 2 moves in $S$. Now, if he makes the move from $S$, he will lose, because we will have the final move. So, if he has an option, he will not make a move from $S$.

By trying some examples I find out that opponent move not from $S$ increases our $S$ by 1 move - exactly the move he made. So, we get back the board with 3 (different) moves in $S$. But, he can't stop us like this forever - finally all his possible moves will be in $S$.

So, my hypothesis is: When you don't make a move from $S$, $|S|$ will increase exactly by 1 and his "destroying" move will be added to the set (so later we can "undo" his "destroying" move).

Which would mean that the opponent can only do -1 (by playing a move from $S$) or +1 (by not playing a move from  $S$), so he can't change the parity of our $|S|$!

So, if all of above is true, it means that players have no influence on the result of the game. We could even play by random and if the initial $|S|$ was odd - we will win, if it was even - we will lose.

It's hard for me to believe in that. What will be the sense of making AI challenge on HackerRank with such a game? Players in the leader-board have different scores and these scores are consistent each time after recalculation (same people in the top 10), suggesting that players actually have an influence on the game result.

Maybe you can see some mistake I'm making?

答:
let's take a simple 1x3 board

101

This is a loss for player 1.

But the loss is faster if the cell on the left is chosen first.

Since there is a move limit, the losing play can squeeze out a draw by stalling as much as possible.

The winning player, when possible needs to make a move that can't be undone.

Let's try a simple board where player 1 wins.

1011

player 1 can take out the right 2, which forces the game to 5 moves.
player 1 can take left for

0111

player 2 then can't take left or right, which lets player 1 win on move 3.

instead, player 2 will take the middle, which recreates the same situation as if player 1 had started with the right.

So now we have the strategy.

solve the board as quickly as possible. making required moves that can't be undone, working from upper left to lower right.  Count the number of moves it took. This is very easy. You simply sweep from upper left to lower right, clicking each 1. This will reveal all the currently required moves, and is an optimal solution always.

If ODD, you are winner. Try to find a move that can't be tagged back (there is always at least one) that doesn't overlap any required cells, but does overlap as many unrequired cells that are 1 as possible, to deny them to your opponent on their turn.  Say you are starting on a full 3x3 board. upper left is the ideal move because it permanently removes the corner 1, and knocks out two non required moves next to it so they can't be selected by the opponent.  Prefer permanent unlights of non required cells, but always pick one that cannot relight. Therefore, you will never directly lengthen the game.

if EVEN, you are loser, and must avoid those required moves if you can. instead, pick a NOT required move that unlights 2 required moves if possible, and if not, unlights one 1. If only required cells ae currently available, then try to pick the one that is closest to the bottom right.  If you are starting with the 3x3 board with the bottom right cell already taken, then picking one of the two unrequired next to the start is a good move, because it tags two required cells, but is not a required move itself. the upper left move is still required, and it will tag your cell you picked again, making it a new required cell.  Hopefully if you keep doing this, you will extend the game past the 100 move mark and salvage a draw.  You may get a chance to force extra clicks on required cells too. take these if you find them.

And here we go.  :)

if you were playing alone, the order would not matter for the required moves. but since you are competing, the important thing to do is permanently remove as many chances for the opponent to extend the game as you can.

ContentLicense: CC BY-SA 3.0


问:
$M$ is a maximal subgroup of $G$ then either $N_G(M)=M$ or $N_G(M)=G$.
Prove that if $M$ is a maximal subgroup of $G$ then either $N_G(M)=M$ or $N_G(M)=G$.

Recall that

$N_G(M)=\{g\in G | gM=Mg\}$

A proper subgroup $M$ of $G$ is called maximal if whenever $M\leq H \leq G$, either $H=M$ or $H=G$.

I think I am overthinking this but let's just make sure. So

By the definition of maximal, it is enough to show $M \leq N_g(M) \leq G$. Well isn't this statement trivial? If not, why?

答:
Proof: Assume $M$ is maximal subgroup of $G$. Then $M\leq H\leq G$, for some $H\leq G$, implies $H=M$ or $H=G$. We need to show that $N_G(M)=H$.

Facts:

If $M$ is a subgroup of $G$, then $M\subseteq N_G(M)$.

$N_G(M)$ is a subgroup of $G$.

What we are missing is that $M$ is a subgroup of $N_G(M)$. i.e. $x,y\in M \Rightarrow x^{-1}y\in M$. But we have this since $M$ is a group. Hence we can claim $H=N_G(M)$ which implies $N_G(M)=M$ or $N_G(M)=G$.


问:
Explanation for Proof of $[T(u)]_{\gamma}=[T]_{\beta}^{\gamma}[u]_{\beta}$
I am struggling with understanding the proof of the following theorem:

Theorem. Let $$ and $$ be finite-dimensional vector spaces over a field $$, having ordered basis $\beta$ and $\gamma$, respectively. Let $: \rightarrow $ be linear. Then for each $u∈$ we have:

$$[T(u)]_{\gamma}=[T]_{\gamma}^{\beta}[u]_{\beta}$$

The proof for this theorem is given as:

Proof: Fix $u∈V$, and define the linear transformations $:F \rightarrow V$ by $(a)=au$ and $g:F \rightarrow W$ by $g(a)=aT(u)$ for all $a∈F$. Let $\alpha=\{1\}$ be the standard ordered basis for $F$. Notice that $g=T$. Identifying column vectors as matrices and using Theorem 2.11, we obtain

$$[T(u)]_{\gamma}=[g(1)]_{\gamma}=[g]_{\alpha}^{\gamma}=[Tf]_{\alpha}^{\gamma}=[T]_{\beta}^{\gamma}[f]_{\alpha}^{\beta}=[T]_{\beta}^{\gamma}[f(1)]_{\beta}=[T]_{\beta}^{\gamma}[u]_{\beta}$$

I am having trouble understanding the logic behind this proof. It seems like it just plugs in 2 linear transformations ($f$ and $g$) that make this proof work. How does it show that there isn't an example of $f$ and $g$ that won't make this theorem work? How does this proof prove the theorem for all possible $T$ and $u$?

答:
The logic behind this proof is that it wants to utilize the already proved similar result, theorem 2.11.: for any conposable linear maps $F,G$ and fixed bases $\alpha,\beta,\gamma$ we have
$$[F\circ G]_\alpha^\gamma=[F]_\beta^\gamma \cdot [G]_\alpha^\beta\,.$$
For that end, we wish to represent the single vector $u$ as some linear map, so we define $f:F\to V$ which just 'picks out' $u$ (all such linear maps are determined by their value at $1$ since $\alpha=(1)$ is a basis for $F$), that is, $f(\lambda)=\lambda u$.

The main observation is that
$$[f]_\alpha^\beta=[f(1)]_\beta=[u]_\beta\,,$$
so when applying the theorem for $T$ and this specific $f$ (which depends on $u$, but $u$ was arbitrary), we get
$$[T\circ f]_\alpha^\gamma=[T]_\beta^\gamma\cdot [f]_\alpha^\beta=[T]_\beta^\gamma\cdot[u]_\beta$$
where the left hand side is $[T(f(1))]_\gamma=[T(u)]_\gamma$, and note that $g=T\circ f$.


问:
Is Skolem arithmetic effectively axiomatizable?
Skolem arithmetic

I found nothing via Google about its
axioms. Wikipedia says nothing about whether it has ever been effectively axiomatized. If it has, then what do its effective axioms look like?

答:
In my initial comment to the OP, I wrote:

A better question would be: "What is an explicit axiomatization of Skolem arithmetic"? Which probably can be best answered by reading the reference for Wikipedia's claim that this theory is decidable.

I finally took my own advice on this. Wikipedia's reference is to A Survey of Arithmetical Definability by Bès. Here one can read that Skolem arithmetic was axiomatized by Cegielski in the paper Theorie elementaire de la multiplication des entiers naturels. One can find a summary of Cegielski's results in (in English) in Chapter III.5 of Logical Number Theory I by Smorynski. I'll give a further summary of what I read in Smorynski's book here.

Let's agree to axiomatize the complete theory of $(\mathbb{N}_+,\cdot)$ here (where $\mathbb{N}_+ = \mathbb{N}\setminus 0$). If you want to include $0$, it's easy to obtain an axiomatization of $(\mathbb{N},\cdot)$ from this by asserting the existence of $0$ ($\exists x \forall  y (xy = x\land yx = x)$) noting it must be unique, and then restricting all the quantifiers in the axioms below to the non-zero elements.

Let's introduce some abbreviations:

$x|y$ is an abbreviation for $\exists z(xz = y)$. It means "$x$ divides $y$".

$x|_n y$ is an abbreviation for $\exists z(xz^n = y)$. It means "$x$ divides $y$  and the quotient is an $n^{\text{th}}$ power". We say "$x$ $n$-divides $y$". Note $0$-dividing is equality and $1$-dividing is ordinary dividing.

$O(x)$ is an abbreviation for $\forall y\, (xy = y)$. It means "$x = 1$".

$P(x)$ is an abbreviation for $\lnot O(x)\land \forall  y(y|x\rightarrow (O(y)\lor y = x))$. It means "$x$ is prime".

$PP(x,y)$ is an abbreviation for $P(x)\land x|y\land \forall z((P(z)\land z\neq x\rightarrow \lnot (z|y))$. It means "$x$ is prime and $y$ is a power of $x$".

$V(x,y,z)$ is an abbreviation for $PP(x,z)\land z|y \land \forall w((PP(x,w)\land w|y)\rightarrow w|z)$. It means "$z$ is the highest power of the prime $x$ which divides $y$".

$T(x,y,z)$ is an abbreviation for $\forall p(P(p)\rightarrow ((p|x\rightarrow \exists w(V(p,y,w) \land V(p,z,w)))\land (\lnot p|x\rightarrow \lnot p|z)))$. It means "$z$ is the number obtained by deleting from the prime factorization of $y$ all primes not dividing $x$".

$I(x,y)$ is an abbreviation for $\forall p(P(p)\rightarrow ((p|x\rightarrow \exists w(V(p,x,w)\land V(p,y,pw))\land (\lnot p|x\rightarrow \lnot p|y))$. It means "$y$ is the number obtained by increasing each prime exponent appearing in the prime factorization of $x$ by $1$".

$S_n(p,x,y)$ is an abbreviation for $P(p)\land p|xy\land \exists z\exists w(V(p,x,z)\land V(p,y,w)\land z|_n w)$. It means "$p$ is a prime which divides $xy$, and the highest power of $p$ dividing $x$ $n$-divides the highest power of $p$ dividing $y$.

