问:
Similar questions in GATE 2021 and different answer keys | Answer Key Challenge

GATE CSE 2021 Set 1 | GA Question: 9

Given below are two statements 1 and 2, and two conclusions I and II

Statement 1: All bacteria are microorganisms.

Statement 2: All pathogens are microorganisms.

Conclusion I: Some pathogens are bacteria.

Conclusion II: All pathogens are not bacteria.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

(A) Only conclusion I is correct

(B) Only conclusion II is correct

(C) Either conclusion I or II is correct

(D) Neither conclusion I nor II is correct

My attempt:

I have to find counter example,

Case-1 is counter example of Conclusion II, because of all pathogens are bacteria in this diagram.
Conclusion I would be true in this case, because all so also some.
(So, Conclusion I is true and II is false.).

Case-2 is counter example of Conclusion I, because no pathogens are bacteria.
Conclusion II would be true in this case, because all so also some.
(So, Conclusion II is false and II is true.).

Case-3: Conclusion I is true, and Conclusion II is false as some pathogens are already bacteria.

Case-4: Conclusion I is true, and Conclusion II is false as some pathogens are already bacteria.

So, I found counter example for both Conclusion. Hence, option (D) is true (Official Key given by GATE 2021.).

I've read this method for finding counter examples for such questions.

Doubt here, is as both given Conclusions are negation of each other. That is,

∃x(p(x) ∧ b(x)) = ¬∀x (p(x) → ¬b(x))

∀x (p(x) → ¬b(x)) = ¬∃x(p(x) ∧ b(x))

Peoples are debating this question and saying that the correct answer should be option (C) Either conclusion I or II is correct.

A exact similar question appeared in the GATE EC branch, which has same above logic.

GATE EC 2021 | GA Question: 6 :

Given below are two statements and two conclusions.

Statement 1: All purple are green.

Statement 2: All black are green.

Conclusion I: Some black are purple.

Conclusion II: No black is purple.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

(A) Only conclusion I is correct

(B) Only conclusion II is correct

(C) Either conclusion I or II is correct

(D) Both conclusion I and II are correct

Here, as above my attempt, we can find counter examples for the both the statements, so both (D) Both conclusion I and II are correct.

But, this time official key is given (C) Either conclusion I or II is correct.

Note these are two different branches of Engineering (Computer Science and Engineering, Electronics and Communication Engineering). Professors who designed these questions for GATE 2021, should not be same person (but, could be as these questions are from General Aptitude which is common (but not same questions) in all branches of GATE Papers).

My Question

What is correct approach to solve such questions and which answer keys are correct which are wrong ? If both keys are correct respective their questions then what is possible difference between these questions ?

Thank you,

Update :

Final answer keys given by GATE official :

For first question (GATE CSE 2021 Set 1 | GA Question: 9) : Both
(C) or (D)

For second question (GATE EC 2021 | GA Question: 6) : Only (C)

Now, why they have given option (D) as answer for the first question
!?

答:
You're correct in that both questions have exactly the same idea, and in both cases the solution is "Either I or II is correct". We're using the law of the excluded middle, so either $P$ is true or $\neg P$ is. The official answer for the first question seems to be mistaken.

You'll notice that in your attempt, in all your cases $C$ was still true, which should indicate to you that $C$ could be valid, and you shouldn't automatically jump to $D$. With your cases you could confirm $A$ and $B$ were false. But you didn't prove $C$ was false. In this case we can prove $C$ is true with the law I mentioned earlier.


问:
Find distribution of $X$
If there are random variables $X$ and $N$ which satisfies
$$N\sim poisson(\lambda) \\ X\mid N\sim B(N, p)$$
$N$ is a poisson distribution with expectation $\lambda$ and $X$ given $N$ is a binomial distribution with N trials and p success probability.

I already get if $X, Y$ are poisson distribution with expectations $\lambda p$ and $\lambda(1- p), N= X+ Y$ then it satisfies above conditions.

But, I cannot explain to myself why it has to be like that.

Could I find out distribution of $X$ and reasonable explanation for why is it ?

答:
The probability mass function of $X$ can be written as
$$
P(X=k) = \sum_{i=0}^{\infty}P(X=k|N=i)P(N=i)
$$
You said, $X|N$ is a binomial with parameter $N$ and $p$, therefore for any non-negative integer $k$

$$
P(X=k) = \sum_{i=k}^{\infty} {i\choose k} p^{k}(1-p)^{i-k} \exp(-\lambda)\frac{\lambda^i}{i!} $$
$$= \exp(-\lambda p)\frac{(\lambda p)^k}{k!}\sum_{i=k}^{\infty}\exp(-\lambda(1-p))\frac{(\lambda(1-p))^{i-k}}{(i-k)!}.
$$
Now write $j=i-k$ and you will see that the last sum is $1$ and thus

$$
P(X=k) =\exp(-\lambda p)\frac{(\lambda p)^k}{k!}
$$
for $k=0,1,\ldots$.

ContentLicense: CC BY-SA 4.0

Title: Find distribution of $X$


问:
Definability of Sets of Truth Assignments
I have some questions about under what conditions a set of truth assignments is the model of some set of sentences.

To be more precise, suppose I'm dealing with only propositional logic. Let $K$ be a set of truth assignments, i.e. those maps assigning truth values to propositional variables.

For any set $\Sigma$ of sentences, let $Mod(\Sigma):=\{v:v\ \mbox{is a truth assignment and v satisfies $\Sigma$}\}$.

Definition. Let $K$ be a set of truth assignments. Call $K$ definable if $K=Mod(\Sigma)$ for some set $\Sigma$ of sentences; if such $\Sigma$ can be chosen finite (or equivalently, containing a single sentence), then $K$ is finitely definable.

I can prove that, in a language with only finitely many propositional variables, any set of truth assignments is definable. Since in such language there are only finitely many truth assignments, what I proved can be rephrased as: any finite set of truth assignments is definable. Now my questions are:

In a language with denumerably many propositional variables, there are continuum many truth assignments. Does what I proved still hold in this case? Is any set of truth assignments, or any finite set of truth assignments, definable? Or finitely definable?

Is there a criterion for a set of truth assignments to be definable or finitely definable in a general setting? What would happen if we work in a language with uncountably many propositional variables?

Thank you in advance for anyone who might help me about this.

答:
A few observations:

Being finitely definable implies there are only finitely many variables used in $S$. Thus, any valid truth assignment for $S$ will remain one if we flip the truth value of any of the remaining variables from true to false or vice versa. If this is not the case with a set of truth assignments (i.e. no finite subset of the variables has this property), it is not finitely definable.

Any finite set of truth assignments is definable (regardless of the cardinality of the set of variables; the set of sentences will actually have the same cardinality).

This is easy to achieve for each truth assignment separately (just add one sentence for each variable, forcing it to be true or false, depending on what the assignment says).

In order to do it for more than one assignment simultaneously, it's sufficient to note that one can find a finite subset of variables which distinguishes them (i.e. every pair of truth assignments will differ in at least one variable from this subset). This set can then be used to build one distinguishing sentence for each truth assignment which says "We're dealing with this truth assignment." (being just a conjunction of the variables and their negations, depending on what the assignment says about them).

Then, for each variable in the assignment, we'll add an implication of the form "distinguishing sentence $\Rightarrow$ variable" (or $\neg$ variable,  depending on what the assignment tells us). Thus, we've essentially added (infinite) implication saying that the distinguishing sentence determines the rest of the truth assignment completely.

Finally, a sentence one formed as a disjunction of all distinguishing sentences guarantees the only satisfying assignments will be the ones we are interested in.

Note that the above construction does not work for infinitely many statements since the "big disjunction" would yield an infinite sentence. I actually suspect that a set of all truth assignments which assign value of "true" to exactly one variable might not be definable -- although that's just a gut feeling for now.

ContentLicense: CC BY-SA 3.0


问:
Equation with $ \sqrt[5]{}$
How to show that this equation $ \sqrt[5]{20-2x} + \sqrt[5]{7-x}+\sqrt[5]{3x+5}=2$ have 3 solutions 9, -6 and -25

Wolframsays  the equation have no solutions !

答:
If $\sqrt[5]{20-2x}= a$  etc.,   $a+b+c=2$

$$a^5+b^5+c^5=32=(a+b+c)^5$$

Use How to factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$? to discover that $(a+b)(b+c)(c+a)=0$ should give three solutions.

Title: Equation with $ \sqrt[5]{}$


问:
Prove or give a counterexample: Over C,$\,$ $0$ is the only eigenvalue $\iff$ $T$ is nilpotent.
Suppose $V$ is a complex vector space and $T$ is a linear map on $V$.

Prove or give a counterexample: $\ $ $0$ is the only eigenvalue of $T$ $\iff$ $T$ is nilpotent.

Two things already known :

(1) $\ $ If $T$ is nilpotent, then $0$ is the only eigenvalue of
$T$.

(2) $\ $ On finite-dimensional complex vector spaces, if $0$
is the only eigenvalue of $T$, then $T$ is nilpotent.

So we only need to consider one direction. The question is equivalent to:

On infinite-dimensional complex vector spaces, $\,$ $0$ is the only eigenvalue $\,$ $\Longrightarrow$  $\,$$T$ is nilpotent $\ ?$

Any insights are much appreciated.

答:
Consider the operator $T:(x_i)_{i\in\mathbb{N}}\mapsto(0,0,x_2,x_3,\ldots)$, that is $T\mathbf{x}=R\mathbf{x}-x_1\mathbf{e}_2$, where $R$ is the right-shift operator.

Then $0$ is an eigenvalue since $T\mathbf{e}_1=\mathbf{0}$.

But there are no other eigenvalues (as in the proof that $R$ has no eigenvalues). Moreover, $T$ is not nilpotent since $T^n\mathbf{e}_2=\mathbf{e}_{2+n}\ne\mathbf{0}$


问:
Interior of intersection of essential closures
Let $X$ be a topological space. Then any two open sets $A,B\subset X$ satisfy
$$
\mathrm{Int}\big(\overline A\cap \overline B\big)=\mathrm{Int}\big(\overline {A\cap B}\big),
$$
as explained in the answer to this question. It is easy to see that the above equality does not hold for arbitrary sets $A$ and $B$. For example, consider  $X=\mathbb R$, $A=\mathbb Q$, $B=\mathbb R\setminus \mathbb Q$. (However, it seems to be somewhat difficult to write down a counterexample in $\mathbb R$ where one of the sets is open).

Now, suppose that there is a measure $\mu$ on $X$ on some $\sigma$-algebra containing the Borel $\sigma$-algebra. Recall that the essential support of a measurable function $f:X\to \mathbb R$ is defined by
$$
\mathrm{ess\,supp}\, f=X\setminus\bigcup\big\{\Omega\subset X \;|\; \Omega \text{ open},\; f=0\text{ almost everywhere in }\Omega\big\}.
$$
Given some measurable set $A\subset X$, let us define its essential closure by
$$
\overline{A}^\mu:=\mathrm{ess\,supp}\, 1_A,
$$
where $1_A$ is the characteristic function of $A$.

I am interested in the question which measurable sets $A,B\subset X$ satisfy the equality
$$
{(\bf I)}\qquad\mathrm{Int}\big(\overline A^\mu\cap \overline B^\mu\big)=\mathrm{Int}\big(\overline {A\cap B}^\mu\big).
$$
If the measure $\mu$ is strictly positive, i.e., $\mu(U)>0$ for every non-empty open set $U\subset X$, then one has $\overline A^\mu=\overline A$ for every open set $A$ and thus ${\bf(I)}$ holds for any two open sets $A,B$.
On the other hand, if the support of the measure $\mu$ has empty interior (e.g. a Dirac measure on a space where points have empty interior), then ${\bf(I)}$ holds trivially for any two measurable sets $A,B$.

I expect that there should be plenty of counterexamples to ${\bf(I)}$ even when $X=\mathbb R$ and $\mu$ is the Lebesgue measure. However, I find it surprisingly difficult to find such counterexamples. For example, the above pair $A=\mathbb Q$, $B=\mathbb R\setminus \mathbb Q$ no longer works. Also $A=C$, the fat Cantor set, and $B =\mathbb R\setminus C$ does not work.

Who knows a counterexample to ${(\bf I)}$ ?

Preferably with $X=\mathbb R$ and $\mu$ the Lebesgue measure, but other topological measure spaces would also be interesting.

A reference where such questions are addressed would be highly appreciated.

