问:
Markov chain exercise 1
I have the transition matrix $$P= \begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
\frac14 & \frac14 & \frac12
\end{pmatrix}$$

and I have to determine the period of each state, compute $f_{2,2}{(n)}$ for each $n\geq 1$ and compute $p_2(X_n=3)$
I think $f_{2,2}{(n)} = p_{(2,3)}p_{(3,2)}=1·\frac14=\frac14$ but I am not sure
and the last part I have $p_i(X_n=j)=p_{i,j}{(n)}$ but I don't know how to do it

thank you

答:
$p_{i,j}^{(n)}$ can be found from the $n^{th}$ power of the transition matrix computed with spectral decomposition of $P=V\Lambda V^{-1}$ s.t., $P^n=V\Lambda^nV^{-1}$,

Here we have

$P
= \begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
\frac14 & \frac14 & \frac12
\end{pmatrix}
$

$V = \begin{pmatrix}
0.5773503 & 0 &  0\\
0.5773503 & 0.7774375 & 0.9554226\\
0.5773503 & 0.6289602 & -0.2952418
\end{pmatrix}
$

$\Lambda = \begin{pmatrix}
1 & 0 &  0\\
0 & 0.809017 & 0\\
0 & 0 & -0.309017
\end{pmatrix}
$

Value of $p_{2,2}^{(n)}$ can be obtained from $P^n(2,2)$
s.t., for n = 2,

$ P^2 = \begin{pmatrix}
1.000 & 0.000 & 0.0 \\
0.250 & 0.250 & 0.5 \\
0.375 & 0.125 & 0.5
\end{pmatrix}
$

s.t.,  $p_{2,2}^{(2)}=\frac1 4$ and $p_{2,3}^{(2)}=\frac1 2$

Note that 1 being the absorbing state of the Markov chain, eventually $p_{i,1}^{(n)}=1, \forall{i}$ and  $p_{i,j}^{(n)}=0, \forall{j \neq 1}$ when $n \to \infty$

The following animation shows how $p_{i,j}^{(n)}$ changes with increasing $n$.

Title: Markov chain exercise 1


问:
A triangle $ABC$ is given with $AB=8,BC=9,AC=13$
A triangle $ABC$ is given with $AB=8,BC=9,AC=13$. The vertex $A$ lies on the plane $\alpha$ and $BC\parallel \alpha$. A line $p$ through the midpoint of $AC$ and $B$ intersects $\alpha$ in $D$. Find the length of $BD$.

 $BD=11$

I am new to solid geometry and it's difficult for me to imagine the figures. I am really used to plane geometry. Can you help me with the diagram? I am not sure if we can use these formulas here, but if $K$ is the midpoint of $AC$, then $BK$ is a median of the triangle $ABC$, so $$m_b=\dfrac{1}{2}\sqrt{2a^2+2c^2-a^2}=\dfrac{11}{2}.$$ What can I do from here?

答:
We know the plane $\alpha$ is parallel to $BC$ and so the line $AD$ is parallel to $BC$ (line $AD$ is in the same plane as $\triangle ABC$).

If $K$ is the midpoint, $\triangle AKD \cong \triangle CKB$ given $BC \parallel AD$ and $CK = AK$.

So $BD = 2 \cdot BK = 11$

EDIT: On your question on why $BC \parallel AD$, say $\triangle ABC$ is in plane $\beta$. The key point is that the line through points $A$ and $D$ are in both planes $\alpha$ and $\beta$.

Now it comes down to using the following properties of Euclidean space  - (i) If a line and a plane are parallel, they never meet. As line through $B, C$ and plane $\alpha$ are parallel, they never meet and this implies lines through $B, C$ and through $A, D$ never meet (ii) If two lines in the same plane never meet, they are parallel. So lines through $B, C$ and through $A, D$ being in the same plane $\beta$ must be parallel.

ContentLicense: CC BY-SA 4.0


答:
I tried to draw it in order to understand it better, so here it is

From here you can deduce that the triangles $KCB$ and $KAD$ are similar because of the criteria $ASA$.Can you finish it?

Body:
A triangle $ABC$ is given with $AB=8,BC=9,AC=13$. The vertex $A$ lies on the plane $\alpha$ and $BC\parallel \alpha$. A line $p$ through the midpoint of $AC$ and $B$ intersects $\alpha$ in $D$. Find the length of $BD$.

 $BD=11$

I am new to solid geometry and it's difficult for me to imagine the figures. I am really used to plane geometry. Can you help me with the diagram? I am not sure if we can use these formulas here, but if $K$ is the midpoint of $AC$, then $BK$ is a median of the triangle $ABC$, so $$m_b=\dfrac{1}{2}\sqrt{2a^2+2c^2-a^2}=\dfrac{11}{2}.$$ What can I do from here?


问:
Graph-theory question regarding degrees and cliques
Suppose we have a graph $G$, where each node in there has a degree of at most $d$. I need to show that it is possible to determine if $G$ has a clique of size $k$ in time of at most $n^c \cdot d^k$ steps where $c$ is some constant.

Any help will be great.

答:
For each vertex $v$, denote by $\Gamma(v)$ the set of neighbors of $v$. For each vertex $v$, set $A_1:=\Gamma(v)$ and $C_1:=\{v\}$. In the following recursion, $A_i$ denotes the set of possible candidates to enter the clique and $C_i$ is the largest clique containing $v$ which we have already found:

For each $w\in A_i$, check whether $C_i\subseteq\Gamma(w)$. If so, then we have found a larger clique $C_{i+1}:=C_i\cup\{w\}$. Also, we can set $A_{i+1}:=\Gamma(w)\cap A_i$ since any vertex in an even larger clique will have to be a neighbor of all vertices in $C_{i+1}$. For performance reasons, we would stop as soon as $|A_i\cup C_i|<k$, of course.

If at some point you reach $i=k$, then $C_k$ is a clique of size $k$. On the other hand, if there is a clique of size $k$, then this procedure will find it by starting at any of its vertices. The running time is $nd^k$.

ContentLicense: CC BY-SA 3.0


问:
Is the local max point also a global max?
Consider the following function $f$ of three variables, defined on $\mathbb{R^{3}}$: $$f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_{x} = 15 + y - 8x = 0,$$ $$f_{y} = x - 4y + 2z = 0,$$ $$f{z}=-2z + 2y = 0.$$

So there's only one critical point $\left ( 2,1,1 \right )$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = \begin{bmatrix}
-f_{xx} &f_{xy}  &f_{xz} \\
 f_{yx}&f_{yy}  &f_{yz} \\
f_{zx} &f_{zy}  &zz
\end{bmatrix}$$

$$ \Rightarrow  H = \begin{bmatrix}
-8 &1  &0 \\
 1&-4  &2 \\
0 &2  &-2
\end{bmatrix}.$$

After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $\left ( 2,1,1 \right )$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z \in \mathbb{R^{3}}$.

So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

答:
The level surfaces of the function $ \ f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7 \ = \ c \ $ prove to be triaxial ellipsoids  (using the determinant method shown here ) with their axes not parallel to the coordinates axes and centered on $ \ (2  ,  1  ,  1 ) \ , $ the critical point found by OGC.  We can show that the maximum value of the function, $ \ f(2,1,1) = 15·2 + 2 - 4·2^2 - 2 - 1 + 2 + 7 \ = \ 22 \ $ is the global maximum of the function by an analysis of these surfaces.

The intersections of a level surface with "constant" planes through the center $ \ (2  ,  1  ,  1 ) \  $ are the ellipses

$$ \mathbf{x  =  2:  } \ \  30 \ + \ 2y \ - \ 16 \ - \ 2y^2 \ - \ z^2 \ + \ 2yz \ + \ 7 \ \ = \  -2y^2 \ + \ 2y \ - \ z^2 \ + \ 2yz \ + \ 21 \ \ = \  \ c \ \ ; $$

$$ \mathbf{y  =  1:  } \ \  15x \ + \ x \ - \ 4x^2 \ - \ 2 \ - \ z^2 \ + \ 2z \ + \ 7 \ \ = \  -4x^2 \ + \ 16x \ - \ z^2 \ + \ 2z \ + \ 5 \ \ = \  \ c \ \ ; $$

$$ \mathbf{z  =  1:  } \ \  15x \ + \ xy \ - \ 4x^2 \ - \ 2y^2 \ - \ 1 \ + \ 2y \ + \ 7 $$ $$ = \  -4x^2 \ + \ 15x \ - \ 2y^2 \ + \ 2y \  + \ xy \ + \ 6 \ \ = \  \ c \ \ . $$

The "cross-section" ellipse in the $ \ y \ = \ 1 \ $ plane does have its axes parallel to the $ \ x-$ and $ \ z-$ axes, so we can simply "complete the squares" to obtain

$$ 4·(x^2 - 4x + 4) \ + \ (z^2  +  2z  +  1) \ \ = \  \ 5 \ - \ c \ + \ 16 \ + \ 1  $$ $$ \Rightarrow \ \ 4·(x-2)^2 \ + \ (z-1)^2 \ \ = \  \ 22 \ - \ c \ . $$

This gives us an ellipse for $ \ c \ < \ 22 \ $ and "shrinks to a point" for $ \ c \ = \ 22 \ . $

The other two cross-sections must be rotated first, applying the standard method (coefficients shown are approximate):

$$ \mathbf{x  =  2:  } \ \ \left[ \tan 2\theta \ = \ \frac{2}{(-2)-(-1)}  =  -2 \right] \   -2.618y^2 \ + \ 1.701y \ - \ 0.382 z^2 \ + \ 1.051z \ + \ 21 \  \ = \ \ c \ \ , $$

$$ \mathbf{z  =  1:  } \ \ \left[ \tan 2\theta \ = \ \frac{1}{(-4)-(-2)}  =  -\frac12 \right]  \  -4.118x^2 \ + \ 14.139x \ - \ 1.882 y^2 \ + \ 5.393y \ + \ 6 \  \ = \ \ c \ \ ; $$

after which we "complete the squares" here (note: the centers of the ellipses are altered by these rotations):

$$ \mathbf{x  =  2:  } \quad 2.618·(y^2 \ - \ 0.650y \ + \ 0.1055) \ + \ 0.382·(z^2 \ - \ 2.751z \ + \ 1.892) $$ $$ = \ \ 21 \ - \ c \ + \ 0.276 \ + \ 0.723 \ \ = \ \ 22 \ - \ c \ \ , $$

$$ \mathbf{z  =  1:  } \quad 4.118·(x^2 \ - \ 3.433x \ + \ 2.947) \ + \ 1.882·(y^2 \ - \ 2.866y \ + \ 2.053) $$ $$ = \ \ 6 \ - \ c \ + \ 12.136 \ + \ 3.863 \ \ = \ \ 22 \ - \ c \ \ . $$

So indeed, all of the cross-sections are ellipses for $ \ c \ < \ 22 \ $ and "collapse" to the point $ \ (2,1,1) \ $ for $ \ c \ = \ 22 \ . $  Hence this is the global maximum for our function.


问:
Continuity of Functions defined from a supremum
Let $\overline{\Omega}\subset\mathbb{R}^{N}$ be a compact set and $T<\infty$. Assume $f : \overline{\Omega}\times(0,T)\to\mathbb{R}$ is continuous and sastifies

(A) $\forall x\in\overline{\Omega}, \sup\limits_{t\in(0,T)}|f(x,t)|<\infty$

(B) $\forall t\in(0,T), \sup\limits_{x\in\overline{\Omega}}|f(x,t)|<\infty$

Is it possible to show that a function defined as $\bar{f}:=\sup\limits_{t\in(0,T)}f(\,\cdot\,,t)$ to be a continuous function in $\overline{\Omega}$? I would like to apply this problem for my PDE research so any help (or counterexample) will be useful. (Remark : I know that $\sup\limits_{x\in\overline{\Omega}}|f(x,\,\cdot\,)|$ is a continuous function in $(0,T)$ since $\overline{\Omega}$ is compact.)

Thank you very much!

答:
No, consider the following counterexample.

Let $f:(0,1)\times [0,1]\longrightarrow \mathbb{R} $ be a function defined as $f(x,y)=\text{arctan}(y/x)$.
Then, $\bar{f}$ is not continous on $0$.


问:
Perturbated sum of independent random variables
It is $(X_{jm}), m\in\mathbb{N}, j\le m$ i.i.d with $E X_{11} = 0$ and $EX_{11}^2=1$ (no higher moments). Let $(v_{jm})_{j\le m}, (e_{lm})_{l\le m} \subset \mathbb{R}^m$ be orthonormal bases and $a_{jm} \in (1\pm \varepsilon)$. Does it hold that

$$\frac{1}{m} \sum_{j=1}^m a_{jm} \left( \left( \sum_{l=1}^m X_{lm} (v_{jm},e_{lm})\right)^2-1\right) \to 0$$

in probability (or in $L^1$)? The statement is obviously true, if either $v_{lm}=e_{lm}$, or $a_{lm}=1$, and I am curious, wether also the general case is true...

答: In the case someone is interested in the answer. The statement is true. One can first consider the special case with finite fourth moments ($E [X_11^4]<\infty$), then the general statement follows with truncation. A proof of a similar statement can be found in https://arxiv.org/pdf/2010.04519.pdf (Proposition 4.8 for the special case, the general case on pages 30-32).


问:
Separate a compact and a closed subset in $\mathbb{R}^n$
Let $K\subset \mathbb{R}^n$ a compact subset. And $F\subset \mathbb{R}^n$ a closed subset such that
$$ K \cap F = \emptyset.$$
How can we deduce that there is $a>0$ such that
$$\| y-x\|>a,$$
for all $x\in K$ and $y\in F$.

答:
Hint

Suppose such $a$ doesn't exist, i.e. for all $n\in \mathbb N$, there is $x_n\in K$ and $y_n\in F$ s.t. $$\|x_n-y_n\|\leq \frac{1}{n}.$$

Prove that $(x_n)$ is bounded.

Prove that $(x_n)$ and $(y_n)$ are convergent (up to a subsequence) and has the same limit.