For $n\geq 0$, $SP_n(x,y,z)$ is an abbreviation for $\forall p(P(p)\rightarrow ((S_n(p,x,y)\rightarrow V(p,z,p))\land (\lnot S_n(p,x,y)\rightarrow \lnot (p|z)))$. It means "$z$ is the product of those primes which divide $xy$ and for which the highest power of $p$ dividing $x$ $n$-divides the highest power of $p$ dividing $y$." You should think of this as a kind of comprehension/separation condition. If we think of an element as coding a (multi)set of primes (its prime factors), then $z$ codes the set of primes $p$ for which a certain modular arithmetic condition holds between the exponent of $p$ in the prime factorizations of $x$ and $y$.

Now we can give our axioms:

$\forall x\forall y\forall z((xy)z = x(yz))$

$\forall x\forall y(xy = yx)$

$\exists x (O(x))$ [Note that uniqueness of $x$ follows from the previous axioms.]

$\forall x\forall y\forall z(xz = yz\rightarrow x = y)$

$\forall x\forall y(O(xy)\rightarrow (O(x)\lor O(y)))$

$\forall x\forall y(x^n = y^n \rightarrow x = y)$ for each natural number $n>0$

$\forall x\exists y\exists z(x = y^nz\land \forall y'\forall z'(x = (y')^nz'\rightarrow z|z'))$ for each natural number $n>0$

$\forall x\exists p (P(p)\land \lnot  p|x)$

$\forall x\forall y\forall z((PP(x,y)\land PP(x,z))\rightarrow (y|z\lor  z|y))$

$\forall p\forall x(P(p)\rightarrow \exists y(V(p,x,y)))$ [Note that uniqueness of $y$ follows from the previous axioms.]

$\forall x\forall y(x = y\leftrightarrow \forall p(P(p)\rightarrow \exists z(V(p,x,z)\land V(p,y,z))))$

$\forall p\forall x\forall y(P(p)\rightarrow (\exists z\exists w (V(p,x,z)\land V(p,y,w)\land V(p,xy,zw))))$

$\forall x\forall y(\forall  p(P(p)\rightarrow \exists z\exists w(V(p,x,z)\land V(p,y,w)\land z|w))\rightarrow x|y)$

$\forall x\forall y\exists z(T(x,y,z))$ [Note that uniqueness of $z$ follows from the previous axioms.]

$\forall x\exists y(I(x,y))$ [Note that uniqueness of $y$ follows from the previous axioms.]

$\forall x\forall y\exists z(SP_n(x,y,z))$ for each natural number $n\geq 0$ [Note that uniqueness of $z$ follows from the previous axioms.]

Further, this theory has quantifier elimination in the definable expansion $\{\cdot,1,t,i,(sp_n)_{n\geq0},(E_n)_{n>0}\}$ where $1$ is a constant symbol defined by $O$ above, $t$ is a binary function symbol defined by $T(x,y,z)$ above, $i$ is a unary function symbol defined by $I(x,y)$ above,  $sp_n$ is a binary function symbol for each $n\geq 0$ defined by $SP_n(x,y,z)$ above, and $E_n(x)$ is a unary relation symbol for each $n>0$ defined by $$\exists p_1\dots\exists p_n(\bigwedge_{1\leq i < j \leq n} (p_i\neq p_j) \land \bigwedge_{1\leq i \leq n} (P(p_i)\land p_i|x)).$$

Disclaimer: I have not read Cegielski's proofs, only the summary of results in Smorynski. Some errors may have arisen in the "game of telephone" by which I summarized Smorynski's summary of Cegielski. But writing out the axiomatization here has given me some intuition for the kinds of complexities that arise in Skolem arithmetic - I hope it does the same for you!


问:
Solving integral equation with Laplace's Transform.
I'm trying to prove the following

$$\int\limits_0^\infty  {\frac{{\cos tu}}{{{u^2} + 1}}\log udu}  =  - \frac{\pi }{2}\int\limits_0^\infty  {\frac{{\sin tu}}{{{u^2} + 1}}du} $$

The original problem suggested the use of the Laplace Transform, thus I have

$$\int\limits_0^\infty  {\frac{s}{{{u^2} + {s^2}}}\frac{{\log u}}{{{u^2} + 1}}du = }  - \frac{\pi }{2}\int\limits_0^\infty  {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$

ADD The RHS transform can be evaluated as follows:

$$- \frac{\pi }{2}\int\limits_0^\infty  {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$

$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$

Now by partial fractions you can separate and get:

$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  =  - \frac{\pi }{4}\frac{1}{{{s^2} - 1}}\left[ {\log \left( {\frac{{m + 1}}{{m + {s^2}}}} \right)} \right]_0^\infty $$

$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  = -\frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$

For the second one I make a similar manipulation:

$$\frac{1}{{{s^2} + {u^2}}}\frac{1}{{{u^2} + 1}} = \frac{1}{{{s^2} - 1}}\left( {\frac{1}{{{u^2} + 1}} - \frac{1}{{{u^2} + {s^2}}}} \right)$$

$$\frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\left( {\frac{{\log u}}{{{u^2} + 1}} - \frac{{\log u}}{{{u^2} + {s^2}}}} \right)du} $$

Substitution to show integral over $\mathbb{R}$ of an odd function is zero so,

$$\frac{s}{{{s^2} - 1}}\left( {\int\limits_{ - \infty }^\infty  {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx}  - \int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}} }du \right)$$

$$ - \frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}}du $$

And again a suitable $u = m s$ produces

$$ - \frac{1}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm =  - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{1}{{{m^2} + 1}}} dm =  - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$


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

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

solution :

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

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

答:
$J\in k[x_1,\ldots,x_n]$

$J$ doesn't vanish on $\overline{k}^n$

Thus $J$ is constant, it is in $k^\times$

For $n=2$ try with $\varphi(x_1,x_2)=(x_1+x_2^2,x_2)$ to see that $Aut(\Bbb{A}^n)$ is much more complicated than $Aut(\Bbb{P}^n)$ where we get "it must be linear".

The Jacobian conjecture is the converse: if $J\in k^\times$ then does $\varphi$ have to be an automorphism?

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


问:
Proof that there is exactly one equivalence relation that forms a partition
So I've just started trying to teach myself some topology and i cant quite grasp how an equivalence class forms a partition more specifically i don't understand the proof  that there is exactly one equivalence relation that forms the partition.

For example in my book it says that a partition $D$ is formed from an equivalence relation $R$ and that equivalence relation  is unique that is there is only one equivalence relation that forms the partition however they do not give a proof of this  i myself cant come up with one so is there a proof of this statement?
Thanks in advance.

答:
Let $\Omega$ be a set. We say that a partition $P$ of $\Omega$ is compatible with an equivalence relation $R$ on $\Omega$ if for all $x, y \in \Omega$,
$$
xRy\iff \exists S\in P \text{ such that } x \in S \text{ and } y \in S.
$$
Claim: For any equivalence relation $R$ on $\Omega$, there exists exactly one partition $P$ of $\Omega$ that is compatible with $R$.

If I understand correctly, you are asking about the proof that there is at most one $P$ compatible with $R$. Suppose $P$ and $P'$ are both compatible with $R$; we shall show that $P = P'$.

Consider an arbitrary $S \in P$. Since $P$ is a partition, $S$ is nonempty; say $x \in S$. Since $P'$ is a partition, there is some set $S' \in P'$ such that $x \in S'$. For any $y \in S$, since $P$ is compatible with $R$, we have $x R y$. Since $P'$ is compatible with $R$, this implies that $y \in S'$. Therefore, $S \subseteq S'$. Similarly, $S' \subseteq S$, i.e., $S = S'$. Thus, $S \in P'$. Since $S$ was an arbitrary element of $P$, this shows that $P \subseteq P'$. By symmetry, we also have $P' \subseteq P$, i.e., $P = P'$.


问:
Trouble understanding notation of symplectic forms, differential forms and differentials
I found the exerpt below in the book "Introduction to Symplectic Topology".
I understand that a 2-form $\omega(p,\zeta, \zeta')$ on $\mathbb R^{2n}$ is a function that maps $\mathbb R^{2n}\times\mathbb R^{2n}\times\mathbb R^{2n}$ to $\mathbb R$. If $\omega$ is independent of $p$, then it essentially maps $\mathbb R^{2n}\times\mathbb R^{2n}$ to $\mathbb R$. That's why I can understand

$$\omega_0(\zeta,\zeta')=\sum_{j=1}^n (\xi_j \eta_j' - \eta_j\xi_j')$$

But what is

$$\omega_0 =\sum_{j=1}^n dx_j\land dy_j$$
and why is it equivalent to the above? What exactly do they mean with $dx_j$ and $dx_j\land dy_j$ here?

I should note that the first $n$ components of $\zeta$ are denoted $\xi$ or $x$ and the last $n$ components of $\zeta$ are denoted $\eta$ or $y$.

The symplectic form

Consider the $2$-form
$$\omega_0:=\sum_{j=1}^n dx_j\wedge dy_j\tag{1.1.20}$$
on $\mathbb{R}^{2n}$. We can think of this either as a differential form on $\mathbb{R}^{2n}$ with constant coefficients, or as a nondegenerate skew-symmetric bilinear form
$$\omega_0:\mathbb{R}^{2n}\times\mathbb{R}^{2n}\longrightarrow\mathbb{R}$$
on the vector space $\mathbb{R}^{2n}$. (The two notions coincide when the vector space $\mathbb{R}^{2n}$ is viewed as the tangent space of the manifold $\mathbb{R}^{2n}$ at a point $z$.) The value of $\omega_0$ on a pair of vectors $\zeta=(\xi,\eta)$ and $\zeta'=(\xi',\eta')$ with $\xi,\eta,\xi',\eta'\in\mathbb{R}$ is given by
$$\omega_0(\zeta,\zeta')=\sum_{j=1}^n(\xi_j\eta_j'-\eta_j\xi_j')=\langle J_0\zeta,\zeta'\rangle=-\zeta^TJ_0\zeta'.\tag{1.1.21}$$

答:
If we define

$$ dx_j(\zeta) = \xi_j $$
$$ dy_j(\zeta) = \eta_j $$

and we define the exterior product like wikipedia:

$$ (f\wedge g)(v_1,\ldots, v_{k+\ell})=\frac{1}{k!\ell!}\sum_{\sigma\in S_{k+\ell}} (sgn(\sigma)) f(v_{\sigma(1)}, \ldots, v_{\sigma(k)})g(v_{\sigma(k+1)}
,\ldots,v_{\sigma(k+\ell)}) $$

then

$$\omega_0(\zeta, \zeta') =\sum_{j=1}^n (dx_j\land dy_j)(\zeta, \zeta')
= \sum_{j=1}^n (dx_j(\zeta) dy_j(\zeta') - dx_j(\zeta') dy_j(\zeta)) = \sum_{j=1}^n (\xi_j \eta_j' - \eta_j\xi_j') $$


问:
If $w: [0, \infty) \to [0, \infty)$ is upper semicontinuous, increasing and $w(r)/r^2 = o(1)$, then $w$ is $C^2$ near $0$

If $w: [0, \infty) \to [0, \infty)$ is upper semicontinuous, increasing, $w(0) = 0$ and $w(r)/r^2 = o(1)$, then $w$ is $C^2$ near $0$.