答:
In the meantime I have found a counterexample myself, inspired by the exercises on p. 118 of Bressoud's A radical approach to Lebesgue's theory of integration. Let me share it with you:

We take $X=[0,1]$, $\mu$ the Lebesgue measure, and let $(\lambda_k)_{k\in \mathbb N}\subset (0,\infty)$ be a sequence of positive real numbers such that
$$
L:=\sum_{k=1}^\infty \lambda_k<\infty.
$$
Put $a_k:=e^{-\lambda_k}(1+2e^{-\lambda_k})$ and denote by $C_{a_k}\subset [0,1]$ the  Smith–Volterra–Cantor set with parameter $a_k$. The latter has the volume $1-e^{-\lambda_k}$. Let $C_1:=C_{a_1}$ and for $n\geq 2$ let $C_n$ be defined by replacing each open interval in $[0,1]\setminus\{C_1\cup\cdots\cup C_{n-1}\}$ with a rescaled version of $C_{a_n}$. Note that $C_n$ is disjoint from $C_k$ for all $k<n$. Then one has
$$
\mu(C_1\cup \cdots \cup C_n)=1-e^{-\sum_{k=1}^N\lambda_k},
$$
which implies that $C:=\bigcup_{k=1}^\infty C_k$ satisfies
$$
\mu(C)=1-e^{-L}>0,\qquad \mu([0,1]\setminus C)=e^{-L}>0.
$$
Now, if $(a,b)\subset [0,1]$ is a non-empty open interval, then since each SVC-set is nowhere dense the above procedure places a small rescaled version of $C_{a_k}$ inside $(a,b)$ for some large enough $k$, so a similar argument as above shows that
$$\mu((a,b)\cap C)>0,\qquad \mu((a,b)\cap ([0,1]\setminus C))>0.$$
This shows that $\mu(\Omega\cap C)>0$ and $\mu(\Omega\cap (X\setminus C))>0$ for every non-empty open set $\Omega\subset [0,1]=X$, so that
$$
\overline{C}^\mu=X,\qquad \overline{X\setminus C}^\mu=X.
$$
However, $\overline{C\cap(X\setminus C)}^\mu=\overline{\emptyset}^\mu=\emptyset$, so $A:=C$ and $B:=X\setminus C$ provide a counterexample to (I).


问:
What is the geometrical meaning or implications of making variable substitution?
Let $x,y,z$ be real numbers and $$F(x,y,z)=ax^3+bx^3+cy^3+dx^2y+ex^2z+fxy^2+gxz^2+hy^2z+iyz^2+jxyz+k$$ $(a,b,c,\cdots$ are nonzero integers)

If I write $z=rx+sy$ where $r,s$ are real numbers and substitute $z$ in $F$, I will obtain a bivariate polynomial of degree 3 with real coefficients. I am curious to know the  geometrical implication  of such variable change? Is this a projection of the curve F on a 2-dimensional plane? Please let me know. Thanks.

答:
It may be helpful to first consider a lower dimensional example.
Suppose $\,F(x,y)\,$ is a real valued function of two real variables.
Then the equation $\,z=F(x,y)\,$ is a surface in three space.
If we make a substitution such as $\,y=rx+s,\,$ now $\,z=F(x,y)\,$
is a curve in three space which is a slice of the original surface.
That is, we find the intersection of the surface with a plane whose
result is a curve embedded in three space.

In your example of $\,F(x,y,z)\,$ and substitution $\, z=rx+sy,\,$ a very similar process happens. The equation $\,w=F(x,y,z)\,$ is a hypersurface in four space. After
the substitution we get the intersection of the hypersurface with a hyperplane. The
result is a two dimensional surface embedded in four space.


问:
Motivation for proof of $(((\phi \to \psi) \to \phi) \to \phi)$
I'm trying to solve this exercise.

If $\phi, \psi$ are propositional formulas, prove $\vdash (((\phi \to \psi) \to \phi) \to \phi)$.

I was wondering if there was a "simple" proof to this (I'm aware of the proof Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$?), and most importantly, how to motivate a proof to this exercise. (For me, the motivation is the most important bit! I already have a proof)

I asked such a question before, with a much simpler exercise, and the answer was "try it out in words first, and then write it symbolically!" which worked very well until coming across this example.  Currently I have a proof to this, which I brute forced; to me it feels extremely mechanical and non-replicable but for luck. I'm not sure if this is relevant to the question, but I'll attach it as a picture.

Edit: I actually don't know what the expression "proof system" means. Sorry; I was only told that this was "natural deduction", and the rules for $\land, \lnot, \rightarrow , \leftrightarrow, \perp$, etc were given.

答:
If $\phi$ is false, then $\phi \to \psi$ is true, but then since $(\phi \to \psi) \to \phi,$ from modus ponens $\phi$ would be true. So that much shows $\phi$ cannot be false, i.e. $\phi$ must be true.

This isn't a formal proof, but could be turned into one in some systems.


答: Given the last instance of the material conditional then the last rule to use must be an instance of the deduction theorem (or ($\to I$)), so for the solution a good strategy is to assume the antecedent: $(\phi \to \psi)\to \phi$, turn it into its equivalent disjunctive version: $\lnot(\phi \to \psi) \lor \phi$; then assume the two disjuncts to derive the formula $\phi$ on both sides and dischard the two assumptions $(\lor E)$. then you can make use of $(\to I)$, dischard the original assumption and derive the desired sequent.

Title: Motivation for proof of $(((\phi \to \psi) \to \phi) \to \phi)$

Body:
I'm trying to solve this exercise.

If $\phi, \psi$ are propositional formulas, prove $\vdash (((\phi \to \psi) \to \phi) \to \phi)$.

I was wondering if there was a "simple" proof to this (I'm aware of the proof Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$?), and most importantly, how to motivate a proof to this exercise. (For me, the motivation is the most important bit! I already have a proof)

I asked such a question before, with a much simpler exercise, and the answer was "try it out in words first, and then write it symbolically!" which worked very well until coming across this example.  Currently I have a proof to this, which I brute forced; to me it feels extremely mechanical and non-replicable but for luck. I'm not sure if this is relevant to the question, but I'll attach it as a picture.

Edit: I actually don't know what the expression "proof system" means. Sorry; I was only told that this was "natural deduction", and the rules for $\land, \lnot, \rightarrow , \leftrightarrow, \perp$, etc were given.


问:
Is the subspace of an eigenspace invariant?
Let $\phi$ be a diagonalizable endomorphism. Let $E_{\lambda}$ be an associated eigenspace and $F_{\lambda}$ is a subspace of $E_{\lambda}$, that is $F_{\lambda}  \subset E_{\lambda}$.

Is $F_{\lambda}$ $\phi$-invariant?

答: If $E_\lambda$ is an eigenspace of $\phi$ associated to an eigenvalue $\lambda$, then (by definition), $\phi(v)=\lambda v$ for all $v\in E_\lambda$.  So if $F_\lambda\subset E_\lambda$ is a subspace, then for all $u\in F_\lambda$, we have $\phi(u)=\lambda u\in F_\lambda$. Hence, $\phi(F_\lambda)\subset F_\lambda$.  So $F_\lambda$ is $\phi$-invariant.


问:
Which sets are by definiton countable/uncountable?
Are there some general rules on whether a set is countable or not? Thus far I only know about $\mathbb{R}$ and $\mathbb{C}$ being uncountable, so I wonder if there are possibly any theorems that ascertain the countability/uncountability for sets, i.e., the cartesian product of a set, the power set of a set... etc. Is there maybe an intuitive way to see that? Other than going by the definition of countability itself.

Thank you very much!

答:
A set non-empty set $X$ is countable iff there is a bijection $f:\mathbb{N}\to X$ iff there is an injection $i:X\to\mathbb{N}$ iff there is a surjection $s:\mathbb{N}\to X$.

Just as another piece of information, if $A$ is a set and $B$ is a proper subset of $A$ and there is a bijection $f:A\to B$, then $A$ is infinite.


问:
Finding argument of complex number and conversion into polar form
How do I find the argument of a complex number, for example $z = 3 + 4i$?

I know the polar form of $z$ is $r(\cos\theta + i\sin\theta)$ where $r$ is the modulus of $z$ ($\sqrt{3^2+4^2}$) which would leave me with $z = 5(\cos\theta+i\sin\theta)$, I'm just unsure how to deduce $\theta$ here.

The textbook says two things: $\theta = \arg(z)$ and $\tan\theta = b/a$, I'm not sure how these help.

答:
Question: Let $z=a+ib.$ Why does the following formula not hold in general?
$$\mathrm{Arg}(z)=\arctan\left(\frac ba\right)$$

Answer: Suppose that $b<0.$

Then $Z \text{ lies in quadrant }2\;\text{or}\;3.$

So $\;\mathrm{Arg}(z)>\frac{\pi}2\quad\text{or}\quad\mathrm{Arg}(z)<-\frac{\pi}2.$

But since $-\frac\pi2<\arctan\left(\frac ba\right)<\frac\pi2,\quad\mathrm{Arg}(z)$ cannot possibly equal $\arctan\left(\frac ba\right).$

There are several compact formulae for $\mathrm{Arg}(z)$, one of which is $$\mathrm{Arg}\left(a+ib\right)=\begin{cases}
 \quad\arccos \left(\frac a{\sqrt{a^2+b^2}}\right) \text{ if } b\geq0;\\
 -\arccos \left(\frac a{\sqrt{a^2+b^2}}\right) \text{ if } b<0.\end{cases}$$

However, it is easiest to just use an Argand diagram:

locate the angle between $OZ$ and its projection on the real axis;

evaluate it by taking $\arctan$ of the corresponding ratio of
lengths;

take the adjacent supplementary angle iff Re$(z)<0;$

assign a negative sign iff the resultant angle is directed clockwise from the positive real axis;

the desired principal argument is thus obtained.

For example, $$\mathrm{Arg}\left(-2-3i\right)\\=-\left[\pi-\arctan\left(\frac32\right)\right]\\=\arctan\left(\frac32\right)-\pi.$$


问:
Is relative growth rate affects the accuracy of Malthusian Model on Certain Interval?
So, namely i have $5$ models. And as we know the Malthusian model has form
$$\frac{dP}{dt} = kt$$
And the solution of the ODE above is
$$P(t) = ce^{kt}$$.
And the formula for Relative Growth Rate (RGR) from Wikipedia is
$$RGR = \frac{\ln(P(t_2)) - \ln(P(t_1))}{t_2-t_1}$$
Where $t_2$ and $t_1$ represent the year of the population. E.g. $t_2=2021$ and $t_1=2000$. We know we need $2$ points to determine the solution of Malthusian model, and the informations $t_2$ and $t_1$ have contribution on this.

And as i said before, i have 5 models means i use different $t_2$ and $t_1$ as their initial conditions. And i realized that the smaller the RGR value, the more accurate it is (at least on the nearest interval, suppose $t_2=2021$ and $t_1=2000$, the nearest interval is between $1980$ and $2025$).  It is evident from the plot data I have drawn (I will upload if it's necessary).  Can anyone confirm this thought, or is this just a coincidence?

Thanks in advance!

Edit:

In this edit section, i'd like to include the calculation of RGR and compare the results with the predicted data plot and the real (actual) data.

Here is my calculation of RGR on several models:

And the data plots:

$RGR=0.0195738$ :

$RGR=0.016577$

$RGR=0.0159$

For convenient reason, i only give 3 of them. And as you can see, the smaller RGR gives me better prediction on given interval (Might not accurate on another interval). Anyone can explain about this if this is just a conincidence or not? Thanks

答:
The wikipedia formula for RGR is actually the calculation of the constant $k$ when you know two points on the exponential curve.

The fact that the fit happens to look best when you use any two particular points to find $k$ is a coincidence. The right way to find the "best" $k$ is to fit a linear regression line to the logarithmic plot of the data. Then $k$ is the slope of that line.

That's what Excel does when you ask it for an "exponential trendline". It's what you should do in this example.


问:
Axiom of Choice question using the "If $f$ is surjective, then $f$ has a right inverse" proof as an example
I've tried to absorb several of the features of the Axiom of Choice, as well as its distinction between existential instantiation , but many deficits remain. The aim of this question is to address at least two of these deficits.

The implication

If $f: S \to T$  is a surjective function, then $f$ has a right inverse $g: T \to S$ such that $f \circ g = \text{Id}_T$, where $\text{Id}_T:=\{\langle t,t \rangle \ |\  t \in T\}$

requires the invocation of the Axiom of Choice (for the general case where $f$ is not also injective).

The need for the AoC occurs when we are trying to construct the function $g$ at the following stage:

Let $s$ be an element in $S$ such that $f(s)=t$. Let $X_s$ be the set of elements in $S$ that are mapped to $t$ through $f$. Equivalently, $f[X_s]=\{t\}$. Now, when it comes to determining how to construct $g$, we run into the issue of how to define $g(t)$ so that $f(g(t))=t$. Specifically, which $x \in X_s$ should $g(t)$ map to?

I believe there are two "bad scenarios" that need to be accounted for (but I have a question on the second scenario):

Firstly, let $X_s$ be uncountably infinite. Based on what I know, this is "bad" because $X_s$ can therefore not be related to the natural numbers $\omega$, which means I cannot select a minimum element in $X_s$. If I were able to select a minimal element from $X_s$, I would gain access to a definable rule (i.e. a rule that I could express using some finite combination of symbols) that I can use to describe $g$...thus necessarily circumventing the need for the Axiom of Choice.

Secondly, there are $3$ different possibilities when it comes to the number of $X_s$-like sets that arise in our function $f$. There could be

i) a finite number of $X_s$-like sets

ii) a countably infinite number of $X_s$-like sets

iii) an uncountably infinite number of $X_s$-like sets

For clarification, by "number of $X_s$-like sets", I mean the number of "instances of non-injectivity". Using "finite" as an example, suppose there are 4 elements in $T$ ($t_1,t_2,t_3, t_4$) that are "non-injectively mapped" into...i.e. $f[X']=\{t_1\}, f[X'']=\{t_2\}, f[X''']=\{t_3\}$, and $f[X'''']=\{t_4\}$.