Find a contradiction.


答:
Assume there exists no such $a$, and use compactness of $K$ to find a solution.

Hint: $K$ is a compact subset of $\mathbb{R}^n$, so every sequence of points in $K$ has a converging subsequence. :) You can use this to find a contradiction.

Title: Separate a compact and a closed subset in $\mathbb{R}^n$

Body:
Let $K\subset \mathbb{R}^n$ a compact subset. And $F\subset \mathbb{R}^n$ a closed subset such that
$$ K \cap F = \emptyset.$$
How can we deduce that there is $a>0$ such that
$$\| y-x\|>a,$$
for all $x\in K$ and $y\in F$.


问:
I don't know how to solve $\sin(x) + \sin(2x) + \sin(3x) = -3$
I have no idea how to solve this equation $\sin(x) + \sin(2x) + \sin(3x) = -3$. I've tried to use the formulas for $\sin(2x) = 2\sin(x)\cos(x)$ and for $\sin(3x) =  3\sin(x) - 4\sin^{3}(x)$, but all for nothing.

答:
If you continue the expansions, let $\sin(x)=t$ and you end with
$$-4 t^3+2 t\sqrt{1-t^2} +4 t=-3$$ After squaring processes, you need to find the roots of
$$16t^6-28t^4-24t^3+12t^2+24t+9=0$$ which cannot be solved (in fact, no real root at all).

Numerically, the solutions are $$t=-0.635632 \pm 0.219414\,i\implies x=-0.608034\pm 0.177981\,i$$

There would be solution if, in absolute value, $\text{rhs} < 2.49960$.


问:
Cauchy-Schwartz inequality in Einstein notation
I am learning about Einstein (index) notation, and I am trying to write Cauchy-Schwartz inequality as such. If we have $(\boldsymbol{a} \cdot \boldsymbol{b})^2 \le ||\boldsymbol{a}||^2 \ ||\boldsymbol{b}||^2$, and knowing that whenever an index is repeated in a term, a summation over this index is implied, we can write:
$$(\boldsymbol{a} \cdot \boldsymbol{b})^2 = (a_1b_1 + a_2b_2 + a_3b_3)^2 = \sum_{i=1}^3 a_ib_i = a_i b_i  $$
$$ ||\boldsymbol{a}||^2 = a_1a_1 + a_2a_2 + a_3a_3 = a_ia_i$$
Equivalently $||\boldsymbol{b}||^2 = b_i b_i$

So, I thought that the answer would be $(a_ib_i)^2 \le (a_ia_i) (b_ib_i)$, yet my textbook gives the answer as $(a_ib_i)^2 \le \boxed{a_j ^2 b_k ^2}$

Can someone explain how can we get to the right hand side of the inequality? (or point out if I have a misunderstanding for the index notation).

答:
With index notation you have to be careful about repeating indices.
First as @ClementC pointed out
$$ \|\mathbf{a}\| = a_1^2 + a_2^2 + a_3^2 = a_ja_j = a_j^2$$
where we understand that even though $a_i^2$ only has one explicit index since it is squared we can write $a_j^2 = a_ja_j$ and so still sum over $j$.
and similarly $\|\mathbf{b}\| = b_k^2$.
Here we choose a different index to avoid issues later.
The Cauchy-Schwartz inequality is then, as stated in your textbook,
$$ (a_ib_i)^2 \le a_j^2b_k^2. $$
The fact that $a$ and $b$ have different indices is important. For example the sum on the right includes terms like
$$ a_1a_1b_2b_2 $$
whereas if we instead had $a_i^2b_i^2$ then $a$ and $b$ always have the same index so this term would not be included.
In general if you have multiple terms with indices and then combine them into some larger term you should make sure not to have the same indices in each term unless you explicitly want no cross terms.


问:
Proof about distribution of $\forall$ universal quantifier over implication
I am struggling with this proof:

$\forall x(A \to B) \vDash A \to \forall x B$, if $x \not \in {free}(A)$

Below is my thoughts:

Proof: Let $\mathfrak{I} =(D,\varphi)$ be an interpretation, and let $\sigma$ be a variable assignment over $\mathfrak{I}$.

Assume   $\mathfrak{I}, \sigma \vDash \forall x(A \to B), $ so for all d $\in$ D,  $\mathfrak{I} , \sigma^{\frac{d}{x}} \vDash A \to B $,

so either $\sigma \nvDash A $ or  $\sigma \vDash B$,

if $\sigma \nvDash A $, then $\sigma \vDash A \to \forall xB$, then I have no idea how to do...

I want to know how to prove this claim. Would anyone help me out please?

答:
By contradiction: assume that the premise holds and the conclusion does not, i.e. that:

$I, \sigma \vDash \forall x (A \to B)$ and $I, \sigma \nvDash A \to \forall xB$.

This means that $I, \sigma \vDash A$ and $I,\sigma \nvDash \forall x B$.

This implies that, for some $d \in D$, we have that $I, \sigma (d|x) \nvDash B$.

But due to the fact that $x \notin \text {FV}(A)$, we have that form $I, \sigma \vDash A$, also: $I,\sigma(d|x) \vDash A$.

In fact, if $x$ does not occur free in $A$, and $\sigma$ satisfies $A$, also the new sequence $\sigma(d|x)$ that assigns $d$ to $x$ will satisfy it.

And thus:

$I, \sigma (d|x) \nvDash (A \to B)$,

contradicting the assumption above.


问:
Let $ K=\mathbb{Z}_{2}$ and $V=K^{4}$. For subspace $W=lin((1,0,1,0),(0,1,1,1))\subset V$ find all subspace layers.
This is the task:

Let $ K=\mathbb{Z}_{2}$ and $V=K^{4}$. For subspace $W=lin((1,0,1,0),(0,1,1,1))\subset V$ find all subspace layers.

I know, that there is 16 vectors in $V$. My initial idea was to check all subspaces $\forall_{\alpha \in W} \ \ \ \alpha + W $, but there should be a clever way.

Please don't use advanced techniques, it's from first year algebra perparation book with tasks

答:
The nonzero vectors in $W$ are  $(1,0,1,0)$, $(0,1,1,1)$, $(1,1,0,1)$. The solution takes about as long as writing it down:

pick a nonzero vector not yet listed, such as $(1,0,0,0)$. Add it to nonzero vectors in $W$ to get $(0,0,1,0)$, $(1,1,1,1)$, $(0,1,0,1)$.

pick a nonzero vector not yet listed, such as $(0,1,0,0)$. Add it to nonzero vectors in $W$ to get  $(1,1,1,0)$, $(0,0,1,1)$, $(1,0,0,1)$

and so on...


问:
What is steady state of differential equation?
I am a beginner in Dynamical Systems and Stability Analysis. The theory starts from the definition of steady-state solution (equilibrium) of differential equations, but I cannot understand how the following two explanations of steady-state solution match:

Given an ordinary differential equation $$\frac{dy}{dt}=f(t)$$

We say $y$ is a steady state solution of the above equation, if $\frac{dy}{dt}=0$.

The steady state is a state that the behavior of the system is unchanging over time. If a system is in a steady state, then the recently observed behavior of the system will continue into the future.

The first explanation (definition) means the critical point of the following curve, but how does it match the second explanation?

Thank you!

答:
You seem to be confused by two things. First about the difference between the derivative "having a zero" and "being zero". The condition 1 means that the function's derivative is zero, as in "it is zero for all time". Having a critical point doesn't satisfy this.

The second thing you seem to be confused by, is that in addition to having a derivative that is zero with respect to time, it also has to be a solution to the differential equation. An arbitrary $y$ with a zero time derivative isn't a steady state solution unless it is also a solution in the first place.


问:
Finding UMVUE of function of poisson parameter
I am to estimate $exp(-\lambda)\lambda^2/2$ from the distribution $Exp(\lambda) \sim \frac{e^{-\lambda}\lambda^x}{x!}$

I used the Indicator function $I_{X_1=2}(X)$ as an initial unbias estimator. statistic $T=\sum{X_i}$. Using Rao-Blackwell, $\hat{W} = E(W|T) = \frac{t(t-1)(n-1)^{t-2}}{2n^t}$

However, when I check $E(\hat{W}) = \sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}e^{-n\lambda}(n\lambda)^t}{2n^tt!}=\frac{1}{2}\exp(-n\lambda)\lambda^2$.

Where is the n coming from and how do i get rid of it? Where did i made a mistake?

答:
your initial unbiased estimator looks fine to me...using standard procedure I get exactly the resutl you will expect.

$$\mathbb{E}[W|T]=\mathbb{P}[W=1|T=t]=\frac{\mathbb{P}[X_1=2]\cdot\mathbb{P}[T=t|X_1=2]}{\mathbb{P}[T=t]}=$$

$$=\frac{\frac{e^{-\lambda}\lambda^2}{2!}\cdot\mathbb{P}[\Sigma_{i=2}^n=t-2]}{\frac{e^{-n\lambda}(n\lambda)^t}{t!}}=$$

$$=\frac{\frac{e^{-\lambda}\lambda^2}{2!}\cdot\frac{e^{-(n-1)\lambda}[(n-1)\lambda]^{t-2}}{(t-2)!}}{\frac{e^{-n\lambda}(n\lambda)^t}{t!}}=\frac{t(t-1)(n-1)^{t-2}}{2\cdot n^t}$$


问:
Prove that $(A × B) ∩ (C × D) ⊇ (A ∩ C) × (B ∩ D)$
Can someone help me with this please? I did find different solutions through the Internet, but they do not say the same:

Prove or show counterexample: $$(A \times B) \cap (C \times D) \supseteq (A \cap C) \times (B \cap D)$$

答: If $(x,y) \in (A \cap C) \times (B \cap D)$ then $x \in A \cap C$ and $y \in B \cap D$. Thus in particular $x \in A$ and $y \in B$, so $(x,y) \in A \times B$, and similarly $(x,y) \in C \times D$, so $(x,y) \in (A \times B) \cap (C \times D)$ as required.


答: Take some $x \in (A\cap C) \times (B\cap D)$. Then $x = (x_1, x_2)$ for some $x_1 \in A \cap C$ and $x_2 \in B \cap D$. However, this means that $x_1 \in A$ and $x_2 \in B$, so $x = (x_1, x_2) \in A \times B$. Try to do the same thing to show it is also in $C\times D$, and you will get your inclusion.

Title: Prove that $(A × B) ∩ (C × D) ⊇ (A ∩ C) × (B ∩ D)$

Body:
Can someone help me with this please? I did find different solutions through the Internet, but they do not say the same:

Prove or show counterexample: $$(A \times B) \cap (C \times D) \supseteq (A \cap C) \times (B \cap D)$$


问:
How is this argument deductively invalid?
I am reading the book An Introduction to Formal Logic by Peter Smith. I was checking the following argument  (Exercise-1 Question no-15) :

Miracles cannot happen. Why? Because, by definition, a miracle is an event incompatible with the laws of nature. And everything that happens is always consistent with the laws of nature.

Now, the author has written the following in the answer sheet:

Invalid. The premisses tell us that in fact no miracle ever happens. But they don’t tell us that miracles cannot happen. Compare:

Crimes cannot happen. Since, by deﬁnition, a crime is an act incompatible with the criminal law. And everything that happens is always consistent with the criminal law.

In a law-abiding community the premisses may be true: it doesn’t follow that criminal activity is impossible!

The author and some people seem to stress that 'And everything that happens is always consistent with laws of nature' means everything that has happened or is happening is consistent with laws of nature. It doesn't eliminate the possibility that anything can happen in future which is incompatible with the laws of Nature.

Similarly, they want to stress that 'No miracle ever happens'
doesn't eliminate the possibility that it can happen in future. However, 'Miracle cannot happen' is the statement that eliminates that possibility. The same thing can be said about the second argument.

Now, Is the author making the right point? I mean we can certainly have counter-examples:

Two plus two equals four.

Zero cannot divide any number.

Now, does the former statement mean that up to now two plus two has been equal to four but it doesn't need to remain the same way in future?
Does the latter statement mean that zero never divided, is not dividing and will not divide any number? I mean is it conceivable that the former statement can be false but still the latter statement is always true?

The way these simple present tenses is interpreted by the author seems wrong. That's why I have cross-posted it on English Language and Literature Stack Exchange. So, shouldn't this argument be deductively valid?

答:
I agree with you that the example can be misleading.

IMO, the key point is the usage of the modal operator expressing possibility, that is not present in the premises of the argument.

If we formalize the premises in predicate logic, what we get is the conclusion:

$\lnot \exists x (\text {Miracle}(x) \land \text {Happen}(x))$,

that reads: "No miracle happens."

This (following the Author's comment) asserts that miracles do not happen, not that it is not possible that... This one need modalities to be expressed:

$\lnot \diamond \exists x (\text {Miracle}(x) \land \text {Happen}(x))$.


答:
Let $(x,y) \in (A\cap C) \times (B\cap D)$.

$\Rightarrow x\in (A\cap C)$ and $y\in (B\cap D)$

$\Rightarrow x\in A$ and $x\in C$, $y\in B$ and $y\in D$

$\Rightarrow (x,y) \in A\times B$ and $(x,y) \in C\times D$

$\Rightarrow (x,y) \in (A\times B) \cap (C\times D)$

You just use definitions of intersection and what you know about sets like $I_1 \times I_2 \times ...$


问:
Question about Noncommutative Rings by I. N. Herstein
Here's an example that I do not quite understand it fully, it's on page 6 of the book. And here's what it says:

Let $F$ be a field, and $F_n$ be a ring of all $n\times n$ matrices over $F$. We consider $F_n$ as the ring of all linear transformations on the vector space $V$ of $n-$tuples of elements of $F$. If $A$ is a subset of $F_n$, let $\overline{A}$ be the sub-algebra generated by $A$ over $F$. Clearly, $V$ is a faithful $F_n-$module, and so, $\overline{A}$-module. $V$ is in addition, both a unitary, and irreducible $F_n-$module.