Why does the above hold?

I understand that, by the definition of derivative, there exists $w'(0) = 0$. However, I cannot proceed further.

Context: Theory of viscosity solutions. More precisely, this is a claim of my teacher when proving that if $(p, X)$ belongs to the superjet of $u$ at $x_0$, then there is a $C^2$ supertangent function $\psi$ such that $D\psi(x_0) = p$ and $D^2\psi(x_0) = X$.

答:
That is not correct. The given conditions do not even imply that $w$ is differentiable near $0$.

As an example, define $f: [0, \infty) \to [0, \infty)$ by $f(0) = 0$,
$$
f(x) = \frac{1}{2^n} \quad\text{for }\frac{1}{2^n} \le x < \frac{1}{2^{n-1}}\, , n = 1, 2, 3, \ldots
$$
and $f(x) = 1$ for $x \ge 1$. $f$ is increasing and upper semicontinuous, but not differentiable in any interval $[0, \epsilon]$.

Then $w(x) = x^2 f(x)$ is increasing and upper semicontinuous with $w(x)/x^2 = f(x) = o(1)$ for $x \to 0$, but $w$ is not differentiable in a neighbourhood of $0$, and $w''(0)$ does not exist.


问:
Convergence of $ \sum_{i =0}^\infty i! \left( \frac{x}{i+1} \right)^i $
One of my assignment questions was to find if the series $$ \sum_{i=0}^\infty S_i = \sum_{i=0}^\infty i!\left( \frac{x}{i+1} \right)^i $$ converges or diverges, using d'Alembert's ratio test and the logarithmic test (if $ l =\lim_{n\to 0} n \ln \frac{S_{n}}{S_{n+1}}$, the series converges if $ l > 1$ and diverges if $l < 1$).

I've determined that it converges if $ x < \frac{1}{\sqrt{e}} $  and diverges if $ x > \frac{1}{\sqrt e} $, but when $ x = \frac{1}{\sqrt e} $, I get $ l = \lim_{n\to\infty} -(n + \frac{1}{2})$, which is always less than one, meaning the series is divergent. But this seems wrong to me (the value of $l$ that I got), and WolframAlpha fails to even tell if the series is convergent or divergent. Is this correct?

I also have a doubt that I got the actual summation wrong. The series was: $$ 1 + \frac{x}{2} + 2!\frac{x^2}{3^2} + 3!\frac{x^3}{4^3} ...$$ WolframAlpha also fails to find the general term of this series. Is my interpretation correct?

答:
$\sum_{i=0}^\infty S_i = \sum_{i=0}^\infty i!( \frac{x}{i+1} )^i
$

Applying the n-th root test,
the term is
$(i!)^{1/i}( \frac{x}{i+1} )
\approx \frac{i}{e} \frac{x}{i+1}
\approx \frac{x}{e}
$

so it converges if
$|x| < e$
and diverges if
$|x| > e$.

I don't know where your
$\sqrt{e}$
came from.


问:
Why is 0 factorial equal to 1? Is there any pure basic mathematical proof?
I just got a question while reading permutation. Why 0 is factorial equal to 1?</p>

答:
One definition of the factorial that is more general than the usual

$$
N! = N\cdot(N-1) \dots 1
$$

is via the gamma function, where

$$
\Gamma(N) = (N-1)! = \int_0^{\infty} x^{N-1}e^{-x} dx
$$

This definition is not limited to positive integers, and in fact can be taken as the definition of the factorial for non-integers. With this definition, you can quite clearly see that

$$
0! = \Gamma(1) = \int_0^{\infty} e^{-x} dx = 1
$$

If you are starting from the "usual" definition of the factorial, in my opinion it is best to take the statement $0! = 1$ as a part of the definition of the factorial function, as anything else would require proofs using the factorial to include special cases for $0!$ and $1!$. It's a definition that is consistent and makes our lives easier.

OwnerDisplayName: Elack

OwnerDisplayName: gabe


答: There is exactly one permutation of a set of zero elements.</p>

Title: Why is 0 factorial equal to 1? Is there any pure basic mathematical proof?

Body: I just got a question while reading permutation. Why 0 is factorial equal to 1?</p>

OwnerDisplayName: WillO


问:
Property of Markov chains
Let $(X_k)_{k\in \mathbb{Z}_{+}}$ be a Markov chain with state space $I$ and let $J_0,J_1, \ldots,J_{n+1}$ be subsets of $I$. Prove that:

$$\mathbb{P}(X_{n+1} \in J_{n+1} | X_n \in J_n,\cdots,X_0 \in J_0) = \mathbb{P}(X_{n+1} \in J_{n+1} | X_n \in J_n) $$

I was thinking of using the Markov property to be able to take it to each element of $J_{n+1}$ but I have not been able to prove anything.

答:
For a simple counterexample just take almost any situation where there are multiple states, $J_n$ is the entire space, and $J_{n+1}$ is not the entire space.

For example $I=\{ 0,1 \},J_0=\{ 0 \},J_1=\{ 0,1 \},J_2 = \{ 0 \}$, the transition matrix is just the identity, and the initial distribution in the right hand side is uniformly distributed on both states. Then on the right side, assuming $X_1 \in J_1$ adds nothing, so $P(X_2 \in J_2 \mid X_1 \in J_1)=P(X_2=0)=1/2$. Meanwhile on the left side, assuming $X_0 \in J_0$ does tell you something: $P(X_2 \in J_2 \mid X_1 \in J_1,X_0 \in J_0)=P(X_2=0 \mid X_0=0)=1$.

The point is that the Markov property says that if you know exactly where the particle is at time $n$, then it doesn't matter how it got there. But if you only have an idea of where it is at time $n$, then knowing something about how it got there may give you a better idea of where it actually is at time $n$, which is what determines its distribution in the future.

Title: Property of Markov chains


问:
Combinatorics Identity about Catalan numbers: $\sum_{k=0}^n \frac{1}{k+1}\binom{2k}k \binom{2n-2k}{n-k}=\binom{2n+1}n$
I need to prove this identity:

$\sum_{k=0}^n \frac{1}{k+1}{2k \choose k}{2n-2k \choose n-k}={2n+1 \choose n}$

without using the identity:

$C_{n+1}=\sum_{k=0}^n C_kC_{n-k}$.

Can't figure out how to.

答:
Rewriting the LHS as:
$$
\sum_{k=0}^n \frac{1}{1+2k}{1+2k \choose k}{2n-2k \choose n-k}
$$
one observes that this is a particular case ($a=1,b=2,c=2n$) of the Rothe–Hagen identity:
$$
\sum_{k=0}^n\frac a{a+bk}\binom{a+bk}k\binom{c-bk}{n-k}=\binom{a+c}n,
$$
valid for any complex numbers $a,b,c$.


问:
How to solve $\int_0^\infty f(x) \sin(wx) \, dx=\frac a {a^2+w^2}$ with using Fourier transform?
$\int_0^\infty f(x) \sin(wx) \, dx=\frac{a}{a^2+w^2}$
I thought it is Fourier sine transform of $f(x)$

since  $\ F_s(f(x))=F_s(k)=\sqrt{\frac{2}{\pi}} \int_0^\infty f(x)\sin(wx) \, dx$

so $\sqrt{\frac{\pi}{2}} F_s(w)=\frac{a}{a^2+w^2}$. by taking invers sin transform of this

$$\sqrt{\frac{\pi}{2}}f(x)=\int_0^\infty \frac{a}{a^2+w^2} \sin(wx)\,dx$$

答:
Upon second thought, there seems to be nothing wrong with this question. In fact, if one restricts the domain of the FST to $\omega>0$, the transform has a unique inverse, which should exist as long as $f$ is absolutely integrable. Indeed it can be shown that the inverse defined by

$$F^{-1}_s\left[\int_{0}^{\infty}dxf(x)\sin{\omega x}\right](\omega)=\frac{2a}{\pi}\int_{0}^{\infty}\frac{\sin\omega t}{t^2+a^2}dt$$

converges absolutely and therefore it is equal to $f(\omega)$. This integral can only be done in terms of special functions. One can compute that

$$\int_{0}^{\infty}\frac{e^{-zx}}{x^2+a^2}dx=\int_{\infty}^{z}\frac{\sin a (z-z')}{z'}dz'=\frac{\sin az}{a}\text{Ci}(az)+\frac{\cos az }{a}\left(\frac{\pi}{2}-\text{Si}(az)\right), ~~\text{Re}~z\geq 0$$

where the last equality is given in terms of the cosine-sine integrals. We can set therefore $z=\pm i\omega $ in the expression above to get the final result

$$\frac{\pi}{2a}f(\omega)=\frac{\cosh a\omega}{a}\frac{\text{Si}(ia\omega)-\text{Si}(-ia\omega)}{2i}-\frac{\sinh a\omega}{a}\frac{\text{Ci}(ia\omega)+\text{Ci}(-ia\omega)}{2}$$

which converges for all $\omega>0$.


答:
$$a_i=\frac{i!}{(i+1)^i}$$
radius of convergence is
$$r=\lim_{i\to\infty}\left|\frac{a_i}{a_{i+1}}\right|$$
that is
$$r=\lim_{i\to\infty}\frac{\frac{i!}{(i+1)^i}}{\frac{(i+1)!}{(i+2)^{i+1}}}=\lim_{i\to\infty}\left(\frac{i+1}{i+2}\right)^{-i-1}=$$
substitute $i+1=n$
$$=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{-n}==\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n}=e$$

Title: Convergence of $ \sum_{i =0}^\infty i! \left( \frac{x}{i+1} \right)^i $

Body:
One of my assignment questions was to find if the series $$ \sum_{i=0}^\infty S_i = \sum_{i=0}^\infty i!\left( \frac{x}{i+1} \right)^i $$ converges or diverges, using d'Alembert's ratio test and the logarithmic test (if $ l =\lim_{n\to 0} n \ln \frac{S_{n}}{S_{n+1}}$, the series converges if $ l > 1$ and diverges if $l < 1$).

