问:
Creating the Liar's Paradox with one truth and one lie (that aren't meta).
Some friends were playing a game where you say 1 truth and 1 lie about yourself, and the others have to guess which is which.

Just for fun, I was wondering if there was a reasonable way to give 2 statements about myself (1 assumed to be truth, and the other a lie) that would generate a Liar's Paradox (i.e. assigning a truth value either way leads to a paradox).

Obviously, I could make the statements:

1.) Statement 2 is False.

2.) Statement 1 is True.

and that would cleanly get me the Liar's paradox, but each of those are "meta" statements (statements that explicitly refer to the other statement). But what I want are more "reasonable" statements (preferably that could be about myself) that would lead to such a paradox. Statements like "I own all 50 state coins", or "I've never eaten waffles", etc..

答:
Since realty is logically consistent (to the best of our knowledge), it is not possible to obtain truly paradoxical statements that are purely factual (without changing our fundamental understanding of the world). However, as you noted we can get such statements using meta references.

However, it is still possible to satisfy the intent of your question.

One solution can be to make factual statements about the kind of statements you make.

Example:

Statement 1: I have not told any lies in this game

Statement 2: I never speak the truth

The statements are fine by themselves if spoken from third person point of view  ("Person A hasn't told any lies in this game" and "Person A never speaks the truth" are individually non-paradoxical statements") but cause a paradox if spoken by the same person.

Assume that statement 1 is true. Then statement 2 is a lie since you have spoken a true statement.

If statement 2 is true, then it is again a contradiction.

The trick is to make meta statements without making explicit references to the statements ;)

Note that statement 2 is actually sufficient by itself here. It is equivalent to the famous self-contradictory statement "This sentence is false".

So, strictly you could say:

The Earth is a planet (Truth)

I never speak the truth (Lie)


问:
Proof by contradiction and invalid reasoning
I am very aware of the several posts that are in this forum, but to be honest I am not quite satisfied with the provided answers. Also, I can tell that the best post is the following:

Proof by Contradiction, Circular Reasoning?

I would love to continue the discussion in that post, however my reputation level does not allow me to add any comments, and I still have some questions. Therefore, I want to continue discussing more about proof by contradiction and circular reasoning.

Suppose we wish to prove $(P \wedge Q) \rightarrow R$. We assume P and Q. Proof by contradiction begins by assuming $\neg R$, and from these two assumptions, a "contradiction" is derived, and sometimes that contradiction is that $\neg P$ or $\neg Q$ (our initial valid assumptions). Therefore, since there is a contradiction with our assumptions, we conclude that it cannot happen that $\neg R$, so it must be the case that $R$.

This proof strategy bothers me, because I have the feeling that there is some kind of circular (or invalid) reasoning going on there i.e. $(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$, so if we already know that the consequent is false, in order for the statement to be true, $\neg R$ has to be false, so R is true, but R is part of the consequent part of the conditional statement (isn't doing this like assuming there is a biconditional i.e. because R is true, then $P \wedge Q$ is true?). I probably don't know how to explain it better, but how/why can we say that this is indeed a proof? Hence, my main question is why it is valid to assume $\neg R$ if our conclusion involves R? I have one explanation that kind of makes sense to me (still a little iffy though):

In more simple terms, where we want to prove that $P \rightarrow Q$, we are trying to do the following: $[(P \wedge \neg Q) \rightarrow Contradiction] \rightarrow (P \rightarrow Q)$ if we can prove that $(P \wedge \neg Q)$ leads to a Contradiction, then we are indeed proving that $P \rightarrow Q$, which was our original goal. Given that this is a tautology, whenever we follow this proof strategy structure, we are always proving $P \rightarrow Q$. However, saying that this is a tautology kind of bothers me for some reason.

I would sincerely appreciate your insights, and if there is anything you would improve in my post, please feel free to do so. I'm starting my formal journey in the proof realm (currently reading How To Prove It), and I have always heard that proofs by contradiction were the easiest strategy. However, I have learned that this strategy seems to be difficult at the beginning for all those who are trying to learn proof strategies.

答:
$(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$

Yes, and the "proof by contradiction" is just proving $\neg R \rightarrow \neg(P \wedge Q)$ instead of $(P \wedge Q) \rightarrow R$.

ContentLicense: CC BY-SA 4.0


问:
mirror image with respect to a parabola
$$P: y^2 = 4x$$
$$L: x+y+4 = 0$$

Find mirror image of P with respect to L and mirror image of L with respect to P.

I know how to find the mirror image of a curve with respect to a line... Mirror image of P with respect to L is $x^2+4y+8x+32=0.$ How to find the mirror image of L with respect to P? Or for that matter mirror image of any curve with respect to a parabola? I'm trying to find some relations for closed form.

EDIT:
Someone suggested that this problem was duplicate as another one: find image of parabola with respect to line. Please note that its not. I've seen the solution for that, and understood it.

答:
A parametric equation for the parabola $y^2 = 4x$ is $P(t) = (t^2, 2t)$

Then the slope of the tangent line to the parabola at the point $P(t)$ is

$$\dfrac{dy}{dx} = \dfrac{2 \, dt}{2t \, dt} = \dfrac 1t$$

Thus the equation of the normal line to the parabola at the point $P(t)$ is

\begin{align}
   y - 2t &= -t(x - t^2) \\
   y-2t &= x -tx + t^3\\
   tx + y &= 2t + t^3
\end{align}

This line intersects the line of reflection, $x+y=-4$, at

$$ L(t) = \left(\frac{4 + 2t + t^3}{t-1}, -\frac{6t+t^3}{t-1} \right)$$

Thus, $L'(t)$, the reflection of the point $L(t)$ about the parabola, $P(t)$ is

$$L'(t) = 2P(t) - L(t) =
\left(\dfrac{t^3 - 2 t^2 - 2 t - 4}{t - 1},
       \dfrac{t^3 + 4 t^2 + 2 t}{t - 1}\right)$$

Title: mirror image with respect to a parabola


问:
Understanding a proposition about $\mathcal R$-section in ordered sets (VERY EDITED)
The correct title must be: Proof of Theorem 90 of Appendix of Kelley's book General Topology or, Proving a theorem about $\mathcal R$-sections or Prove of this prove that proof aboutTheorem 90 of Appendix of Kelley's book General Topology .

Proposition. Let $(X,\mathcal R)$ be an ordered set (or class). If $Y\subset X$ is an $\mathcal R$-section and every element $y\in Y$ is also an $\mathcal R$-section of $X$, then $\bigcup Y$ and $\bigcap Y$ are $\mathcal R$-sections of $X$.

EDIT 1: The statement was misunderstood by me. The real porposition is:

Let $I$ non-empty (possibly $I$ is a set). If every element $y\in I$ is an $\mathcal R$-section of $X$, then $\bigcup I$ and $\bigcap I$ are also.

So the comment of @WilliamElliot is unnecesary, because in any momment we have said $I\subset X$. And probably it may be false. In my opinion, $I$ plays the role of an index set.

Addendum 1. This statement corresponds to the Appendix of J. L. Kelley General Topology, Theorem 90 (p. 264).

EDIT 2: Below is the rest of my question. Whit the new remarks it has no sense.
My problem is that I can't understand why (or how) an element $y\in Y$ can be an $\mathcal R$-section of $X$ if $y$ is not a subset of $X$ (as least as far as I know). It would have sense for me if the text said $\{y\}$, which is actually a subset of $X$.

The same applies to $\bigcup Y$ and $\bigcap Y$: by DEFINITION, both are sets (classes) formed by elements of sets in $Y$:

$$
\bigcup Y= \{x:\exists y\in Y \mbox{ such that } x\in y\}\\
\bigcap Y= \{x|x\in y \;\forall y\in Y\}.
$$

Again, to be an $\mathcal R$-section, we should consider $\{\bigcup Y\}$ and $\{\bigcap Y\}$.

Addendum 2. (Real questions) Once I have understood the proposition, the result seems me evident. However, I don't know how how to prove it rigorously. I know I have a collection of sets $\{y:y\in I\}$ such that

$$ \mbox{if } y_1,y_2\in Y \mbox{ with } y_1\neq y_2 \Longrightarrow y_1\subset y_2 \mbox{ or } y_2\subset y_1 , $$

So I actually have a descendent chain of sets

$$ y\supset y' \supset y'' \supset \cdots $$

1.- Proof for $\bigcup I$. With the above in mind, I think $bigcup I = y$ which is an $\mathcal R$-section by hypothesis.

2.- Proof for $\bigcap I$. I think the definition of $\mathcal R$-section avoids there is some $y\in I$ such that $y=\emptyset$. So I think all them has a common element, the least element $y_0$, and thus $\bigcap I$ is non-empty. Now, let $x\in X$ and $y\in\bigcap I$ with $x\mathcal R y$ and suppose $x\notin \bigcap I$. That implies that $x$ is less than the least element $y_0$. But then, $x$ wouldn't below either to $y\in I$, which is a contradiction, because $y$ was an $\mathcal R$-section$. So $\bigcap I$ is an $\mathcal R$-section.

Questions about this proof:

1.- Is the proof of $bigcup I$ correct?

2.- Is the proof of $\bigcap I$ correct?

3.- In 2., do you think I need to proove $y_0\in \bigcap I$? Is there a shorter proof?

4.- If all the inclusions are proper, then is it possible to show that $\bigcap I=\{y_0\}$?

Definition. Let $\mathcal R$ be a well-order in $X$. An $\mathcal R$-section is $Y\subset X$ such that if $x\in X$ and $y\in Y$ with $x\mathcal R y$, then $x\in Y$.

Informally, a set (class) $Y$ is said to be an $\mathcal R$-section if there is no element in $X\setminus Y$ that precedes the elements of $Y$.

Final Remark. I have prefered keep the old question and add the edits and corrections, because I think that is more proper. Sorry if it is a bad idea.

Thanks a lot for your patience.

答:
Well I think the notations in the book have been misunderstood. I have the 1964 edition, in which the theorem is,

$90$ THEOREM If $n\not=0$ and each member of $n$ is an $r$-section of $x$ , then $\bigcup n$ and $\bigcap n$ are $r$-sections of $x$.

Here $0$ is the void class, and $n$ is a class-of-classes; this may be understood from the definition of the union $\bigcup$ and the intersection $\bigcap$ operators.

Let $n=\{\{1,2\},\{1,2,5\},\{1,2,3,4,5\} \}$. Then, $\bigcup n=\{1,2,3,4,5\}$ and $\bigcap n = \{1,2\}$.

In this sense, the theorem can be reworded in terms of ordinary set theory as:

If $Y_i$, where $i=1,2,3,...$ be $r$-sections of a set $X$, so are $\cup_iY_i$ and $\cap_iY_i$.

A subset $Y\subset X$ is an $r$-section of $X$ iff the relation $r$ well orders $X$ and there is no $u\in (X-Y)$ such that $u\ r\ v$ for any $v\in Y$.

Using the definition alone the theorem appears logically obvious. A formal proof I leave on the author.

ContentLicense: CC BY-SA 3.0


问:
Uniform continuity of $f(x)=(1-\cos(x))/\sin(x)$ on the interval $(0,1)$
I got this question:

Prove that the function $f(x)=\frac{1-\cos(x)}{\sin(x)}$ is uniformly continuous on the interval $(0,1)$

I tried to prove it directly using the definition of uniform continuity but I failed this way. Then I tried to prove it using the fact that the sum of two uniformly continuous functions is uniformly continuous by writing $f(x)=\frac{1}{\sin(x)} - \cot(x)$ and then I tried to show that both $\frac{1}{\sin(x)}$ and $cot(x)$ are uniformly continuous on (0,1) but I wasn't managed to proceed that much.

Some hints will be helpful. Thanks.

答:
Extension to Simon's answer:
$$\lim\limits_{x\to 0}{\frac{1-\cos\,x}{\sin\,x}}=\lim_{x\to 0}{\frac{\frac{1}{2}x^2}{x}}=0$$
the limit exists, so you can extend $f$ from $(0,1)$ to $[0,1]$


问:
Finite-dimensional unitary operators
I need help in proving the following

Let $A$ and $B$ be finite-dimensional unitary operators. Then either
$$ {\left\|\left(\frac{A+B}{2} \right)^k\  \right\|}<1 $$
for all $k \geq 1$ , or else for some vector $f \ne 0$ we have
$Af=Bf; \ A^2f=ABf=BAf=B^2 f;\ A^3f=A^2Bf=ABAf=···=B^3f;\ ... $

I know that the norm of unitary operator is $1$,and if ${\left\|f+g  \right\|}={\left\|f  \right\|}+{\left\|g  \right\|}$ with ${\left\|f  \right\|}={\left\|g  \right\|}=1$ then $ f=g$ (the equality case of the triangle inequality ),but I cannot see the proof. If any one can help!