I assume that any three of these possibilities (in conjunction with the first bad scenario) require the invocation of the Axiom of Choice...and perhaps the manner in which AoC is invoked is the same regardless of which three is encountered. Is this correct?

The equivalent variant of the Axiom of Choice that I have seen used in the above context of the surjectivity proof is the following:

Every relation contains a function having the same domain...symbolically: $\forall x \in T \exists y \in S \big[R(x,y)\big] \rightarrow \exists g \big[g:T \to S \land \forall x \in T [R(x,g(x))]\big]$

In essence, this is saying, "$f$ may not be injective, so $f^{-1}$ may not qualify as a function...but by AoC, we claim that there exists a function-variant of $f^{-1}$, call it $g$" (as a side note, this $g$ must be injective by default, because $f$ was defined as a function). Importantly, my understanding is that AoC does not give us a "rule" for the assignment choice $g$ makes when dealing with the $X_s$-like sets. This $g$ is then used to complete the surjectivity proof.

My confusion with this invocation of the AoC is that I cannot see why it is any different from the claim that the set $S^T$ exists. And thus, because $f:S \to T$ is surjective (which says something about the size of $T$ relative to $S$), there must necessarily be a $g \in S^T$ that satisfies our criteria for $f\circ g = \text{Id}_T$. In the same way as with the AoC, we are not given a "rule" for this $g$. But we know it exists.

So why do I need to Axiom of Choice here? Don't I effectively use the $g$ from $S^T$ as blindly as I do the $g$ from the AoC (blind in the sense that I do not know the specific assignment rule associated with $g$)?

答:
First of all, you're falling into the naive pitfall of separating into cases, or at least the wrong cases, when that is not necessary. There's no need to distinguish between countable and uncountable cases. Finite and infinite cases are the correct distinction, and even that could be argued to be wrong.

The correct cases is "$\sf ZF$ proves the existence of a choice function" and its negation. And yes, this would depend on $S$ and $T$. So for example, if $S$ is well-ordered (which can be so much more than just $\omega$), then we can choose the minimum from each $X_s$. Or, if $T$ is finite, then we can prove by induction that a choice function exists.

Starting to break into cases is natural, it's a good approach of sorting out "what do I know is possible", but it's not always the right one, and it's easy to make mistakes that will just pull you deeper into this knot of not understanding.

The difference between using $\sf AC$ and using Existential Instantiation is obvious: EI can be used finitely many times. So if $T$ is not finite1, we cannot repeatedly use EI to exhaust the choices necessary for the inverse.

The axiom of choice, however, simply posits the existence of functions with the desired property: since $f$ is surjective, for all $t\in T$, $f^{-1}(t)$ is non-empty, so by $\sf AC$ a choice function from $\{f^{-1}(t)\mid t\in T\}$ exists. Now we use EI once and we have such a choice function in our hands.

Your question has been addressed before on the site. The existence of functions, or the existence of a relation $R\subseteq T\times S$ whose domain is $T$ (in this case, $f^{-1}$) is not enough to conclude that a function with the wanted properties exists. To prove it exists, you need to be able to use the axioms for a proof. The obvious candidate is Separation or Replacement. But to invoke those you need to have a defining property (which may or may not already be a functional one). So, what is that property?

If $S=\{x,y\}$ and $T=\{0\}$, consider $f(x)=f(y)=0$. Now to construct $g$ we need to choose an element for which $g$ maps $0$. Is it $x$? Is it $y$? Both options are good. But there's no preference for either one here, so we have to "leave it to chance": let the EI work its magic and choose one of them. But if we have to make infinitely many choices, how can you do that without already appealing to the axiom of choice?

When I say finite here, I mean meta-finite. That is, finite in the sense of the meta-theory, where the proofs live. Internally it might be that there are non-standard integers, and so "internally finite" could be longer than any valid proof. This is a subtle point when it comes to "finitely many choices" in $\sf ZF$, and it is the difference between proving internally that "$T$ is finite implies $f\colon S\to T$ has an inverse" and constructing that inverse by repeated use of EI.


答:
Case 1

Statement 1: All B are M.

Statement 2: All P are M.

Conclusion I: Some P are B.

Conclusion II: All P are not B.

Counterexample to I): All Triangles are Plane figures. All Quadrilateral are Plane figures.

Therefore: Some Quadrilateral are Triangles. The conclusion is False.

Counterexample to II): All multiples of 2 are integers. All multiple of 3 are integers.

Therefore: All multiple of 3 is not a multiple of 2. The conclusion is False (consider 6).

We are speaking of logical (or: deductive) validity: an argument is valid by virtue of the "logical form".

The "factual" truth of the purported conclusion does not mean that the argument is formally valid.

What happens with Case 2 ?

We have to note that option (D) is different from the previous one. It reads "Both conclusion I and II are correct."

But I and II are contradictory: thus, they cannot be both true.

Thus, if (C) is the correct answer, the issue is: I or II? "Either" is usually used in logic in the inclusive form. But we have already noted that they cannot be both true.

Thus, "Either I or II" amounts to choosing between (A) and (B).

Title: Similar questions in GATE 2021 and different answer keys | Answer Key Challenge

Body:
GATE CSE 2021 Set 1 | GA Question: 9

Given below are two statements 1 and 2, and two conclusions I and II

Statement 1: All bacteria are microorganisms.

Statement 2: All pathogens are microorganisms.

Conclusion I: Some pathogens are bacteria.

Conclusion II: All pathogens are not bacteria.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

(A) Only conclusion I is correct

(B) Only conclusion II is correct

(C) Either conclusion I or II is correct

(D) Neither conclusion I nor II is correct

My attempt:

I have to find counter example,

Case-1 is counter example of Conclusion II, because of all pathogens are bacteria in this diagram.
Conclusion I would be true in this case, because all so also some.
(So, Conclusion I is true and II is false.).

Case-2 is counter example of Conclusion I, because no pathogens are bacteria.
Conclusion II would be true in this case, because all so also some.
(So, Conclusion II is false and II is true.).

Case-3: Conclusion I is true, and Conclusion II is false as some pathogens are already bacteria.

Case-4: Conclusion I is true, and Conclusion II is false as some pathogens are already bacteria.

So, I found counter example for both Conclusion. Hence, option (D) is true (Official Key given by GATE 2021.).

I've read this method for finding counter examples for such questions.

Doubt here, is as both given Conclusions are negation of each other. That is,

∃x(p(x) ∧ b(x)) = ¬∀x (p(x) → ¬b(x))

∀x (p(x) → ¬b(x)) = ¬∃x(p(x) ∧ b(x))

Peoples are debating this question and saying that the correct answer should be option (C) Either conclusion I or II is correct.

A exact similar question appeared in the GATE EC branch, which has same above logic.

GATE EC 2021 | GA Question: 6 :

Given below are two statements and two conclusions.

Statement 1: All purple are green.

Statement 2: All black are green.

Conclusion I: Some black are purple.

Conclusion II: No black is purple.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

(A) Only conclusion I is correct

(B) Only conclusion II is correct

(C) Either conclusion I or II is correct

(D) Both conclusion I and II are correct

Here, as above my attempt, we can find counter examples for the both the statements, so both (D) Both conclusion I and II are correct.

But, this time official key is given (C) Either conclusion I or II is correct.

Note these are two different branches of Engineering (Computer Science and Engineering, Electronics and Communication Engineering). Professors who designed these questions for GATE 2021, should not be same person (but, could be as these questions are from General Aptitude which is common (but not same questions) in all branches of GATE Papers).

My Question

What is correct approach to solve such questions and which answer keys are correct which are wrong ? If both keys are correct respective their questions then what is possible difference between these questions ?

Thank you,

Update :

Final answer keys given by GATE official :

For first question (GATE CSE 2021 Set 1 | GA Question: 9) : Both
(C) or (D)

For second question (GATE EC 2021 | GA Question: 6) : Only (C)

Now, why they have given option (D) as answer for the first question
!?


问:
Derivative chainrule on khanacadamy ignoring some terms
I watched the chain rule series on khanacademy.org and decided to do the "questions". One of the questions is:

Let $y = \sin(6x^2−4x−1+3x^{−1}−5x^{−2})$

$dy/dx=?$

The answer is $dy/dx=(\cos(6x^2−4x−1+3x^{−1}−5x^{−2}))(12x−4)$

Since

$dy/dx [f(g(x))] = {\rm D}f(g(x))*{\rm D}g(x)$

I figure that it should be

$dy/dx = \cos(6x^2−4x−1+3x^{−1}−5x^{−2})(12x-4-3x^{-2}+10x^{-3})$

Why aren't they deriving the negative exponent terms in the inner function "$g(x)$"?

Thank you in advance for your help!

答:
It seems that there is some error in presenting the solution. That being said, the answer that you have provided is correct. The following is a step-by-step solution for future users.

$$f(x) = \sin(6x^2−4x−1+3x^{−1}−5x^{−2})$$

In order to find the derivative, we must use the chain rule. Therefore:
$$f'(g(x)) * g'(x)$$ Where $f(x)$ is $\sin(x)$ and $g(x)$ is $6x^2−4x−1+3x^{−1}−5x^{−2}$. Therefore, the solution would be:

$$\begin{align}
\cos (6x^2−4x−1+3x^{−1}−5x^{−2}) * 12x-4+3x^{-2}-5x^{-3}
\end{align}$$

because the derivative of  $\sin(x)$ is $\cos(x)$ and derivative of $g(x)$ can be found through the general forumla $\frac{dy}{dx} (x^n) = nx^{n-1}$.


答:
For the Cartesian product of two sets, you basically multiply the cardinalities.  In the case of infinite cardinality, you get the larger cardinality back.  I have jumped up to infinite cardinality, since the distinction between countable and  uncountable is really just the beginning of the story.

This is just a special case of Cantor's "infinite paradise", and its cardinal arithmetic.

For instance, in the case of power sets, the cardinality of the power set of a set is always greater than the cardinality of the original set:  $|\mathcal P(A)|\gt|A|$, by Cantor's famous diagonal argument.  Particularly salient is the proof that the real numbers are not countable, since if you claim you could put them in a list, it is possible by the diagonal argument to construct a real not in the list by changing the diagonal entry on each element.

Title: Which sets are by definiton countable/uncountable?

Body:
Are there some general rules on whether a set is countable or not? Thus far I only know about $\mathbb{R}$ and $\mathbb{C}$ being uncountable, so I wonder if there are possibly any theorems that ascertain the countability/uncountability for sets, i.e., the cartesian product of a set, the power set of a set... etc. Is there maybe an intuitive way to see that? Other than going by the definition of countability itself.

Thank you very much!


问:
If $a(u,u) \leq \liminf a(u_n, u_n)$, is $a(u,u) \leq \liminf a(u_n, u_{n-1})$?
Let $a\colon H \times H \to \mathbb{R}$ be some bilinear form and $H$ is a Hilbert space. If it is weak lsc, i.e.,
$$a(u,u) \leq \liminf a(u_n, u_n)$$
for a sequence $u_n \rightharpoonup u$ in $H$, then is $$a(u,u) \leq \liminf a(u_n, u_{n-1})$$
true? Or is there some name for this kind of property?

答:
No, this is not true. Take $a$ to be the inner product and consider the sequence
$$
e_1, -e_1, e_2, -e_2, \ldots,
$$
where $(e_i)$ is an infinite orthonormal system.

I do not think that many bilinear forms $a$ have this property (unless there is some compactness hidden in $a$).


问:
Solve transcendental equation / maximum likelihood estimation
Consider $(X_1,\dots X_n)$ with $$\mathbb P[X=k]=\begin{cases}p_0,&k=0\\\frac{(1-p_0)\lambda^k}{k!(e^\lambda)-1},&k\geq1\end{cases}$$

I want to determine the maximum likelihood estimator of $\lambda$

I arrive at $$\frac{\partial}{\partial \lambda}\log L((p_0,\lambda))=\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)\big(\frac{X_i}{\lambda}-\frac{e^\lambda}{e^\lambda-1}\big)\stackrel{!}{=}0$$

Which is equivalent to
$$\frac{1}{\lambda}\frac{e^\lambda-1}{e^\lambda}=\frac{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)}{\sum_{i=1}^{n}1_{\{X_i\geq 1\}}(X_i)X_i}$$

How do I solve this equation? When can it be solved? (If I didn't make any mistakes this is a transcendental equation)

答:
your distribution function is well known, it is a "Zero modified Poisson distribution" that is a generalization of the "Zero Truncated Poisson distribution".

For the "Zero Truncated Poisson distribution", there is no a closed form for $\hat{\lambda}_{MLE}$, which is the solution of the following equation

$$\overline{X}_n=\frac{\lambda}{1-e^{-\lambda}}$$

Estimators of $\lambda$ for the zero truncated Poisson distribution are studied in


问:
How do I compute $\Delta_x\frac{1}{|x-y|}$ when $x$ has arbitrary $n$ components?
How do I compute $\Delta_x\frac{1}{|x|}$ when $x$ has arbitrary $n$ components?

Is there any good way to simply this?

