问:
Determine regularity of transition matrix from diagram
Hi I am doing an exercise where I have to sketch a diagram from a Markov process and then only from that sketch argue if the matrix is regular or not. The process is a happening on a square where clockwise transitions have the probability of a and anti clockwise transitions has zero probability. Also no transitions occurs on the diagonal, only between neighboring sites. My sketch looked something like this:

```<code>             a
       2 ----->->----   3
       |                    |
  a   |                    |
       ^                   v   a
       |                    |
      1  ---<--<------4
             a
</code>```

Only by looking at this sketch how am I supposed to determine if the transition matrix is regular or not?

答:
Hint:

As I pointed out in a comment, the diagram can represent a Markov chain only if $\ a=1\ $.

By definition, a Markov chain is regular if there is some power $\ P^n\ $ of its transition matrix $\ P\ $ whose entries are all strictly positive.  This means that if the chain starts off in state $1$ (say) with probability $1$ at a time $\ t=0\ $, then at time $\ t=n\ $ it has a positive probability of being in any of its states, since the probability of its being in state $\ i\ $ at time $\ t=n\ $ is just the $\ i^\text{th}\ $ entry in the first row of $\ P^n\ $ (assuming you're using row-stochastic transition matrices).

If the Markov chain represented by your diagram starts off in state $\ 1\ $ with probabiliity $1$ at time $\ t=0\ $, what possible states can it be in at time $\ t=4k+j\ $ when $\ j=1,2,3\ $ or $\ 4\ $?  Is there any time $\ t=n\ $ when it could have a positive probability of being in every one of its four states?


问:
Intersection of ranges of projections
Let $A$ be a unital $\mathrm{C}^*$-algebra and $p,\,q$ non-zero projections in $A$. Consider two faithful representations of $A$ on Hilbert spaces $\mathsf{H}_1,\,\mathsf{H}_2$.

Is it possible for the intersections of the ranges of the projections to be zero in one representation and non-zero in the other?

答:
Let $H$ be a Hilbert space with orthonormal basis $(e_n)$.  Let $p\in B(H)$ be the projection onto the span of the vectors $e_{2n}$ and let $q\in B(H)$ be the projection onto the span of the vectors $e_{2n}+2^{-n}e_{2n+1}$.

Note now that $p-q$ is a compact operator (it can be written as the norm limit of the finite rank operators $p_n-q_n$ where $p_n$ and $q_n$ are the corresponding projections onto the spans of just the first $n$ of the basis vectors).  So, $p$ and $q$ are equal and nonzero projections in the quotient $B(H)/K(H)$.  Now take a faithful representation of $B(H)/K(H)$ on some Hilbert space $H'$, so $p$ and $q$ act by the same nonzero projection on $H'$.  Then $B(H)$ has a natural faithful representation on $H\oplus H'$, and the ranges of $p$ and $q$ have nontrivial intersection in this representation.  However, in the faithful representation on just $H$, the ranges of $p$ and $q$ have trivial intersection.

ContentLicense: CC BY-SA 4.0

Title: Intersection of ranges of projections


问:
$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$
Suppose that $X$ is an infinite set of cardinality $\alpha$.  Also, suppose that, for some $A \subseteq X$, we have that $|A| = |X\backslash A|$.  I want to show that $|A| = |X|$.

When, for example, $X = \mathbb{N} = \{1, 2, 3, \dots \}$, and $A = \{2n: n \in \mathbb{N}\}$, the requisite bijection $X\rightarrow A$ is about as obvious as it can be (since the definition of $A$ already provides one such bijection; in fact, in this case, the fact $|A| = |X\backslash A|$ is not even used.)  But I have a hard time pinning down the proof of the more general case stated above.

答:
I work in ZFC. $X=A\cup (X\setminus A)$ with $A$ and $X\setminus A$ disjoint sets. Therefore $\vert X\vert =\vert A\vert +\vert X\setminus A\vert =\vert A\vert +\vert A\vert$. Now, since this equation holds and $\vert X\vert\geq \omega$ also $\vert A\vert \geq \omega$ (otherwise $\vert A\vert$ would be finite and then $\vert A\vert +\vert A\vert$ could not be equal to $\vert X\vert$). Hence $\vert X\vert=\vert A\vert +\vert A\vert=\vert A\vert$, since $\vert A\vert$ is an infinite cardinal.

To prove that if $\kappa$ is an infinite cardinal $\kappa +\kappa =\kappa$, one can argues as follows. It suffices to show that $\kappa\times 2=\kappa$. Take $F:=\{f\colon \alpha\longrightarrow \alpha\times 2:\ 0\neq \alpha\subseteq \kappa\ \text{and f is bijective}\}$ ($0=\emptyset$) with the partial order given by set theoretic inclusion. Notice that $F\neq\emptyset$ since $\omega\leq \kappa\implies \omega \subseteq \kappa$ and $\omega\approx \omega\times 2$ in an obvious manner. It is not difficult to show that $(F,\subseteq)$ satisfies the assumptions of Zorn's Lemma (easy exercise). Therefore we get a maximal element $f\in F$, so that, in particular, $f\colon \alpha\longrightarrow \alpha\times 2$ with $0\neq\alpha\subseteq \kappa$ and $f$ bijective. To conclude it suffices to prove that $\kappa\setminus \alpha$ is a finite set (if $\kappa$ is an infinite cardinal and $n\in\omega$ then $\kappa +n\approx \kappa$). But this is clear: if $\vert\kappa\setminus \alpha\vert\geq\omega$, there is a bijection between $\omega$ and a subset $X\neq\emptyset$ of $\kappa\setminus \alpha$, i.e there is a bijection $g\colon X\longrightarrow X\times 2$ and therefore $f\cup g\colon \alpha\cup X\longrightarrow (\alpha\cup X)\times 2$ is a bijection such that $f\subsetneq f\cup g$, contradicting maximality of $f$.

ContentLicense: CC BY-SA 3.0


问:
Differentiating $\frac{2t-4}{(t^2+2)^2}$ by the quotient rule?
The quotient rule $\frac{u}{v} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

There comes a point where I need to differentiate my $v$ which is $(t^2+2)^2$, so when it comes to $\frac{dv}{dx}$ how do I go about this? Do I just expand the brackets like this $(t^2+2)(t^2+2) = t^4+2t^2+2t^2+4$ and then differentiate those terms and plug them back in the formula?

I found that the result is quite messy.

答:
You can expand it that way if you want, or you can use the chain rule
$$\frac d{dt}(t^2+2)^2=2(t^2+2)\frac d{dt}(t^2+2)=2(t^2+2)\cdot 2t$$
which is the same as you got another way.  This approach is much easier for more complicated compositions.


问:
Find generating function for linear recurrence
Let $a_0=0, a_1=1,$ and $a_{n+2}=2a_{n+1}+a_n.$ Find the closed form for this sequence using generating functions.

I'm not sure what to do, I tried getting some cancellation like when solving for Lucas/Fibonacci numbers but I'm not sure how to.

Body:
$$a_0=0, a_1=1\\a_{n+2}=2a_{n+1}+a_n\tag{1}$$
Let $$f(x)=\sum_{n=0}^{\infty}a_nx^n$$
multiply both sides of $(1)$ by $x^n$ and sum from $n=0$ to $\infty$

$$\sum_{n=0}^{\infty}a_{n+2}x^n=2\sum_{n=0}^{\infty}a_{n+1}x^n+\sum_{n=0}^{\infty}a_nx^n\tag{2}$$
We have
$$\sum_{n=0}^{\infty}a_{n+2}x^n=\frac{1}{x^2}\sum_{n=0}^{\infty}a_{n+2}x^{n+2}=\frac{1}{x^2}\left(f(x)-a_0-a_1x\right)=\frac{1}{x^2}\left(f(x)-x\right)$$

$$\sum_{n=0}^{\infty}a_{n+1}x^n=\frac{1}{x}\sum_{n=0}^{\infty}a_{n+1}x^{n+1}=\frac{1}{x}\left(f(x)-a_0\right)=\frac{f(x)}{x}$$

Thus $(2)$ can be written as
$$\frac{1}{x^2}\left(f(x)-x\right)=2\frac{f(x)}{x}+f(x)$$
and then
$$f(x)=\frac{x}{1-2x-x^2}$$


问:
$Var(X_1 X_2)=Var(X_1) Var(X_2)$
Let : $X_1$ and $X_2$ two independent variables, square-integrable et non constant.

At which necessary and sufficient condition, do we have $ Var(X_1 X_2)= Var(X_1) Var(X_2)$ ?

Attempt :
$V(Z)= V( E(Z|X_1) ) + E(V(Z|X_1))$ if $Z$ is a random variable.

Title: $Var(X_1 X_2)=Var(X_1) Var(X_2)$

Body:
With $Z=X_1 X_2$ your attempt leads to
\begin{align}
V(X_1 X_2)
&= V(X_1 E[X_2 \mid X_1]) + E[X_1^2 V(X_2 \mid X_1)]
\\
&= E[X_2]^2 V(X_1) + V(X_2) E[X_1^2]
\\
&= V(X_1)V(X_2) + V(X_1)E[X_2]^2 + V(X_2) E[X_1]^2
\end{align}


问:
Does the limit exist if a function approaches a limit where it is discontinuous??

If the function is discontinuous at $x=0$ and$f(0)=4.5$ . And as $x$ approaches $0$, the value of function nears $3$.

Does the limit in this case exist?

Imho ,it does and equals $3$ but one of my friend disagreed.

I'd be grateful if someone even hints at my error.

Thank you.

答: The limit exists and is 3. As I recall form my first year of calculus (Analysis?):  If a function f is real-valued, then the limit of f at p is L if and only if both the right-handed limit and left-handed limit of f at p exist and are equal to L.    $lim_{x\rightarrow0^+}f(x)=lim_{x\rightarrow 0^-}f(x)=3$


问:
Value of $t$ minimizing the distance $\lVert a-tb \rVert$
The question is

Let $a$ and $b$ be nonzero vectors in $\mathbb R^n$.

(a) What value of $t \in \mathbb R$ minimizes the distance $\lVert a-tb \rVert$? (Hint. It’s easier
to minimize the value of $\lVert a-tb \rVert ^2$.)

(b) What is the minimum distance in (a)?

For part a),

$$\lVert a-tb \rVert ^2 = \lVert a \rVert ^2  -  2t(a \cdot b)  +  t^2\lVert b \rVert ^2.$$

To minimize it, we set the derivative of it with respect to $t$ to $0$. That is,

$$-2(a \cdot b) + 2t\lVert b \rVert ^2=0 \implies t=\frac{a \cdot b}{\lVert b \rVert ^2}.$$

For part b), we substitute the found $t$ value into the original distance formula and get

$$\lVert a - \frac{a \cdot b}{\lVert b \rVert ^2} \cdot b \rVert .\tag{1}$$

But the answer to part b) is given as,

$$\sqrt{\frac{\lVert a \rVert ^2 \lVert b \rVert ^2 - (a \cdot b)^2}{\lVert b \rVert ^2}}.\tag{2}$$

Can you please explain how do we get from $(1)$ to $(2)$?

答:
But the two expressions are the same !

Actually when you replace $t$ by your expression $\frac{a\cdot b}{||b||^2}$ in the developped expression $\lVert a-tb \rVert ^2 = \lVert a \rVert ^2  -  2t(a \cdot b)  +  t^2\lVert b \rVert ^2$, you get the answer $\sqrt{\frac{\lVert a \rVert ^2 \lVert b \rVert ^2 - (a \cdot b)^2}{\lVert b \rVert ^2}}$.

If you really want to find again this answer with yours ($\lVert a - \frac{a \cdot b}{\lVert b \rVert ^2} \cdot b \rVert$), you just need to do the same thing you did at first, which is taking the square and developping, then taking the square root.

Hope it helped.


Title: $Var(X_1 X_2)=Var(X_1) Var(X_2)$

Body:
Let : $X_1$ and $X_2$ two independent variables, square-integrable et non constant.

At which necessary and sufficient condition, do we have $ Var(X_1 X_2)= Var(X_1) Var(X_2)$ ?

Attempt :
$V(Z)= V( E(Z|X_1) ) + E(V(Z|X_1))$ if $Z$ is a random variable.

Body:
$$var(X_1 X_2) = E(var(X_1 X_2|X_2))+var(E(X_1 X_2 | X_2)\\
=E(X_2^2 var(X_1)) + var(X_2 E(X_1))\\
= var(X_1) (E(X_2)^2 + var(X_2)) + E(X_1)^2 var(X_2)\\
= var(X_1) var(X_2) + E(X_2)^2 var(X_1) + E(X_1)^2 var(X_2)$$

Since $X_1, X_2$ are not constants then we must have $E(X_1)=E(X_2)=0$ which is necessary and sufficient.


问:
What does the notation $S^{n} \rightarrow S$ mean?
If S is a set and an operation on $S$ is denoted by $S^{n} \rightarrow S$, does the n in the exponent of S mean that the function's inputs are tuples of the set $S\times S \times S  \times S \space  \times \space ... \space \times S  $   (n-times)?

答: Yes. This is true in general. For example for $n=3$ and $S=\mathbb{Z}$, the function $f(x,y,z) = xyz$; $x,y,z \in \mathbb{Z}$, is a function from $\mathbb{Z}^3$ to $\mathbb{Z}$ and can be written $$f: \mathbb{Z}^3 \rightarrow \mathbb{Z}.$$


问:
Order of operations (BODMAS)
$$40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5$$

I'm studying and this is from an example.

In BODMAS, aren't addition and subtraction have same level?

So, in the 3rd line, it should be from left to right, correct?

答: As @GerryMyerson noted, there was indeed an error. The last two terms were grouped, which changed the expression to $40 - (10 + 22.5)$, which is entirely different from the correct expression. This is equivalent to $40 + (-1)(10 + 22.5)$, or $40 - 10 - 22.5$. Note how the second plus sign has become a minus sign.</p>

Title: Order of operations (BODMAS)


问:
How would you geometrically describe a Span consisting of the linear combinations of more than $2$ vectors in $\mathbb{R^3}$?