Thanks in advance.

答:
$$\left\|\left(\frac{A+B}{2}\right)^k\right\|\le\left\|\frac{A+B}{2}\right\|^k\le\left(\frac{\|A\|+\|B\|}{2}\right)^k=1$$

If the left-hand term is $1$ for some $k$, then all the terms must be equal to $1$. But the equation $\|A+B\|=2$ holds iff there is a sequence of unit vectors $x_n$ such that  $$2\leftarrow\|(A+B)x_n\|\le\|Ax_n\|+\|Bx_n\|\le2$$ Since $Ax_n$ is bounded in $\mathbb{R}^m$, it must have some convergent subsequence, $Ax_{n_i}\to y$. Let $f:=A^{-1}y$, so that $x_{n_i}\to f$, and thus $\|f\|=1$, $Bx_{n_i}\to Bf$, etc. The above set of inequalities now become, for the convergent subsequence, $$\|(A+B)f\|=\|Af\|+\|Bf\|=2$$ But as the question observes, this can only hold iff $Af=\lambda Bf$. It thus follows that $$ABf=\lambda B^2f=BAf$$ and so also the rest of the equations.

Title: Finite-dimensional unitary operators


问:
What is the difference between finiteness for a module versus an algebra?
I encountered the following line on page 60 in section 1.5.3 of Shavarevich's Basic Algebraic Geometry I:

"A ring $B$ that is finitely generated as an $A$-algebra is integral over $A$ if and only if it is finite as a module over $A$."

What do they mean here by "finite as a module over $A$" and how is that different from $B$ being finitely generated as an $A$-algebra?

答:
Suppose $A$ is a commutative ring, and $B$ is an $A$-algebra. We say $B$ is finitely generated as an $A$-algebra if $B$ admits a surjection $A[X_{1}, \ldots, X_{n}] \to B$ of $A$-algebras for some $n \in \mathbb{N}$. We say $B$ is finitely generated as an $A$-module if $B$ admits a surjection $A^{n} \to B$ of $A$-modules for some $n \in \mathbb{N}$.

The key difference in the above definitions is in what operations are "allowed". Less formally, if $B$ is finitely generated as an $A$-algebra, then that means that there are finitely many elements $b_{1}, \ldots, b_{n} \in B$ such that every element of $B$ can be expressed as a polynomial in $b_{1}, \ldots, b_{n}$ with coefficients in $A$. Likewise, $B$ is finitely generated as an $A$-module if there are $b_{1}, \ldots, b_{n} \in B$ such that every element of $B$ is an $A$-linear combination of $b_{1}, \ldots, b_{n}$. You can use the multiplication in $B$ in the first definition, but not the second.

Examples abound of finitely generated algebras which are not module finite. A nice example to illustrate the difference between the two: for any ring $A$, the polynomial ring $A[X]$ is finitely generated as an $A$-algebra, but not as an $A$-module. Indeed, $A[X]$ is a free $A$-module with infinite basis $\{1,X,X^{2}, \ldots\}$, and admits no $A$-basis with finite cardinality.


问:
Calculate the radius and intervall of convergence
We have the function $f(x)=x\cdot \ln (x)$.

I have calculated the Taylor series \begin{align*}T(x)&=\sum_{n=0}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n\\ & =\frac{f^{(0)}(1)}{0!}(x-1)^0+\frac{f^{(1)}(1)}{1!}(x-1)^1+\sum_{n=2}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n  \\ & =0+(x-1)+\sum_{n=2}^{\infty}\frac{(-1)^n(n-2)!}{n!}(x-1)^n \\ & =(x-1)+\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n-1)}(x-1)^n\end{align*} To calculate the radius and intervall of convergence do we have to write this term as one sum? Or is it possible to calculate that also in that form?

答: The first term $x-1$ has no effect on the radius of convergence. By root test the radius is $1$. So the sreis convers for $0 <x<2$ it is also absolutely convergent for $x=0$ and $x=2$ by comparison with $\sum \frac 1 {n^{2}}$.


问:
Let $c$, $d$ $\in$ $\mathbb{N}^+$ be positive integers which are coprime. Show that $cd \mid a$, for an integer $a$.
I wrote a proof for this, not sure if it's correct though.

Let $c$, $d$ $\in$ $\mathbb{N}^+$ be positive integers which are coprime. Show that $cd \mid a$, for an integer $a$.

Assume $cd \mid a$

By the assumption, $a$ is some multiple of $cd$

So $a$ can be written as $cda$

Hence $cd \mid a$.

I'm not sure if this is correct because the question mentions that $c$ and $d$ are coprime. Any help appreciated.

答: This question seems like it has to be missing something, I would double check to make sure the problem is correctly formulated. If you let $c,d$ be any integers then $cd \mid cd$ so set $a=cd$ and you're done.


答:
It is well-known that

$$\log(z+1)=\sum_{n=1}^\infty(-1)^{n+1}\frac{z^n}{n}$$ has the radius of convergence $1$. Multiplying by the polynomial $z+1$ has no effect on that radius.

By the way, this establishes that the general term of the development is, for $n>1$,

$$(-1)^{n+1}\frac{z^n}{n}+(-1)^{n}\frac{z^{n-1}}{n-1}z=(-1)^{n}\frac{z^n}{n(n-1)}.$$

Title: Calculate the radius and intervall of convergence

Body:
We have the function $f(x)=x\cdot \ln (x)$.

I have calculated the Taylor series \begin{align*}T(x)&=\sum_{n=0}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n\\ & =\frac{f^{(0)}(1)}{0!}(x-1)^0+\frac{f^{(1)}(1)}{1!}(x-1)^1+\sum_{n=2}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n  \\ & =0+(x-1)+\sum_{n=2}^{\infty}\frac{(-1)^n(n-2)!}{n!}(x-1)^n \\ & =(x-1)+\sum_{n=2}^{\infty}\frac{(-1)^n}{n(n-1)}(x-1)^n\end{align*} To calculate the radius and intervall of convergence do we have to write this term as one sum? Or is it possible to calculate that also in that form?


问:
Fundamental perspectives on subtraction and if how it is viewed/applied changes based on other context
A common error in reasoning is the off by one error which I found that has a more "formal" name as "Fence post error".

The error is that if we do e.g. $11 - 8$ while subtraction gives $3$ in the context of where we try to count items including $8$ the correct answer is actually $4$ i.e. $N - i + 1$.

E.g. clean floors $8$ to $11$ that is $4$ floors and not $3$.

I have been bitten by this error occasionally and I am always curious how these logical falacies arise in the thought process. So looking online I found this post which I first thought seemed not to make sense to me: How to avoid fence post error

Basically it states:

Subtraction is a span between numbers, not a count. The equation 11 - 8 = 3 means there are 3 "spans" between 8 and 11, but four numbers in that range! A span is a distance measure, like time from 8am and 11am (3 hours) or the distance between 8 and 11 inches (3 inches).

To me this whole explanation seemed weird at first reading and wrong. Because the core of the problem is when we say $11 - 8$ we mean that $8$ is removed too exactly like you have $11$ apples and remove $8$ to end up with $3$ which you can count (apple $9th$, $10th$ and $11th$).

But then I thought may be there is something about the basic operation of subtraction that escapes me and this post might be alluding to something I am not getting.

Are there multiple perspectives of what is subtraction fundamentally and if it is related to the context of counting or something different changes?

答:
When I am confronted by problems like this. I try to think of a very simple example, figure that out, and then try to generalize.

So suppose we have three circles lined up on the ground that are each exactly one step apart from the previous circle.

If we want to measure the distance from the first circle to the third,

We stand on the first circle. Step count = $0$.

We step to the second circle. Step count = $1$.

We step to the third circle. Step count = $2$.

So the distance between the $3$ circles is $2$ steps.

The important concept is that you start counting at $0$.

If we want to plant a fence post in each circle.

We stand on the first circle and plant a fence post. Fence post count = $1$.

We step to the second circle and plant a fence post. Fence post count = $2$.

We step to the third circle and plant a fence post. Fence post count = $3$.

So the number of fence posts planted is $3$.

The important concept is that you start counting at $1$.

This isn't about subtraction. It's about counting. Are you going to start at $0$ or are you going to start at $1$? Figure that out and choose your number line accordingly.


问:
What is "exclusive neighborhoods"?
I am trying to read a proof but don't understand what does this mean. Thank you very much!

So here is the question I am trying to prove: "Show that for a Lefschetz map f on a compact manifold X, there can be only ﬁnitely many ﬁxed points." And the proof is: "The graph of f is transversal to the diagonal inside X×X." Then I get lost here: "Now we take the preimage of the diagonal under the graph map from X to X×X. This is a 0-manifold inside X, and being a manifold can’t have limit points."

And "exclusive neighborhoods" appears in the followed proof:

Then (a) we can put an open set around each one containing no others, (b) the union of them is a closed set. Take the following list of open sets: the exclusive neighborhoods, plus the complement of the set of ﬁxed points. This is an open cover of X, so by compactness it has a ﬁnite subcover, so there must have been ﬁnitely many of them. QED.

答: I would guess two neighbourhoods $M$ and $N$ with $M\cap N=\emptyset$.</p>

Title: What is "exclusive neighborhoods"?


问:
Universal elimination when $\forall$ is not the main operator
I was asked to apply universal elimination on $\forall x A(x) \rightarrow B,$ but the $\forall$ here is not the main operator, what should I do?
here's the question and what I have for now

答:
Universal elimination when $∀$ is not the main operator.

You cannot apply UE when the quantifier is not the main operator.

You have to derive $\forall x A(x)$ in order to use it to "detach" $B$ using ($\to$-E):

$\forall x A(x) \to B$ --- premise

$\lnot \exists x (A(x) \to B)$ --- assumed [a]

$\lnot A(y)$ --- assumed [b]

$A(y)$ --- assumed [c]

$\bot$

$B$ --- from 5)

$A(y) \to B$ --- from 4) and 6) by ($\to$-I), discharging [c]

$\exists x (A(x) \to B)$ --- from 7) by ($\exists$-I)

$\bot$

$\lnot \lnot A(y)$ --- from 3) and 9) by ($\to$-I), discharging [b]

$A(y)$ --- from 10) by ($\lnot \lnot$-E)

$\forall x A(x)$ --- from 11) by ($\forall$-I)

$B$ --- from 1) and 12) by ($\to$-E)

$A(y) \to B$ --- from 13) by ($\to$-I)

$\exists x (A(x) \to B)$ --- from 14) by ($\exists$-I)

$\bot$

$\exists x (A(x) \to B)$ --- from 2) and 16) by ($\to$-I), discharging [a], followed by ($\lnot \lnot$-E)


问:
Calculation in Routh's theorem
The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:

Option 1

The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.

That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.

Option 2

This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.

The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.

The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.

答: A variant on my option 2: as @RiverLi notes, the numerator is of at most degree $2$ in $z$, so is of the form $(c-1)(Aa+Bb+Cc+D)$. That $A=B=0,\,C=D=1$ follows from the case $x=y=z$.

Title: Calculation in Routh's theorem


问:
$A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements.
$(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$

I know I have to start with $A = ...$ to get to $B$. I incorporated $A = A \cap (A \cup C)$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $(A \cup C) = (B \cup C)$? I've looked at the set identities, and think that I have to use absorption law again to show that $A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$. Can you guys please help me solve this problem, I'm stuck.

答:
You can use absorption and distributivity:
\begin{align}
A&=A\cap(A\cup C)&&\text{absorption}\\
&=A\cap(B\cup C)&&\text{assumption}\\
&=(A\cap B)\cup(A\cap C)&&\text{distributivity}\\
&=(A\cap B)\cup(B\cap C)&&\text{assumption}\\
&=(B\cap A)\cup(B\cap C)&&\text{commutativity}\\
&=B\cap(A\cup C)&&\text{distributivity}\\
&=B\cap(B\cup C)&&\text{assumption}\\
&=B&&\text{absorption}
\end{align}
You can see that this holds in every distributive lattice.


问:
Proof in FOL with no CQ rule

I had the first completed but when doing its converse, since there is no CQ rule, I do not know where to proceed. Please help

答:
$\lnot \exists x (P(x) \land Q(x))$ --- premise

$P(a)$ --- assumed [a]

$Q(a)$ --- assumed [b]

$\exists x (P(x) \land Q(x))$

$\bot$

$\lnot Q(a)$

$P(a) \to \lnot Q(a)$

$\forall x (P(x) \to \lnot Q(x))$

Title: Proof in FOL with no CQ rule


问:
An application (question) of Tietze Extension Theorem
Below is a question, followed by a partial solution that I am confused about.