We say the set of matrices $A \subset F_n$ is irreducible if $V$ is an irreducible $\overline{A}-$module. In matrix terms, it merely says that there is no invertible matrix $S$ in $F_n$ so that

$$S^{-1}aS = \left( \begin{array}{c|c} a_1 & 0 \\ \hline * & a_2  \end{array} \right)$$ for all $a \in A$.

The bolded parts are what I don't understand

Firstly, why must $V$ be a faithfull $F_n-$module in order to be an $\overline{A}-$module? What will happen if $V$ is not faithful?

Secondly, I don't understand the bolded part start with "in matrix terms...", say I have $V$ is an irreducible $\overline{A}-$module. How can I deduce the last part? And I don't get the star in the expression $S^{-1}aS$ as well, what does it stand for, is it a wild-card? I think this has to do with some matrix knowledge that I'm missing, so if the explanation is pretty long, please just give me a reference on what I should read.

Thank you guys very much,

And have a good day,

答:
I don't see that you need faithful in order to have an $\overline{A}$-module. Whenever you have an algebra homomorphism $i: \overline{A}\to F_n$ (such as inclusion in your case) then any $F_n$-module is also an $\overline{A}$-module by restriction, i.e. $a\cdot m:= i(a)m$. The author probably means that $V$ is also a faithful $\overline{A}$-module, which is quite clear since $i$ is the inclusion (although that $V$ is faithful is not used in the paragraph you provided).

For the last part, you are right $*$ stands for a wildcard. The explanation is not really long, the intuition you should have is that block matrices correspond to direct sum decompositions and block matrices with a wild card instead of a zero correspond to submodules. This is how you proceed. Suppose $V$ is not irreducible. Then it has a submodule $U$. This means that $a\cdot U\subset U$ for all $a\in A$. Now doing a base change from $U$ to the standard basis this means that $S^{-1}aS V'\subset V'$ where $V'$ denotes the vector space with the first $r=n-\dim U$ entries equal to zero (and the others arbitrary). And if you right this down in matrix language, this means that $S^{-1}aS$ has the claimed form with $a_1$ a matrix of size $r$. For the reverse direction view $S$ as such a base change matrix, then you have your submodule (in the same way as above) by changing back $V'$. I'll leave the details for you. Feel free to ask if I should explain something in more detail.


问:
If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$.
Question from Math Olympiad:

Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.

My approach:

$a\sqrt{a} + b\sqrt {b} = 183$

$a\sqrt{b} + b\sqrt {a} = 182$

Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$

I only see two integral possibilities:

$1 \times 365 $  and $5 \times 73 $

And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.

Please help me with the general method to go about the same.

This is the picture from the source.

Body:
Let $\sqrt a=x,\sqrt b=y$

$$\implies\dfrac{183}{182}=\dfrac{x^3+y^3}{xy(x+y)}$$

As $x+y\ne0,$  $$182 x^2-365xy+182y^2=0\implies\dfrac xy=\dfrac{13}{14}\text{ or }\dfrac{14}{13}$$

If $\dfrac xy=\dfrac{13}{14},\dfrac x{13}=\dfrac y{14}=z$(say)

So, we have $$182=z^3(13\cdot14)(13+14)\implies z^3=?$$

As $z$ is real, $z=\sqrt[3]{\dfrac1{27}}=\dfrac13\implies x=?,y=?$

We need $$\dfrac{9(a+b)}5=\dfrac{9(x^2+y^2)}5=?$$


问:
Number of positive integer solutions of a linear equation with maximal value of two summands being restricted.
My question is how many possible solutions there are for the equation subject to the condition that $x_1 \leq k_1$ and $x_r \leq k_r$, where $x_i$ are positive integers.

The equation is
$$
x_1 + x_2 + \ldots + x_r = n.
$$
I have tried to find the number of solutions so that if L = number of solutions we can obtain the combinatorial number but i haven't reached any conclusion.

Any tip would be much appreciated, thanks.

Edit:
I solved it using the inclusion-exclusion principle.

答:
Let us consider the general case where all $x_i$ are restricted: $0<x_i\le k_i$.

In terms of generating functions we are looking for the coefficient at $x^n$ in the product:
$$\begin{align}
x^r\prod_{i=1}^r\sum_{j=0}^{k_i-1}x^j
&=x^r\prod_{i=1}^r\frac{1-x^{k_i}}{1-x}\\
&=\frac{x^r\prod_{i=1}^r(1-x^{k_i})}{(1-x)^{r}}\\
&=x^r\prod_{i=1}^r(1-x^{k_i})\sum_{j=0}^\infty\binom{-r}{j}(-x)^j\\
&=x^r\prod_{i=1}^r(1-x^{k_i})\sum_{j=0}^\infty\binom{r+j-1}{j}x^{j}\\
&=\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}x^{\boldsymbol{\mu}\cdot\mathbf{k}}
\sum_{j=0}^\infty\binom{r+j-1}{r-1}x^{r+j},
\end{align}$$
which is
$$
\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}\binom{n-1-\boldsymbol{\mu}\cdot\mathbf{k}}{r-1},\tag1
$$
where $\boldsymbol{\mu}$ are all $2^r$ binary vectors of the length $r$, $|\boldsymbol{\mu}|=\sum_{i=1}^r\mu_i$ and $\mathbf{k}=(k_1,k_2,\dots,k_r)$.

In (1) the equality $\binom pq=0$ is assumed for $p<q$.

If some $x_i$ are unrestricted one can simply ignore them while using (1). For example if (as in your question) only two values are restricted the answer is:
$$
\binom{n-1}{r-1}-\binom{n-1-k_1}{r-1}-\binom{n-1-k_r}{r-1}+\binom{n-1-k_1-k_r}{r-1}.
$$

This can be obtained also using inclusion-exclusion principle.


答:
Sorry but following your procedure (that is correct)

$$W=\mathbb{1}_{\{2\}}(X_1)$$

thus using the law of total expectation

$$\mathbb{E}[\hat{W}]=\mathbb{E}[\mathbb{E}[W|T]]=\mathbb{E}[W]=1\cdot \frac{e^{-\lambda}\lambda^2}{2}+0\cdot\left[1-\frac{e^{-\lambda}\lambda^2}{2}\right]=\frac{e^{-\lambda}\lambda^2}{2}$$

Direct calculation

$$\mathbb{E}[\hat{W}]=\sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}}{2\cdot n^t}\cdot\frac{e^{-n\lambda}(n\lambda)^t}{t!}=$$

$$=\frac{e^{-n\lambda}\lambda^2}{2}\sum_{t-2=0}^{n}\frac{[(n-1)\lambda]^{t-2}}{(t-2)!}=\frac{e^{-n\lambda}\lambda^2}{2}\cdot e^{(n-1)\lambda}=\frac{e^{-\lambda}\lambda^2}{2}$$

Title: Finding UMVUE of function of poisson parameter

Body:
I am to estimate $exp(-\lambda)\lambda^2/2$ from the distribution $Exp(\lambda) \sim \frac{e^{-\lambda}\lambda^x}{x!}$

I used the Indicator function $I_{X_1=2}(X)$ as an initial unbias estimator. statistic $T=\sum{X_i}$. Using Rao-Blackwell, $\hat{W} = E(W|T) = \frac{t(t-1)(n-1)^{t-2}}{2n^t}$

However, when I check $E(\hat{W}) = \sum_{t=0}^{n}\frac{t(t-1)(n-1)^{t-2}e^{-n\lambda}(n\lambda)^t}{2n^tt!}=\frac{1}{2}\exp(-n\lambda)\lambda^2$.

Where is the n coming from and how do i get rid of it? Where did i made a mistake?


问:
Example of differential equation with two general solutions
This is a teaching-related question.

When teaching ordinary differential equations, we typically distinguish between the general solution (which includes unspecified integration constants) and particular solutions (where the integration constants are fixed by, say, initial values).

This makes sense for linear ordinary differential equations. But I can imagine that some nonlinear ordinary differential equations have two different solutions that look nothing like each other, that is, their difference is not an integration constant.

Do you know an example for such differential equations (and solutions)?

答:
You can also just multiply together two linear homogeneous ODEs, to form a nonlinear ODE, e.g.

(y'+ay) (y'+by) = 0

This shows explicitly how nonlinear ODEs can have separate solution branches (if a and b are different).


问:
If terms of $\left\langle Tu,w\right\rangle$ are of the form $\left\langle Tv,v\right\rangle$, then $\left\langle Tu,w\right\rangle=0$. Why?
In order to prove following theory, my textbook gives reasoning that $\left\langle Tu,w\right\rangle =0, \forall u,w$ implies $T=0$.
But I am unable to understand where is the fact used in this proof, that it is strictly a complex inner product space (not real), because it is false for real inner product space.

Over $\mathbb{C}$, $T v$ is orthogonal to $v$ for all $v$ only for the $0$ operator.

We have $\left\langle Tu,w\right\rangle=\frac{\left\langle T(u+w),u+w\right\rangle-\left\langle T(u-w),u-w\right\rangle}{4}+i\frac{\left\langle T(u+iw),u+iw\right\rangle-\left\langle T(u-iw),u-iw\right\rangle}{4}.$

Each term on the RHS are of the form $\left\langle Tv,v\right\rangle$. Thus our hypothesis implies that $\left\langle Tu,w\right\rangle=0$, $\forall u,v\in V$. This implies that $T=0$ (taking $w=Tu$).

But real space is a subspace of a complex space. So, the proof should be valid for real space as well, which is not true?

答:
In that proof, multiplication by $i$ is used. Therefore, the proof makes no sense in a real vector space.

On the other hand, your sentence “But real space is a subspace of a complex space.” is strange. Are you saying that every real vector space is a subspace of a complex vector space? What happens is that, for every real vector space $V$, there is a complex vector space such that $V$ is isomorphic to a real vector subspace of $W$. But that's a quite heavy approach for this problem.


问:
A function with the following properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function with the following property: Given $\epsilon > 0$, there exists a positive real number $r > 0$ and a set $D \subset \mathbb{R}$ such that

Any interval $I$ with length $r >0$, is such that $D \cap I \ne \emptyset$

$|f(x+t) - f(x)| < \epsilon$ for any $t \in D$ and $x\in \mathbb{R}$

Show that a function with those properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.

I've been stuck trying to prove this for a while. I think that any $f$ is a periodic function, like $\sin$, and that would imply that it is bounded and that if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant. But I don't know how to show that $f$ is periodic, nor if it is the best approach for this proof.

答:
This should be a comment except that it is much too long. I will answer the sub-(sub-)question of whether $f$ is necessarily periodic in the negative; that is, we will construct an $f$ that satisfies the constraints but is not periodic.

First, define $f_0 : \mathbb Z \to \mathbb R$ by setting
$$
f_0(n) = \begin{cases} 0 & \text{if $n = 0$;}\\
2^{-k} & \text{if $n \neq 0$ and $2^k$ is the largest power of two dividing $n$}.
\end{cases}
$$

Then we extend $f_0$ to a function $f: \mathbb R \to \mathbb R$ by linear interpolation on each interval $[k, k+1]$. Note that $f$ is not periodic: $f(x) = 0 \iff x = 0$.

Now let us prove that it satisfies the hypotheses of the problem. Let $\epsilon > 0$ be given, and take $K$ so that $2^{-K} < \epsilon$. Then I claim that taking $D = \{m2^K \mid m \in \mathbb Z\}$ and $r$ any number greater than $2^K$ works. Clearly $r$ is large enough, so it remains to prove that for $x \in \mathbb R$ and $t \in D$, we have $|f(x + t) - f(x)| < \epsilon$. Since $t$ is an integer, the fractional parts of $x + t$ and $x$ are identical. Since $f$ is defined for non-integral arguments by linear interpolation, if $|f(x+t) - f(x)|$ exceeds $\epsilon$, then so does at least one of $|f(\lfloor x + t\rfloor) - f(\lfloor x \rfloor)|$ and $|f(\lceil x + t\rceil) - f(\lceil x \rceil)|$, so without loss of generality we can take $x$ to be an integer, too.

Let $x = j2^k$ with $j$ odd. Then the largest power of two dividing $x + t = j2^k + 2^K$ is the minimum of $k$ and $K$. If it is equal to $k$, then $f(x) = f(x + t)$. If it is equal to $K$, then $K < k$, so $2^{-K} > 2^{-k}$, and
$$
|f(x + t) - f(x)| = |2^{-K} - 2^{-k}| \leq 2^{-K} < \epsilon.
$$


问:
$\sigma(\delta)_{A}=\sigma(\delta_{A})$
Let $A\subset X$ a non-empty set, $\emptyset\not=\delta\subset \mathcal{P}(X)$ and $\delta_{A}:=\{A\cap B: B \in \delta$}.
Si $\sigma(\delta_{A})$ is is the σ-algebra in $A$ generated by $\delta_{A}$, proof that $\sigma(\delta)_{A}=\sigma(\delta_{A})$

since $ A\cap B\subset B \in \delta$ then $A \cap B \in \delta$, this shows that $\delta_{A} \subset \delta$ then $\sigma(\delta_{A})\subset \sigma(\delta)_{A}$

I have tried to demonstrate the other restraint without success. Which way do you advise me to follow?

答: Consider $\{B\in \sigma(\delta): B \cap A \in \sigma(\delta_A)\}$. Verify that it is a sigma algebra and that it contains $\delta$. Hence, it contains $\sigma(\delta)$. This is what you want to prove.

Title: $\sigma(\delta)_{A}=\sigma(\delta_{A})$


问:
Help with Combination
Guys need help to solve this one..
How will we arrange  Red balls in '$N$' places , so that if you choose any '$M$' consecutive places, there should be at least '$K$' Red balls among this '$M$' chosen places.And we should use minimum number of Red balls.

Now, If $N = 6$ and $M = 3$ and $K = 2$ then, one combination is like '$1\,\; R\,\; R\,\; 1\; R\; R\; 1$' Here if we choose any $3$ consecutive places, there will be $2$ Red Balls. So like that how many ways we can do it..

There could be a larger numbers also.
Please help..