In $\mathbb{R^3}$, when we are given the span of $2$ vectors, we can describe the Span geometrically by

-a line if one of these vectors is a multiple of the other (since a line is the Span of all scalar multiples of a single vector.)

-a plane if they are not scalar multiples of each other.

Q) Now if we have $3$  vectors, could we describe their span geometrically (if the set of vectors making up this span are linearly independent) as a sphere, that is with infinite volume?

If we go to $4$ linear independent vectors in $\mathbb{R_4}$, would their spanning set occupy the entirety of the four dimensions, and so on? It makes sense to me, but I might be making an error, and I don't want to get the wrong ideas.

Please correct me if this question is not clear enough, nevertheless, thank you.

I edited the question as I made a stupid mistake, which I noticed from the answer, I failed to move on to $\mathbb{R_4}$ when extending to $4$ vectors, that wasn't the meaning of my question at all so I just edited it.

答:
As Tanner commented, The span of three linearly independent vectors in $\mathbb{R^3}$ is the entirety of the vector space $\mathbb{R^3}$. Geometrically, this means that the vectors span all of three-dimensional Euclidian space. A sphere of infinite volume is technically the same, but I think it's best to think of it just as the entire space.

If the vectors are not linearly independent, then the span will just be a plane or a line. There are no sets of four linearly independent vectors in $\mathbb{R^3}$, but in $\mathbb{R^4}$, four linearly independent vectors would span the entirety of that (four-dimensional) space, as you guessed.


问:
Identification of ends of Bruhat-Tits Tree
I am trying to understand why a canonical identification exists between the "ends" or "rays" of the Bruhat-Tits tree defined on $\mathbb{Z}_p$-lattices (with metric invariant under the action of $PGL(2,\mathbb{Q}_p)$) and $\mathbb{P}^1(\mathbb{Q}_p)=\mathbb{Q}_p\cup\{\infty\}$. I have consulted several online references on the Bruhat-Tits tree, all of which more or less avoid constructing the actual identification.

Intuitively, I see that since any vertex in the tree has $p+1$ neighbors, a path that travels outwards without backtracking should correspond to a $p$-adic number - for instance, if $p=2$, you could label a path and end up with some binary string such as $0101010011...$, which can be converted into a $2$-adic by using the 0s and 1s as coefficients of a formal Laurent series. I am unsure, however, of what the end that corresponds to $\infty$ should look like, as well as how the exact identification between a path and a $p$-adic would work - given a binary string, for instance, how would we know which power of $p$ would be the starting point? I am also unsure of what the implications of there being $p+1$ choices for the first outgoing edge are.

Edit: After thinking further on the matter, I've realized that the $p+1$-th edge (i.e. the "extra" one at your starting point) can be used to denote elements of $\mathbb{Q}_p\setminus\mathbb{Z}_p$, by letting "extra edge followed by $n$ zeroes" denote multiplication of the subsequent binary sequence by $p^{-n-1}$. So any path that does not travel through the extra edge initially simply belongs to $\mathbb{Z}_p$, and I can identify $\infty$ with the path that goes through the extra edge and is followed by infinite 0s (since this one seems to be degenerate/lack interpretation via the above). Feedback on this identification and whether it is correct/can be made more precise would be appreciated.

答:
I am more or less sure that following those lines you'll get your result, after proving that the Bruhat-Tits graph is a tree, ie. that my sequence $(L_n)$ is determined by any subsequence $(L_{k_j})$.

Let $A_n$ be the set of lattices $$A_n= \{ L\subset Z_p^2, [Z_p^2:L] = p^n\}$$

Write $$L = (a,b)Z_p+(c,d)Z_p$$ Wlog $r=\min(v(a),v(b))\le \min(v(c),v(d))$.   If $v(a)=r$, with $d' = d-\frac{c}ab$ then $$L = (a,b)Z_p+(0,d')Z_p= (a,b)Z_p+(0,p^{v(d')})Z_p=(a,b)Z_p+p^{v(d')}Z_p^2$$

Since $$[(a,b)Z_p+p^{v(d')}Z_p^2:p^{v(d')}Z_p^2]= p^{v(d')-r}$$ we get that $$p^n=[Z_p^2:(a,b)Z_p+p^{v(d')}Z_p^2]=p^{2 v(d')}/p^{v(d')-r}=p^{v(d')+r}$$
Thus in fact $L$ is determined by $(a,b)$. Obviously it works the same way when $v(b)=r$ so $$A_n = \{ (a,b)Z_p+p^e Z_p^2,(a,b)\in Z_p^2,  \min(v(a),v(b))\le n/2,e= n-\min(v(a),v(b))\}$$

Then let's look at the sequences $(L_n), L_n\in A_n$ such that $L_{n+1}\subset L_n$ and $p^k L_n \ne L_{n+2k}$.

Write $$L_n = (a_n,b_n)Z_p+p^{e_n} Z_p^2, \qquad e_n\ge r_n = \min( v(a_n),v(b_n)), n=r_n+e_n$$

The $p^k L_n \ne L_{n+2k}$ condition imposes that $e_n-r_n\to \infty$. This is because once $n$ and $e_n-r_n$ are known then there are finitely many choices for $L_n$.

Thus in fact there is $(a,b)\in Z_p^2-(0,0)$ such that for all $n\ge N$ large enough $$L_n = (a,b) Z_p+p^{n-\min(v(a),v(b))} Z_p^2$$

We impose $p^k L_n \ne L_{n+2k}$ because we are going to say that two lattices are equivalent if $L=\lambda L'$, so our $(a,b)$ becomes an element of $(\Bbb{Q}_p^2-(0,0))/\Bbb{Q}_p^* =\Bbb{P^1(Q_p)}$.


问:
Suppose $G$ is a group and $|G|=p^k$ , and $X$ is finite set. Let $G$ act on $X$ or the cardinality $|X|$
While studying group theory I came across this question:

Suppose $G$ is a group and $|G|=p^k$ for some $p$ prime number and $k$ integer, and $X$ is finite set. Let $G$ act on $X$ if $p$ does not divide $|X|$ see if $G_{\alpha}=G$ for some $\alpha \in X$ true or not.

I need a hint the relation between the cardinality if $X$ and order of the group $G$. In other words why does the number of elements in $X$ matter?

答:
Hint:

Use the orbit-stabilizer theorem.

If $G$ acts on $X$, then every point $x \in X$ has some orbit. Moreover, since every element is in exactly one orbit we see

$$|X| = \sum_{\text{orbits}} |\text{orbit}|$$

By the orbit-stabilizer theorem, though, we know each orbit satisfies
$|\text{orbit}| \cdot |\text{stabilizer}| = |G|$.

If $|X|$ is not a multiple of $p$, but each $|\text{orbit}|$ must divide $|G| = p^k$, what does this tell you about the size of some orbits? It might be helpful to work mod $p$.

I hope this helps ^_^


问:
Number of leaves in a tree with maximum degree $k$
Let $G$ be a tree with a maximum degree of the vertices equal to $k$.
At least how many vertices with a degree of $1$ can be in $G$ and why?

I think the answer must be $k$ but I don't know how to prove it.

答:
For $d\in\{1,\dots,k\}$, let $n_d$ be the number of nodes of degree $d$.  By the handshake lemma, we have
$$\sum_{d=1}^k d n_d = 2\left(\sum_{d=1}^k n_d - 1\right),$$
which implies that
$$n_1 = 2 + \sum_{d=3}^k (d-2) n_d \ge 2 + (k-2) n_k \ge 2 + (k-2) 1 = k.$$
This lower bound is attained by a star with $k+1$ nodes.

Alternatively, perform a depth-first search rooted at a node with degree $k$.  Each branch yields at least one leaf.


问:
How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$?
Consider Numbers:

\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is congruent to $10^m \bmod 41$ for some $m$?

答:
Since $1..1$ with $n$ ones is equal to $\frac{10^n-1}{9}$, we can write the general term of the sequence as
$$
a_n = 245 \cdot \Big(200 \cdot \frac{10^n-1}{9} + 11\Big) - 239.
$$
Looking modulo 41 we have $245 \equiv -1$, $200 \equiv -5$, $-239 \equiv 7$ and $9^{-1} \equiv -9$ (since $9 \times (-9) = -81 \equiv 1$). We can write
\begin{align*}
a_n
&\equiv (-1) \cdot 9^{-1} (-5) (10^{n}- 1) - 11 + 7 \\
&\equiv 9 \cdot  (-5 \cdot 10^{n} + 5) -4 \\
&\equiv -45 \cdot 10^{n} + 45 - 4 \\
&\equiv -4 \cdot 10^n \\
&\equiv -40 \cdot 10^{n-1}\\
&\equiv 10^{n-1}.
\end{align*}


问:
How to solve $\tan (a)\cdot \tan (b)>1$
Considering my answer for the question Solve $\cos(a)\cos(b)\cos(a+b) = -1/8$ for $0 < a,b < \pi/2$ I have thought to put

$$x=\tan(a), \quad y=\tan(b)$$

in the inequality

$$\tan (a)\cdot \tan (b)>1 \tag 1$$

I will have

$$xy>1 \tag 2$$

that the $(2)$ it is an equilateral hyperbola when $xy=1$. The crop zone done with Desmos is:

How can I find $a=b=\frac\pi3$?

Using instead Wolfram Alfa I will have many steps https://www.wolframalpha.com/input/?i=%5Ctan+%28a%29%5Ctan+%28b%29%3E1

Is it possible to find with my approach

$$a=b=\frac\pi3\quad ?$$

答:
First of all, we need both $\tan A,\tan B$ of the same sign

If both are $>0,$ which is true in the given range,

$$\tan A>\cot B=\tan(\pi/2-B)$$

As $\tan(x) $ is increasing in the given range of $A,B$
we need $$A>\pi/2-B$$

We find
$$-8(1-xy)=(x^2+1)(y^2+1)$$

$$\iff(x-y)^2+(xy-3)^2=0$$

Can you take it from here?

Title: How to solve $\tan (a)\cdot \tan (b)>1$


答:
$$40-10+22.5=52.5$$

You're correct, subtraction and addition have the same level of priority, so when both exist, then you operate from left to right. You are entirely correct about that.  If you were puzzled when you saw the "answer" $\,7.5$, good for you, since $\;\;40-10+22.5\neq 7.5!$

Now, note: $$40-(10+22.5)=7.5$$

because operations in "brackets" (or parentheses) have greatest priority.

So let's be forgiving: perhaps the author of the problem statement/review notes forgot parentheses!

Title: Order of operations (BODMAS)

Body:
$$40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5$$

I'm studying and this is from an example.

In BODMAS, aren't addition and subtraction have same level?

So, in the 3rd line, it should be from left to right, correct?


问:
Number of configurations of dots and loops without mirrors

The equation for number of ways to put $b$ loops around $b$ of $a$ dots is
$$a \choose b$$

How can one modify this so that all palindromes/mirrors are excluded?

This is a continuation of this question.  Imagine you have $a$ amount of dots and $b$ amount of loops, where $a>b$. Now imagine you put those loops around the dots, and see how many permutations/combinations you get. Is there an equation to give you the answer if both $a$ and $b$ are known? Here is a picture that shows what I mean:

Here the dots are squares, and the loops are the red squares.

答:
We'll have to work for odd, even $a,b$ separately.

The mirrors can be counted by looking at one half of the string, since the other half is identical.

For a even, b odd, there won't be any mirrors. So following is enough $$\binom{a}{b}$$

For a even, b even, $a=2n$, $b=2m$ there'll be $m$ loops in each half, hence there'll be $\binom{n}{m}$ mirrors

$$\bigcirc \bullet \bullet \boxed{\bullet \bullet \bigcirc}$$

For a odd, b odd, $a=2n+1$, $b=2m+1$ there'll always be a loop in the center. So no. of mirrors $=\binom{n}{m}$

$$\bullet \bigcirc \bullet \boxed{\bigcirc} \boxed{\bullet \bigcirc \bullet } $$

For a odd, b even, $a=2n+1$, $b=2m$ there'll never be a loop on central dot. So no. of mirrors $=\binom{n}{m}$

$$\bigcirc \bigcirc \bullet \boxed{\bullet} \boxed{\bullet \bigcirc \bigcirc } $$

Desired equation is $$\binom{a}{b} - \binom{n}{m}$$
for three of the cases.


答:
We have
$$\tan a > \frac{1}{\tan b} = \cot b = \tan (\frac{\pi}2 - b),$$ hence  $a > \frac{\pi}2 - b$ and $a+b > \frac{\pi}2$.

Title: How to solve $\tan (a)\cdot \tan (b)>1$

Body:
Considering my answer for the question Solve $\cos(a)\cos(b)\cos(a+b) = -1/8$ for $0 < a,b < \pi/2$ I have thought to put

$$x=\tan(a), \quad y=\tan(b)$$

in the inequality

$$\tan (a)\cdot \tan (b)>1 \tag 1$$

I will have

$$xy>1 \tag 2$$

that the $(2)$ it is an equilateral hyperbola when $xy=1$. The crop zone done with Desmos is:

How can I find $a=b=\frac\pi3$?

Using instead Wolfram Alfa I will have many steps https://www.wolframalpha.com/input/?i=%5Ctan+%28a%29%5Ctan+%28b%29%3E1

Is it possible to find with my approach

$$a=b=\frac\pi3\quad ?$$


答:
HINT:

You can organize differentiation of a quotient/fraction in the following convenient way:

In a raster plug in  derivatives of numerator and denominator

$$\boxed{\left(\dfrac{u}{v}\right)'=\left(\dfrac{u}{v}-\dfrac{u'}{v'}\right) \cdot \dfrac{-v'}{v}}$$

Let $ t^2+2 =[]$.  Take out $2$ as common factor and plug into Raster

$$\left(\dfrac{t-2}{[]^2}- \dfrac{1}{4t}[] /[]^2\right)\cdot \left(\dfrac{-4t}{[]}\right)$$

