问:
Help proving a simple tautology $(p\land(p\to\lnot q))\to(\lnot q\lor p)$
Specifically this  $$(p\land(p\to\lnot q))\to(\lnot q\lor p)$$-

spent a lot of time going down the wrong path

答:
You were almost done, just two more steps are missing in your caluculation to reach tautologicalitly:

$
\vdots\\
\equiv q \lor (\neg q \lor p)\\
\equiv (q \lor \neg q) \lor p \quad\text{(associativity)}\\
\equiv \top \lor p \quad \text{(complement)}\\
\equiv \top \quad \text{(domination)}$


问:
Prove $ A \vdash \neg \neg A $ (Natural Deduction)
From this set of rules:

$ (\wedge E) $  $A \wedge B \vdash A $

$ (\wedge I) $ $A, B \vdash A \wedge B $

$ (\vee E) $ $ A \vee B, A \rightarrow C, B \rightarrow C \vdash C $

$ (\vee I) $ $ A \vdash A \vee B $

$ (\neg E) $ $ \neg A \rightarrow B, \neg A \rightarrow \neg B \vdash A $

$ (\neg I) $ $ A \rightarrow B, A \rightarrow \neg B \vdash \neg A $

$ (\rightarrow E) $ $ A, A \rightarrow B \vdash B $

$ (\rightarrow I) $ $ Premises \vdash A \rightarrow B $

With the last rule ($ \rightarrow I $), you can introduce any implication you like as long as you prove in a sub-proof the consequent of the implication under the assumption of the hypothesis of the implication e.g.

$ P \rightarrow Q \quad $ Premise

$ \neg Q \quad\quad\space\space $ Premise

$ P \rightarrow \neg Q \quad \rightarrow I,$ subproof below

3.1 $ P \quad\space\space\space $ Assumption

3.2 $ \neg Q \quad $ from (2)

$ \neg P \quad $ from (1)(3), $ \neg I $

Prove $ A \vdash \neg \neg A $

$ A \quad Premise $

$ \neg (\neg \neg A) \rightarrow A \quad \rightarrow I, subproof $

2.1 $ \neg (\neg \neg A) \quad Assumption $

2.2 $ A \quad\quad\quad (1) $

$ \neg (\neg \neg A) \rightarrow \neg (\neg \neg A) \quad \rightarrow I, subproof $

3.1 $ \neg (\neg \neg A) \quad Assumption $

3.2 $ \neg (\neg \neg A) \quad (3.1) $

$ \neg \neg A \quad (2)(3), \neg E $

This is the proof I have currently and I'm unsure about it because of line 3 (specificially, the consequent of the implication which would usually just be $\neg A$) and subsequently line 4 which uses line 3 as a premise.

Is my proof correct?

答:
No. As you pointed out yourself, the $\lnot E$ rule requires two premises of the form $\lnot P\rightarrow Q$ and $\lnot P\rightarrow \lnot Q$. Your pair of premises $\lnot (\lnot \lnot A)\rightarrow A$ and $\lnot (\lnot \lnot A)\rightarrow \lnot (\lnot \lnot A)$ do not have this form: $Q$ would have to be both $A$ and $\lnot \lnot A$.

However, a correct proof can be given with essentially the same form as yours. The idea is simple.  We want to prove $\lnot \lnot A$. This sentence begins with $\lnot$, so we can try to prove it using the $\lnot$ introduction rule - this is a bit more natural than working with $\lnot \lnot \lnot A$ and trying to use $\lnot$ elimination! To apply $(\lnot I)$, we need to prove $\lnot A\rightarrow B$ and $\lnot A \rightarrow \lnot B$ for some choice of $B$. Can you see a $B$ that works?

Complete proof hidden in the spoiler block below.

 \begin{align*}(1) &\quad A &\quad \text{Premise}\\(2) &\quad \lnot A\rightarrow A &\quad (\rightarrow I)\\&\quad (2.1) \quad \lnot A &\quad \text{Assumption}\\&\quad (2.2) \quad  A &\quad \text{from }(1)\\(3) &\quad \lnot A \rightarrow \lnot A&\quad (\rightarrow I)\\&\quad (2.1) \quad \lnot A &\quad \text{Assumption}\\&\quad (2.2) \quad \lnot A &\quad \text{from }(2.1)\\(4) &\quad \lnot \lnot A &\quad \text{from }(2), (3), (\lnot I)\end{align*}

ContentLicense: CC BY-SA 4.0


问:
References for Linear Algebra (on infinite dimensional vector spaces) preferably with an eye on Functional Analysis
I am taking a self study graduate course on Functional Analysis and the notes are quite terse. I am struggling in particular with the linear algebra background part, including dealing with infinite dimensional vector spaces.

So I was looking for a good reference to complement the notes. It would be particularly good if the book involved solved exercises and detailed proofs.

Thank you.

答:
I'm not aware of any such reference that isn't either a text on functional analysis or a text on commutative algebra. It would help if you were more specific about what you wanted to know about infinite-dimensional vector spaces. Most of what is worth knowing about them is a list of things that don't generalize from finite-dimensional vector spaces. If $V$ is infinite-dimensional:

In general you can't define the determinant, trace, or characteristic polynomial of an endomorphism $T : V \to V$.

It's not true that $T : V \to V$ must have an eigenvalue even over an algebraically closed field.

It's not true that $V$ has the same dimension as $V^{\ast}$.

It's not true that an injection $V \to V$ is automatically a bijection, and similarly for surjections.

Then there's some stuff that relies on the axiom of choice:

$V$ still has a basis, but only if you assume the axiom of choice (and in fact it's a famous result due to Blass that AC is equivalent to the assertion that every vector space has a basis). Due to the independence of AC it follows that it's consistent with ZF that there are infinite-dimensional vector spaces without bases.

Linear functionals still separate points, but again only if you assume the axiom of choice. It's consistent with ZF that there are infinite-dimensional vector spaces with no nonzero linear functionals (see this MO thread).


问:
Trying to delete in a sum of ideals: P+I=Q+I so P=Q?
Let be $(A,M)$ a local ring (if it needs noetherian ring), $P_1$ and $P_2$ two prime ideals with $P_1\subseteq P_2$, $\ a_1\in M\setminus P_2$. I don't know if it suffices to conclude this sentence:

If $P_1+(a)=P_2+(a)$ so $P_1=P_2$.

Someone can help me?
Maybe in a more general setting it is possible to say that if $P+I=Q+I$ so $P=Q$.

Thank you.

答:
I assume $A$ is commutative since you have commutative algebra as a tag, and I do use a hypothesis that $A$ is noetherian. I am not sure whether the result holds in more general settings.

Lemma: if $R$ is a noetherian integral domain, and $x\in R$ is a non-unit, then the only prime ideal strictly contained in $xR$ is $\{0\}$.

Proof: Let $P\subset xR$ a prime ideal strictly contained in $xR$. Certainly $(xR)P=xP=P$: indeed, given any $p\in P$, since $p\in xR$ there is $r\in R$ such that $p=xr$, whence – because $P$ is prime and $x\notin P$ – we must have $r\in P$. Now, $R$ is noetherian, so $P$ is a finitely generated ideal – ie a finitely generated $R$-module – and thus we may apply Nakayama's lemma to obtain an element $xy\in xR$ such $(1-xy)P=\{0\}$. Now, if $P$ is non-zero – say $p\in P\setminus\{0\}$ – then, because $(1-xy)p=0$ and $R$ is an integral domain, we have that $xy=1$ and so $x$ is a unit, a contradiction. Thus $P$ must indeed be the zero ideal.

Proof of main result: We consider the quotient ring $R:=A/P_1$. Note, since $P_1$ is prime, $R$ is an integral domain. Also, since $(a)+P_1\subseteq M$ is a proper ideal of $A$, the principal ideal $(a+P_1)R$ is certainly a proper ideal of $R$.

Now note, $P_2/P_1\subseteq (P_2+(a))/P_1=(P_1+(a))/P_1=(a+P_1)R$. We claim this inclusion is proper; indeed, otherwise we have $a+P_1\in P_2/P_1$, whence there are $p_2\in P_2$ and $p_1\in P_1$ such that $a=p_2+p_1\in P_2$, a contradiction.

So $P_2/P_1$ is a strict subset of the proper principal ideal $(a+P_1)R$. It is also a prime ideal, since $P_2$ is, and hence we may apply our lemma above to obtain $P_2/P_1=0$. But this means precisely that $P_2=P_1$, as desired.


问:
Is there any discernible pattern for the number list $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$
I was looking at the notes of a calculus' student and one problem their teacher stated was to find a way to algebraically define the list of numbers $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$ into a sequence. Is there a "logical" answer for this? To me, one could add literally any other number into the list and find a 4-degree polynomial arbitrarily. This could be done with any finite list of 4 numbers however there are cases where 'logic' or 'common sense' is appealed (e.g. finding the next term of the sequence $1, 2, 3, 4, ...$).

I tried to analyze their differences in a spreadsheet and found nothing, as well as the ratio between the consecutive terms. Any ideas?

Edit: Here's an image of the notes (they're from a class in spanish).

It was part of a task where they were asked to find the $5311th$ term, this can only be doable by finding an algebraic expression.

答: A bit of thought process: I notice that the denominators seem to be jumping around a little bit, which seems unlikely — so my guess would be that several of these are 'artificially' reduced. A bit more pondering gives that $3|15$, $4|16$, and $9|18$, so that $1$ in the sequence is probably masking $\frac{17}{17}$. Writing the terms this way gives $\frac{35}{15}$, $\frac{20}{16}$, $\frac{17}{17}$, and $\frac{16}{18}$. Unfortunately, that sequence of numerators isn't much more compelling; first differences give 15, 3, 1, which fits far too many sequences. My urge would be to say $\frac{16}{19}$ based on a read of the differences as $2^{a_n}-1$, but I'd be hard-pressed to defend that well.


问:
Conditional expectation with 3 random variables
If we know $E(Y) = E( E( Y \mid X ) )$, and $X$ and $Y$ are independent, I am wondering whether $E(Y)$ is still equal to $E( E( Y \mid X, f(X) ) )$, where $f(X)$ is an approximation of $Y$?

答: Yes. If you are familiar with conditional expectaions given a sigma field then $EY=E(Y|\sigma (X))$ which implies that $EY=E(Y|X,f(X))$ for any measurable function $f$.


答:
Assume $A$ and $\neg A.$ From $A$ and $\neg A$ you get $\bot$ by ($\rightarrow$-E). Thus $\neg A \rightarrow \bot$ by ($\rightarrow$-I), i.e. $\neg\neg A.$

As a diagram:
$$
\dfrac
{
  \dfrac
  {
    A
    \quad
    [\neg A]
  }
  {
    \neg
  }
  (\rightarrow\text{-E})
}
{
  \neg\neg A
}
(\rightarrow\text{-I})
$$

Title: Prove $ A \vdash \neg \neg A $ (Natural Deduction)

Body:
From this set of rules:

$ (\wedge E) $  $A \wedge B \vdash A $

$ (\wedge I) $ $A, B \vdash A \wedge B $

$ (\vee E) $ $ A \vee B, A \rightarrow C, B \rightarrow C \vdash C $

$ (\vee I) $ $ A \vdash A \vee B $

$ (\neg E) $ $ \neg A \rightarrow B, \neg A \rightarrow \neg B \vdash A $

$ (\neg I) $ $ A \rightarrow B, A \rightarrow \neg B \vdash \neg A $

$ (\rightarrow E) $ $ A, A \rightarrow B \vdash B $

$ (\rightarrow I) $ $ Premises \vdash A \rightarrow B $

With the last rule ($ \rightarrow I $), you can introduce any implication you like as long as you prove in a sub-proof the consequent of the implication under the assumption of the hypothesis of the implication e.g.

$ P \rightarrow Q \quad $ Premise

$ \neg Q \quad\quad\space\space $ Premise

$ P \rightarrow \neg Q \quad \rightarrow I,$ subproof below

3.1 $ P \quad\space\space\space $ Assumption

3.2 $ \neg Q \quad $ from (2)

$ \neg P \quad $ from (1)(3), $ \neg I $

Prove $ A \vdash \neg \neg A $

$ A \quad Premise $

$ \neg (\neg \neg A) \rightarrow A \quad \rightarrow I, subproof $

2.1 $ \neg (\neg \neg A) \quad Assumption $

2.2 $ A \quad\quad\quad (1) $

$ \neg (\neg \neg A) \rightarrow \neg (\neg \neg A) \quad \rightarrow I, subproof $

3.1 $ \neg (\neg \neg A) \quad Assumption $

3.2 $ \neg (\neg \neg A) \quad (3.1) $

$ \neg \neg A \quad (2)(3), \neg E $

This is the proof I have currently and I'm unsure about it because of line 3 (specificially, the consequent of the implication which would usually just be $\neg A$) and subsequently line 4 which uses line 3 as a premise.

Is my proof correct?