答:
Hint: after making the substitution $\boldsymbol r := \boldsymbol x - \boldsymbol y $, switch to hyperspherical coordinates on $\mathbb R^n\setminus\{0\}$, for example using the following convention:
$$
\begin{split}
r_{1}&=\rho \cos \theta \prod_{j=1}^{n-2} \cos \varphi_{j} \\
r_{2}&=\rho \sin \theta \prod_{j=1}^{n-2} \cos \varphi_{j} \\
r_{k}&=\rho \sin \varphi_{k-2} \prod_{j=k-1}^{n-2} \cos \varphi_{j}, \quad 3 \leq k \leq n,
\end{split}
$$
with $r \in (0,\infty)$, $\theta \in (0,2\pi)$, and $\varphi_j \in (-\pi/2,\pi/2)$. Write the laplacian $\Delta_\boldsymbol r$ in these coordinates, then apply it to the spherically-symmetric function $(r,\theta,\dots)\mapsto 1/r$.


问:
Proof of Astroid?
How can I prove that an astroid is an envelope of all line segments of length 1 from the x-axis to the y-axis?

I read one proof of this online at the link Link
but I don't understand how this proof works.

Therefore, it would be much appreciated if someone could show me another proof or make the online proof more understandable. For example, for the first step, why is it x**/t^2**+y/(a^2-t^2)? I mean, I get that this is because of the Pythagorean theorem, but why are these the denominators of x and y?

答:
For a reverse proof, compute the tangent of the astroid, find the intersection with the axis and show they distance is 1:

Start with the parametric equation $(x,y)=(\cos^3\theta, \sin^3\theta)$, whose derivative relative to $\theta$ is $(x',y')=(3 \cos^2\theta \sin\theta, -3 \sin^3\theta \cos\theta)$. The tangent at $\theta$ has equation $\frac {x - \cos^3\theta} {3 \cos^2\theta \sin\theta} = \frac {x - \sin^3\theta} {-3 \cos\theta \sin^2\theta}$, which simplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$. The tangent cuts the axis in $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$. As $PQ=1$, this proves that the tangent of the astroid at any point is a segment of length 1 from the x-axis to the y-axis.

For the constructive proof, take the segment computes its equation $y=F_\theta(x)$, and, accordingly to the books,  solve the system $y=\frac {\partial F_\theta} {\partial \theta} (x)$, which express that you choose the point on the segment in such y way that the segment is tangent to the envelope.

Take $P=(0, \cos\theta)$ and $Q=(\sin\theta, 0)$, the segment equation is $\frac x {\cos\theta} + \frac y {\sin\theta}=1$, which symplifies in $x \cos\theta + y \sin\theta= \cos\theta \sin\theta$ (as above). Derivating (relative to $\theta$!) leads to $-x \sin\theta + y \cos\theta= \cos^2\theta - \sin^2\theta$. So, the equation of the envelope is the solution of the system:

$\begin{cases} +x \cos\theta + y \sin\theta &= \cos\theta \sin\theta\\ -x \sin\theta + y \cos\theta &= \cos^2\theta - \sin^2\theta \end{cases}$

To solve the system multiply the first line by $\sin\theta$, the second by $\cos\theta$ and add to eliminate $x$, which leaves $y=\sin^3\theta$. Then multiply the first line by $\cos\theta$, the second by $\sin\theta$ and add to eliminate $y$, which leaves $x=\cos^3\theta$. So the envelope parametric equation is $(x,y)=(\cos^3\theta, \sin^3\theta)$. You can eliminate the parameter by replacing $(\cos\theta, \sin\theta)=(x^\frac 1 3, y^\frac 1 3)$ in $\cos^2 \theta+\sin^2\theta=1$ to get the familiar $x^\frac 2 3 + y^\frac 2 3 =1$.

I may seem quick with mental calculations, but a lot of simplifications will appear a long the way, as long as you keep your formulas as symmetric as possible.

Title: Proof of Astroid?


问:
K-derivations and Lie algebras
I had a question regarding a statement that was on the Wikipedia page for Derivations. It states that $\text{Der}_k(A)$ is a Lie algebra with the Lie bracket being the commutator. Here $\text{Der}_k(A)$ is the $K$-module consisting of $K$ linear mappings $D: A \rightarrow A$ that satisfies the Leibniz rule. Also, here $A$ is a commutative $K$-algebra.

If I wanted to show this for my self, would I just need to show that $[ \cdot , \cdot]$ is a bilinear mapping that is skew symmetric and satisfies the Jacobi identity?

For instance for skew symmetry, would it be as simple as taking a derivation $D \in \text{Der}_k(A)$ and stating that $[D,D] = DD - DD = 0?

Also, would I have to show that the commutator of 2 derivations is a derivation? If so, I'm not sure how to conclude that.

答:
Yes, obviously $[D,D]=DD-DD=0$ is satisfied. For the Leibniz rule, we have for all $x,y\in A$,
\begin{align*}
[D_1,D_2](x\cdot y) & = (D_1D_2)(x\cdot y)-(D_2D_1)(x\cdot y)\\
 & = D_1(D_2(x)\cdot y+x\cdot D_2(y))-  D_2(D_1(x)\cdot y+x\cdot D_1(y))\\
 & = (D_1D_2(x))\cdot y +x\cdot (D_1D_2(y)))+D_2(x)\cdot D_1(y) + D_1(x)\cdot D_2(y)\\
 & - (D_2D_1(x))\cdot y-x\cdot (D_2D_1(y))-D_1(x)\cdot D_2(y)-D_2(x)\cdot D_1(y) \\
 & = [D_1,D_2](x)\cdot y+x\cdot [D_1,D_2](y).
\end{align*}

Here we have added and subtracted the terms $D_1(x)\cdot D_2(y)$ and $D_2(x)\cdot D_1(y)$.

So we have shown that for $D_1,D_2 \in {\rm Der}(A)$ we have $[D_1,D_2]\in {\rm Der}(A)$, so that ${\rm Der}(A)$ is a Lie subalgebra of $\mathfrak{gl}(A)$. Moreover we haven't used that $A$ is commutative, or associative.

Title: K-derivations and Lie algebras


问:
expectation on different steps
$f_1,f_2,f_3: \mathbb R\to \mathbb R$, starting from a deterministic $X_0=x\in \mathbb R$, let $X_1=f_1(X_0)$ with probability $p_1=1/3$, $X_1=f_2(X_0)$ with probability $p_2=1/6$, $X_1=f_3(X_0)$ with probability $p_3=1/2$.

I observe $X_1,\dots, X_{10}$.

Now, starting from $X_{10}$ which I understand not deterministic, let  $X_{11}=f_1(X_{10})$ with probability $q_1=1/2$, $X_{11}=f_2(X_{10})$ with probability $q_2=1/5$, $X_{11}=f_3(X_{10})$ with probability $q_3=3/10$.

I see that $\mathbb E[X_1\mid X_0=x]=\sum\limits_{k=1}^{3}p_kf_k(x)$? I am not sure, as after every 10 steps $p_i$ changes to $q_i$..making me confuse about the expectation. Thanks for any help.

答:
Well, I mean, the situation was always going to be a little complicated. Let's define $f \circ g(x) = f(g(x))$ (the composition of functions). Let's first try to work out the situation when $X_0$ is given, then go for the general $\mathbb E[f(X_n) | X_{n-k}]$ for $k \leq n$.

First, $\mathbb E[f(X_1) | X_0 =a]$. That is $\sum_{k=1}^3 p_k f(f_k(a))$, since $X_1$ is $f_i(a)$ with probability $p_i$, $i=1,2,3$.

Now what about $\mathbb E[X_2 | X_0 = a]$? Well, for $i,j = 1,2,3$ , with probability $p_ip_j$, we have $X_2 = (f_i \circ f_j)(a)$. Hence, $$
\mathbb E[X_2 | X_0 = a] = \sum_{i,j = 1}^3 p_ip_j( f \circ f_i \circ f_j)(a)
$$

in similar fashion, you can see that
$$
\mathbb E[X_{10} | X_0 = a] = \sum_{k_1,...,k_{10} = 1}^3 p_{k_1}...p_{k_{10}} f \circ f_{k_1} \circ ... f_{k_{10}}(a)
$$

once we go to a number $11 \leq n \leq 20$ we have to add the $q_{i}$ into play. But it is easy to see that for example :
$$
\mathbb E[f(X_{17}) | X_0 = a] = \sum_{k_1,...,k_{17} = 1}^{3} q_{k_{17}}...q_{k_{11}}p_{k_{10}} ... p_{k_1} (f \circ f_{k_{17}} \circ ... \circ f_{k_1})(a)
$$

now what happens if we go to a very large quantity? Say $n = 46$. Then for $1 \to 10, 21 \to 30 , 41 \to 46$ we use $p$, for $11 \to 20, 31 \to 40$ we use $q$. So , we have :
$$
\mathbb E[X_{46} | X_{0} = a] = \sum_{k_{1},...,k_{46} = 1}^{3} p_{k_{46}}p_{k_{45}}...p_{k_{41}}p_{k_{30}}...p_{k_{21}}p_{k_{10}}...p_{k_1} q_{k_{40}}...q_{k_{31}}q_{k_{20}}...q_{k_{11}} (f \circ f_{k_{46}} \circ ... \circ f_{k_1})(a)
$$

so we are slowly warming to our task. In particular, let us see how this works. What does not change is the composition of functions, which has a clear pattern. Instead, we need to look at when $p$ is used and when $q$ is used. Then again, isn't that pattern clear? See the remainders of the numbers which sit beneath a $p$ when divided by $20$ : it's all between $1$ and $10$. On the other hand, the numbers which sit beneath a $q$ have remainders between $11$ and $19$, and $0$. Let $i \% 20$ be the remainder upon division by $20$ (note : the percentage sign may seem unusuual, but in a lot of programming languages it is considered standard notation for the remainder), and let $P = \{1,2,...,10\}$ and $Q= \{0,11,12,...,19\}$. Then we have the general expression :

$$
\mathbb E[f(X_n) | X_0 = a] = \sum_{k_i=1 , 1 \leq i \leq n}^3 \left(\prod_{1 \leq j \leq n, j \% 20 \in P} p_{k_j}\right) \left(\prod_{1 \leq l \leq n, l \% 20 \in Q} q_{k_l}\right)
(f \circ f_{k_{n}} \circ ... \circ f_{k_1})(a)
$$

Generalizing this is not going to be very difficult : indeed, once we have the remainders written down we are in great shape.

For example, let's take $\mathbb E[X_{54} | X_{16} = a]$. Then for the transitions at times $21 \to 30,41 \to 50$ we have to use $p$ and $17,18,19,20,31\to 40, 51 \to 54$ we have to use $q$. With this in mind, when we index $l_1,...,l_{k}$ in the sum, then we must actually look at whether $(n-k)+1, (n-k)+2,...,(n-k)+k$ belong in $P$ or $Q$ and perform accordingly.

That kind of analysis leads to the following expression for $n \geq k$ :
$$
\mathbb E[f(X_{n}) | X_{n-k}] = \sum_{l_i = 1 , 1 \leq i \leq k}^3 \left(\prod_{1 \leq j \leq k , ((n-k)+j) \% 20 \in P} p_{l_{j}}\right)\left(\prod_{1 \leq m \leq k , ((n-k)+m)  \% 20 \in Q} q_{l_{m}}\right)(f \circ f_{l_k} ... \circ f_{l_1})(a)
$$

In general, this is a difficult expression to work with because compositions of functions are difficult to understand. However, this is comfortably the most general one can get.

Note : one can write membership in $P,Q$ using the greatest integer function. For $N$ let $L = \left[\frac N{20}\right]$. Then $N - 20L = N \% 20$. With that, we note that $N \% 20 \in P$ means that $1 \leq N - 20\left[\frac N{20}\right] \leq 10$, and if $N \% 20 \in Q$ then note that $10 \leq N - 20 \left[\frac N{20}\right] \leq 19$ or $ N - 20 \left[\frac N{20}\right]  = 0$. Thus we can write the conditions that way as well.

Title: expectation on different steps


问:
Finding limits of integration for triple integral
I'm trying to get the volumne of the solid bounded by the surfaces $z=x^2+y^2$, $z=3(x^2+y^2)$ and $z=1$, $z=3$.
I'm doing this example but I can't get the limits of integration. Can you give me a hand? I know how the graph looks like but I can't figure it out..

EDIT: I've tried using polar coordinates, taking the disk from z=1 to z=3. So, I can see that r (in polar) changes with the two equations (the radius gets bigger). And I can't find the limits for r. I know it can be simple but I'm new into polar and change of variables. (I get that r should be in polar coordinates, between 3r^2 and r^2... but I think i've made a nonsense)

答:
We are finding volume bound between paraboloid surfaces $z = x^2+y^2, z = 3(x^2+y^2)$ and between $1 \leq z \leq 3$.

So in cylindrical coordinates, we have

$\displaystyle \frac{z}{3} \leq r^2 \leq z$ and $1 \leq z \leq 3$.

Cross section area of the solid at any given $z$ is the difference of cross section areas of the given paraboloids.

So, $V = \displaystyle \int_1^3\pi \ (z - \frac{z}{3}) \ dz$

If you are doing triple integral, use $\sqrt{z/3} \leq r \leq \sqrt z$.


问:
Equation with $ \sqrt[5]{}$
How to show that this equation $ \sqrt[5]{20-2x} + \sqrt[5]{7-x}+\sqrt[5]{3x+5}=2$ have 3 solutions 9, -6 and -25

Wolframsays  the equation have no solutions !