with $[]^3$ in denominator, the  numerator is

$$ -(4t^2 -8t -t^2-2 )$$

results in

$$-2\dfrac{(3t^2-8t-2)}{[]^3}$$

Title: Differentiating $\frac{2t-4}{(t^2+2)^2}$ by the quotient rule?

Body:
The quotient rule $\frac{u}{v} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

There comes a point where I need to differentiate my $v$ which is $(t^2+2)^2$, so when it comes to $\frac{dv}{dx}$ how do I go about this? Do I just expand the brackets like this $(t^2+2)(t^2+2) = t^4+2t^2+2t^2+4$ and then differentiate those terms and plug them back in the formula?

I found that the result is quite messy.


问:
Integration over a triangular region using barycentric coordinantes?
Example problem: Compute $\int _A 3x^2+y \, \mathrm{d}A$, where $A$ is a triangle with vertices $(0,0),(2,0),(1,1)$.

Can this be computed using barycentric coordinates? I found a Wikipedia article on that matter, but I don't really understand it. Could you work through this example? I'd really appreciate that!

答:
Your integral is $\displaystyle \int_0^{1} \int_y^{2-y} (3x^2 + y) \, dx \, dy = \frac{23}{6}$

Now we can always use change of variable to any two barycentric coordinates such that limits of our integral over a triangular region becomes

$0 \leq \lambda_1 \leq 1 - \lambda_2, 0 \leq \lambda_2 \leq 1$

In barycentric coordinates,

$\lambda_1 + \lambda_2 + \lambda_3 = 1 \,$ so $\lambda_3 = 1 - \lambda_1 - \lambda_2$

Given the vertices of the triangle $(0, 0), (2, 0), (1, 1)$, we have

$x = \lambda_1 x_1 + \lambda_2 x_2 + \lambda_3 x_3 = 1 - \lambda_1 + \lambda_2$

Similarly, $y = 1 - \lambda_1 - \lambda_2$

So our integral becomes

$\displaystyle I = 2 A \int_0^1 \int_0^{1-\lambda_2} [3 (1 - \lambda_1 + \lambda_2)^2 + 1 - \lambda_1 - \lambda_2] \, d \lambda_1 \, d\lambda_2 = \frac{23}{6}$

$A = 1$ is the area of the triangle (which is easy to find as it is isosceles triangle with base $2$ and altitude $1$).

While in this case finding the integral is easier with the regular method, it is not always the case and we have to split our integral into two. Barycentric coordinates can be easier in such cases.


问:
Find constant term in the expansion of $\left (1+\frac x2 -\frac 2x \right)^4$
Finding the constant term for

$$\left (1+\frac x2 -\frac 2x \right)^4$$

is easy, but that would require converting the expression into a binomial.

However, I have no idea about how to do that.   Completing squares doesn’t work because we have a negative sign on $\frac 2x$.

答:
We are looking for constant products of four factors taken from $\{1, x/2, -2/x\}$. If you pick one of $x/2$ or $-2/x$ you have to pick the other one as well to cancel $x$. Hence, there's just three ways to form constant products:

All four factors are $1$, so the product is $1$. There's only one way to achieve this.

Two factors are $1$ and the other two are $x/2$ and $-2/x$, so the product is $-1$. There are $\binom{4}{2,1,1} = 12$ ways to do this.

Two factors are $x/2$ and the other two are $-2/x$, so the product is $1$. There are $\binom{4}{2,2} = 6$ ways to do this.

Hence, the sum of all these products is
$$
1 - 12 + 6 = -5.
$$


问:
Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$
How do I prove that:

$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$

Do I use induction?

答:
You can use integral:

$$\frac { 1 }{ \sqrt { 1 }  } +\frac { 1 }{ \sqrt { 2 }  } +\dots +\frac { 1 }{ \sqrt { 100 }  } <\int _{ 0 }^{ 100 }{ \frac { 1 }{ \sqrt { x }  }  } dx=20$$

You can imagine approximating the integral with rectangles of side $\frac { 1 }{ \sqrt { n }  }$ and $1$, will give less area than the integral because of the behaviour of the curve.


问:
Dual spaces, Am I doing right?

Let $\mu$ be the counting measure on the measurable space $(N,P(N))$.
$1 <p,q< \infty$ , $\frac{1}{p}+\frac{1}{q}=1$.
Define: $\phi: l_q\to (l_p)^*$ such as for all $(a_n)\in l_p$ and $(b_n)\in l_q$:
$\phi((b_n)_n)((a_n)_n)=\sum_{n=1}^{\infty} a_n b_n$.

A.Show that $\phi$ is well-defined, linear, injective and $\|\phi(b_n)_n\|\leq \|(b_n)_n\|_q$.

B.Show that $\phi$ is surjective and $\|\phi(b_n)_n\|=\|(b_n)_n\|_q$.
Hints for part B:

1.Let $f\in (l_p)^*$, $\forall n\in N$, denote $e_n(k)$ as:
$e_n(k)$ = 1 if n=k and 0 otherwise.
So $e_n\in l_p$ (why?). And look at $b_n:=f(e_n)$.

$\forall n\in N$ define:
$\lambda_n= |b_n|^q/ b_n$ if $b_n \neq 0$ and 0 otherwise.
Let $N_0\in N$ show that $\sum_{n=1}^{N_0} (|b_n|^q)^{1/q} \leq \|f\|$.

This what I did (according to my understanding of this question).

$\phi$ is well defined:
It means to make sure that the given sum is in $(l_p)^*$ which is equivalent to say that the defined sum is a function from $l_p$ to $C$.
$|\phi(b_n)(a_n)|=\left|\sum_n a_n b_n\right|\leq \sum_n |a_n b_n| \leq \|a_n\|_p \|b_n\|_q < \infty$, by using the Holder inequality.

$\phi$ is Linear:
Let $b_n , c_n \in l_q$ , $a_n \in l_p$ , $b,c \in C$.
Then, $\phi(b(b_n)+c(c_n))(a_n)= \sum_n (a_n)(b(b_n)+c(c_n))=\sum_n (b(a_n)(b_n)+c(a_n)(c_n))=b\sum_n (a_n)(b_n)+ c\sum_n (a_n)(c_n)= b \phi(b_n)(a_n)+ c \phi (c_n)(a_n)$.

$\phi$ is injective:
Let $(a_n)\in l_p$ , $(b_n), (c_n) \in l_q$.
Assume that :  $\phi(b_n)(a_n)=\sum_n (a_n)(b_n)=\phi(c_n)(a_n)=\sum_n (a_n)(c_n)$, so $\sum_n (a_n)((b_n)-(c_n))=0$ and since it is true for every $a_n \in l_p$, we have $b_n=c_n$ ,$\forall n$ thus $(c_n)_n=(b_n)_n$.

And the last thing, $\|\phi(b_n)\| =\|\sum_n (a_n)(b_n)\| \leq \sum_n \|(a_n)(b_n)\| \leq \|a_n\|_p \|b_n\|_q$ ,here i used the holder inequality, so $\|\phi(b_n)\|=\sup_{\|a_n\|_p=1}  |\phi(b_n)| \leq \|b_n\|$ .

Part B:
$\phi$ is surjective:
Is is easy to see that $e_n \in l_p$ since
$\|e_n\|_p=(\sum_n |e_n|^p)^{1/p}=1 < \infty$ so $e_n \in l_p$.
(Here we use the p-norm definition), we'll need this later!.
We aim to find a sequence $(b_n)_n\in l_q$ such as $\phi((b_n)_n)=f$. Look at:
$b_n=f(e_n)$ so we get $\phi((b_n)_n)(e_n)=\sum_n e_n*b_n=b_k=f(e_k)=f(e_n)$ (using $e_n$ definition).
We want to show that $b_n\in l_q$:
$\|b_n\|_q= \|f(e_n)\|\leq \|f\|_{l_p^*} \|e_n\|_p = \|f\|_{l_p^*}<\infty$ since we assume that $f\in l_p^*$.

Is what I did fine?
Can you please help me with the other part in 2?

答:
As for the part A, for $\| \phi(b_n)\|$ what is meant is the norm in $\ell_p^*$, the standard dual norm in a Banach space, so, if you apply the definition of dual norm, with your computation you've almost finished.

As for the party B, I didnt understand what you write. The idea of the proof is that you want to prove that there exists $\mathbf{b}=(b_n)_n$ such that $f=\phi(\mathbf{b})$, but of such a sequence exists, it is determined by the fact that $\phi(\mathbf{b})(e_k)= b_k$, so it must necessarily be $f(e_k)= b_k$. Then you have to prove that $\mathbf{b}\in\ell_q$ and that $\phi(\mathbf{b})= f$. The latter is easier, the former a bit trickier

Title: Dual spaces, Am I doing right?


问:
Give some example of quotient topology?
Give some easy example of Quotient topology ?

My attempt : I got some example here But im facing difficulties in understanding the  example

I need some example that is easy to understand

I know the definition of

Quotient topology : let $X$ be a topological space and $~$ an equivalence relation on $X$. For every $x \in X$ , denote by $[x]$ its equivalence class .The quotient space  of $X$ modulo $ \sim$ is given by the set

$X/\sim~ =\{[x] : x \in X \}$

we have the projection map $p: X \to /\sim ,x \to [x]$
and we equib $X/\sim$ by the topology

$U\subseteq$ $X/ \sim~$ is open if and only if $p^{-1}(u)$ is an open subset  of $X$

答:
Here is a nice example:  Start with the compact interval $[0,1]$.  "Identify the endpoints" to get a (topological) circle.  This means your equivalence relation is $x \sim y$ iff either $x=y$ or else $\{x,y\} = \{0,1\}$.

Other common ones start with a square $[0,1] \times [0,1]$ and identify the boundaries in certain ways.  Depending on how you identify the points of the boundary, you could get a (topological) sphere, a cylinder, a Möbius band, a torus, a Klein bottle, or other things.

Title: Give some example of quotient topology?


问:
Separable ODE and assumptions on the solution
Suppose I have an ODE-problem of the form:
\begin{equation}
\begin{cases}
x'(t)= g(t) h(x(t))\\
x(t_{0})=x_{0},
\end{cases}
\end{equation}
where both functions $g$ and $h$ are smooth. Suppose further that $h(t)=0$ if and only if $t=0$.

Now, if $x_{0} =0$, then clearly the solution of the problem is the constant function $x=0$. Else, if $x_{0} \neq 0$, then there exists an open interval containing $x_{0}$ where $h$ is non-zero. Thus, dividing by $h(x(t))$, we can solve the separable ODE using the standard method.

Suppose that, when $x_{0} \neq 0$, the solution of
\begin{equation}
\begin{cases}
x'(t)/h(x(t))= g(t) \\
x(t_{0})=x_{0},
\end{cases}
\end{equation}
is well-defined and non-zero for all $t \in \mathbb{R}$. Since this solution also solves the first system, by uniqueness, it must be the solution of the first system.

My question is: Is this reasoning correct? I am a bit confused because it seems to me like we are finding the solution under the assumption that it is non-zero. What is the best way of handling this problem?

答:
Your argument is wrong. Consider the following IVP:
$$ x'=x-1,x(0)=0. $$
Clearly it has the solution
$$ x=1-e^t $$
which is not a zero solution.


问:
Intuitive logic behind the following equivalences in propositional logic
I was going through the text Discrete mathematics and its applications by Kenneth H. Rosen where I came across few logical equivalences as follows:

$1. (p→q)∧(p→r)≡p→(q∧r)$

$2. (p→r)∧(q→r)≡(p∨q)→r$

$3. (p→q)∨(p→r)≡p→(q∨r)$

$4. (p→r)∨(q→r)≡(p∧q)→r$

Now the thing is that the text does not deal with any intuition behind the same, though there is the obvious method of proof by truth table or using boolean laws, I was thinking about some intuitive logic, which would help me understand the actual essence behind each of these equivalences. So I came up with an intuition for each equivalence, going from left to right, in each case.

$1. (p→q)∧(p→r)≡p→(q∧r)$ : Since, we have $p$ as the hypothesis in each of the ANDed terms on the LHS, I assume that $p$ is true, if $p$ is true, then both $q$ as well as $r$ have to be true. So we have the equivalence of the RHS as: if $p$ is true then both $q$ and $r$ are true.

$2. (p→r)∧(q→r)≡(p∨q)→r$  : Since we have $r$ as the conclusion in each of the ANDed terms on the LHS, I assume that $r$ is false, then $\lnot r$ is true. Now if $r$ is false, then both $p$ as well as $q$ have to be false, so have $(\lnot r \rightarrow (\lnot p \land \lnot q)) \equiv (\lnot r \rightarrow \lnot (p \lor q)) \equiv ((p∨q)→r)$

$3. (p→q)∨(p→r)≡p→(q∨r)$ : [Similar to 1.] Since, we have $p$ as the hypothesis in each of the ORed terms on the LHS, I assume that $p$ is true, if $p$ is true, then either $q$ or $r$ or both have to be true. So we have the equivalence of on the RHS as: if $p$ is true then ($q$ or $r$) is true.

$4. (p→r)∨(q→r)≡(p∧q)→r$ : [Similar to 2] Since we have $r$ as the conclusion in each of the ORed terms on the LHS, I assume that $r$ is false, then $\lnot r$ is true. Now if $r$ is false, then either $p$ or $q$ or both have to be false, so have $(\lnot r \rightarrow (\lnot p \lor \lnot q)) \equiv (\lnot r \rightarrow \lnot (p \land q)) \equiv ((p \land q)→r)$

I just made an attempt to convince myself intuitively, I might be wrong, or a better intuitive approach might exist, if so please guide me. Thank you.

答:
I believe there is a more concise and intuitive explanation for parts $2$ and $4$.

For $2$, if both $p$ and $q$ imply $r$ independently of each other, then if either are true (the OR), then $r$ is true.

For $4$, if one of $p$ and $q$ implies $r$, then if both are true, then the one that truly implies $r$ must be true, meaning $r$ is true.


问:
Estimating Poisson rate and using the resulting distribution.
The following table shows the frequency of arrival of patients at a health clinic ina 15-minute time interval. What is the probability that more than 6 patients will arrive in next half an hour?