问:
Linear Algebra - Dual vector Space
My question is:

''In this problem, we shall explore some if the concepts surrounding dual vector spaces for the specific case of $V= P(3)$, the vector space of polynomials with real coeficients of degree less than or equal to $3$. Assume through that $V=P(3)$

a) Describe a basis for $V^*$ (dual vector space) and express the linear functional $f:V\to R$ given by: $$f(a_3x^3+a_2x^2+a_1x+a_0)=2a_3+a_1-5a_0$$ as a linear combination of the basis you found for $V^*$.

b) Find a basis for the annihilator $W$ of the subspace $W= \text{span}(1+x^2, x+x^3)$ which belongs to $C$. What's dimension of annihilator of $W$?

c) We know that the evaluation map $E:V\to V^{**}$ is a vector space isomorphism. Then $E(x^2+x^3)$  is a linear functional on $V^*$. What is the value of the linear functional $E(x^2+x^3)$ on the linear functional $f$ from part (a) ? (I.e, what is $E(x^2+x^3)(f)$?''
Please help me ! I even don't know how to start !

答:
I'll give some hints for you to get started:

If $V$ is a finite dimensional vector space with basis $B$, then there's a natural choice for a basis $B^\ast$ in the space $V^\ast$ which is the dual basis how can you use this?

Recall that the annihilator of $ W \subset V$ is the set of all $f \in V^\ast$ such that $f(w)=0$ for all $w \in W$. It seems you know a basis for $W$ agree? So, use the definition!

Write down explicitly the evaluation map and apply it. Example: if we take $x \in P(3)$, then the evaluation $E: V \to V^{\ast\ast}$ is $E(x)(f) = f(x)$.

If you need further aid ask in comment.

ContentLicense: CC BY-SA 3.0

Title: Linear Algebra - Dual vector Space


答: Here is an example to show that this need not be true if $A$ is not Noetherian.  Let $A$ be a valuation ring with value group $\mathbb{Z}^2$ with the lexicographic order.  Let $P_1=0$ and let $P_2$ be the prime ideal consisting of elements whose valuation has positive first coordinate.  Let $a\in A$ be any element with valuation $(0,1)$.  Then $P_1\subset P_2\subset (a)$ so $P_1+(a)=P_2+(a)=(a)$ but $P_1\neq P_2$.

Title: Trying to delete in a sum of ideals: P+I=Q+I so P=Q?

Body:
Let be $(A,M)$ a local ring (if it needs noetherian ring), $P_1$ and $P_2$ two prime ideals with $P_1\subseteq P_2$, $\ a_1\in M\setminus P_2$. I don't know if it suffices to conclude this sentence:

If $P_1+(a)=P_2+(a)$ so $P_1=P_2$.

Someone can help me?
Maybe in a more general setting it is possible to say that if $P+I=Q+I$ so $P=Q$.

Thank you.


问:
Path independence of complex integral
I'm working through my Complex Analysis notes independently (because of lockdown mode) and I've stumbled across something I don't completely understand. It says here that complex functions $f: \mathbb{C} \to \mathbb{C}$ which have a primitive $F: \mathbb{C} \to \mathbb{C}$ such that $F'(z) = f(z)$ are path independent if the endpoints remain constant, which makes sense. As a counter-example, it gives $f(z)=f(x+iy)=x$, which is shown not to be path independent. However, $F(x+iy)=\frac{x^2}{2}$ seems to me like a perfectly fine primitive function. Where do I go wrong?

答: Existence of a primitive means existence of an analytic function $F$ such that $F'(z)=f(z)$ for all $z$. (The derivative is in the complex analytic sense). $F(x)=\frac {x^{2}} 2$ is not differentiable in this sense.

Title: Path independence of complex integral


问:
Comparing functions using asymptotic notations
I am learning algorithmic analysis and I want to know how to compare functions...I am trying to apply the definitions of Big Oh, Omega and theta yet unable to find the correct answer.

For example, how to know if $f(n) = 100n^{5.001} + n^5\log(n)^{10}$ is

(a) $O(n^{10})?$

(b) $\Theta(n^5\log(n))?$

(c) $\Theta(n^{5.001}\log(n))?$

(d) $\Omega(n^{5.001} + n^5\log(n))?$

(e) $\Omega(\sqrt n)?$

(f) $o(n^{5.1})?$

Body: Use this:$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\begin{cases}\infty,&g(n)\in o(f(n))\\c\ne0,&g(n)\in\Theta(f(n))\\0,&g(n)\in\omega(f(n))\end{cases}$$


问:
Van Kampen on $\mathbb{R}P^2$ with a "bridge"
Consider $X = \mathbb{R}P^2 \sqcup [0, 1]/ \sim$ where, for some pair of distinct points $x, y \in \mathbb{R}P^2$, we have the identifications $x \sim 0$ and $y \sim 1$.

The goal is to find the fundamental group of $X$ using Van Kampen.

So I was thinking of expressing it in terms of two open sets. Call the first one $A_1$. I was thinking of just letting this be $\mathbb{R}P^2$ itself (is that a problem?). Then let the other one, $A_2$ be $[0, 1]$ along with a path connecting $x$ and $y$ (i.e, $1$ and $0$). So the intersection would be the path between $x$ and $y$ which is certainly path connected, and (I believe) contractible; it's basically $[0 , 1]$ (I am dubious of this however). Then the only other unknown is $A_2$ which I want to say is just $S^1$, but I feel like that's wrong. In which case I would obtain $\pi_1(X) = \mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}$ which is probably incorrect (that's the impression I get from the follow up question).

I expect that I have misclassified $A_1 \cap A_2$ and $A_2$ so I would appreciate some help.

Edit: I just realized that $A_2$ isn't open. That's very dumb of me. Would it be fixed if I thicken the path into some open strip? This deformation retracts into a single path anyway so I'm not sure if it changes the story.

答:
Your argument sounds good to me. You can pull back everything to $S^2$ to picture things better, then the situation looks like this:

As long as you thicken your $A_1$ to include some open neighborhoods of 0 and 1 (up to the blue lines in the picture), $A_1$ is open in the space, and as long as you thicken your $A_2$ to include an open (purple) neighborhood of the path between $x$ and $y$ in $S^2$, $A_2$ is open in the space. Also we have to pick a simple path from $x$ to $y$, e.g. a path passing through the hemisphere would probably yield complications down in $\Bbb{R}P^2$.


问:
Weird $3^n$ in an identity to be combinatorially proved
Give a combinatorial proof of the following identity:

$$3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}$$

I can't see any counting argument that would yield $3^n$, and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course.

答:
Suppose you have $n$ balls to be put into 3 boxes. You may keep any or all boxes empty.Then the number of ways of putting $n$ balls into 3 boxes is $3^n$.
For the right hand side of the equation, you may fix a box $C$. The box $C$ may contain $0,1,2,\dots n$ balls. If the box $C$ has $i$ balls, you may choose the $i-\text{balls}$ to be put into $C$ in $\binom{n}{i}$ ways and the remaining $n-i$ balls may be put into boxes A or B in $2^{n-i}$ ways. Hence the number of ways of putting $n$ balls into 3 boxes is $\displaystyle \sum_{i=0}^n \binom{n}{i}2^{n-i}$ ways and hence $$3^n=\displaystyle \sum_{i=0}^n \binom{n}{i}2^{n-i}.$$ Please let me know if my solution is incorrect.</p>


问:
Extreme Value Theorem
For any function, given an interval with closed/open brackets, does a function have to have a min and max for the Extreme Value Theorem to apply? For example, does the extreme value theorem apply for the function f(x) = |x|?

In my opinion, I believe not because this function doesn't have a discrete maximum value. Is this thinking correct?

In addition, if at some point of any function is discontinuous within a interval, does it automatically mean that the application of the EVT is invalid? For example, take a simple floor function, since it's not continous, then it doesn't apply.

答:
The EVT applies exactly when the hypothesis I mentioned in my comment exist, and there is no redundancy, meaning all of them must apply in order for the theorem to work. Importantly, it has nothing to do with differentiability.

$f(x)=\lvert x\rvert$  is indeed continuous so, pick a bounded, closed interval (say, $[a,b]$) then indeed, the EVT applies. Namely, the extreme values are $0$ if $ab<0$ or $min(\lvert a\rvert,\lvert b\rvert)$ else, for the minimum; $max(\lvert a\rvert,\lvert b\rvert)$ for the maximum.

Again, for a discontinuous function such as the floor function the EVT does not apply, which doesn't mean that, by chance, the extreme values are not attained, which indeed they are, for any bounded interval.

Title: Extreme Value Theorem


问:
Simple question in verifying Cauchy's integral theorem
I'm trying to verify Cauchy's integral theorem "by hand" for
$$
I = \oint_C\frac{z}{2z-5}dz\,,
$$
where $C$ is the circle $\vert z-3\vert=2$. From Cauchy's integral theorem, the answer is $5\pi i /2$ although I'm not quite sure how to finish the last step of my verification.

My attempt is as follows.
\begin{equation*}
\begin{aligned}
I &= \oint_C \frac{z-3}{2(z-3)+1}dz+\oint_C\frac{3}{2(z-3)+1}dz \\
&= \int_0^{2\pi i} \frac{e^{2x}}{e^x+1/4}dx + \frac{3}{2}\int_0^{2\pi i} \frac{e^{x}}{e^{x}+1/4}dx \\
&= \int_0^{2\pi i} \frac{e^x(e^x+1/4-1/4)}{e^x+1/4}dx + \frac{3}{2}\left[\log{\left(e^{ix}+1/4\right)}\right]_0^{2\pi i} \\
&= \left[e^{x} + \frac{5}{4}\log{\left(e^{x}+1/4\right)}\right]_0^{2\pi i} \\
&= \frac{5}{4}\log{\left(e^x+1/4\right)}\bigg\vert_0^{2\pi i}
\end{aligned}
\end{equation*}
How do I show that the evaluation of the log term is $2\pi i$? The factor of $1/4$ in the $\log$ is throwing me off. I suspect that I am to use
$$
\log z = \log r + i(\theta+2\pi n)
$$
here, although if I am, I feel silly because I'm not seeing a simple way to employ it.

答:
Write $I = \oint_C\frac{z}{2z-5}dz=\frac12\oint_C \frac{z-\frac52+\frac52}{z-\frac52}dz=\frac12 \underbrace{\oint_Cdz}_{=0}+\frac54 \oint_C\frac{1}{z-\frac52}dz$. Hence, the problem really amounts to solving $$II=\frac54\oint_C \frac{1}{z-\frac52}dz,$$
which can be done only using elementary methods. The trick in the picture below, we see that the integral along $C_1+C_2$ is equivalent to the integral along $C$ plus the integral along the clockwise circle, centered at $z=\frac52$ of radius, say, $r$.
Hence, $\oint_{C_1}+\oint_{C_2}=\oint_C-\oint_{C_r}$, where $C_r$ is the positively oriented counterclockwise circle. Rearranging, we see that
$$
II=\frac54 \oint_{C_1} \frac{1}{z-\frac52}dz+\frac54\oint_{C_2} \frac{1}{z-\frac52}dz+\frac54\oint_{C_r} \frac{1}{z-\frac52}dz.
$$
Now the key for the first two integrals is that since we are avoiding the singularity, there exists an analytic logarithm such that  $\frac{d}{dz}\log(z-\frac52)=\frac{1}{z-\frac52}$. Hence, by the fundamental theorem, the integral around a closed loop is zero. This leaves the last integral, which is easy to solve with parameterization, since the circle is centered at $z=\frac52$.


问:
Surface Integral Question
I have this question:

Evaluate $\int r . dS$ over the surface of a sphere, radius a, centred at the origin.

I'm not really sure what '$r$' is supposed to be? I would guess a position vector? If so, I would have $r . dS$ as $(asin\theta cos\phi, a sin\theta sin\phi, acos\theta) . (a^2sin\theta d\theta d\phi)$ which doesn't seem right. Any pointers appreciated, thanks!

答:
There's two ways to look at this.  For both ways, though, the notation $r \cdot dS$ is a shorthand for $r \cdot n dS$, where $n$ is the normal vector to $S$, pointed outward.  In this case, the normal vector is just the position vector $r$, normalized to unit length.  Thus the surface integral is simply

$$a \int_S dS = 4 \pi a^3$$

This may also be analyzed using the divergence theorem.  Because $\nabla \cdot r = 3$, we have the surface integral equal to

$$3 \int_{\text{sphere}} dV = 4 \pi a^3$$

Title: Surface Integral Question


问:
An integer $n \geq 2$ is called square-positive- proof?
An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.

So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.

Here is my answer:

2 : 4 + 5 = 9  which is 3^2

3 : 2 + 3 + 4 = 9 which is 3^2

5: 18 + 19 + 20 + 21 + 22 = 100  which is 10^2

6: 35 + 36 + 37 + 38 + 39 + 40 = 225  which is 15^2

Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :

(I show it in an example because I still can't write it algebraically):

For example: the sum of 95 subsequent numbers is 5n + 10

(10^2 - 10) /5 = 18

So your first number is 18

And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.

答:
I don't know what tools you are supposed to use to demonstrate that 4 is not a square-positive number. I guess you can do it in different ways. Here's one:

Take four consecutive numbers.
Now take a unit from the last one and put it to the second.
The last three numbers are now equal. If you sum them all, you get: $3n + (n - 2) = 4n - 2$.