I've determined that it converges if $ x < \frac{1}{\sqrt{e}} $  and diverges if $ x > \frac{1}{\sqrt e} $, but when $ x = \frac{1}{\sqrt e} $, I get $ l = \lim_{n\to\infty} -(n + \frac{1}{2})$, which is always less than one, meaning the series is divergent. But this seems wrong to me (the value of $l$ that I got), and WolframAlpha fails to even tell if the series is convergent or divergent. Is this correct?

I also have a doubt that I got the actual summation wrong. The series was: $$ 1 + \frac{x}{2} + 2!\frac{x^2}{3^2} + 3!\frac{x^3}{4^3} ...$$ WolframAlpha also fails to find the general term of this series. Is my interpretation correct?


问:
If $A^{n-1}$ is not diagonalizable but $A^n$ is, show $A^n$ is zero
Let $n$ be an integer greater than or equal to 2 and let $A$ be a $n^{th}$ order complex square matrix. Show that if $A^{n-1}$ is not diagonalizable and  $A^n$ is diagonalizable, then $A^n=0$.

I have tried to think of a way to prove that but couldn't. I know that, If a matrix is diagonalizable over a field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$. Can we use this to prove it or what is the proper way to think about it?

答:
a nice way to do this is to exploit the inequality

(i) $\text{rank}\big(A B\big) \leq \text{rank}\big( A\big)$

this immediately implies

$\text{rank}\big(A^n\big) \leq \text{rank}\big( A^{n-1}\big)\leq ... \leq \text{rank}\big(A^2\big)\leq \text{rank}\big(A\big)$

checking the equality conditions of (i), we see that
$\text{rank}\big(A^k\big) = \text{rank}\big( A^{k-1}\big)$ implies $A$ is injective on the vector space $A^{k-1}\cdot V$, which implies  $\text{rank}\big(A^{k+1}\big)= \text{rank}\big(A^k\big) = \text{rank}\big( A^{k-1}\big)$ and so on.

Thus we have 2 cases to consider

1.)  $\text{rank}\big(A^n\big) \lt \text{rank}\big( A^{n-1}\big)$

$\implies \text{rank}\big(A^n\big) \lt \text{rank}\big( A^{n-1}\big)\lt ... \lt \text{rank}\big(A^2\big)\lt \text{rank}\big(A\big)\leq n-1$

$\implies \text{rank}\big(A^n\big) = 0 \implies A^n =\mathbf 0$

2.)  $\text{rank}\big(A^n\big) = \text{rank}\big( A^{n-1}\big)$

this implies that the eigenvalue 0 is semi-simple for $A^{n-1}$ and $A^n$

$A^{n-1} \text{ defective } \implies A \text{ has some non-zero eigenvalue that is not semi-simple }$

call this eigenvalue $\lambda_1$ and via re-scaling $A$ we can assume WLOG that $\lambda_1=1$. This has algebraic multiplicity $r$.

via any triangularization scheme (Jordan, Schur, or others) we have

$S^{-1} A S = \displaystyle \left[\begin{matrix}I_r + N & * & \\ \mathbf 0 & *\end{matrix}\right]$

and
$S^{-1} A^n S = \displaystyle \left[\begin{matrix}(I_r + N)^n & * & \\ \mathbf 0 & *\end{matrix}\right]=\displaystyle \left[\begin{matrix} I_r & * & \\ \mathbf 0 & *\end{matrix}\right]$

where $N$ is some non-zero nilpotent matrix

(book-keeping note: $N$ is $r\times r$ so argument 1 tells us $N^r=\mathbf 0$)

Thus we have

$I_r = (I_r + N)^n = I_r + \binom{n}{1}N^1+ \binom{n}{2}N^2+....+ \binom{n}{r-1}N^{r-1}$

$\implies \mathbf 0 = \binom{n}{1}N^1+ \binom{n}{2}N^2+....+ \binom{n}{r-1}N^{r-1}$

which shows that these $(r-1)$ powers of $N$ are linearly dependent, a contradiction.

note 1: $N^r=\mathbf 0$ but $N\neq \mathbf 0$ implies there is a min positive integer $k\geq 2$ such that $N^k=\mathbf 0$, multiply the above linear relation by $N^{k-2}$ to verify the contradiction

note 2: This highlights the role of our field being of characteristic zero (i.e. the scalars given by the binomial coefficients are necessarily non-zero).  The above argument fails for fields of positive (prime) characteristic -- repeatedly applying the Frobenius Homomorphism reveals the existence of unipotent elements in such a case.


答:
Suppose $R_1$ and $R_2$ on a set S induce the same partition of S but are not identical.

Since they are not identical there is ( at lest) one  couple $(a,b)\in S\times S$ that belongs to one of them but does not belong to the other. [ Relations are sets - of couples - and two different sets must differ by at least one element, by the extensonnality principle]

So, one relation will produce a partition having as "cell"  an equivalence class with $a$ and $b$ as elements; while the other relation will not produce such a partition.

The reason is that in the other relation there is no couple $( a,b)$ , meaning that there is no equivalence class having  $a$ and $b$ as members. Since the partition is a set of equivalence classes, and since the two partitions differ by ( at least) one equivalence class, they are not identical.

The fact that the partitions induced are not identical contradicts our assumption.

So $R_1 = R_2$.

Title: Proof that there is exactly one equivalence relation that forms a partition

Body:
So I've just started trying to teach myself some topology and i cant quite grasp how an equivalence class forms a partition more specifically i don't understand the proof  that there is exactly one equivalence relation that forms the partition.

For example in my book it says that a partition $D$ is formed from an equivalence relation $R$ and that equivalence relation  is unique that is there is only one equivalence relation that forms the partition however they do not give a proof of this  i myself cant come up with one so is there a proof of this statement?
Thanks in advance.


问:
Help me solve this recurrence relation
I'm trying to solve the recurrence relation

$$a_n = (\lambda +\mu)a_{n+2}+\mu a_{n+3}.$$

with initial relationships of:

$\lambda a_1 = \mu a_2$

$(\lambda +\mu)a_2 = \mu a_3.$

I found a site online which suggests using the roots of a, in this case, cubic polynomial to solve it, but the roots are of course really complex and since I don't know what $\lambda$ and $\mu$ are they don't simplify very much.  Is there another way to go about this?  Thanks.

答:
The site online probably wanted you to write
$$
\mu \left[
\begin{array}{c}
a_{n+3} \\
a_{n+2} \\
a_{n+1}
\end{array}
\right]
=
\left[
\begin{array}{ccc}
-\lambda - \mu & 0 & 1 \\
\mu & 0 & 0 \\
0 & \mu & 0
\end{array}
\right]
\left[
\begin{array}{c}
a_{n+2} \\
a_{n+1} \\
a_{n}
\end{array}
\right]
$$
and then diagonalize the matrix using eigenvalues to compute a general formula.  This is extremely gross, but I don't see any better way.</p>

Title: Help me solve this recurrence relation


答:
Let $\sim$ and $\sim'$ be equivalence relations on $S\ne\emptyset.$ For $x\in S$ let $[x]_{\sim}=\{y\in S:y\sim x\}$ and let $[x]_{\sim'}=\{y\in S:y\sim' x\}.$ Let $P=\{[x]_{\sim}:x\in S\}$ and let $P'=\{[x]_{\sim'}:x\in S\}.$

Now for any $x\in S,$ the only member of $P$ that contains $x$ (as a member) is $[x]_{\sim}$ and the only member of $P'$ that contains $x$ is $[x]_{\sim'}\,.$

So if $P=P'$ then $\forall x\in S\,([x]_{\sim}=[x]_{\sim'}).$ That is, if $P=P'$ then $\forall x\in S\,(\{y\in S:y\sim x\}=\{y\in S:y\sim' x\}).$

So if $P=P'$ then $\sim$ and $\sim'$ are the same thing.

Equivalently, if $\sim$ and $\sim'$ are not equal then $P\ne P'.$


问:
Quotients of finite C-star algebras: are they finite?
Edit: to whomever downvoted this post: please explain why. Seriously, I'd like to know why because I cannot think of any reason. This is a fine question, it got 3 upvotes, in my humble opinion it is interesting and it deserves to be here. There are also two beautiful answers by Ruy and PStheman presented here to address this question. So why the downvote?

Edit 2: Now I see someone retracted their upvote. Please explain why you changed your mind. What is it that you saw and decided that, on second thought, this post does not deserve your encouragement and support? I would really like to see some explanations for all this drama.

Edit 3: Another downvote. Makes one wonder, why is this happening?

Original post

As I was trying to answer this question, I thought of the following: If $A$ is a unital, finite $C^*$-algebra and $I$ is an ideal in $A$, then is $A/I$ a finite $C^*$-algebra?

Recall that a unital $C^*$-algebra is called finite when the implication
$$p\sim_{M-vN}1_A\implies p=1_A$$
holds, i.e. every isometry in $A$ is a unitary. Besides the main question, I realized something: I do not know what $C^*$-algebras are finite or not. For example, I know that abelian $C^*$-algebras are stably finite, as it is (briefly) discussed in this post, and finite dimensional $C^*$-algebras are (stably) finite too, but that's all I know. So a bonus question is: what are some interesting examples of finite $C^*$-algebras besides abelian and finite dimensional ones?

Note that in the answer of the linked question I am using a particular feature of the ideal $\sum_nA_n$ in $\prod_nA_n$, something that we cannot do in the general case.

答:
There is probably an obvious example laying around, but I can't think of one! Here is an overkill answer showing that quotients of finite C*-algebras need not be finite.

There is a characterization of separable exact C*-algebras as subquotients (quotient of a subalgebra) of the CAR algebra, i.e., the UHF algebra $M_{2^{\infty}}$. This algebra is finite and so is any subalgebra since the faithful trace on the CAR algebra restricts to a faithful trace on any smaller algebra. So take $B \subseteq M_{2^{\infty}}$ ($B$ unital) and $I \lhd B$ such that $B/I \simeq \mathcal{O}_2$, the Cuntz algebra generated by 2 isometries. The Cuntz algebra is (purely) infinite.