答:
The claim that $9,-6,-25$ are the solutions of the equation does not hold.

Substituting for example $x=9$ into the equation one obtains:
$$
(2)^\frac15+(-2)^\frac15+(2^5)^\frac15\ne2
$$
because
$$
(2)^\frac15+(-2)^\frac15\ne0
$$
for any pair of the fifth order roots of $2$ and $-2$ belonging to the same branch.

Title: Equation with $ \sqrt[5]{}$


问:
Fourier expression for infinite recurring binary sequence
We have infinite binary sequences of type
$$\langle g_n \rangle_{j=4}=\{0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,...\} \,;\, n\to\infty$$
where $j$ indicates the length of a period.

we try to express them with a Fourier in a shape similar the spectrum:

$$G(x)=\sum_{k=1}^\infty a_k\,e^{(i\,2\pi\,k\,x / j)}$$

None of us is good in Fourier at all. May someone help us with the right expression.

Many thanks

答:
The inverse Fourier in its most elegant form is for such sequences:

$$G(x)=\sum_{k=-j}^j \frac{2 \pi}{j}\,e^{(i\,2\pi\,(k-1)\,x / j)}$$

while $i=\sqrt{-1}$ and angular frequency:

$$\omega_j=\frac{2 \pi}{j}$$


问:
Saddle-node behaviour of planar systems
I'm teaching a course on ODEs using Lawrence Perko's book. Having assigned Q3(c) of section 2.11, and both working out the solution using theorems in the sections and consulting the solution manual, I get an answer that seems to conflict with other calculations. The details are as follows:

The system is:
$$
\dot{x} = y, \qquad \dot{y} = x^4 + xy.
$$

According to a Thm 2.11.3 in Perko (see screenshot of Thms 2.11.2 and 2.11.3 below), taken from Andronov--Leontovich--Gordon--Maier, the critical point $(0,0)$ is a saddle-node, "because" the lowest order monomial of the form $x^k$ in the $\dot{y}$ equation is of even degree, the lowest order monomial of the form $x^ny$ has $n = 1$, with coefficient $b_1 = 1 \not= 0$, and $4/2 > 1$. The solution manual concurs.

Now from the equation, $\dot{x} > 0$ on $\{y > 0\}$, $\dot{x} < 0$ on $\{y < 0\}$. Moreover $\dot{y} = xy + O(|x,y|^4)$, which means $\dot{y} > 0$ in the 1st and 3rd quadrants, and $\dot{y} < 0$ in the 2nd and 4th quadrants close to $(0,0)$. And any way I draw this, it looks like a cusp near the critical point.

Finally, winplot confirms this: .

So my question is, is it known that Thm 2.11.3 is in fact not a theorem, or have I missed something?

答:
Let's take the scalar equation
$$
\ddot x=x\dot x+x^4
$$
and divide it up into a slightly different first order system with $v=\dot x-\frac12x^2$ so that
\begin{align}
\dot x&=v+\frac12x^2\\
\dot v&=x^4
\end{align}
A minimal perturbation of the second equation to $\dot v=x^4-\varepsilon$ results in two close-by critical points $(\pm\sqrt[4]ε,-\frac12\sqrtε)$ with Jacobians
$$
J=\pmatrix{x&1\\4x^3&0}=\pmatrix{\pm\sqrt[4]ε&1\\\pm4\sqrt[4]ε^3&0}
$$
with eigenvalues $\frac{x}2(1\pm\sqrt{1+16x})$, so that for $ε<2^{-16}$ one eigenvalue is around $x=\pm\sqrt[4]ε$ and the other $-4x^2=-4\sqrtε$. For $ε>2^{-16}$ the eigenvalues for the left point become a complex conjugate pair with real part $-\frac12\sqrt[4]ε$.

This gives one saddle point at $(+\sqrt[4]ε,-\frac12\sqrtε)$ and one attracting node at $(-\sqrt[4]ε,-\frac12\sqrtε)$. For $ε\to 0$ they melt together, giving a saddle-node.

Title: Saddle-node behaviour of planar systems


问:
Applying Cantor's diagonal argument
I understand how Cantor's diagonal argument can be used to prove that the real numbers are uncountable. But I should be able to use this same argument to prove two additional claims: (1) that there is no bijection $X \to \mathcal{P}(X)$ and (2) that there are arbitrarily large cardinal numbers.

I don't understand how to apply the diagonal argument, or a variation of its logic, in this case. I know how to prove Cantor's theorem. I suppose there exists a surjection $f: X \to P(X)$, consider the set $B = \{x \in X \mid x \not \in f(x)\}$, argue that there has to exist $y \in X$ such that $f(y) = b$, and then show that $y \in B$ and $y \not \in B$ both give contradictions. To show that there are arbitrarily large cardinals, I take $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$, $\mathcal{P}(\mathcal{P}(\mathbb{N}))$, and continue ad infinitum.

Is there a way that I'm silently using the diagonal argument here? Or is there a different way to think of these arguments?

答:
Your solution to (2) is wrong. To show that there are arbitrarily large natural numbers divisible by $3$ we don't start with $0$, apply the successor function (or any increasing function), and hope for the best. We show that if $n$ is a number then one of $n+1,n+2,n+3$ is divisible by $3$ and all are strictly larger than $n$.

The same here. You're not supposed to iterate power sets from $\Bbb N$. How do you know that the sequence doesn't have an upper bound which is the largest cardinal? Instead, show that if $X$ is any set whatsoever, there is another set which has a strictly larger cardinality. For example, apply (1), but note that "there is no surjection from $X$ to $Y$" does not necessarily mean that $Y$ is larger, we also need to exhibit an injection from $X$ into $Y$.

Title: Applying Cantor's diagonal argument


问:
Finding out the argument a complex number
So this is something I'm still having issues with. Lets say I have the following:
$$z^6 = -8$$
Now if I take $\frac{1}{6}$ power on both sides, I get $z = \sqrt{2}i$. Now, if I look at the point that this is in, I mean the value for the real number is $0$ so it must be at angle $\frac{\pi}{2}$ right? For some reason though, it should be the $arg(z) = \frac{\pi}{6} + \frac{\pi}{3}k$ which I don't quite understand. Would appreciate some help, thanks in advance!

答:
Yes, your calculation is correct, as we can see by calculating: $(\sqrt(2) i)^6 = (-2)^3 = -8$

However, when we calculate roots of complex numbers the number of roots, there are always $n$ $n$th roots (This result is related to the Fundamental Theorem of Algebra). That is there are $2$ square roots, $3$ cube roots and so on.

In thus case, there are 6 roots and you haver successfully calculated one of them. In particluar, $\sqrt{2} i $ is a root lying on the X-axis with argyment $\pi /2 $.

The other roots all lie at intervals of $\pi /3$ or $60$ degrees from this root.

To calculate them all, note thta we are trying to solve

$z^6 = -8$

Now, $ z  =  (-8)^{\frac{1}{6}}  =  (-1)^{1/6} 8^{1/6} = \sqrt{2} \cdot (e^{(2n+1)\pi i})^{1/6} = \sqrt{2} \cdot e^{i \frac{(2n+1) \pi}{6}} $

Thus, $\arg (z) = {(2n + 1) \pi}{6} = \frac{\pi}{6} + n \cdot \frac{\pi}{3}$ as required.

Note that, we can set $n = 0,1, \dots , 5$ and get the required solutions:

$z = \sqrt{2} e^{i \frac{\pi}{6}}, \sqrt{2} e^{ i \frac{ 3\pi }{6} }, \sqrt{2} e^{i \frac{5\pi}{6}}, \sqrt{2} e^{i \frac{9\pi}{6}}, - \sqrt{2} i, \sqrt{2} e^{i \frac{11\pi}{6}}$

You can see the $6$ sextic roots of $(-1)$ graphically using the complex plane:


问:
An application of the Mean Value Theorem in lemma 2.1 of Gidas, Ni and Nirenberg

Let $\Omega$ be a bounded domain with $C^2$ boundary and let $x_0 \in \partial \Omega$ such that $\nu_1(x_0) > 0$, $\nu$ outer normal unit vector. For $\varepsilon > 0$ let $\Omega_\varepsilon = \Omega \cap B_\varepsilon(x_0)$, $u \in C^2(\overline{\Omega_\varepsilon})$, $u > 0$, $u = 0$ on $\partial \Omega \cap B_\varepsilon(x_0)$. Then $\exists \delta > 0$ such that $u_1 < 0$ in $\Omega \cap B_\delta(x_0)$

This is a lemma in the Famous Paper of Gidas, Ni and Nirenberg about symmetry results. The beginning of the proof is as follows:

Denote $S = \partial \Omega \cap B_\varepsilon(x_0)$. We can assume that $\nu_1 > 0$ on $S$, decreasing $\varepsilon$ if necessary. Since $u > 0$ in $\Omega$ and $u = 0$ on $S$, then $u_1 \leq 0$ on $S$. Indeed, if this was not the case, it would mean that $u$ arrives at $S$ ascending, that is, increasing towards $S$, which is absurd.

We proceed by contradiction.

If the lemma was false, there would be a sequence $(x_j)$ such that $x_j \to x_0$ and $u_1(x_j) \geq 0$. By continuity, $u_1(x_0) \geq 0$. On the other hand, we already know that $u_1 \leq 0$ on $S$, and thus
$$
u_1(x_0) = 0
$$
Moreover, if $j$ is sufficiently large, the interval in the $e_1$ direction going from $x_j$ to $\partial \Omega$ hits $S$ at a point $y_j$, where we know that $u_1(y_j) \leq 0$. It is clear that $u_1(y_j) \nearrow 0 \swarrow u_1(x_j)$, so $u_1(x_0) = 0$.

By the Mean Value Theorem,
$$
u_{11}(x_0) = 0.
$$

Why does this claim hold? How is the Mean Value Theorem used?

I'm really clueless.

Thanks in advance and kind regards.

EDIT:

Precise reference:

B. Gidas. Wei Ming Ni. L. Nirenberg. "Symmetry and related properties via the maximum principle." Comm. Math. Phys. 68 (3) 209 - 243, 1979.

https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-68/issue-3/Symmetry-and-related-properties-via-the-maximum-principle/cmp/1103905359.full?tab=ArticleLink

答:
This answer only addresses the claim $u_{11}(x_0)$, not the main claim.

Let us say that we already know $u_1(x_0) = 0$. There are three possibilities:

If $u_{11}(x_0) < 0$, then (by continuity) $u_{11} < 0$ on some neighborhood. In particular, for the one-variable function $g(t) = u(x_0 + t \mathbf{e_1})$ we have that $g''(t) < 0$. This means that $g'(t)$ is descreasing, but since $g'(0) = u_1(x_0) = 0$, we have $g'(t) > 0$ for $t<0$, which means that $g$ itself is increasing. But this is a contradiction with $u > 0$ inside $\Omega$ and $u = 0$ on $\partial \Omega$.

If $u_{11}(x_0)$, then $u_{11} > 0$ on some neighborhood. For any $y \in \partial \Omega$ close to $x_0$ the function $g(y+t \mathbf{e_1})$ now has $g'' > 0$ (for small $t < 0$). Hence, $g'$ is increasing, but $g'(0) \le 0$, so $g' < 0$ for $t < 0$, and this is exactly the claim of the theorem.

Finally, it is possible that $u_{11}(x_0) = 0$.


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
The square of this is $2x+2\sqrt{x^2-y^2}$.  One could find this.

Regarding $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$, the modulus of this number $35+18\sqrt{-3}$ is 2197, which is $13^3$.  This means that we are looking for a number whose cube is $35+18\sqrt{-3}$.  An example of such a number is $3.5 + 0.5\sqrt{-3}$, which we cube to get the expression of $35+18\sqrt{-3}$.

The conjucate gives $3.5-0.5\sqrt{-3}$ gives a cube of $35-18\sqrt{-3}$.

The process here is to multiply the two numbers together, to establish the modulus.  Here, the product gives $2197$, a known cube.  In general, it needs to be a cube to fall on an integer real.  One could also divide the two numbers, in the form $\sqrt[3]{35 + 18i\sqrt{3}} / \sqrt[3]{35 - 18i\sqrt{3}}$, to get a value on the unit circle.

However, it is easier to trace around a circle of radius $\sqrt{13}$, by looking for solutions to $x^2 + 3y^2 = 52$ for solutions of the type $(x+y\sqrt{-3})/2$.  One finds numbers like $7,3$, and then $5,3$ and $2,4$, all of which are likely suspects.

It's a similar method that i have employed in more complex integer systems.


问:
What does "#" mean here? "[these structures] are diffeomorphic to $\#^kS^1\times B^3$"
I am reading The topology of 4-manifolds by Kirby.

At page 7 the author uses the symbol # what does it mean?

The sentence for context is this:

However, the 3-handles and 4-handle of a closed $M^4$ together are diffeomorphic to $\#^kS^1\times B^3$...

答:
In this particular passage, notation $\#$  is for the boundary connected sum, see my answer here for the detailed definition and comparison to the usual connected sum.

Incidentally, one should be very proficient in basic algebraic topology to have any chance succeeding in reading Kirby's book.