Now for reduction ad absurdum you want that expression to give you the square of an integer: $4n - 2 = x^2$ which is equivalent to: $n = ( x^2 + 2) / 4$. Now you can say that n is an integer only if $( x^2 + 2 )$ is multiple of 4, but this can be true only if x is an even number. Now the square of any even number is of the form $(2^2n *...)$ i.e. $(4^n *...)$ So it is always a multiple of 4. If $x^2$ needs to be always divisible for 4, then $(x^2 + 2)$ can never be a multiple of 4 and therefore n can never be an integer. But $n$ is of course an integer, it is the third of the four consecutive numbers chosen.


问:
Proving regular expression with induction - $(a \cup ba)^{\ast}b^{\ast}$
Let $L$ be the language $\{w \in \{a, b\}^{*} : bba$ is not a substring of $w\}$. We claim that  $L$ is a regular language, and, for such, we will prove that the following regular expression $R = (a \cup ba)^{\ast}b^{\ast}$ describes $L$.

Proof. We will prove that $L(R) = L$.

Proof of $L(R) \subseteq L$.

Suppose that $w \in L(R)$, we will prove that $w \in L$ by induction on $|w|$. If $|w| = 0$ or $|w| = 1$ or $|w| = 2$, then, $w \in \{\epsilon\} \cup \{a, b\} \cup \{aa, ab, ba, bb\} \subseteq L$,  therefore, since $bba$ cannot be a substring of $w$, it follows that $w \in L$.

If $|w| = 3$, then, we have the following possibilities: $w = aaa$, $w = aba$, $w = bab$, $w = bbb$, $w = baa$, $w = aab$, or $w = abb$. Since $bba$ does not occur in any of these possibilities, we have that $w\in L$.

If $|w| \geq 4$, then, we have the following possibilities: $w = aaaxy$, $w = abaxy$, $w = babxy$, $w = bbbxy$, $w = baaxy$, $w = aabxy$, or $w = abbxy$, where $x \in L((a \cup ba)^{\ast})$ e $y \in L(b^{\ast})$. Thus, we have the follwing cases:

Case $w = bbbxy$ or $w = abbxy$: Note that, the sufix of both possibilities is $bbxy$, thus, we have that $|bbxy| < |w|$, hence by the inductive step follows that $bbxy \in L$. Then, whenever $w = bbbxy$ or $w = abbxy$, it follows that $x = \epsilon$ to not have $bba$ as substring. Thus, $bba$ cannot be a substring of $w = bbbxy$ and
cannot be a substring of $w = abbxy$, therefore, we can conclude that $bbbxy \in L$ e $abbxy \in L$.

Case $w = aabxy$ ou $w = babxy$: In a similar way to the previous case, we have that the suffix $abxy \in L$ by the inductive hypothesis, since $|abxy| < |w|$. Thus, the string $x$ cannot begin with $ba$, so none of the possibilities considered here will have $bba$ as substring, hence, we can conclude that $aabxy \in L$ e $babxy \in L$.

Case $w = aaaxy$ ou $w = abaxy$ ou $w = baaxy$: For each of these cases, we have $w \in L$
if and only if $xy \in L$, but since $xy \in L((a \cup ba)^{\ast}b^{\ast})$ and $|xy| < |w|$, we have $xy \in L$ by induction hypothesis. Thus, $w$ does not have substring $bba$, therefore, $w \in L$.

Since we exhausted all possibilities, given that we take an arbitrary $w$ of $L(R)$, then we can conclude that $L(R) \subseteq L$.

Proof of $L \subseteq L(R)$.

Suppose that $w \in L$, we will prove that $w \in L(R)$ by induction on $|w|$. If $|w| = 0$,
or $|w| = 1$, or $|w| = 2$, by the definition of $L$ we have that $w \in \{a, b\}^{\ast}$, then,
it follows that $w \in \{\epsilon\} \cup \{a, b\} \cup \{aa, bb, ab, ba\} \subseteq L(R)$.
Hence, for any of these possibilities, we can conclude that $w \in L(R)$.

If $|w| \geq 3$, since $w \in \{a, b\}^{\ast}$, then we have the following possibilities:
$w = aax$ or $w = bbx$ or $w = abx$ or $w = bax$, with $x \in L$. Since $w$ does not have
substring $bba$, note that, when $w = bbx$, $w \in L$ if and only if $x$ ndoes not begin with $ba$ or $x$ does not begin with $a$. Similarly, when $w = ab$, $w \in L$
if and only if, $x$ does not begin with $ba$. Since, $|x| < |w|$, by the inductive hypothesis we have $x \in L(R)$ and therefore $x = uv$ with $u \in L((a \cup ba)^{\ast})$
and $v \in L(b^{\ast})$. Since $aau$, $ bbu$, $abu$ and $bau$ are in $L(R)$, we see that
$w \in L(R)$, as we wanted to show.

Since we explored all possibilities, given that we take an arbitrary $w$ of $L$, then we can conclude that $L \subseteq L(R)$.

Therefore, since we showed that $L(R) \subseteq L$ e $L \subseteq L(R)$, we can conclude that $L(R) = L$. $\square$

Is my proof okay? I am always getting confused in the inductive hypothesis, if I always using the correct number of symbols when I assume that $|w| \geq k$ for some $k \in \mathbb{N}$.

答:
The first half of the proof is okay, but there’s a problem with the second half: in the last sentence of the $|w|\ge 3$ paragraph it’s not true that $bbu$ and $abu$ are in $L(R)$ unless $u=\epsilon$.

In this kind of argument I find it easier to to use the least counterexample style of induction argument, like this:

Suppose that there is some word in $L\big((a\cup ba)^*\big)$ that contains the substring $bb$, and let $w$ be one of minimal length. There is an $x\in L\big((a\cup ba)^*\big)$ such that $w=xa$ or $w=xba$, and clearly no word in $L\big((a\cup ba)^*\big)$ ends in $b$, so any $bb$ substring of $w$ must be contained in $x$. However, $|x|<|w|$, so by hypothesis $x$ does not contain the substring $bb$. This contradiction shows that in fact no word in $L\big((a\cup ba)^*\big)$ has a substring $bb$.

Now $w\in L(R)$; then $w=xy$ for some $x\in L\big((a\cup ba)^*\big)$ and $y\in L(b^*)$. Clearly there is no $a$ in $y$, so $x$ must have a substring $bba$. But then $x$ has a substring $bb$, which is impossible. Thus, no word in $L(R)$ contains a substring $bba$, and therefore $L(R)\subseteq L$.

If $L\setminus L(R)\ne\varnothing$, let $w$ be word in $L\setminus L(R)$ of minimal length. Clearly $w\ne\epsilon$, so we can write $w=x\sigma$, where $\sigma\in\{a,b\}$ and $x\in L$. (Clearly every substring of $w$ belongs to $L$.) The minimality of $|w|$ implies that $x\in L(R)$, so $x=uv$ for some $u\in L\big((a\cup ba)^*\big)$ and $v\in L(b^*)$. There are now several possibilities.

If $\sigma=b$, then $v\sigma\in L(b^*)$, and $w=uv\sigma\in L(R)$, contrary to hypothesis.

If $\sigma=a$, $x$ cannot end in $bb$, so either $x=\epsilon$, $x=b$, $x$ ends in $a$, or $x$ ends in $ab$. The first two cases are ruled out: $w$ would be $a$ or $ba$, both of which are in $L(R)$. If $x$ ends in $a$, then $v=\epsilon$, so $$w=xa=ua\in L\big((a\cup ba)^*\big)\subseteq L(R)\,,$$ again contrary to hypothesis.

The only remaining possibility is that $x$ ends in $ab$, i.e., that $w$ ends in $aba$. In that case let $w=zaba$ (i.e., $x=zab$). Then $za\in L$, and $|za|<|w|$, so $za\in L(R)$. Clearly $za\in L\big((a\cup ba)^*\big)$, so $$w=zaba\in L\big((a\cup ba)^*\big)\subseteq L(R)\,,$$ yet again contrary to hypothesis. It follows that no such word $w$ exists, so $L\subseteq L(R)$, and therefore $L=L(R)$.


问:
Describe finite structure up to isomorphism as model of two sentences.
I'm trying to solve the following exercise:

Let $A$ be a finite structure.

Find $σ_1\in$ Th$(A)$ such that any model of $σ_1$ has universe of the same
cardinality as $A$.

Now assume $L$ (the language we're using) has finitely many symbols.  Find a single existential sentence $σ_2$ (where an existential sentence is one of the form $\exists x_1 ... \exists x_n\varphi  $ where $\varphi$ is quantifier free) such that any model of $\lbrace σ_1, σ_2\rbrace$ is isomorphic to $A$.

The first part seems simple - if we assume $ |A|=n$, then we can just take a formula

$$ σ_1 = \exists x_1 ... \exists x_n(x_1\neq x_2 \wedge x_1\neq x_3 \wedge ... \wedge x_{n-1}\neq x_n \wedge \forall y (y=x_1 \lor y=x_2 \lor ... \lor y=x_n ))$$

saying that there are at least $n$ elements and at most $n$ elements.

I have no idea what form $σ_2$ would take - I tried thinking about a simple case where $ |A|=1$ and the language contains only a unary relation symbol $R$, but even then I don't know what you could do. Clearly in this case either $R^A = \emptyset$ or $R^A =A$, but then without knowing what interpretation $A$ gives to $R$ I don't see how you'd settle on the sentence. In this case it'd be easy enough to distinguish cases and just say e.g. that if $R^A =A$ then let $σ_2 = \exists x Rx$ but I struggle to see how you'd generalise to any arbitrary finite language and size of $A$.

I have no idea where to begin, so any help you could offer would be really appreciated.

答:
Well let's start by observing that the idea (where our language consists of a single unary relation symbol $R$)

if $R^A=A$ then let $\sigma_2=\exists xRx$.

does not work if there is more than one element in $A$: all this $\sigma_2$ says is that $R^A$ has at least one element. Already there are two (up to isomorphism) structures with two elements which satisfy $\sigma_2$ but are not isomorphic (one in which $R$ holds once, and one in which $R$ holds twice).

"$\forall xRx$" would do the job but of course we don't want that. So how can we do things with just existential quantifiers?

Well, let's think about a structure with two elements $u,v$ where $R$ holds of $u$ but not $v$. We can describe this as follows:

There are two elements.

There is one element where $R$ holds.

There is one element where $R$ fails.

The first point is taken care of by $\sigma_1$. The second and third can be written together as $$(\exists xRx)\wedge (\exists x\neg Rx),$$ which in turn can be put in the form desired as

 $\exists x,y(Rx\wedge \neg Ry)$.

This last form suggests what we should be doing in general: to describe an $n$-element structure, we first fix some enumeration $a_1,...,a_n$ of its elements and then

 consider a sentence of the form $$\exists x_1,...,x_n([\mathsf{stuff}])$$ where "$\mathsf{stuff}$" describes how the $a_i$s interact - with $x_i$ interpreted as $a_i$.

This does not work, however: consider, in the language with a single unary relation $R$, a structure with three elements where $R$ holds on exactly one element versus a structure with three elements where $R$ holds on exactly two elements. If we look at the corresponding sentences of the type above, we get equivalent sentences (which we definitely don't want) since

 nothing is keeping the variables distinct.

We can fix this by

 folding appropriate distinctness clauses into the idea above, so that e.g. in the $R$-structure with two elements where $R$ holds once we'd get $$\exists x,y(x\not=y\wedge Rx\wedge\neg Ry).$$

At this point there are a couple questions to ask yourself:

Where did we use the assumption of a finite language?

Why do we need $\sigma_1$ as well as this $\sigma_2$?


问:
Is there a symbol for potential equality?
Is there a symbol for potential equality? Essentially I'd like to condense:

$$
(a = b) \lor (a \ne b)
$$

so that I can express the phrase "a may or may not be equal to b". Apologies if my syntax is not entirely correct; I come from a computer science background.

答:
≟

The Unicode symbol ≟ can be used for questionable equality.

Unicode Reference: https://www.compart.com/en/unicode/U+225F


问:
Regular average calculated accumulatively
is it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1.

for example: 2 3 10
the average is of course: 5

but in the last step to calculate i only have access to the previous average of 2 and 3: 2.5
the next number: 10
and the count of numbers: 3

if this is possible, how?

答:
What you are asking for is commonly called sequential estimation.
A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400–407.]