There is a nice description of finiteness in unital C*-algebras (see Left Invertibility Implies Right Invertibility in Certain $C^*$-algebras). Some examples are UHF algebras (like the CAR algebra), the irrational rotation algebra, the Jiang-Su algebra, etc.

References:
This was originally proved in "On subalgebras of the CAR-algebra" by Kirchberg, and later Wassermann gave a "simpler" (take with a grain of salt) proof in "Subquotients of UHF C*-algebras".


问:
How does one show that two functors are *not* isomorphic?
Let $C$ be the category of finite-dimensional vector spaces over some field. It is easy to construct pairs of endofunctors $F, G$ of $C$, of the same variance, such that $F(V)$ and $G(V)$ have the same dimension for every $V \in C$, yet are not naturally isomorphic.

For example, we could take $G(V) := (V^*)^{\otimes 2}$, and $F(V) := (V^{\otimes 2})^*$, where $*$ denotes the dual. Or, I'm sure many of us have, at some time, tried showing that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, before realizing that it is false.  A convincing reason why they are not canonically isomorphic is that there is no obvious reason, in terms of universal properties, why they should be. In other words: if you try proving that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, you're going to have a bad time.

However, how does one explicitly show that $F$ and $G$ are not isomorphic? Is there a slick, systematic way of doing this?

Edit: Well, my examples couldn't be worse, as Mariano points out. But the gist of my question still stands (perhaps just get rid of the finite-dimensionality assumption ;))

答:
The functors $FV=(V^*)^{\otimes 2}$ and $G(V)=(V^{\otimes2})^*$ on the category of finite fimensional vector spaces are isomorphic. The map $\phi:(V^*)^{\otimes2}\to(V^{\otimes2})^*$ such that $\phi(f\otimes g)(v\otimes w)=f(v)g(w)$ is a natural transformation (defined on the whole category of vector spaces) which in the category of f.d. vector spaces is an isomorphism of functors.

Similarly, $\Lambda(V^*)$ and $(\Lambda V)^*$ are canonically isomorphic on the category of f.d. vector spaces.


问:
Integrating $\int_{0}^{\infty} x^{-k} e^{-a(x+d)^2} dx$ where $k \in \mathbb{Z}_{+}$
I want to solve the following integral:

$$I = \int_{0}^{\infty} x^{-k}  e^{-a(x+d)^2} dx$$

where $k \in \mathbb{Z}_{+}$, and $a, d \in \mathbb{R}_{++}$.

When $k =0$, the integral is a standard Gaussian integral. However, when $k > 0$, the problem is not so easy. We see that when $x \rightarrow 0$, the integrand blows up.

We cannot set $I = \underset{\epsilon \rightarrow 0^{+}}{\text{lim}} \int_{0}^{\infty} x^{-k}  e^{-a(x+d)^2} dx $ and then expand $x^{-k}$ into its Maclaurian series because we are integrating over all $x > 0$ and the series converges only within a finite region. Similarly, we cannot put a contour that goes around the singularity because the pole is of higher order. Also, solving as if $k < 0$ and then using analytic continuation of the Gamma function won't work, because $k$ is an integer.

Any ideas? Or does this integral simply not exist?

Thanks

答:
$$I=\int_0^\infty x^{-k}e^{-[\sqrt{a}(x+d)]^2}dx$$
$u=\sqrt{a}(x+d)\Rightarrow dx=du/\sqrt{a}$ so we have:
$$I=\frac{1}{\sqrt{a}}\int_{\sqrt{a}d}^\infty\left(\frac{u}{\sqrt{a}}-d\right)^{-k}e^{-u^2}du$$
$$=a^{(k-1)/2}\int_{\sqrt{a}d}^\infty(u-\sqrt{a}d)^{-k}e^{-u^2}du$$
now try and expand $(u-\sqrt{a}d)^{-k}$ to get this as a series of integrals and then use the incomplete gamma function


答: This integral does not exist.  Consider the case when $k=1$.  (I think other cases reduce to this via integration by parts.)  We know $\int_0^n\frac{dx}{x}$ diverges.  But our integral $\int_0^n\frac{e^{-a(x+d)^2}dx}{x}\ge\int_0^n\frac{mdx}{x}$ where $m$ is the minimum of $e^{-a(x-d)^2}$ on $[0,n]$.

Title: Integrating $\int_{0}^{\infty} x^{-k} e^{-a(x+d)^2} dx$ where $k \in \mathbb{Z}_{+}$

Body:
I want to solve the following integral:

$$I = \int_{0}^{\infty} x^{-k}  e^{-a(x+d)^2} dx$$

where $k \in \mathbb{Z}_{+}$, and $a, d \in \mathbb{R}_{++}$.

When $k =0$, the integral is a standard Gaussian integral. However, when $k > 0$, the problem is not so easy. We see that when $x \rightarrow 0$, the integrand blows up.

We cannot set $I = \underset{\epsilon \rightarrow 0^{+}}{\text{lim}} \int_{0}^{\infty} x^{-k}  e^{-a(x+d)^2} dx $ and then expand $x^{-k}$ into its Maclaurian series because we are integrating over all $x > 0$ and the series converges only within a finite region. Similarly, we cannot put a contour that goes around the singularity because the pole is of higher order. Also, solving as if $k < 0$ and then using analytic continuation of the Gamma function won't work, because $k$ is an integer.

Any ideas? Or does this integral simply not exist?

Thanks


问:
Struggling with proof of obvious statement regarding an inequality
Given that $n_1(n_1-1)\leq n_2(n_2-1)$ for $n_1, n_2 \in \{1, ..., n\}$ is true. How can I prove that $n_1 \leq n_2$ must also be true?

I thought about saying if $n_1 \leq n_2$ than $n_1^2 \leq n_2^2$ and therefore $n_1^2-n_1 \leq n_2^2 - n_2$, but here I'm already using the solution.

Body:
hint

$$n_1^2-n_1=(n_1-\frac 12)^2-\frac 14$$


问:
Tangent vector of geodesic on sphere trough points $(a,b,c)$ and $(d,e,f)$
I am seeking the unit vector $v=(x,y,z)$ tangent at $p$ to the geodesic on sphere connecting $p=(a,b,c)$ and $q=(d,e,f)$. The curve is orientated from $p$ to $q$.

Once the tangent vector is ortogonal to the point vectors, so I must have

$$\begin{cases}ax+by+cz=0\\
x^2+y^2+z^2=1\end{cases}$$

I need more two equations to get the vector, could you help me?

The aim is to prove that the application $(p,q)\mapsto v$ is continuous.

Thank you in advance.

答: HINT: There is no calculus or differential geometry here. If you want the great circle through $p$ passing through $q$ ($q\ne -p$), it lies in the plane with normal vector $p\times q$. The tangent vector to a great circle will lie in that plane and be orthogonal to the position vector. How do you find a vector in that plane orthogonal to $p$?


答:
Expanding, we get $n_1^2-n_1\leq n_2^2-n_2$.

Now we complete the square: $(n_1-\frac{1}{2})^2-\frac{1}{4}\leq (n_2-\frac{1}{2})^2-\frac{1}{4}$. Rearrange to obtain $(n_1-\frac{1}{2})^2-(n_2-\frac{1}{2})^2\leq 0$.

By the difference of two squares factorisation, $(n_1-\frac{1}{2}+n_2-\frac{1}{2})(n_1-\frac{1}{2}-n_2+\frac{1}{2})\leq 0$, and so $(n_1+n_2-1)(n_1-n_2)\leq 0$.

Now, using the fact that $n_i\geq 1(\forall i\in\mathbb{N})$ we gave that $n_1\leq n_2$ from the second bracket, or $n_1+n_2\leq 1$ from the first, which is impossible.

Hence $n_1\leq n_2$ as we wanted to prove.

Title: Struggling with proof of obvious statement regarding an inequality

Body:
Given that $n_1(n_1-1)\leq n_2(n_2-1)$ for $n_1, n_2 \in \{1, ..., n\}$ is true. How can I prove that $n_1 \leq n_2$ must also be true?

I thought about saying if $n_1 \leq n_2$ than $n_1^2 \leq n_2^2$ and therefore $n_1^2-n_1 \leq n_2^2 - n_2$, but here I'm already using the solution.


问:
Vector Spaces - Understanding counter exmaple
I am trying to understand a counter-example to say that this statement is false:

Let $U,W$ be subspaces of a vector space $V$, than  $U ∪W$ is a subspace of $V$.

Counter Example:

Let $V= \mathbb{R}^2$ and let $U=span \left \{ \begin{bmatrix} 0\\ 1 \end{bmatrix} \right \}$ and let $W=span \left \{ \begin{bmatrix} 1\\ 0 \end{bmatrix} \right \}$

So $\begin{bmatrix} 1\\ 1 \end{bmatrix} \notin U ∪ W$ .

I do not understan why is  $\begin{bmatrix} 1\\ 1 \end{bmatrix} \notin U ∪ W$?

Or first should I ask, what is $U ∪ W$ equal in this case?

Thanks a lot!

答:
$$ U\cup V = \{a\cdot (0\;1), a\in\mathbb{R}\} \cup\{a\cdot (1\;0), a\in\mathbb{R}\}$$
So there are no vectors with both coordinates different than zero


问:
If exist a extension of $f$ to a map $D^2\to X$, then $f: S^1\to X$ is homotopic to a constant map

Every map $f: S^1\to X$ is homotopic to a constant map, if and only if, exist a extension of $f$ to a map $D^2\to X$.

Here, $X$ is an arbitrary topological space. I managed to prove the direction $\Rightarrow$, for reference I will also leave some links for similar answers. (1), (2), (3), (4), (5).

I know that there is a list of equivalences, in this case:

The following three conditions are equivalent:

Every map $S^1\to X$ is homotopic to a constant map, with image a point.

Every map $S^1\to X$ extends to a map $D^2\to X$.

$\pi_1(X, x_0) = 0$ for all $x_0\in X$.

As I said, the direction I prove is $1\Rightarrow 2$, and the direction I'm having trouble proving is $2\Rightarrow 1$. From the list of equivalences, it would be enough to prove $2\Rightarrow 3$ and then $3\Rightarrow 1$, but I would like to know the direct demonstration of $2\Rightarrow 1$.