Let $X$ be a T$_4$ space and $M\subseteq X$ be closed. Let $\mathbb{S}^1$ be the unit circle with the subspace topology. Suppose that $f\colon M\to\mathbb{S}^1$ is a continuous such that $f[M]$ is a closed proper subset of $\mathbb{S}^1$. Show that there exists a continous function $\hat{f}\colon X\to\mathbb{S}^1$ such that $\hat{f}\vert_{M}=f$.

So, I figured I could just directly apply Tietze Extension Theorem here, but when I checked the solution this was not the case. My question is, why can't I just apply Tietze Extension Theorem here? Is it because Tietze Extension Theorem requires $f$ to map to $\mathbb{R}$? For reference, the given solution shows that there exists an open arc $\gamma$ in $\mathbb{S}^1$ such that $\gamma\subseteq\mathbb{S}^1\setminus f[M]$ hence $f[M]\subseteq\mathbb{S}^1\setminus\gamma$. Then goes on to say, $\mathbb{S}^1\setminus\gamma$ is a non-empty closed arc in $\mathbb{S}^1$, thus homeoremorphic to the interval $[0,1]$. Hence, by Tietze Extension Theorem there exists a continuous function $\hat{f}\colon X\to\mathbb{S}^1\setminus\gamma\subseteq \mathbb{S}^1$ such that $\hat{f}\vert_{M}=f$.

To add, I though I could simply say: "By Tietze Extension Theorem, there is a continuous function $\hat{f}\colon X\to\mathbb{S}^1$ such that $\hat{f}\vert_M=f$." However, based on the given solution, I can't just say this. So again, my question really is, why can't I just say that?

答:
Tietze extension theorem states:

Let $X$ be a $T_4$ space, $M\subseteq X$ a closed subset and $f:M\to\mathbb{R}$ a continuous function. Then there exists a continuous function $F:X\to\mathbb{R}$ such that $F_{|M}=f$. Moreover $F$ can be chosen in such a way that $sup(|f(m)|:\ m\in M)=sup(|F(x)|:\ x\in X)$.

Note that there's "$\mathbb{R}$" on the right side. This is no accident. It doesn't work for an arbitrary space. It does work for some but not all.

The first observation is that the supremum property implies that we can replace $\mathbb{R}$ with $[0,1]$ in the theorem. But we cannot replace $\mathbb{R}$ with $S^1$ without some additional assumptions.

Here's a counterexample: consider $D=\{v\in\mathbb{R}^2 :\ \lVert v\rVert\leq 1\}$. Note that $S^1\subseteq D$ is a closed subset. Then consider the identity $id:S^1\to S^1$, $id(x)=x$. Can this $id$ be extended to a $D\to S^1$ map? This is the famous "is $S^1$ a retract of $D$?" question and the answer is "no", which is a consequence of Brouwer Fixed Point Theorem.

The reason why it works in your case is because $f[M]$ is a closed proper subset of $S^1$. This is the additional assumption that makes a difference. Such subsets are always contained in some arc $A\subseteq S^1$. And arcs are homeomorphic to $[0,1]$. By using such a homeomorphism, its inverse and Tietze for $[0,1]$ we can construct the extension we are looking for.


问:
Doubt in Factoring quadratic equations
I was factorising the equation $x^2 - 3x + 10$ and realised that I can do it as $x^2 -5x + 2x + 10$, since $-5x +2x = -3x$ and $5 \cdot 2 = 10$. However, this doesn't work. As we don't get the common terms in the brackets $x(x-5) + 2(x+5)$

Is this mistake because we need $-10$ instead of $10$?
That is, while checking out the multiplication of factors of $a$ and $c$, we need to make sure that it equals $ac$ along with its sign ($+$ or $-$)?

答:
You are right that this doesn't work, but this is because you have $x-5$ in one bracket and $x+5$ in the other. Recall the distributive law, shown in this image with numbers:

Multiplying two numbers can be thought of the area of a rectangle having the numbers as side lengths. When we have rectangles of different dimensions, there is no way to combine them to make a larger rectangle with the same width or height, unless the sides have a common factor.

The same applies for variables, so that there is no way to combine a rectangle with sides $x, x-5$ and another one with sides $2, x+5$ as in your example.

Title: Doubt in Factoring quadratic equations


答:
It works but you can no more guess. So or the roots of $Ax^2+Bx+C=0$
$$x_{1,2}=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

Also see:
Getting back to the quadratic equation again

Title: Doubt in Factoring quadratic equations

Body:
I was factorising the equation $x^2 - 3x + 10$ and realised that I can do it as $x^2 -5x + 2x + 10$, since $-5x +2x = -3x$ and $5 \cdot 2 = 10$. However, this doesn't work. As we don't get the common terms in the brackets $x(x-5) + 2(x+5)$

Is this mistake because we need $-10$ instead of $10$?
That is, while checking out the multiplication of factors of $a$ and $c$, we need to make sure that it equals $ac$ along with its sign ($+$ or $-$)?


问:
Riemann zeta function at zero
Can the value of Riemann zeta function at 0, $\zeta(0)=-1/2$, be deduced from the identity $E(z)=E(1-z)$, where
$$E(z)=\pi^{-z/2}\Gamma(z/2)\zeta(z)?$$</p>

答:
The functional equation under consideration yields:
$$
\zeta(s)=\frac{\pi^s}{\sqrt{\pi}}\frac{\Gamma\left(\frac{1-s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta(1-s).
$$
So the fact that $\zeta(0)=\lim_0\zeta(s)=-\frac{1}{2}$ follows from the following three ingredients:
$$
\zeta(s)\sim_1\frac{1}{s-1}\qquad \Gamma(s)\sim_0\frac{1}{s}\qquad \sqrt{\pi}=\Gamma\left(\frac{1}{2}\right)=\lim_{1/2} \,\Gamma(s).
$$
I assume you know the first two ones. The last one can be computed directly from the integral definition of $\Gamma$ by variable change $u=\sqrt{t}$, using $\int_0^{+\infty}e^{-u^2}du=\frac{\sqrt{\pi}}{2}$.

The so-called functional equation of Riemann zeta function is:
$$
\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).
$$
Using
$$
\zeta(s)\sim_1\frac{1}{s-1}\qquad\sin\left(\frac{\pi s}{2}\right)\sim_0\frac{\pi s}{2}\qquad1=\Gamma(1)=\lim_1\,\Gamma(s)
$$
it follows that
$$
\zeta(s)\sim_02^0\pi^{-1}\frac{\pi s}{2}\Gamma(1)\left(\frac{-1}{s}\right)=-\frac{1}{2}.
$$

Title: Riemann zeta function at zero


问:
"Summing" the series $\sin(x)+\frac{1}{2}\sin(2x)+\frac{1}{3}\sin(3x)+\frac{1}{4}\sin(4x)+...$
"Summing" the series $\sin(x)+\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)+...$

Pose $$S=\sin(x)+\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)+...$$
$$C=\cos(x)+\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)+\dfrac{1}{4}\cos(4x)+...$$

$$C+iS = e^{ix}+\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3+\dfrac{1}{4}(e^{ix})^4+...$$

Let $t$ = $e^{ix}$

Then we have a series of $t+\dfrac{t}{2}+\dfrac{t}{3}+\dfrac{t}{4}+...=\log(1+t)$

Which is $-\log(1-e^{ix})=\log(1-\cos(x)-i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$

$-\log([1-\cos(x)]-i\sin(x))=-\dfrac{1}{2}\log([1-\cos(x)]^2-i\sin^2(x))-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right))$. Since we are interested only in the imaginary part, we have the sum for $S$ is:

$$-\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$

Which is $-\arctan\left(-\cot(\dfrac{x}{2})\right)$

I don't know what to do next.

The "sum" of this series should be $\dfrac{\pi-x}{2}$, according to Euler.

答:
First, the series are
$$
\log(1+x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}k
$$
and
$$
-\log(1-x)=\sum_{k=1}^\infty\frac{x^k}k
$$
The series is $2\pi$-periodic and the sum for $-\pi\le x\le\pi$ is
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(kx)}k
&=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ikx}-e^{-ikx}}k\\
&=-\frac1{2i}\left(\log(1-e^{ix})-\log(1-e^{-ix})\right)\\
&=\frac1{2i}\log\left(\frac{1-e^{-ix}}{1-e^{ix}}\right)\\
&=\frac1{2i}\log\left(-e^{-ix}\right)\\[3pt]
&=\operatorname{sgn}(x)\frac\pi2-\frac x2
\end{align}
$$

Note about $\boldsymbol{\operatorname{sgn}(x)}$

Since $\sin(x)$ is odd, the series should be odd. However, $\frac{\pi-x}2$ is not odd. The limit as $x\to0^+$ is $\frac\pi2$, but at $x=0$, all the terms of the series are $0$, so the sum is $0$. So that should clue us into the fact that there is a discontinuity there (this is because the convergence is not uniform near $0$).

Taking the oddness into account we get $\operatorname{sgn}(x)\frac\pi2-\frac x2$. This is also shown by looking at $\frac1{2i}\log\left(-e^{-ix}\right)$: when $x$ is increasing from $0$, $-e^{-ix}$ is moving clockwise from $-1$; thus, $\frac1{2i}\log\left(-e^{-ix}\right)$ is decreasing from $\frac\pi2$. When $x$ is decreasing from $0$, $-e^{-ix}$ is moving counter-clockwise from $-1$; thus, $\frac1{2i}\log\left(-e^{-ix}\right)$ is increasing from $-\frac\pi2$.


问:
contour integral with hyperbola branch cuts
I am reading Karl Graff's book on wave motion and stuck by a contour integral where the branch cuts are made of two hyperbolas in the 2nd and 4th quadrants. See figure attached Contour integral figure.

Here only the branch cut in the 2nd quadrant is shown as the only interest is in the upper half of the complex plane. The hyperbola degrades into a path along the positive imaginary axis and a section of the real axis on the negative side close to the origin after some manipulation as shown in the figure.

If two new polar coordinate systems are defined at the two branch points on the real axis, which are symmetric to the imaginary axis and define the ends of the cut on the real axis , as done in the book, the variable now takes the form of the product of coordinates redefined in the two new coordinate systems, say, $\gamma$ =
$\rho_1e^{i\theta_1}*\rho_2e^{i\theta_2}$. In other words, the argument of $\gamma$ is $\theta_1+\theta_2$.

I can work out the argument by calculating the limit along each of the paths AO, OB, BO and ON but this is tedious. I am not able to read the argument out directly from the defined $\theta_1$ and $\theta_2$ as stated in the book. It seems to me something needs to be considered whereas the path is on one side of the complex plane when the coordinate is defined on the other. I do not know how this is done.

I would like to know how this works. Any advice that can help me improve my understanding on the technique, be it a book, a paper or a direct illustration, is appreciated.

答:
I am here to answer my own question in case other people may face a similar problem. I have to acknowledge the help of a math Senior Lecturer in our university who solved this for me.

AO: $\theta_1+\theta_2=\pi$ (evident)

OB: $\theta_1 (=\pi)+\theta_2(=0) = \pi$

BO: $\theta_1 (=\pi)$ + $\theta_2(=-2\pi$, rotates clockwise one circle) = $-\pi$

OA: $\theta_1 + \theta_2 = -\pi $($\theta_2$ is negative)


问:
With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100$ by using Karamata's inequality

Given $n$ real numbers $a_{1}, a_{2}\cdots a_{n}$ so that
$$a_{1}\geq a_{2}\geq\cdots\geq a_{n}, a_{1}+ a_{2}+ \cdots+ a_{n}= 300, a_{1}^{2}+ a_{2}^{2}+ \cdots+ a_{n}^{2}> 10000$$
With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100.$

I wanna create a contractdiction. So we may assume that
$$a_{1}+ a_{2}+ a_{3}< 100\Rightarrow\sum_{k= 1}^{n}a_{k}^{2}\leq a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+ a_{3}\sum_{k= 4}^{n}a_{k}= a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+$$
$$+ a_{3}\left ( 300- a_{1}- a_{2}- a_{3} \right )\leq 100\left ( a_{1}- a_{3} \right )+ 300a_{3}\leq 100\left ( a_{1}+ a_{2}+ a_{3} \right )= 10000$$
I tried to solve this inequality by using Karamata but unsuccessfully, who can help me with a way ?? Thanks a real lot !

答: You are very close to the solution, just change the last equality sign to a $<$ sign because of your assumption $a_1+a_2+a_3<100$, and this contradicts the second part of the hypothesis.


问:
Proof by contradiction and invalid reasoning
I am very aware of the several posts that are in this forum, but to be honest I am not quite satisfied with the provided answers. Also, I can tell that the best post is the following:

Proof by Contradiction, Circular Reasoning?

I would love to continue the discussion in that post, however my reputation level does not allow me to add any comments, and I still have some questions. Therefore, I want to continue discussing more about proof by contradiction and circular reasoning.