To add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember).
$$\mu_n = \mu_{n-1} + \frac{1}{n}(x_n - \mu_{n-1}) $$

```<code># algorithmically: sequential average computation
avg += (x_n - avg)/n
</code>```

Body:
What you are asking for is commonly called sequential estimation.
A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400–407.]

To add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember).
$$\mu_n = \mu_{n-1} + \frac{1}{n}(x_n - \mu_{n-1}) $$

# algorithmically: sequential average computation
avg += (x_n - avg)/n


问:
Non isomorphic principal $G$-bundles
Remark: throughout this post we work in the smooth category, so that all manifolds, bundles, maps etc are assumed to be smooth.

An exercise asks me to show that there exists no principal $S^1$-bundle isomorphism between the Hopf bundle $S^1 \to S^3 \xrightarrow\pi S^2$ (where $S^1$ acts on $S^3$ by multiplication as usual), and the bundle with the same fiber structure but equipped with the "inverted" action $(w_1,w_2) \cdot z = (w_1z^{-1}, w_2z^{-1})$.

Aiming to reach a contradiction I suppose an $S^1$-bundle isomorphism between these two bundles exists. This is a diffeomorphism $\Phi: S^3 \to S^3$ satisfying:
\begin{array}{rc}
\pi \circ \Phi = \pi, & (*)\\[0.2cm]
\Phi(w_1 z,w_2 z) = \Phi(w_1,w_2)z^{-1}. & (**)
\end{array}

As far as I can see, the conditions $(*)$ and $(**)$ alone do not violate each other, since $\pi$ is invariant under multiplication by elements of $S^1$ (and hence their inverses) anyway. So I guess there should be issues with the diffeomorphy of $\Phi$. More particularly, I would guess smoothness/continuity of $\Phi$ (or its inverse) is an issue, since again I don't see why the equivariance condition $(**)$ would contradict the bijectivity of $\Phi$.

But I'm failing to see how to derive a contradiction here. Am I missing something simpler or am I on the right track?

EDIT:  I just realized that the book has solutions to the exercises in the appendix (I don't remember seeing this in an advanced math book so I didn't even notice they were there). I will try to add an outline of the solution to this post once I figure the details out.

EDIT 2: As promised, here is an outline of a solution. Firstly, since $\Phi$ respects the fibers by $(*)$, it must be of the form
$$ \Phi(w) = w \, \phi(w) $$
for some smooth $\phi:S^3 \to S^1$. By $(**)$ we then see that
$$ w\,\phi(wz)z = \Phi(wz) = \Phi(w)z^{-1} = w\,\phi(w)z^{-1}, $$
or in other words,
$$ \phi(w) = \phi(wz)z^2 $$
since the action is free. This is the condition that we are going to use to derive a contradiction. Since $S^3$ is simply connected, we may write
$$ \phi = e^{i\psi} $$
for some smooth $\psi : S^3 \to \mathbb R$. Thus, taking $z = z(\alpha) = e^{i\alpha}$ with $\alpha \in \mathbb R$, we get
$$ e^{i\psi(w)} = \phi(w) = \phi(wz)z^2 = e^{i\psi(we^{i\alpha})+2i\alpha}. $$
This implies that
$$ \psi(we^{i\alpha}) - \psi(w) + 2\alpha = 2\pi n(\alpha) $$
for some integer $n = n(\alpha)$. But the left hand side depends smoothly on $\alpha$, so the right hand side must as well, i.e. $n(\alpha)$ must be constant (since it is integer-valued). In fact, $n(\alpha) = n(0) = 0$. But then
$$ \psi(we^{i\alpha}) - \psi(w) = -2\alpha, $$
which is absurd, since the left hand side is bounded (since $\psi(S^3)$ is compact), while the right hand side is unbounded.

答: I think the "standard" argument would be that the Hopf fibration is classified by a nontrivial element of $H^1(S^2, S^1) \cong H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$ (I think more specifically by twice a generator but don't quote me on that) so the inverted bundle is presumably classified by the negative of this element, and since $\mathbb{Z}$ has no $2$-torsion no such bundle is equivalent to its inverse. Probably there's something to learn from trying to show that the diffeomorphism $\Phi$ doesn't exist more concretely though.

Title: Non isomorphic principal $G$-bundles


问:
Show that $|G| = mn$ implies $G$ cyclic.

Let $m$, $n$ be primes with $m < n$, $m \neq n \text{ (mod $1$)}$, and suppose $|G| = mn$. Prove that $G$ is cyclic.

The first thing I need to show is that there is only one Sylow $m$-subgroup and one Sylow $n$-subgroup. By the third Sylow theorem, we have:
\begin{align*}
n_m \equiv 1 \text{ (mod $m$)} \\
n_m \mid n
\end{align*}
The second condition implies that $n_m = n$ or $n_m = 1$. I should be able to rule out $n_m = n$, but I cannot figure out how. If $n_m = n$, then $n \equiv 1 \text{ (mod $m$)}$, so $n = 1 + mk$ for integer $k$. I doI qm wn't see how this produces a contradiction.

On the flip side:
\begin{align*}
n_n \equiv 1 \text{ (mod $m$)} \\
n_n \mid m.
\end{align*}
I am still not able to conclude $n_m = n_n = 1$.

Help would be appreciated.

答:
It's obviously false, as $S_3$ is a non-cyclic group of order $6=2\cdot 3$.

You need another condition for this to be true, such as $m\nmid n-1$ for example.

In this case, you can prove there is a unique $n$-Sylow (you are almost there). If $H$ is the unique $n$-Sylow subgroup and $K$ is a $m$-Sylow subgroup, the next step is to prove that $G=H\rtimes K$.
Now the corresponding morphism $K\to Aut(H)$ is trivial because $m$ is prime to $n-1$ (why ???). Then $G=H \odot K\simeq H\times K$, and CRT (for example) will lead to the desired conlusion.

Details are left to you.


问:
Complete orthonormal sequence, Hilbert Space, Kronecker Delta
Let $H$ be a Hilbert space and $(e_n)_{n=1,2,\ldots}$ be a complete orthonormal sequence in $H$. We want to show that if  $a_{np}=(e_n,f_p)$ then $\sum_{p=1}^{\infty}a_{np} \overline{a_{mp}}=\delta_{nm}$ and $\sum_{n=1}^{\infty}a_{np}\overline{a_{nq}}=\delta_{pq}$ where $\delta_{ij}$ is the Kronecker delta and $(f_p)_{p=1,2,\dots}$ is an other complete orthonormal sequence in $H$..

I've been thinking about this for a while now - any thoughts / hints about where to start? Thanks!

答:
I guess $(f_p)_{p\ge1}$ must be another complete orthonormal sequence.

In that case, you ought to be able to show that $\sum_{p=1}^{\infty}a_{np} \overline{a_{mp}}=(e_n,e_m)$ and $\sum_{n=1}^{\infty}a_{np}\overline{a_{nq}}=(f_q,f_p)$. By orthonormality, these inner products are $\delta_{nm}$ and $\delta_{pq}$.

Hint: you can use the fact that $x=\sum_{n=1}^\infty (x,e_n)e_n=\sum_{p=1}^\infty (x,f_p)f_p$ for any vector $x\in H$.


问:
Is this sequence and/or function monotone?
I know that from what I am learning, Monotone is where a sequence and/or function is neither increasing or decreasing at a consistent rate.

Take for example, the function: $$P(t): 2+\frac{10}{2t}$$

Is this function monotone or decreasing? I feel as if this function is decreasing at all times according to the math, but at a certain point the function just flatlines. What is the correct answer to this?

答:
A function is increasing if $f(x)\le f(y)$ whenever $x<y$.

A function is decreasing if $f(x)\ge f(y)$ whenever $x<y$.

A function is monotone if it is increasing or decreasing.

Some authors use $f(x)<f(y)$ and $f(x)>f(y)$ in the above definitions.

Any function which is increasing at a constant rate is thus monotone. The function $P(t)=2+\dfrac5t$ is decreasing on $(0,\infty)$ because, if $0<x<y$, then $\dfrac1y<\dfrac1x$, which means $2+\dfrac1y<2+\dfrac1x$, so $P(y)<P(x)$.


问:
Find a linear transformation $ f: \mathbb{R} ^ {3} \rightarrow \mathbb{R}^{3} $ such that the unit cube becomes the parallelepiped
Find a linear transformation $ f: \mathbb{R} ^ {3} \rightarrow \mathbb{R}^{3} $ such that the unit cube becomes the parallelepiped with vertices $(0,0,0), (2 , 0,0), (2,2,0), (0,2,0), (2,1,3), (0,1,3), (2, -1,3)$ and $(0, -1,3)$

I think the unit cube is $(0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1), and (0, 1, 1).$

I don't know how to do it I think I can do

$(2,1,2)=a(1,0,0)+b(0,1,0)+c(0,0,1)$ and

$(2,-1,2)=a(1,0,0)+b(0,1,0)+c(0,0,1)$

but it's absurd

答:
The unit cube has vertices $\{(1,1,1),(-1,1,1),(-1,-1,1),(-1,-1,-1),(1,-1,-1),(1,1,-1),(1,1,-1)\}$, and the transformation in question may be an affine transformation in general:
$$
f(x) = Ax + b,
$$
or more compactly:
$$
f(x) = \begin{pmatrix}A&b\end{pmatrix}\begin{pmatrix}x\\1\end{pmatrix}.
$$

Lets first pick 4 linearly independent vertices from the unit cube and stack them in a matrix including an additional row of ones to account for $b$, say:
$$
X = \begin{pmatrix}-1&-1&-1&1\\-1&-1&1&1\\-1&1&1&1\\1&1&1&1\end{pmatrix}.
$$

Now, pick 4 affinely independent vertices from the output parallelepiped, and stack them in a matrix $V$, say:
$$
V = \begin{pmatrix}2&2&2&0\\0&2&1&1\\0&0&3&3\end{pmatrix}.
$$

Then, you can solve for $A$ and $b$ as:

$$
\begin{pmatrix}A & b\end{pmatrix} = VX^{-1},
$$
which gives:
$$
A = \begin{pmatrix}
         0    &     0 &  -1  \\
         1 & -0.5 & 0 \\
         0 & 1.5 & 0\end{pmatrix},\qquad b = \begin{pmatrix}1 \\ 0.5 \\ 1.5\end{pmatrix}.
$$

EDIT: Actually, in your case it turns out that $f$ is indeed linear, and the vertices that you have are outputs of $f$ on the unit cube according to your definition. So, finding the columns of $A$ indeed amounts to stacking all of the vertices horizontally in a matrix $V$ and performing Gaussian elimination. The columns of $A$ are the columns of $V$ which become standard basis vectors after elimination. However, the approach above is more general.


问:
Bijection between continuous maps $C(X\times A,Y)$ and continuous maps $C(X,Y^A)$

If $A$ is a discrete space, and $X,Y$ are any spaces, I want to show there is a bijection between the sets of continuous maps $C(X\times A,Y)$ and $C(X,Y^A)$. Here, $Y^A$ is the set of all continuous maps from $A$ to $Y$ (which is all maps from $A$ to $Y$ since $A$ is discrete), and we topologize $Y^A$ with the product topology.

Here is what I've tried.

Define the map $\Psi:C(X\times A,Y)\to C(X,Y^A)$ via $f\mapsto\Psi(f)$ where for each $x\in X$ the map $\Psi(f)(x)$ is defined by $(\Psi(f)(x))(a)=f(x,a)$. It was easy to show that $\Psi$ is injective. I'm stuck on surjective. My intuition was to pick $h\in C(X,Y^A)$ and define the map $\tilde{h}:X\times A\to Y$ via $(x,a)\mapsto h(x)(a)$. Clearly $\Psi(\tilde{h})=h$, so I'm pretty sure this is what I want to show surjectivity, but I cannot figure out why $\tilde{h}$ is continuous. I tried picking an open $U$ in $Y$ and showing $\tilde{h}^{-1}(U)$ is open in $X\times A$, but all I can say about this is that $U^A$ is open in $Y^A$, so $h^{-1}(U^A)$ is open in $X$. I'm not sure where to go from here.

Maybe I can use the fact that since $A$ is discrete, we have a homeomorphism between the cartesian product and the coproduct $X\times A\cong\coprod_AX$? Any suggestions? By the way, I want to show this directly, i.e. by exhibiting an actual bijection, rather than hitting it with any sledgehammers.

答:
You give $Y^A$ the product topology, i.e. you take $Y^A = \prod_{a \in A} Y_a$ with $Y_a = Y$. As a set, the product $\prod_{a \in A} Y_a$ is defined as the set of all functions $\varphi : A \to Y$ between the sets $A, Y$. But since $A$ is discrete, the set of all these functions is nothing else than the sets all continuous maps $f : A \to Y$ between the spaces $A, Y$.

Let $p_a : Y^A \to Y$ denote the projection on the factor $Y = Y_a$. It is given by $p_a(f) = f(a)$.

The universal property of the product says that a function $h : X \to Y^A$ is continuous iff all $p_a \circ h : X \to Y$ are continuous. Note that $(p_a \circ h)(x) = p_a(h(x)) = h(x)(a)$.

So let $h : X \to Y^A$ be continuous and $U \subset Y$ be open. Then all $(p_a \circ h)^{-1}(U)$ are open in $X$ and, because $A$ is discrete, all $(p_a \circ h)^{-1}(U) \times \{a\}$ are open in $X \times A$. Thus
$$(\overline h)^{-1}(U) = \{(x,a) \in X \times A \mid \overline h(x,a) = h(x)(a)  = (p_a \circ h)(x) \in U \} \\ = \{(x,a) \in X \times A \mid x \in (p_a \circ h)^{-1}(U) \}  \\ = \bigcup_{a \in A} (p_a \circ h)^{-1}(U) \times \{a\}$$
which is open in $X \times A$.

To get a more general picture, I recommend to learn about the compact-open topology on arbitrary function spaces $Y^Z$ = set of all continous maps $f : Z \to Y$ . You will see that if $A$ is discrete, then the compact-open topology on $Y^A$ agrees with the product topology. Moreover, if $Z$ is locally compact (note that discrete spaces have this property), then you have a canonical bijection

$$e : C(X \times Z,Y) \to C(X,Y^Z), e(\phi)(x)(z) = \phi(x,z) .$$

This is known as the exponential map. In this forum you will find a lot of questions about it.


问:
How can you prove that the square root of two is irrational?
I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

答:
Let me give a proof based on (partly) Newton's method and (mainly) Pell's equation. The proof technique is also similar to that of continued fractions.