问:
Prove that the space is totally-disconnected.
Given.

$$D \text{ - some dense set in ℝ} $$
$$P(D) - \text{power set over D} $$
$$T_{d_E} - \text{canonical topology on ℝ} $$
$$T_D = T_{d_E}\cup P(D) $$
$$T - \text{ topology generated from the subbasis } T_D$$
$$\underline X = (ℝ, T) $$

Prove.

$\underline X$ is totally-disconnected.

Attempt.

If $A \subseteq R, \ card(A) > 1$ and $A$ is open, then it contains some $d \in D$ by the definition of density.

Since $\underline X$ is a $T_1$ space, then $\{d\}$ is closed but it's also open in $A$ due to the way $T_D$ is defined. So $\{d\} \neq A$ is "clopen" set in $A$ and hence $A$ is disconnected.

Same goes by the above logic for the case if $A$ is not open but is a neighbourhood containing an open set.

If the above is right then one can handle the cases: $[ a, b], $[a, b), $(a, b]$ and $(a,b)$.

But how can one show that e.g. non-interval set is disconnected: $B = (a,b)\cup(c,d)$?

If the topology defined on $\underline X$ were $T(\le)$ Topology, then one could argue that since $B$ is not an interval then it can't be connected. But $T_D$ is not an order topology, so how can I prove that $B$ type sets are disconnected too?

答:
$T_D$ is not a topology but it is a subbase for a topology, which is probably what's meant here. Taking $D= \Bbb P$ gives us the Michael line, and this question is meant to generalise this situation, I suppose.

To see that $\underline{X}$ is totally disconnected, take a subspace $A$ having $2$ or more points. If $d \in D \cap A \neq \emptyset$ then $\{d\}$ is a clopen subset of $A$, so $A$ is not connected. Otherwise $A$ has empty interior in the Euclidean topology and $A \subseteq \Bbb R\setminus D$. In the usual topology any $A$ that has empty interior and has two or more points is disconnected (as it's not an interval of any type) and so the same holds for $A$ (which has the Euclidean topology, been disjoint from $D$). QED


问:
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$?
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$

$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$

$3x+5=x-1$

$x=-3$

But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$

So I tried to do other things, such as making sure the base is larger than $0$.

$x^2-9>0$

$x<-3$ or $x>3$

But I still didn't obtain the $x$.
So what is $x$?

答:
Hint

I assume you are looking for real solution.

First, check the cases:

a) $(x^2-9)\in \{-1,0,1\}$. Test each case in the inital equation;

b) When $(x^2-9)>0$ you do $3x+5=x-1$, and take the intersection.

An observation, if you try $x=-3$ in the first equation, the term $(x-3)^{x-1}$ is equal to $(-6)^{-4}$. Is it a solution?

EDIT:
Plotting both functions (RHS and LHS) it seems we have two intersection. One at $x=3$ an other close to it. @user showed how to find the second solution.


问:
What is the expectation of $(\langle x, w \rangle - \langle y, w \rangle)^2$, where $x,y,w$ are independent Bernoulli random vectors?
I'm stuck on the following problem.

Consider three independent Bernoulli random vectors $x,y,w$ of length $n$, where each entry follows the Bernoulli distribution $B$ with $P(B=0)=P(B=1)=\frac{1}{2}$.

$\langle  x,w \rangle$ denotes the standard scalar product and, in this case, just counts the number of positions i = 1,...,n, where both entries are 1.

Let $X := \langle  x,w \rangle$ and $Y := \langle  y,w \rangle$.
Clearly, $\mathbb{E}(X) = \mathbb{E}(Y) = \frac{n}{4}$.

Now, I'm interested in the random variable $Z := (X-Y)^2$ and in particular in the value of $\mathbb{E}(Z)$ as a function of $n$.

I have trouble evaluating this, because $X$ and $Y$ are not independent. I have tried to condition on $w$ and using the normal approximation of the binomial distribution, but didn't get anywhere.

答:
You can compute it as follows. Note first that $X= \sum_{i=1}^n x_iw_i$, therefore $x_iw_i$'s are Bernoulli$(1/4)$ and you said in the comment that they are independent therefore $X$ is binomial$(n,1/4)$. Same for $Y$. Now you can compute $\mathbb{E}(X^2)$ and $\mathbb{E}(Y^2)$, right?

For $\mathbb{E}(XY)$,

$$
\mathbb{E}(XY) = \mathbb{E}(\sum_{i=1}^{n}x_iw_i\sum_{j=1}^{n}y_iw_i) = \mathbb{E}(\sum_{i=1}^nx_iw_i^2y_i + \sum_{i\neq j} x_iw_i y_jw_j)
$$
$$
=\sum_{i=1}^n\mathbb{E}(x_i)\mathbb{E}(w_i^2)\mathbb{E}(y_i) + \sum_{i\neq j} \mathbb{E}(x_i)\mathbb{E}(w_i) \mathbb{E}(y_j)\mathbb{E}(w_j)
$$

Now plug-in the expectations of the Bernoulli random variable.


问:
What is the combination of Complex, Split-Complex and Dual Numbers
If $a+bi:i^2=-1$ is a complex number, $a+cj:j^2=+1$ is a split-complex number, and $a+d\epsilon:\epsilon^2=0$ is a dual number; what is the term for the combination $a+bi+cj+d\epsilon:i^2=-1,j^2=+1,\epsilon^2=0$?</p>

答:
Dual tessarines. Combination of complex and split-complex numbers is tessarines, and when we add dual unit we usually add "dual" to the algebra's name.

UPDATE

One can implement this in Mathematica using the following code:

```<code>$Post=(Normal[Series[#,{ε,0,1}]]/.j->{-1,1})/.{x_,y_}->(x+y)/2+(j(y-x))/2&;
</code>```

Test:

```<code>Log[j + ε]
</code>```

Output:

```<code>(I π)/2 + 1/2 j (-I π + 2 ε)
</code>```

UPDATE2

The following code also works (and works better):

```<code>$Post=((#/.ε->0)+ε(D[#,ε]/.ε->0)/.j->{-1,1})/.{x_,y_}->(x+y)/2+(j(y-x))/2&;
</code>```

Body:
Dual tessarines. Combination of complex and split-complex numbers is tessarines, and when we add dual unit we usually add "dual" to the algebra's name.

UPDATE

One can implement this in Mathematica using the following code:

$Post=(Normal[Series[#,{ε,0,1}]]/.j->{-1,1})/.{x_,y_}->(x+y)/2+(j(y-x))/2&;

Test:

Log[j + ε]

Output:

(I π)/2 + 1/2 j (-I π + 2 ε)

UPDATE2

The following code also works (and works better):

$Post=((#/.ε->0)+ε(D[#,ε]/.ε->0)/.j->{-1,1})/.{x_,y_}->(x+y)/2+(j(y-x))/2&;


答:
When $n=1$ the expectation is $\frac14$ as there are $8$ equally likely combinations, two of which give $1$ and the other six give $0$

For $n>1$ each index's contribution is independent of the others, as the expectation of the difference is $0$ and they are identically distributed, and there are $n$ values of the index. So this is like adding variances.

So the overall expectation is $\dfrac n4$.

Title: What is the expectation of $(\langle x, w \rangle - \langle y, w \rangle)^2$, where $x,y,w$ are independent Bernoulli random vectors?

Body:
I'm stuck on the following problem.

Consider three independent Bernoulli random vectors $x,y,w$ of length $n$, where each entry follows the Bernoulli distribution $B$ with $P(B=0)=P(B=1)=\frac{1}{2}$.

$\langle  x,w \rangle$ denotes the standard scalar product and, in this case, just counts the number of positions i = 1,...,n, where both entries are 1.

Let $X := \langle  x,w \rangle$ and $Y := \langle  y,w \rangle$.
Clearly, $\mathbb{E}(X) = \mathbb{E}(Y) = \frac{n}{4}$.

Now, I'm interested in the random variable $Z := (X-Y)^2$ and in particular in the value of $\mathbb{E}(Z)$ as a function of $n$.

I have trouble evaluating this, because $X$ and $Y$ are not independent. I have tried to condition on $w$ and using the normal approximation of the binomial distribution, but didn't get anywhere.


问:
Motivation for proof of $(((\phi \to \psi) \to \phi) \to \phi)$
I'm trying to solve this exercise.

If $\phi, \psi$ are propositional formulas, prove $\vdash (((\phi \to \psi) \to \phi) \to \phi)$.

I was wondering if there was a "simple" proof to this (I'm aware of the proof Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$?), and most importantly, how to motivate a proof to this exercise. (For me, the motivation is the most important bit! I already have a proof)

I asked such a question before, with a much simpler exercise, and the answer was "try it out in words first, and then write it symbolically!" which worked very well until coming across this example.  Currently I have a proof to this, which I brute forced; to me it feels extremely mechanical and non-replicable but for luck. I'm not sure if this is relevant to the question, but I'll attach it as a picture.

Edit: I actually don't know what the expression "proof system" means. Sorry; I was only told that this was "natural deduction", and the rules for $\land, \lnot, \rightarrow , \leftrightarrow, \perp$, etc were given.

答:
Your proof has a few redundant steps.  Let us rebuild it.

You have the miniproof: $$\dfrac{\dfrac{\dfrac{\lnot\phi\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}$$

And you know that you want to end with a conditional introduction:

$$\dfrac{\dfrac{[(\phi\to\psi)\to\phi]^1\\\qquad\quad\vdots}{\phi}{}}{((\phi\to\psi)\to\phi)\to\phi}{\small{\to}\mathsf I^1}$$

Well simply combining them to use a conditional elimination will derive $\phi$ but ...

$$\dfrac{\lower{1.5ex}{(\phi\to\psi)\to\phi}\quad\dfrac{\dfrac{\dfrac{\overset{\color{red}\downarrow}{\lnot\phi}\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}}{\phi}{\small{\to}\mathsf E}$$

Uh oh! You cannot yet introduce the conditional unless you can discharge that undischarged $\lnot\phi$ and still derive $\phi$.

Fortunately, if you could derive a contradiction you may then use RAA to do that.

Okay, can you derive that contradiction?

 Uh, yes, obviously: the assumption $\lnot\phi$ and sub-conclusion $\phi$ do contradict.

So...

$$\dfrac{\dfrac{\dfrac{\dfrac{\lower{1.5ex}{[(\phi\to\psi)\to\phi]^1}\quad\dfrac{\dfrac{\dfrac{[\lnot\phi]^2\quad[\phi]^3}{\bot}{\small\lnot\mathsf E}}{\psi}{\small\bot}}{\phi\to\psi}{\small{\to}\mathsf I^3}}{\phi}{\small{\to}\mathsf E}}{\bot}{\small\lnot\mathsf E}}{\phi}{\small\textsf{RAA}^2}}{((\phi\to\psi)\to\phi)\to\phi}{\small{\to}\mathsf I^1}$$


问:
Nonnegativity of solution of ODE if $y'(0)=0$
It's known that, if I have positive initial condition and $f(t,0)>0$, so $y(t)$, solution of ODE
$$y'(t)=f(t,y),$$ is nonnegative, because if $T$ is the first time s.t. $y(T)=0$  so $y'(t)>0$ for $t\in (T-\epsilon,T+\epsilon)$, contradiction.

I'd like to know more about the case $f(t,0)=0$. If the initial condition is positive and $f(t,0)=0$, what can I conclude about positiveness?

Thank you so much.

答: $y(t)=0$ is a solution, and if you have the uniqueness property, then no other solution can cross or branch off or in to this solution. Meaning every other solution has one definite sign.


问:
Multiplicativity of maximum norm
Let $f\in C([a,b], \mathbb{K})$ and $\| f\|_\infty:= \max_{x\in[a,b]} |f(x)|$ be the maximum norm.

I was wondering if $\|f^n\|_\infty = (\|f\|_\infty)^n$ holds for $n\in\mathbb{N}$.

By submultiplicativity of $\|.\|_\infty$, we clearly have $\|f^n\|_\infty \leq (\|f\|_\infty)^n$.

Is the reverse estimate also correct?

答:
That is correct, because $t \mapsto t^n$ is increasing on $[0, \infty)$.

For all $x \in [a,b]$ is
$$
|f(x)^n| = |f(x)|^n \le (\Vert f \Vert_\infty)^n
$$
with equality for the $x$ where the continuous function $|f|$ attains its maximum on $[a, b]$. It follows that $\Vert f^n \Vert = (\Vert f \Vert_\infty)^n$.

By the same argument is
$$
 \Vert h \circ f \Vert_\infty = h(\Vert f \Vert_\infty)
$$
if the function $h: \Bbb K \to \Bbb R$ is increasing on $[0, \infty)$ and satisfies $h(|t|) = |h(t)|$ for all $t$.

Title: Multiplicativity of maximum norm


问:
Various possible modification of fixed point property
I was thinking about of possible variation of the fixed point property, as in the title.
A space $X$ satisfies the fixed point property (FPP) if every continuous function $f:X\to X$ has a fixed point, i.e. $\exists x\in X,f(x)=x$.
I realized that, for example, the space $X=[0,1)$ doesn't clearly satisfies FPP, BUT every surjective continuous function from $X$ to itself has a fixed point ($\exists xf(x)=0\leq x$ and $f(0)>0$).
We can call this property SFPP (surjective FPP) and, more in general, if we select a class $\mathcal{A}$ of continuous functions from $X$ to itself, let's define $\mathcal{A}$-FPP the property: "every $f\in \mathcal{A}$ has a fixed point".