答: Consider a continuous function $f : S^1 \to X$, and let $\widetilde{f\,} : D^2 \to X$ one of its extensions. Now note that if $z \in S^1$ and $t \in [0,1]$, $$|(1-t)z+t| \leq (1-t)|z|+t = 1$$ and so we can define $H : S^1 \times [0,1] \to X$ by $H(z,t) := \widetilde{f\,}((1-t)z+t)$. Finally, observe that $H(-,0) = f$ and that $H(-,1)$ is the constant map $z \mapsto f(1)$.


问:
Convert with variable change the form Sturm- Liouville to $y^{\prime\prime}+(\lambda \rho_{1}(x)+q_{1}(x))y=0 $
I have the problem of Sturm Liouville in $[a,b]:$
$$
\left(p(x) y^{\prime}\right)^{\prime}+(\lambda \rho(x)-q(x))y=0,
$$
developing the expression I got:
$$
p(x)y^{\prime\prime}+p(x)^{\prime}.y^{\prime}+(\lambda \rho(x)-q(x))y=0
$$
by changing the variable I must arrive at the expression:
$$
y^{\prime\prime}+(\lambda \rho_{1}(x)+q_{1}(x))y=0
$$

the problem is not about reducing but making a change of variable,any help will be appreciated.

答:
Let $y=h(x)u$, do your product rules and collect all your terms to get an ODE of the form
\begin{align}
f_1(x)u''+f_2(x)u'+f_3(x)u=0.
\end{align}
Set $f_2(x)$ to zero. This will give you an easy ODE to solve for $h(x)$ in terms of $p(x)$ which will simplify it to the form you desire.


问:
Results that came out of nowhere.
Most big results in mathematics are built on years and years of groundwork by the author and other mathematicians, such as Wiles' proof of FLT or the classification of finite simple groups.  Nevertheless, there is this big romantic idea in pop culture where a young genius comes up with a brilliant proof of a major conjecture, based on some creative new idea which flies in the face of the mathematical community.

What are some examples where this has actually happened?  That is, which results stem from independent work by a mathematician who came along "out of nowhere" and solved a huge problem by surprise through nonstandard techniques?

I read that the recent proposed proof of the ABC conjecture comes from years of autonomous theory by Prof Mochizuki, most of which ventures far outside of the current literature; this would seem to qualify as one example provided the proof turns out to be true.

答:
Perhaps the recent proof that there are infinitely many primes with a (fixed) bounded gap qualifies? (Although this story is not necessarily a happy ending for all, it has still caused quite the buzz!)

For those interested, I may as well add the actual work (subscription/campus IP/proxy required)

Title: Results that came out of nowhere.


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

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

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

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

答:
The binomial identity
\begin{align*}
\color{blue}{\binom{n+1}{k+1}=\binom{n}{k+1}+\binom{n}{k}}\tag{1}
\end{align*}
is not only valid for natural numbers $n$ but also for $n\in\mathbb{Z}$ and $k\geq 0$ a natural number. We can use this relationship to extend the Pascal triangle to negative numbers $n$ shown in the table below. The numbers of the Pascal triangle for $n,k\geq 0$ follow by setting $\binom{n}{0}=1, \binom{0}{k}=\delta_{k,0}$. The entries with negative $n$ follow by extending the rule $\binom{n}{0}=1$ for $n\in\mathbb{Z}$.
\begin{align*}
\begin{array}{r|rrrrrrrrl}
k&0&1&2&3&4&5&6&&\\
\hline
n&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&&\\
-4&1&\color{blue}{-4}&\color{blue}{10}&-20&35&-56&84&&(1+x)^{-4}\\
-3&1&-3&\color{blue}{6}&-10&15&-21&28&&(1+x)^{-3}\\
-2&1&-2&3&-4&5&-6&7&&(1+x)^{-2}\\
-1&1&-1&1&-1&1&-1&1&&(1+x)^{-1}\\
0&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^0\\
1&1&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^1\\
2&1&2&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^2\\
3&1&\color{blue}{3}&\color{blue}{3}&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^3\\
4&1&4&\color{blue}{6}&4&1&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^4\\
\end{array}
\end{align*}

When looking for instance at the last line of the table above we see the number $6$ is according to (1) the sum of the entry $3$ above $6$ with its left neighbor $3$.
\begin{align*}
6=3+3\qquad\qquad \binom{4}{2}=\binom{3}{1}+\binom{3}{2}
\end{align*}
The algebraic connection is given by
\begin{align*}
\color{blue}{(1+x)^4=(1+x)(1+x)^3}
\end{align*}
or more detailed with $[x^k]$ denoting the coefficient of $x^k$:
\begin{align*}
\binom{4}{2}&=[x^2](1+x)^4\\
&=[x^2](1+x)(1+x)^3\\
&=\left([x^2]+[x^1]\right)(1+x)^3\\
&=\binom{3}{2}+\binom{3}{1}\tag{2}
\end{align*}
The algebraic connection holds also for negative $n$, for instance:
\begin{align*}
\color{blue}{(1+x)^{-3}=(1+x)(1+x)^{-4}}
\end{align*}
and we obtain similarly to (2):
\begin{align*}
6=10+\left(-4\right)\qquad\qquad \binom{-3}{2}=\binom{-4}{2}+\binom{-4}{1}
\end{align*}
or more detailed with $[x^k]$ denoting the coefficient of $x^k$:
\begin{align*}
\binom{-3}{2}&=[x^2](1+x)^{-3}\\
&=[x^2](1+x)(1+x)^{-4}\\
&=\left([x^2]+[x^1]\right)(1+x)^{-4}\\
&=\binom{-4}{2}+\binom{-4}{1}\tag{2}
\end{align*}

We can also see in the table the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k
\end{align*}
is given as rotation of a somewhat modified Pascal triangle.
\begin{align*}
\begin{array}{r|rrrrrrrrl}
k&0&1&2&3&4&5&6&&\\
\hline
n&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&\color{white}{-0}&&\\
-4&\color{blue}{1}&-4&10&-20&35&-56&84&&(1+x)^{-4}\\
-3&\color{blue}{1}&\color{blue}{-3}&6&-10&15&-21&28&&(1+x)^{-3}\\
-2&\color{blue}{1}&\color{blue}{-2}&\color{blue}{3}&-4&5&-6&7&&(1+x)^{-2}\\
-1&\color{blue}{1}&\color{blue}{-1}&\color{blue}{1}&\color{blue}{-1}&1&-1&1&&(1+x)^{-1}\\
0&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^0\\
1&1&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^1\\
2&1&2&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^2\\
3&1&3&3&1&\color{lightgrey}{0}&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^3\\
4&1&4&6&4&1&\color{lightgrey}{0}&\color{lightgrey}{0}&&(1+x)^4\\
\end{array}
\end{align*}


答:
Consider
$$
  A = \{f\in C([0, 1], \mathcal T): f(0)\in  \mathbb C\cdot 1\},
  $$
where  $\mathcal T$ is the Toeplitz algebra.  Every isometry $f$ in $A$ gives rise to a path of isometries in
$\mathcal T$ beginning with a scalar.  Observing that  all isometries in the Toeplitz algebra have finite Fredholm
index, and also that the index is continuous, we can easily show that $A$ is finite.  However
the map
$$
  f\in  A\mapsto  f(1)\in \mathcal T
  $$
is a surjection from $A$ to a non-finite algebra.

EDIT. Here is an argument to show the finiteness of $A$ which does not rely on the theory of  Fredholm operators:

Lemma.  Let $B$ be a unital C*-algebra.  Denote by   $\mathscr I(B)$  the set of all isometries in $B$, and by   $\mathscr U(B)$ the subset
formed by the unitary elements.  Then $\mathscr U(B)$ is clopen in $\mathscr I(B)$.

Proof.  It is well known that the set  $\text{GL}(B)$ formed by all invertible elements is open.  Since
$$
  \mathscr U(B) = \mathscr I(B) \cap \text{GL}(B),
  $$
it follows  that   $\mathscr U(B)$ is open in $\mathscr I(B)$.  To show that $\mathscr U(B)$ is also closed,  suppose by
contradiction that the
sequence $\{u_n\}_n$  of unitaries converges to a proper isometry $s$ (meaning that $ss^*\neq 1$).
The element
$$
  p = 1-ss^*
  $$
is then a non-trivial projection and,  clearly, $ps=0$.  We then deduce that
$$
  0 = ps = \lim_{n\to \infty }pu_n,
  $$
but this is a contradiction since $\|pu_n\|=\|p\| = 1$.  QED

This said,  suppose that $f$ is an isometry in $A$,  so we have that   $f(t)$ lies in $\mathscr I(\mathcal T)$, for all
$t$ in $[0, 1]$.

Since $f(0)$ is necessary a scalar, we have that $f(0)\in \mathscr U(\mathcal T)$, but then
$f(t)\in \mathscr U(\mathcal T)$, for every $t$ because  $[0,1]$ is connected.

This shows that $f$ is unitary and hence that
$A$ is finite.

Title: Quotients of finite C-star algebras: are they finite?

Body:
Edit: to whomever downvoted this post: please explain why. Seriously, I'd like to know why because I cannot think of any reason. This is a fine question, it got 3 upvotes, in my humble opinion it is interesting and it deserves to be here. There are also two beautiful answers by Ruy and PStheman presented here to address this question. So why the downvote?

Edit 2: Now I see someone retracted their upvote. Please explain why you changed your mind. What is it that you saw and decided that, on second thought, this post does not deserve your encouragement and support? I would really like to see some explanations for all this drama.

Edit 3: Another downvote. Makes one wonder, why is this happening?

Original post

As I was trying to answer this question, I thought of the following: If $A$ is a unital, finite $C^*$-algebra and $I$ is an ideal in $A$, then is $A/I$ a finite $C^*$-algebra?

Recall that a unital $C^*$-algebra is called finite when the implication
$$p\sim_{M-vN}1_A\implies p=1_A$$
holds, i.e. every isometry in $A$ is a unitary. Besides the main question, I realized something: I do not know what $C^*$-algebras are finite or not. For example, I know that abelian $C^*$-algebras are stably finite, as it is (briefly) discussed in this post, and finite dimensional $C^*$-algebras are (stably) finite too, but that's all I know. So a bonus question is: what are some interesting examples of finite $C^*$-algebras besides abelian and finite dimensional ones?

Note that in the answer of the linked question I am using a particular feature of the ideal $\sum_nA_n$ in $\prod_nA_n$, something that we cannot do in the general case.