Consider the iterations: $x_{n+1} = \frac{x_n}{2}+\frac{1}{x_n}$. This is the formula given by Newton's method for the function $f(x) = x^2 -2$. Let's use a special initial value: $x_0 = \frac{3}{2}$.

Denote $x_n = \frac{p_n}{q_n}$. We immediately get the formula:
\begin{equation}
p_{n+1} = p_n^2 + 2q_n^2, q_{n+1} = 2p_n q_n.
\end{equation}

Furthermore, you might notice that $p_n, q_n$ are all the solutions of the Pell's equation $x^2-2y^2 = 1$. But we only need the following calculation:
\begin{equation}
p_{n+1}^2-2q_{n+1}^2 = (p_n^2 -2q_n^2)^2,
\end{equation}
\begin{equation}
p_0^2-2q_0^2 = 3^2-2\cdot 2^2 = 1.
\end{equation}

From the above formula, we have the relation: $\frac{2}{x_n}< \sqrt{2} < x_n$. Now assume $\sqrt{2} = \frac{s}{t}$. We can get the estimate:
\begin{equation}
|\frac{p_n}{q_n}-\frac{s}{t}| = |x_n - \sqrt{2}| < |x_n - \frac{2}{x_n}| = \frac{1}{p_n q_n}.
\end{equation}

So $|tp_n-sq_n| < \frac{t}{p_n}$. Because $p_n$ grows to infinity, we have $|tp_n - sq_n|<1$ for some $n$. Then $tp_n = sq_n$ and $x_n = 2$. Contradiction!

If you know more about Pell's equation ($x^2-ny^2 = 1$ has solution for all positive nonsquare integer $n$), you can prove that $\sqrt{n}$ is irrational.

ContentLicense: CC BY-SA 2.5


问:
On how many ways can the teacher assign his grades
The question goes by:

A class of 40 students are getting the grades A,B,C,D and F. The teacher fails half of his class, that is, 20 of the students get grade F. On how many possible ways can the teacher assign grades to his class?

My idea is that the order doesn't matter, so we have a combinations with repetitions problem, that is, we will be using the formula $$\binom{k + n-1}{k}$$ where $k$ is the combinations (amount of students to be distributed) and $n$ is the objects (grades to distribute the students over).

So first we have 20 students to give F, and since the order does not matter, we get $$\binom{20 + 1-1}{20} = 1.$$

and for the remaining 20 students, we get $$\binom{20 + 4-1}{20} = \binom{23}{20} = 1771$$

so the final answer would be $1\cdot 1771 = 1771$.

But I'm not sure if this is correct... Anyone who can confirm or explain how this problem should be solved?

答:
First, you choose which 20 students will be given an F. This is combinations without repetitions, because the same student can't be given an F twice, and order does not matter. So, there are $\binom{40}{20}$ ways to assign half of the class a grade of F.
Then, for each of the 20 remaining students, you assign them an A, B, C or D. There are 4 different choices for each student. This is permutation with repetition (as multiple students can have the same grade) and where order matters. So, there are $4^{20}$ ways to grade the rest of the class.
The total answer should be $\binom{40}{20}\times 4^{20}$.


问:
Contrapositive of statement regarding open sets
I am taking a Real Analysis class and i am having problems with the following:

"A set $A \subset \mathbb{R}$ is open $\Leftrightarrow$ the following holds: if a sequence $\left(x_{n}\right)$ converges to a point $a \in A$, then $x_{n} \in A$ for $n \in \mathbb{N}$ sufficiently large"

I am trying to prove $(\Leftarrow)$. I think i can prove the contrapositive statement. However, i am not sure how to properly write the statement in the contrapositive form.

Attempt 1 to write the contrapositive:
If $A \in \mathbb{R}$ is not open, then there exists a sequence $\left(x_{n}\right)$ such that $\lim x_{n}=a$ and $\mathrm{x}_{\mathrm{n}} \notin \mathrm{A}$.

Attempt 2 to write the contrapositive:
If $A \in \mathbb{R}$ is not open, then there exists a sequence $\left(x_{n}\right)$ such that $\lim x_{n}$ is not equal to $a$ and $\mathrm{x}_{\mathrm{n}} \notin \mathrm{A}$.

Why should (or should not) the statement $\lim x_{n}=a$ be negated?

Thanks in advance!

答:
Neither of them is quite right, even apart from the use of $\in$ instead of $\subseteq$ at the beginning of each of them. A correct statement of the contrapositive is:

If $A\subseteq\Bbb R$ is not open, then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $\Bbb R$ that converges to a point $a\in A$, but for each $m\in\Bbb N$ there is an $n\ge m$ such that $x_n\notin A$.

Equivalently, one could say:

If $A\subseteq\Bbb R$ is not open, then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $\Bbb R$ that converges to a point $a\in A$ but has infinitely many terms that are not in $A$.


问:
Conservative Vector Field question
Given $F(x,y) = \langle 2xy^2+5, 2x^2y-3y^2 \rangle$, compute $\int_{C} F\cdot dr$ where $C$ is the ellipse given by $\frac{(x-4)^2}{5} + \frac{y^2}{4} = 1$.

I know that $F$ is conservative, and that $\int_{C} F\cdot dr$ is independent of path. It is just the remainder of the integral that I am confused in attempting. I converted $r(t)$ from $C$ into polar, and tried a u-sub integral with what I got from $r'(t)$, but I didn't get anywhere. I feel like I've missed something trivial. Could someone point me in the direction to start with the integration?

答:
If $F$ is conservative, then the integral is zero. This is because $C$ is a closed curve (it begins and ends at the same point).

As you said, the integral is independent of path. If you go all the way around the ellipse, then you end at the same point you started. You could instead take a different path, which just does nothing (I mean the path just sits at that point for all values of $t$, without moving), and you would begin and end at the same point. The integral over the path that just does nothing is clearly zero.

Title: Conservative Vector Field question


问:
How can the Bergman space $L^2_a(G)$ of a complex open set $G$ be a subspace of $L^2(\mu)$?
How can the Bergman space $L^2_a(G)$ of a complex open set $G$ be a subspace of $L^2(\mu)$ where $\mu$ is the area measure on G? If $L^2(\mu)$ is a space of equivalence classes of functions then how is $L^2_a(G)$, a space of functions, contained within it? Are we using shorthand for some sort of representation of $L^2(\mu)$ where we're only considering representatives of each equivalence class as those that are continuous? Is there a unique element of each equivalence class of $L^2(\mu)$ that is continuous?

答:
Say $f\sim g$ if $f=g$ ae. Say $[f]$ is the corresponding equivalence class. It's not literally true that $L^2_a\subset L^2$, but it's clear that $$L^2_a/\sim\,\,:=\{[f]:f\in L^2_a\}\subset L^2.$$

And there's no real difference between $L^2_a$ and $L^2_a/\sim$.

The Point: It happens a lot that people say something about $X$ when they really mean $X/\sim$. You just have to get used to it...


问:
Is existence of antiderivative equivalent to analyticity and path independence?
I'm working through my Complex Analysis notes independently (because of lockdown mode) and would like to draw a general conclusion from some seemingly disjointed theorems.

One theorem states that functions having an antiderivative are path independent. Another (Cauchy-Goursat) implies path independence of analytic functions on simply connected domains. It is also proven that functions having an antiderivative are analytic. Furthermore, an analytic function is continuous, which, together with path independence, let's us construct an antiderivative. So it would seem analyticity and existence of antiderivative are equivalent. Is this true? Am I missing a subtlety?

答:
Analyticity is equivalent to the existence of an antiderivative for functions on a simply connected domain.  It is not true on arbitrary domains: for instance, the function $f(z)=1/z$ is analytic on $\mathbb{C}\setminus\{0\}$ but does not have an antiderivative (locally it has an antiderivative given by a branch of a logarithm, but you can't define a consistent global antiderivative).

Another formulation is that analyticity is equivalent to locally having an antiderivative (i.e., every point has a neighborhood on which the function has an antiderivative).  For the forward direction, just note that since a disk is simply connected, an analytic function has an antiderivative on a small disk around each point of its domain.  For the reverse direction, locally having an antiderivative implies being locally analytic, but analyticity is a local property so this just means it is analytic.


问:
What does it mean for a Vector field to be tangent at every point of S?
As background, I am in Vector Calculus. My question is giving me some "short cuts" to use to help with evaluating flux integrals. The hint says "If F (Vector Field) is tangent at every pointof S, then the integral of S of F dot dA =0. What does this mean exactly? How can I know if this vector field is tangent?

The second hint is "If F is perpendicular at every point of S and has a constant magnitude on S, then the integral of S of F dot dA = +/- ||F|| dot Area of the Surface (choose the positive sign if F points in the same direction s the orientation of S, Choose negative if F points in the direction opposite the orientation of S)

I am having trouble understand what these hints mean and when/ how to use them.

The Flux integral I have to solve is: F=e^(y^2+z^2)i through the disk of radius 2 in the yz-plane, centered at the origin and oriented in the positive x direction

答:
Presumably, you learned in your class about the normal vector to a surface at a point. Saying the vector field is "tangent to the surface" means that the vector field is perpendicular to the normal vector of the surface. Geometrically/visually, this means the vector points in a direction within the tangent plane at that point.

So if $F$ is the vector field, and $dA$ is the normal to the surface, then $F \cdot dA = 0$, because they are perpendicular. If you integrate zero, you get zero.

The second hint is similar. Saying the vector field is "perpendicular to the surface" means that it points in the same (or opposite) direction of the normal vector everywhere. I guess the notation you are using probably means $dA$ is the unit normal vector. Then $F \cdot dA = \pm |F|$. Remember that if two vectors $v$ and $w$ are parallel, then $|v \cdot w| = |v| |w|$.

So if $|F|$ is the same length everywhere, then $F \cdot dA = \pm |F|$ is a constant function on the surface (it's the same at every point). Integrating a constant function is that constant times the area (this is just a basic principle about double integrals).


问:
Uniform continuity of functions and restrictions.
Hi I have this exercise on my real analysis course, I haven´t seen topology course:

Let $A_{i} \subset \mathbb{R}$ an interval for all $i \in \mathbb{N}$. Is it true or false?

a) If $f_{1}: A_{1} \longrightarrow \mathbb{R}$ and $f_{2}: A_{2} \longrightarrow \mathbb{R}$ are uniformly continous and $f: A_{1} \cup A_{2} \longrightarrow \mathbb{R}$ is continous, then $f$ is uniformly continous.

b) Let $f_{i}: A_{i} \longrightarrow \mathbb{R}$ uniformly continous for all $i \in \mathbb{N}$ and $f: \displaystyle{\bigcup_{i \in \mathbb{N}}{A_{i}}} \longrightarrow \mathbb{R}$ continous, then $f$ is uniformly continous.

Since they seem very similar I suppose I need induction on $i \in \mathbb{N}^{\geq 2}$ for both of them. However I am not abble to prove any case. For the fisrst $i = 2$ I do not have any theorem to use or a definition wich implies union of intervals on my textbook. Do you have any hint?

答:
Assuming that you define $f$ to be $f(x) = f_i(x) \iff x\in A_i$ and that $A_i \cap A_j = \varnothing $, then b) is false. Consider $f_i(x) = \frac 1x \; \forall x$ and $A_1 = (1,2], A_i = (1/i,\inf {A_{i-1}}]$ (Edit: notice the domain of $f$ is an interval. I previously defined it erroneously and made it disconnected). Then $f_i$ is uniformly continuous but $f$ is not.

You have that each $f_i$ is uniformly continuous, but $f$ is not bounded and therefore not uniformly continuous.


问:
Does $(L-\lambda I )=0$ mean that $\lambda$ must be an eigenvalue?
I am working on a proof that if $\lambda$ is the root of a linear operator's minimal polynomial, then it must be an eigenvalue of the linear operator.

I reach a point in my proof where I have, $(L-\lambda I) = 0$. Is this sufficient to show it is an eigenvalue? This is what is tripping me up: if $(L-\lambda I)= 0$ then for any nonzero vector, wouldn't it be true that $Lv-\lambda v = 0$ meaning any nonzero vector would be an eigenvector rather than specific vectors?

答: $L - \lambda I = 0$ if and only if $L = \lambda I$ for some $\lambda$ in the desired field. You are correct. If $L$ is as defined above, then any vector would be an eigenvector. I think the criterion you were looking for is that $\lambda$ is an eigenvalue if $\det(L - \lambda I) = 0$.


问:
Let $A$ and $B$ be sets. Prove that $A\cup B = (A-B)\cup (B-A)\cup (A\cap B)$