答:
Completely revised to answer the general question:

Assume first that $n=3m$, so that we may segment the string into $m$ blocks of three balls. Each block must be of one of the forms RRW, RWR, and WRR, and a string of $m$ RRW blocks is certainly acceptable, so there must be $2m$ red and $m$ white balls.

Let $s_m$ be the number of acceptable strings of $3m$ balls. Suppose that I have such a string. If it ends with a WRR block, I can append any of the three kinds of block to get an acceptable string of length $3(m+1)$. If it ends with an RWR block, I can append either an RWR or an RRW block, but not a WRR block. And if it ends with an RRW block, I can only append another RRW block. To turn it around, a WRR block can only follow another WRR block; an RWR block can follow a WRR block or another RWR block; and an RRW block can follow anything.

Let $r_m$ be the number of acceptable strings of length $3m$ that end with an RRW block, $c_m$ the number that end with an RWR block, and $\ell_m$ the number that end with a WRR block. Then from the last paragraph we see that

$$\begin{align*}
s_m&=r_m+c_m+\ell_m\;,\\
r_m&=s_{m-1}\;,\\
c_m&=c_{m-1}+\ell_{m-1}\;,\text{ and}\\
\ell_m&=\ell_{m-1}\;.
\end{align*}\tag{1}$$

Clearly $r_1=c_1=\ell_1=1$, so $\ell_m=1$ for all $m\ge 1$, and $(1)$ reduces to

$$\begin{align*}
s_m&=r_m+c_m+1\;,\\
r_m&=s_{m-1}\;,\text{ and}\\
c_m&=c_{m-1}+1\;.\\
\end{align*}\tag{2}$$

From the last line of $(2)$ it’s clear that $c_m=m$ for all $m\ge 1$, so $$s_m=s_{m-1}+m+1\;.\tag{3}$$ If we set $s_0=1$ and $r_0=c_0=0$, the recurrences are all valid for $m\ge 1$, and it’s not hard to verify that

$$s_m=\sum_{k=1}^{m+1}=\frac12(m+1)(m+2)=\binom{m+2}2\;.$$

This completely answers the question when $n$ is a multiple of $3$.

Now suppose that $n=3m+1$. There is an acceptable string of length $3m$ ending in WRR, namely, a string of $m$ WRR blocks; we can append a white ball to this string, so there is an acceptable string of length $n$ with just $2m$ red balls. However, this is the only such string, since, as we saw before, a WRR block must follow another WRR block, and so on back to the beginning of the string. Thus, there is only one acceptable string of length $n=3m+1$ that uses the minimum possible number of red balls: $(\text{WRR})^m\text{W}$.

Finally, suppose that $n=3m+2$. Clearly at least one of the last two balls must be red, and one suffices, since we can append a red ball to the string $(\text{WRR})^m\text{W}$, so the minimum possible number of red balls in this case is $2m+1$. We can append WR to an acceptable string of length $3m$ if and only if the string is $(\text{WRR})^m$; that accounts for one acceptable string of length $n$. We can append RW to any acceptable string of length $3m$ that ends in RWR or WRR. There is just one of the latter, and there are $c_m=m$ of the former, so this case accounts for another $m+1$ acceptable strings of length $n$. Thus, there are $m+2$ acceptable strings of length $n=3m+2$.

Title: Help with Combination


问:
$11$-adic expansion of 1/10
I am working through Koblitz's book on $p$-adic numbers and in one of the exercises, he asks us to find the $11$-adic expansion of $\frac{1}{10}$. My idea was to use the fact that $$\frac{1}{10}=(-1)\frac{1}{1-11^1}=(-1)\sum_{n=0}^\infty 11^n.$$

But now I'm confused. The series on the right looks exactly like a multiplicative inverse (it's $\dots 1111_{11}$). I know $(-1)=\sum_{n=0}^\infty (p-1)p^n$ for all $p$-adics but then it seems to me that the multiplication returns $-1$. What am I missing?

答:
You have shown that $-\frac{1}{10} = \ldots\!111_{11}$. Which is true, no problems here! It sounds like you are worried that $(-1)\times \ldots\!111_{11} = \ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11}$ gives $-1$? Well, it doesn't... maybe you are multiplying pointwise (term by term) by mistake?

Instead of working out the result of $\ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11}$, it's easier to do something like:
$$
-\frac{9}{10} = 9\cdot \left(-\frac{1}{10}\right) = \ldots\!999_{11}
$$
$$
\implies \frac{1}{10} = -\frac{9}{10} + 1 = \ldots\!999(10)_{11}
$$
It should also be possible to verify directly that $\ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11} = \ldots\!999(10)_{11}$.

Title: $11$-adic expansion of 1/10


答:
Your argument for the inclusion $\sigma (\delta _A)\subset \sigma (\delta )_A$ is not correct. Take for example $X=\{1,2,3,4\}$ and $\delta =\{1,2\}$. Then $\{1\}\subset \{1,2\}\in \sigma (\delta )$, but $\{1\}\notin \sigma (\delta )$.

The way to proceed is the following. You have that $$\sigma (\delta )_A=\{A\cap B\mid B\in \sigma (\delta )\}.$$ In particular, it's a $\sigma -$algebra that contains $\delta _A$. Therefore $\sigma (\delta _A)\subset \sigma (\delta )_A$. For the converse inclusion, let $$\mathcal S:=\{S\in \mathcal P(X)\mid A\cap S\in \sigma (\delta _A)\}.$$

Then $\mathcal S$ is a $\sigma -$algebra that contains $\delta $. Therefore, $\sigma (\delta )\subset \mathcal S$, and thus $\sigma (\delta )_A\subset \sigma (\delta _A)$.

Title: $\sigma(\delta)_{A}=\sigma(\delta_{A})$

Body:
Let $A\subset X$ a non-empty set, $\emptyset\not=\delta\subset \mathcal{P}(X)$ and $\delta_{A}:=\{A\cap B: B \in \delta$}.
Si $\sigma(\delta_{A})$ is is the σ-algebra in $A$ generated by $\delta_{A}$, proof that $\sigma(\delta)_{A}=\sigma(\delta_{A})$

since $ A\cap B\subset B \in \delta$ then $A \cap B \in \delta$, this shows that $\delta_{A} \subset \delta$ then $\sigma(\delta_{A})\subset \sigma(\delta)_{A}$

I have tried to demonstrate the other restraint without success. Which way do you advise me to follow?


问:
Finding the common ratio of an arithmetic and geometrical series
I am working through a maths text book as a hobby and am stuck on the following problem:

An arithmetic series and a geometric series have r as a common difference and the common ratio respectively. The first term of the arithmetic series is 1 and the first term of the geometric series is 2. If the fourth term of the arithmetic series is equal to the sum of the third and fourth terms of the geometric series, find the three possible values of r.

This is how I have calculated:

Let $u_{n}$ refer to the nth member of a series. Then for the arithmetic series the 4th term is $1 + 3r$.

If $S_{n}$ refers to the sum to n of a series then $S_{4} - S_{2}$ will give $u_{3} + u_{4}$ of the geometric series.

$S_{4} = \frac{2(1 - r^4)}{1 -r}$ and $S_{2} = \frac{2(1 - r^2)}{1 -r}$

$S_{4} - S_{2} = \frac{2r^2 - 2r^4}{1-r}$

$(1 + 3r) = \frac{2r^2 - 2r^4}{1-r}$

$(1 + 3r)(1-r) = 2r^2 - 2r^4$

$\implies 2r^4 - 5r^2 + 2r + 1 = 0$

Now my textbook hasn't got as far as solving polynomials higher than 3 yet - and I cannot do it - and I'm wondering if there is another way of approaching this problem.

答:
The terms of GP are $2,2r,2r^2,2r^3,\ldots$ So we can directly write

$$1+3r=2r^2+2r^3$$
$$\Rightarrow 2r^3+2r^2-3r-1=0$$

which indicates that there is mistake made in simplification in OP. EDIT : There is no mistake in simplification as pointed by @DavidK below in comments since
$$2r^4-5r^2+2r+1=(r-1)(2r^3+2r^2-3r-1)$$

$\vdots$

Clearly $1$ is a root, so the above factorizes as
$$(r-1)(2r^2+4r+1)=0$$

The second factor is a quadratic whose roots can be found using quadratic formula.


答:
Suppose we have a chess tournament.  After the tournament ends, we analyze the results of all of the games.  We come to the conclusion that:

"Every move that gets made does not involve a pawn promoting into a knight."

In other words we might say that:

"Knight promotion does not happen."

However, in the rules of chess, it's still possible for a pawn to promote into a knight.  In other words, knight promotion can happen.

Thus, the argument "knight promotion does not happen implies that knight promotion cannot happen" is an invalid argument.

Perhaps as a better example there may exist some positions in chess or other games where a position of pieces on the board can happen in accordance with the ordinary rules of chess, but has never, in fact, happened in a game that people have played.

Also, to say that something cannot happen in the real world implies that such couldn't happen at any future events.  Everything that happens being consistent with the laws of nature applies to past and present events.  The quote doesn't say that everything that will happen will be consistent with the laws of nature.  Thus, the premise of the argument that everything that is happening and has happened doesn't apply to future events, while the claim that something cannot happen does apply to future events.  So, by that account also, the argument is invalid... at least as stated.

Title: How is this argument deductively invalid?

Body:
I am reading the book An Introduction to Formal Logic by Peter Smith. I was checking the following argument  (Exercise-1 Question no-15) :

Miracles cannot happen. Why? Because, by definition, a miracle is an event incompatible with the laws of nature. And everything that happens is always consistent with the laws of nature.

Now, the author has written the following in the answer sheet:

Invalid. The premisses tell us that in fact no miracle ever happens. But they don’t tell us that miracles cannot happen. Compare:

Crimes cannot happen. Since, by deﬁnition, a crime is an act incompatible with the criminal law. And everything that happens is always consistent with the criminal law.

In a law-abiding community the premisses may be true: it doesn’t follow that criminal activity is impossible!

The author and some people seem to stress that 'And everything that happens is always consistent with laws of nature' means everything that has happened or is happening is consistent with laws of nature. It doesn't eliminate the possibility that anything can happen in future which is incompatible with the laws of Nature.

Similarly, they want to stress that 'No miracle ever happens'
doesn't eliminate the possibility that it can happen in future. However, 'Miracle cannot happen' is the statement that eliminates that possibility. The same thing can be said about the second argument.

Now, Is the author making the right point? I mean we can certainly have counter-examples:

Two plus two equals four.

Zero cannot divide any number.

Now, does the former statement mean that up to now two plus two has been equal to four but it doesn't need to remain the same way in future?
Does the latter statement mean that zero never divided, is not dividing and will not divide any number? I mean is it conceivable that the former statement can be false but still the latter statement is always true?

The way these simple present tenses is interpreted by the author seems wrong. That's why I have cross-posted it on English Language and Literature Stack Exchange. So, shouldn't this argument be deductively valid?


答:
I disagree with the author on both counts:

By the principle of charity, the crime argument is valid, albeit unsound
(its 2nd premise is false, whichever way one interprets “everything
that happens is always consistent with the criminal law”).

On the other hand, the miracles argument also is valid; however its 2nd
premise (“everything that happens is always consistent with the laws
of nature”) may or may not be true.

The author rightly observes that the premises together say that no
miracle ever happens, i.e., miracles never happen. However, whether
“never happens” necessarily implies “cannot happen” really is a
matter of semantics: the author says nay, while I say aye (so, when I
state that speed is never negative, I mean it in the sense that speed
can never be negative).

(If the premises had instead been along the lines of “miracles do not happen”, then yes, reasonably, that wouldn't have necessarily implied that “miracles cannot happen”.)


问:
If $B_n$ is compact and $B_n\subseteq B_{n+1}$, is there a $n_0$ with $d(x,B_n)\ge d(x,\bigcap_nB_n)/2$ for all $n\ge n_0$?
Let $(E,d)$ be a metric space, $x\in E$ and $$d(x,B):=\inf_{y\in B}d(x,y)\;\;\;\text{for }B\subseteq E.$$

Let $(B_n)_{n\in\mathbb N}$ be nonincreasing and $B:=\bigcap_{n\in\mathbb N}B_n$. It seems like that in the proof on page 9 in this paper, it s claimed that if $x\not\in B$ (hence $\delta:=d(x,B)>0$), there is a $n_0\in\mathbb N$ with $d(x,B_n)\ge\frac{d(x,B)}2$ for all $n\ge n_0$.

As the example in this answer shows, this is not true in general. In the paper, every $B_n$ (hence $B$) is compact, but I don't see how this makes any difference. So, is there anything else I'm missing or is the proof in the paper wrong?

答: There exists $b_n \in B_n$ such that $d(x,B_n)+\frac 1 n >d(x,b_n)$ $\, \, (1)$. There is  a subsequence $(b_{n_k})$ of $(b_n)$ converging to some $b$. Let $N \geq 1$.  Since $b_{n_k} \in B_{n_k} \subseteq B_N$ for all $k$ sufficiently large it follows that $b \in B_N$. This is true for all $N$, so $x \in B$. Now $d(x,B) \leq d(x,b)=\lim d(x,b_{n_k})\leq \lim  d(x,B_n)$ by (1).  Of course, $d(x,B_n) \leq d(x,B)$ for all $n$. Conclusion: $d(x,B_n) \to d(x,B)$ as $ n \to \infty$.


问:
$\Omega$ be a set $(A_n)_{n \in \mathbb{N}}$
$\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$ a sigma algebra.

How can I show that a measurable function $f: (\Omega, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ is constant on each $A_n$?

I know that since $f$ is measurable, we have $f^{-1}(\mathcal{B}) \subset \mathcal{A}$. Then I tried to come up with a contradiction. I assumed that $f$ is not constant, meaning that $\exists a,b \in A_n$ such that $f(a) \neq f(b).$ I think that both $f(a)$ and $f(b)$ are elements of $\mathcal{B}(\mathbb{R})$ since they're just points, so open, and the Borel-Algebra contains all open sets. I'm stuck here.