问:
submaximal space

A space X is said to be submaximal if every dense set of X is open in X (i.e., every preopen set of X is open in X. A subset $S$ of $(X, \tau)$ is said to be "preopen" if $S \subset \text{int}(\text{cl}(S)$.

A space X is said to be extremally disconnected (in brief , E.D.,) space if Cl(G) is open set for each open set G of X.

Is every submaximal space, extremally disconnected? If not, can you tell me a counterexample?

答:
Let $X=\{p\}\cup(\Bbb N\times\Bbb N)$, where $p$ is some point not in $\Bbb N\times\Bbb N$. Points of $\Bbb N\times\Bbb N$ are isolated, and the sets

$$B_n=\{p\}\cup\{\langle k,\ell\rangle\in\Bbb N\times\Bbb N:k\ge n\}$$

form a local base at $p$. The only dense sets in $X$ are $X$ and $X\setminus\{p\}$, both of which are open, so $X$ is submaximal.

Let $G=\Bbb N\times\{0\}$. Then $G$ is open, but $\operatorname{cl}G=G\cup\{p\}$ is not open, so $X$ is not extremally disconnected.

Title: submaximal space


问:
How to minimize objective function?
Fit a line with zero $y$-intercept ($\hat{y} = ax)$ on the curve $y=x^2+x$. Instead of minimizing the sum of squares of the errors, minimize the following objective function:

$$\sum_i \left[ \left(\frac{y^i}{\hat{y}^i}\right)^2 + \left(\frac{\hat{y}^i}{y^i}\right)^2 \right]$$

We are allowed to assume that the distribution of $x$ is uniform on $[0,1]$. What is the optimal value for $a$?

I know how to minimize using sum of squares but I do not know how to minimize using the following function. Could someone please show how to solve this problem?

Title: How to minimize objective function?

Body:
$$f(a)=\int_0^1 \left(\left(\frac{x^2+x}{a x}\right)^2+\left(\frac{a x}{x^2+x}\right)^2\right) \, dx=\frac{a^2}{2}+\frac{7}{3 a^2}$$
$$f'(a)=a-\frac{14}{3 a^3}$$
$$f'(a)=0\to 3a^4-14=0\to a=\sqrt[4]{\frac{14}{3}}\approx 1.47$$
as $f''(a)=\frac{14}{a^4}+1$ it is positive at $a=1.47$ which is actually a minimum.


问:
Notation for existential qualifier without repeats
Apologies for any incorrect terminology, my background is in CS, not math. I'm trying to figure out how to notate that in a list of sets (of numbers), the sets can be paired such that in each pair of sets, the sum of the members of each set are equal. What I currently have is:
$$\forall_A \exists_B (\sum_{x \in A} x = \sum_{y \in B} y)$$
where A and B are sets in the list. The problem (as I see it) is that this would be true for a list of three sets which all have equal sums, while what I am trying to write is that once a set is used in one pair, it cannot be in another (the list would have to include an even number of sets), and in the list of three sets, the statement would be false, because one set would have no match. Please let me know if what I'm asking is unclear, and I can clarify.

答:
In mathematics, a list is really a function from some indexing set - e.g. the list $$[\{1,2\},\{2\}, \{3,4\}, \{2\}]$$ (using Python notation) would correspond to the function with domain $\{0,1,2,3\}$ given by $$0\mapsto \{1,2\},\quad 1\mapsto \{2\},\quad 2\mapsto \{3,4\},\quad 3\mapsto\{2\}.$$ Per your comment to William Elliot, you're interested in the following property of a function $F$ whose codomain is the set of finite sets of natural numbers (I'm guessing based on your examples that you're not interested in non-natural numbers or infinite sets here):

There is a partition of $dom(F)$  into pairs such that for each pair $\langle A,B\rangle$ in the partition we have $\sum_{x\in A}x=\sum_{y\in B}y$.

(Note that this makes perfect sense regardless of what $dom(F)$ is; for example, it need not be ordered, or finite, or anything else nice.)

Now this is not entirely symbolic, but - barring unusual context - that's a feature rather than a bug: it's rarely the case in mathematics that a symbols-only expression is the best option for expressing a given notion.


问:
How can I show that the function $ h(a)$ is continuous, if I don’t know it?
Prove the existence and unicity of the equation $e^x + x^3 + x + cos(x) = a$ for all
$a \in R$. If $g: R → R$ is the function that maps a real number a to the unique solution $x(a)$  of
$e^x + x^3 + x + cos(x) = a$, is $g$ continuous?.

Using Bolzano's theorem I have shown that the function has at least one solution. The uniqueness, I proved it by monotony, in fact, the function is strictly increasing, therefore, the function has a unique solution for $ a $ fixed, I also proved this by assuming two solutions and using Rolle's theorem.

But, that unique solution depends on $ a $, let's say $ x = x(a) $, now I need to show that $ x(a) $ is continuous.

Somebody could help me? thanks!

答:
Let $$f(x)=e^x+x^3+x+\cos(x)$$

$ f $ is clearly continuous and differentiable at $ \Bbb R $ and

$$(\forall x\in \Bbb R)$$
$$\;f'(x)=e^x+3x^2+1-\sin(x)>0$$
because
$$e^x+3x^2>0 \text{ and } 1-\sin(x)\ge0$$

thus, $ f $ is a bijection from $ \Bbb R $ to $ (\lim_{-\infty}f,\lim_{+\infty}f)=\Bbb R$.
So
$$(\forall a\in \Bbb R)\;(\exists ! x\in \Bbb R)\;:\; f(x)=a$$
and

$g=f^{-1} :a\mapsto g(a)=x(a) $ is also a continuous bijection from R to R.


问:
Every element of a tensor product space can be expressed as a linear combination of tensor products
Hi. I'm studying the book written by John M. Lee on smooth manifolds, and I got a problem that is bothering me so much. Let me introduce first the definition of a tensor product space in this book:

Now let $V_1,\ldots,V_k$ be real vector spaces. We begin by forming the free vector space $\mathscr{F}(V_1\times\cdots\times V_k)$, which is the set of all finite formal linear combinations of $k$-tuples $(v_1,\ldots,v_k$) with $v_i\in V_i$ for $i=1,\ldots,k$. Let $\mathscr{R}$ be the subspace of $\mathscr{F}(V_1\times\cdots\times V_k)$ spanned by all elements of the following forms:
$$(v_1,\ldots,av_i,\ldots,v_k)-a(v_1,\ldots,v_i,\ldots,v_k),$$
$$(v_1,\ldots,v_i+v_i',\ldots,v_k)-(v_1,\ldots,v_i,\ldots,v_k)-(v_1,\ldots,v_i',\ldots,v_k),$$
with $v_j,v_j'\in V_j$, $i\in\{1,\ldots,k\}$, and $a\in\mathbb{R}$.
Define the tensor product of the spaces $V_1,\ldots,V_k$, denoted by $V_1\otimes\cdots\otimes V_k$, to be the following quotient vector space:
$$V_1\otimes\cdots\otimes V_k=\mathscr{F}(V_1\times\cdots\times V_k)/\mathscr{R},$$
and let $\Pi:\mathscr{F}(V_1\times\cdots\times V_k)\longrightarrow V_1\otimes\cdots\otimes V_k$ be the natural projection. The equivalence class of an element $(v_1,\ldots,v_k)$ in $V_1\otimes\cdots\otimes V_k$ is denoted by
$$v_1\otimes\cdots\otimes v_k=\Pi(v_1,\ldots,v_k),$$
and is called the (abstract) tensor product of $v_1,\ldots,v_k$. It follows from the definition that abstract tensor products satisfy
\begin{align*}
v_1\otimes\cdots\otimes av_i\otimes\cdots\otimes v_k= & a(v_1\otimes\cdots\otimes v_i\otimes\cdots\otimes v_k),\\
v_1\otimes\cdots\otimes(v_i+v_i')\otimes\cdots\otimes v_k= & (v_1\otimes\cdots\otimes v_i\otimes\cdots\otimes v_k)+\\
& +(v_1\otimes\cdots\otimes v_i'\otimes\cdots\otimes v_k).
\end{align*}

After defining the tensor product of vector spaces, Prof. Lee said that every element of a tensor product space can be expressed as a linear combination of tensor products:

Note that the definition implies that every element of $V_1\otimes\cdots\otimes V_k$ can be expressed as a linear combination of elements of the form $v_1\otimes\cdots\otimes v_k$ for $v_i\in V_i$; but it is not true in general that every element of the tensor product space is of the form $v_1\otimes\cdots\otimes v_k$ (see Problem 12-1).

I don't know how to arrive at this conclusion. Could someone please do me a favor? Thank you.

答:
Since $\Pi: \mathscr{F}(V_1\times\cdots\times V_k) \to V_1\otimes\cdots\otimes V_k$ is surjective, for every $z \in V_1\otimes\cdots\otimes V_k$ there is $v \in \mathscr{F}(V_1\times\cdots\times V_k)$ such that $z=\Pi(v)$. Now, as every element of $\mathscr{F}(V_1\times\cdots\times V_k)$ is a finite (formal) linear combination of elements in $V_1\times\cdots\times V_k$, there exists $m \in \mathbb Z^+$, $a_1,\dots,a_m \in \mathbb R$ and $(v_{i1},\dots,v_{ik}) \in V_1 \times \cdots \times V_k$ for every $i \in \{1,\dots,m\}$ such that
$$v = \sum_{i=1}^m a_i(v_{i1},\dots,v_{ik}).$$
Then, using the linearity of $\Pi$ we get that $$z = \Pi(v) = \sum_{i=1}^m a_i \Pi(v_{i1},\dots,v_{ik}) = \sum_{i=1}^m a_i(v_{i1} \otimes \dots \otimes v_{ik}).$$


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

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

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

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

答:
In the numbers $0\ldots (10^k-1)$, each digit appears in the ones place in one-tenth of those numbers: $\frac{1}{10}\cdot10^{k}=10^{k-1}$. So just adding the digits that originally come from the ones place, you have $(0+1+2+\cdots+9) 10^{k-1}=45\cdot 10^{k-1}$.

Nothing changes as you consider the tens place. In the numbers $1\ldots (10^k-1)$, each digit appears in the tens place one-tenth of the time. So just adding the digits that originally come from the tens place, you have $(0+1+2+\cdots+9) 10^{k-1}=45\cdot 10^{k-1}$.