So we came to the question: what implications hold between FPP, SFPP, IFPP (injective FPP), BFPP (bijective FPP) (note that it's different from SFFP $\cap$ IFPP, at least in principle) and HFPP (homeomorphism FPP)?

Clearly FPP $\Rightarrow$ SFPP/IFPP $\Rightarrow$ BFPP $\Rightarrow$ HFPP, but what other implications hold? In particular between SFPP and IFPP.

And, even more important, also why the others doesn't hold? The previous example shows SFPP $\nRightarrow$ FPP, and also a "8"-shape (or two intersecting straight line) space  shows BFPP $\nRightarrow$ FPP. Clearly an example that shows HFPP $\nRightarrow$ BFPP (if it's exists at all) cannot be compact.

EDIT. I would also like to know if BFPP=SFFP $\cap$ IFPP (probably not, but I can't think to a counterexample).

答:
Partial answer covering some cases, I'll think about the rest and update it if I come up with more examples.

$SFPP\not\Rightarrow IFPP$ by your $[0,1)$ example.

$HFPP\not\Rightarrow BFPP$ by the following example: Consider the subspace of $\Bbb R^2$ composed of the $x$ axis and, for every integer $z\leq 0$ attach to $(z,0)$ a vertical half open segment, while for $z>0$ attach to $(z,0)$ a circle. This space has a fixed point free continuous bijection, namely shift every vertical segment and every circle to the right by $1$, and wrap the vertical segment over $(0,0)$ around the circle over $(1,0)$. However homeomorphisms of this space must preserve the basepoints of segments and circles.


问:
Formula for the sum of the digits of a number in base b
In an answer to Cyclic numbers are characterized by the reciprocals of full reptend primes?
 a user states that the sum of the digits of a number in base b is given as below:

$\frac{b^d-1}{b-1}s=
\frac{d(d+1)}{2}n
\ \text{where} \
s=d\frac{b-1}{2}\frac{(d+1)n}{b^d-1}\in\mathbb{N}$

I am not able to see how to arrive at this formula for $s$. Kindly provide me some insight.

Thanks!

Body:
Using the example from the original question, the argument is that

$$\begin{aligned}
1 \times 142857 &= 142857 &= 1\times 10^5 + 4\times 10^4 + 2\times 10^3 + 8\times 10^2 + 5\times 10 + 7\times 1\\
+2 \times 142857 &= 285714 &= 2\times 10^5 + 8\times 10^4 + 5\times 10^3 + 7\times 10^2 + 1\times 10 + 4\times 1\\
+3 \times 142857 &= 428571 &= 4\times 10^5 + 2\times 10^4 + 8\times 10^3 + 5\times 10^2 + 7\times 10 + 1\times 1\\
\vdots & \quad & \vdots\\
\downarrow & \quad & \downarrow\\
=\color{blue}{\frac{d(d+1)}{2}n} &= 2999997 &= (1+4+2+8+5+7) \times 111111 = 27 \frac{10^6-1}{10-1} = \color{blue}{s \frac{b^d-1}{b-1}}
\end{aligned}
$$


问:
On the definition of the affine subspace and complex lines
I'm reading "Algebraic Geometry: A problem solving approach" and on page 197 the author mentions:

For a field $k$, the affine n-space over k is the set
$$\mathbb{A}^n(k) = \{(a_1,a_2,...,a_n) : a_i \in k \text{ for } i = 1,...,n\}
$$

Then it is mentioned that $\mathbb{A}^1(\mathbb{C})$ is referred to as the complex line. But it isn't this just the complex plane? In other words, isn't $\mathbb{A}^1 = \{(a) : a \in \mathbb{C}\} = \mathbb{C}$?

What is the definition of affine n-space saying and what is the difference between $\mathbb{A}^1(k)$ and $k$?

答:
In a first approach, there seems to be very little difference between $k$ and $\mathbb{A}^1(k)$ – or even between $\mathbb{A}^n(k)$ and $k^n$. And in many situations you can indeed think of them as similar (as a guide for intuition, if nothing else). The interest of the notation $\mathbb{A}^n$ is that it turns a set of “scalars” ($k$) into a “variety”, ie some form of “shape”. In this way, $\mathbb{A}^n$ has the same “type” as, say, $\mathbb{P}^n$, or any set of zeroes of any family of polynomials.

Very soon in algebraic geometry, it is understood that the notion of zero set of a family of polynomials is insufficient of itself, or troublesome to manipulate. And so such a set is enriched with a ring (and, in more modern versions, a sheaf of rings because some phenomena need to be studied locally to be understood) of regular functions. Again, the notation $\mathbb{A}^n$ makes the distinction between the scalar values of functions and the topological space made up with the elements of $k^n$, and polynomial functions as regular functions.

When $k=\mathbb{C}$, we then realize, through the Nullstellensatz, that the points of $\mathbb{A}^n$ can be paired up with an invariant directly defined in terms of its ring of regular functions, instead of conflating “points” (geometric) and “values” (algebraic) – every maximal ideal of $k[x_1,\ldots,x_n]$ is some $(x_1-a_1,\ldots,x_n-a_n)$ for some scalars $a_i$. So we can just forget the “point characterization” of $\mathbb{A}^n$ to just remember that it is the space of maximal ideals (this has been outdated since 1960-ish and schemes but let’s ignore that) with the ring of regular functions $k[x_1,\ldots,x_n]$ – and recall that shorthand for “the maximal ideal $(x_1-a_1,\ldots,x_n-a_n)$” is “the point $(a_1,\ldots,a_n)$”.

There’s an issue with this definition, though. A huge issue: this creates “phantom” points when the base field isn’t algebraically closed. For instance, consider $\mathbb{A}^1$ on $k=\mathbb{R}$ (and try and think of the same on $\mathbb{Q}$ if you’re brave enough). With our new definitions, the points are the maximal ideals of $\mathbb{R}[x]$, so they’re essentially the irreducible polynomials of $\mathbb{R}$. They are of two kinds: the first kind is that of polynomials $x-a$ – the points $a$ – and then the second-degree irreducible polynomials, for instance $x^2+1$... but what kind of point is that?

Wait a second... we said that $a$ was the scalar corresponding to the ideal $(x-a)$ – because (more or less) the polynomial $x-a$ vanishes at $a$... so the “scalar” corresponding to $x^2+1$ would be $i$? Or perhaps $-i$? It’s both at once, actually...

So $\mathbb{A}^1$ knows. $\mathbb{A}^1$ remembers. Even if you consider it on a non-algebraically closed field, $\mathbb{A}^1$ has some points for algebraic numbers.

And $\mathbb{A}^n(k)$? Well, sometimes we really don’t want to consider these “appearing points” for algebraic numbers that aren’t even numbers. So $\mathbb{A}^n(k)$ is the set of points of $\mathbb{A}^n$ which correspond to actual scalars, that is, the maximal ideals of the form $(x_1-a_1,\ldots,x_n-a_n)$ where the $a_i$ are in $k$.

TL; DR: actually $\mathbb{A}^n$ is a very big, very rich object of nature different from $k^n$, and $\mathbb{A}^n(k)$ simply is the subset of points of $\mathbb{A}^n$ that are nice enough to have “coordinates” in $k$.

As for your first question, it’s one basic thing in algebraic geometry to make the concept of dimension field-invariant: for instance, if the real line had dimension one but $\mathbb{A}^1(\mathbb{C})$ had dimension two, what would be the dimension of $\mathbb{A}^1(\mathbb{Q})$? Or $\mathbb{A}^1(\mathbb{F}_p)$? Intuitively, these objects are “similar” over different fields (not unlike linear algebra)... so it’s normal that they be given the same algebraic  dimension.


问:
Joint PDF with graph and variable, problem
I was given this function:

$\displaystyle f_{X,Y}( x,y) =\begin{cases}
c & in\ the\ area\ of\ the\ picture\\
0 & o.w.
\end{cases}$

The picture is the following:
plot

The question is to find $\displaystyle c$, using of course the fact that $\displaystyle \int \int c=1$.

Their solution:

They caculated the area geometrically, and said $\displaystyle A=0.5$.

Then, $\displaystyle c=\frac{1}{A} =2$.

I don't undestand too much their solution, why is $\displaystyle c=\frac{1}{A}$? For what reason?

Anyway, my solution was longer, and I occurend into a strange situation.

My solution:

Let's divide the ranges of the picures into two areas.

$\displaystyle 0\leqslant x\leqslant 1$ and $\displaystyle -x+0.5\leqslant y\leqslant -x+1$

$\displaystyle 0.5\leqslant x\leqslant 1$ and $\displaystyle -x+1.5\leqslant y\leqslant 1$

I then calculated:

$\displaystyle \int ^{1}_{0}\left(\int ^{-x+1}_{-x+0.5} c\ dy\right) dx\ +\int ^{1}_{0.5}\left(\int ^{1}_{-x+1.5} c\ dy\right) dx\ =\ 1$.

My result was $\displaystyle c=1.6$.

The strange thing is that when I do (notice the difference of $\displaystyle x$ bound in second integral):

$\displaystyle \int ^{1}_{0}\left(\int ^{-x+1}_{-x+0.5} c\ dy\right) dx\ +\int ^{1}_{0}\left(\int ^{1}_{-x+1.5} c\ dy\right) dx\ =\ 1$.

I also get $\displaystyle c=2$.

Same thing as if I did:

$\displaystyle \int ^{1}_{0}\left(\int ^{-x+1}_{-x+0.5} c\ dy+\int ^{1}_{-x+1.5} c\ dy\right) dx=1$.

I don't undestand the difference between these results. It seems to me that in both cases, no matter the change of the $\displaystyle x$ for $\displaystyle 1$ to $\displaystyle 0.5$, the area still represents the same chunk in the picture.

Could anyone explain both the answer in the book, and both the strange behaviour I occured?

Thanks a lot

答:
You are given an uniform density (in fact you have $f(x,y)=c$. a constant) in a specific area, that is the colored area in the picture. Thus the density is exactly the reciprocal of the area.

This because your joint distribution is a solid with Base as the area $A$ in the picture and Height constant $c$.

The probability is the volume of the solid figure, thus you have

$$A\times c=1 \rightarrow c=\frac{1}{A}$$

You can integrate the constant in the colored area finding the same result

$$\int_0^{1/2}\left[\int_{1/2-x}^{1-x}dy  \right]dx+\int_{1/2}^1\left[\int_{0}^{1-x}dy  \right]dx+\int_{1/2}^1\left[\int_{3/2-x}^{1}dy  \right]dx=\frac{1}{c}$$

You get $c=2$


问:
Multiplication by a simple function and compactly supported inverse Fourier transform
Let $f\in C_c^\infty$ be a compactly supported smooth function function with support say in $(0,1)$.  It is clear that when I multiply $\tau$ with $\hat{f}(\tau)$ (Fourier transform of $f$), then inverse Fourier transform of $\tau\hat{f}(\tau)$ is equivalent to $g(x)=id/dx(f(x))$ which is clearly compactly supported with the same support at the largest. How about other powers of $\tau$. Let's say $F(\tau)=\tau^r\hat{f}(\tau)$ where $0<r<1$. Is it true that inverse Fourier transform of  $F$ has compact support. If so, how big would be the support? Many thanks!

答:
For $k\ge 0$, the inverse Fourier transform of $ (i\tau)^k \hat{f}$ is $ f^{(k)}$ which is $C^\infty_c$ with a support at most as large as $f$.

This is because $(i\tau)^k$ is the Fourier transform of the distribution $ \delta^{(k)}$ with support $\{0\}$ so the support of $f\ast \delta^{(k)}=f^{(k)}$ is included in that of $f$.

The Fourier transform of a smooth compactly supported function is entire, so $\hat{f}(\tau)$ is entire. For $f\ne 0$ and $k\ge 0$ then $\hat{f}(\tau)\tau^k$ is entire iff $k\in \Bbb{Z}_{\ge 0}$, ie. its inverse Fourier transform is compactly supported iff $k\in \Bbb{Z}_{\ge 0}$.


问:
$C^1[0, 1]$ is complete with the following norm.
$C^1[0, 1]$ is complete with the $C_1$ norm. Consider the following norm,
$$||f||=|f(0)|+||f^\prime||_\infty$$
This norm is equivalent to $C_1$ norm. But in the book they asked to prove the space is complete using fundamental theorem of calculus and convergence theorem of Riemann integration.

How to prove the result, without using the fact that the norms are equivalent?

答:
Let $(f_n)$ be a Cauchy sequence of $(C^1, ||.||)$, then the sequence of derivatives $(f_n')$ is a Cauchy sequence in $(C^0,||.||_{\infty})$. This last space is complete so $(f_n')$ converges to a function $g$ that is continuous over $[0,1]$. Now we only need to show that $(f_n)$ converges to an antiderivative of $g$ to have that $(f_n)$ converges in $C^1$. Using again the Cauchy property satisfied by $(f_n)$, $(f_n(0))$ is a Cauchy sequence of $\mathbb{R}$ that is complete and thus this last sequence converges to a real number $a$. Now there is no choice for the candidate limit of $(f_n)$, it is an antiderivative of $g$ taking the value $a$ at $0$. We introduce
$$ f : x \longmapsto a + \int_0^xg(t)dt $$
that is a $C^1$ function over $[0,1]$ by the fundamental theorem of calculus, let us check that $(f_n)$ goes to $f$,
$$ ||f_n-f|| = |f_n(0)-f(0)| + ||f_n'-f'||_{\infty} = |f_n(0)-a| + ||f_n'-g||_{\infty} $$

where the two last terms converge to $0$ by construction.