Let $A$ and $B$ be sets. Prove that $A\cup B = (A-B)\cup (B-A)\cup (A\cap B)$.

My proof: To prove $A\cup B \subseteq (A-B)\cup(B-A)\cup(A\cap B)$ we let $x\in A\cup B$.
This means $x\in A$ or $x\in B$. Without loss of generality, assume $x\in A$ so we must have that $x\in B$, or $x\notin B$.

Case $1$: Assume $x\in B$. This means since $x\in A$, and $x\in B$, then $x\in A\cap B$.

Case $2$: Assume $x\notin B$. This means since $x\in A$ and $x\notin B$, then $x\in A-B$.

So we have $A\cup B\subseteq (A-B)\cup(B-A)\cup(A\cap B)$.

Now to prove $(A-B)\cup(B-A)\cup(A\cap B) \subseteq A\cup B$, we let $y\in (A-B)\cup(B-A)\cup(A\cap B)$. This means $y\in (A-B)$, or $y\in (B-A)$, or $y\in (A\cap B)$.

Case $1$: Let $y\in (A-B)$. This means $y\in A$ and $y\notin B$.

From here I'm not sure how to get $A\cup B$.

答: Since $A \subseteq A \cup B, y \in A \Rightarrow y \in A \cup B$.


问:
Application of complex number s for solutions to a quadratic example from Spivak - Ch.25 - pg 522 - 2nd ed
I had a question of some algebra the author used to explain an idea. As stated before, it comes from Spivak's Calculus - Ch 25 - pg. 522 - 2nd ed. The example in question revolves around the use of complex numbers and is the following:

In particular I can't seem to figure out the algebra to arrive at the expressions to the right of the word "then". I thought to solve for $x$ the author used the typical formula for solving a quadratic equation, namely:

$$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$

So using the assumptions of $u^{2} - v^{2} = b^{2} - 4ac$ and $uv = 0$, I obtained:

$$x = \frac{-b \pm \sqrt{u^{2} - v^{2}}}{2a}$$

But I don't see how this is going to lead to

$$x = \frac{-b + u}{2a}$$

unless we assume $v = 0$, but in doing that I have no idea how the $x^{2}$ would even be possible of coming about.

答:
If you are wondering about the "then" - that is, why are those formulas true, given that $u^2 - v^2 = b^2 - 4ac, uv = 0$, then that is a matter of noting that $uv=0$ means either $$\begin{align}u^2 &= b^2 - 4ac\\v&=0\end{align}$$ or $$\begin{align}u &= 0\\v^2 &= 4ac - b^2\end{align}$$
If you set either $u = 0$ or $v=0$ in the two expressions, you can expand them out and cancel to get an expression only in the square of the other variable, and it turns out that substituting the above expressions for the squares will make both equations true. Well, almost: Spivak or the publisher made a typo in setting the equations. It will have to be corrected to get the right result.

On the other hand, if you are wondering how Spivak managed to pull those particular expressions out of thin air:

Move $c$ to the other side and multiply by $4a$: $$4a^2x^2 +4ab x = -4ac$$

Note that $(2ax + b)^2 = 4a^2x^2 + 4abx+b^2$. So $$(2ax + b)^2 = b^2 - 4ac$$ or making the substitution $z = 2ax+b$, $$z^2 = b^2 - 4ac$$

$z$ is some complex number, so write $z = u + iv$, then $z^2 = u^2 - v^2 + 2uvi$, so $$u^2 - v^2 + 2uvi = b^2 - 4ac$$

Equating the real and imaginary parts of the above equation,
$$\begin{align}u^2 - v^2 &= b^2 - 4ac\\2uv &= 0\end{align}$$ and of course the last equation can be divided by $2$.

$u+iv = z = 2ax+b$, so $$x = \frac{u-b}{2a} + i\frac v{2a}$$

$$x^2 = \left[\left(\frac{u-b}{2a}\right)^2 - \left(\frac v{2a}\right)^2\right] + \left[2\left(\frac{u-b}{2a}\right)\left(\frac v{2a}\right)\right]i$$
so
$$\begin{align}ax^2 + bx + c = &\left\{a\left[\left(\frac{u-b}{2a}\right)^2 - \left(\frac v{2a}\right)^2\right]+ b\left[\frac{u-b}{2a}\right] + c\right\}\\&+ \left\{a\left[2\left(\frac{u-b}{2a}\right)\left(\frac v{2a}\right)\right]+b\left[\frac v{2a}\right]\right\}i\end{align}$$

and to be $0$, both its real and imaginary parts must be $0$.


答: This is similar to the characteristic equation, which tells us the eigenvalues of any given operator(matrix). The characteristic equation for a matrix $L$ is $$\det(L-\lambda I)=0$$ if $$(L-\lambda I)=0$$ then all that we have is the zero matrix, which just tells us that $$L=\lambda I$$ i.e., our matrix is just a scaled up identity matrix. It does seem to be true that any vector $\nu$ is and eigenvector of $L$ if $$L=\lambda I$$ but that is just a trivial thing since $$L=\lambda I \Rightarrow L\nu = \lambda I\nu =\lambda \nu$$ So, you are right, but only in the special case. In the more general case, you must compute the values of $\lambda$ satisfying $$\det(L-\lambda I)=0$$

Title: Does $(L-\lambda I )=0$ mean that $\lambda$ must be an eigenvalue?

Body:
I am working on a proof that if $\lambda$ is the root of a linear operator's minimal polynomial, then it must be an eigenvalue of the linear operator.

I reach a point in my proof where I have, $(L-\lambda I) = 0$. Is this sufficient to show it is an eigenvalue? This is what is tripping me up: if $(L-\lambda I)= 0$ then for any nonzero vector, wouldn't it be true that $Lv-\lambda v = 0$ meaning any nonzero vector would be an eigenvector rather than specific vectors?


问:
Help proving a simple tautology $(p\land(p\to\lnot q))\to(\lnot q\lor p)$
Specifically this  $$(p\land(p\to\lnot q))\to(\lnot q\lor p)$$-

spent a lot of time going down the wrong path

答:
Your work until line 6. is correct, but there is no Associative law as $a\lor(b\land c)\equiv(a\lor b)\land c$, so Complement law doesn't apply on line $7$, as when we see $a\lor b\land c$, the logical connective "$\land$" always connect first.

\begin{align}
\unicode{x2714}\hspace{6.3ex}~&\vdots\\
\unicode{x2714}\hspace{2ex}6.~\equiv&\lnot p\lor (p\land q)\lor (\lnot q\lor p)\tag*{Double negation}\\
\unicode{x2718}\hspace{2ex}7.~\equiv&\top\land q\lor(\lnot q\lor p)\tag*{Complement law}
\end{align}

To fix this, we can apply Distributive law instead as following

\begin{align}
&\vdots\\
\equiv&\lnot p\lor (p\land q)\lor (\lnot q\lor p)\tag*{Line $6.$}\\
\equiv&(\lnot p\lor p)\land(\lnot p\lor q)\lor (\lnot q\lor p)\tag*{Distributive law}\\
\equiv&(\top\land(\lnot p\lor q))\lor (\lnot q\lor p)\tag*{Complement law}\\
\equiv&(\lnot p\lor q)\lor (\lnot q\lor p)\tag*{Identity law}\\
\equiv&\lnot p\lor (q\lor \lnot q)\lor p\tag*{Associative law}\\
\equiv&\lnot p\lor \top\lor p\tag*{Complement law}\\
\equiv&\top\tag*{Domination law}
\end{align}


问:
General form of centralizer
If we say $B = J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) ... \oplus  J_{a_n}(\lambda_n)$ I am trying to find the general form of the centralizer of $B$. First when our lambdas are distinct but then also when they aren't and what the dimension is on each of the cases.

So I know B should be an $\sum a_i \times \sum a_i $ matrix with entries $\lambda_i$ on the diagonals but then I am not sure how to find the centralizer for each case?

I am not really sure how to start and if someone could point me in a good direction that would be appreciated.

答:
I will use the language of modules.

A choice of linear operator $A$ on a vector space $V$ is equivalent to making $V$ a $\mathbb{C}[T]$-module, where $T$ acts by the operator $A$. Since $\mathbb{C}[T]$ is a PID, the fundamental theorem of finitely-generated modules over PIDs applies, so we can say $V$ is isomorphic as a $\mathbb{C}[T]$-module to a direct sum of cyclic submodules, and cyclic modules (by Chinese Remainder Theorem) are WLOG of the form $\mathbb{C}[T]/(T-\lambda)^d$. If we write the matrix for multiplication-by-$T$ on such a module with respect to the power basis $\{1,T,\cdots,T^{d-1}\}$ we get the $d\times d$ Jordan block associated to the generalized eigenvalue $\lambda$. The component containing all direct summands (or equivalently all cyclic submodules) with a given value of $\lambda$ forms the generalized eigenspace of $V$ associated to $\lambda$.

The centralizer of $A$ corresponds to finding the endomorphisms of $V$ as a $\mathbb{C}[T]$-module.

Note that $\hom$ is distributive, any element of $\hom(\bigoplus V_i,\bigoplus V_j)$ may be written as a matrix whose $ij$-entry is a homomorphism $V_i\to V_j$. Thus, to find the module endomorphisms of $V$, we simply need to determine all possible module homomorphisms $\mathbb{C}[T]/(T-\lambda)^d\to \mathbb{C}[T]/(T-\mu)^k$. Such a map is $T$-equivariant and its range is the cyclic submodule generated by the image of $1$. The only condition is that $1$ must be sent to an element annihilated by $(T-\lambda)^d$. Either $k\le d$ or this means $1$ is sent to something in the cyclic submodule $(T-\lambda)^{k-d}$ which has dimension $d$ within $(T-\lambda)^k$. Thus, the vector space of module homomorphisms is $\min\{d,k\}$ if $\lambda=\mu$, else $0$.

So if the Jordan blocks associated to $\lambda$ have sizes $d_1\ge d_2\ge\cdots\ge d_n$ then the dimension of the endomorphism algebra / centralizer is $\sum_{i,j}\min\{d_i,d_j\}$ which is the sum of terms in the array

$$ \begin{array}{cccc}
d_1 & d_2 & \cdots & d_n \\
d_2 & d_2 & \cdots & d_n \\
\vdots & \vdots & \ddots & \vdots \\
d_n & d_n & \cdots & d_n
 \end{array} $$

which equals $\sum_{j=1}^n (2j-1)d_j$.

In general, we take this sum for each generalized eigenvalue $\lambda$ and combine.

To describe the centralizer / endomorphism algebra as an algebra, first decompose it as a direct sum of subalgebras associated to each $\lambda$, then write elements of these subalgebras as matrices whose entries are elements of quotient modules $(T-\lambda)^{k-d}/(T-\lambda)^k$ (interpreting $k-d$ as $0$ if $k\le d$) for the various choices of $d$ and $k$.

Here is an example. Consider a matrix with Jordan decomposition

$$ \color{Red}{J_1(0)\oplus J_1(0)\oplus J_2(0)}\oplus \color{Green}{J_3(i)\oplus J_4(i)} \oplus \color{Blue}{J_5(\omega)} $$

Associated to $\color{Red}{\lambda=0}$ is $2\ge1\ge1$ with sum $\sum_{j=1}^3 (2j-1)d_j=(1)2+(3)1+(5)1=\color{Red}{10}$.

Associated to $\color{Green}{\lambda=i}$ is $4\ge3$ with sum $\sum_{j=1}^2(2j-1)d_j= (1)4+(3)3=\color{Green}{13}$.

Associated to $\color{Blue}{\lambda=\omega}$ is $5$ with sum $\sum_{j=1}^1 (2j-1)d_j=(1)5=\color{Blue}{5}$.

Thus the full centralizer has dimension $\color{Red}{10}+\color{Green}{13}+\color{Blue}{5}=28$.

Title: General form of centralizer


问:
What are the last 3 digits of $3^{100}$
How do I find the last 3 digits of $3^{100}$. I know that the last digit will be 1. But how do I find the digits occupying the ten's and hundred's places. Please give a hint. Thanks</p>

答:
We can break the problem down into several steps.

Step 1: Show that the set of $\text{modulo-}200$ residue numbers,

$\quad [1],[21],[41],[61],[81],[101],[121],[141],[161],[181]$

forms a multiplicative cyclic group of order $10$ and $81^5 \equiv 1 \pmod{200}$.

Step 2: Show that $3^{20} \equiv 1 \pmod{200}$ (observe that $3^4 = 81$).

Step 3: Show that the set of $\text{modulo-}1000$ residue numbers,

$\quad [1],[201],[401],[601],[801]$

forms a multiplicative cyclic group of order $5$.

Step 4: Using step 2 and step 3 show that

$\quad 3^{100} \equiv 1 \pmod{1000}$

Title: What are the last 3 digits of $3^{100}$


答:
Since $y \in A$, this guarantees that $y \in A \cup B$ because $y$ satisfies at least one "parameter" of the union.

By the union definition, if $x \in A \cup B$, then $x$ is an element of at least one set, $A$ or $B$.

Body:
Let $A$ and $B$ be sets. Prove that $A\cup B = (A-B)\cup (B-A)\cup (A\cap B)$.

My proof: To prove $A\cup B \subseteq (A-B)\cup(B-A)\cup(A\cap B)$ we let $x\in A\cup B$.
This means $x\in A$ or $x\in B$. Without loss of generality, assume $x\in A$ so we must have that $x\in B$, or $x\notin B$.

Case $1$: Assume $x\in B$. This means since $x\in A$, and $x\in B$, then $x\in A\cap B$.

Case $2$: Assume $x\notin B$. This means since $x\in A$ and $x\notin B$, then $x\in A-B$.

So we have $A\cup B\subseteq (A-B)\cup(B-A)\cup(A\cap B)$.

Now to prove $(A-B)\cup(B-A)\cup(A\cap B) \subseteq A\cup B$, we let $y\in (A-B)\cup(B-A)\cup(A\cap B)$. This means $y\in (A-B)$, or $y\in (B-A)$, or $y\in (A\cap B)$.

Case $1$: Let $y\in (A-B)$. This means $y\in A$ and $y\notin B$.

From here I'm not sure how to get $A\cup B$.


问:
Prove that $(-1)^n \text{Laguerre}_n(2) \leq 1$.
I would like to prove the following inequalities on Laguerre polynomials evaluated at point 2:
$$
(-1)^n \text{Laguerre}_n(2) \leq 1
$$
This seems to hold numerically.
I tried to use the recurrence relation between Laguerre polynomials but I was not successful. Any ideas? I also tried expression via Bessel function $J_0$ and then contour integral but it was not successful.

答: This is a graph of $(-1)^n L_{n}^{(0)}(2)$ where $0 \leq n \leq 2500$. In this range $|(-1)^n L_{n}^{(0)}(2)| \leq 1$. This may be of interest. The Digital Library of Mathematical Functions, NIST gives $| L_{n}^{(0)}(2)| \leq e$ in equation 18.14.8 as above. Proving a better bound is probably difficult.


问:
Chromatic polynomial of a grid graph
I have the following graph with $nm$ vertices:

```<code>1 --- 2 --- 3 --- (n)
|     |     |
4 --- 5 --- 6 --- (n)
|     |     |
7 --- 8 --- 9 --- (n)
|     |     |
|     |     |
(m)   (m)   (m)
</code>```

In other words, this is a graph of a grid with $m$ rows and $n$ columns.  How can I determine the chromatic polynomial of this graph?  Say, for $n=4$ and $m=4$? Is there a way to form a general formula for such a graph,?

答:
Let $\chi_{m,n}(q)$ denote the chromatic polynomial of an $m\times n$ grid graph. An application of the transfer-matrix method shows that for fixed $m$, the generating function $\sum_{n\geq 1}\chi_{m,n}(q)x^n$ is a rational function of $q$ and $x$. For $m=3$ we get
$$ \sum_{n\geq 1}\chi_{3,n}(q)x^n = \frac{q(1-q)(1-q+(1-3q+q^2)x)x}
    {1+(10-11q+5q^2-q^3)x+(11-24q+19q^2-7q^3+q^4)x^2}. $$
It is also easy to see directly that
$$ \chi_{2,n}(q) = q(q-1)(q^2-3q+3)^{n-1}. $$

Title: Chromatic polynomial of a grid graph


答: We obviously have $L^2_a(G)\subset L^2_c(G)$, where $L^2_c(G)$ is the space of continuous square integrable functions. Then $L^2_c(G)\subset L^2(\mu)$ in the following sense: There is an injection $j:L^2_c(G)\to L^2(\mu)$ such that for any $u\in L^2_c(G)$ the only continuous representative of $ju\in L^2(\mu)$ is $u$.

Title: How can the Bergman space $L^2_a(G)$ of a complex open set $G$ be a subspace of $L^2(\mu)$?

Body: How can the Bergman space $L^2_a(G)$ of a complex open set $G$ be a subspace of $L^2(\mu)$ where $\mu$ is the area measure on G? If $L^2(\mu)$ is a space of equivalence classes of functions then how is $L^2_a(G)$, a space of functions, contained within it? Are we using shorthand for some sort of representation of $L^2(\mu)$ where we're only considering representatives of each equivalence class as those that are continuous? Is there a unique element of each equivalence class of $L^2(\mu)$ that is continuous?


答:
Here it is a proof based on the properties of the set operations:
\begin{align*}
(A - B)\cup (A\cap B) = (A\cap B^{c})\cup(A\cap B) = A\cap(B\cup B^{c}) = A
\end{align*}

Consequently, one has that
\begin{align*}
(A - B)\cup(B-A)\cup(A\cap B) = A\cup(B - A) = A\cup(B\cap A^{c}) = (A\cup B)\cap(A\cup A^{c}) = A\cup B
\end{align*}

Hopefully this helps.

Title: Let $A$ and $B$ be sets. Prove that $A\cup B = (A-B)\cup (B-A)\cup (A\cap B)$


问:
How is changing the $t$ to a $s$ in $v(t) = \frac{ds}{dt} \implies \frac{dt}{ds} = \frac{1}{v(s)}$ justified?
Sorry if the title wasn't descriptive enough as I didn't found a better way to formulate the question.

So I found this article that introduces you to calculus of variations by using it to solve the brachistochrone problem.

In page 2 lies the problem, as the article gets velocity in terms of position $x$ ($v(x) = \sqrt{2gy(x)}$).

Then the distance traveled (so the arclength) of $y(x)$ is defined as $s(x) = \int_0^x{\sqrt{1+y'(\hat{x})^2}}d\hat{x}$ as expected so its derivative over $x$ is $\frac{ds}{dx} = \sqrt{1+y'(x)^2}$.

Then by definition velocity is $v(t) = \frac{ds}{dt}$ (note it now depends on time).

Now here its the troublesome part, as the article rearranges the equation above in the following way: $\frac{dt}{ds} = \frac{1}{v(s)}$ and later on $v(s)$ is replaced somewhere with $\sqrt{2gy(x)}$ (so $v(x)$) and my question is:

Why can we change the function from depending on $t$ to depending on the distance traveled $s$? Doesn't the function take a time and returns you the velocity? why you can just input $s$ like that?

答:
There's a joke that if you show $f(x,y)=\sqrt{x^2+y^2}$ to a mathematician and a physicist and ask them what $f(r,\theta)$ is, the mathematician will say $f(r,\theta)=\sqrt{r^2+\theta^2}$ and the physicist will say $f(r,\theta)=r$.

Do you get the joke?

The mathematician is viewing $f(\cdot,\cdot)$ formally as notation that says to plug two things into a formula, while the physicist is viewing $f$ as a variable that depends on the two things in parentheses. If $f=\sqrt{x^2+y^2}$, then we can say $f$ depends on $x$ and $y$ and call it $f(x,y)$, but since this is also $r$ we can say $f=r$ and that $f$ depends on $r$ and $\theta$ and call it $f(r,\theta)$, at least in the physicist's mind.

Thus, velocity $v$ may very well depend on $t$, but that would also depend on $s$ then too! Because you can write $s$ as a function of $t$ and vice-versa, that means you can write $v$ as a function of $s$ or of $t$ and take the derivative with respect to either variable.

Let's look at an example with different letters. Say $z=\sqrt{1-y}$ and $y=x^2$. Then a physicist would say $z(y)=\sqrt{1-y}$ but they might also say $z(x)=\sqrt{1-x^2}$, assuming $x,y,z$ represent physical quantities. Obviously this is at odds with the usual function notation that mathematicians understand.


问:
Prove A is symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$
Prove A is a symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$. (for all $x,y \in \mathbb{R}^n$)

Going from the assumption that it is symmetric to the two conditions is fairly straightforward.

However, going the other way, I am stuck at proving $A^T = A$ using the second and condition, being stuck at $X^T (A-A^T)y=0$.

Note T is for transpose!

答:
The key part, alas missing, from $x^TAy=(Ax)^Ty$ is that this holds for all vectors $x,y$.  If $A-A^T$ had a nonzero entry, you could find an $x,y$ that would extract that entry.

Hint: try standard basis vectors.

Double Hint: If $x=e_i, y=e_j$, then $x^TAy=(A)_{i,j}$.


问:
Continuous proof question
Let f : [0, ∞) → R be a continuous function and let g : [0, ∞) → R be the function
defined by
g(x) = sup{f(y) | y ∈ [0, x]}.
Prove that g is also continuous.

I tried using the Intro to Real Analysis text to complete this proof but I am unsure of how to manipulate the supermum.
I know that I need to show that g is increasing, but then I do not know where to go from there.

Any help would be appreciated.

答:
I would prove the contrapositive: if $g$ is discontinuous at $x_0$, then so is $f$. Indeed, showing $g$ is increasing is of great help because discontinuities in increasing functions need to be jump discontinuities. In particular (with the exception of the special case $x_0 = 0$) we must have
$$\lim_{x \to x_0^-} g(x) < \lim_{x \to x_0^+} g(x).$$
These one-sided limits exist essentially by the continuous version of the monotone convergence theorem. From the left, the function is increasing and bounded above by $g(x_0 + 1)$. From the right, the function is decreasing, bounded below by $g(x_0/2)$ (I didn't choose $x_0 - 1$, just in case $x_0 \in (0, 1)$).

On the other hand if $x_0 = 0$, then the only way that we can have a discontinuity is if
$$f(0) = g(0) < \lim_{x \to 0^-} g(0).$$
This needs to be dealt with as a separate case.

If $x_0 > 0$, then let the left limit of $g$ be $L$ and the right limit be $R$. I would suggest proving that $f(x) \le L$ for all $x \in [0, x_0]$, implying that $\lim_{x \to x_0} f(x) \le L$ if such a limit exists. But, on the other hand, for any $n \in \Bbb{N}$, there must exist some $x_n \in [0, x_0 + 1/n]$ such that $f(x_n) \ge R$. This implies that $\lim_{x \to x_0} f(x) \ge R$ if it exists. Thus, the limit cannot exist, hence $f$ cannot be continuous at $x_0$.

I'll leave the $x_0 = 0$ case to you, as well as filling in the blanks! Let me know if you need more help.

Title: Continuous proof question


问:
Growth-rate vs totality
How can one prove the statement, "If a function grows fast enough, it cant be proven total in PA, unless PA is inconsistent"? How fast must it grow to be not provably total? </p>

答:
There is a hierarchy of fast-growing functions $f_\alpha:\mathbb N\to\mathbb N$ indexed by ordinals $\alpha<\epsilon_0$, where $\epsilon_0$ is the first ordinal fixed point of the ordinal exponentiation map $\tau\mapsto\omega^\tau$, that is,
 $$ \epsilon_0=\omega^{\omega^{\dots}}. $$
(Note that this is a countable ordinal.) Let $\zeta_0=1$ and $\zeta_{k+1}=\omega^{\zeta_k}$ for $k\in\mathbb N$.

For each $k\ge0$, let $I\Sigma_{k+1}$ denote the restriction of Peano Arithmetic where the induction axiom is only stated for $\Sigma_{k+1}$-formulas, so $\mathsf{PA}$ is the union of all the $I\Sigma_{k+1}$ for $k\in\mathbb N$.

Theorem. (Wainer) If $f$ is a recursive function, provably total in $I\Sigma_{k+1}$ , then $f$ is eventually dominated by some $f_\alpha$, $\alpha<\zeta_{k+1}$. In particular, any recursive $f$ provably total in $\mathsf{PA}$ is eventually dominated by some $f_\alpha$, $\alpha<\epsilon_0$.

Note that the function $f$ must be recursive. There are non-recursive functions that only take the values $0$ and $1$, so this restriction is essential. The specific hierarchy of functions we pick is not so important. All natural hierarchies that have been considered give the same result that $\epsilon_0$ is the limit. The specific hierarchy mentioned in the theorem is defined as follows:

Any ordinal $\alpha<\epsilon_0$ can be written in a unique way as $\alpha=\omega^\beta(\gamma+1)$ where $\beta<\alpha$ (note that $\beta$ could be $0$). By transfinite recursion, define for limit $\alpha<\epsilon_0$ an increasing sequence $(d(\alpha, n)\mid n<\omega)$ cofinal in $\alpha$ by setting
$$ d(\alpha, n) = \omega^\beta\gamma+\left\{\begin{array}{cl}\omega^\delta n&\mbox{ if }\beta=\delta+ 1,\\ \omega^{d(\beta,n)}&\mbox{ if }\beta \mbox{ is limit.}\end{array}\right. $$
The fast growing hierarchy $(f_\alpha)_{\alpha<\epsilon_0}$ of functions $f:\mathbb N\to\mathbb N$, due to Löb and Wainer, can now be defined as follows:

$f_0 (n) = n + 1$.

For $\alpha<\epsilon_0$, $f_{\alpha+1} (n) = f^n_\alpha(n)$, where the superindex indicates that $f_\alpha$ is iterated $n$ times.

For limit $\alpha<\epsilon_0$, $f_\alpha(n) = f_{d(\alpha,n)} (n)$.

Note that these functions grow very fast indeed. One can prove that these functions are strictly increasing, and that whenever $\alpha<\beta<\epsilon_0$, there is a number $m$ such that $f_\alpha(n)<f_\beta(n)$ for all $n\ge m$. We say that $f_\beta$ eventually dominates $f_\alpha$.