Continuing in this way, summing digits from the $10^0$ place, the $10^1$ place, up to the $10^{k-1}$ place, the total sum would be $k$ copies of the same number: $$k\cdot45\cdot10^{k-1}$$

Title: challenging digit sum problem


问:
Do "Parabolic Trigonometric Functions" exist?
The parametric equation

$$\begin{align*}
x(t) &= \cos t\\
y(t) &= \sin t
\end{align*}$$

traces the unit circle centered at the origin ($x^2+y^2=1$).   Similarly,

$$\begin{align*}
x(t) &= \cosh t\\
y(t) &= \sinh t
\end{align*}$$

draws the right part of a regular hyperbola ($x^2-y^2=1$).  The hyperbolic trigonometric functions are very similar to the standard trigonometric function.

Do similar functions exist that trace parabolas (because it is another conic section) when set up as parametric equations like the above functions?  If so, are they also similar to the standard and hyperbolic trigonometric functions?

答:
One place where hyperbolic functions appear is, of course, hyperbolic geometry. There is a phenomenon that formulas of hyperbolic geometry are similar to the formulas of the spherical (elliptic) geometry, except that the hyperbolic formulas use sinh and cosh, while the spherical formulas use sin and cos, and some signs are changed (because of the opposite curvature).

In particular:

spherical geometry:

curvature is $K=1$

the circumference of a circle of radius $r$ is $2\pi \sin(r)$

Trigonometric identity: $\cos^2(x)−K\sin^2(x)=1$

spherical Pythagorean theorem: $\sin(a)\sin(b) = \sin(c)$

differential equations: $\cos(0)=1, \sin(0)=0, \cos′(x)=−K⋅\sin(x),\sin′(x)=\cos(x)$

hyperbolic geometry:

curvature is $K=−1$

the circumference of a circle of radius $r$ is $2π\sinh(r)$

Trigonometric identity: $\cosh^2(x)−K\sinh^2(x)=1$

hyperbolic Pythagorean theorem: $\sinh(a)\sinh(b) = \sinh(c)$

differential equations: $\cosh(0)=1, \sinh(0)=0, \cosh′(x)=−K⋅\sinh(x),\sinh′(x)=\cosh(x)$

parabolic geometry:

The parabolic geometry would be the Euclidean geometry (but we usually just say Euclidean geometry). In the Euclidean geometry we have $K=0$. So the parabolic sine and the parabolic cosine would be the functions which make the similar formulas true in Euclidean geometry.

The only choice is: sinp$(x)=x$, cosp $(x)=1$. Not very interesting, but this is what you should be using if you want to use the same formulas for spherical, parabolic, and hyperbolic geometry (see e.g. functions sin_auto and cos_auto here). The formulas above work, just as many others.

Not very formal, but you could also say that the curvature is infinitesimal ($K=\epsilon$) -- then you get cosp$(x)=1+\epsilon x^2/2$, correct and usable Pythagorean theorem, and the graph (cosp($x$), sinp($x$)) is actually a parabola.


问:
Determinant Identity: Elegant Solution Please.
Can anyone provide a proof of this identity:
$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}=2abc(a+b+c)^3 ?$$

Ideally I would like a series of row column operations resulting in the formula.
No brute force calculations. Note that letting $a=0$ we get two rows equal. So $abc$ is a factor. But I have unable to produce this by a simple series of operations, I mean to get one row or column all of whose entries are divisible by $a$. Getting divisibility by $(a+b+c)^2$ is also not hard just subtract one column from the other two. After that however I have been unable to make any further progress. Hopefully someone here is smarter than I.

Body:
$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}= \begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b)^2&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix}$$

$$\begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b^2)&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix} =(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
c-a-b& c-a-b&(a+b)^2\\
\end{vmatrix}$$

$$=(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
-2b& -2a& 2ab\\
\end{vmatrix} (R_3 \to R_3 -(R_1+R_2))$$

$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
a(b+c-a) &0&a^2\\
0&b(a+c-b)&b^2\\
-2ab& -2ab& 2ab\\
\end{vmatrix}$$

After this just do $R_{1} \to R_1 + R_3$ and $R_{2} \to R_2 + R_3$

$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
ab+ac &a^2&a^2\\
b^2&ba+ac&b^2\\
0& 0& 2ab\\
\end{vmatrix} ={(a+b+c)^2}\begin{vmatrix}
b+c &a&a\\
b&a+c&b\\
0& 0& 2ab\\
\end{vmatrix}$$

$$={(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
1&a+c&b\\
0& 0& 2ab\\
\end{vmatrix} = {(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
0&c&b-a\\
0& 0& 2ab\\
\end{vmatrix}$$

This gives the required value of $2abc(a+b+c)^3$.


问:
A consistent notation for algebraic numbers
I'm working on a simple CAS and I have run into the issue of number representation, or more specifically, number comparison.

while programming with integers, the CPU compares them by checking if their bits have the same values and the same order.

Now given these integers, you can create a data structure (a number representation), such as a fraction, and ensure that each possible value that it can represent will only ever be represented in one way. For a fraction, we do this by removing all factors shared by the numerator and the denominator. This means that we can just compare each element of the fraction individually, and we never need to worry about accidentally comparing 12/4 with 3/1.

We also have the ability, however, to do the simplification at the time of the comparison. Again with the fraction example, we can also find the common denominator of the two and scale them so that we can compare them (or something like that).

Now how can we create some sort of list-based data structure (or rather number notation) that allows us to compare algebraic numbers with either of these methods? This might be asking too much because this is all under the assumption that we will still be able to perform operations.

答:
I am not an expert here.  So please take this with a grain of salt or, even better, point out I should fix things.  Now, with the apology out of the way:

If we are focussed on the equality test (which is what the bulk of the OP was about) then we can proceed as follows.  We specify an algebraic number $\alpha$ by giving two pieces of data

the minimal polynomial $P(x)$ of $\alpha$ (monic by definition) and

a bounding box $B$ about $\alpha$ in the complex plane.

We arrange matters so that

bounding boxes are computed deterministically from $P$ and

boxes for distinct roots of $P$ are disjoint.

Given algebraic $\alpha$ and $\beta$ in this way, it is easy to decide if they are equal.  (It is also easy to decide if $\alpha$ is real.)  It should not be too painful to compute $-\alpha$ and $1/\beta$, as well.

However, given algebraic $\alpha$ and $\beta$ in this way, it is quite a bit of work to compute representations for $\alpha + \beta$ or $\alpha \cdot \beta$ (or, if both are real, to decide if $\alpha < \beta$).  In addition to computing new minimal polynomials we will have to shrink the given boxes.

There is another method with different trade-offs.  To specify $\alpha$ we find a number field $K$, equipped with a place, that contains $\alpha$.  We think of $K$ as a $\mathbb{Q}$-vector space and "represent" $\alpha$ as a rational vector.  Of course, the place has to be given to sufficient precision to separate $\alpha$ from its Galois conjugates.  If $\beta$ also lives in $K$ then arithmetic is much faster.  (Comparisons as real numbers is still expensive.)  However, if $\beta$ lives in a different number field $L$ then we cannot ask about equality until we find a third number field that contains both, and so on.  So the faster arithmetic is paid for by a slower equality test.

(I guess that any representation will have to have trade-offs of this kind.  However, lower bounds in complexity theory are famously difficult, so please don't ask me to prove my guess is correct.)

ADDED: After writing the answer, I looked around a bit and found the paper "Thom's Lemma, the Coding of Real Algebraic Numbers and the Computation of the Topology of Semi-algebraic Sets" by Coste and Roy.  They reduce the size of the representation of $\alpha$ (assumed real) considerably (but the simpler representation appears to be harder to compute).


问:
Question about orthogonal projection and some properties
Let be $V$ a vector space with an inner product, $W \subset V$ a subspace with $\dim W < \infty$ and $P:V \to V$ a projection s.t $Im P = W$ and $\vert \vert P(v) \vert \vert \leq \vert \vert v \vert \vert$, for all $v \in V$. then $P=P_W$, where $P_W$ is the orthogonal projection of $W$.

I know that if $\{w_1,w_2,\dots,w_n\}$ is an orthonormal basis of $W$ then $P_W(v)= \sum_{k=1}^n \langle v,w_k\rangle w_k$. I was thinking about using that and triangle inequality to show some way that $\vert \vert P(v) - P_W(v) \vert \vert = 0$, can you give me a hint?

答:
Suppose that $P\ne P_W$; I will prove that there is some $v\in V$ such that $\|v\|<\bigl\|P(v)\bigr\|$.

Since $P\ne P_W$, $\ker P\ne W^\perp$ and so, since $\ker P$ and $W^\perp$ are distinct subspaces of $V$ with the same dimension, there is some $v_0\in V\setminus W^\perp$ such that $P(v_0)=0$. The vector $v_0$ can be written as $w+v_1$, with $w\in W$ and $v_1\in W^\perp$, and then$$P(v_1)=P(v_0-w)=-w,$$since $P(v_0)=0$ and $P(w)=w$.

If $\lambda\in\Bbb R$, let $v_\lambda=v_1+\lambda v_0$. Then, since $P(v_0)=0$, $P(v_\lambda)=P(v_1)=-w$. On the other hand\begin{align}\|v_\lambda\|&=\|v_1+\lambda v\|\\&=\bigl\|v_1+\lambda(w+v_1)\bigr\|\\&=\bigl\|(\lambda+1)v_1+\lambda w\bigr\|\\&=\sqrt{(\lambda+1)^2\|v_1\|^2+\lambda^2\|w\|^2},\end{align}since $v_1$ and $w$ are orthogonal. Since $P(v_\lambda)=-w$, if I am able to find a $\lambda$ such that $\|v_\lambda\|<\|-w\|=\|w\|$, then $\|v_\lambda\|<\bigl\|P(v_\lambda)\bigr\|$.

Consider the function$$\begin{array}{rccc}q\colon&\Bbb R&\longrightarrow&\Bbb R\\&\lambda&\mapsto&\sqrt{(\lambda+1)^2\|v_1\|^2+\lambda^2\|w\|^2}.\end{array}$$Then $q(-1)=\|w\|$. Besides, $q'(-1)=-\|w\|\ne0$ (if $w=0$, then $v_0=v_1\in W^\perp$). So, $-1$ is not a local minimum of $q$ and therefore there is some $\lambda\in\Bbb R$ such that $q(\lambda)<\|w\|=q(-1)$. In other words, there is some $\lambda\in\Bbb R$ such that $\|v_\lambda\|<\|w\|$.