Can anyone help?

答:
Remark that $f$ is $\mathcal A-\mathcal B(\mathbb R)-$measurable if and only if $$\forall \omega \in \Omega , f^{-1}(\{f(\omega )\})\cap A_i\in \{A_i,\varnothing \}.$$

Suppose that $f$ is not constant on some $A_i$, i.e. there are $a,b\in A_i$ s.t. $f(a)\neq f(b)$. In particular, $f^{-1}(\{f(a)\})\cap A_i\subsetneq A_i$, and thus, $f^{-1}(\{f(a)\})\notin \mathcal A$.

Edit

a set $S$ is $\mathcal A-$measurable if and only if there is a sequence $\{i_n\}_{n\in\mathbb N}\subset \mathbb N$ s.t. $S=\bigcup_{n=1}^\infty A_{i_n}$. Since $\{f(\omega )\}$ is a Borel set for all $\omega \in \Omega $, if $f^{-1}(\{f(\omega )\})\notin \mathcal A$ for at least one $\omega \in \Omega $, then your function won't be measurable. Now, you'll have that $f^{-1}(\{f(\omega )\})\cap A_i$ is either included in $A_i$ or empty. So, if $f^{-1}(\{f(\omega )\})\cap A_i$ is not in $\{A_i,\varnothing \}$ for at least one $\omega $ and one $i$, then $f^{-1}(\{f(\omega )\})$ won't be measurable, and thus $f$ won't be measurable either.


问:
Is the restriction of a function to its set of continuity points continuous? i.e. is $h|_{\mathcal C(h)}$ continuous?
Let $h: S \to S'$ be a function between topological spaces $(S, \tau)$ and $(S', \tau')$.

Let $\mathcal C(h)$ denote the continuity points of $h$.

Is then the restriction $h|_{\mathcal C(h)} : \mathcal C(h) \to S'$ a continuous function? [That is, with respect to the subspace topology $(\mathcal C(h), \tau_{\mathcal C(h)})$.]

I think the answer is yes. Here is my proof (attempt). Perhaps someone could check it.

Proof: If $\mathcal C(h) = \emptyset$, then $h|_{\mathcal C(h)}$ is the empty function to $S'$ and trivially continuous.

Now let $\mathcal C(h) \neq \emptyset$ and $x \in \mathcal C(h)$. Let $V$ be a neighborhood of $h(x)$. Since $x$ is a continuity point of $h$, there exists a neighborhood $U$ of $x$ such that $h(U) \subseteq V$. Then $U_0 := U \cap \mathcal C(h) \in \tau_{\mathcal C(h)}$ is a neighborhood of $x$ in the subspace topology $\tau_{\mathcal C(h)}$. Furthermore $h|_{\mathcal C(h)}(U_0) = h(U) \setminus h\big(\mathcal C(h)^c\big) \subseteq h(U) \subseteq V$.

Thus, for any neighborhood $V$ of $h|_{\mathcal C(h)}$, we can find a neighborhood $U_0 \in \tau_{\mathcal C(h)}$ of $x$ such that $h(U_0) \subseteq V$. Hence, $h|_{\mathcal C(h)}$ is continuous. $\blacksquare$

答:
It's enough to directly note the trivial inclusion $h\restriction_{\mathcal{C}(h)}[U_0] \subseteq h[U]$, the intermediate equality is not needed.

Or note:

If $x$ is a point of continuity of $h$, then it remains so in the restriction of $h$ to any subspace that contains $x$. So all points of $\mathcal{C}(h)$ are points of continity of $h\restriction_{\mathcal{C}(h)}$. So it is continuous on that subspace.

The proof of the first fact goes just as yours for the specific $\mathcal{C}(h)$:

Given $V \ni h(x)$ pick open $U \ni x$ that $h[U] \subseteq V$. But then $(h\restriction_A)[U \cap A]=h[A \cap U] \subseteq h[U] \subseteq V$ so $U \cap A \ni x$ witnesses the continuity at $x$ of $h\restriction_A$.


问:
Presheaves over sieves and posets
I'm looking for a proof of the claim that given a poset P, the topos of presheaves over P is equivalent to the topos of presheaves over the complete Heyting algebra of sieves on elements of P. I found this claim in John Bell's paper "causal sets and frame valued set theory". I get the impression that the proof is probably pretty trivial, but I can't quite see it at the moment. If anyone could either outline the proof or give a suitable reference, that would be great. </p>

答:
The claim is false. Let $P$ be the trivial poset with one element; then presheaves on $P$ are just sets. But the complete Heyting algebra of sieves in $P$ is the poset $\{ 0 < 1 \}$, and presheaves on this constitute a non-boolean topos. Here is the correct statement:

For any poset $P$, the topos of presheaves on $P$ is equivalent to the topos of sheaves on the complete Heyting algebra of sieves in $P$.

It is a consequence of standard facts in topos theory:

The presheaf topos on a poset is localic. [Sketches of an elephant, Example A4.6.2(d)]

The subobjects of $1$ in $[P^\mathrm{op}, \mathbf{Set}]$ are the sieves in $P$. [Exercise]

A localic topos is the topos of sheaves on the complete Heyting algebra $\mathrm{Sub}(1)$. [Sketches of an elephant, Theorem C1.4.7]

Title: Presheaves over sieves and posets


问:
Separate a compact and a closed subset in $\mathbb{R}^n$
Let $K\subset \mathbb{R}^n$ a compact subset. And $F\subset \mathbb{R}^n$ a closed subset such that
$$ K \cap F = \emptyset.$$
How can we deduce that there is $a>0$ such that
$$\| y-x\|>a,$$
for all $x\in K$ and $y\in F$.

答: On $K$, the function $x \to d(x,A)$ is continuous and real-valued (values in $(0,+\infty)$. It assumes a minimum by compacntess of $K$. QED.


问:
Riemann Zeta function: Integral representation in terms of Gamma function, interchange integration and sum?
It is well known, that the Zeta function has this integral formula for $\Re z>1$:

$\Gamma(z)\zeta(z)=\int _0^{\infty}\frac{t^{z-1}}{e^t-1} \ dt$

In the proof (attached in the link), we can interchange the integration and the sum:
$\sum_{n=1}^{\infty}n^{-z}\Gamma(z)=\sum_{n=1}^{\infty}\big(\int _0^{\infty}t^{z-1}e^{-nt} \ dt \big)=\int _0^{\infty}\big(\sum_{n=1}^{\infty} t^{z-1}e^{-nt}  \big) \ dt$.
Can anybody tell me why we can interchange the integration and the sum? How can we prove this? Which theorem justifies this? (Maybe Lebesgue's dominated convergence theorem?)

The proof from Garling book

Body:
For $\Re(s) > 1$
$$\sum_{n=1}^\infty n^{-s}\Gamma(s) = \lim_{N\to \infty}\sum_{n=1}^N \int_0^\infty x^{s-1}e^{-nx}dx= \lim_{N\to \infty}\int_0^\infty x^{s-1}\frac{1-e^{-Nx}}{e^x-1}dx$$

$$= \lim_{N\to \infty}\int_0^\infty \frac{x^{s-1}}{e^x-1}dx+O(e^{-N^{1/2}}\int_{N^{-1/2}}^\infty \frac{|x^{s-1}|}{e^x-1}dx)+O(\int_0^{N^{-1/2}} \frac{|x^{s-1}|}{e^x-1}dx)$$
$$ = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$


问:
Equilateral Triangle built from Pearls
Can someone find a solution to this problem:

You try to build an equilateral triangle with a number of pearls. Your pearls are placed the following way in order to solve the triangle:

$1$ pearl in the first row, $2$ in the second row, $3$ in the next, and so on, with as many rows as you may need to form an equilateral triangle.

In the first round you get $2008$ pearls to try and form this equilateral triangle and you see that it doesn‘t work. In the next round you get one more pearl, so $2009$, and again it does not work. Then you get two pearls, so $2011$, and it still doesn‘t work.
Every new round you get the amount of stones you got in the round before and $1$ more than that. (so after the try with $2011$ we get the $2+1=3$ pearls added to $2011$ for the next try).

How many equilateral triangles can you build this way, if you continue to get more and more pearls until the end of time.

So far I know that the amount of pearls we gain in addition to $2008$ is an arithmetic series with the difference of $1$ between its consecutive terms and so after $n$ rounds we have $2008 +\dfrac{n(n+1)}{2}$ pearls.

Where do I go from here?

答:
We can form a complete equilateral triangle only when the total number of pearls is a Triangular number (that is, expressible as $1+2+3+\ldots$).

So we're looking for solutions of
$$2008 + \frac{n(n+1)}{2}=\frac{m(m+1)}{2}$$
for positive integers $m,n$. On rearranging,
$$m^2-n^2+m-n=2008\cdot 2$$
$$\Rightarrow (m-n)(m+n+1)=4016=2^4 \cdot 251$$

which is easily solvable by observing that $m-n$ and $m+n$ have same parity for all integers $m,n$.

One solution would be $m=133$, $n=\ldots$

And there would be others.

Title: Equilateral Triangle built from Pearls


问:
Derivation of the identity $\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}=4\pi\delta^3(\vec{r})$
Let $\vec r=x\hat i+y\hat j+z\hat k$ a position vector and $\delta^3(\vec{r})=\delta (x)\delta (y)\delta (z)$ is the tree-dimensional Dirac's delta function, undertood in the sense of distributions (it is zero at all points except at the origin, where it equals infinite). I'm looking for a proof of the following identity,

$$\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}=4\pi\delta^3(\vec{r})$$

How could it be derived?

答: Since $\phi\vec{\nabla}\cdot\vec{A}=\vec{\nabla}\cdot(\phi\vec{A})-\vec{\nabla}\phi\cdot\vec{A}$, we want to prove scalar test functions $f$ satisfy$$\int_\Omega\frac{\vec{r}}{r^3}\cdot\vec{\nabla} f(\vec{r})d^3\vec{r}=-4\pi f(\vec{0}),$$where $\Omega$ denotes $3$-dimensional space, which we'll parameterize with spherical polar coordinates so the integral is$$4\pi\int_0^\infty\frac{\hat{r}}{r^2}\cdot\vec{\nabla} f(\vec{r})r^2dr=4\pi\int_0^\infty\frac{\partial f}{\partial r}dr=-4\pi f(\vec{0}).$$


问:
Why $(nkn^{-1})k^{-1}=n(kn^{-1}k^{-1})\in N\cap K$?
I'm reading Hungerford's Algebra and trying to understand the proof of $(iv)$:

I'm a bit confused, in the end of the proof, why $(nkn^{-1})k^{-1}=n(kn^{-1}k^{-1})\in N\cap K$? I understand this looks like the condition for normality but it's shifted in a weird way.

答:
$(nkn^{−1})k^{−1}$ lies in $K$ since $K$ is normal, and

$n(kn^{−1}k^{−1})$ lies in $N$ since $N$ is normal.

Since both expressions are the same, they lie in the intersection $N\cap K$.


问:
Gaussian estimate
I am trying to find some $a\gt 0$ such that $$\int_{-2}^{2}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt\leq e^{-a}$$ Any good method to do so?
I'm looking for an analytical argument rather than a numerical solution.
Edit: there was a typo in the integral.

答:
To make the problem general, you search for $a$ such that, for a given $x$,
$$\int_{-x}^{+x}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt\leq e^{-a}$$ that is to say
$$\text{erf}\left(\frac{x}{\sqrt{2}}\right)\leq e^{-a}$$ First, we can solve the equation using some approximation. For example
$$\text{erf}(t)\sim \sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha\, t^2}{1+\beta \,t^2}\,t^2 \Big)}$$ where
$$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi } \qquad \text{and} \qquad \beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$ so
$$a > -\frac 12 \log \Bigg[1-\exp\Bigg(-\frac{2 x^2 \left(\alpha  x^2+2\right)}{\pi  \left(\beta  x^2+2\right)} \Bigg)\Bigg]$$

Using $x=2$, this approximation gives $a >0.04673$ while the exact solution would be $a >0.04657$.

We can have much better using the second approximation I gave here.

Title: Gaussian estimate


问:
Using mathematical induction to show that a binary tree of height $h$ has no more than $2^h$ leaf nodes

Use mathematical induction to show that a binary tree of height $h$ has no more than $2^h$ leaf nodes.

I'm familiar with mathematical induction proofs, but I haven't encountered one like this. I'm not sure how to set up the base case.

I thought about letting $n$ = the number of leaf nodes, then showing by induction that $n \le 2^h$. This didn't seem right though, because all of the proofs I have done by induction have been in terms of one variable. Any hints?

答:
I thought about letting $n =$ the number of leaf nodes, then showing by induction that $n≤2^h$. This didn't seem right though, because all of the proofs I have done by induction have been in terms of one variable.

Indeed, that does not seem right. Generally, the variable of induction is something you can grasp already, not something you are trying to understand (as the number of nodes here). The correct approach is to induct on $h$, starting with the base case $h=0$ and proceeding with the observation that adding another step can at most double the number of leaves.


问:
Can you project unto closed subspaces of normed spaces that are not necessarily pre-Hilbert?
I'm working through some notes for my signal processing class and they introduce the whole notion of pre-Hilbert spaces (inner product spaces) essentially only in order to be able to project elements onto closed subspaces.

My question is: suppose $V$ is a general normed space (not necessarily derived from an inner product) and $U \subset V$ is a complete subset, can we not use the same proof to show that $\forall v \in V, \exists u \in U$ such that $||u-v||$ is minimal? In that case, the only thing the inner product does is to make the projection orthogonal, right?

答:
You can always project onto any subspace at all. Namely, given any $W\leq V$, you can choose an algebraic basis $B_1$ of $W$ and extend it to a basis $B=B_1\sqcup B_2$ of $V$. Then for any $v\in V, v=\sum_{b_1\in B_1} a_{b_1}b_1+\sum_{b_2\in B_2}a_{b_2}b_2$, you can put $F(v)=\sum_{b_1\in B_1} a_{b_1}b_1$, which gives you a projection $V\to W$, and any projection is of this form.