Suppose we wish to prove $(P \wedge Q) \rightarrow R$. We assume P and Q. Proof by contradiction begins by assuming $\neg R$, and from these two assumptions, a "contradiction" is derived, and sometimes that contradiction is that $\neg P$ or $\neg Q$ (our initial valid assumptions). Therefore, since there is a contradiction with our assumptions, we conclude that it cannot happen that $\neg R$, so it must be the case that $R$.

This proof strategy bothers me, because I have the feeling that there is some kind of circular (or invalid) reasoning going on there i.e. $(P \wedge Q) \rightarrow R$ is equivalent to $\neg R \rightarrow \neg(P \wedge Q)$, so if we already know that the consequent is false, in order for the statement to be true, $\neg R$ has to be false, so R is true, but R is part of the consequent part of the conditional statement (isn't doing this like assuming there is a biconditional i.e. because R is true, then $P \wedge Q$ is true?). I probably don't know how to explain it better, but how/why can we say that this is indeed a proof? Hence, my main question is why it is valid to assume $\neg R$ if our conclusion involves R? I have one explanation that kind of makes sense to me (still a little iffy though):

In more simple terms, where we want to prove that $P \rightarrow Q$, we are trying to do the following: $[(P \wedge \neg Q) \rightarrow Contradiction] \rightarrow (P \rightarrow Q)$ if we can prove that $(P \wedge \neg Q)$ leads to a Contradiction, then we are indeed proving that $P \rightarrow Q$, which was our original goal. Given that this is a tautology, whenever we follow this proof strategy structure, we are always proving $P \rightarrow Q$. However, saying that this is a tautology kind of bothers me for some reason.

I would sincerely appreciate your insights, and if there is anything you would improve in my post, please feel free to do so. I'm starting my formal journey in the proof realm (currently reading How To Prove It), and I have always heard that proofs by contradiction were the easiest strategy. However, I have learned that this strategy seems to be difficult at the beginning for all those who are trying to learn proof strategies.

答:
It might help to reframe the concept of how the proof works: Instead of assuming ¬R, you consider the hypothetical possibility of ¬R.

You work out the consequences of what would be true if R is false. You prove that this hypothetical situation includes a contradiction, and therefore cannot be real.

After considering that hypothetical, you know that it is impossible for R to be false. In symbolic terms, you know that ¬(¬R) is true. This is equivalent to simply R being true, so you have proven R.


问:
If $X_n\to X$ a.e. it does not follow that $\mu_n(P)\to \mu(P)$
If $X_n\to X$ a.e. and $\mu_n$ and $\mu$ are the p.m.'s of $X_n$ and $X$, it does not follow that $\mu_n(P)\to \mu(P)$ even for all intervals $P$.

I am having trouble coming up with an example that illustrates this. Why does Egoroff's theorem not guarantee that $\mu_n(P)\to \mu(P)$?

答:
Let's look at the measures we have in this problem.
We are working on $(\Omega, \mathcal{F},\mathbb{P})$ so when we say $X_n \to X$ a.s., we mean  $\mathbb{P}$ almost surely. Then we have the laws $\mu_n$ of $X_n$ and $\mu$ of $X$.

Egorov's Theorem tells us we may find a set of $\mathbb{P}$ measure at least $1-\epsilon$ on which $X_n \to X$ uniformly. For sake of argument, suppose we could find a set of $\mathbb{P}$ measure $1$ on which $X_n \to X$ uniformly. Great, now what? The point is that the measure $\mathbb{P}$ isn't really related to $\mu_n = \mathbb{P}\circ X_n^{-1}$ since $X_n^{-1}$ can do whatever it pleases with sets before $\mathbb{P}$ sees them.

Here is a trivial example which demonstrates the conclusion may fail for an interval: If $X_n = 1/n$ for all $n$ with probability $1$, and $P=(0,1]$ then $\mu_n(P)=1 \not\to 0 = \mu(P)$.


问:
Measurable function definition
In my lecture notes for stochastic processes we define measurable functions as

Here $\mathcal{A}$ is a $\sigma-$algebra on $\Omega$. However, when I read on other sites (e.g. Wolfram), the set that should be measurable is defined in another way, specifically the inequality sign should be flipped. Is the definition of measurable function incorrect in my picture? Or is there some detail that I've missed?

答:
$\{\omega: X(\omega) \leq x \} \in \mathcal A$ for every real number $x$ if and only if $\{\omega: X(\omega) \geq x \} \in \mathcal A$ for every real number $x$. So the definitions are equivalent.

Proof is based on the following: $\{\omega: X(\omega) \geq x \}=(\bigcup_n  \{\omega: X(\omega) \leq x -\frac 1 n\})^{c}$ and $\{\omega: X(\omega) \leq x \}=(\bigcup_n  \{\omega: X(\omega) \geq x +\frac 1 n)^{c}$.

Title: Measurable function definition


问:
Homomorphism of Rings.
How to prove this one?

Let $f: R \longrightarrow S$ be a homomorphism of rings $R$ and $S$. If $M$ is a maximal ideal of $S$, then $f^{-1}(M)$ is a maximal iseal of $R$.

答:
What you have written is not correct. Consider the ring homomorphism $f : \Bbb Z \to \Bbb Q$ given by $f(x) = x$.

Then, $\{0\}$ is a maximal ideal of $\Bbb Q$ but its preimage $\{0\}$ is not maximal in $\Bbb Z$.

What you have written is true if $f$ is onto. (At least with the convention that $R$ and $S$ are commutative with unity and $f$ maps $1_R$ to $1_S$.) An answer to that can be found here.

Title: Homomorphism of Rings.


问:
How to prove this :$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$?

$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$

I saw this in a math book. I tried to prove it but I'm unable to. Any hints or solutions would be most appreciated.

答:
Nothing to prove, we know $n!=\Gamma(n+1)$ where:
$$\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-z}dt$$

Now we can "deduce" $\big(\frac{1}{2}\big)!=\frac{\sqrt\pi}{2}$


问:
An inequality for sums
If I have two non-negative series $f(k)$ and $g(k)$ such that
$$\sum_{k=1}^{n+1}f(k)\leq \sum_{k=1}^{n+1}g(k),\tag1$$
Does this imply that

$$\sum_{k=1}^{n}f(k)\leq \sum_{k=1}^{n}g(k)\tag2$$

I have tried to prove it as follows
$$\sum_{k=1}^{n}f(k)\leq \sum_{k=1}^{n+1}f(k)\leq \sum_{k=1}^{n+1}g(k)\nleq
\sum_{k=1}^{n}g(k)$$
which don't give (2) !!

Any alternative proof or counterexample to this question is very appreciated.

答:
Welcome to MSE!

Consider

$f(1) = 1$, $f(2) = 1000000$

$g(1) = 10$, $g(2) = 0$

Then $\sum_{k=1}^1 f \leq \sum_{k=1}^1 g$, but
$\sum_{k=1}^2 f \not \leq \sum_{k=1}^2 g$.

Can you see how to find a similar example for larger $n$?

I hope this helps ^_^

Title: An inequality for sums


Title: An inequality for sums

Body:
If I have two non-negative series $f(k)$ and $g(k)$ such that
$$\sum_{k=1}^{n+1}f(k)\leq \sum_{k=1}^{n+1}g(k),\tag1$$
Does this imply that

$$\sum_{k=1}^{n}f(k)\leq \sum_{k=1}^{n}g(k)\tag2$$

I have tried to prove it as follows
$$\sum_{k=1}^{n}f(k)\leq \sum_{k=1}^{n+1}f(k)\leq \sum_{k=1}^{n+1}g(k)\nleq
\sum_{k=1}^{n}g(k)$$
which don't give (2) !!

Any alternative proof or counterexample to this question is very appreciated.

Body: $f(k)=0$ for $2 \leq k \leq n$, $f(1)=1$, $g(k)=0$ for $1 \leq k \leq n$, $f(n+1)=0, g(n+1)=2$ is a counter-example.


答: Please also keep in mind that this is nothing more than an analytic continuation of the factorial. In other words, when you take the factorial of a fraction, it loses its initial meaning of multiplying the number with a number lower until you reach $1$. The Gamma function is, in a sense, an interpolation between the integer values of the factorial that do make sense with the original meaning. So it wouldn't have been possible proving this with the ''ordinary'' definition of the factorial.

Title: How to prove this :$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$?

Body:
$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$

I saw this in a math book. I tried to prove it but I'm unable to. Any hints or solutions would be most appreciated.


问:
Normal subgroup and non-equivalent representations
I am looking for an example to illustrate that to representations are not equivalent in a particular situation.

If H is a normal subgroup of G, $\pi$ a representation of $H$, and the second representation is defined as follows: For $g \in G$ define $\pi^g : G \to GL(V)$ by $\pi^g (h)=\pi (g^{-1}hg)$. For $g \in G$ we define the representation $\pi^g$ of H as before.

So now I have to find an example of G, H a representation $\pi$ of H and $g \in G$ such that $\pi$ and $\pi^g$ are not equivalent.

However, I am struggling a bit to find a nice and simple example to show this.
I've tried with $G= (\mathbb{C}, +)$ and $H=(\mathbb{R}, +)$ as this is a normal subgroup of G (as it is abelian) and as the representation of H has been given in the lectures. However, it seems as if this doesn't work

答:
Since $\pi(g^{-1}hg) = \pi(g)^{-1}\pi(h)\pi(g)$, you're just conjugating the outputs, i.e. you're using $\pi(g)$ to change bases. So, they're equivalent representations.

Edit: As the comments noted, you intended to write $\pi^g \colon H \to GL(V)$, in which case they don't need to be equivalent.

For instance, take $G = S_3$, the symmetric group on 3 elements, and $H = A_3 = \{1, (123), (132)\}$, the alternating group on 3 elements, written in cycle notation. Use $V = \mathbb{C}$ and $\pi \colon A_3 \to \mathbb{C}^\times$ by $\pi((123)^k) = \exp(2\pi i k/3)$.

Set $g = (12)$. Easy calculations give $\pi^g((123)) = \pi((123)^2) = \exp(2\pi i \cdot 2/3) = \overline{\pi((123))}$, the complex conjugate. Hence they're inequivalent representations.


答:
Your idea of applying Tietze is OK, but needs a small tweak:

Realise that for any $p \in \Bbb S^1$, we have that $\Bbb S^1\setminus \{p\} \simeq \Bbb R$ (using the stereographic projection from $p$) and if we have $f: M \to \Bbb S^1$ non-surjective we can pick any $p\in \Bbb S^1 \setminus f[M]$, and a homeomorphism $h: \Bbb S^1 \setminus \{p\} \to \Bbb R$ and consider $h \circ f: M \to \Bbb R$ instead, which is wel-defined (as $p$ is not in $f[M]$) and continuous.

By Tietze, this has a continuous extension $g: X \to \Bbb R$ (so $g = h \circ f$ on $M$) and then we can define $\bar{f}: X \to  \Bbb S^1$ by $\bar{f} = h^{-1} \circ g$, which is of course continuous as a composition of continuous maps.

This extends $f$ as $$\forall x \in M: \bar{f}(x) = h^{-1}(g(x)) = h^{-1}(h(f(x))) = f(x)$$

You cannot expect that all maps into $\Bbb S^1$ can be extended from a closed subspace to the whole space, this turns out to be equivalent to $\dim(X) \le 1$, where dim is the covering dimension of $X$. And we cannot extend the identity from $\Bbb S^1 \subseteq \Bbb D^2$ to the whole disk $\Bbb D^2$, because that would contradict the so-called non-retraction theorem.

Spaces $Y$ as a codomain for which we can always extend from a closed subspace to the whole space are called AR’s (absolute retracts), and these include all spaces $\Bbb R^n$ and many other linear spaces and convex subsets of them.

Title: An application (question) of Tietze Extension Theorem

Body:
Below is a question, followed by a partial solution that I am confused about.

Let $X$ be a T$_4$ space and $M\subseteq X$ be closed. Let $\mathbb{S}^1$ be the unit circle with the subspace topology. Suppose that $f\colon M\to\mathbb{S}^1$ is a continuous such that $f[M]$ is a closed proper subset of $\mathbb{S}^1$. Show that there exists a continous function $\hat{f}\colon X\to\mathbb{S}^1$ such that $\hat{f}\vert_{M}=f$.