问:
Using the idea of heron's problem to find focal property of ellipse
I'm trying to follow this proof for reflection property of ellipse, I understood the idea of reflection and using it to minimize distances because it's the same concept which we use to solve heron's problem.

However, I don't get the second part of his arguments where they dtldarek actually speaks of the ellipse. How did he argue that the point on tangent is the one which minimizes the "heron's distance" between F' and F?

Note: By heron's distance I mean shortest distance with two points on the constraint that you must first move from one of the point to a line and then from the line you must move to theother  point

答: By definition, all points $Q$ on the tangent, different from tangency point $T$, are external to the ellipse. Hence $FQ+F'Q>FT+F'T$, that is $T$ minimises "heron's distance".


问:
Multiple ultrafilter quantifiers
I'm reading Todorcevic's book Introduction to Ramsey spaces and I stumble on the following use of ultrafilter quantifiers: given a formula $\phi$ with variables $x_1,\dots,x_k$, $\mathcal{U}x_1\dots\mathcal{U}x_k\phi(x_1,\dots,x_k)$. I understood the definition for one variable: $\mathcal{U}x\phi(x)$ iff $\{x:\phi(x)\}\in\mathcal{U}$, but I can't write down the precise definition for the quantifier in two or more variables.

答:
$\mathcal Ux_1\mathcal Ux_2\phi(x_1,x_2)$ means that $\{y\mid\mathcal Ux_2 \phi(y,x_2)\}\in\mathcal U$, which means that $\{y\mid\{x\mid\varphi(y,x)\}\in\mathcal U\}\in\mathcal U$.

I think the clarifying example is the definition of the semigroup operation on $\beta\omega$, extending addition of natural numbers we set $$\mathcal U_1+\mathcal U_2=\{A\subseteq\omega\mid \{x\in\omega\mid\{y\in\omega\mid x+y\in A\}\in\mathcal U_2\}\in\mathcal U_1\},$$ or equivalently $$\mathcal U_1+\mathcal U_2=\{A\subseteq\omega\mid\mathcal U_1x\mathcal U_2y\,\, x+y\in A\}.$$

Title: Multiple ultrafilter quantifiers


答:
Mindlack already gave a complete answer, so I just want to underscore something in the complex case. If we work over $\Bbb{C}$, we can choose an underlying real structure to see $\Bbb{C}$ as a $2-$dimensional vector space over $\Bbb{R}$. Topologically, we think of this as a real plane. However, when we work purely over $\Bbb{C}$, it is actually more correct to think of $\Bbb{C}$ (by extension $\Bbb{A}^1_{\Bbb{C}})$ as being $1-$dimensional, i.e. as a "line".

At first, this does not really make sense. However, when we think of the intersection theoretic properties of copies of $\Bbb{C}$ in $\Bbb{C}^2$ it becomes clear why this is more correct. Even though a $1-$dimensional subspace $V(\cong \Bbb{C})$ of $\Bbb{C}^2$ is like a "real plane", any two such $V$ and $V'$ which intersect in a point besides the origin are identical. So, as far as complex/algebraic geometry is concerned, copies of $\Bbb{C}$ really behave more like lines than planes.

Actually, this phenomenon persists when one thinks about algebraic curves. A nonsingular algebraic curve over $\Bbb{C}$ can be thought of as a Riemann surface also, which we think of as topologically a real surface. However, if these are "plane" curves (in the sense that they embed in $\Bbb{P}^2)$, they satisfy Bézout's Theorem:

If $X$ and $Y$ are plane curves in $\Bbb{P}^2$ the number of points at which they intersect (counted with multiplicity) is $\deg(X)\cdot \deg(Y)$.

Here, degree is the degree of the polynomial cutting out the curve. So, in spite of being (classically) topologically $2-$dimensional as $\Bbb{R}-$manifolds, these curves intersect in discrete sets of points unless they are equal. This is quite confusing at first, but coming to terms with this makes us realize that $1-$dimensional complex varieties look more like "curves" than they do like "surfaces."

Title: On the definition of the affine subspace and complex lines

Body:
I'm reading "Algebraic Geometry: A problem solving approach" and on page 197 the author mentions:

For a field $k$, the affine n-space over k is the set
$$\mathbb{A}^n(k) = \{(a_1,a_2,...,a_n) : a_i \in k \text{ for } i = 1,...,n\}
$$

Then it is mentioned that $\mathbb{A}^1(\mathbb{C})$ is referred to as the complex line. But it isn't this just the complex plane? In other words, isn't $\mathbb{A}^1 = \{(a) : a \in \mathbb{C}\} = \mathbb{C}$?

What is the definition of affine n-space saying and what is the difference between $\mathbb{A}^1(k)$ and $k$?


问:
Confusion over continuity on surfaces and L'Hospital's rule
I am really confused and was hoping someone could help clear up my understanding.

Suppose I have two continuous differentiable surfaces $f,g:\mathbb{R}^N\rightarrow\mathbb{R}$ and there exists a point $x_0$ such that $f(x_0)=g(x_0)=0$. Further $g(x)$ is non-zero everywhere else in the domain. I am interested in the surface $\frac{f}{g}$, which must be continuous everywhere where $g$ is non-zero.

Suppose I find a curve $C_1:[0,1]\rightarrow\mathbb{R}^N$ such that $x_0=C_1(0)$ and $x_1=C_1(1)$ for some $x_1 \in \mathbb{R}^N$ and I am able to show that:
$$\lim_{y \rightarrow 0}\frac{f(C_1(y))}{g(C_1(y))}=k$$

The question which is confusing me is this:

Does the existence of the above limit imply that if I were to find any curve function, say $C_2:[0,1]\rightarrow\mathbb{R}^N$ such that $x_0=C_2(0)$ and $x_2=C_2(1)$ for some $x_2 \in \mathbb{R}^N$, it would be true that:

$$\lim_{y \rightarrow 0}\frac{f(C_2(y))}{g(C_2(y))}=k$$

i.e. is the limit now the same along all curves? Or in other words, if I defined $h(x)=\frac{f(x)}{g(x)}$ for all $x\neq x_0$ and $h(x_0)=k$ would $h$ be continuous?

My intuition for why the answer to the above should be "yes, $h$ would be continuous" is this:

$x_0$ should be a "removable" discontinuous point for $\frac{f}{g}$ and therefore it seems it should have a well-defined limit at $x_0$ from all sides and that limit should be equal.

My intuition for why the answer to the above should be "no, $h$ is not continuous" is this:

Suppose additionally that $C_1$ and $C_2$ were differentiable everywhere and that $f$ and $g$ had a non-zero directional derivative along $C_1$ and $C_2$. By L'hopitals rule, we would have:
$$k=\lim_{y\rightarrow 0}\frac{(f \circ C_1) (y)}{(g \circ C_1) (y)}=\lim_{y\rightarrow 0}\frac{(f \circ C_1)' (y)}{(g \circ C_1)' (y)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_1} f(x)}{\nabla_{C_1} g(x)}$$
and:
$$k=\lim_{y\rightarrow 0}\frac{(f \circ C_2) (y)}{(g \circ C_2) (y)}=\lim_{y\rightarrow 0}\frac{(f \circ C_2)' (y)}{(g \circ C_2)' (y)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_2} f(x)}{\nabla_{C_2} g(x)}$$
Thus:
$$\lim_{x\rightarrow x_0}\frac{\nabla_{C_1} f(x)}{\nabla_{C_1} g(x)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_2} f(x)}{\nabla_{C_2} g(x)}$$
This seems like quite a strong statement and its hard for me to believe it is true for every curve $C_2$ such that $g$ has non-zero directional derivative along $C_2$...

Apologies for the very long post but I am really very confused and any help clearing up my understanding here would be very appreciated!

答:
If I understand your question correctly, the claim is clearly false. Just take $f(x,y)=xy$ and $g(x,y)=x^2+y^2$ (with $N=2$, obviously), so that
$$
h(x,y) = \frac{xy}{x^2+y^2}
.
$$
For $C_1(t)=(t,0)$ the limit of $h(t,0)$ as $t \to 0$ is zero, and for $C_2(t)=(t,t)$ the limit of $h(t,t)$ is $1/2$. The confusion seems to lie in your idea about a removable singularity, which isn't correct.


问:
Prime 3-Manifolds
I'm trying to understand what the prime 3-manifolds are. Milnor's Prime Decomposition Theorem states that any compact, orientable 3-manifold is a connected sum of a unique finite collection of prime 3-manifolds.

I've read that these prime manifolds will be one of the following:

$S^3$ or any quotient of $S^3$

$S^2 \times S^1$

Some $K(\pi, 1)$ manifold

I'm wondering where this statement comes from. Why these are the only prime 3-manifolds. My thoughts are it could be something to do with $\pi_2$, because Milnor's paper states a 3 manifold is prime $\iff$ $S^2 \times S^1$ or irreducible. The latter condition $\implies$ $\pi_2 =0$.

Thus the prime 3 manifolds are $S^2 \times S^1$ and those with trivial second homotopy group. It is here where I become stuck since I have not formally studied second homotopy groups before.

My question is how do we conclude that the the only 3 options are those stated above?

答:
Here are the details, which use a lot of additional tools. You can find all these arguments (except for the application of Perelman's theorem) in older 3-manifolds textbooks such as Hempel's "3-manifolds".

Let $M$ be a closed, orientable 3-manifold.

First, the sphere theorem of Papakyriakopolous says that if $\pi_2(M)$ is nontrivial then there exists an embedding $S^2 \approx S \subset M$ which represents a nontrivial element of $\pi_2(M)$. Nontriviality implies that $S$ does not bound a 3-ball. It follows that if $S$ is separating (or indeed if any other separating sphere exists in $M$ that does not bound a 3-ball), then $M$ is not prime. Finally, if every separating sphere in $M$ bounds a 3-ball, and if $S$ is nonseparating, then a pretty easy separate argument shows $M$ is homeomorphic to $S^2 \times S^1$. Therefore, the conclusion that you desire is true in the case that $\pi_2(M)$ is nontrivial.

So now let's assume that $\pi_2(M)$ is trivial.

Consider the universal covering space $\widetilde M$.

If $\pi_1(M)$ is finite then $\widetilde M$ is a closed, simply connected 3-manifold. Before the work of Perelman, the best conclusion one could make here was that $\widetilde M$ is homotopy equivalent to $S^3$. But nowadays, by applying Perelman's Theorem, which says that the Poincare conjecture is true, it follows that $\widetilde M$ is homeomorphic to $S^3$, and therefore $M$ is a quotient of $S^3$. (A broader statement of Perelman's theorem says that Thurston's whole geometrization conjecture is true, and therefore $M$ is a spherical space form, meaning that $M$ is a quotient of $S^3$ with a deck transformation group consising of isometries of $S^3$).

If $\pi_1(M)$ is infinite, then $\widetilde M$ is noncompact. Applying the long exact homotopy sequence of a fibration to the universal covering map $\widetilde M \to M$, it follows that $\pi_2(\widetilde M)$ is homeomorphic to $\pi_2(M)$ and hence is trivial. Since $\pi_1(\widetilde M)$ and $\pi_2(\widetilde M)$ are both trivial, it follows from the Hurewicz theorem that $\pi_3(\widetilde M)$ and $H_3(\widetilde M)$ are isomorphic. But noncompactness of $\widetilde M$ implies that $H_3(\widetilde M)$ is trivial; this is one of the side-results accompanying the Poincare duality theorem. Using that $M$ has a 3-dimensional triangulation, and hence so does $\widetilde M$, it follows that all of the higher homology groups $H_n(\widetilde M)$ are also trivial. Again, by applying Hurewicz inductively, it follows that all of the higher homotopy groups $\pi_n(\widetilde M)$ are trivial. Therefore, $\widetilde M$ is contractible and $M$ is a $K(\pi,1)$ space.

Title: Prime 3-Manifolds


答:
The equation:
$$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}
$$
obviously holds if $x^2-9=0$ provided that the exponents are non-negative. This is the case for $x=3$.

If $x^2-9\ne0$ we can divide both sides by $(x^2-9)^{(x-1)}$ obtaining:

$$(x^2-9)^{(2x+6)}=1.$$

This equation holds if $2x+6=0$ or $x^2-9=1$. The roots of the equations are $x=-3$ and $x=\pm\sqrt{10}$, respectively.

The substitution in the original equation shows that
$x=3,\sqrt{10}$ satisfy the equation whereas the roots $x=-3,-\sqrt{10}$ do not suit it.

Title: $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$?

Body:
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$

$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$

$3x+5=x-1$

$x=-3$

But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$

So I tried to do other things, such as making sure the base is larger than $0$.

$x^2-9>0$

$x<-3$ or $x>3$

But I still didn't obtain the $x$.
So what is $x$?