```<code>Number of customers  : 0 1 2 3 4 5 6
Observed frequency   : 1 4 5 6 4 3 2
</code>```

The problem I face is here random variable is defined for number of intervals but we have to find the probability for number of customers How is that possible?

I am new to Statistics Please help me with ideas or examples

答:
Using data to estimate Poisson rate per 15 min. Let $k = 0, 1, \dots 6,$ and $f_k = 1, 4, \dots, 2.$
Then there are $n = \sum_{i=k}^6 f_k = 25$ fifteen minute periods with an average of $\bar X = \frac 1n\sum_{k=1}^6 kf_k
= 3.$ So the average fifteen minute period has $\hat \lambda_{15} = \bar X = 3$ arrivals.  (See @heropup's Comment on estimation. Computations using R as a calculator.)

```<code>k = 0:6
f = c(1, 4, 5, 6, 4, 3, 2)
n = sum(f); n
[1] 25
sum(k*f)/n
[1] 3
</code>```

Poisson rate for 30 min. Then the number $Y$ of arrivals over half an hour (30 minutes) can be modeled as a Poisson random variable with arrival rate $6.$ That is,
$\lambda_{30} = 2\lambda_{15} = 6.$
Now, we can assume $Y\sim\mathsf{Pois}(\lambda=6).$

Using Poisson PDF. You seek

$$P(Y > 6) = 1 - P(Y \le 6) = 1-\sum_{k=0}^6 P(Y=k)\\
= 1-\sum_{k=0}^6 e^{-\lambda}\frac{\lambda^k}{k!} = 0.3937,$$

where the next-to-last step uses the Poisson PDF (or PMF).

You can use a calculator to find and sum the required six terms. A statistical calculator might be able to find
the sum automatically. In R, where ppois stands for
the CDF of a Poisson distribution, the computation
is shown below:

```<code>1 - ppois(6, 6)
[1] 0.3936972
</code>```

Rough normal approximation. Sometimes, if $\lambda$ is large enough, Poisson probabilities can be found using a normal approximation.
I don't know whether you are expected to use this method, because $\lambda = 6$ is a little too small for a good
approximation.

If so, here is a start:

$$P(Y > 6) = P(Y>6.5) = P\left(\frac{Y-\lambda}{\sqrt{\lambda}} > \frac{6.5 - 6}{\sqrt{6}}\right)\\
\approx P(Z > 0.2041) \approx 0.42,$$
where the numerical value of the probability is from printed normal CDF tables. (Look in the margins of the table for z-values $0.20$ and $0.21.)$

The first equality amounts to a 'continuity correction'; the last line uses the normal approximation. If you use
this method, you should (a) justify each step and (b) comment that it is only a rough approximation.

Because $E(Y) = 6, Var(Y) = 6,$ and $SD(Y) = \sqrt{6},$ we can compute the
normal approximation directly in R without standardization as shown below:

```<code>1 - pnorm(6.5, 6, sqrt(6))
[1] 0.4191282
</code>```

In the figure below the black vertical bars show the PDF of $\mathsf{Pois}(\lambda=6),$ the red curve is the density function of $\mathsf{Norm}(\mu=6, \sigma=\sqrt{6}),$ and the vertical dotted line is at
$Y = 6.5.$ The exact answer is the sum of the heights of the bars to the right of that dotted line. Finally, an approximate answer (a little too big) is the area under
the density curve to the right of the vertical dotted line.

R code for figure:

```<code>k = 0:15;  PDF = dpois(k, 6)
plot(k, PDF, type="h", lwd=3, main="PDF of POIS(6)")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 curve(dnorm(x, 6, sqrt(6)), add=T, col="red")
 abline(v = 6.5, lwd=2, lty="dotted", col="blue")
</code>```

Body:
Using data to estimate Poisson rate per 15 min. Let $k = 0, 1, \dots 6,$ and $f_k = 1, 4, \dots, 2.$
Then there are $n = \sum_{i=k}^6 f_k = 25$ fifteen minute periods with an average of $\bar X = \frac 1n\sum_{k=1}^6 kf_k
= 3.$ So the average fifteen minute period has $\hat \lambda_{15} = \bar X = 3$ arrivals.  (See @heropup's Comment on estimation. Computations using R as a calculator.)

k = 0:6
f = c(1, 4, 5, 6, 4, 3, 2)
n = sum(f); n
[1] 25
sum(k*f)/n
[1] 3

Poisson rate for 30 min. Then the number $Y$ of arrivals over half an hour (30 minutes) can be modeled as a Poisson random variable with arrival rate $6.$ That is,
$\lambda_{30} = 2\lambda_{15} = 6.$
Now, we can assume $Y\sim\mathsf{Pois}(\lambda=6).$

Using Poisson PDF. You seek

$$P(Y > 6) = 1 - P(Y \le 6) = 1-\sum_{k=0}^6 P(Y=k)\\
= 1-\sum_{k=0}^6 e^{-\lambda}\frac{\lambda^k}{k!} = 0.3937,$$

where the next-to-last step uses the Poisson PDF (or PMF).

You can use a calculator to find and sum the required six terms. A statistical calculator might be able to find
the sum automatically. In R, where ppois stands for
the CDF of a Poisson distribution, the computation
is shown below:

1 - ppois(6, 6)
[1] 0.3936972

Rough normal approximation. Sometimes, if $\lambda$ is large enough, Poisson probabilities can be found using a normal approximation.
I don't know whether you are expected to use this method, because $\lambda = 6$ is a little too small for a good
approximation.

If so, here is a start:

$$P(Y > 6) = P(Y>6.5) = P\left(\frac{Y-\lambda}{\sqrt{\lambda}} > \frac{6.5 - 6}{\sqrt{6}}\right)\\
\approx P(Z > 0.2041) \approx 0.42,$$
where the numerical value of the probability is from printed normal CDF tables. (Look in the margins of the table for z-values $0.20$ and $0.21.)$

The first equality amounts to a 'continuity correction'; the last line uses the normal approximation. If you use
this method, you should (a) justify each step and (b) comment that it is only a rough approximation.

Because $E(Y) = 6, Var(Y) = 6,$ and $SD(Y) = \sqrt{6},$ we can compute the
normal approximation directly in R without standardization as shown below:

1 - pnorm(6.5, 6, sqrt(6))
[1] 0.4191282

In the figure below the black vertical bars show the PDF of $\mathsf{Pois}(\lambda=6),$ the red curve is the density function of $\mathsf{Norm}(\mu=6, \sigma=\sqrt{6}),$ and the vertical dotted line is at
$Y = 6.5.$ The exact answer is the sum of the heights of the bars to the right of that dotted line. Finally, an approximate answer (a little too big) is the area under
the density curve to the right of the vertical dotted line.

R code for figure:

k = 0:15;  PDF = dpois(k, 6)
plot(k, PDF, type="h", lwd=3, main="PDF of POIS(6)")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 curve(dnorm(x, 6, sqrt(6)), add=T, col="red")
 abline(v = 6.5, lwd=2, lty="dotted", col="blue")


答:
One issue is that in all your explanations, you only show that the RHS follows from the LHS, and not that the LHS follows from the RHS.

Also, I am not sure that your explanations are really all that more 'intuitive' than going through a truth-table or using boolean algebra. What might help more, is to come up with some concrete examples.

For example, for the second equivalence $(p \to r) \land (q \to r) \equiv  (p \lor q) \to r $ you could say that the LHS says that if you are a fruit ($p$) then you are delicious ($r$), and also being a vegetable ($q$) makes you delicious ($r$), whereas the RHS says that if you are either a fruit of a vegetable ($p \lor q)$ then you are delicious ($r$). And these two claims are really saying the same thing: both fruits and vegetables are delicious!

Title: Intuitive logic behind the following equivalences in propositional logic

Body:
I was going through the text Discrete mathematics and its applications by Kenneth H. Rosen where I came across few logical equivalences as follows:

$1. (p→q)∧(p→r)≡p→(q∧r)$

$2. (p→r)∧(q→r)≡(p∨q)→r$

$3. (p→q)∨(p→r)≡p→(q∨r)$

$4. (p→r)∨(q→r)≡(p∧q)→r$

Now the thing is that the text does not deal with any intuition behind the same, though there is the obvious method of proof by truth table or using boolean laws, I was thinking about some intuitive logic, which would help me understand the actual essence behind each of these equivalences. So I came up with an intuition for each equivalence, going from left to right, in each case.

$1. (p→q)∧(p→r)≡p→(q∧r)$ : Since, we have $p$ as the hypothesis in each of the ANDed terms on the LHS, I assume that $p$ is true, if $p$ is true, then both $q$ as well as $r$ have to be true. So we have the equivalence of the RHS as: if $p$ is true then both $q$ and $r$ are true.

$2. (p→r)∧(q→r)≡(p∨q)→r$  : Since we have $r$ as the conclusion in each of the ANDed terms on the LHS, I assume that $r$ is false, then $\lnot r$ is true. Now if $r$ is false, then both $p$ as well as $q$ have to be false, so have $(\lnot r \rightarrow (\lnot p \land \lnot q)) \equiv (\lnot r \rightarrow \lnot (p \lor q)) \equiv ((p∨q)→r)$

$3. (p→q)∨(p→r)≡p→(q∨r)$ : [Similar to 1.] Since, we have $p$ as the hypothesis in each of the ORed terms on the LHS, I assume that $p$ is true, if $p$ is true, then either $q$ or $r$ or both have to be true. So we have the equivalence of on the RHS as: if $p$ is true then ($q$ or $r$) is true.

$4. (p→r)∨(q→r)≡(p∧q)→r$ : [Similar to 2] Since we have $r$ as the conclusion in each of the ORed terms on the LHS, I assume that $r$ is false, then $\lnot r$ is true. Now if $r$ is false, then either $p$ or $q$ or both have to be false, so have $(\lnot r \rightarrow (\lnot p \lor \lnot q)) \equiv (\lnot r \rightarrow \lnot (p \land q)) \equiv ((p \land q)→r)$

I just made an attempt to convince myself intuitively, I might be wrong, or a better intuitive approach might exist, if so please guide me. Thank you.


问:
A basic doubt on Lebesgue integration
Can anyone tell me at a high level (I am not aware of measure theory much) about Lebesgue integration and why measure is needed in case of Lebesgue integration? How the measure is used to calculate the horizontal strip mapped for a particular range? </p>

答:
Imagine a cashier who is in-charge of counting coins at a bank and thereby report the total money collected everyday to the bank authorities. Also, let us assume that the coins can only be of denomination $1$, $2$, $5$ and $10$. Now say he receives the coins in the following order:
$$5,2,1,2,2,1,5,10,1,10,10,5,2,1,2,5,10,2,1,1,1$$

Now he has two different ways to count.

$1$. The first way is to count the coins as and when they come, i.e., he does
$$5+2+1+2+2+1+5+10+1+10+10+5+2+1+2+5+10+2+1+1+1$$
which gives $79$.

$2$. The second way is as follows. He has $4$ boxes, one box for each denomination, i.e., the first box is for coins with denomination $1$, the second box is for coins with denomination $2$, the third box is for coins with denomination $5$ and the last box is for coins with denomination $10$. He drops the coins in the corresponding box as and when it comes. At the end of the day, he counts the coins in each box, i.e., he counts that there are $7$ coins with denomination $1$, $6$ coins with denomination $2$, $4$ coins with denomination $5$ and $4$ coins with denomination $10$. He hence finally reports the total money as
$$7 \times 1 + 6 \times 2 + 4 \times 5 + 4 \times 10 = 79$$

$\color{red}{\text{The first method is the Riemann way of summing}}$ the total money, while $\color{blue}{\text{the second method is the Lebesgue way of summing}}$ the same money.

In the second way, note that there are $4$ sets, i.e., the boxes for denominations $1$, $2$, $5$ and $10$. The measure of each of these sets/boxes is nothing but the denomination of each of these boxes, i.e., the measure of each of these sets is $1$, $2$, $5$ and $10$ respectively and the functional value on each of these sets is nothing but the number of coins in that particular denomination.

Title: A basic doubt on Lebesgue integration


问:
What does the notation $S^{n} \rightarrow S$ mean?
If S is a set and an operation on $S$ is denoted by $S^{n} \rightarrow S$, does the n in the exponent of S mean that the function's inputs are tuples of the set $S\times S \times S  \times S \space  \times \space ... \space \times S  $   (n-times)?

答:
Indeed, $S^n$ denotes the set $\underbrace{S\times S\times\cdots\times S}_{n\text{ times}}$, and until some level in math education, no further explanation is usually given.

There is quite a profound extension of this notation.

Often, with $A^B$ we denote the set of all functions $f:B\to A$. The reason we call it $A^B$ and not $B^A$ is that, in the cases where $A$ and $B$ are finite, the number of those functions ($|A^B|$) is $|A|^{|B|}$, as for every element $b\in B$ we have $|A|$ possible choices in $A$ to map $b$ into.

This lets us define:

$$|A|^{|B|}:=|A^B|$$

even for infinite sets, prove that this is a well-defined operation, and this lets us use exponentiation on cardinal numbers.

On the other hand, in set theory it is often taken that:

$$n:=\{0,1,2,\ldots,n-1\}$$

(with $0:=\emptyset$), so $S^n$ may be taken as the set of all the functions of the set (of indices) $\{0,1,\ldots,n-1\}$ into $S$, i.e. as the set of all $n$-sized sequences $(s_0,s_1,\ldots,s_{n-1})$, where $s_i\in S$ for $i=0,1,\ldots,n-1$.

This may not be exactly the same as $\underbrace{S\times S\times\cdots\times S}_{n\text{ times}}$ (in fact, this depends on how you define the latter), but there is an obvious bijection between those sets, so you may just as well identify $\underbrace{S\times S\times\cdots\times S}_{n\text{ times}}$ with $S^n$.


问:
Derivation of Euler-Lagrange equations for Brox Optical Flow
I'm having trouble understanding how the Euler Lagrange equations are derived for the optical flow equation in the paper: https://scihub.wikicn.top/10.1007/978-3-540-24673-2_3
Eq 7 on page 4 of the pdf.

$$
E(\mathit{u,v}) = \int \psi (|I(\mathbf{x+w}) - I(\mathbf{x})|^2 + \gamma |\nabla I(\mathbf{x+w}) - \nabla I(\mathbf{x})|^2) + \alpha \int \psi (|\nabla u|^2 + |\nabla v|^2)
$$

Euler lagrange equations will be :

$$
\dfrac{\partial L}{\partial u} - \dfrac{d}{dx} \dfrac{\partial L}{\partial u_{x}} - \dfrac{d}{dy} \dfrac{\partial L}{\partial u_{y}} = 0
$$

$$
\dfrac{\partial L}{\partial v} - \dfrac{d}{dx} \dfrac{\partial L}{\partial v_{x}} - \dfrac{d}{dy} \dfrac{\partial L}{\partial v_{y}} = 0
$$

I would really appreciate it if someone could walk me through the u partial derivative part. Apologies if this is straightforward. I'm new to both optical flow and variational calculus.

Thanks in advance.

答:
Given that $\mathbf{w}=(u, v, 1)^T$, so you follow the definition to compute the first term through the chain rule:
$$\frac{\partial L}{\partial u}=\psi'(|I(\mathbf{x}+\mathbf{w})-I(\mathbf{x})|^2+\gamma|\nabla I(\mathbf{x}+\mathbf{w})-\nabla I(\mathbf{x})|^2)\ 2[(I(\mathbf{x}+\mathbf{w})-I(\mathbf{x})\frac{\partial I}{\partial x}(\mathbf{x+w})+\gamma (\nabla I(\mathbf{x+w})-\nabla I(\mathbf{x}))\cdot \nabla \frac{\partial I}{\partial x}(\mathbf{x+w})].$$

Next you can calculate the dependence on $\nabla u$:
$$\frac{\partial L}{\partial \nabla u}=\alpha\psi'(|\nabla u|^2+|\nabla v|^2)\ 2\nabla u.$$

You can assemble the last two term (it probably should have been three terms with the one on $\frac{\partial}{\partial t}$) by
$$\nabla\cdot\frac{\partial L}{\partial \nabla u}=\nabla\cdot(\alpha\psi'(|\nabla u|^2+|\nabla v|^2)\ 2\nabla u).$$

Note that in that paper, they put divergence instead $\rm{div}=\nabla\cdot$


问:
Formula on the sum of all the elements of $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$
Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where:

$p_1\dots,p_n$ are primes;

$\alpha_1,\dots,\alpha_n$ are nonnegative integers;

$k:=\prod_{i=1}^np_i^{\alpha_i}$.

As an example, let's take:

\begin{alignat}{1}
\Bbb Z_{2^2} \times \Bbb Z_3\times \Bbb Z_{2}= &\{(0,0,0),(0,1,0),(0,2,0),(1,0,0),(1,1,0),(1,2,0),\\
&(2,0,0),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(3,2,0),\\
&(0,0,1),(0,1,1),(0,2,1),(1,0,1),(1,1,1),(1,2,1),\\
&(2,0,1),(2,1,1),(2,2,1),(3,0,1),(3,1,1),(3,2,1)\} \\
\end{alignat}

Then we get:

\begin{alignat}{1}
\sum_{i=1}^{24}a_i &= \bigl(3\cdot 2\cdot(0+1+2+3),\space\space 2^2\cdot2\cdot (0+1+2)),\space\space 2^2\cdot 3\cdot (0+1)\bigr) \\
&= \biggl(3\cdot 2\cdot\sum_{i=0}^{2^{2}-1}i,\space\space 2^2\cdot2\cdot\sum_{i=0}^{3-1}i,\space\space 2^2\cdot 3\cdot \sum_{i=0}^{2-1}i\biggr) \\
\end{alignat}

This suggests the following:

Claim. Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where $k:=\prod_{i=1}^np_i^{\alpha_i}$. Then:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i= \biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr) \\
\tag 1
\end{alignat}

I've tried to prove $(1)$ by induction on $(\alpha_1,\dots,\alpha_n)$. The claim holds for $(\alpha_1,\dots,\alpha_n)=(0,\dots,0)$; in fact, in this case $k=1$ and $G=\{a_1\}=\{(\underbrace{0,\dots,0}_{n\space slots})\}$, whence:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i= a_1=(\underbrace{0,\dots,0}_{n\space slots})
\end{alignat}

and

\begin{alignat}{1}
&\biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr)= \\
&\biggl(1\dots 1\sum_{i=0}^{0}i,\space\space\dots,\space\space 1\dots 1\sum_{i=0}^{0}i\biggr)= \\
&(1\dots 1\cdot 0,\space\space\dots,\space\space 1\dots 1\cdot 0)= \\
&(\underbrace{0,\dots,0}_{n\space slots}) \\
\end{alignat}

As inductive step, I've assumed $(1)$ to hold for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j,\alpha_{j+1},\dots,\alpha_n\}$ and checked what happens for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j+1,\alpha_{j+1},\dots,\alpha_n\}$, for an arbitrary $j\in\{1,\dots,n\}$. This is where I get stuck, and I don't know whether I get lost in the algebra, or I'm applying induction incorrectly, or simply the claim is false.

答:
I think you should use induction over the number of groups in the crossproduct. I will simply write $G = G_1 \times G_2 \times \cdots \times G_n$.

$q_k := |G_k|$ and $G_k = \{a^k_1 , \dots , a^k_{q_k}\}$.

For $n = 1$ the proposition is trivially true.

Now proof $n$ assuming the claim is true for $n-1$. The sum looks like this:

$$\sum_{a\in G} a = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_n=1}^{q_n}   (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n)$$

Because of the way addition is defined on product groups it follows:
\begin{equation}
\sum_{j_n=1}^{q_n}   (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n) = (q_n a_{j_1}^1, q_n a_{j_2}^2, \dots, q_n a_{j_{n-1}}^{n-1} ,\sum_{j_n=1}^{q_n} a_{j_n}^n)
\end{equation}

If we take this result and plug it back into the original equation we get:

$$\sum_{a\in G} a = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_n=1}^{q_n}   (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n) = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_{n-1}=1}^{q_{n-1}} (q_n a_{j_1}^1, q_n a_{j_2}^2, \dots, q_n a_{j_{n-1}}^{n-1} ,\sum_{j_n=1}^{q_n} a_{j_n}^n) $$

But as you may realize now  we can use our original induction result, as it is now only a sum over $n_1$ summation signs. $\square$

Remark: We did neither use that we were using the integer groups nor that our subsets over which we add even are groups. In fact the claim in your questions holds for the product of subsets of groups. All we needed was the way addition is defined on products.

Hope it helps.


问:
Present Value of Perpetuity
Really really struggling with the following question:

Suppose you receive $\$44000$ per year, but get a $\$1000$ raise every $k = 9$ years. That is, you are paid $\$44000$ in years $1, 2, ..., k$, and paid $\$44000 + \$1000$ in years $k+1, k+2, ..., 2k$, and paid $\$44000+\$2000$ in years $2k+1, 2k+2, ..., 3k$, and so on into perpetuity.  Given an interest rate $2.7\%$, what is the present value of these cash flows?  (to nearest $\$0.01$) (Hint: work out a formula rather than just calculating numerically. Can you convert this question to the PV of an annuity of length 9 years?)

The answer should be $\$1,766,315.72$.

I tried calculating annuity for $n=9$ first and get the annuity of $\$44,000+\$1000$ and $\$44,000+\$2000$ separately. Is that what I do? If yes, how?

Any help would be appreciated! Thanks.

Title: Present Value of Perpetuity

Body:
The present value can be written as an infinite sum:
\begin{align}PV \,&=\, \color{red}{44000} + \color{blue}{\frac{44000}{1.027}}+\color{green}{\frac{44000}{1.027^2}} +\color{purple}{\frac{44000}{1.027^3}} + \cdots + \frac{44000}{1.027^8}\\\\ &+\,\color{red}{\frac{45000}{1.027^9}} +\color{blue}{\frac{45000}{1.027^{10}}} +\color{green}{\frac{45000}{1.027^{11}}} +\color{purple}{\frac{45000}{1.027^{12}}}+ \cdots +\frac{45000}{1.027^{17}}\\\\
&+\,\color{red}{\frac{46000}{1.027^{18}}} +\color{blue}{\frac{46000}{1.027^{19}}} +\color{green}{\frac{46000}{1.027^{20}}} +\color{purple}{\frac{46000}{1.027^{21}}}+ \cdots +\frac{46000}{1.027^{26}}\\\\
&+ \,\,\,\, \cdots \\\\
\end{align}

We split this sum into $9$ sums, grouping terms as follows:

\begin{align}
PV \,&=\, \color{red}{44000} + \color{red}{\frac{45000}{1.027^9}} + \color{red}{\frac{46000}{1.027^{18}}} + \cdots \\\\
&+ \, \color{blue}{\frac{44000}{1.027}} + \color{blue}{\frac{45000}{1.027^{10}}} + \color{blue}{\frac{46000}{1.027^{19}}} + \cdots \\\\
&+ \, \color{green}{\frac{44000}{1.027^2}} + \color{green}{\frac{45000}{1.027^{11}}} + \color{green}{\frac{46000}{1.027^{20}}} + \cdots \\\\
&+ \, \color{purple}{\frac{44000}{1.027^3}} + \color{purple}{\frac{45000}{1.027^{12}}} + \color{purple}{\frac{46000}{1.027^{21}}} + \cdots \\\\
&\phantom{+} \qquad \vdots \\\\
&+ \, \frac{44000}{1.027^8} + \frac{45000}{1.027^{17}}+\frac{46000}{1.027^{26}} + \cdots
 \\\\ \tag{1}\end{align}

Each of these series is what is known as an arithmetico-geometric series. In general, series of this form can be reduced to a geometric series by multiplying by the common ratio and subtracting the result. In one of my previous answers, I've described this reduction, for a simpler arithmetico-geometric series.

In this case, dividing both sides of $(1)$ by $1.027^{9}$ solves the problem:

\begin{align}
 \frac{PV}{1.027^9} \,&=\,  \color{red}{\frac{44000}{1.027^9}} + \color{red}{\frac{45000}{1.027^{18}}} + \color{red}{\frac{46000}{1.027^{27}}} + \color{red}{\frac{47000}{1.027^{36}}} + \cdots \\\\
&+ \, \color{blue}{\frac{44000}{1.027^{10}}} +  \color{blue}{\frac{45000}{1.027^{19}}} + \color{blue}{\frac{46000}{1.027^{28}}} + \color{blue}{\frac{47000}{1.027^{37}}} + \cdots \\\\
&+ \, \color{green}{\frac{44000}{1.027^{11}}} + \color{green}{\frac{45000}{1.027^{20}}} + \color{green}{\frac{46000}{1.027^{29}}} + \color{green}{\frac{47000}{1.027^{38}}} + \cdots \\\\
&+ \, \color{purple}{\frac{44000}{1.027^{12}}} + \color{purple}{\frac{45000}{1.027^{21}}} + \color{purple}{\frac{46000}{1.027^{30}}} + \color{purple}{\frac{47000}{1.027^{39}}} + \cdots \\\\
&\phantom{+} \qquad \vdots \\\\
&+ \, \frac{44000}{1.027^{17}} + \frac{45000}{1.027^{26}} + \frac{46000}{1.027^{35}}+\frac{47000}{1.027^{44}} + \cdots \\\\ \tag{2}
\end{align}

Subtracting $(2)$ from $(1)$ gives:

\begin{align}
PV - \left(\frac{PV}{1.027^9} \right) \, &= \,  \color{red}{44000} + \left(\, \color{red}{\frac{1000}{1.027^9}} + \color{red}{\frac{1000}{1.027^{18}}} + \color{red}{\frac{1000}{1.027^{27}}} + \cdots \,\right)\\\\
&+\, \color{blue}{\frac{44000}{1.027}}  + \left(\, \color{blue}{\frac{1000}{1.027^{10}}} + \color{blue}{\frac{1000}{1.027^{19}}} + \color{blue}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\
&+\, \color{green}{\frac{44000}{1.027^2}}  + \left(\, \color{green}{\frac{1000}{1.027^{10}}} + \color{green}{\frac{1000}{1.027^{19}}} + \color{green}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\
&+\, \color{purple}{\frac{44000}{1.027^3}}  + \left(\, \color{purple}{\frac{1000}{1.027^{10}}} + \color{purple}{\frac{1000}{1.027^{19}}} + \color{purple}{\frac{1000}{1.027^{28}}} + \cdots \,\right)\\\\
&\phantom{+} \qquad \vdots \\\\
&+ \frac{44000}{1.027^8} + \left(\,\frac{1000}{1.027^{17}} + \frac{1000}{1.027^{26}} + \frac{1000}{1.027^{35}} + \cdots \,\right)\\\\ \tag{3}
\end{align}

Each of the nine series in parentheses on the right hand side of $(3)$ are infinite geometric series with common ratio $1/1.027^9 \approx 0.786803$. If the
first term is $a$, and the common ratio is $r$, then the sum of each infinite geometric series is $$a + ar + ar^2 + \cdots = \frac{a}{1 - r}\tag{4}$$

Applying $(4)$ to each of the nine infinite geometric series on the right hand side of $(3)$ gives:

\begin{align}
PV \left(1 - \frac{1}{1.027^9}\right) \, &=\, \left(\color{red}{44000} + \color{red}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\\\\
&+\, \left(\color{blue}{44000} + \color{blue}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{blue}{\frac{1}{1.027}}\right)\\\\
&+\, \left(\color{green}{44000} + \color{green}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{green}{\frac{1}{1.027^2}}\right)\\\\
&+\, \left(\color{purple}{44000} + \color{purple}{\frac{1000000000000000000000000000000}{270966174450725282733813987}}\right)\left(\color{purple}{\frac{1}{1.027^3}}\right)\\\\
&\phantom{+} \qquad \vdots \\\\
&+ \, \left(44000 + \frac{1000000000000000000000000000000}{270966174450725282733813987}\right)\left(\frac{1}{1.027^8}\right)\\\\
\end{align}

This is, at last, equivalent to the present value of an annuity of length $9$ years.

The present value is:

$$PV \, = \, \frac{13271419491079374076175586444556000}{7316086710169582633812977649} \, \approx  \, \$ 1\text{,}814\text{,}005.22\, $$

This calculation assumes that the first payment occurs right now.

If, instead, the first payment of $\$44000$ occurs one year from now, then every term is discounted by an additional factor of $\frac{1}{1.027}$. In that case, the present value is:

$$PV \, = \, \frac{12922511675831912440287815428000000}{7316086710169582633812977649} \, \approx \, \boxed{ \, \$ 1\text{,}766\text{,}314.72\, }$$


问:
Does the exists invertible matrix $S$ such that $S^{-1}\operatorname{diag}(A,B)S=\operatorname{diag}(B,A)$ with $\det(S)=1.$
Let $A$ be an $m$ by $m$ matrix, and $B$ be an $n$ by $n$ matrix. I know that
$$\begin{bmatrix} 0&I_n\\I_m&0 \end{bmatrix}\begin{bmatrix} A&0\\0&B \end{bmatrix}\begin{bmatrix} 0&I_m\\I_n&0 \end{bmatrix}=\begin{bmatrix} B&0\\0&A \end{bmatrix}.$$
But $\det\biggl(\begin{bmatrix} 0&I_m\\I_n&0 \end{bmatrix}\biggr)$ may not equal to $1$. I want to know whether there is an invertible matrix $S$ such that $S^{-1}\operatorname{diag}(A,B)S=\operatorname{diag}(B,A)$ with $\det(S)=1.$

答: Since $\det\pmatrix{0&I_n\\ I_m&0}=(-1)^{mn}$, if $mn$ is even, the matrix already has determinant $1$. If $mn$ is odd, then both $m$ and $n$ are odd and you may consider $S=\pmatrix{0&-I_n\\ I_m&0}$ instead.


问:
Likelihood of the 'closest' vector having the same feature?
Suppose I have vectors $\mathbf{x_1},\mathbf{x_2},\ldots,\mathbf{x_n}$, each sampled uniformly and independently from the set of vertices of the unit hypercube of dimension $d$, and that I'm then given a new vertex vector $\mathbf{v}$. If $\mathbf{x_c}$ denotes the closest of my $\mathbf{x}$ vertices to $\mathbf{v}$, does an expression exist for the probability $p$ that $\mathbf{x_c}_1 \neq \mathbf{v}_1$ (i.e. they differ in some given dimension)?

答:
Let $\mathbf D$ be the random variable representing the number of different coordinates (or the squared distance) between $\boldsymbol v$ and $\boldsymbol{x}_c$. Then by the law of total probability
$$
    \Pr[\boldsymbol v_1 \ne (\boldsymbol x_c)_1] = \sum_{k=0}^d \Pr[\boldsymbol v_1 \ne (\boldsymbol x_c)_1 \mid \mathbf D=k] \cdot \Pr[\mathbf D=k].
$$
By symmetry, $\Pr[\boldsymbol v_1 \ne (\boldsymbol x_c)_1 \mid \mathbf D=k] = \frac kd$: given that $\boldsymbol v$ and $\boldsymbol x_c$ differ in exactly $k$ coordinates, the probability is $\frac kd$ that the first coordinate is one of those $k$. So we get
$$
   \Pr[\boldsymbol v_1 \ne (\boldsymbol x_c)_1] = \sum_{k=0}^d \frac kd \cdot \Pr[\mathbf D = k] = \frac1d \sum_{k=0}^d k \cdot \Pr[\mathbf D =k] = \frac1d \mathbb E[\mathbf D].
$$

Now, we are going to compute $\mathbb E[\mathbf D]$ in a different way. For any nonnegative integer-valued random variable $\mathbf X$, the identity
$$
   \mathbb E[\mathbf X] = \sum_{k=1}^\infty \Pr[\mathbf X \ge k]
$$
holds, because $\Pr[\mathbf X=k]$ is included in the $k$ probabilities $\Pr[\mathbf X \ge 1], \Pr[\mathbf X\ge 2], \dots, \Pr[\mathbf X \ge k]$.

In the case of $\mathbf D$, $\Pr[\mathbf D \ge k] = \left(1 - \frac1{2^d} \sum_{i=0}^{k-1} \binom di\right)^n$, which is just the probability that none of $\boldsymbol x_1, \dots, \boldsymbol x_n$ are closer than that to $\mathbf v$. So we get
$$
   \Pr[\boldsymbol v_1 \ne (\boldsymbol x_c)_1] = \frac1d \sum_{k=1}^d \left(1 - \frac1{2^d} \sum_{i=0}^{k-1} \binom di\right)^n.
$$
Unfortunately, that's maybe as simple as it gets. We've gotten a pretty nice expression for the probability as a function of $n$, when $d$ is fixed and not too large; when $d$ can also vary, this sum probably has no nice closed form, because the partial sum of binomial coefficients inside it doesn't.


答:
I think you are applying induction incorretly, by applying induction on $(\alpha_1,\dots,\alpha_n)$ you are assuming that $\alpha_1=\dots=\alpha_n$ but it can happen otherwise. I think it would be easier to prove the claim for all natural numbers by induction instead of only for powers of primes.

New Claim. Let $G=\Bbb Z_{l_1}\times \dots \times \Bbb Z_{l_n}$ be listed as $G=\{a_1,\dots,a_k\}$, where $k:=\prod_{i=1}^nl_i$. Then:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i= \biggl(l_2\dots l_n\sum_{i=0}^{l_1-1}i,\space\space\dots,\space\space l_1\dots l_{n-1}\sum_{i=0}^{l_n-1}i\biggr) \\
\tag 2
\end{alignat}
Proof:
\begin{alignat}{2}
\sum_{i=1}^{k}a_i= \biggl(\sum_{i=1}^{k}i,...,\sum_{i=1}^{k}i\biggr)
\end{alignat}, where the $i$ is an element of $ \Bbb Z_{l_j}$.

therefore it suffices to prove that for each coordinate j, $\sum_{i=1}^{k}i = l_1...l_{j-1}l_{j+1}... l_n\sum_{i=0}^{l_j-1}i$.

For all $1\le j\le n$,

there are $l_1\times...l_{j-1}\times l_{j+1}...\times l_n$ elements in $\Bbb Z_{l_1}\times ... \times \Bbb Z_{l_j-1} \times \Bbb Z_{l_j+1} \times ... \times \Bbb Z_{l_n}$
therefore for each $i$ in $\Bbb Z_{l_j}$ there are $l_1\times...l_{j-1}\times l_{j+1}...\times l_n$ elements with $j-th$ coordinate $i$ in $G=\Bbb Z_{l_1}\times \dots \times \Bbb Z_{l_n}$.

Therefore  $\sum_{i=1}^{k}i = (1\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) + (2\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) + ... + (l_{j-1}\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) = (l_1\times...l_{j-1}\times l_{j+1}...\times l_n) \times (1+2+...+l_j-1) = l_1...l_{j-1}l_{j+1}... l_n\sum_{i=0}^{l_j-1}i$

Since (2) holds for all numbers it also holds for powers of primes.

Title: Formula on the sum of all the elements of $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$

Body:
Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where:

$p_1\dots,p_n$ are primes;

$\alpha_1,\dots,\alpha_n$ are nonnegative integers;

$k:=\prod_{i=1}^np_i^{\alpha_i}$.

As an example, let's take:

\begin{alignat}{1}
\Bbb Z_{2^2} \times \Bbb Z_3\times \Bbb Z_{2}= &\{(0,0,0),(0,1,0),(0,2,0),(1,0,0),(1,1,0),(1,2,0),\\
&(2,0,0),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(3,2,0),\\
&(0,0,1),(0,1,1),(0,2,1),(1,0,1),(1,1,1),(1,2,1),\\
&(2,0,1),(2,1,1),(2,2,1),(3,0,1),(3,1,1),(3,2,1)\} \\
\end{alignat}

Then we get:

\begin{alignat}{1}
\sum_{i=1}^{24}a_i &= \bigl(3\cdot 2\cdot(0+1+2+3),\space\space 2^2\cdot2\cdot (0+1+2)),\space\space 2^2\cdot 3\cdot (0+1)\bigr) \\
&= \biggl(3\cdot 2\cdot\sum_{i=0}^{2^{2}-1}i,\space\space 2^2\cdot2\cdot\sum_{i=0}^{3-1}i,\space\space 2^2\cdot 3\cdot \sum_{i=0}^{2-1}i\biggr) \\
\end{alignat}

This suggests the following:

Claim. Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where $k:=\prod_{i=1}^np_i^{\alpha_i}$. Then:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i= \biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr) \\
\tag 1
\end{alignat}

I've tried to prove $(1)$ by induction on $(\alpha_1,\dots,\alpha_n)$. The claim holds for $(\alpha_1,\dots,\alpha_n)=(0,\dots,0)$; in fact, in this case $k=1$ and $G=\{a_1\}=\{(\underbrace{0,\dots,0}_{n\space slots})\}$, whence:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i= a_1=(\underbrace{0,\dots,0}_{n\space slots})
\end{alignat}

and

\begin{alignat}{1}
&\biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr)= \\
&\biggl(1\dots 1\sum_{i=0}^{0}i,\space\space\dots,\space\space 1\dots 1\sum_{i=0}^{0}i\biggr)= \\
&(1\dots 1\cdot 0,\space\space\dots,\space\space 1\dots 1\cdot 0)= \\
&(\underbrace{0,\dots,0}_{n\space slots}) \\
\end{alignat}

As inductive step, I've assumed $(1)$ to hold for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j,\alpha_{j+1},\dots,\alpha_n\}$ and checked what happens for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j+1,\alpha_{j+1},\dots,\alpha_n\}$, for an arbitrary $j\in\{1,\dots,n\}$. This is where I get stuck, and I don't know whether I get lost in the algebra, or I'm applying induction incorrectly, or simply the claim is false.


问:
How do the 2 descriptions of column space describe the same subspace of a matrix?
From Gilbert Strang's textbook, the augmented matrix
$$
\begin{bmatrix}
1 & 2 & 3 & 5 & \mathbf{b_1}\\
2 & 4 & 8 & 12 & \mathbf{b_2}\\
3 & 6 & 7 & 13 & \mathbf{b_3}
\end{bmatrix}
$$
reduces to upper-triangular
$$
\begin{bmatrix}
1 & 2 & 3 & 5 & \mathbf{b_1}\\
0 & 0 & 2 & 2 & \mathbf{b_2-2b_1}\\
0 & 0 & 0 & 0 & \mathbf{b_3+b_2-5b_1}
\end{bmatrix}
$$

As a first description of the column space, he writes that the column space is all linear combinations of the pivot columns $(1 ,2 ,3)$ and $(3 ,8 ,7)$. As a second description. he writes that the column space is all vectors that satisfy the bottom equation $0=b_3+b_2-5b_1$.

I believe that those pivot columns are a basis for the column space, and I also checked that they satisfy that bottom equality. I just don't understand why the two pivot columns satisfy that equation.

答: Well the two vectors above define a basis for the column space and the second condition is basically giving what type of vectors are in the column space. Hence eventually the basis columns will also satisfy that condition.</p>


问:
Graph of a convex function and normal vectors
Recall that the graph of a function $f:\mathbb R^n\to\mathbb R$ is
$$
G(f)=\{ (x,y)\in\mathbb R^n\times \mathbb R \mid y =f(x)\}
$$
Suppose $f$ is a piece-wise smooth convex function.

Fix two points on the graph $P,Q\in G(f)$. We pick up two inner normal vectors $\overrightarrow N_P$ and $\overrightarrow N_Q$ at $P$ and $Q$ contained in the subset $\mathbb R^n\times \{1\}\subset \mathbb R^{n+1}$.

The normal vectors mean that $\overrightarrow N_P$ or $\overrightarrow N_Q$ is orthogonal to the tangent vector at $P$ or $Q$)

Question: Do we always have $\langle \overrightarrow N_Q -\overrightarrow{N_P}, \overrightarrow{PQ}\rangle <0$?

I ask this because of the following example. If $f(x)=x^2$ and $P=(x_0,x_0^2), Q=(x_1,x_1^2)$, then
$\overrightarrow N_P=(-2x_0,1)$ and $\overrightarrow N_Q=(-2x_1,1)$. Also, we have $\overrightarrow N_Q -\overrightarrow{N_P}=(-2(x_1-x_0),0)$ and $\overrightarrow{PQ}=(x_1-x_0, x_1^2-x_0^2)$, implying that $\langle \overrightarrow N_Q -\overrightarrow{N_P}, \overrightarrow{PQ}\rangle=-2(x_1-x_0)^2<0$

答:
First let us take the case $n=1$, i.e., $f\colon\mathbb{R}\to\mathbb{R}$.

Up to some stretching and translating, let us say that $P=(0,0)$, $Q=(1,f(1))$, $N_P=(-f'(0),1)$, $N_Q=(-f'(1),1)$. Then
$$N_Q-N_P=(f'(0)-f'(1),0)$$
and
$$Q-P=Q=(1,f(1))$$
so
$$\langle N_Q-N_P,Q-P\rangle=f'(0)-f'(1)$$

Now you should know that the derivative of a convex function is non-decreasing (see details below, as I could not find the proof of this here at MSE), so $f'(0)-f'(1)\leq 0$.

For general $f\colon\mathbb{R}\to\mathbb{R}$, a similar argument will yield the result. Or you can "stretch and translate", as in the beginning of the solution: If $P=(x_0,f(x_0))$ and $Q=(x_1,f(x_1))$, apply the case above to $g(x)=f((x_1-x_0)x+x_0)-f(x_0)$ and rewrite the normals of $f$ in terms of the normals of $g$.

As for the fact I mentioned about derivatives of convex functions, we need the following general fact (see Prob. 23, Chap. 4 in Baby Rudin: Every convex function is continuous and every increasing convex function of a convex function is convex):

if $f$ is convex on an interval containing $x_1<x_2<x_3$, then
$$\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq\frac{f(x_3)-f(x_1)}{x_3-x_1}\leq\frac{f(x_3)-f(x_2)}{x_3-x_2}$$
(i.e., the slope of the secans increase when you move them to the right)

Then given $x_0<x_1$, we have, for all $x_0<z<w<x_1$,
\begin{align*}
\frac{f(z)-f(x_0)}{z-x_0}
&\leq\frac{f(x_1)-f(x_0)}{x_1-x_0}\\
&\leq\frac{f(x_1)-f(w)}{x_1-w}\\
&=\frac{f(w)-f(x_1)}{w-x_1}
\end{align*}
Letting $z\to x_0^+$ and $w\to x_1^-$, we conclude
$$f'(x_0)\leq f'(x_1)$$

For general $n$, let $f\colon\mathbb{R}^n\to\mathbb{R}$ convex. The tangent space at a point $(x_0,f(x_0))$ of the graph is given by vectors of the form $(x,\nabla f(x_0)\cdot x)$ (where $\nabla f$ is the gradient of $f$ and "$\cdot$" denotes the usual inner product). It follows that the inner normal vectors are given by $(-\nabla f(x_0),1)$.

Let $P=(x_0,f(x_0))$ and $Q=(x_1,f(x_1))$ be given, with $x_0,x_1\in\mathbb{R}^n$. Then
$$\langle N_Q-N_P,Q-P\rangle=(\nabla f(x_1)-\nabla f(x_0))\cdot (x_0-x_1)\tag{$\bigstar$}$$

The following argument is a standard way of transforming a multivariable problem into a single-variable problem: Restrict your function to a line!

Consider the function $g\colon\mathbb{R}\to\mathbb{R}$ given by $g(\lambda)=f((1-\lambda)x_0+\lambda x_1)$. Then $g$ is convex. The single variable case yields $g'(1)\geq g'(0)$.

Let us use the chain rule to calculate $g'$ in terms of $\nabla f$ (I can never remember the formulas, so skip this if you want). I'll denote the derivative of a function $F\colon\mathbb{R}^N\to\mathbb{R}^M$ at a point $x_0$ by $dF_{x_0}$.

Let $h\colon\mathbb{R}\to\mathbb{R}^n$, $h(\lambda)=(1-\lambda)x_0+\lambda x_1=x_0+\lambda(x_1-x_0)$. Then

$dh_\lambda(t)=t(x_1-x_0)$ for any $\lambda$

$df_{x_0}(x)=\nabla f(x_0)\cdot x$.

$dg_\lambda(t)=d(f\circ h)_\lambda(t)=df_{h(\lambda)}(dh_{\lambda}(t))=\nabla f(h(\lambda))\cdot(x_1-x_0)$

$g'(\lambda)=dg_\lambda(1)$

So in particular
$$g'(1)=\nabla f(h(1))\cdot(x_1-x_0)=\nabla f(x_1)\cdot(x_1-x_0)$$
and similarly
$$g'(0)=\nabla f(x_0)\cdot (x_1-x_0)$$

Using these equalities along with $g'(0)\leq g'(1)$ and equation ($\bigstar$), we conclude that $\langle N_Q-N_P,Q-P\rangle\leq 0$, as we wanted.


答:
Suppose we plug in the first pivot column $1,2,3$ for $b_1, b_2, b_3$. We get the matrix $$\begin{bmatrix}1 & 2 & 3 & 5 & 1 \\ 2 & 4 & 8 & 12 & 2 \\ 3 & 6 & 7 & 13 & 3\end{bmatrix}$$ and as we row-reduce it, it will always be true that the first column is the same as the last. (Row operations cannot change this property!) Therefore we must end with $$\begin{bmatrix}1 & 2 & 3 & 5 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0\end{bmatrix}.$$ In other words: when we start with $(b_1, b_2, b_3) = (1,2,3)$, we get $(b_1, b_2-2b_1, b_3+b_2-5b_1) = (1,0,0)$. In particular, the condition that the last row is $0$ must be true of the column space.

The same will be true if we set $(b_1, b_2, b_3)$ equal to the other pivot column $(3,8,7)$. Then the last column will always equal the third column as we row-reduce. So when we're done, the last entry of the third column ($0$) will equal the last entry of the last column ($b_3+b_2-5b_1$).

Title: How do the 2 descriptions of column space describe the same subspace of a matrix?

Body:
From Gilbert Strang's textbook, the augmented matrix
$$
\begin{bmatrix}
1 & 2 & 3 & 5 & \mathbf{b_1}\\
2 & 4 & 8 & 12 & \mathbf{b_2}\\
3 & 6 & 7 & 13 & \mathbf{b_3}
\end{bmatrix}
$$
reduces to upper-triangular
$$
\begin{bmatrix}
1 & 2 & 3 & 5 & \mathbf{b_1}\\
0 & 0 & 2 & 2 & \mathbf{b_2-2b_1}\\
0 & 0 & 0 & 0 & \mathbf{b_3+b_2-5b_1}
\end{bmatrix}
$$

As a first description of the column space, he writes that the column space is all linear combinations of the pivot columns $(1 ,2 ,3)$ and $(3 ,8 ,7)$. As a second description. he writes that the column space is all vectors that satisfy the bottom equation $0=b_3+b_2-5b_1$.

I believe that those pivot columns are a basis for the column space, and I also checked that they satisfy that bottom equality. I just don't understand why the two pivot columns satisfy that equation.


问:
Question about integer congruence modulo positive primes.
I am going through a book on abstract algebra over Christmas break, and I've come across this exercise that I'm having trouble cracking. This is in the introductory section on modular arithmetic and congruence classes, so it's more relevant to number theory than algebra. Here it is:

If $a$ and $b$ are integers such that $a\equiv b\,(\mbox{mod }p)$ for every positive prime $p$, prove that $a=b$.

Here's what I've tried:

Direct proof. I've tried beginning under the assumption that $a\equiv b\,(\mbox{mod }p)$ for every prime $p$. By definition, this means $a-b=pk$ for some $k\in\mathbb{Z}$. To show that $a=b$, I'd have to show that $k=0$. Right? I can't come up with a way to get to this conclusion.

Contrapositive Proof. I've also tried beginning under the assumption that $a\neq b$. For this method, I'd need to show that there exists a positive prime $p$ such that $a\not\equiv b\,(\mbox{mod }p)$. I like beginning here since it only requires that I find one such positive prime, but it's complicated in the sense that I need to show noncongruence, which I know is more difficult than showing congruence.

I have a feeling that one of these two proof methods will work. I thought about doing a proof by contradiction, but that lost momentum pretty quickly. Any hints as to how I should proceed would be greatly appreciated. This is just for fun, so no rush. Thanks.

答:
Hint:

The congruence class of an integer $x$ has a unique representative $\in [\![0 ..p-1]\!]$, denoted $x\bmod p$.

If $p>x$, what is the value of $x\bmod p$?


问:
The parametric representation of a line
I have a lecture for my linear algebra and geometry module that focuses on the cross product and line and planes within Euclidean geometry. The lecture notes talk about the parametric representation of a line which is something I covered in my A-levels, however I can't make the connection in my head between what I learned at A-level and what was in the lecture leading to me not being able to understand this content. The part that I am struggling on is as follows:

The straight line through the point $P\in\Bbb R^n$ in direction of vector $\vec a$ has the parametric form:
$$\vec x (t)=P+\vec a t\quad\forall t\in\Bbb R$$
Similarly, given two points $P, Q\in\Bbb R^n$, the line passing through $P$ and $Q$ has parametric form:
$$\vec x (t)=P+t(Q-P)\quad\forall t\in\Bbb R$$
I was wondering if anyone could explain how this links back to earlier content that I have studied, hopefully allowing me to understand this?

答: Have a look at 3D vectors online to understand how your teacher maps this onto a graph it will help a lot with your understanding</p>

Title: The parametric representation of a line


问:
Simplify the expression: $A'BC' + A'BC + AB'C + ABC' + ABC$
The online calculator(not posting the name because I don't know if its's allowed) is giving me a different result and I can't find what I'm doing wrong:

Mine:

$A'BC' + A'BC + AB'C + ABC' + ABC=
BC'(A'+A) + BC(A' + A) + AB'C=
BC' + BC + AB'C=
B(C' + C) + AB'C=
B + AB'C$

The calculator's:

$=AC + BC' + BA'$

答:
The two answers are equivalent:

$$\begin{align*}
AC+BC'+BA'&=AC+B(A'+C')\\
&=AC+B(AC)'\\
&=AC(B+B')+B(AC)'\\
&=ACB+ACB'+B(AC)'\\
&=B\big(AC+(AC)'\big)+AB'C\\
&=B+AB'C
\end{align*}$$


问:
Embedding linear orders into the reals such that infima/suprema are preserved
Let $(A,\leq^A)$ be a countable linear (or total) order. It is well-known that $(A,\leq^A)$ order-embeds into the rationals $(\mathbb Q,\leq)$.

What I am wondering is whether there also is an embedding of $(A,\leq^A)$ into the reals $(\mathbb R,\leq)$ such that, whenever the infima/suprema
$$\bigwedge_i a_{i}\text{ or }\bigvee_{i}b_i$$
exist in $(A,\leq^A)$, then these are preserved by the embedding.

I think this should be true and my proof would be akin to the embedding of $(A,\leq^A)$ into the rationals (for simplicity, I embed the order into the open unit interval): enumerate the elements of $A$ by $(a_n)_{n\in\mathbb N}$ and then define the mapping
$$q_0:a_0\mapsto 1/2$$
and from $q_n$, define
$$q_{n+1}:=q_n\cup\left(a_{n+1}\mapsto\frac{q_n(x)+q_n(y)}{2}\right)$$
where $x,y\in\mathrm{dom}(q_{n})=\{a_0,\dots,a_n\}$ are such that they fulfill
$$x<^Aa_{n+1}<^Ay$$
and $x$ is maximal with this property and $y$ minimal with this property in $\mathrm{dom}(q_n)$. The embedding $q$ is given by the union of all the $q_n$.

My question is whether the above is correct and whether there is maybe a simpler proof? Could it be possible to show that the completion of $(A,\leq^A)$ embeds into $(\mathbb R,\leq)$ while preserving infima/suprema (the completion of $(\mathbb Q,\leq)$)?

答:
Your proof isn't complete, since you haven't justified that $q$ in fact has the desired properties. At a glance I suspect that it (or rather, its modification to account for the case where $a_{n+1}>a_i$ for all $i\le n$ and similarly with $<$) does indeed work, but I also think it will be tedious to verify this.
To avoid doing work, here's another approach, which ultimately uses similar machinery but hides it "under the hood" via an application of Cantor's theorem:

First, we introduce a construction. For a linear order $S$, let $S'$ be the linear order gotten from $S$ by adding a new point $p$ between every pair of points $x,y\in S$ such that $y$ is the immediate successor of $x$ in $S$. We now have the following observation:

Suppose $X\subseteq S$ and $y\in S$ is the least upper bound of $X$ in $S$. Then $y$ remains the least upper bound of $X$ in $S'$.

Proof. There are two cases. If $X$ has a greatest element, that element must be $y$, and trivially $y$ will remain the greatest element of $X$ no matter how we add points to $S$. If $X$ does not have a greatest element, then every new point $a\in S'\setminus S$ which is $<y$ is also $<x$ for some $x\in X$, and so no point in $S'$ below $y$ is greater than every element of $X$. $\quad\Box$

The analogous result holds for infima:

Suppose $X\subseteq S$ and $y\in S$ is the greatest lower bound of $X$ in $S$. Then $y$ remains the greatest lower bound of $X$ in $S'$.

Using these facts, we can proceed as follows. Let $A$ be a countable linear order; WLOG $A$ is nonempty and without endpoints. Let $A_0=A$, and for $i\in\mathbb{N}$ let $A_{i+1}=A_i'$. Setting $B=\bigcup_{i\in\mathbb{N}} A_i$, we see that $B$ is a dense linear order without endpoints. Now fix an isomorphism $f:B\cong\mathbb{Q}$ and let $g=f\upharpoonright A_0$; $g$ is an embedding of $A$ into $\mathbb{Q}$, and by the facts above we have that $g$ preserves arbitrary sups and infs.


答:
This is more about Lebesgue integration in general, and not the the horizontal strip business.

I imagine it something like this: the Riemann integral is only able to approximate functions by rectangles. Rectangles basically only use a single set that we "know" the length of: the interval. It's easy to compute the length of an interval, and so if $\chi_{[a,b]}$ is the function that is 1 on the interval $[a,b]$ and 0 off of it, we can easily compute the integral $$\int c \chi_{[a,b]} = c(b-a).$$ To compute the integrals of other functions, we approximate them by functions like these. This gives the Riemann integral.

But the Riemann integral has a few issues. It doesn't behave well with limits and there are lots of functions that "ought" to be integrable but aren't. So what we do is replace boring intervals $[a,b]$ as our "basic integration set" with a much larger class: measurable sets. Think of these sets as being a very, very large collection of sets that we can find the length of. A measure is an assignment of a number to each of these sets in a way that is compatible with our notion of area. Let's call $\mu(A)$ the measure of a set, given by the measure $\mu$. This is basically the area of $A$, or perhaps the length of $A$.

Now, we think about our function $\chi_A$ again, that takes the value of $1$ on $A$ and $0$ off $A$. If our integral does anything like what is should, then we had better have $$\int \chi_A d\mu = \mu(A).$$ How do we compute the integrals over other functions? We essentially approximate them by finite sums of functions like the ones above and that tells us what the area should be.

There are, of course, lots of details not mentioned here, but in short: the Lebesgue integral allows more flexibility in approximation by letting us approximate by a much richer collection of sets.

Title: A basic doubt on Lebesgue integration

Body: Can anyone tell me at a high level (I am not aware of measure theory much) about Lebesgue integration and why measure is needed in case of Lebesgue integration? How the measure is used to calculate the horizontal strip mapped for a particular range? </p>


问:
Find constant term in the expansion of $\left (1+\frac x2 -\frac 2x \right)^4$
Finding the constant term for

$$\left (1+\frac x2 -\frac 2x \right)^4$$

is easy, but that would require converting the expression into a binomial.

However, I have no idea about how to do that.   Completing squares doesn’t work because we have a negative sign on $\frac 2x$.

答:
The combinatorial approach from @Christoph is elegant and efficient. Here we do it the same way using just a slightly different notation. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. We also use consequently multinomial coefficients which is convenient in the current situation.

We obtain
\begin{align*}
\color{blue}{[x^0]}&\color{blue}{\left(1+\frac{x}{2}-\frac{2}{x}\right)^4}\\
&=\binom{4}{4,0,0}1^4\left(\frac{1}{2}\right)^0\left(-2\right)^0+\binom{4}{2,1,1}1^2\left(\frac{1}{2}\right)^1(-2 )^1
+\binom{4}{0,2,2}1^0\left(\frac{1}{2}\right)^2(-2)^2\\
&=\frac{4!}{4!0!0!}\,1\cdot 1\cdot 1+\frac{4!}{2!1!1!}\,1\cdot\frac{1}{2}(-2)+\frac{4!}{0!2!2!}\,1\cdot\frac{1}{4}\cdot4\\
&=1-12+6\\
&\,\,\color{blue}{=-5}
\end{align*}


问:
Differentiating $\frac{2t-4}{(t^2+2)^2}$ by the quotient rule?
The quotient rule $\frac{u}{v} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

There comes a point where I need to differentiate my $v$ which is $(t^2+2)^2$, so when it comes to $\frac{dv}{dx}$ how do I go about this? Do I just expand the brackets like this $(t^2+2)(t^2+2) = t^4+2t^2+2t^2+4$ and then differentiate those terms and plug them back in the formula?

I found that the result is quite messy.

答: Hope this helps-Final answer as followed


问:
Extension field of degree 2 contains an element whose square is in the field
Little question.
Let $F$ be a field with $\operatorname{char}(F) \neq 2$. Let $E$ be a field extension such that $[E:F]=2$

I need to show there exists $y \in E$ such that $y^2 \in F$ and $E=F[y]$.

Let $x \in E$ and not in $F$. If $[E:F]=2$, then $\deg(m_x(T))=2$ and so $m_x(T)=T^2+bT+c$ for some $b,c \in F$. So for all $x$ not in $F$: $x^2+bx+c=0$ (still for some $b(x),c(x) \in F$ )

The idea now is to set $y=x-\frac{b}{2}$ which
$\textbf{if} y \notin F$ leads to $y^2=\frac{b^2}{4}-c \in F$ and clearly $y$ satisfies $E=F[y]$

My questions is: why (if true) may I assume $y$ is not in $F$?

Thank you.

答: Since $m_x(T)$ has degree $2$, $\{1,x\}$ is an $F$-basis for $F[x]$, that is, $F[x] = \{a+bx : a,b \in F\}$. So, an element of $F[x]$ is in $F$ if and only if can be written as $a+bx$ with $b=0$. In particular $y \notin F$.


问:
Visualising the Minkowski sum of two triangles
Consider the figure given below. I am told that the Minkowski sum of the two triangles $A,B \subset \mathbb{R}^2$ is the regular hexagon, as shown right next to them.

The Minkowski sum is defined as $$A+ B = \{a+b: a\in A, b\in B\}$$

Note that $A$ and $B$ are such that $A$ is the reflection of $B$ about the origin (and vice versa). So, knowing that the sum is a regular hexagon, I can somewhat make sense of it, but I'm not really able to see why it is a regular hexagon in the first place! Is there an intuitive way of looking at this? Finding $A+ B = \{a+b: a\in A, b\in B\}$ manually using every $a\in A, b\in B$ is obviously impossible, so there must be some special $a_1,a_2,...,a_p \in A$ and $b_1, b_2,...,b_q \in B$ we should use to construct the figure (extreme cases of some sort). Which ones are these?

As an addendum, I am wondering if there is some general intuition (at least in $\mathbb{R}$,$\mathbb{R}^2$, and $\mathbb{R}^3$) I can carry with me, when finding the Minkowski sum of two figures (sets)?

P.S.

I am not looking for algorithms to find Minkowski sums.

答:
Here is a GIF I made that might help you visualize the Minkowski sum; the animation shows translates of the downward-facing triangle by points on the upward-facing triangle as they cycle around the perimeter, and how those translates fit onto the hexagon.

Of course, the Minkowski sum also includes translates of the downward-facing triangle by interior points of the other triangle, so this picture is not quite complete - if the second triangle were larger, the red triangles in the animation would only trace out the outer part of the resulting hexagon.

In your notation, we're looking at translates of $a$ in the boundary of $A$ by $b$ in the boundary of $B$, where each frame shows a single value of $b$. When the shapes are convex, the Minkowski sum is given by taking the convex hull of these "extreme" sums. In fact, in the case of convex polygons (or polyhedra, in the 3-dimensional case), it suffices to take just the vertices of $A$ and $B$.


问:
Double integral of a function
I am trying to calculate:

$$J = \int\limits_0^1 \int\limits_0^1 \dfrac{x-y}{(x+y)^3} \, dy \, dx .$$

I tried the substitutions $u=x-y$ and $v=x+y$ and found $J=1/2$, $0 \leq v \leq 2$ and $-1 \leq u \leq 1$ but found $J=0$.  Is my way of thinking correct?

答:
This double integral does not converge absolutely, and so the answer will depend upon how we interpret and calculate it.

If we treat is as literally the iterated integral in that order, we obtain
$$
\int\limits_0^1 \bigg( \int\limits_0^1 \dfrac{x-y}{(x+y)^3} \, dy \bigg) \, dx = \int\limits_0^1 \bigg( \dfrac{y}{(x+y)^2} \bigg|_0^1 \bigg) \, dx = \int\limits_0^1 \dfrac{1}{(x+1)^2} \, dx = \frac{-1}{x+1} \bigg|_0^1 = \frac12.
$$
However, if we try to calculate the iterated integral in the opposite order, we get a different answer:
$$
\int\limits_0^1 \bigg( \int\limits_0^1 \dfrac{x-y}{(x+y)^3} \, dx \bigg) \, dy = \int\limits_0^1 \bigg( \dfrac{-x}{(x+y)^2} \bigg|_0^1 \bigg) \, dy = \int\limits_0^1 \dfrac{-1}{(y+1)^2} \, dy = \frac{1}{y+1} \bigg|_0^1 = -\frac12.
$$
And other ways of evaluating the integral would lead to other answers as well—for example, interchanging the variables $x$ and $y$ preserves the region of integration but negates the integrand, which (if valid) would lead to $J=-J$ and thus $J=0$.

So in this case, we don't have free reign to use any technique we want to evaluate the integral—we have to evaluate it as given to us (the first calculation above).

Title: Double integral of a function


问:
Expansion of conditional probability with $X$ and $Y$
Question
$\mathbb{E}(X^2 +Y | Y =2)  $

Whereas the answer is
$\mathbb{E}(X^2|Y =2)+2$

Where did I go wrong? What would your approach be?

答:
I do not see a difference of two squares here, and also how a factorization of $X^2 + Y$ could help here. On the other hand, every conditional expectation $\mathbf E[- \mid \mathfrak A]$ is linear. Hence
$$ \def\E#1{\mathbf E\left[#1\right]} \E{X^2 + Y \mid Y = 2} = \E{X^2\mid Y = 2} + \E{Y \mid Y =2} = \E{X^2\mid Y = 2} + 2 $$
To discuss your solution here, you should write down your calculations.


问:
Are stationary states of infinite irreducible Markov chains unique?
Suppose I have a continuous-time Markov chain with a countable state space, and suppose that it is irreducible, in the sense that every state is accessible from every other state.

Since the state space is infinite it need not have a stationary state, but if it does have one, is it guaranteed to be unique? If so, how can I prove this?

In the case of a finite state space the existence and uniqueness of the stationary state follow straightforwardly from the Perron-Frobenius theorem, but the countably infinite case seems not so obvious.

答:
Answer: Yes, if a stationary state exists, then it is unique.

Explanation: There is indeed an infinite dimensional version of Perron Frobenius theory which yields the result you are seeking for.

More precisely, a continuous-time Markov chain on a countable state space can be described by a stochastic $C_0$-semigroup $T$ on the space $\ell^1$. The Markov chain is irreducible if and only if the semigroup $T$ is irreducible in the sense of positive $C_0$-semigroups on Banach lattices. (That latter definition of irreducibility can, for instance, be found in [1, Definition C-III-3.1 on page 306] or in [2, Definition 14.9 on page 222]).

The fact that irreducibility of the semigroup implies that its fixed space is at most one-dimensional is, for instance, proved in [1, Proposition C-III-3.5(c) on page 310] or in [2, Proposition 14.12(c) on page 224].

Remarks:

(a) I'm pretty sure that there are also more stochastic approaches to the theory; but personally, I'm more familiar with the operator theoretic approach described above.

(b) Concerning the definition of irreducibility in [1, Definition C-III-3.1 on page 306], one has to be a bit careful. In contrast to what is claimed in this reference, assertion (iii) there is not equivalent to the other assertions in this definition (and thus, (iii) is not equivalent to irreducibility). However, if the space $E$ under consideration has order continuous norm (as is, for instance, the case for $\ell^1$), then (iii) is indeed equivalent to the other assertions in the definition.

References:

[1] Arendt, Grabosch, Greiner, Groh, Lotz, Moustakas, Nagel, Neubrander, Schlotterbeck: One-parameter Semigroups of Positive Operators, 1986 (zbMATH and link to publisher; the book is also freely available on the homepage of Rainer Nagel)

[2] Bátkai, Kramar Fijavž, Rhandi: Positive Operator Semigroups - From Finite to Infinite Dimensions, 2007 (zbMATH and link to publisher)


问:
Order of operations (BODMAS)
$$40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5$$

I'm studying and this is from an example.

In BODMAS, aren't addition and subtraction have same level?

So, in the 3rd line, it should be from left to right, correct?

答:
I think you have your answer but still I want to clear this thing to you.
$$40-10+22.5$$
 you are writing it as    $40-(10+22.5)$ but if you try to remove this you will get $40-10-22.5$ which is not same as previous line so if you want to do add and subtraction in a line (they have same priority) you can think as $10$ and $22.5$ have opposite sign so we write their difference and put the sign of bigger number to the difference so here we get

$-10+22.5=12.5$

then we solve this

$40+12.5=52.5$

Title: Order of operations (BODMAS)


问:
On the join of simplicial sets as a dependent product
Prop. 3.5 of Joyal notes in quasicategories gives a description of $X\star Y$ as $i_*(X,Y)$, where $i^*\dashv i_*$ is the adjunction
$$
i^*\colon \mathbf{sSet}/\Delta[1] \leftrightarrows \mathbf{sSet}/\partial\Delta[1]
$$
where $i^*$ is "pullback along $\partial\Delta[1]\to \Delta[1]$ and $i_*$ is the associated dependent product, which being $\partial\Delta[1]$ discrete should morally be a mere product; this baffles me a lot, since $X\star Y$ seems much more complicated than what I get if I blind myself and I try to write what $i_*(A,B)$ is for $A,B\in \bf sSet$ (confusing ${\bf sSet}/\Delta[1]$ with ${\bf sSet}^{\Delta[1]}$ it chooses $A\times B\to B$).

Here are the words of Joyal:

any clue?

I have some observations to make: somewhere there must be a typo, since either Joyal's symbol for adjointness is reversed, or he uses the left adjoint of pullback, $i_!$, which simply composes $A\to \partial\Delta[1]$ with $i$ (but this is pretty strange too, since then I am dealing with the mere coproduct map $X\sqcup Y\to \partial\Delta[1]\to\Delta[1]$).

I have absolutely no clue about what he does in the proof so I tried something different: my alternative proof strategy was to simply prove that the unit $X\star Y\to i_*(X,Y)$ is a monic (only in this component); it's always a split epi since $i^*$ is a full functor. And, alas, I'm not able to prove this either!

答:
This is coming way late, but I ran into similar trouble with this and figured I would expand on the answer given, as it is difficult for us beginners and there are never too many details to be shared.

We know that the pullback functor $i^*$ has a left and right adjoint, this is common to locally Cartesian closed categories, see this answer for more details. We typically denote the adjoints as $i_! \dashv i^* \dashv i_*$ or in the answer as $\Sigma_i \dashv i^* \dashv \Pi_i$. So yes Joyal did mess up some notation where he writes $i^* \vdash i_*$, but maybe that is just how he chooses his notation, idk. Now $i_!$ is easy to compute, it is basically post-composition, but by the answer I linked to, it is comparably difficult to compute $i_*$ directly. So in this solution we probe $i_*(X,Y)$ with simplices instead.

So suppose we have a morphism $\Delta^n \xrightarrow{f} \Delta^1$. We know that $\Delta^1 \cong \Delta^0 * \Delta^0$, and the nerve functor is fully faithful, so $\mathsf{sSet}(\Delta^n, \Delta^1) \cong \mathsf{Cat}([n], [1])$, so by the discussion in section 23.5 by Rezk, we know that $f$ decomposes uniquely into a map $\Delta^{i_f} * \Delta^{j_f} \xrightarrow{f_1 * f_2} \Delta^0 * \Delta^0$. Notice that the number of ways of decomposing $\Delta^n$ in this form is the same as the number of components in $$(X*Y)_n = X_n \sqcup X_{n-1} \times Y_0 \sqcup \dots \sqcup X_i \times Y_j \sqcup \dots \sqcup X_0 \times Y_{n-1} \sqcup Y_n$$
Now a morphism in $(\mathsf{sSet}/\Delta^1)(\Delta^n \xrightarrow{f} \Delta^1, i_*(X \sqcup Y) \xrightarrow{p} \Delta^1)$ is equivalent to a choice of simplex $x \in i_*(X \sqcup Y)$. However, by the adjunction this is also equivalent to a map in $(\mathsf{sSet}/\partial \Delta^1)(i^*(\Delta^n \xrightarrow{f} \Delta^1), X \sqcup Y)$, but by Lemma 3.4 in Joyal's notes linked above, $i^*(\Delta^n \xrightarrow{f} \Delta^0)$ is then the map $\Delta^{i_f} \sqcup \Delta^{j_f} \to \Delta^0 \sqcup \Delta^0$. Thus a map $i^*(\Delta^n \xrightarrow{f} \Delta^1) \to X \sqcup Y$ is just a choice of a $i_f$-simplex in $X$ and a $j_f$-simplex in $Y$, which we recognize as a choice of an $n$-simplex in $X * Y$. This provides us with our desired bijection. Since this bijection holds for all simplices, it tells us that there is an isomorphism $X * Y \cong i_*(X\sqcup Y)$.