So, I figured I could just directly apply Tietze Extension Theorem here, but when I checked the solution this was not the case. My question is, why can't I just apply Tietze Extension Theorem here? Is it because Tietze Extension Theorem requires $f$ to map to $\mathbb{R}$? For reference, the given solution shows that there exists an open arc $\gamma$ in $\mathbb{S}^1$ such that $\gamma\subseteq\mathbb{S}^1\setminus f[M]$ hence $f[M]\subseteq\mathbb{S}^1\setminus\gamma$. Then goes on to say, $\mathbb{S}^1\setminus\gamma$ is a non-empty closed arc in $\mathbb{S}^1$, thus homeoremorphic to the interval $[0,1]$. Hence, by Tietze Extension Theorem there exists a continuous function $\hat{f}\colon X\to\mathbb{S}^1\setminus\gamma\subseteq \mathbb{S}^1$ such that $\hat{f}\vert_{M}=f$.

To add, I though I could simply say: "By Tietze Extension Theorem, there is a continuous function $\hat{f}\colon X\to\mathbb{S}^1$ such that $\hat{f}\vert_M=f$." However, based on the given solution, I can't just say this. So again, my question really is, why can't I just say that?


问:
Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$
I'm reading Lang's Undergraduate Analysis:

Let ${n}\choose{k}$ denote the binomial coefficient,

$${n\choose k}=\frac{n!}{k!(n-k)!}$$

where $n,k$ are integers $\geq0,0\leq k\leq n$, and $0!$ is defined to be $1$. Prove the following assertion:

$${n\choose k}={n\choose n-k}$$

I proceded by making the adequate substitutions:

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!}$$

And then I simplified and achieved:

$$\frac{n!}{(n-k)!-k!}$$

But I'm not sure on how to proceed from here, I've noticed that this result is very similar to $\frac{n!}{k!(n-k)!}$. What should I do? I guess it has something to do with the statement about the nature of $n$ and $k$:

$n,k$ are integers $\geq0,0\leq k\leq n$

So should I just change the minus sign to plus sign and think of it as a product of $(n-k)!$?

$$\frac{n!}{(n-k)!-k!}\Rightarrow\frac{n!}{(n-k)!+k!}\Rightarrow \frac{n!}{k!(n-k)!}$$

I'm in doubt because I've obtained the third result on Mathematica, but obtained the first with paper and pencil. I'm not sure if there are different rules for simplification with factorials. I'm not sure if this $(n-k)!+k!$ mean a sum or a product in this case.

Body: $ n - (n-k) = k $, not $ -k $.


问:
For which $n$ the first player has a strategy so that he wins no matter how the other player plays?

The Question

From Junior O-level Tournament of the Towns paper, Fall 2020:

There are $n$ stones in a heap. Two players play the game by
alternatively taking either $1$ stone from the heap or a prime number
of stones which divides the current number of stones in the heap. The
player who takes the last stone wins. For which $n$ the first player has
a strategy so that he wins no matter how the other player plays?

My Understanding

Let player $A$ play first and player $B$ second. Also, let a heap with $n$ stones be an $n$-heap.

My initial observations:

$A$ wins any $p$-heap for a prime number $p$.

If an $n$-heap is
winning for $B$, then the $(n + 1)$-heap is winning for $A$, since
$A$ can remove $1$ stone and then use $B$'s winning strategy for the
$n$-heap (since for $n$-heap $A$ moves second).

I couldn't think of anything else so I tried finding a pattern with some examples.

\begin{array}{|c|c|c|c|}
\hline
n& \text{winner} \\ \hline
 1 &A \\ \hline
 2  & A\\ \hline
 3  & A\\ \hline
 4 & B\\ \hline
 5  & A\\ \hline
 6  & A\\ \hline
 7 & A\\ \hline
 8  & B\\ \hline
 9  & A\\ \hline
 10  & A\\ \hline
 11 & A\\ \hline
 12  & B\\ \hline
\end{array}

Since it seems $B$ wins exactly when $n = 4k$, I tried to prove this by induction on $k$.

Basis: the claim holds for $k \leq 3$ as shown in the examples.

Inductive hypothesis: Suppose that $B$ wins exactly when $n = 4k$ for arbitrary $k$.

Now I try to prove that $4k + 1$, $4k + 2$, $4k + 3$ have winning strategies for $A$ and that $4k + 4$ has a winning strategy for $B$.

$\textbf{(4k + 1)-heap:}$ $A$ wins by observation 2.

$\textbf{(4k + 2)-heap:}$ $4k + 2 = 2(2k + 1)$ which is even. $A$ can remove 2 stones to get a $4k$-heap which is winning for $A$ since now $B$ moves first.

Now I am stuck for the $(4k + 3)$-heap and $(4k + 4)$-heap. I don't want a complete solution, instead I just want a hint and some confirmation that this is a way to solve this problem. Also if you have an easier way, I would like a hint to discover it!

答:
I have used lonza leggiera's hints to complete my solution.

$4k + 3$ must have a prime factor $p \not = 2$ since $4k + 3$ is odd. Then either

$p \equiv 0 \pmod{4}$, then $p = 4x$ which is not prime,

$p \equiv 1 \pmod{4}$, then $p = 4x + 1$,

$p \equiv 2 \pmod{4}$, then $p = 4x + 2$ which implies $p = 2$, but have already established $p \not = 2$,

$p \equiv 3 \pmod{4}$, then $p = 4x + 3$.

If all prime factors satisfy $p \equiv 1 \pmod{4}$, then $4k + 3$ would satify $4k + 3 \equiv 1 \pmod{4}$ which is false. Hence at least one $p$ must satisfy $p \equiv 3 \pmod{4}$.

$A$ can then remove $4x + 3$ stones from the heap which means the heap has $4k + 3 - (4x + 3) = 4(k - x)$ stones. By the inductive hypothesis a $4(k - x)$-heap is winning for $B$. Hence the $(4k + 3)$-heap is winning for $A$.

Now for the $(4k + 4)$-heap. Again let $p$ be a prime factor of $4(k + 1)$, then either

$p \equiv 0 \pmod{4}$, then $p = 4x$ which is not prime,

$p \equiv 1 \pmod{4}$, then $p = 4x + 1$,

$p \equiv 2 \pmod{4}$, then $p = 2$, or

$p \equiv 3 \pmod{4}$, then $p = 4x + 3$.

If $p = 2$ then $A$ can reduce the heap to a $(4k + 2)$-heap which we have established is always winning for $A$. No good.

If $p = 4x + 1$, then $4k + 4 - (4x + 1) = 4(k - x) + 3$ which we have established is always winning for $A$. No good.

If $p = 4x + 3$, then $4k + 4 - (4x + 3) = 4(k - x) + 1$ which we have established is always winning for $A$, hence $(4k + 4)$-heap is always winning for $B$.


问:
Understanding algorithm to show $(x+\partial_x)e^{-x^2/2}=0$ using python?
I am given the following Python code which is supposed to verify numerically that

$$(\partial_x + x)e^{-x^2/2}=0.$$

The algorithm does this, by transforming everything into a Fourier basis and then verifying it using the Fourier transform. However, I have some difficulties understanding the mathematical basis of this algorithm. In short, my question is:

When the final result is computed as np.dot( D, yl ), what is that actually in mathematical terms that we are computing here?

```<code> import numpy as np

 ## non-normalized gaussian with sigma=1
 def gauss( x ):
return np.exp( -x**2 / 2 )

  ## interval on which the gaussian is evaluated
 L = 10
 ## number of sampling points
 N = 21
 ## sample rate
 dl = L / N
 ## highest frequency detectable
 kmax= 1 / ( 2 * dl )

 ## array of x values
 xl = np.linspace( -L/2, L/2, N )
 ## array of k values
 kl = np.linspace( -kmax, kmax, N )

 ## matrix of exponents
## the Fourier transform is defined via sum f * exp( -2 pi j k x)
  ## i.e. the 2 pi is in the exponent
 ## normalization is sqrt(N) where n is the number of sampling points
 ## this definition makes it forward-backward symmetric
 ## "outer" also exists in Matlab and basically does the same
exponent = np.outer( -1j * 2 * np.pi * kl, xl )
## linear operator for the standard Fourier transformation
 A = np.exp( exponent ) / np.sqrt( N )

 ## nth derivative is given via partial integration as  ( 2 pi j k)^n f(k)
 ## every row needs to be multiplied by the according k
 B = np.array( [ 1j * 2 * np.pi * kk * An for kk, An in zip( kl, A ) ] )

 ## for the part with the linear term, every column needs to be multiplied
## by the according x or--as here---every row is multiplied element
 ## wise with the x-vector
C = np.array( [ xl * An for An in  A ] )

 ## thats the according linear operator
 D = B + C

 ## the gaussian
 yl = gauss( xl )

 ## the transformation with the linear operator
 print(  np.dot( D, yl ).round( decimals=9 ) )
## ...results in a zero-vector, as expected
</code>```

答:
The line

print(  np.dot( D, yl ).round( decimals=9 ) )

calculates the integrals of the Fourier transform at set of frequencies using the rectangle integration method:
\begin{equation}
F(k) = \int\limits^{\infty}_{-\infty}e^{-ikx}\left(\partial_x + x\right)e^{-\frac{x^2}{2}}dx=\left.e^{-ikx}e^{-\frac{x^2}{2}}\right\vert^{\infty}_{-\infty} +\int^{\infty}_{-\infty}(ik + x)e^{-ikx}e^{-\frac{x^2}{2}}dx=\int^{\infty}_{-\infty}(ik + x)e^{-ikx}e^{-\frac{x^2}{2}}dx\approx\int\limits^{5}_{-5}\left(ik + x\right)e^{-ikx}e^{-\frac{x^2}{2}}dx\approx\sum\limits^{N}_{j = 0}(ik + x_j)e^{-ikx_j}\Delta{x}e^{-\frac{x^2_j}{2}}=\sum\limits^{N}_{j = 0}v^{(k)}_jf_j=(\vec{v}^{(k)}, \vec{f}),
\end{equation}
where $\Delta{x}$ is the coordinate step, $v^{(k)}_j = (ik + x_j)e^{-ikx_j}\Delta{x}$, $f_j = e^{-\frac{x^2_j}{2}}$. Since we want to calculate the following ressult vector $\vec{r}$: $r_p = F(k_p)$, we can to introduce the matrix $D$:
\begin{equation}
D =
\begin{pmatrix}
(\vec{v}^{(k_0)})^{\intercal}\\
(\vec{v}^{(k_1)})^{\intercal}\\
\vdots\\
(\vec{v}^{(k_n)})^{\intercal}
\end{pmatrix}
\end{equation}
or
\begin{equation}
D =
\begin{pmatrix}
v^{(k_0)}_0 & v^{(k_0)}_1 &\ldots & v^{(k_0)}_N\\
v^{(k_1)}_0 & v^{(k_1)}_1 & \ldots & v^{(k_1)}_N\\
& &\ldots &\\
v^{(k_n)}_0 & v^{(k_n)}_1 & \ldots & v^{(k_n)}_N
\end{pmatrix},
\end{equation}
so the vector $\vec{r}$ can be calculated by the following way
\begin{equation}
\vec{r} = D\vec{f}.
\end{equation}
The last equation corresponds to np.dot( D, yl ) in the code.

It is also important to note that according to the Kotelnikov-Nyquist–Shannon theorem, there will be no loss of information only if (spectral density) the highest frequency of the useful signal is equal to half or less than the sampling frequency. Half of the sampling frequency is called the Nyquist frequency:

kmax= 1 / ( 2 * dl )


问:
Why axes should be 90 degrees to each other?
So why we can't (or we actually can?) draw coordinate system like this:

We can add more than 3 dimensions!

Are there any problems in this coordinate system with describing n-dimentional bodies?

PS: I've heard something about affine coordinates and parallel coordinates but I can't put it all together and clarify those ideas.

答:
IMO, this is kind of a confused question, since obviously the title of it does not match its description. The title asks about the $90^{\circ}$ angle between axes, while the description asks whether the proposed picture has represented an n-dimensional space. Here, I want to disprove the latter case clearly:

Regarding the axes you've shown in your picture, you also have to make sure the basis vectors in those directions will be linearly independent(You can't, read $2$ below).

You won't be able to do $1$. Because it can be proved that every possible Basis for a vector space will have the same cardinality as the other ones. I.e., Since we know that x, y, and z dimensions can cover all the 3D space, so no matter if you add more axes like the picture in your question, some of them will definitely be superfluous(wherever they could go, could be covered with x, y and z only as well), i.e., you have not added more "dimensions" in your picture, at all.


问:
Find a function whose domain and range do not contain any positive numbers
Find a function whose domain and range do not contain any positive numbers. Explain why your function satisfied.</p>

答:
\begin{gather*}
y=-( 1+\mid \ln( -1-x) \mid )\\
The\ domain\ is\ ( -\infty ,-1)\\
and\ the\ range\ is\ ( -\infty ,-1) .
\end{gather*}
Does this answer your question?


问:
multiply two Fibonacci numbers
As you know this is Fibonacci sequence:

$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots$

As I noticed if you multiply $T_{n-1}$ and $T_{n+1}$ from Fibonacci numbers, you almost always get $(T_{n})^2 \pm  1$.
for example:

$2 \times 5 = 3^2 + 1$

$3 \times 8 = 5^2 - 1$

$5 \times 13 = 8^2 + 1$

$13 \times 34 = 21^2 + 1$

So I was wondering, is there any algebraic proof for this?

答:
This is a known identity:

$F_{n-1}F_{n+1} = F_n^2 + (-1)^n$.

Title: multiply two Fibonacci numbers


答:
Yes. A way to see this identity:

Let $Q = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.
Then, $Q^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$, where $F_i$ is the $i$-th Fibonacci number (a fun exercise would be to check this - it's a fairly straightforward induction argument).

Then, we notice $\det(Q) = -1$, so $\det(Q^n)={(-1)}^n$, but also we explicitly see that $\det(Q^n) = F_{n+1}F_{n-1} - {F_n}^2$.

So: $F_{n+1}F_{n-1} - {F_n}^2 = {(-1)}^n$.

Title: multiply two Fibonacci numbers

Body:
As you know this is Fibonacci sequence:

$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots$

As I noticed if you multiply $T_{n-1}$ and $T_{n+1}$ from Fibonacci numbers, you almost always get $(T_{n})^2 \pm  1$.
for example:

$2 \times 5 = 3^2 + 1$

$3 \times 8 = 5^2 - 1$

$5 \times 13 = 8^2 + 1$

$13 \times 34 = 21^2 + 1$

So I was wondering, is there any algebraic proof for this?


问:
Approximating nonlinear first order recurrence with differential equation
I have been investigating the following recurrence that pops out in describing the probability a particular Monte Carlo algorithm for min cuts succeeds:

$$P_k = P_{k-1} - P_{k-1}^2 / 4$$

where $P_0 = c$ for some $0 < c < 1$. In my case, all I need is a lower bound for $P_k$ but ultimately I want it to be asymptotically as tight as I can get because this success probability plays a role in how many times I will re-run the algorithm to make it much less likely the algorithm fails. A naive approach uses the fact that $0 \leq P_{k-1} \leq 1$ to make the recurrence become

$$P_{k} \geq P_{k-1}\left(1 - 1/4\right) = \frac{3}{4} P_{k-1} \geq \left(3/4\right)^{k} P_{0}$$

The depth of the recursion tree in the algorithm is $d = c \log(n)$ for some constant $c$ which makes the above bound state that $P_{d} = \Omega(1/n^b)$ for some constant $b$. This is not great because it can be shown that $P_{d} = \Omega(1/\log(n))$. From the original paper on the min cut algorithm, they tackled this recurrence by making $P_k = 4/Z_k$ for some new variable $Z_k$. With this, one can then find that $P_k = \Omega(1/k)$ and so the desired result falls out.

I asked a question recently here to try and get some intuition into this transformation because it was not intuitive to me. One of the simpler approaches mentioned in the answer effectively treated the recurrence as a finite difference approximation to some differential equation and used that idea to spit out a differential equation we could work on instead. By this approach, one can find that the related diffeq is $\dot{p} = -p/4$ with $p(0) = c$ and it gives us that $p(k) = 4/(4c^{-1} + k)$. Interestingly, this $p(k)$ is asymptotically correct when you compare to the $P_k = \Omega(1/k)$ result.

My question is, is there any rigorous way to relate the answer to a recurrence of the form $x_{k} = x_{k-1} + f(x_{k-1})$ for some function $f$ to the solution you get from the differential equation $\dot{x}(t) = f(x(t))$? In particular, are there any well known results that give inequalities that relate the two so I can find lower bounds or upper bounds of a recurrence using the corresponding differential equation (and vice versa)?

Edit

More specifically, if we have the recurrence $X_{k} = X_{k-1} + f(X_{k-1})$ with $X_{0} = c$ and differential equation $\dot{x}(t) = f(x(t))$ with $x(0) = c$, do we have any bound of the form $K_1 x(k) + d_1 \leq X_{k} \leq K_2 x(k) + d_2$ for all $k \geq 0$ with constants $K_1, d_1, K_2, d_2$?

答:
Such recursions that look like Bernoulli equations can often be solved using the "Bernoulli trick". That is, consider the corresponding negative power of $P$,
$$
P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1}
=P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+\tfrac1{64}P_{k-1}^2+...\ge P_{k-1}^{-1}+\tfrac14
$$
so that
$$
P_k\ge \frac{k}4+P_0^{-1}\implies P_k\le \frac{4P_0}{4+P_0k}
$$

On the other side, use the observation that $P_k<P_{k-1}<...<P_1<P_0\le 1$ is strictly falling to get
$$
P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1}
=P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+...\le P_{k-1}^{-1}+\tfrac14\frac{1}{1-\frac14P_0}=P_{k-1}^{-1}+\frac1{4-P_0}
$$
From this one concludes
$$
P_k^{-1}\le\frac{k}{4-P_0}+P_0^{-1}\implies P_k\ge \frac{4P_1}{4+P_0(k-1)}
$$
So the result is the same as that of Henry, only with a more direct derivation.


问:
An interesting gambling problem in probability
Back in high school we used to play a simple gambling game with cards. Two people would randomly choose a card from a deck of $52$ cards. The person with a higher value card (Ace being the strongest and two the weakest) would then be the winner.

In this seemingly simple scenario is it possible to calculate the probability of a win? That is the first person choosing a higher value card. (I believe a similar line of thought would hold for the second person winning?)

Although appearing quite straightforward at first glance I couldn't figure out how to even get started.

Any help or ideas would be appreciated. Thank you!

答:
Let $X_{i}$ denote the value of the $i$-th draw (from 1 to 13). Conditionally on $X_1$,
$$
\mathsf{P}(X_2>X_1\mid X_1=v)=\frac{4(13-v)}{51}.
$$
Thus, the probability that the second player wins is
$$
\sum_{v=1}^{13}\mathsf{P}(X_2>X_1\mid X_1=v)\mathsf{P}(X_1=v)=\sum_{v=1}^{13} \frac{4(13-v)}{51}\times \frac{4}{52}=\frac{8}{17}.
$$
Similarly,
$$
\mathsf{P}(X_2<X_1\mid X_1=v)=\frac{4(v-1)}{51},
$$
and so, the probability that the first player wins is also $8/17$.


问:
Table of separation properties of various topological spaces
I was attempting to make a collection of some of the separation properties of various topological spaces, can anyone tell me which of the properties I got incorrect? Also does anyone know of any source with a collection like this with more topological spaces and more separation properties?

$$\begin{array}{cc|c}
 & \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\
\hline
\mathbb{R}&Y & Y & Y&Y&Y\\
\mathbb{R}^n&Y & Y & Y&Y&Y\\
\mbox{indiscrete}&N & N & Y&Y&Y\\
\mbox{discrete}&Y & Y & Y&Y&Y\\
\mbox{Cofinite}& Y& N&N &N&Y \\
\mbox{Cocountable}&Y&N&N&N&Y\\
\mathbb{R}_l&Y &Y &Y &Y&Y\\
\mbox{line w  2 origins} &Y &N &N &N&Y\\
\mbox{ordered  square} & Y&Y &Y &Y&N \\
\mathbb{R}_k &Y &Y &N &N&Y\\
\{0,1\}^A &Y &Y &Y &Y&Y
\end{array}$$

Comment:$\{0,1\}^A$ is the set of all functions from a set $A$ to $\{0,1\}$. $\mathbb{R}$ and $\mathbb{R}^n$ have the standard topology.$\mathbb{R}_l$ is $\mathbb{R}$ with the lower limit topology and $\mathbb{R}_k$ is $\mathbb{R}$ with the $k$ topology. I am not assuming regular/normal necessarily implies $T_1$.

答:
As a source of more of such tables the classic book “Counterexamples in Topology” is still very nice (it has such tables in the back for many more such properties), and a modern adaptation (taking that and other sources and making it searchable in a modern way) is $\pi$-base where you can search for a space and find its properties or look for a combination of properties, and find a space for it.

As to your table: separable is not a separation property, despite the name. It’s more of a notion of size IMO.

You cannot say that the cocountable topology is not Hausdorff, because that’s only the case if the underlying set is uncountable. And on a countable or finite set, any topology is separable, so you should specify the exact underlying set in the case of cofinite, discrete, cocountable topologies. Maybe fix the set on $\Bbb R$ for definiteness and contrastive value.

$\{0,1\}^A$ is separable iff $|A|\le |\Bbb R|$, so that needs an extra specification too.


问:
Complete local ring which is not complete valuation ring
Complete valuation ring is obviously complete local, because valuation ring is always local ring.

I would like to know the example of

Complete local ring which is not complete valuation ring

Any examples? I could find example of local ring which is not valuation ring, but the example ring is not complete.