If $V$ is a topological vector space, e.g. a normed space, then such a projection will typically not be continuous, however, and $F(v)$ may not minimize the distance from $v$ to $W$ (even if $F$ happens to be continuous with respect to some given metric). In general, there may be no continuous projection at all, not even if $W$ is a closed (or complete, when it makes sense) subspace of $V$.

If $V$ is an arbitrary normed space, and you try to mimic the proof you mention, it does does not work: given any $v\in V$, even if you choose a sequence $w_n\in W$ such that $\lvert v-w_n\rvert<1/n+\inf_{w\in W}\lvert v-w\rvert$, the sequence $w_n$ need not be Cauchy. For example, if $V$ is the plane with the $L^\infty$ norm and $W$ is spanned by $(0,1)$, then for $v=(1,0)$, you can choose $w_n=(0,(-1)^n)$ and this is clearly not Cauchy, even though $\lvert v-w_n\rvert=1=\inf_{w\in W}\lvert v-w\rvert$. In this case, a distance-minimizing projection does exist, but it is not unique.


问:
Why do we "plug" the approaching value when solving limits?
This is a question regarding basic limits.

If we have

$\lim_{x\to3} (2x + 2)$

then,

$\lim_{x\to3} (2x + 2) = 2(3) + 2 = 8$

Now, we know that x approaches 3, but never actually gets to 3 (correct?). So, if x never gets to 3, then why do we use the approaching value itself when solving the limit, when it's just close to 3, but not equal?

答:
When we say that $8$ is the limit of the function $x \mapsto (2x + 2)$ when $x$ approaches 3, what we mean is that the value of $2x + 2$ gets closer and closer to $8$ as we let $x$ move closer and closer to $3$. More specifically, it says that if we pick any sequence $x_1, x_2, x_3, \dots$ of real numbers converging to $3$, the sequence $2x_1 + 2, 2x_2 + 2, 2x_3 + 3, \dots$ converges to $8$. The whole point of taking a limit rather than just evaluating some function at a point is that the number $8$ does not actually need to appear anywhere in the sequence. For example, the limit of the sequence
$$ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \dots$$ is 0, even though 0 never appears in this sequence.

Nevertheless, as pointed out in the comments, it frequently happens that we are taking the limit $\lim_{x \to a}f(x)$ of some continuous function $f: \mathbb{R} \to \mathbb{R}$, where $a$ is some fixed real number at which we want to know the limit (so $a = 3$ in your case.) In this case, it follows essentially by definition of continuity that if we choose $x$ to be closer and closer to $a$, then $f(x)$ gets closer and closer to $f(a)$. So in this case, the limit can be computed very easily by simply plugging in $a$ into your function $f$.

In your case, the function $x \mapsto 2x + 2$ is continuous (since if you change $x$ by a small amount $\epsilon$, the value of $2x + 2$ will change by at most $2\epsilon$) and thus $\lim_{x \to 3} 2x + 2 = 2\cdot 3 + 2 = 8$.


问:
Help with recursive definition for satisfaction in FOL
In Enderton pgs 83-84, he provides a recursive definition for satisfaction in FOL. I am having trouble understanding the clause regarding quantified sentences in FOL.

Let $s:\mathcal{V}\longmapsto|\mathfrak{A}|$, where $\mathcal{V}$ is the set of all variables and $|\mathfrak{A}|$ is the domain of the structure $\mathfrak{A}$.

The clause relating to quantified sentences runs as follows:

$\vDash_\mathfrak{A}\forall x\phi[s]$ iff for every $d\in|\mathfrak{A}|,\vDash_\mathfrak{A}\phi s(x|d) $

And this is how Enderton explains $s(x|d)$:

$s(x|d)$ is exactly like $s$ except for one thing: At the variable $x$, $s(x|d)$ assumes the value $d$.

Also, for clarity, earlier he says informally:

$\vDash_\mathfrak{A}\phi[s]$ iff the translation of $\phi$ determined by $\mathfrak{A}$, where the variable $x$ is translated as $s(x)$ wherever it occurs free, is true.

Intuitively, I understand perfectly what satisfaction requires when it comes to universally quantified sentences. Where I am struggling is on the role played here by $s(x|d)$ in the recursive definition. $s$ is a function that maps all of the variables onto members of the domain. So, the value of $s(x)$ is a member of the domain. And according to Enderton $s(x|d)$ does the same thing. That is, for all variables $y$, the value of $s(x|d)(y)$ is also a member of the domain. Indeed the same member $s(y)$ returns for that variable, unless the variable is $x$, then $s(x|d)(x)=d$.

Part of my confusion is that I don't really understand what $s(x|d)(x)=d$ means. Is $d$ one of the variables here, or is it one of members of the domain? And if the latter, which one? And lastly, I don't see how this move mechanically achieves the intuitive notion of satisfaction for quantified sentences in FOL.

Thanks for all the help. I really appreciate it!

答:
I don't really understand what $s(x|d)(x)=d$ means.

Recall that $s : \text {Var} ⟼ |\mathfrak A|$, where $\text {Var}$ is the set of all variables and $|\mathfrak A|$ is the domain of the structure $\mathfrak A$.

Thjus, $s$ is a function that maps variables into objects of the domain.

What is $d$? An object of the domain: $d \in |\mathfrak A|$.

Thus, if we have a variable assignment function $s$, we will have that:

$s(x|d)(y)=s(y)$ for $y \ne x$ and $s(x|d)(x)=d$.

This is the gist of: "$s(x|d)$ is exactly like $s$ except for one thing: at the variable $x$ it assumes the value $d$."

How this move achieves the intuitive notion of satisfaction for quantified sentences in FOL?

Exactly because we want that $\forall x P(x)$ is true in a structure when the "meaning" of the predicate $P$ holds of every object in the domain of the structure.

Thus, whatever is the value of $s$ for variable $x$, we want that $s$ satisfies $P(x)$ in the structure.


问:
For any metric $g_{ab}$ show that $g^{ab}g_{ab}=N$ where $N$ is the dimension of the manifold
I'm bugged by this question of my physics book... As far as I know, shouldn't the product $g^{ab}g_{ab}$ be precisely the $N\times N$ identity matrix (given the fact that, expressing the contravariant and covariant metric tensor as matrices, they are the inverse matrix of each other)?

答:
If you don't use the Einstein notation, you'll see that
$$ g^{ab}g_{ab}=\displaystyle\sum_{a=1}^{\dim M}\sum_{b=1}^{\dim M}g^{ab}g_{ab},$$
where $M$ is your manifold. As you said, $g^{ab}$ are the coefficients of the inverse of $(g_{ab})$. In other words,
$$\sum_{k=1}^{\dim M}g^{ak}g_{bk}=\delta^a_{\;b},$$
where $\delta^a_{\;b}$ is the Kronecker delta. It follows that
$$ g^{ab}g_{ab}=\sum_{a=1}^{\dim M}\delta^a_{\;a}=\sum_{a=1}^{\dim M} 1=\dim M.$$


问:
Residues of Elliptic functions and non-isolated singularity
Consider a function $f(z,q)$ ($|q| < 1$) meromorphic (at least) in the region $|z| < 2$, with a periodicity $f(zq,q) = f(z,q)$. So if one replace $a = e^{2\pi i \mathfrak{a}}$, $f$ is an elliptic function with modulus $\tau$ given by $q = e^{2\pi i \tau}$.

Suppose $f(z,q)$ has a simple pole at $z = z_0$ (and therefore an infinite series of poles $z_0q^k$, and an additional pole to cancel the residue) leading to a non-isolated singularity at $z = 0$.

My question is: is there any relation between $\operatorname{Res}_{z = z_0}z^{-1}f(z,q)$ and the "$\operatorname{Res}_{z = 0}z^{-1}f(z,q)$ " defined by first expanding $f(z,q) = \sum_{n} f_n(z)q^n$ and
$$
\begin{align}
  \operatorname{Res}_{z = 0} \frac{ 1 }{ z }f(z,q) \equiv \sum_n \Bigg[   \operatorname{Res}_{z = 0}\frac{ 1 }{ z }f_n(z)  \Bigg]q^n \ ?
\end{align}
$$

In certain physical problem, we encounter such a situation where the two residues satisfy common  differential equations, and seem to be related by modular $S$-transformation. But we have no idea why it is the case.

答:
There no residues for non-isolated singularities. If $f$ is meromorphic on $\Bbb{C}^*$ and $f(z)=f(zq)$ with $|q|<1$ then for any $p,r\in (|q|,1),k\in \Bbb{Z}$, $$I=\int_{|s|=p} f(s)\frac{ds}{is}=\int_{|s|=r q^k} f(s)\frac{ds}{is}$$
This is telling the $z^{-1}$ coefficient of the Laurent exansion of $f$ on any annulus $a<|z|<A$ not containing a pole.

Let $F(u)=f(e^{2i\pi u})$ which is meromorphic $\Bbb{Z+\frac{\log q}{2i\pi} Z}$ periodic, then $I=\int_0^1 F(u) du$, in general there is no closed-form for it, take a look at Weierstrass_eta_function which is $I$ when $F=\wp$.

We can find $I$ from the negative Laurent coefficients $c_{j,l}$ at the poles $b_j$ plus one constant Laurent coefficient and some $\zeta(b_i-b_j;\Lambda)$ constants, this is because $F-\sum_j \sum_{l=1}^{e_j} c_{j,l}\zeta(z-b_j;\Lambda)$ is entire and doubly periodic thus constant.

When the $b_j \in \Bbb{Q}\Lambda$ there are some algebraic relations and $I$ is algebraic in the $c_{j,l}$ and the $g_2,g_3$ parameters of the lattice.

Feel free to clarify what is your $f$ and what you'd like to do with $I$.


问:
injectivity radius of hyperbolic manifolds
It is known that one can construct hyperbolic surfaces that have arbitrarily large injectivity radii. Is this true for higher dimensional hyperbolic manifolds? In particular I'm interested in the case of compact hyperbolic manifolds of dimension $n>4$. </p>

答: Yes, you can do it just by rescaling. Let $(M,g)$ be a compact  hyperbolic manifold of dimension $n$. By compactness, it has a positive injectivity radius $r>0$. Given $R>0$, consider the manifold $(M,(R/r)g)$. Its injectivity radius is $R$.</p>


问:
Probability of Brownian Motion touching a circle
Currently, I am reading a paper about planar Brownian Motion. However, I am just getting to know Brownian Motion so I am not familiar with all well-known properties of BM or stochastic processes.

I am looking at a Brownian Motion that is somewhere in a neighborhood of several circles $\partial B_{r_i}(z_0), z_0\in\mathbb R^2, \ i=1,\dots,n,$ such that $r_1<\dots<r_n.$ The author says that when $BM$ starts on a circle $\partial B_{r_i},$ the probability BM hits $\partial B_{r_{i+1}}$ before $\partial B_{r_{i-1}}$ is $\frac 1 2$.

Although this seems like quite a simple statement, I don't really why this is. My intuition says that if the distances i.e. the $r_i$ were the same, then this statement is clear. However, the $r_i$ don't necessarily have to be equal.

If you could point out a few arguments for this, I'd be really glad. Any help is appreciated.

Edit: Upon request; the values are given as $r_i = R (\frac{\epsilon}{R})^\frac{i}{K}$ where $R\in(0, \frac 1 2)$, $K \geq 1$ and $i = 0,1,\dots,K.$

答:
Let $0<y<|z|<w<\infty$ with $z\in\mathbb{R}^2$, and $z_0=0$. Using the fact that $u(x)=\ln|x|$ is harmonic on $\mathbb{R}^2\setminus \{0\}$ and Ito's lemma, one can show that
$$
N_t:=u\!\left(B_{t\wedge T_y}\right)-u(z)=\sum_{i=1}^2\int_0^t\frac{\partial}{\partial x_i} u(B_{s\wedge T_y})\,d B_{s\wedge T_y}^i,
$$
where $T_r:=\inf\{t\ge 0:|B_t|=r\}$, is a local martingale null at 0. Stopping $N_t$ at $T_w$, one gets a uniformly integrable martingale null at 0 which converges in $L_1$. Therefore,
$$
u(z)=\mathsf{E}^z[u(B_{T_y\wedge T_w})]=\mathsf{P}^z(T_y<T_w)\ln y+\mathsf{P}^z(T_y\ge T_w)\ln w,
$$

and so, when $B_t$ starts at $z$,
$$
\mathsf{P}^z(T_y<T_w)=\frac{\ln w-\ln|z|}{\ln w-\ln y}.
$$

In your case, $r_i=R(\varepsilon/R)^{i/K}$ for $R\in(0,1/2)$ and some $K\in\mathbb{N}$ (see Equation (5) on page 2). So
$$
\frac{\ln r_{i+1}-\ln r_i}{\ln r_{i+1}-\ln r_{i-1}}=\frac{1}{2},
$$
as required.


问:
$\int \frac{1}{x^2\sqrt{x-1}}\; dx$
I am trying to do the following integral
$$
\int \frac{1}{x^2\sqrt{x-1}}\; dx
$$
My first thought is to do a substitution $u = x - 1$ so that I get
$$
\int \frac{1}{(u + 1)^2\sqrt{u}}\; du
$$
The maybe another substitution $t = \sqrt{u}$, so that I get
$$
\int \frac{1}{(t^2 + 1)^2} 2\;dt
$$
This might then be a trig. sub., but I don't think it is supposed to be this hard. Help please.

答: After your calculations, you just have to do the substitution $t = \tan(s)$ and it is straightforward

Title: $\int \frac{1}{x^2\sqrt{x-1}}\; dx$


问:
Find all similar matrices to (alpha)I
Well the question is in the header - find all the similar matrices to skalar * I.
I just can't see what i am supposed to do and i didn't find a similar question over the internet so i thought i will ask here maybe someone know the method to solve this type of question.</p>

答:
Two matrices $A,B\in k^{n\times n}$ ( $n\in\mathbb{N}$ and say $k$ a field) are similar if there exists an invertible matrix $P\in k^{n\times n}$ such that $A=P^{-1}BP$. Now for $\alpha\in k$ and $I$ the identity matrix this means every similar matrix is of the form:
$$P^{-1}\alpha I P\overset{*}=\alpha P^{-1}IP=\alpha P^{-1}P=\alpha I,$$
where $*$ holds by the compatibilty with scalars rule, which holds since $k^{n\times n}$ is an algebra over $k$.

Title: Find all similar matrices to (alpha)I


答:
You can combine your two substitutions into one by setting $x-1=t^2$.

$$2\int \frac{1}{(t^2+1)^2} dt$$

Now letting $t=\tan \theta$, the above simplifies to
$$2\int \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta =\int 2\cos^2 \theta \, d\theta =\int 1+\cos 2\theta \, d\theta$$
$$=\theta + \frac{\sin 2\theta}{2}+C$$
It remains to convert the above into an expression of $x$.

Title: $\int \frac{1}{x^2\sqrt{x-1}}\; dx$

Body:
I am trying to do the following integral
$$
\int \frac{1}{x^2\sqrt{x-1}}\; dx
$$
My first thought is to do a substitution $u = x - 1$ so that I get
$$
\int \frac{1}{(u + 1)^2\sqrt{u}}\; du
$$
The maybe another substitution $t = \sqrt{u}$, so that I get
$$
\int \frac{1}{(t^2 + 1)^2} 2\;dt
$$
This might then be a trig. sub., but I don't think it is supposed to be this hard. Help please.


问:
proving that a certain function defined using integrals is continuous
How can I prove that the function

$$f(x)=\int_{0}^{\pi}\frac{\sin(xt)}{t}dt $$

I'm very confused here, because the variable "x" appears inside the integral, and not in the form $\int_{0}^{x}$ as always, I don't know how can I do here. Please help me!

答:
Hint:
$$
\int\limits_0^\pi\frac{\sin(xt)}{t}dt
=\int\limits_0^\pi\frac{\sin(xt)}{xt}d(xt)
=\int\limits_0^{\pi x}\frac{\sin(s)}{s}ds
$$


问:
How to solve following recurrence relation
I need to solve the following recurrence relation:

$$
\begin{align*}
T(1) &= c_1\\
T(2) &= c_2\\
T(n) &= T(\lfloor\frac{n+2}{2}\rfloor) + c_3,
\end{align*}
$$

where $c_1$, $c_2$ and $c_3$ are constants. The thing is, I really don't know how to battle the $T(\lfloor\frac{n+2}{2}\rfloor)$ part. I don't need an explicit answer, I just need to know wherever it will be logarithmic (i.e. can be upper bounded by $\mathcal{O}(\log{}n)$). I'll be glad for any ideas how to prove/disprove it.

答:
For simplicity assume $n=2^k,k\in \mathbb \{3,4,\cdots \}$. Then $$\left\lfloor\frac{n+2}{2}\right\rfloor=\left\lfloor\frac{n}{2}+\frac12\right\rfloor=\left\lfloor\frac{2^k}{2}+\frac12\right\rfloor=\left\lfloor2^{k-1}+\frac12\right\rfloor=\lfloor2^{k-1}\rfloor$$
So \begin{align*}
T(n)&=T(2^{k-1})+c_3=T(2^{k-2})+2c_3=T(2^{k-3})+3c_3+\cdots =T(2)+kc_3\\
&= c_2+kc_3=c_2+c_3\log n=O(\log n)
\end{align*}


问:
Question about large deviation bounds
Suppose $X$ is some random variable. This question may be vague , but is it always possible to find a large deviation bound of the following form that would work for ALL $s\geq 0$.

i.e $P(X \geq \mathbb{E}[X] + s) \leq \left( \right)$

More importantly, the bound $\left( \right)$ must be something that decays exponentially (either sub-exponential or sub-Gaussian).

I have seen in literature that for certain distributions, existence of such a bound is possible only for $s \geq \mathbb{E}[X]$ (especially Chernoff bounds) so I'm not sure why those bounds can't be extended for all $s\geq 0$.

Any comments on this will be appreciated!

答:
No, it is not always possible to get such a bound. Example: $X$ is a random variable with density

$$
f(x) = \frac{2}{x^3}1(x>1).
$$

You can see for any $t >1$
$$
P(X\geq t) = \frac{1}{t^2}.
$$
You can choose $t=s+\mathbb{E}(X)$ for any $s\geq 0$.

Title: Question about large deviation bounds


问:
Local field extension and its residue extension
If an extension $K/L$ of nonarchimedean local fields is finite Galois, then we have an extension of their residue fields $\kappa/\lambda$. Then we can define the extension to be unramified if their degrees coincide. Do we always have an extension of residue fields $\kappa/\lambda$ for any (finite) extension of nonarchimedean local fields?

答: If $K$ is an extension of $L$, then the rings of integers $\mathcal O_K$ is an extension of $\mathcal O_L$, so we have an embedding $\lambda = \mathcal O_L/\mathfrak m_L \to \mathcal O_K/\mathfrak m_K = \kappa$.


答:
As we have $u(x, 2x) = x$, the chain rule gives
$$ u_x(x,2x) + 2u_y(x,2x) = 1 $$
As $u_x(x,2x) = x^2$, we have $u_y(x, 2x) = \frac{1-x^2}{2}$. Starting with $u_x(x,2x) = x^2$, the chain rule gives
$$ u_{xx}(x,2x) + 2u_{xy}(x,2x) = 2x \tag 1 $$
and starting with $u_y(x, 2x) = \frac{1-x^2}{2}$, we get
$$ u_{xy}(x,2x) + 2u_{yy}(x,2x) = -x \tag 2$$
Subtracting (2) twice from the equation (1), we get
$$ u_{xx}(x,2x) - 4u_{yy}(x,2x) = 4x $$
As $u_{xx} = u_{yy}$, we have
$$ u_{xx}(x,2x) = u_{yy}(x,2x) = -\frac 43x $$
And hence, by (2)
$$ u_{xy}(x,2x) = -x + \frac 83 x = \frac 53 x. $$</p>

Title: the partial derivatives computing

Body:
Given that:
$u_{xx}^{''}=u_{yy}^{''} \;,  \;
u(x,2x)=x \;,  \;
u_{x}^{'}(x,2x)=x^{2}\;,  \;
$
How to find the following values?
$u_{xx}^{''}(x,2x)=?\;
u_{xy}^{''}(x,2x)=? \;
u_{yy}^{''}(x,2x)=?\; $

Thanks a lot.


问:
Shouldn't the probability of sampling a point from a continuous distribution be 0?
Hey I was reading about Gaussian EM algorithm in which you first calculate the likelihood of data points being sampled from a Gaussian and then adjust your mean and variance to maximize it. To calculate the probability of a point being sampled from a distribution we do
$$
P(x_i\mid\theta) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-(x_i-\mu)^2}{2\sigma^2}\right)
$$
But how can you sample a point from a continuous distribution? Shouldn't this be zero? Moreover I read many times about sampling a point/data from some continuous distribution but I can't understand how could you do that as for continuous random variable X the $P(X=x_1) = 0 $, so how could yo sample data point(s) from a continuous distribution? Or does this sampling have some other meaning. I've seen many questions on this platform but I couldn't get my answer.

答:
On digital computers, it is certainly impossible to sample from a truly continuous distribution; i.e. a distribution that contains infinite possible "discrete" outcomes. One would have to wait an infinite amount of time to compute every one of the infinite decimals. In practice, we chop up the continuous distribution into a discrete number of possibilities and sample from those, and make a case that the discrete distribution adequately represents the continuous distribution.

The simplest example might be a random sampling from a continuous, uniform distribution between values 0 and 1. If we split the interval (0,1) uniformly (chunks of dx) into the infinite number of possibilities, the probability of picking a particular $X$ from $X \in (0,1)$ is

$P(X) = \lim_{dx \to 0} dx=0$

But there's an integral hidden in the denominator:

$\int_{0}^{1}dx = 1$

The integral adds up an infinite number of $dx \to 0$ to get probability 1 of picking any number between 0 and 1.

Same thing happens in the numerator for the infinite number of $dx \to 0$ in a finite subinterval, resulting in a non-zero probability for finite subintervals.


问:
Necessary condition to be closed Irreducible subspace

Let $X=\text{Spec}(R)$ where $R$ is a commutative ring with unity. If $Y\subset X$ is a closed subspace that is irreducible, then $Y = V(P)$ for some $P\in X$.

Proof. Since $Y$ is closed, $Y = V(I)$ for some ideal $I\subset R$. Need to show $I$ is prime ideal. Let $f,g\notin I$. Then consider $D(f)\cap Y$ and $D(g)\cap Y$. Assume they are both nonempty. Then they are nonempty open sets in $Y$ and as $Y$ is irreducible, $D(f)\cap D(g)\cap Y = D(fg)\cap Y$ is nonempty. But if $P\in D(fg)\cap Y$ then $I\subset P$ and $fg\notin P$ which implies $fg\notin I$.

The problem is I don't know how prove  $D(f)\cap Y$ is nonempty. I don't know if it's helpful information but I found that such $f,g$ are non-units.
But still don't know. Could you give any hint?

答: You need to assume $I$ to be a radical ideal (since $V(I) = V(\sqrt{I})$). Then notice that $\sqrt{I} = \bigcap_{\mathfrak p \in V(I)} \mathfrak p$, therefore $f\notin I$ iff there is a $\mathfrak p\in V(I)$ with $f\notin \mathfrak p$.


问:
Spherical coordinate system
I can easily write $z$ axis value is $r\cos\theta$ but what will be for $x$ and $y$ axis, explain a bit please.

From the above how can I write the area element as $d\vec{a} = r^2\sin\theta d\theta d\phi\hat{r}$?

答: The piece of the sphere with radius between $r$ and $\mathrm{d}r$ etc. has a shape tends towards a cuboid as the size of the piece cut decreases, with height $\mathrm{d}r$, width $\mathrm{d}\theta$ and length $r\sin\theta\mathrm{d}\phi$. Its volume is $r^2\sin\theta \mathrm{d}\theta \ \mathrm{d}\phi\mathrm{d}r$. </p>

Title: Spherical coordinate system


问:
function in $\mathbb{Q}$ with positive derivative
A function $f: \mathbb{Q} \to \mathbb{Q}$ is differentiable over $\mathbb{Q}$ with $f'(q)$>$0$ for every $q$ in $\mathbb{Q}$.
Normaly when the derivative is positive for everything in the domain the function is rising. I'm looking for an counter example that this isn't always for a function defined in $\mathbb{Q}$.

答:
Consider the function

$$f(q)=\begin{cases}q, & \textrm{if }q^3<2 \\ q-1, & \textrm{if }q^3\ge2.\end{cases}$$

We have
$$f'(q)=\lim_{h\to 0}\frac{f(q+h)-f(q)}{h}=1 \quad \forall q \in \mathbb{Q}.$$

However, $f(q)$ is not monotonically increasing.


答:
Hint:

Set $t=\sqrt{x-1}$, so that $x^2=1+t^2$ and therefore $\;\mathrm dx=2t\,\mathrm dt$.

The integral becomes $$\int\frac{2t\,\mathrm dt}{(1+t^2)^2 t}=2\int\frac{\mathrm dt}{(1+t^2)^2}.$$

Now the integrals $\;I_n=\displaystyle\int\frac{\mathrm dt}{(1+t^2)^n} $ are standard and computed recursively in the following way:

$I_1=\arctan t$.

For the inductive step, use integration by parts with $I_{n-1}$, setting
$$u=\frac1{(1+t^2)^{n-1}}\enspace\mathrm dv=\mathrm dt,\enspace\text{whence }\quad\mathrm du=-\frac{2(n-1)t\,\mathrm dt}{(1+t^2)^n},\quad v=t,$$
and ultimately
\begin{align}
I_{n-1}&=\frac t{(1+t^2)^{n-1}}+2(n-1)\int\frac {t^2\,\mathrm dt}{(1+t^2)^n}=\frac t{(1+t^2)^n}+2(n-1)\int\frac {(1+t^2)-1}{(1+t^2)^n}\,\mathrm dt \\
&=\frac t{(1+t^2)^{n-1}} +2(n-1)(I_{n-1}-I_n) \\
\implies I_n&= \frac t{2(n-1)(1+t^2)^n}+\frac{2n-3}{2n-2}I_{n-1}.
\end{align}

Title: $\int \frac{1}{x^2\sqrt{x-1}}\; dx$


问:
De re/dicto knowledge
Define $\mathcal{K}$ as a knowledge operator characterized by a S4 modal system. There is a distinction between de re and de dicto expressions of knowledge:

$\exists x\mathcal{K}A(x)$ is a de re expression of knowledge: there exists $x$ such that the agent knows $A(x)$.

$\mathcal{K}\exists xA(x)$ is a de dicto expression of knowledge: the agent knows that there exists $x$ such that $A(x)$.

As is usually assumed in the literature, de re knowledge entails de dicto knowledge (but usually not the other way around):
$$\exists x\mathcal{K}A(x)\rightarrow\mathcal{K}\exists xA(x)\tag{$*$}$$

My question is the following: given the above definitions and $(*)$, can we prove $(**)$?
$$\mathcal{K}(\exists xA(x)\rightarrow\exists yB(y))\rightarrow(\exists x\mathcal{K}A(x)\rightarrow\exists y\mathcal{K}B(y))\tag{$**$}$$

I tried to use the distribution axiom for $\mathcal{K}$ which is available from the S4 system, but I cannot see how to proceed. Can anyone help? Is $(**)$ provable or not?

答:
given the above... can we prove $(**)$?

No, nor can we prove its negation.

Here are the countermodels, generated using this tool, courtesy of one Wolfgang Schwarz:

Let $W$ be the set of worlds, $D$ the set of individuals, $R$ the accessibility relation on $W$, and $A'$ and $B'$ the interpretations of the relation symbols $A$ and $B$, respectively.

Then, for

$$\begin{array}
\ W=\{w_0,w_1\}\\
D = \{0,1\}\\
A'=\{(0,w_0),(0,w_1)\}\\
B'=\{(1,w_0),(0,w_1)\}\\
R = W\times W
\end{array}$$

we have $w_0\Vdash\neg(**)$, and for

$$\begin{array}
\ W=\{w_0\}\\
D = \{0\}\\
A'=\{(0,w_0)\}\\
B'=\{(0,w_0)\}\\
R = W\times W
\end{array}$$

we have $w_1\Vdash(**)$.

However, $(**)$ is provable with the addition of the Buridan formula ($\mathsf{BuF}$) to the axioms. Here is the proof:

$$\def\knows{\mathcal K}
\begin{array}
\ 1. & \knows\exists xAx\to\exists x\knows Ax &\mathsf{BuF}\\
2. & \knows\exists yBy\to\exists y\knows By & \mathsf{BuF}\\
3. & \knows(\exists xAx\to\exists yBy)\to(\knows\exists xAx\to\knows\exists yBy) & \mathsf{K}\\
4. & \quad \knows(\exists xAx\to\exists yBy) & \text{assumption}\\
5. & \quad \knows\exists xAx\to\knows\exists yBy &\mathsf{MP}\ 3,4\\
6. & \quad \quad \knows\exists x Ax &\text{assumption}\\
7. & \quad \quad \knows\exists yBy &\mathsf{MP}\ 5,6\\
8. & \quad \quad\exists y\knows By &\mathsf{MP}\ 2,7\\
9. & \quad \knows \exists x Ax\to\exists y\knows By & \mathsf{DT}\ 6-8\\
10. & \quad \exists x\knows Ax\to\knows\exists xAx & (*)\\
11. & \quad \quad \exists x\knows Ax & \text{assumption}\\
12. & \quad \quad \knows\exists x Ax & \mathsf{MP}\ 10,11\\
13. & \quad \quad\exists y\knows By & \mathsf{MP}\ 9,13\\
14. & \quad\exists x\knows Ax\to\exists y\knows By & \mathsf{DT}\ 11-13\\
15. & \knows(\exists xAx\to\knows\exists yBy)\to(\exists x\knows Ax\to\exists y\knows By) & \mathsf{DT}\ 4-14
\end{array}$$

Title: De re/dicto knowledge


问:
Existence of a cut-off function
I have a question about the existence of a "cut-off" function with a specific purpose:

Question. Let $g:\mathbb R\to \mathbb R$ be a locally bounded function. Does there exist a sequence of functions $(f_k)_{k\geq 1}\subset C^\infty_c(\mathbb R)$ such that $f_k\geq 0$ and $f_k\to 1$ almost everywhere and $g f'_k\to 0 $ uniformly on $\mathbb R$?

Alternatively, it is enough to find a bound of the form $|gf'_k|\leq a_k$ with $a_k\to 0$. In my attempt below, I try $a_k=Ck^{-1}$, but I fail.

My attempt.
One can easily construct a sequence $f_k\to 1$. Indeed, let $\varphi:\mathbb R\to\mathbb R$ be such that $\varphi(x)=1$ for all $|x|\leq \frac 1 2$ and $0$ if $|x|\geq 1$ and for $\frac 1 2 \leq |x|\leq 1$ a smooth transition from $1$ to $0$. So I tried to adapt this as follows:
$$f_k(x)=\varphi(x/k)\exp\left(-b_kx^2\right)$$
for some sequence $(b_k)_{k\geq 1}$  to work against the supremum of $g$ on $\{|x|\leq k\}$. Let us differentiate and see what happens
$$ f_k'(x)=\left(\frac 1 k  \varphi'(x/k) -2b_kx \right)\exp\left(-b_kx^2\right)$$
Of course $\| \varphi'\|_\infty<\infty$. So for instance to ensure $|g f_k'|\leq k^{-1}C$, I thought maybe I can start with trying to get
$$\frac 1 k \|\varphi'\|_\infty |g(x)|\exp(-b_kx^2)\leq Ck^{-1}, \ \  \ \text{ on } \{k/2\leq |x|\leq k\},$$
becaue $\varphi'$ is zero on $k/2\leq |x|$. But  at the same time I also need to have
$$2|b_kxg(x)|\exp\left(-b_kx^2\right)\leq Ck^{-1}, \ \  \ \text{ on } \{|x|\leq k\}.$$
I think it is difficult. Even when $g$ has linear growth, like $|g(x)|\leq M(1+|x|)$. The problem is, if you ensure the first inequality, you cannot ensure the other inequality.

Other than this, I know there are many localization functions in PDE theory, but I could not find the one for my purpose. I would appreciate if one has a reference for an example. Also, if it is difficult, can we do it for $g$ with linear growth?

Note. My approach works if $|g(x)|\leq M(1+|x|^\alpha)$ with $\alpha\in [0,1)$ by taking $b_k\equiv 0$. This is, unfortunately, the best I could achieve.

Short motivation. This problem arises in localizing coefficients from diffusion operator to prove a result on equality of one-dimensional distributions of SDEs with locally bounded coefficients. It is very long proof and I am stuck with this little technical result.

答:
I claim this is not true in general. For this consider
$$ g(x) = (1+|x|^{1+\varepsilon}) $$
for any $\varepsilon > 0.$ The main point is that
$$ A(\varepsilon) := \int_0^{\infty} \frac1{g(x)} \,\mathrm{d}x < \infty. $$
As you will see, this argument works more generally for any non-negative $g$ whose reciprocal is integrable.

Suppose there is a sequence $f_k \in C^{\infty}_c(\Bbb R)$ such that $f_k \to 1$ almost everywhere and $gf_k' \to 0$ uniformly in $\Bbb R.$ Then we know there is some point $x_0 \geq 0$ where $f_k(x_0) \to 1$ as $k \to \infty.$ Then there is $k_0$ such that for all $k \geq k_0$ we have $f_k(x_0) \geq \frac12$ and
$$ g(x) |f_k'(x)| \leq \frac1{4A(\varepsilon)} $$
for all $x \in \Bbb R$ by uniform convergence. Then by the fundamental theorem of calculus, for all $x>x_0$ we have
\begin{align*} f_k(x) &= f_k(x_0) + \int_{x_0}^x f'_k(x) \,\mathrm{d}x \\
&\geq \frac12 - \int_{x_0}^x |f'_k(x)| \,\mathrm{d}x \\
&\geq \frac12 - \frac1{4A(\varepsilon)}\int_{x_0}^x \frac1{g(x)} \,\mathrm{d}x \\
&\geq  \frac12 -  \frac1{4A(\varepsilon)}\int_{0}^{\infty} \frac1{g(x)}\,\mathrm{d}x\\
&= \frac14.
\end{align*}
Hence $f_k(x) \geq \frac14$ for all $x \geq x_0,$ so $f_k$ is not compactly supported. Hence no such sequence $(f_k)$ exists.

This shows that the result only holds if $g$ does not grow too quickly, and this is consistent with your working that shows it suffices to assume $|g(x)| \leq (1+|x|^{\alpha})$ for $\alpha \in (0,1).$ In the critical case of linear growth the result is also true; for this assume $|g(x)| \leq C(1+|x|).$

We will exploit the fact that
$$ \int_0^{\infty} \frac1{1+x} \,\mathrm{d}x = \int_1^{\infty} \frac{\mathrm{d}}{\mathrm{d}x} (\log (1+x)) \,\mathrm{d}x = \infty,$$
using the fundamental theorem of calculus. Then we can construct a sequence of piecewise smooth functions by setting
$$ \varphi_k(x) = \begin{cases}
  1 & \text{ if } |x| \leq k \\
  1+\frac1k\left(\log(2+k) - \log(2+|x|)\right) & \text{ if } k < |x| \leq e^k(2+k) -2 \\
  0 & \text{ if} |x| > e^k(2+k) -2. \end{cases}$$
Note that $\varphi_k$ is continuous and $|\varphi_k'(x)| \leq \frac1{k (|x|+2)}$ whenever $k < |x| < e^k(2+k)-2.$ We need to regularise $\varphi_k,$ so let $\rho$ be a mollifier supported in $(-1,1)$ and set
$$ f_k = \varphi_k \ast \rho \in C^{\infty}_c(\Bbb R).$$
We claim this sequence works; evidently $f_k \equiv 1$ on each $[-(k-1),k-1]$ so $f_k \to 1$ pointwise as $k \to \infty.$ Moreover since
$$|f_k'(x)| \leq C\, \operatorname*{ess \, sup}_{|y|\leq 1} |\varphi_k'(x+y)| \leq \frac{C}{k(|x|+1)} \chi_{k-1 < |x| < e^k(2+k)-2} $$
(by mollifying the $L^{\infty}$ function $\varphi_k'$) for $|x|>k-1$ we can estimate
$$ |g(x)f_k'(x)| \leq \frac{C(1+|x|)}{k(1+|x|)} = \frac{C}{k}. $$
Since $f_k'(x)$ vanishes for $|x| \leq k - 1,$ uniform convergence follows.

From the above we see there is a dichotomy dependent on the integrability of $1/|g(x)|;$ for $g(x) \sim (1+|x|^{\alpha})$ we see this occurs if and only if $\alpha \leq 1,$ but the criticality is not occuring at linear growth. For instance the above can be refined to hold if $g(x) = (1+|x| \log(1+|x|))$ by noting that $\frac{\mathrm{d}}{\mathrm{d}x}(\log \log x) = \frac1{x \log x},$ but I will omit the details here.

Title: Existence of a cut-off function


问:
Decomposition of matrix NxN into product of three matrices of size NxN each
Let F be the group of n × n invertible matrices over field X. Let $$G \in F$$

Is there any way to decompose a given matrix of size NxN into product of 3 matrices of size NxN each where I want my middle matrix to be some fixed known matrix.
For example - M = AGB where G is already known to me  and A,B should belong to F such that $$AB \neq BA$$

答:
I believe one answer could be to take (assuming $G$ non-singular)
$$
M=G^{-1}GM.
$$
Where we see that $A=G^{-1}$ and $B=M$ (corresponding to your original form $M=AGB$).

This is rather trivial though and I am not sure if this was the answer you are looking for. Maybe indicate any other constraints you might have to result in a less trivial answer?


问:
A few questions on nonstandard analysis
I know that nonstandard analysis is analysis plus the existence of infinitesimal numbers. Does it mean that nonstandard analysis is the same theory as $ZF+\exists$infinitesimal numbers?

From what I read about it on Wikipedia there seem to be a few different approaches to it. There is "Robinson's method" which is "based on studying models". There is "Nelson's method" which is "Internal set theory" which is an extension of ZF. Are they the same? Or are there theorems that are true in one but not in the other?

If IST is an extension of ZF how much stronger than ZF is the "theory of nonstandard analysis"?

答: $\mathsf{IST}$ is a conservative extension of $\mathsf{ZFC}$: any internal formula that can be proved within internal set theory can be proved in $\mathsf{ZFC}$. This is proved in the Appendix to Edward Nelson’s original $1977$ paper in the Bulletin of the American Mathematical Society, ‘Internal Set Theory: A New Approach to Nonstandard Analysis’.

Title: A few questions on nonstandard analysis


问:
Doubt regarding definition of a real computable number
I saw in Wikipedia that a real number $a$ is computable if there exists a computable sequence of rational  numbers $q_n$ converging to $a$ such that $|q_n - q_{n+1}| < 2^{-n}$, for each $n$.

I just want to know if this is correct, because I couldn't find anywhere else such definition.

答:
Yes, it is correct. Not sure what is the reference you are looking at, but I would say that this definition is rather widespread. Wikipedia cites Weihrauch's book [1] in the end. I may add Pour-El and Richard's book [2,definition 3].

[1]: Weihrauch, Klaus, Computable analysis. An introduction, Texts in Theoretical Computer Science. An EATCS Series. Berlin: Springer. x, 281 p. (2000). ZBL0956.68056.

[2]: Pour-El, Marian B.; Richards, J. Ian, Computability in analysis and physics, Perspectives in Mathematical Logic, Berlin etc.: Springer-Verlag. x, 206 p. DM 128.00 (1989). ZBL0678.03027.


问:
Verify if $f$ is injective and/or surjective, and then describe the inverse function which is bijective, also verify if $f$ is monotomic
My professor give this exercise for homework, but due to distance learnig protocols, the class with this content was very unclear to me. I understand how to verify if a function is injective but can't verify if it is surjective, let alone if it is monotonic.

$f : N → N, f(n) = 3n − 1$

Verifying if $f$ is injective:
$f(n_1)=f(n_2)$

So

$3n_1 - 1 = 3n_2 - 1$

$3n_1 = 3n_2$

$n_1 = n_2$

Now, how can I verify if $f$ is surjective and monotonic??

答:
It is not surjective since you can't find an antiimage for any natural in the form 3m or 3m+1. For example, imagine you want to find and antiimage for 1, then there would exist an $n \in \mathbb{N}$ that 3n-1=1, so $n=2/3 \notin \mathbb{N}$.

For the other part, if $n_1<n_2$ then $3n_1-1<3n_2-1$, so this function is monotone increasing.


问:
Integral using gamma function $\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(1 + a) \Gamma(1 + b)}{\Gamma(2 + a + b)} $
Does anyone knows how to show

$$\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(a+b-2)}{\Gamma( a -1)\Gamma(b-1)} $$
for $a,b \gt -1?$

Also why it doesn't work for $a$ or $b\lt -1$ ?

答:
Does anyone knows how to show...

I cannot, because it is wrong. Using the definition of Beta function you have that

$$\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

that in your case gives the following result

$$\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$

to realize that your result does not make sense, set $a,b=1$ and you get

$$\int_0^1 dx=\frac{\Gamma(-2)}{\Gamma(-1)\Gamma(-1)}$$

??????

While using my result you get

$$\int_0^1 dx=\frac{\Gamma(1)\Gamma(1)}{\Gamma(2)}=\frac{0!\cdot0!}{1!}=1$$