In Ramsey theory one sometimes encounters the first four or five levels $f_0$–$f_4$ and rarely one needs to go up to $f_\omega$: Note that $f_1(n)=2n$, $f_2(n)=n2^n$, $f_3(n)$ is larger than a stack of $n$ powers of $2$, etc. The function $f_\omega$ grows roughly as fast as the diagonal of Ackermann's function, so it eventually dominates all primitive recursiev functions. Goodstein's problem requires all the functions up to $\epsilon_0$ and therefore the totality of Goodstein's function is not provable in $\mathsf{PA}$. Several famous unprovability results have been established the same way. Typically, unprovability in arithmetic considers $\Pi^0_2$ sentences, statements $\phi$ of the form "For all $n$ there is an $m$ such that $\psi(n,m)$" where $\psi(n,m)$ is an easily verifiable (recursive) statement about $n,m$. This naturally gives rise to a function, $f(n)=m$ iff $m$ is the least number such that $\psi(n,m)$ holds. If $\phi$ is true, then $f$ is a recursive function: To compute $f(n)$, you consider in succession all natural numbers, verifying for each $k$ whether $\psi(n,k)$ holds. As soon as we find one such $k$, that is the value of $f$. This search succeeds, since $\phi$ is true, so $\psi(n,m)$ holds for some $m$, and therefore there is a least such $m$. The issue is whether the formal statement $\phi$ is provable from $\mathsf{PA}$, which is equivalent to the question of whether $\mathsf{PA}$ can prove that $f$ is a total function. In view of Wainer's theorem, this means that we should verify whether $f$ is eventually dominated by some $f_\alpha$, $\alpha<\epsilon_0$. In practice, when $\phi$ is not provable, we usually show that in fact $f$ eventually dominates each $f_\alpha$, $\alpha<\epsilon_0$. This is the case, for example, for Goodstein's function.

I list now some references, taken from my paper on Goodstein's function, available at my papers page. To get an idea of how fast these functions grow, I suggest you look at the few values of Goodstein's function I compute in that paper.

Fairtlough, M., and Wainer, S. Handbook of Proof Theory. Elsevier-North Holland,
  1998, ch. Hierarchies of provably recursive functions, pp. 149–207.

Löb, M., and Wainer, S. Hierarchies of number theoretic functions. I. Arch. Math.
  Logik Grundlagenforsch. 14 (1970), 39–51.

Wainer, S. A classification of the ordinal recursive functions. Arch. Math. Logik Grundlagenforsch. 13 (1970), 136–153.

Note that the hierarchy can be extended significantly past $\epsilon_0$ (maintaining that the functions are increasing, and each new one eventually dominates its predecessors), and unprovability results in stronger systems are sometimes obtained by appropriate generalizations of Wainer's results to these other systems. The question of how to extend these functions is always the question of how to pick for (smallish) countable ordinals $\alpha$ a cofinal sequence in an explicit fashion. This leads to the delicate topic of natural well-orderings, see

John N. Crossley, Jane Bridge Kister. Natural well-orderings, Archiv für mathematische Logik und Grundlagenforschung, 26 (1), (1987), 57-76.

Title: Growth-rate vs totality


问:
Prove that : $x_{n}=\sqrt[n]{n^{2}+2}$ is convergent.
Prove that : $x_{n}=\sqrt[n]{n^{2}+2}$ is convergent.

I intended using inductive method. From inductive step, I supposed $x_{n}<x_{n-1}$, and I need to prove $x_{n+1}<x_{n}$. However, with $x_{n}<x_{n-1}$, I could only show that $x_{n-1} >1$. Please help me, thank you so much!

答:
Just a hint: You can alternatively try using the Sandwich Theorem

We have, $\forall n\in\mathbb{N}$, $n^2<n^2+2\implies n^{\frac{2}{n}}<(n^2+2)^{\frac{1}{n}}$

Also, $(n^2+2)<n^3$ $\forall n\geq2\implies (n^2+2)^{\frac{1}{n}}<n^{\frac{3}{n}}$.

So, combining these two, we get,
$$n^{\frac{2}{n}}\leq(n^2+2)^{\frac{1}{n}}\leq n^{\frac{3}{n}}$$ $$\forall n\geq2$$

Can you complete this now?


问:
What is the motivation behind defining $[\alpha] + [\beta] = [(\alpha \land \lnot \beta) \lor (\lnot \alpha \land \beta)]$?

Consider $\mathcal{F}$ to be the set of all well-formed formulae (wffs), and $\mathcal{F}/\equiv$ denotes the set of equivalence classes under the equivalence relation induced by $\equiv$ (logical equivalence). Then, $(\mathcal{F}/\equiv, +, \land, 0, 1)$ is a boolean algebra (defined as a ring), where the operation $+$ is defined as $[\alpha] + [\beta] = [(\alpha \land \lnot \beta) \lor (\lnot \alpha \land \beta)]$. $[\alpha]$ refers to the equivalence class of wff $\alpha$. Clearly, $0$ refers to all contradictions and $1$ refers to all tautologies.

What is the motivation behind defining $[\alpha] + [\beta] = [(\alpha \land \lnot \beta) \lor (\lnot \alpha \land \beta)]$? This surely works, i.e. under this definition of $+$, $(\mathcal{F}/\equiv, +, \land, 0, 1)$ is indeed a boolean algebra, but this definition looks very unintuitive.

答:
It corresponds to symmetric difference,

$$A\triangle B=(A\setminus B)\cup(B\setminus A)=(A\cap B^c)\cup(B\cap A^c)\,,$$

in the algebra of sets and gives you not just a Boolean algebra, but a Boolean ring, just as $\langle\wp(X),\triangle,\cap,\varnothing,X\rangle$ is a Boolean ring of sets.


问:
Can there be generalization of Monty Hall Problem?
Monty Hall problem basically is: there are three doors, behind one of these doors is a brand new car, and behind the other two are goats, one picks a door that one thinks has a car behind, then the host opens one of the doors one didn't pick to reveal one of the goats, and the host asks the person whether one wishes to switch his choice. By the nature of probability, it is always better to switch.

I would like to generalize this case to the following variation:

Assume there are many finite doors (cardinality $n$). $t$ doors have cars behind. As in the original formulation, one picks a door. Then, the host reveals $s$ doors that have goats. Then, the host as in the original formulation asks whether one wishes to switch. For any possible $n$ bigger than 3, and given $t$, how is one able to find minimum $s$ that makes switching the choice always a better choice?

答:
It has been done here - The Monty Hall Paradox Probability Equations.

A quick summary below.
It will greatly simplify the calculations if we start with these two ideas:

The probability values for $t$ doors having cars behind them are just $t$ times the original values (when there is only one door having a car). This is simply because the winning chances of all the strategies and subsets are simply inflated by $t$ by having $t$ winning doors. So let's start with $t=1$.

We assume that the host momentarily identified a set $y > s$ doors (just in his mind) and opened $s$ doors from this $y$ set. Later we'll just make $y=n-1$.

In the beginning, the winning probability of group $y$ as a whole is simply $y \over n$. And, if $y$ wins, the winning probability of each door inside it is $1 \over y$. With $s$ non-winning doors now opened from $y$, the winning probability of each door in $y$ becomes $1 \over y-s$, provided $y$ wins. But $y$ doesn't always win.

Hence, the absolute winning probability of each door within $y$ is simply ${y \over n} \times {1 \over y-s}$, or ${1 \over n} \times {y \over y-s}$. Since ${y \over y-s} >1$, switching to any door within $\boldsymbol{y}$ is always advantageous - even for $\boldsymbol{s=1}$.

Finally, on getting rid of $y$ by making it as large as $n-1$, i.e. it having every door except the player's original choice, and having $t$ doors instead of $1$, we get the winning probability of switching which is: $$ {t \over n} \times {y \over y-s} = {t \over n} \times {n-1 \over n-1-s} $$

Have a look at these two pages for comparing the real world results with the predictions from the equation - Monty Hall Paradox Test Tool and The Monty Hall Paradox Probability Equations.


问:
How to determine a value of a variable in a polynomial that is not x in order to determine the amount of real roots that the polynomial has?
Sorry for such a terribly worded question, that is just how I came to verbalize it. The real question and it's answer is provided below:

My problem comes with understanding the solution to 4bi. and 4biii. Why does setting the polynomial less than zero then determining it's positive regions ensure that the polynomial will have no zeros (with the whole polynomial having one real zero because of the x-1)? Why does doing the opposite ensure 2 real zeroes?

答:
$1$ is certainly a root to the cubic equation. The remaining roots come from the quadratic part and we have proven that $1$ is not a root for the quadratic part.

Hence if the discriminant is negative, then the quadratic part has $0$ real root, hence the cubic equation will have $1$ real root.

If the quadratic part has $r$ real root, then the cubic equation has $r+1$ roots.


问:
order of stabilizer
$G$ is a group which acts transitively on a set $S$. $G_s$ is the stabilizer of $s$. How to see the map from $G/G_s$ to $S$ given by $g*G_s \to g(s)$ is one to one?</p>

答:
Let $g,h\in G$. If $g(s)=h(s)$, then $$g^{-1}(g(s))=g^{-1}(h(s)),$$ which becomes $$(g^{-1}g)(s)=e(s)=s=(g^{-1}h)(s).$$
Thus $g^{-1}h\in G_s$ because $g^{-1}h$ stabilizes $s$. Thus $g^{-1}hG_s=G_s$, and thus $hG_s=gG_s$.

Therefore, given any $gG_s$ and $hG_s$, if $g(s)=h(s)$ then $gG_s=hG_s$. Thus the map is injective (which is one possible meaning of one-to-one).

Some people instead use "one-to-one" to mean bijective. Under this meaning, the map $gG_s\mapsto g(s)$ is still one-to-one. We have already shown the map is injective above; so all that is left is to prove it is surjective, and then we will have shown it is bijective.

Given any $t\in S$, there is a $g\in G$ such that $g(s)=t$ (this is the definition of transitive). Thus given any $t\in S$, there is an element of $G/G_s$ that maps to $t$, namely $gG_s$ where $g$ is any element that sends $g(s)=t$ (and we know at least one must exist).

Title: order of stabilizer


问:
Limit of a log function using expansion of $\log (1+x)$
I am trying to find the limit of

$$\lim_{n\to \infty} n\log \left(1+\left(\frac{f(x)}{n}\right)^\alpha\right)$$

where $f$ is some function $f:X\to[0,\infty]$ and $1\leq \alpha < \infty$ in some other problem I am looking at.

I cannot use the expansion $\log (1+x)=x-\frac{x^2}{2}+\cdots$ because we don't know if $|x|<1$. I know that the answer is $f(x)$ for $\alpha=1$ and $0$ for $\alpha>1$ but I'm not sure how to show it?

答:
HINT:

For any fixed $x$ and fixed $\alpha\ge1$, there exists an $N$ such that whenever $n>N$, $\left(\frac{f(x)}{n}\right)^\alpha<1$.

Then note that for $t\ge0$ we have

$$\log(1+t)=t+O(t^2)$$

Can you finish?


答: $|\frac{f(x)}{n}|\lt 1$ for fixed $x$ and sufficiently large $n$,  so you use the approximation for the log ~ $(\frac{f(x)}{n})^\alpha$.  For large n the expression is then ~ $\frac{f(x)^\alpha}{n^{\alpha-1}} \to 0$, as $n\to \infty$ for $\alpha \gt 1$.

Title: Limit of a log function using expansion of $\log (1+x)$

Body:
I am trying to find the limit of

$$\lim_{n\to \infty} n\log \left(1+\left(\frac{f(x)}{n}\right)^\alpha\right)$$

where $f$ is some function $f:X\to[0,\infty]$ and $1\leq \alpha < \infty$ in some other problem I am looking at.

I cannot use the expansion $\log (1+x)=x-\frac{x^2}{2}+\cdots$ because we don't know if $|x|<1$. I know that the answer is $f(x)$ for $\alpha=1$ and $0$ for $\alpha>1$ but I'm not sure how to show it?


问:
Determine the length of the projection of PQ onto the x-axis, with position vectors of P <3,4> and Q <7,8>

Determine the length of the projection of PQ onto the x-axis, with
position vectors of P <3,4> and Q <7,8>

My working out so far:

$PQ = PO + OQ$

$= -(3i + 4j) + (7i + 8j)$

$ = 4i + 4j$

Since $PQ$ is being projected onto the $x$-axis, do you choose a random vector lying on the $x$-axis, like a unit vector $(1i + 0j)$ to calculate the scalar projection with the formula?

Or the answer key provided suggests that the scalar is just the x-component of $PQ$ - so in this case it would just be $4$ since that is the $i$ component, but I don't understand why. Could someone please explain this to me?

Thanks in advance!

答:
Both methods work. Using the projection formula we get the projected vector length as
$$(4,4)\cdot(1,0)=4$$
Note that $(1,0)$ is already a unit vector.

Since we're projecting on the $x$-axis, it suffices to throw away the $y$-component, so we still get $4$. Indeed, using the projection formula shows that we can make this simplification.