Thank you for your help.

答:
$k[x]/(x^2)$ is a complete local ring, it is not a valuation ring because it is not an integral domain.

$R=k[[x^2,x^3]]$ is a local ring, it is complete because $R\to \underset{n\to \infty}\varprojlim R/\mathfrak{m}^n$ is an isomorphism, it is not a valuation ring because $x,x^{-1}\in Frac(R),\not \in R$.

Equivalently its principal ideals are not totally ordered: $(x^3)\not\subset (x^2)$, $(x^2)\not\subset (x^3)$.

The completion at the maximal ideal doesn't have to be the same as the completion for the valuation: $S= \{ \sum_{j\ge 1} c_j x^{b_j},c_j\in k, b_j\in \Bbb{Q}_{\ge 0}, \lim_{j\to \infty} b_j=\infty\}$, then $\mathfrak{m}^n = \mathfrak{m}$ so $\underset{n\to \infty}\varprojlim S/\mathfrak{m}^n=S/\mathfrak{m}\cong k$ while $S$ is complete for the valuation (every Cauchy sequence converges in $S$).


答:
Remember,
$$n-(n-k)=n-n+k=k.$$
Now just use the fact that multiplication is commutative to switch the order of the terms in the denominator.</p>

Title: Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$

Body:
I'm reading Lang's Undergraduate Analysis:

Let ${n}\choose{k}$ denote the binomial coefficient,

$${n\choose k}=\frac{n!}{k!(n-k)!}$$

where $n,k$ are integers $\geq0,0\leq k\leq n$, and $0!$ is defined to be $1$. Prove the following assertion:

$${n\choose k}={n\choose n-k}$$

I proceded by making the adequate substitutions:

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!}$$

And then I simplified and achieved:

$$\frac{n!}{(n-k)!-k!}$$

But I'm not sure on how to proceed from here, I've noticed that this result is very similar to $\frac{n!}{k!(n-k)!}$. What should I do? I guess it has something to do with the statement about the nature of $n$ and $k$:

$n,k$ are integers $\geq0,0\leq k\leq n$

So should I just change the minus sign to plus sign and think of it as a product of $(n-k)!$?

$$\frac{n!}{(n-k)!-k!}\Rightarrow\frac{n!}{(n-k)!+k!}\Rightarrow \frac{n!}{k!(n-k)!}$$

I'm in doubt because I've obtained the third result on Mathematica, but obtained the first with paper and pencil. I'm not sure if there are different rules for simplification with factorials. I'm not sure if this $(n-k)!+k!$ mean a sum or a product in this case.


问:
Given 4 numbers $a, b, c, d> 0,$ show $16\max\limits_{\bigcirc}\left \{ a^{3}+ 3bcd \right \}\!\geq\!\left ( a+ b+ c+ d \right )^{3}$

Given four positive numbers $a, b, c, d.$ Prove that
$$16\max\left \{ a^{3}+ 3bcd, b^{3}+ 3cda, c^{3}+ 3dab, d^{3}+ 3abc \right \}\geq\left ( a+ b+ c+ d \right )^{3}$$

the way I think is using the inequality $\max\left \{ m, n \right \}\geq\dfrac{x}{x+ y}m+ \dfrac{y}{x+ y}n$ with 4 positive numbers $x, y, m, n.$ So we must find $x, y:=f\left ( a, b, c, d \right ), f\left ( b, c, d, a \right )$ here, I need to the help.

答: Hint: Essentially, you need just $4\,\max\{A, B, C, D\}\ge A+B+C+D$. Expand $(a+b+c+d)^2$, and use $a^2b+ab^2\le a^3+b^3$ etc, the latter follows from $$(a^2-b^2)(a-b)=(a-b)^2(a+b)\ge0.$$ Is this sufficient, or do I have to elaborate?


问:
10 arbitrary points inside a rectangle with dimensions 5 and 2
Prove that if we put $10$ arbitrary points inside a rectangle with dimensions $5$ and $2$, there exist $2$ points $a,b$ such that $d(a,b)<\sqrt{2}$ where $d(a,b)$ is the distance of $a$ and $b$.
I hope the problem is clear.
I don't have major results to show.

答:
This question is equivalent to the following question:

What is the maximum number of circles with radius $r = \sqrt{2}/2$ which can be packed into a rectangle of side lengths $l = 5+2\sqrt{2}$ and $w = 2 + 2\sqrt{2}$ ?

In 1940 it has been proven that the densest of all possible equal circle packings in the plane is the hexagonal packing.

Source: https://en.wikipedia.org/wiki/Circle_packing


问:
Inequality involving product of iid random variables
Let $a>0$, $b>1$. Let $(X_k)$ be a sequence of iid random variables, with density $f(t) =\frac{a}{b} \left(\frac{b}{t}\right)^{a+1}$ for $t>b$ and 0 otherwise.

I can work out that the CDF is $F_{X_k} (x) =1-(b/x)^a$.

Let $Z_n:=\prod_{k=1}^nX_k$, and let $Z$ be any random variable. I'd like to show the inequality:

$\mathbb{P} (X_{n+1}>3/2) < \mathbb{P} (|Z_{n+1}-Z|>1/4)+\mathbb{P}(|Z_n-Z|>1/4)$.

The only thing I can think of is work out the lhs to be $(2b/3)^a$, and use sub additivity for the rhs, and hope that $\mathbb{P} ([|Z_{n+1}-Z|>1/4] \cup [|Z_n-Z|>1/4)] $ will already be greater than the lhs, but to no avail.

Presumably I should use both indepedence and the definition of $Z_{n+1}=Z_n X_{n+1}$, and perhaps $b>1$, but cannot find how.

答:
First note that $x_k >b>1$ for all $k$ so $Z_n >1$ for all $n$.

Let $X_{n+1} >\frac 3  2 $. Then $ \frac {Z_{n+1}} {Z_n} >\frac  3 2 $ which gives $ \frac {Z_{n+1}-Z_n} {Z_n} >\frac  3 2-1=\frac 1 2 $. Hence, $Z_{n+1}-Z_n >\frac 1 2Z_n >\frac 1  2$. Can you see now why we must have either $|Z_{n+1}-Z| >\frac 1  4$ or $|Z_n-Z| >\frac 1  4$?


问:
How to compute the unknown power x in $0.15^x + 0.36^x = 1$
More generally, I have two known variables $a$ and $b$ and one unknown variable $x,$ where:

$$a^x + b^x = 1, a,b \in [0, 1]$$

Is there a way to compute the value of $x$ analytically?

答:
Consider that we look for the zero of function
$$f(x)=a^x+b^x-1$$ As already said in answers, for the most general case, there is no analytical solution.

However, we can make very good approximations considering instead that we look for the zero of function
$$g(x)=\log(a^x+b^x)$$ which much closer to linearity than $f(x)$. We have
$$g'(x)=\frac{a^x \log (a)+b^x \log (b)}{a^x+b^x}$$
$$ g''(x)=\frac{a^x \log ^2(a)+b^x \log ^2(b)}{a^x+b^x}-\frac{\left(a^x \log (a)+b^x \log
   (b)\right)^2}{\left(a^x+b^x\right)^2}$$

What is interesting to notice is that
$$g(0)=\log(2) >0 \qquad \text{and}\qquad f''(0)=\frac{1}{4} \log ^2\left(\frac{a}{b}\right)>0$$ By Darboux theorem, this means that Newton method would converge without never overshooting the solution.

We could use Halley method and have as an estimate
$$x=-\frac{4 \log (2) \log (a b)}{(2-\log (2)) \log ^2\left(\frac{a}{b}\right)+8 \log
   (a) \log (b)}$$

Applied to the case $a=0.15$, $b=0.36$ this would give
$x=0.490244$ while the "exact" solution is $x=0.490642$.

Not exact but so close.

Doing the same with Householder method would give $x=0.490752$.

Edit

Please, notice that all the above has been obtained starting at $\color{red}{x=0}$. Assuming, without loss of generalization $a<b$, we can improve the process noticing that the solution is such that
$$\alpha=-\frac{\log (2)}{\log (a)} < x < -\frac{\log (2)}{\log (b)}=\beta$$

Taking into the quasi-lineartion of function $g(x)$, using the secant, you can generate a better estimate
$$\gamma=\frac{\beta \, g(\alpha )-\alpha \, g(\beta )}{g(\alpha )-g(\beta )}$$ and, using Halley method with $x_0=\gamma$, obtain the approximation
$$x=\gamma-\frac{2 g(\gamma )\, g'(\gamma )}{2 g'(\gamma )^2-g(\gamma )\, g''(\gamma )}$$

Applied to the working case, this would give
$$\alpha=0.365368\cdots \qquad \beta=0.678458\cdots \qquad \gamma=0.492216\cdots$$ from which the estimate
$$x=0.4906415695461$$ while the "exact" solution is
$$x=0.4906415695447$$


问:
$\alpha$ and $\beta$ are the roots of the quadratic equation.
$ω≠1$ and $ω^{13}=1$. If $\alpha=ω+ω^3+ω^4+ω^{−4}+ω^{−3}+ω^{−1}$ and $\beta=ω^2+ω^5+ω^6+ω^{−6}+ω^{−5}+ω^{−2}$, then quadratic equation, whose roots are $\alpha$ and $\beta$ is:

I tried finding the sum and product of roots and placing in the equation $x^2-(\alpha+\beta)x+(\alpha\cdot\beta)=0$ but I was not able to solve for it.

答:
By the formula for a finite geometric series,

$$\alpha + \beta = \sum_{i=-6}^{-1} \omega^i + \sum_{i=1}^6 \omega^i =  \sum_{i=7}^{12} \omega^i + \sum_{i=1}^{6} \omega^i =\frac{(1-\omega^{13})}{1 - \omega} -1 =  -1 \tag{$\omega^{13} =1$}$$

For the product $\alpha \beta$, note that $\omega^n = \overline{\omega^{-n}}$. Taking $\omega$ as $e^{2 i \pi /13} = \cos(2\pi /13) + i \sin(2\pi /13)$ for example, $\alpha = 2 \left(\cos(2 \pi/13) + \cos(6 \pi/13)+\cos(8 \pi/13)\right)$ and you can do the same for $\beta$. Then you can use the product-to-sum formula $\cos A \cos B = \frac{1}{2} \left(\cos(A-B) + \cos(A+B) \right)$: by expanding the brackets, there are nine terms in this form.


问:
If a two variable smooth function has two global minima, will it necessarily have a third critical point?
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a  $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$, and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.

Those two points are connected by a compact line segment $K$. As a continuous function, $f$ attains a maximum value $M$ on the set $K$. This means that when the water level has reached $M$, the two lakes will have been merged.

The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$.

It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$. But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ($S^1\times S^1$) or the surface of a sphere ($S^2$). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?

Similarly, what if we have local minima instead of global?

If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).

答:
Morse theory says that every Morse function $f$ (that is all critical points were non-degenerate and distinct critical points take distinct critical values) satisfies $$\#\min+\#\max-\#\mathrm{saddle}=\chi(M).$$ So, in the case of torus (that its Euler char is $0$), the functions must have $\#\mathrm{saddle}=\#\max+2$. By the fact that  all continuous functions on a compact domain attaint  at least a max and a min therefore we should have at least a max point then at least 3 saddle point for torus. in the case of sphere is similar.
$\#\min+\#\max-\#\mathrm{saddle}=\chi(\Bbb S^2)=2$ so having two global minima we must have $\#\max=\#\mathrm{saddle}\neq 0$. In any case we have at least a saddle point.

Note that these are non-degenerate critical points (that means Hessian is nonsingular at that points) and there is a function on torus with 3 critical points i.e. a min, a max and a degenerated saddle point.


问:
Is it always possible to extend a homeomorphism from a closed ball in the plane onto a closed disk in a surface to their neighbourhoods?
Let $S$ be a two dimensional surface and $D$ a closed disk contained in $S$. Is it true that if $f$ is a homeomorphism from a closed ball $B$ in the plane onto $D$, then $f$ extends to a homeomorphism from a neighbourhood of $B$ in the plane onto a neighbourhood of $D$ in the surface?

答:
Assuming that you don't mean manifold with boundary (in which case it's not true, since $D$ might already be the whole space!) then yes, it's true.  One way to do it is with the Tubular Neighborhood Theorem by switching to the smooth category.

Let $J$ be the boundary curve of your disc in $S$.  Then it has a tubular neighborhood, which is diffeomorphic to a neighborhood of $J$ in its normal tangent bundle.  For a simple closed curve $J$ that's just an open annulus.  This neighborhood is separated into two components by $J$, one of which will meet the interior of your disc.  Keep the other one and glue your annulus homeomorphism and disc homeomorphism together along $J$.

If it's annoying having to go into the smooth category, just be aware that these sorts of theorems get VERY hard without appealing to smooth, algebraic/PL or complex-analytic techniques.


问:
2-dimensional representation of modular group
My question probably follows from something basic in character theory, but I don't see what I'm missing. Let $\rho: \Gamma \to \mathrm{GL}(2,\mathbb C)$ be a two-dimensional representation of the modular group $\Gamma$, assumed to be indecomposable. Then for the identity $I$ one has $\rho(-I) = \pm I$. Now I am confused by:

Because $\det \rho(-I) = 1$, $\det \rho$ is a character of order
dividing $6$.

I do not see how this follows. I assume here the determinant character means if $\rho$ affords a character $\chi$ then the determinant character $\det\chi$ is defined as $\det \rho(\cdot)$ and here the author means order in the sense as an element of the group of characters.

I have read that the determinant character has order dividing $|G|$, but in this case the modular group $C_2 *C_3$ has infinite cardinality. What am I missing?

答:
Let $\chi$ be a character $\in Hom(SL_2(\Bbb{Z}),\Bbb{C}^\times)$.

Here $\chi(-I)=1$ thus $\chi$ is a character of
$SL_2(\Bbb{Z})/\pm I= G=PSL_2(\Bbb{Z})$ whence of $G/[G,G]$.

Then $G= \langle S\rangle \ast \langle ST\rangle \cong C_2 \ast C_3$ (free product of two cyclic groups) so $G/[G,G]\cong C_2 \times C_3$ ie. $\chi$ has order dividing $6$.


问:
Analytic Properties of $f(x) = x^x$
The function $f(x) = x^x, x > 0$ can be plotted on graphing software and inspected to see a local minimum around .367. The function is convex, decreasing from 0 to its minimum, and increasing thereafter. The derivative of the function can be found by implicit differentiation to be $f'(x) = x^x(\ln(x)+1)$.

Can the exact value of the local minimum be found?

Can someone explain intuitively why the function first decreases and then increases?

Is there anything interesting about the class of functions $\{ \, f(x) \, | \, f'(x) = f(x) \cdot g(x) \, \}$

Edit: I forgot, I'll be subject to a firing squad if I don't explain what I have tried!

Setting the derivative equal to 0 doesn't do much for me. If this can be solved exactly without optimization algorithms, I suspect something really clever will need to happen.

I tried thinking about what is happing for different types of inputs (irrational $x$, rational $x$, natural $x$). Still didn't get anywhere.

Nothing to try really, does this class of functions come up anywhere in math?

答:
Among the properties of the function $x^x$, don't forget

The antiderivative $\quad \text{Sphd}(x)=\int_0^x t^{\,t}\, dt\quad$ namely the Sophomore's dream function (Note below).

The inverse function of $\quad x^x=y(x)\quad$ which is
$$x(y)=e^{W\big(\ln(y)\big)}$$

$W$ is the Lambert W function (graph below)

The graph is a copy of page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function .

Note : In this paper one can find many properties of the antiderivative of $x^x$ wich is a special function (not standardized).

Title: Analytic Properties of $f(x) = x^x$


问:
Evaluating a double integral containing absolute value squared of sum of two functions.
I need to evaluate integrals of this form

$\int\limits_{\phi = 0}^{2\pi }\ \int\limits_{\theta = 0}^{\pi} \left | f_1(\theta, \phi)e^{ix} + f_2(\theta, \phi)e^{ix'} \right |^2\sin{\theta} \ d\theta d\phi$

Where,

$f_1$ and $f_2$ are harmonic functions of $\theta$ and $\phi$. Mostly, $\cos^2{\frac{\theta}{2}}$ or $\sin^2{\frac{\theta}{2}}$ and in general, $f_1 \neq f_2$. $x$ and $x'$ are some real constants.

Any help regarding how I can evaluate these integrals is highly appreciated.

Body:
Hint: If $f_1$ and $f_2$ are real, then
$$|f_1 e^{i x}+f_2 e^{i x'}|^2 = f_1^2+2f_1 f_2\cos(x-x') + f_2^2.$$


问:
Learning roadmap and prerequisites for Isbell duality
I'm looking for a roadmap to learning about Isbell duality. I know a reasonable amount about several of the "specific" dualities (Gelfand duality, AffSch - CRing, frames - locales, etc), especially affine schemes. However, I'm having trouble finding anything that amounts to a coherent presentation of the general case (with clear prerequisites), and how it ties back to the specific cases.

The nlab article on space and quantity I can follow the "idea" part of, but when they get to the actual Isbell duality, I get lost at symmetric monoidal categories, $V$-enriched categories, ends, coends... and what's more problematic is, I can't find a source that cleanly ties the general case back to the specific cases, or that discuss interesting examples where you have just an adjunction (instead of a full equivalence).

Any guidance or resources are helpful, especially those with prerequisites clearly spelled out. I currently know basic category theory (first few chapters of Categories Work, chapter 1 of Vakil Algebraic Geometry).

答:
Since you are unfamiliar with enriched categories, let me describe the special case of ordinary categories first.

Let $\mathcal{C}$ be an essentially small category, e.g. the opposite of the category of finitely presentable commutative rings.
The category $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ has a universal property: it is the free completion of $\mathcal{C}$ by colimits of small diagrams.
Dually, $[\mathcal{C}, \textbf{Set}]^\textrm{op}$ is the free completion of $\mathcal{C}$ by limits.
It turns out that $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ has limits and $[\mathcal{C}, \textbf{Set}]^\textrm{op}$ has colimits, so the two universal properties applied to the two Yoneda embeddings $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ and $\mathcal{C} \to [\mathcal{C}, \textbf{Set}]^\textrm{op}$ yield a limit-preserving functor $\operatorname{Spec} : [\mathcal{C}, \textbf{Set}]^\textrm{op} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ and a colimit-preserving functor $O : [\mathcal{C}^\textrm{op}, \textbf{Set}] \to [\mathcal{C}, \textbf{Set}]^\textrm{op}$.
Isbell duality is the observation that this constitutes an adjunction:
$$[\mathcal{C}, \textbf{Set}]^\textrm{op} (O (X), B) \cong [\mathcal{C}^\textrm{op}, \textbf{Set}](X, \operatorname{Spec} B)$$
(This is basically an application of the Yoneda lemma.)
Being "free", colimits in $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ behave very nicely and resemble what we expect of colimits in a category of spaces.
We can think of $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ as a category of spaces locally modelled by objects in $\mathcal{C}$.
On the other hand, we may think of $[\mathcal{C}, \textbf{Set}]$ as a category of algebras modelled by objects in $\mathcal{C}^\textrm{op}$, but the justification for this is not as good, in my opinion.

As a general fact, any adjunction restricts to an equivalence of categories between certain subcategories.
By construction, the representable functors on both sides are in those subcategories, but there may be more objects.
In general, this adjunction is "just" an adjunction.

One can produce many variations on this theme: you could replace $\textbf{Set}$ with some other category $\mathcal{V}$ and look at $\mathcal{V}$-enriched categories instead of ordinary categories; you could restrict to specified subcategories of $[\mathcal{C}^\textrm{op}, \textbf{Set}]$ or $[\mathcal{C}, \textbf{Set}]^\textrm{op}$, etc.
Unfortunately, none of them really capture the classical dualities on the nose – the size restrictions are an unavoidable nuisance, and unless you rig the situation by redefining the objects under study eventually you do have to do some non-trivial work to show that two prima facie different things are the same.
The case of affine schemes and commutative rings is especially an example of this – if you use the functor of points definition of scheme the duality is almost a triviality, but if you use the locally ringed space definition of scheme you get a remarkable representation theorem.

If, nonetheless, you want to pursue the subject, I suppose you could first finish reading the rest of CWM, and then read [Kelly, Basic concepts of enriched category theory].


问:
Can $\sf AC_\omega$ and $\sf DC$ be generalized?
When one come across axiom of choice, one reads about weaker forms like countable choice $\sf AC_\omega$ and dependent choice $\sf DC$.

Are there known uses of restrictions to higher cardinals like $\sf AC_{\omega_1}$ for example, or more generally $\sf AC_{\omega_\alpha}$ or a restriction to a stage of the cumulative hierarchy $\sf AC_{V_\alpha}$?

Along the same lines, are there higher forms of $\sf DC$ that use longer sequences, for example $\sf DC_{\omega_1}$ that uses $(x_n)_{n \in \omega_1}$ instead of $(x_n)_{n \in \mathbb N}$, so we may generally have $\sf DC_{\omega_\alpha}$.

What is the comparison in consistency strength between the two lines? For example how $\sf DC$ compare with the $\sf AC_{\omega_\alpha}$'s.

答:
Yes, there are general variants of both $\sf DC$ and $\sf AC_\omega$.

Since $\sf AC$ is about choice functions from a family, we can give it three parameters. I follow the convention1 that $\sf AC_X^Y(Z)$ means "Every family of subsets of $\sf Z$ which is in bijection with $\sf X$, and every member is in bijection with $\sf Y$ admits a choice function". When we omit one of the parameters, we replace it with a universal quantification. So $\sf AC^{\rm fin}(\Bbb R)$ would be the axiom of choice for arbitrarily families of finite subset of $\Bbb R$, for example; and $\sf AC_{WO}$ is the axiom of choice for arbitrarily families that can be well-ordered.

$\sf DC$, in contrast, talks about iterations of a function, and sequences, so it is inherently well-ordered. So when we write $\sf DC_\kappa$ it only makes sense that $\kappa$ is an $\aleph$ number. We can still restrict to a set with $\sf DC_\kappa(X)$,2 and while the formulations using relations and sequences is a bit awkward, we have a simpler formulation in terms of trees: $\sf DC_\kappa(X)$ means that every $\kappa$-closed tree on $X$ has a branch of length $\kappa$. Again, removing $X$ means a universal quantification, and we can write $\sf DC_{<\kappa}$ for $\forall\lambda<\kappa\,\sf DC_\lambda$.

Now, $\sf AC_{WO}$ implies $\sf DC$, but not $\sf DC_{\omega_1}$. And $\sf DC_\kappa$ does not imply $\sf AC_{WO}$ for any $\kappa$. If you also look closely, you will see that $X$ can be well-ordered if and only if $\mathcal P(X)\setminus\{\varnothing\}$ admits a choice function. So $\mathsf{AC}_{V_\alpha}$ implies that for all $\beta<\alpha$, $V_\beta$ can be well-ordered. It is not equivalent, of course, since $\sf AC_\omega$ can fail for the first time in an arbitrarily high rank.

It is worth pointing out that $\sf AC_X$ does not come up very often. One of the reasons is that we don't really understand sets that cannot be well-ordered, so it's hard to work with them. We can prove a few small things, e.g. that $\sf AC_X$ holds when every surjection onto $X$ admits an inverse, but not a whole lot more.

One of the reasons is that we never really had good tools for preserving $\sf AC_X$ for an arbitrary set $X$. Although in one of my recent papers I provide a crude preservation theorem for symmetric extensions that allows us to prove $\sf AC_X$ holds for some sets $X$ which cannot be well-ordered.

As for the consistency strength, all these are of course equivalent, since they all follow from $\sf ZFC$, which is equiconsistent with $\sf ZF$. You can look at Jech's "Axiom of Choice" book, where Chapter 8 is dedicated for these principles; and in the Howard–Rubin dictionary of choice principles as well.

Footnotes.

Some people use the opposite notation for superscript and subscripts, i.e. $\sf AC^{WO}$ for what I'd write $\sf AC_{WO}$. This is confusing, yes. Other people use something like $X$-$\sf AC$.

Here too there is a discrepancy in notation, some people use $\sf DC_X$ to denote $\sf DC(X)$, and $\kappa$-$\sf DC$ to denote $\sf DC_\kappa$. Again, confusing.


问:
Expanding $\boldsymbol \nabla\phi(\mathbf{x},t)$ where $\mathbf{x} = \mathbf{x}(\mathbf{a},t)$
How does $\boldsymbol \nabla\phi(\mathbf{x},t)$ expand where $\mathbf{x} = \mathbf{x}(\mathbf{a},t))$ and $\mathbf{x} =(x_1,x_2,x_3) $ and $\mathbf{a} = (a_1,a_2,a_3)$ where $\boldsymbol \nabla = (\partial_{a_1}, \partial_{a_2}, \partial_{a_3})$?

Here $x_1 = x_1(a_1)$ etc.

$$\boldsymbol \nabla \phi = (\partial_{a_1}, \partial_{a_2}, \partial_{a_3})\phi = (\partial_{a_1}\phi, \partial_{a_2}\phi, \partial_{a_3}\phi)$$

The first component: $$\partial_{a_1}\phi = \frac{\partial \phi}{\partial \mathbf{x}}\frac{\partial \mathbf{x}}{\partial a_1}$$

but writing derivative wrt $\mathbf{x}$ is a little confusing for me.

Could anyone please clarify?

答:
Here's an answer taken from my comment:

If this makes it clearer, we have $\phi(\mathbf{x},t)≡\phi(x_1,x_2,x_3,t)$. Use the chain rule to this function that takes 4 arguments. For example,
$$\partial_{a_1}\phi=\partial_t ϕ\ \partial_{a_1}t+\sum_{i=1}^3\partial_{x_i}\phi\ \partial_{a_1}x_i$$

If only $x_1$ depends on $a_1$, then this expression simplifies into
$$\partial_{a_1}\phi=\partial_{x_1}\phi\ \partial_{a_1}x_1$$


问:
Exact sequence of abelian Groups with unknown groups
I have a exact sequence of abelian groups
$$0 \rightarrow Y \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow X \rightarrow 0$$
where $X$ and $Y$ can be any abelian groups and $\mathbb{Z}$ is the group of Integers.

My question is can I calcuclate $X,Y$? They should be Zero but I have no proof for this.

I found out that if one of them is Zero it follows that the other one is Zero.
If one could show that they have to be isomorphic then it follows that they have to be Zero as well.

答: They don't indeed both have to be zero.  For we could have the map $\phi$ in the middle being zero, the trivial homomorphism.  Consider $0\to\Bbb Z\stackrel{\text{id}}\to\Bbb Z\stackrel{\phi}\to\Bbb Z\stackrel{\text{id}}\to\Bbb Z\to0$.


问:
Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$
I'm reading Lang's Undergraduate Analysis:

Let ${n}\choose{k}$ denote the binomial coefficient,

$${n\choose k}=\frac{n!}{k!(n-k)!}$$

where $n,k$ are integers $\geq0,0\leq k\leq n$, and $0!$ is defined to be $1$. Prove the following assertion:

$${n\choose k}={n\choose n-k}$$

I proceded by making the adequate substitutions:

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!}$$

And then I simplified and achieved:

$$\frac{n!}{(n-k)!-k!}$$

But I'm not sure on how to proceed from here, I've noticed that this result is very similar to $\frac{n!}{k!(n-k)!}$. What should I do? I guess it has something to do with the statement about the nature of $n$ and $k$:

$n,k$ are integers $\geq0,0\leq k\leq n$

So should I just change the minus sign to plus sign and think of it as a product of $(n-k)!$?

$$\frac{n!}{(n-k)!-k!}\Rightarrow\frac{n!}{(n-k)!+k!}\Rightarrow \frac{n!}{k!(n-k)!}$$

I'm in doubt because I've obtained the third result on Mathematica, but obtained the first with paper and pencil. I'm not sure if there are different rules for simplification with factorials. I'm not sure if this $(n-k)!+k!$ mean a sum or a product in this case.

答:
You simply failed to distribute the negative sign (i.e., multiplying the difference by $-1$: $$\begin{align} n - (n - k) &= n + -1(n -k) \\ \\& = n + - 1\cdot n - (-1)\cdot k \\ \\
& = n - n + k \\ \\ &= k\end{align}$$

(I.e., Negation distributes over sums and differences.)

$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!} =\frac{n!}{(n-k)!(n - n + k))!} =\frac{n!}{(n-k)!(k)!}= \frac{n!}{k!(n-k)!}$$


问:
How to visualize a Markov chain in the following way?
Nice to meet you, this is my first question in this site.

I now can draw $A\to B\to C$ but stuck at $A\to B\to C\to D$. Let me show you my graph: $A\to B\to C$.

It's easy to see that $A$, $B$ and $C$ are not independent to each other, but conditional on $B$, then $A$ and $C$ would be independent, which can deduce the Markov property: $P(C|B,A)=P(C|B)$.

So, how to draw $A\to B\to C\to D$ in a similar way, or more generally, draw $X_1\to X_2\to \dots\to X_n$? Or is it possible?

答:
Just found the answer by myself:

And here is $X_1\to X_2\to\dots\to X_{10}$: