问:
How to formulate continuum hypothesis without the axiom of choice?
Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers :

1) The definition of $\aleph_1$ makes sense even without choice as $\aleph_1$ is an ordinal number (whose construction doesnt depend on the axiom of choice) with some minimal property. With or without choice, there is no cardinal number $\mathfrak{a}$ such that $\aleph_0 < \mathfrak{a} < \aleph_1$.

2) In ZFC, all cardinal numbers are $\aleph$ and are comparable (by the trichotomic property of the ordinals). The continuum hypothesis states that $\aleph_1 = 2^{\aleph_0}$.

3) In ZF, $2^{\aleph_0}$ need not be an $\aleph$. However, I don't know if talking about  $2^{\aleph_0}$ in ZFC makes sense or not.

Does it make sense in ZF to define CH to be the statement that if a set is larger than the natural numbers, then it must contain a copy of the reals (up to a relabeling of the elements) no matter what the cardinality of the reals is ?

答:
You are correct that without the axiom of choice $2^{\aleph_0}\newcommand{\CH}{\mathsf{CH}}$ may not be an $\aleph$. Therefore the continuum hypothesis split into two inequivalent statements:

$(\CH_1)$ $\aleph_0<\mathfrak p\leq2^{\aleph_0}\rightarrow2^{\aleph_0}=\frak p$.

$(\CH_2)$ $\aleph_1=2^{\aleph_0}$.

Whereas the second variant implies that the continuum is well-ordered, the first one does not.

You suggested a third variant:

$(\CH_3)$ $\aleph_0<\mathfrak b\rightarrow 2^{\aleph_0}\leq\mathfrak b$.

Let's see why $\CH_3\implies\CH_2\implies\CH_1$, and that none of the implications are reversible.

Note that if we assume $\CH_3$, then it has to be that $2^{\aleph_0}\leq\aleph_1$ and therefore must be equal to $\aleph_1$. If we assume that $\CH_2$ holds, then every cardinal less or equal to the continuum is finite or an $\aleph$, so $\CH_1$ holds as well.

On the other hand, there are models of $\sf ZF+\lnot AC$, such that $\CH_1$ holds and $\CH_2$ fails. For example, Solovay's model in which all sets are Lebesgue measurable is such model.

But $\CH_2$ does not imply $\CH_3$ either, because it is consistent that $2^{\aleph_0}=\aleph_1$, and there is some infinite Dedekind-finite set $X$, that is to say $\aleph_0\nleq |X|$. Therefore we have that $\aleph_0<|X|+\aleph_0$. Assuming $\CH_3$ would mean that if $X$ is infinite, then either $\aleph_0=|X|$ or $2^{\aleph_0}\leq|X|$. This is certainly false for infinite Dedekind-finite sets (one can make things stronger, and use sets that have no subset of size $\aleph_1$, while being Dedekind-infinite).

One can also think of the continuum hypothesis as a statement saying that the continuum is a certain kind of successor to $\aleph_0$. As luck would have it, there are $3$ types of successorship between cardinals in models of $\sf ZF$, and you can find the definitions in my answer here.

It is easy to see that $\CH_1$ states "$2^{\aleph_0}$ is a $1$-successor or $3$-successor of $\aleph_0$", and $\CH_3$ states that "$2^{\aleph_0}$ is a $2$-successor of $\aleph_0$" -- while not explicitly, it follows from the fact that I used to prove $\CH_3\implies\CH_2$.

So where does $\CH_2$ gets here? It doesn't exactly get here. Where $\CH_1$ and $\CH_3$ are statements about all cardinals, $\CH_2$ is a statement only about the cardinality of the continuum and $\aleph_1$. So in order to subsume it into the $i$-successor classification we need to add an assumption on the cardinals in the universe, for example every cardinal is comparable with $\aleph_1$ (which is really the statement "$\aleph_1$ is a $2$-successor of $\aleph_0$").

All in all, the continuum hypothesis can be phrased and stated in many different ways and not all of them are going to be equivalent in $\sf ZF$, or even in slightly stronger theories (e.g. $\sf ZF+AC_\omega$).

Without the axiom of choice we can have two notions of ordering on the cardinals, $\leq$ which is defined by injections and $\leq^*$ which is defined by surjections, that is to say, $A\leq^* B$ if there is a surjection from $B$ onto $A$, or if $A$ is empty. These notions are clearly the same when assuming the axiom of choice but often become different without it (often because we do not know if the equivalence of the two orders imply the axiom of choice, although evidence suggest it should -- all the models we know violate this).

So we can formulate $\CH$ in a few other ways. An important fact is that $\aleph_1\leq^*2^{\aleph_0}$ in $\sf ZF$, so we may formulate $\CH_4$ as $\aleph_2\not\leq^*2^{\aleph_0}$. This formulation fails in some models while $\CH_1$ holds, e.g. in models of the axiom of determinacy, as mentioned by Andres Caicedo in the comments.

On the other hand, it is quite easy to come up with models where $\CH_4$ holds, but all three formulations above fail. For example the first Cohen model has this property.

All in all, there are many many many ways to formulate $\CH$ in $\sf ZF$, which can end up being inequivalent without some form of the axiom of choice. I believe that the correct way is $\CH_1$, as it captures the essence of Cantor's question.

Interesting links:

What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...

Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF


问:
Lipschitz continuity implies continuity
Say $\vert f(x)-f(y)\vert \le L\vert x-y\vert$. How to prove the following: $\forall \lim_{n \to \infty} x_n = x_0 \wedge x_0\in \Bbb{R}$: $\lim_{n \to \infty}f(x_n) = f(x_0)$?

In other words: how to prove that Lipschitz-continuity implies regular continuity(without using differentiation or similar methods)?

答:
Assume that the $f$ is not continuous. Hence, it must be that for some $c$:

$$\lim _{n\to \infty} |f(x_n) - f(c)| \not = 0 $$

where $x_n \to c$ as $n\to \infty$ . Hence it must be:

$$\lim _{n\to \infty} \dfrac{|f(x_n) - f(c)|}{|x_n - c|}  = \infty $$

However this contradicts that $f$ is Lipschitz. Because for all $x_n$, it must be $|f(x_n) - f(c)| \leq L | x_n-c|$ for some real constant $L\ge 0$. As $n\to\infty$, $L$ converges to $\infty$, hence it is not constant.

Contradiction implies that $f$ must be continuous.

ContentLicense: CC BY-SA 4.0

Title: Lipschitz continuity implies continuity


问:
Locally principal Ideal in a Noetherian semilocal domain
Let $R$ be a commutative Noetherian domain with only finitely many maximal ideals $\mathfrak m_1,...,\mathfrak m_n$. If $I$ is an ideal of $R$ such that $I_{\mathfrak m_i}$ is a principal ideal of $R_{\mathfrak m_i}$ for every maximal ideal $\mathfrak m_i$ , then is it true that $I$ itself is a principal ideal?

答:
Yes, this is true. Since $R$ is Noetherian, $I$ is finitely generated, and hence finitely presented. The condition that $I_{\mathfrak{m}_{i}}$ is a principal ideal of $R_{\mathfrak{m}_{i}}$ for each $i$ implies that $I$ is a projective $R$-module. (See Bourbaki's "Commutative Algebra", II.5.2 Theorem 1.)

In fact, $I$ has constant rank $1$, since $R$ is a domain, and the rank function $\mathrm{Spec}(R) \to \mathbb{Z}$ sending $\mathfrak{p}$ to $\mathrm{dim}_{R_{\mathfrak{p}}}(I_{\mathfrak{p}})$ is locally constant. (I'll note that one does not need to use the fact that $R$ is a domain here - Theorem 2 of II.5.3 of Bourbaki's "Commutative Algebra" settles this in greater generality.) But $R$ is semi-local by assumption, which shows that $I$ is principal by Proposition 5 of II.5.3 of Bourbaki. (I also wrote about this here in a previous question you asked.)


问:
Question on sheafification of a presheaf
In chapter 2 of GTM 52 by Robin Hartshone there are definition of presheaf and the associated sheaf of a given presheaf.

I found that the definition of the sheafification is rather less natural and too rigorous. Harthshone did not give any non trivial concrete presheaf and its sheafification.

My questions are :

From the definition of a presheaf $\mathcal{F}$(as Hartshone defined) how can one think about its sheafification $\mathcal{F}^{+}$ as a collection of map : $s: U\rightarrow \cup \mathcal{F}_{p}$ for each open subset $U$ and why is $\cup \mathcal{F}_{p}$ rather than other sets ?

Could you please show me a nontrivial, concrete example of a presheaf(that is not a sheaf itself) and its sheafification ?

Thanks !

答:
First I give an example of sheafification which I think it is important: Let $X=\mathbb{R}^n$, for each open subset $U$ we define $\mathcal{F}(U)$ be the set of all constant functions from $U$ to $\mathbb{R}$. Then easy to see $\mathcal{F}$ is a presheaf on $X$. Now given $f_1\in\mathcal{F}(U)$ and $f_2\in\mathcal{F}(V)$ where $U\cap V=\emptyset$, there is now constant function $f$ on $U\cup V$ such that $f_{|U}=f_1$ and $f_{|V}=f_2$. So $\mathcal{F}$ is not a sheaf!. Now we define the sheaf $\mathcal{F}^+$ which is associated to $\mathcal{F}$: for each $U$, $\mathcal{F}^+(U)=$ the set of all locally constant functions on from $U$ to $\mathbb{R}$. Now you can observe that the sections of $\mathcal{F}^+$ on $U$ are created by glueing the sections of $\mathcal{F}$ in smaller open subset. This is the key point of thinking about sheafification, because for presheaf we can not glue the sections together.

Now for your question (1): in general the sections of a presheaf are not functions, so you can not glue them together. In order to glue them together you must somehow transfer them to functions, the canonically way to do this is Identify each $s\in\mathcal{F}(U)$ with the function $\bar{s}: U\rightarrow \cup \mathcal{F}_{p}$. $\bar{s}$ capture all behaviors of $s$. And now you can glue together $\bar{s}$ to have the sheaf $\mathcal{F}+$. You can read chapter 2 in the very readable book "Sheaf theory" written by Tennison to get more details.

ContentLicense: CC BY-SA 3.0


问:
How many combinations to combine letters with certain restrictions
I have the following letters:

A, B, C, D and E

and these have to be combined with:

F, G, H, I and J.

An example of a possible combination is (A, F), (B, H), (C, G), (D, J), (E, I).

Now, the following restrictions are given:

F can't be with A and E, G can't be with C and D, H can't be with A and E, I cant' be with B and J cant't be with C. I want to calculate the amount of possible combinations and I think I have to use the principle of inclusion and exclusion.

The total number of pairs $T$ is $5!$, I think, because I first want to permutate A-E.

Now my idea is this: $T - F(A, E) - G(C,D)-H(A,E)-I(B)-J(C)$, where $F(A, E)$ is the total number of pairs where $F$ is paired with $A$ or $E$. I'm obiously counting way too much right now, so could anybody point out how I can keep track of what I'm counting? That would help me solve this problem.

Thanks in advance!

答:
F can't be with A and E, G can't be with C and D, H can't be with A and E, I cant' be with B and J cant't be with C. I want to calculate the amount of possible combinations and I think I have to use the principle of inclusion and exclusion.

The total number of pairs $T$ is $5!$, I think, because I first want to permutate $A-E$.

No question that Inclusion-Exclusion, which generalizes well, is the 1st port of call, for a problem like this.  Here, I reject Inclusion-Exclusion, in favor of the direct approach, for two reasons:

Ignoring permutations, there are only $5!$ candidate combinations to consider.  Further, once the number of satisfying combinations are computed, it is a simple matter to multiply this number by $5!$ to obtain the number of satisfying permutations.

The restrictions are convoluted.

Thus, we have to consider : [C-k denotes Constraint #k]

C-1 - (F:BCD)

C-2 - (G:ABE)

C-3 - (H:BCD)

C-4 - (I:ACDE)

C-5 - (J:ABDE)

This is the turning point.  Inclusion-Exclusion might still be the best way to attack this problem.  It depends on the interactions between the constraints.

Here, I will go with the direct approach, rather than Inclusion-Exclusion, specifically because of the dovetailing interaction between C-1 and C-3.

Case-1

(F,H) = (B,C) or (F,H) = (C,B)

What the above syntax means is that F:B, H:C or vice-versa.  The reason that these two possibilities are examined together is because, it is immediate, from the constraints, that the number of satisfying combinations that involve (F,H)=(B,C) will equal the number of satisfying combinations that involve (F,H) = (C,B).

The remainder of the analysis of Case-1 will assume (F,H) = (B,C).

Then, at the end of Case 1, I will simply take the computation and multiply by $2$.

At this point, you know [within Case 1], by C-2, that (G:AE).

If (G:A), then (I,J) must link with (D,E), with C-4,C-5 obeyed.

2 ways.

If (G:E), then (I,J) must link with (A,D), with C-4,C-5 obeyed.

2 ways.

Total # of ways = 4, assuming (F,H) = (B,C).

Total # of ways for Case 1 $= (4 \times 2) = 8.$

Case-2

(F,H) = (B,D) or (F,H) = (D,B)

Assume (F,H) = (B,D).

By C-2, (G:AE)

If (G:A), then (I,J) must link with (C,E), with C-4,C-5 obeyed.

Only 1 way, since J can't link with C.

If (G:E), then (I,J) must link with (A,C), with C-4,C-5 obeyed.

Similar to (G:A), only 1 way.

Final case 2 computation : $(1 + 1) \times 2 = 4.$

Case-3

(F,H) = (C,D) or (F,H) = (D,C)

With (C,D) removed, C-2,C-4,C-5 are adjusted to:

C-2 (G:ABE)

C-4 (I:AE)

C-5 (J:ABE)

You can tell at a glance that there are two ways of satisfying the adjusted C-4, and for each way, there are then two ways of satisfying the adjusted C-2,C-5.

Therefore, in Case 3, (F,H) = (C,D) will generate 4 possibilities.

Thus, for Case-3: $(4\times 2) = 8.$

Adding: the total number of combinations of Cases 1 through 3 combined are $(8 + 4 + 8) = 20.$

Since each combination yields $(5!)$ permutations, there are
$20 \times (5!)$ satisfying permutations.

Addendum

I should probably examine the challenges inherent in applying Inclusion-Exclusion to this problem.  This will explain that even absent the dovetailing of constraints C-1, C-3, for this problem, the alternative of Inclusion-Exclusion is no walk in the park.

This Addendum will be much easier to understand, if you first read this Inclusion-Exclusion article.

Consider the following chart:

D-1 - (F:AE)

D-2 - (G:CD)

D-3 - (H:AE)

D-4 - (I:B)

D-5 - (J:C)

The above chart represents the violations of the constraints.  For example, if constraint C-1 is violated, then D-1 occurs, which means that either (F:A) or (F:E).

Then, suppose that you wanted to examine the situation where at least two constraints are being violated.  This is distinct from the situation where exactly two constraints are being violated.  There are $\binom{5}{2} = 10$ ways of selecting 2 of the constraints out of 5, and then computing the number of ways (i.e. combinations) that represent that at least these two specific constraints are being violated.

As a further illustration, suppose that you are examining the specific situation where constraints C-1 and C-4 were violated.  This means that you are examining how many different ways that D-1 and D-4 could have simultaneously occurred.  There are two ways that D-1 could have occurred, either (F:A) or (F:E).  Coupled with D-4, you have (F:A, I:B) or (F:E, I:B).

Examining (F:A, I:B) you observe that this makes no restriction on how (G,H,J) are to be coupled with (C,D,E).  Clearly, there are $(3!)$ such couplings.  This means, that the number of ways that D-1,D-4 could simultaneously occur is $(3!) \times 2.$

This type of analysis leads to the following chart:

$(T_0) ~:~ (5!) \times (1).$

$(T_1) ~:~ (4!) \times (2 + 2 + 2 + 1 + 1).$

$(T_2) ~:~ (3!) \times (4 + 2 + 2 + 2 + 4 + 2 + 1 + 2 + 2 + 1).$

$(T_3) ~:~ (2!) \times (4 + 4 + 2 + 2 + 2 + 2 + 4 + 2 + 1 + 2).$

$(T_4) ~:~ (1!) \times (4 + 2 + 2 + 2 + 2).$

$(T_5) ~:~ (0!) \times (2).$

It is not a coincidence that

$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 = 20.$

Things to note about the above chart:

$k \in \{0,1,2,3,4,5\}.$

$T_k$ signifies situations where at least $k$ constraints are being violated.

$T_k$ will therefore have $\binom{5}{k}$ entries.

$T_k$ examines couplings of $k$ entries from [F,G,H,I,J] with $k$ entries from [A,B,C,D,E].

When you examine one specific coupling of $k$ entries from [F,G,H,I,J] with $k$ entries from [A,B,C,D,E], you are then leaving unspecified how the other $(5-k)$ entries from [F,G,H,I,J] are to be coupled with the other $(5 - k)$ entries from [A,B,C,D,E].

This explains why each $T_k$ contains the factor $(5-k)!$.

All of the above analysis is merely a backdrop for the real issue: for any such problem, what is the best approach?  Is the best approach the direct approach that I actually used, or is it the indirect approach of Inclusion-Exclusion?

Examining the Inclusion-Exclusion chart of $T_0, \cdots, T_5$, since $\binom{5}{0} + \cdots + \binom{5}{5} = 2^5$, the 5 constraints force the computation of $(2^5)$ separate terms.  This means that when there are $n$ constraints, Inclusion-Exclusion is going to force the computation of $(2^n)$ terms.

Personally, when considering Inclusion-Exclusion for a problem of this nature, which has $(n)$ constraints, I would regard $(n=4)$ as iffy, $(n > 4)$ as perhaps too much, and $(n < 4)$ as do-able.  An exception might be if the constraints had symmetrical properties, so that the $(2^n)$ terms did not have to be computed 1-at-a-time, but instead, some of the computations could be inferred from some of the other computations.

Now you have to examine the other side of the ledger.  When considering the direct approach, how difficult will it be to take off your shoes and count out each possibility?  How unwieldy will it be to manage the $(n)$ constraints?  In the specific problem, with the $(n=5)$ constraints rendering Inclusion-Exclusion problematic, I fortunately was able to capitalize on the dovetailing of constraints C-1,C-3.


问:
Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$
So I came across this step when finding an expontential generating function using this recurrence relation,
$$
T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}
$$
Since $T_{0}=T_{1}=1$
For $n=0$
$$
\sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1
$$
For $n=1$
$$
\sum_{n \geq 1} \frac{T_{1} x^{1}}{1 !}=x
$$
Therefore,
$$
f(x)=\sum_{n \geq 0} \frac{T_{n} x^{n}}{n !}=1+x+\sum_{n \geq 2} \frac{T_{n} x^{n}}{n !}
$$
From the recurrence relation, $T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}$. Hence,
$$
\Rightarrow 1+x+\sum_{n \geq 2} \frac{T_{n-1}+(n-1) T_{n-2}}{n !} x^{n}
$$
Distributing the terms using the commutative law of addition, we get,
$$
\begin{gathered}
\therefore 1+x+\sum_{n \geq 2} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \\
\Rightarrow 1+\sum_{n \geq 1} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n}
\end{gathered}
$$
Where did the ' $x$ ' go in the last step above? And how did the lower limit change from $n \geq 2$ to $n \geq 1$ ?

Could anyone help explain? Any help would be highly appreciated.

答:
Observing that $T_1=T_0=1$, now
$$x=\frac{T_1 x}{1!}=\frac{T_0 x}{1!}$$
so you can absorb this term as $n=1$ of your second last summand.


问:
Does a formula exist that gives a function of only solutions that are multiples of any given integer m up to n multiples?
I feel like this is a stupid question but does a formula exist that gives a continuous function of x such that its only solutions are at integer multiples of m up to n? The domain of the function is any real value x and the range is any real value y for the sake of argument, but could be a wave function, polynomial, etc or any continuous function. This is the working graph I have at m = 3 except that solutions should lie under each bell curve at y = 0 instead of at y = 1, I for some reason I lost the formula that does the former. I have an interesting zeta form of the equations that yield actual solutions at t = m but I'm still playing around with it.

I started playing around with a set of formulas one day after finding a strange relationship with Young modulus, Ramanujan congruencies, Collatz function, and logistic chaos equations I was working with at the time--as far as I could tell what I was looking at was a kernel for the zeta function for the given s expression I had, but I'm no mathematician so don't crucify me, ha. I would share the formulas but I don't know math jax or whatever that is, I only know latex. But other than what my question suggests I don't think the answer will be surprising, it's just that I've seen some weird stuff with this math which is simple so why would it do such as prime factorization and prime maps I didn't learn in school, so I've toyed with it for over a year on occasions, but mostly can't make sense of it.

答:
Have a look at the function $f(x)=-\sin^2( \pi x /m).$ It is negative except at integer multiples of $m$ where it is zero. Is that the behavior you want? If not you can play with what's inside the sine squared function.

Unfortunately I don't think it could be that simple if it was required only to have some finite number of multiples of $m$ as zeros.


问:
Partial derivatives where implicit equations are involved
I have a function:
$$f\bigl(P,R\bigr)=h\bigl(P,R,N(P,R)\bigr)$$
where $P$ and $R$ are independent variables and $N$ is the solution of the equation:
$$P-g(R,N)=0$$
I am trying to find $\frac{\partial f}{\partial P}$ and $\frac{\partial f}{\partial R}$.

答:
$N(P,R)$ is not an independent variable, rather a function (of $P$ and $R$).  Hence partial derivatives of $f$ with respect to $P$ and $R$ exist.

The solution is to obtain $\frac{\partial g}{\partial R}$ and $\frac{\partial g}{\partial N}$ from the solution of $P-g(R,N)=0$.

At the solution, P = g, and hence:
$$\frac{\partial N}{\partial P}=\frac{1}{\frac{\partial g}{\partial N}}$$
$$\frac{\partial N}{\partial R}=-\frac{\frac{\partial g}{\partial R}}{\frac{\partial g}{\partial N}}$$
Use these derivatives with the chain rule in the evaluation of
$$\frac{\partial}{\partial P}f\bigl(P,R\bigr)$$and
$$\frac{\partial}{\partial R}f\bigl(P,R\bigr)$$


问:
Is a quadratic term inside a square root equivalent to a linear term?
For a function such as
$$f(x) = \sqrt{x^2+a^2}, $$
is it correct to say it is linear in $x$?

Can I assume it can be written as $f(x) = \alpha x + \beta$ for some $\alpha$ and $\beta$?

EDIT: what if $x$, $\alpha$ and $\beta$ may be matrices?

答: No, it is not linear in $x$.  A glance at a graph should convince you of that.


问:
Given $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, find truth/falsity

Given $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, where $a,b$ are integers, with $a \geq 1$, $b \geq 1$. Is the following true: $I(2,2)=2I(1,1)$?

I understood this to be whether this integral is linear, like $I(c(a,b))=c(I(a,b))$, where $c\in \mathbb{R}$. I declare this to be false. If you substitute $a=1, b=1$, then the integral evaluates to $2\int_{0}^{1}0dx$, which is clearly not the case with $a=2, b=2$.

Next, is $I(a,b)=I(b,a)$? This is of course a property of symmetry, if there ever was one. This time there's no way to verify whether it holds. Shall I try a few examples to see if it disproves it?

答:
Let $I(a,b)$ be given by the integral

$$I(a,b)=\int_0^1 x^a (1-x)^b\,dx$$

for $a>-1$ and $b>-1$.  Enforcing the substitution $x\mapsto 1-x$, we find that

$$I(a,b)=\int_0^1 (1-x)^a x^b\,dx=I(b,a)$$

as was to be shown!

Now let's look at $I(2,2)$.  Then, we have

$$I(2,2)=\int_0^1 x^2(1-x)^2\,dx=\frac1{30}$$

However,

$$I(1,1)=\int_0^1 x(1-x)\,dx=\frac16$$

So, $I(2,2)=\frac15 I(1,1)$


答: No not exactly. Though the function you have becomes more and more linear as $|x| \to \infty$. This is because as $|x| \to \infty$ the $+a^2$ becomes more and more "irrelevant". In particular, the function approaches $f(x) = |x|$ as $|x| \to \infty$.

Title: Is a quadratic term inside a square root equivalent to a linear term?

Body:
For a function such as
$$f(x) = \sqrt{x^2+a^2}, $$
is it correct to say it is linear in $x$?

Can I assume it can be written as $f(x) = \alpha x + \beta$ for some $\alpha$ and $\beta$?

EDIT: what if $x$, $\alpha$ and $\beta$ may be matrices?


答:
HINT

Does it hold that $f(x + y) = f(x) + f(y)$?


问:
Derivative of $\sec^{-1}(x)$
I'm struggling with problem below which I eagerly want to solve. Let me know from where this problem is, if possible (the origin source of textbook). your answers might really helpful to get through the struggles with which I confront.

Question

One way of defining $\sec^{-1}(x)$ that is sometimes used is to say that $y=\sec^{-1}(x)$ $\iff$ $\sec(y)=x$ and $y \in (0,\pi]$ (alternatively, y is between 0 and pi and is the same 0 and pi, y not 0)

show that, with this definition. $\frac{d}{dx}\sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}$

答:
We already know:

$$\sec \left(\sec^{-1}\left(x\right)\right)=x$$

We want to take a derivative from both sides of this with respect to $x$(obviously the right hand side will give 1). To take the derivative I will use the Chain rule.

$$\frac{d}{dx} \sec \left(\sec^{-1}\left(x\right)\right)=\frac{d\sec \left(\sec^{-1}\left(x\right)\right)}{d\sec^{-1}\left(x\right)}\frac{d\sec^{-1}\left(x\right)}{dx}=1$$

We already know how to calculate the derivative $\frac{d\sec{y}}{dy}$:

$$\frac{d\sec{y}}{dy}=\frac{d}{dy}\frac{1}{\cos{y}}=\frac{\sin{y}}{\cos^2{y}}=\sin{y}\sec^2{y}$$

Let's use this in our derivative(note $\frac{d\sec^{-1}\left(x\right)}{dx}$ is what we are looking for):

$$\frac{d\sec \left(\sec^{-1}\left(x\right)\right)}{d\sec^{-1}\left(x\right)}=\sin\left(\sec^{-1}\left(x\right)\right)\sec^2\left( \sec^{-1}\left(x\right)\right)=x^2\sin\left(\sec^{-1}\left(x\right)\right)$$

Now, how are we supposed to calculate $\sin\left(\sec^{-1}\left(x\right)\right)$? I usually draw a triangle for these type of trigonometry identities.

Looking at the image we figure out: $\sin\left(\sec^{-1}\left(x\right)\right)=\frac{\sqrt{x^2-1}}{x}$

Now putting everything back into our first equation, we find out:

$$x \sqrt{x^2-1}\frac{d\sec^{-1}\left(x\right)}{dx}=1 \\
\Rightarrow \frac{d\sec^{-1}\left(x\right)}{dx}=\frac{1}{x \sqrt{x^2-1}}$$

Title: Derivative of $\sec^{-1}(x)$


问:
Using Lagrange multipliers without a given constraint?

I've solved the problem without the use of Lagrange multipliers, but am unsure on how to solve this problem using lagrange multipliers like was intended?

Is there some way to form my own constraint? I wanted to use distance as a formula and have the ellipse formula act as the constraint, but was unsure on how to go about it?

Does anyone have any insight on how to do this problem? I've scoured the internet for tips, but was unsuccessful.

答:
This looked like a good problem to illustrate how the method from first-semester calculus is related to the use of Lagrange multipliers for answering these questions.  The curve $ \ 4x^2 \ + \ 7xy \ + \ 8y^2 \ = \ 60 \ $ is a rotated ellipse centered on the origin, so it has symmetry about the origin. This means that the point $ \ (X \ , \ Y) \ $ on the ellipse for which the $ \ x-$ coordinate is largest is "mirrored" in the opposite quadrant by the point $ \ (-X \ , \ -Y) \ , $ which is the point with the smallest $ \ x \ \ ; $ a similar statement can be made for the points with the largest and smallest $ \ y-$ values.

Determining these points is equivalent to finding the tangent points on the vertical and horizontal tangent lines to the ellipse (as mentioned by RyRy the Fly Guy), a problem commonly set in single-variable calculus.  We apply implicit differentiation to the curve equation to produce

$$ \frac{d}{dy}[ \ 4x^2 \ + \ 7xy \ + \ 8y^2 \ ] \ = \ \frac{d}{dy}[60] \ \ \Rightarrow \ \ 8x \ + \ 7y \ + \ 7x·\frac{dy}{dx} \ + \ 16y·\frac{dy}{dx} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{dy}{dx} \ \ = \ \ -\frac{8x \ + \ 7y}{7x \ + \ 16y} \ \ . $$

The expressions in the numerator and denominator of this implicit derivative are the same as those which appear in the calculation Claude Leibovici presents.

The horizontal tangent lines are those for which this expression is zero, and so for which the numerator is zero (since the denominator is not also zero for those coordinates).  We learn to insert $ \ 8x \ + \ 7y \ = \ 0 \ \ \Rightarrow \ \ y \ = \ -\frac87 \ x \ $ into the curve equation, which gives us the tangent points
$ \left( \pm \ 7\sqrt{\frac{15}{79}} \ , \ \mp \ 8\sqrt{\frac{15}{79}} \right)  \ \approx \ (\pm  \ 3.05 \ , \ \mp  \ 3.49) \ \ . $  This line  through the origin (in orange) and the  horizontal tangent points are seen in the graph below.  [In the Lagrangian analysis, this will come from
$$ F \ = \ \mathbf{y} \ + \ \lambda(4x^2+7xy+8y^2-60) \ \ \Rightarrow \ \
 \frac{\partial F}{\partial x} \ = \ \lambda  (8 x+7 y) \ = \ 0 \ \ . ] $$

By the same token, we can differentiate the curve equation implicitly with respect to $ \ x \ $ (or simply use the relation $ \frac{dx}{dy} \ \ = \frac{1}{\frac{dy}{dx}} \ $ ) to obtain $ \ \frac{dx}{dy} \  =  \ -\frac{7x \ + \ 16y}{8x \ + \ 7y} \ . $  Setting this ratio equal to zero permits us to locate the vertical tangent points from the condition $ \ 7x \ + \ 16y \ = \ 0 \ \ \Rightarrow \ \ y \ = \ -\frac{7}{16} \ x \ $ (the line in violet in the graph).  The vertical tangent points are then $ \left( \pm \ 8\sqrt{\frac{30}{79}} \ , \ \mp \ \frac72\sqrt{\frac{30}{79}} \right)  \ \approx \ (\pm  \ 4.93 \ , \ \mp 2.16) \ \ . $

The horizontal and vertical tangent lines we have found then form a box which circumscribes the given rotated ellipse.


问:
If the mirror image of $P(a,6,9)$ with respect to line $\frac{x-3}{7} =\frac{y-2}{5}=\frac{z-1}{-9}$..

If the mirror image of $P(a,6,9)$ with respect to line $\frac{x-3}{7} =\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20,b,-a-9)$, then find $|a+b|$

The general point of the line is $(3+7k,2+5k,1-9k)$

The direction ratio line joining this point to P is
$$(3+7k-a,5k-4, -8-9k)$$

So the dot product of the two direction ratios must be zero

So
$$k=\frac{7a-73}{155}$$

Now using the fact that
$$3+7k=\frac{20+a}{2}$$

Gives $a=\frac{-3192}{57}$

Beyond which all the values turn into a random mess. Now I don’t know whether I am making computation error or not (even though I tried it many times), or whether my concept is wrong. But I am not getting the answer with this. Where am I going wrong?

答:
For some $k$, $(3 + 7k, 2 + 5k, 1-9k)$ is the mid-point of $(a,6,9)$ and $(20, b, -a-9$), hence we have:

\begin{cases}
\dfrac {20+a}2 &= 3+7k\\
\dfrac {b+6}2 &= 2 + 5k\\
\dfrac  {9-a-9}2 &= 1-9k
\end{cases}

Solving the system above gives $a=-56, b=-32, k = -3$.


答:
Take the first derivative of $f(x)$:

$$ f'(x) = \dfrac{2x}{2(x^2+a^2)^{0.5}}$$

If $f$ is linear in $x$, it implies that the first derivative must be a constant. We see that $f'(x)$ is not a constant, depends on $x$.


答:
Motivation

$2$ can be written as $1 + 1$, or in fact, in any number of other ways, right?

Idea

Great!

Now that we are motivated, let us take a closer look at $$1 + x + \sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n + \sum_{n \geq 2} \frac {(n - 1) T_{n - 2}} {n!} x^n.$$ In particular, zoom in on $$\sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n.$$ Although the index starts from $n = 2$, there is no harm in trying to see what happens when $n = 1$.

When $n = 1$, observe that $$\frac {T_{n-1}} {n!} x^n = \frac {T_{1 - 1}} {1!} x^1 = T_{0} x = x.$$ This is why the '$x$' (i.e. second term) in $$1 + x + \sum_{n \geq 2} \frac {T_{n - 1}} {n!} x^n + \sum_{n \geq 2} \frac {(n - 1) T_{n - 2}} {n!} x^n$$ has seemingly "disappeared". It has simply been re-written and included into the first summation (i.e. third term).

Title: Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$

Body:
So I came across this step when finding an expontential generating function using this recurrence relation,
$$
T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}
$$
Since $T_{0}=T_{1}=1$
For $n=0$
$$
\sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1
$$
For $n=1$
$$
\sum_{n \geq 1} \frac{T_{1} x^{1}}{1 !}=x
$$
Therefore,
$$
f(x)=\sum_{n \geq 0} \frac{T_{n} x^{n}}{n !}=1+x+\sum_{n \geq 2} \frac{T_{n} x^{n}}{n !}
$$
From the recurrence relation, $T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}$. Hence,
$$
\Rightarrow 1+x+\sum_{n \geq 2} \frac{T_{n-1}+(n-1) T_{n-2}}{n !} x^{n}
$$
Distributing the terms using the commutative law of addition, we get,
$$
\begin{gathered}
\therefore 1+x+\sum_{n \geq 2} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \\
\Rightarrow 1+\sum_{n \geq 1} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n}
\end{gathered}
$$
Where did the ' $x$ ' go in the last step above? And how did the lower limit change from $n \geq 2$ to $n \geq 1$ ?

Could anyone help explain? Any help would be highly appreciated.


问:
How do I show that a group G is the semidirect product of two cyclic subgroups, one of order p^2 and the other of order p?

Let $p$ be an odd prime and $G$ be a group generated by elements $x,y$. Given $$x^{p^{2}}=1, x^{p} = y^{p}, yxy^{-1} = x^{p+1}, (yx^{-1})^{p} = 1,$$
show that $G$ is the semidirect product of a two cyclic subgroups of order $p^{2}$ and $p$, respectively.

The given conditions imply that $|x| = p^{2}, |yx^{-1}|=p$. I think we can pick $x$ and $yx^{-1}$ to be the generator of $\mathbb{Z}/p^{2}\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$, respectively. However, it isn't clear to me where this is leading. Any help is appreciated.

答: Hint:  Try to show that it is the split extension of $\Bbb Z_p$ by $\Bbb Z_{p^2}$ with the homomorphism $\varphi:\Bbb Z_p\hookrightarrow\Bbb Z_{p^2-p}\cong\rm{Aut}(\Bbb Z_{p^2})$ given by the inclusion.


问:
Determine the values of $k$ so that the following linear system has unique, infinite or no solutions.
Determine the values of $k$ so that the following linear system has a unique solution, a infinite solutions and no solution.

$$
x + y -z + w = 1 \\ 2y + z + w = 0 \\ kz + (k-1)w = 2 \\ (k^2-k)w = k-1
$$

I've written the augmented matrix , but I'm stuck.

\begin{bmatrix}
1 & 1 & -1 & 1 & 1 \\
0 & 2 & 1 & 1 & 0  \\
0 & 0 & k & k-1 &2 \\
0 & 0 & 0 & k²-k & k-1 \end{bmatrix}

I've gotten the following solution from a sotware:

$$ w = 1/k+1\\ x = 2k^2 + 4k + 9 / 2k^2+ 2k\\ y = -2k - / 2k² + 2k \\ z = k + 3 / k^2 + k$$ when $$(k \neq 1, k \neq 0, k \neq -1 )$$

But I start getting confused about how to solve the augmented matrix.

And I saw on internet similar problems beeing solved in diferent ways, but I'm not getting what I really should do here.

答:
You don't need the determinant. Note that the 4th row of the REF you computed says $k(k-1)z = (k-1)$, and think about which $k$ will give you in the 4th row:

$0z = 0$ which is infinite solution of $z$,

or $0z = b$, for some $b \neq 0$ which is no solution for $z$,

or $az = b$ where $a \neq 0$, which gives a unique solution of $z$.

Try going from there.


问:
Negative factors of a number
Can a factor of a number be negative?

Is $-5$ a factor of $25$ or $-25$?

A number is said to be prime if it has two factors : $1$ and the number itself. So if $-5$ can be a factor of $5$, how to define the prime numbers?

Thanks.

答:
Indeed, when we enlarge from positive integers to all integers then we have to deal with the fact that this introduces a nontrivial unit (invertible) element $\,-1,$ which may spoil the uniqueness of prime factorizations (up to order), since, e.g. $\,pq = (-p)(-q)\,$ by $\,(-1)(-1) = 1.$  Similarly for polynomials with rational coefficients. Here every nonzero rational number $\,r\,$ is a unit  (invertible) $\ r r^{-1} = 1,\,$ so $\,p(x) q(x) =  (r p(x)) (r^{-1} q(x)).\,$  There are various ways to remedy this.

First, from  the set of all unit multiples of a prime, we could choose some "natural" representative, and call only that element prime. For integers, a natural choice is the positive representative, and for polynomials one that is monic  (leading coefficient $= 1).$ Such representatives are sometimes called unit-normalized representatives.

Alternatively, we may permit all unit multiples to be prime, then alter our statement of uniqueness of prime factorizations to say: prime factorizations are unique up to order and unit multiples.

If you study abstract algebra you will learn that there is a natural algebraic way of "ignoring" these pesky nontrivial units when studying factorization. Namely, since the units form a group $\,U,\,$ we can mod out by $\,U.\,$ This is the algebraic way of working multiplicatively "up to unit multiples".

Title: Negative factors of a number


问:
Proving an object is terminal given that the object has a unique global element
A global element of an object $a$ is a morphism $1 \rightarrow a$ where $1$ is a terminal object. If the object $c$ in a Category has a unique global element, can we prove that $c$ is a terminal object?

答: No. For example, in the category of groups, every object has a unique global element.</p>


答:
Just compare the two lines, focusing on

$$\sum_{n\geq 1} \frac{T_{n-1}}{n!}x^n = \color{red}{\underbrace{\frac{T_{1-1}}{1!}x^1}_{n=1}} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$

$$x + \sum_{n\geq 2} \frac{T_{n-1}}{n!}x^n = \color{red}{x} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$

so it all comes down to $$\frac{T_{1-1}}{1!}x^1 \stackrel{?}{=} x$$ and this is true since $T_0 = 1$.


问:
Removing Implication in the Conversion of WFF to Clausal Form
I understand the conversion algorithm, my concern is regarding the first step of 'Removing Implication'.

Example: In the following figure when one removes implication, the negation comes before ∀z.

However, in other example (given below) the negation when removing implies comes after ∀x and before ∃z.

```<code>∀x [Roman(x) ∧ know(x, Marcus)] → [hate(x, Caesar) ∨ (∀y ∃z hate(y, z) → thinkCrazy(x, y))]

∀x ~[Roman(x) ∧ know(x, Marcus)] ∨ [hate(x, Caesar)  ∨ (∀y ~(∃z hate(y, z)) ∨ thinkCrazy(x, y))]
</code>```

I found a similar example here where ~ is placed after ∀.

I am confused about the placement of ~ while performing the first step during the conversion. What is the criteria of negating the implication when there exists either of the two quantifiers?

答:
It depends on the order of operations by which you parse $\forall x~\phi\to \psi$

$$\color{crimson}(\forall x~\phi\color{crimson})~\to~\psi\iff \lnot\color{crimson}(\forall x~\phi\color{crimson})~\lor~\psi\\\forall x~~\color{crimson}(\phi\to \psi\color{crimson})\iff \forall x~~\color{crimson}(\lnot \phi\lor \psi\color{crimson})$$

The former is the usual convention, but it looks like your other source uses the later.

Unfortunately different authors use different conventions, so it can often be confusing if they don't clearly establish which operation has priority.

When it is ambiguous, use parenthesis.


问:
Non-infinite geometric sum; does not start at 0 or 1
It's bee a long time since I've worked with sums and series, so even simple examples like this one are giving me trouble:

$\sum_{i=4}^N \left(5\right)^i$

Can I get some guidance on series like this? I'm finding different methods online but not sure which to use. I know that starting at a non-zero number also changes things.

My original thought was to do (sum from 0 to N of 5^i) - (sum from 0 to 3 of 5^i) but I'm not sure that's right.

答:
The problem asks for a closed-form solution to:

$$\sum_{i=4}^{N} 5^i = 5^4 + 5^5 + ... + 5^N$$

The OP's original intuition was correct:
$$\sum_{i=4}^{N} = \sum_{i=0}^{N} 5^i - \sum_{i=0}^{3} 5^i$$

More generally, for summing a geometric series starting at an arbitrary index $m$:
$$
\sum_{i=m}^{N} r^i = \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
$$

To get a closed form for the above expression, let's start with the closed-form equation for a geometric series:
$$
\sum_{i=0}^{N} r^i = \frac{r^{N+1}-1}{r-1}
$$

So:

$$
\begin{align*}
\sum_{i=m}^{N} r^i &= \sum_{i=0}^{N} r^i - \sum_{i=0}^{m-1} r^i \\
  &= (\frac{r^{N+1}-1}{r-1}) - (\frac{r^{m-1+1}-1}{r-1}) \\
&= \frac{r^{N+1} - r^m}{r-1}
\end{align*}
$$

An alternative and equivalent form can be found if we multiply the top and bottom by $-1$:
$$
\sum_{i=m}^{N} r^i = \frac{r^m - r^{N+1}}{1-r}
$$


问:
Understanding a colorization algorithm using sparse matrix equation solvers
I am trying to understand a colorization algorithm (converting a gray scale image to color image), available in this paper Colorization using Optimization
https://www.researchgate.net/publication/2896183_Colorization_using_Optimization

For a still image the algorithm inputs a gray-scale image $Y(x,y)$ and finds estimates of the chrominance channels, as in the YUV color space, $U(x,y)$ and $V(x,y)$, given the constraints that a user has assigned some subset of the $U(x,y)$ and $V(x,y)$ pixels as fixed.

Consider the $U$ channel. Then the goal is, for each pixel to minimize the difference between the color $U(x,y)$ at each pixel $(x,y)$ and a weighted average of the colors at neighboring pixels (implemented as a $3 \times 3$ window with appropriate truncation at the image borders) according to the cost function
$$C(U) = \sum \limits_{r}(U(r) - \sum \limits_{s \in N(r)} w_{rs} \cdot U(s))^2$$
where $r$ ranges overall the pixels in the image, and $N(r)$ is the neighbourhood of $r$. The cost function for $V$ is identical. The choice of affinity function $w_{rs}$ measuring how similar the gray-scale values of pixels $r,s$ is up to the implementation, where the authors choose $w_{rs} = \frac {d_{rs}} {\sum_{t \in N(r)} d_{rt}}$ for $s \in N(R)$ where $d_{rs} = e^ \frac{(Y(s)-Y(r))^2}{\sigma_r^2}$ and  $\sigma_r^2$ is the sample variance of intensity values in the window around $r$. For $s \notin N(R)$, $w_{rs}=0$.

The authors state:

Since the cost functions are quadratic and the constraints are linear, this optimization problem yields a large, sparse system of linear equations, which may be solved using a number of standard methods.

So then, I believe for the images $U(x,y), V(x,y)$ that minimizing the cost function can be seen as solving for a vector $x$ that minimizes $||Ax-b||_2$ where $A = D-W$ and

$W$ is the affinity matrix with $W_{rs} = w_{rs}$. Note $W$ is a sparse matrix.

$D$ is a diagonal matrix with $1$'s on the main diagonal

$x$ is the flattened image vector we are solving for

$W, D$ are of size $p \times p$ where $p$ is the number of pixels in the image. $x$ is a vector of size $p \times 1$, and $b$ is the same size as $x$.

My question is what is $b$? I am imagining $b$ enforces the constraints, but I am not sure how. The pixel constraint values are floats in the range $[0,1]$. I would be happy to clarify anything, any insights appreciated.

One possible solution for $b$ is that $b$ consists of a flattened image vector where if at a particular pixel the user has assigned a $U$ or $V$ channel value, then $b_{x,y}$ (before flattening) $=U(x,y)$ and $b_{x,y} = 0$ otherwise.

答:
The linear system that the authors refer to is this system. So instead of minimizing $||Ax-b||$, you minimize $x^TQx + c^Tx$ (you can transfer one objective into the other so they are equivalent, but the latter has an easier expression for the derivative, and the matrix $Q$ is sparse while $A$ might not be sparse).

The dimension of $x$ is the number of pixels. Let me take your cost function and express it in $x$:
$$C(x) = \sum_i(x_i - \sum_j w_{ij} \cdot x_j)^2=\sum_i x_i^2 - \sum_i \sum_j w_{ij}x_i x_j$$
so indeed $Q=D-W$ and $c=0$. In fact, this is why in the paper they mention that the cost function is $x^T(D-W)x$ (in the right column next to formula (1)).

The constraints set the certain values of $x$, which can be modeled as $Ex=d$ where the number of rows in $E$ is the number of colored pixels and each row of $E$ has exactly one nonzero component. An alternative approach is to treat the fixed values as constants, meaning the dimensions of $Q$ and $c$ are a bit smaller, and you need $c \neq 0$ corresponding to the terms $\sum_i \sum_j w_{ij}x_i x_j$ where $x_i$ is fixed.

The only tricky part are the constraints $0 \leq x_i \leq 1$. One way of handling those is via an interior point method. It goes beyond the scope of this answer to fully explain the details, but in essence it adds the term $-\sum_i \mu \log(x_i)$ to the objective and gradually lets $\mu \downarrow 0$. For each $\mu$, you solve:
$$\begin{pmatrix}Q+S & E^T\\ E & O\end{pmatrix}\begin{pmatrix}x \\ \lambda \end{pmatrix} = \begin{pmatrix}e \\ d\end{pmatrix}$$
where $S$ is a diagonal matrix and $S$ and $e$ are based on the current iterate of $x$.


问:
Missing a logic argument
I have to prove
$$(w ⇒ s ∨ p) ∧ (t ⇒ ¬ s) ⇒ (s ⇒ ¬ q)$$

Is there a way to make it right?

答:
The original steps shown in the proof are correct, but I think you're proving the wrong thing. The conclusion $t \land \lnot r \to q$ is provable given the premises. It also uses all the premises and is slightly more difficult to prove, so I think it's the intended problem.

Here's a proof of $t \land \lnot r \to q$

Your premises are equivalent to the following conditional tautologies

$$ \begin{align}
\textbf{(I)} & & q \vdash s, p \;\; \text{is equivalent to $q \to s \lor p$} \\
 \textbf{(II)} & & s, t \vdash \;\; \text{is equivalent to $s \to \lnot t$} \\
\textbf{(III)} & & t, p \vdash r \;\; \text{is equivalent to $t \land p \to r$} \end{align} $$

We can conclude via cut elimination from (I) and (II) that

$$ \begin{align} \textbf{(IIII)} & & q, t \vdash p \end{align} $$

We can use cut elimination again and conclude from (III) and (IIII) that

$$ \begin{align} \textbf{(V)} & & q, t \vdash r \end{align} $$

We can use the negation introduction rule twice

$$ t \vdash r, \lnot q $$
$$ t, \lnot r \vdash \lnot q $$

Note that this is equivalent to the desired conclusion, shown below

$$ \vdash t \land \lnot r \to \lnot q $$

Proof of $t \land \lnot p \to \lnot q$.

I'll use the nonstandard notion of a covariant and contravariant position below. A covariant position is a position like $\square$ in $A \to \square$ or $A \land \square$. Replacing a subexpression $\alpha$ by a subexpression $\beta$ where $\alpha \to \beta$ holds yields a valid conclusion if $\alpha$ is in a covariant position. By contrast, $\square$ is a contravariant position in $\lnot \square$ and $\square \to A$. In a contravariant position, $\beta$ can be replaced with $\alpha$ given that $\alpha \to \beta$ and the conclusion will be a valid inference.

This property is called upward and downward entailment in semantics, I don't know a good name for it in logic but it is a useful concept. Upward entailment is equivalent to what I'm calling a covariant position.

Given your hypotheses

$$ q \to s \lor p \;\;\text{and}\;\; s \to \lnot p \;\;\text{and}\;\; t \land p \to r $$

You can conclude the following by contrapositive, as you did

$$ \lnot(s \lor p) \to \lnot q $$

The De Morgan step is also correct

$$ (\lnot s \land \lnot p) \to \lnot q $$

However, given $s \to \lnot t$, the conclusion $ (\lnot\lnot t \land \lnot p) \to q $ does follow. The position $\square$ in $(\lnot \square \land \lnot p) \to \lnot q$ is a covariant position, so you can replace a subexpression appearing in $\square$ with one it implies. The double elimination step is also valid.

$$ (t \land \lnot p) \to \lnot q $$

Title: Missing a logic argument


问:
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

There is a slight difference in case of $\sin$, which is:
$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$

How do we prove the above two identities?

答:
There's also a nice proof of the cosine formula by Michael Knapp in Mathematics Magazine (vol 82, number 5, December 2009, p. 371-372). Download a free version here: https://www.maa.org/sites/default/files/Knapp200941575.pdf

Knapp mentions in his article that a proof for the sine formula was given by Samuel Greitzer in the obscure student-oriented math journal Arbelos from 1986.


问:
What is the Turing degree of truth in the second-order theory of real numbers?
Let $X$ be the set of Godel numbers of sentences in the second-order language of ordered fields which are true in $\mathbb{R}$.  Then my question is, what is the Turing degree of $X$?

In particular, how does it compare to the Turing degree of truth in second-order arithmetic?  One datapoint is that the second-order arithmetic is interpretable in the second-order theory of real numbers.  Another datapoint is second-order theory of real numbers is interpretable in second-order arithmetic with third-order parameters.  Yet another datapoint is that both second-order arithmetic and the second-order theory of real numbers are categorical theories.  I'm not sure whether these facts imply that the Turing degree of $X$ is the same as the Turing degree of truth in second-order arithmetic, but it least seems intuitively plausible that the two Turing degrees are of the same "order of magnitude".

答:
Since $\mathbb{N}$ is second-order-definable in $\mathbb{R}$, the true second-order theory of $\mathbb{R}$ is basically the same as (= computably isomorphic to) true third-order arithmetic. And the gap between true third-order arithmetic and true second-order arithmetic is gigantic. For example, true second-order arithmetic (construed as a set of natural numbers via Godelization) is definable over $\mathbb{N}$ by a single third-order formula.

Incidentally, there's a methodological point here: what would a "satisfying description" of the relevant Turing degree consist of? For sets of this complexity we're way beyond the point of well-behaved degree constructions, and I think the best we can hope for is descriptions of the form "The Turing degree of the $\mathfrak{L}$-theory of $\mathcal{M}$" for some natural logic $\mathfrak{L}$ and natural structure $\mathcal{M}$. It's often convenient to look for descriptions of this form with $\mathcal{M}$ a nice transitive set, so in this case "The third-order theory of $L_\omega$" would be my suggestion of an answer.

(This point about "naming" Turing degrees is similar to the point about "naming" large ordinals I made here, and I believe in a couple other related answers as well. In each case, unless one has a specific form of description in mind, I think the initial description is actually optimal-ish.)


问:
stability of closed-loop nonlinear system
I would like to understand if it is possible to demonstrate the stability of this closed-loop nonlinear system

$$a b -K_1 u = a K_2 \dot{a}$$

where $a$ is the variable I am trying to control, $u$ is the control law, and the nonlinearity is represented by the fact that $b=f(a)$, and by the multiplication of $a$ with $\dot{a}$ on the right hand side of the equation.

The control law I am using is

$$ u= \frac{[ k_p(a^*-a) + k_i\psi+b] a}{K_1} \\ \dot{\psi}=a^*-a$$

where $k_p$ and $k_i$ are the coefficients of a proportional-integral controller.

I am not a control expert, so I apologise in advance for any imprecision or mistake I might have made in formulating the question.

答:
The proposed control law makes the resulting dynamics linear. Namely, when substituting the control law in the dynamics one can factor out $a$ which gives

$$
\dot{a} = -\frac{k_p}{K_2}(a^*-a) - \frac{k_i}{K_2} \psi.
$$

By using that $\dot{\psi}=a^*-a$ and that $a^*$ is constant, it follows that the second derivative of $\psi$ with respect to time would be $\ddot{\psi}=-\dot{a}$. Substituting the expression for $\dot{a}$ in yields

$$
\ddot{\psi} = \frac{k_p}{K_2} \dot{\psi} + \frac{k_i}{K_2} \psi,
$$

which is linear and can be shown to be stable if $\frac{k_p}{K_2},\frac{k_i}{K_2}<0$.


问:
The sum of a product of two random variables: how do I know if each summand is independent?
So I have a question that involves using SLLN and almost sure convergence:

I'm given $Y_{1},Y_{2},Y_{3}\ldots$ as iid r.v.'s with a finite mean $k$.

I'm also given this sum
$$
\frac{1}{n}\sum_{i=1}^{n}Y_{i}Y_{i+1}\
\overset{\rm a.s.}{\rightarrow}\ k^{2}
$$
and asked if the SLLN applies here.

My question is:

Is there a  surefire way to tell if each summand $\left(~Y_{1}Y_{2},Y_{2}Y_{3}, \mbox{etc.}~\right)$ is iid with finite mean $?$.

What would be the case for a different sum, say,  $$
\frac{1}{n}\sum_{i = 1}^{n}Y_{2i - 01}\ Y_{2i}\ ?
$$

答:
As mentioned by Clement C, SLLN does not apply directly. However, you can apply it indirectly as follows: whenever $n$ is even,
$$
\frac1n\sum_{i=1}^n Y_iY_{i+1}=\frac12\left(\frac{1}{n/2}\sum_{i=1}^{n/2}Y_{2i-1}Y_{2i}+\frac1{n/2}\sum_{i=1}^{n/2}Y_{2i}Y_{2i+1}\right)
$$
Also as mentioned by Clement C, the variables $Y_{2i-1}Y_{2i}$ are iid, so you can apply SLLN two show that the first sum converges to $E[Y_1Y_2]=k^2$, and same for the second. Therefore, the whole sum converges to $\frac12\left(k^2+k^2\right)=k^2$.

(This technically only proves the limit as $n\to\infty$ through the even integers, but you can deal with odd $n$ with a little more work).


问:
Does $I(a,b)=\frac{a}{b+1}I(a-1,b+1)$?
If $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, where $a \geq 1, b \geq 1$, does $I(a,b)=\frac{a}{b+1}I(a-1,b+1)$ for $a \geq 2$? I've tried numbers like $a=2, b=2$, and it seems to be true so far. Any proof that this is true?

答:
My previous answer had a silly calculation error that I had made; the identity is in fact true.

As suggested in the comments, we proceed with integration by parts.

$$\int_0^1 x^a(1-x)^b dx = \left[\frac{-x^a(1-x)^{b+1}}{b+1}\right]_0^1 + \int_0^1\frac{ax^{a-1}(1-x)^b}{b+1}dx$$ $$= 0 + \frac{a}{b+1}\int_0^1 x^{a-1}(1-x)^{b+1} = \frac{a}{b+1}I(a-1, b+1)$$

Title: Does $I(a,b)=\frac{a}{b+1}I(a-1,b+1)$?


问:
reconcile two different cohomolgies
I am in the process of convincing myself of certain results:

I see that in the compact support cohomology mayer vietoris has opposite rows compared to the one in singular topology however it is said that those two cohomologies are the same in case the space is say a compact closed connected manifold. How is that?

I know I am very confused :) I hope I am not confusing you either ;)

Thank you in advance

答:
Let me do it with de Rham cohomology, for variety:

One defines the de  Rham cohomology $H^\bullet(M)$ of a manifold $M$ starting from differential forms on $M$, and one defines the compactly supported de  Rham cohomology $H_c^\bullet(M)$ of a manifold $M$ starting from differential forms on $M$ which have compact support.

Now, if $M$ is a compact manifold, every differential form on $M$ has compact support (because the support of such a thing is a closed subset of $M$, which is automatically compact in this case) so in this case the qualification
«with compact suppot» does not, in fact, exclude any form. It follows immediately that $H^\bullet(M)$ and $H^\bullet_c(M)$ coincide.

Title: reconcile two different cohomolgies


问:
How to get the list price of a tv when a discount has been made?
The problem is as follows:

A seller has made a discount of a $15\%$ from the list price of a tv.
Then he makes another discount of 20 usd and still gets a profit of 12
usd. If the cost of that tv was 206 usd. What was the list price of
the tv?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{280 usd}\\
2.&\textrm{238 usd}\\
3.&\textrm{216 usd}\\
4.&\textrm{286 usd}\\
\end{array}$

What I tried to do was as follows: Assuming that the initial price of the tv is $x$.

$x-\frac{15}{100}x-20-206=12$

Solving this yields.

$x=280$

Which I assume is the initial cost of that tv.
But is this the right answer?. It would help a lot if an answer could indicate the adequate interpretation of this word problem.

答: Yes, that is the right answer. Assume that $x$ is the selling price(list price) of the TV. The seller then gives a discount of $15$%, meaning that only $85$% of the price is left, so it becomes $0.85x$ which equals the $x-\frac{15}{100}x$ in your equation. Now, the seller decreases the price by $20$, so the price is now $0.85x-20$. Now, we subtract the expense of the TV to get the profit of $12$, making our equation $0.85x-20-206=12$. Solving, we get $x= 280$, so yes, you are right.


问:
How to compute the E(x) given its cdf function?
How could I compute the expected value E(x) given that its cdf is,

$F(x) = \dfrac{\exp(x)}{1+\exp(x)}$.

I know the method to derive E(x), as in this post. I still find it very hard to derive the result. Thanks.

答:
I prefer to refer to this question.

We have
$$F(x)=\frac{e^x}{e^x+1}\implies  f(x) = \frac{dF(x)}{dx}=\frac{e^x}{\left(e^x+1\right)^2}$$
$$E[X] = \int_{1}^{\infty}  x f(x)\,dx$$
$$I = \int  x f(x)\,dx=\frac{x \,e^x }{e^x+1}-\log \left(e^x+1\right)$$ Just use the bounds.


Title: Does $I(a,b)=\frac{a}{b+1}I(a-1,b+1)$?

Body: If $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, where $a \geq 1, b \geq 1$, does $I(a,b)=\frac{a}{b+1}I(a-1,b+1)$ for $a \geq 2$? I've tried numbers like $a=2, b=2$, and it seems to be true so far. Any proof that this is true?

Body:
$$I(a,b)=\int_{0}^{1}x^a(1-x)^b\,dx=\frac{\Gamma (a+1) \Gamma (b+1)}{\Gamma (a+b+2)}$$
$$I(a-1,b+1)=\int_{0}^{1}x^{a-1} (1-x)^{b+1}\,dx=\frac{\Gamma (a) \Gamma (b+2)}{\Gamma (a+b+2)}$$
$$\frac{I(a,b) } {I(a-1,b+1)}=\frac{a}{b+1}$$ just as you wrote.


问:
Problem on filters
$\mathcal{F}$ is filter on $\mathcal{I}$, but not ultrafilter.

Prove that $\exists$$\mathcal{X, Y}\notin\mathcal{F}$ | $\forall\mathcal{Z}\in\mathcal{F}$ $\mathcal{X}\cap\mathcal{Z}\neq$ $\mathcal{Y}\cap\mathcal{Z}$

Thank you for your time.

答:
If you are familiar with filter germs this can be a solution:

Suppose the converse:
$$(\forall A,B\subseteq I:A,B\notin \mathcal F)(\exists C\in \mathcal F)(A\cap C=B\cap C)\quad\quad (1)$$

For each $A,B\subseteq I$ define:
$$A \sim  B \quad \leftrightarrow\quad (\exists C\in \mathcal F)(A\cap C=B\cap C)$$
$\sim$ is an equivalence relation on $\mathcal P(I)$. Each equivalence class of $\sim$ is called a germ.

Now your question is answered by these propositions:

$\mathcal F$ is a germ.

$\mathcal F$ is an ultrafilter iff there are exactly 2 germs.

By $(1)$:
$$(\forall A,B\subseteq I:A,B\notin \mathcal F)(A\sim B)$$
which shows there are exaclty  2 germs: $[\emptyset]$ and $\mathcal F=[X]$. So $\mathcal F$ is an ultrafilter. A contradiction.

Title: Problem on filters


问:
Showing that two random variables are standard normal but are not bivariate normal
Question:

If $X\sim \mathcal{N}(0,1)$ and we define $Y$ such that $$ Y = \begin{cases} X& \text{ if }|X|\\ -X& \text{ if }|X|\geq a. \end{cases} $$ Show that $Y\sim \mathcal{N}(0,1)$ but $(X,Y)$ is not bivariate normal.

I'm mainly unsure on how to demonstrate the first part of the problem. I think that all $Y$ does is flip the tails of $X$ but my attempts in rigorously proving that $Y\sim \mathcal{N}(0,1)$ have been fruitless. I've tried computing the moment generating function of $Y$ but I don't know how deal with the limits of the resultant integral; where I wish to split it up in such a way as to recover the mgf of $X$.

The second part of the problem I think I've solved. I considered the distribution $X-Y$ which is clearly defined by $$ \begin{cases} 0 &\text{if }|X|<a, \\ 2X &\text{if }|X|\geq a. \end{cases} $$ Since $\mathbb{P}(|X|<a)\neq 0$, then we have that $X-Y$ cannot be continuous. Hence, they're not bivariate normal.

答: $P(Y \leq y)=P(X \leq y, |X| <a)+P(-X \leq y, |X| \geq a)$. Since $X$ and $-X$ have the same distribution  we can change $X$ to $-X$ in the second term.  So $P(Y \leq y)=P(X \leq y, |X| <a)+P(X \leq y, |X| \geq a)=P(X \leq y)$.


问:
Is the size of the set of injections from a set to another is bigger than the size of the second one?
Specifically, consider the set $X:=\{f:{M_1\times...\times M_n}\to B|f \,\,{\rm is}\,\, 1-1\}$, where all of the sets involved are at least $\aleph_0$ in size. Is $|X|>|B|$?

答:
Since $\vert M_1\times ...\times M_n\vert=\max\{\vert M_1\vert,...,\vert M_n\vert\}$ (all sets involved being infinite), we can just consider a single $M$:

Given infinite sets $M,B$, what can we say about the cardinality of the set $X$ of injections from $M$ to $B$?

There is one trivial case, one easy cases, and then the remaining case is surprisingly complicated.

The trivial case is if $\vert M\vert>\vert B\vert$. Then of course $X=\emptyset$.

The easy case is if $\vert M\vert=\vert B\vert$. It's a standard exercise that the set of self-bijections of an infinite set has the same cardinality as the set of subsets of that set, so now $\vert X\vert=2^{\vert B\vert}$.

Now we're left with the case $\vert M\vert<\vert B\vert$. At this point it's useful to prove the following: if $A,B$ are infinite sets with $\vert A\vert\le\vert B\vert$, there are as many injections from $A$ to $B$ as there are functions from $A$ to $B$ altogether. (This is a good exercise if you haven't seen it before.) So basically we're now just asking:

Suppose $\kappa<\lambda$ are infinite cardinals. What can we say about $\lambda^\kappa$ (= the cardinality of the set of functions from a set of size $\kappa$ to a set of size $\lambda$)?

While "nice" examples tend to be boring (= come out to just $\lambda$ itself), in generality things suddenly explode. Cardinal exponentiation is surprisingly nuanced. For one thing, many (in fact the vast majority of) basic questions about it are undecidable in $\mathsf{ZFC}$. One of the few things we can prove is a connection with cofinality: for all infinite $\alpha$ we have $\alpha^{cf(\alpha)}>\alpha$. So, for example, we could have $M=\mathbb{N}$ and $B$ be "very uncountable" but still have $\vert X\vert>\vert B\vert$.

(Of course I've tacitly applied choice several times in the above. Cardinal arithmetic becomes hideous without choice.)


问:
How to prove the action of the symmetry group of a cube on its pairs of opposite faces defines a surjective homomorphism from $S_4$ to $S_3$?

Show that the action of the symmetry group of a cube on pairs of opposite faces of the cube defines a surjective homomorphism from the symmetric group $S_4$ to $S_3$.

My attempt/logic:

I know that the rigid symmetries of a cube is isomorphic to $S_4$. I began by identifying the vertices of the cube with points {($\pm 1$,$\pm 1$,$\pm 1$) $\in$ $\mathbb{R}^3$}. Then, it should be clear that rotations of angle $\pi$ about the axises $x,y,z$ fix any pair of opposite faces. The number of elements in the set {$\pi_1$,$\pi_2$,$\pi_3$} of pairs of opposite faces is 3. Thus, the action defines a homomorphism $f:$ $S_4$ $\rightarrow$ $S_3$.

I am not sure if my proof is clear enough. I feel like I jumped a step or two since the proof seems a little too short/simple. How can I improve my proof? Any help or suggestion is appreciated. Thank you.

答:
Your proof starts out well: you have correctly shown that the map is a homomorphism from $S_4$ to $S_3$. (I might add a little more detail justifying why the map respects composition, but what you have seems all right.)

However, you have not yet shown that the map is surjective, i.e., that it maps to every element of $S_3$.

As a hint for how to show this, recall that the symmetric group $S_n$ is generated by the transpositions on $n$ elements - if you can find a way to locate each transposition in $S_3$ as an image of some rotation of the cube, that will suffice.


问:
Fourier series of a measurable function
Let $f: [0,2\pi] \to \mathbb{R}$ be a bounded Lebesgue measurable function that satisfies $\int_{0}^{2\pi}f(x)\sin( kx) = 0$ and $\int_{0}^{2\pi}f(x)\cos( kx) = 0$ for every $k\in \mathbb{N}$. It is my guess that this implies $f$ is constant almost everywhere.

However, I have much trouble proving it or finding an counter-example.  The only relevant result which I can think of is that the Fourier series of a continuous function $f$ on closed interval converges pointwise almost everywhere to $f$. So if we make the stronger assumption of $f$ being continuous (instead of just measurable), the statement should be true.

I would appreciate any thoughts whether the original guess is indeed true, and if so, how to go about proving it.

答:
The answer is "$f$ is constant almost everywhere".
Define $a_0 = \int_{0}^{2\pi} f(t) dt$.
Define $g:[0,2\pi] \rightarrow \mathbb{R}$ by $g(t)= f(t) - a_0$.
Then all Fourier coefficients of $g$ are zero.
Since Fourier-Stieltjes transform is injective, so $g =0$ almost everywhere, and hence $f$ is constant almost everywhere.

It was a difficult question asked by a lot of mathematicians in the 19th century: If the Fourier transform of $g$ is zero (for your case, i.e., on the circle group $\mathbb{T}$, all Fourier coefficients are zero), must $g$ be zero (in some sense)?

Obviously, if we modify $g$ on a set of measure zero, all Fourier coefficients are unaffected because they are defined by integrals. Therefore, the question only makes sense if two functions are identified iff they are equal almost everywhere.

If I remember correctly, this problem was solved by Cantor (or Lebesgue?).
When Cantor tried to solve this problem, he considered various subsets of $[0,2\pi]$ on which $g$ can be non-zero but all Fourier coefficients are zero. This sparked the development of set theory and point-set topology.

For a rigorous proof, you may consult any textbooks about Fourier Analysis, search for the theorem which states that "Fourier-Stieltjes transform is injective". A rigorous proof requires a lot of preparation, so there is no few lines proof.

Title: Fourier series of a measurable function


问:
How do I calculate the odds for a raffle when I know three specific facts?
I'm not sure how to calculate the odds for this:

There's a small raffle with 10 tickets sold. 3 of the tickets will win a prize. I've bought exactly 5 tickets. What are my odds of winning at least one prize? What is the formula?

It's not simply 1 in 2 (5 out of 10) because there's more than one prize.

Or more generally, what are my odds of winning an N ticket raffle, with M prizes offered when I buy X tickets?

答:
Your chance of not winning on the first ticket is $\frac 5{10}$ because there are five other tickets than yours.  Assuming you lose the first, your chance of losing the second is $\frac 49$ and assuming you lose that your chance of losing the third is $\frac 38$.  Your chance of winning at least one prize is $1$ minus the product of these
$$1-\frac 5{10}\cdot \frac 49 \cdot \frac 38=\frac {11}{12}$$


问:
$\forall a, b \in \mathbb{R}, \exists M, N \in \mathbb{Z}^*$ such that $Ma=Nb$
Is this true including irrational numbers for $a$ and $b$ (excluding the trivial case of $M = N = 0$)?

答:
There's a trick.

In general it wouldn't be true.  If, for example, of $a$ is rational and $b$ is irrational.  Then $Ma$ would be rational, and $Nb$ would not be rational, if $N \ne 0$.

ANd if $a,b,N,M$ are not zero then $Ma = Nb \iff \frac ab = \frac NM$ and the RHS must be rational and $a,b$ are under no such obligation.

But the question never said we can't have $M, N$ be $0$.

And if $M=0; N=0$ then $Ma = Nb = 0$ for all $a,b$.  So that certainly makes it true.


问:
A function is continuously differentiable in an open neighborhood of $x_o$ and differentiable at $x_o$, is it continuously differentiable at $x_o$?
Until I thought about this, I would have confidently answered yes to my question.

Let $f:\mathbb{R\to\mathbb{R}}$ be differentiable at $x_{0},$ and continuously differentiable on some open neighborhood $\mathcal{N}$ of $x_{o}$. It follows that $f$ is continuous at $x_{o}.$ But, since $\mathcal{N}$ is open, $x_{o}$ is not necessarily in $\mathcal{N}$. So is it possible that

$$\lim_{\delta \to 0} f^{\prime}\left(x_{o}+\delta\right)\ne f^{\prime}\left(x_{o}\right)?$$

For $\Delta x\ne0$ and $x_{o}+\delta+\Delta x\in\mathcal{N},$ the continuity of $f$ gives us,

$$\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\mathit{f}\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}=\frac{\mathit{f}\left(x_{o}+\Delta x\right)-f\left(x_{o}\right)}{\Delta x},$$

and therefore

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\left[\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\mathit{f}\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}\right]=f^{\prime}\left(x_{o}\right).$$

We also have

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\frac{f\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}=f^{\prime}\left(x_{o}+\delta\right).$$

But can we conclude $\underset{\delta \to\mathfrak{0}}{\text{Lim}}\mathit{f}^{\prime}\left(x_{o}+\delta\right)=\mathit{f}^{\prime}\left(x_{o}\right)?$  That is, for all $\epsilon>0$ can we find $\delta$ and $\Delta x$ such that $x_{o}+\delta+\Delta x\in\mathcal{N}$ and

$$\epsilon>\left|\frac{f\left(x_{o}+\delta+\Delta x\right)-f\left(x_{o}+\delta\right)}{\Delta x}-\mathit{f}^{\prime}\left(x_{o}\right)\right|?$$

Another way of stating the question is; can we reverse the order of the limit operation so that

$$\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\left[\underset{\delta\to\mathfrak{0}}{\text{Lim}}\frac{\Delta\mathit{f}_{x_{o}}\left(\delta+\Delta x\right)}{\Delta x}\right]=\underset{\delta\to\mathfrak{0}}{\text{Lim}}\left[\underset{\Delta x\to\mathfrak{0}}{\text{Lim}}\frac{\Delta\mathit{f}_{x_{o}}\left(\delta+\Delta x\right)}{\Delta x}\right]?$$

答: If your $\mathcal N$ is  deleted neighborhhod of $x_0$ then there is a well known counter-example: $f(x)=x^{2} \sin (\frac 1  x)$ for $x \neq 0$ and $f(0)=0$.  In this case $\lim_{x \to 0} f'(x)$ does not even exist.


问:
How to find the equation of common chord between a circle and a parabola?

Question:  Let $S_1$ be the focus of the parabola $$y^2 = 8x$$ and $PQ$ be the common chord of the circle $x^2 + y^ 2-2x-4y= 0$ and the given parabola. The area of the triangle PQS is?

I circumvented this problem knowing that the circle touches the origin as it has no constant value and so does the standard parabola. I assumed (accidentally and also correctly) that the chord was the diameter, knowing the centre was $(1,2)$ and I found the other vertex as $(2,4)$ and solved the question getting the correct answer.

Is there perhaps a generalised method to find the equation of the parabola and the circle? I know that the chord of a parabola is given by the formula $y(t_1+1_2)=2at_1t_1$, but I haven't figured how to use it to find actual equations of chord apart the formula

答:
To find the common chord of the parabola and the circle, we must solve their equations simultaneously. Substituting $x = \frac{y^2}{8}$ in the equation for the circle:
$$ \frac{y^4}{64} + y^2- 2\frac{y^2}{8} - 4y = 0 \qquad \implies \qquad y^4 + 48y^2 - 256y = 0$$
$\therefore$ Either $y = 0, \implies x = 0,$ or $y^3 -48y -256 = 0.$ By rational root theorem, $y=4$ satisfies the cubic equation. $y = 4 \implies x = 2$. Thus $(0,0)$ and $(2,4)$ are the points of intersection.
Let $P = (0,0)$ and $Q = (2,4)$. We know that $S_1 = (2,0)$. Now we can use Heron's formula to calculate the area.


问:
Positioning problem optimal control
Consider the positioning problem:
$\dot x_1 = x_2$,
$\dot x_2 = u$ with $x_1(0)=0,x_2(0)=X, X>0$. show that the bang-bang control switch can be employed to steer the system to the origin. Find the time taken and the position where the switch must be made. </p>

答:
I am going to do an outline and let you fill in the blanks.

For bang-bang, what are the different choices we should investigate for $u$? They are $\pm 1$. You should note why.

Case 1: $u = 1$

$$x'_1 = x_2$$

$$x_2' = 1$$

with $x_1(0)=0,x_2(0)=X, X>0$.

For this system, we have:

$\tag 1 x_2 x'_2 = x'_1, ~~\text{so}~~ \dfrac{1}{2}\left(x_2^2\right)' = x_1$

If we let $t_0$ belong to the time interval where $u = 1$, and integrate from $t_0$ to $t$, we get:

$$\dfrac{x_2^2(t)}{2} - \dfrac{x_2^2(t_0)}{2} = x_1(t) - x_1(t_0)$$

$$x_2^2(t) = 2 x_1(t) + \left(x_2^2(t_0) - 2x_1(t_0)\right)$$

Using the initial conditions, we have $t_0 = 0$ and this reduces to:

$$\tag 2 x_2^2(t) = 2x_1(t) + X^2 \rightarrow x_2^2(t) = 2x_1(t) + w$$

This tells us that we have a parabola and that as long as $u=1$, the trajectory stays on the curve $(2)$ for some constant $w$.

If we draw the trajectories for this, we have:

Case 2: $u = -1$

Repeat the steps for Case 1, but $u = -1$.

The trajectories look like:

Now that we have analyzed the system for the appropriate values of $u$, we see that the solutions $(x_1(t), x_2(t))$, lie on two different parabolas.

To design an optimal control, we start on one parabola and steer to the origin by starting with $u = -1$ and our trajectory is moving down. We just switch control to $u=1$, which causes us to move to the first parabola and we end up at the origin.

You can work the time portion since you have two parabolas and need to find the time when the switch should occur.

Title: Positioning problem optimal control


问:
$Ind_H^G(\pi\oplus \mu) \simeq Ind_H^G(\pi)\oplus Ind_H^G(\mu)$
Let $\pi$ and $\mu$ be representations of a subgorup $H\leq G$. ($G$ a finite group.) My question is whether or not it is true that
$$
Ind_H^G(\pi\oplus \mu) \simeq Ind_H^G(\pi)\oplus Ind_H^G(\mu)
$$
That is, is the induction of the direct sum of two representations equal to the direct sum of the induced representations?

I have been working with inducing irreducible representations, but I want to understand how how induce representations that are not irreducible (so are direct sums of irreducibles). I was hoping the above would hold.

答: The "best" definition of induction is $\mathrm{Ind}_H^G M := kG \otimes_{kH} M$. Once you know $kG$ is a free $kH$-module, the coset-based definition follows immediately. In any case, tensor products are naturally bilinear, giving your result immediately.


问:
Newton's Law example
A cup of tea starts at 80 degres, and moved to outside at -10 degrees. After 7 minutes, tea cools to 55 degrees. How do I make a equation for it? I use Newton's law of cooling, Temperature = (Teperature of original - Temperature of environment)$e^{kt}$+Temperature of environment. So I made equation

$T(7)=(80-(-10))e^{-7k}-10$. Is check to be correct?

答:
Weird, I thought Newton's law of cooling was "the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings" from https://en.wikipedia.org/wiki/Newton's_law_of_cooling.

Let the temperature of the cup be $y$. Let $t$ be time. The surrounding temperature is $-10$, thus the difference in temperatures is $(y+10)$. The rate of heat loss $-\frac{dy} {dt} = k(y+10)$, where $k$ is the constant of proportionality. Rearrange this to get $$\frac{dy}{y+10} = -k \space dt$$. We may integrate both sides to get $$ln(y+10) = -kt + C$$, or $$y = Ce^{-kt}-10$$. We know that $y= 80$ when $t=0$, so we have $$80 = C-10$$ or $$C=90$$. So our equation is now $$y=90e^{-kt}-10$$. We also know that $y=55$ when $t=7$. Thus, we get $$55=90e^{-7k}-10 \implies .72222222 =e^{-7k} \implies -.325 = -7k \implies k=.046$$. Thus, your equation should actually be $$y = 90e^{-.046t}-10$$. Now you can plug in $7$ to get your equation, but you know $k$ now.

Title: Newton's Law example


问:
Calculating $\int \frac{x}{\sqrt{x^{2}-36}} \, {\rm d} x$
I wish to calculate

$$\int \frac{x}{\sqrt{x^{2}-36}} \, {\rm d} x$$

I use a substitution $x=6\sec\theta$ and get $\int \csc\theta d\theta$ as intermediate.. Then, after some calculation

$$-\ln\left|\frac{x}{\sqrt{x^2-36}}+\frac{6}{x^2-36}\right|$$

Is it correct?

答:
Go easy on it. Take

$\displaystyle x^2 - 36 = y \ \implies 2x \ \mathrm {d}x = \mathrm {d}y$

$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1/2} \ \mathrm {d}y = \sqrt {y} + c .$$

Substitute the value of $y$ and you get :

$$\displaystyle \int \frac {x}{\sqrt {x^2 -36}} \ \mathrm {d}x = \sqrt {x^2 -36} + c .$$

But as mentioned by @integration brainstorm : A typo in the Question, using same substitution :

$$\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {1}{2}\int y^{-1} \ \mathrm {d}y = \frac {\ln {y}}{2} + c . $$

Thus $\displaystyle \int \frac {x}{{x^2 -36}} \ \mathrm {d}x = \frac {\ln (x^2 -36)}{2} + c . $


问:
How can we represent set $S= \{z:|z|>|z-1|\}$ graphically.
Here I have plotted graph of $|z|$ and $|z-1|$ but I am not getting the required condition. I have to plot this using divs in a plane for which general set is $\{z:|z-z_0|<r\}$ where $r$ is radius centered at $z_0$.

答: Your condition reads: the distance from $z$ to the origin is greater than the distance to the real number $1$. If you draw the bisector of the segment joining $0$ and $1$, your region is the right half-plane.


问:
About Regulated Functions
Definition. Let $X$ be a Banach space. A mapping $f:[a,b]\to X$ is called regulated if it has one sided limits.

In the setting of a Hausdorff topological vector space $X$, can we still define regulated functions?

答: As in a Hausdorff topological vector space the one sided limit of a function is a defined object, you surely can do this, exactly as in the case of Banach spaces. (You even do not need the Hausdorff condition or vector space structure, as "having a limit" does not need this). </p>

Title: About Regulated Functions


答:
Let
$$
S_{n}(x)=\sum_{m=-n}^{n} \hat{f}(m)\left[\frac{1}{\sqrt{2 \pi}} e^{i m x}\right]
$$

Note $\{\frac{e^{i m x}}{\sqrt{2 \pi}}\}_{m\in \mathbb{Z}} $ is an equivalent base to cos/sen form.

Then it can be proved that Fejer sums $f_N = \frac{1}{2 N+1} \sum_{n=-N}^{N} S_{n}(x)$ converge almost uniformly to $f$ on $[0,2 \pi] .$ If only $\hat{f}(0)\neq0$, then it follows that

$$
f_N \xrightarrow{a.u.} f = S_0 = \frac{\hat{f}(0)}{\sqrt{2 \pi}} \equiv constant
$$

Title: Fourier series of a measurable function

Body:
Let $f: [0,2\pi] \to \mathbb{R}$ be a bounded Lebesgue measurable function that satisfies $\int_{0}^{2\pi}f(x)\sin( kx) = 0$ and $\int_{0}^{2\pi}f(x)\cos( kx) = 0$ for every $k\in \mathbb{N}$. It is my guess that this implies $f$ is constant almost everywhere.

However, I have much trouble proving it or finding an counter-example.  The only relevant result which I can think of is that the Fourier series of a continuous function $f$ on closed interval converges pointwise almost everywhere to $f$. So if we make the stronger assumption of $f$ being continuous (instead of just measurable), the statement should be true.

I would appreciate any thoughts whether the original guess is indeed true, and if so, how to go about proving it.


答: $|z| > |z-1|$ basically translates to, "Find the points $z$ such that the distance from $z$ to the origin is greater than the distance between $z$ & $1$". This is what the graph looks like, please pretend the $y$-axis is the complex axis.

Title: How can we represent set $S= \{z:|z|>|z-1|\}$ graphically.

Body: Here I have plotted graph of $|z|$ and $|z-1|$ but I am not getting the required condition. I have to plot this using divs in a plane for which general set is $\{z:|z-z_0|<r\}$ where $r$ is radius centered at $z_0$.


问:
Find integrating factor for $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$
There's differential equation $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$. Then what is integrating factor. First I convert it to standard form $\frac{dy}{dx}+P(x)y=Q(x)$. And integrating factor is $e^{\int P(x)dx}$. But first to find $P(x)$ I should rearrange it.

答: Yes, You should divided the equation by the coefficient of $\frac{\mathrm{d}y}{\mathrm{d}x}$, The integrating factor is root over $1+\sin^2 x$.


答: Either divide by $2(1+\sin^2 x)$ to find $P(x)=\sin(2x)/[2(1+\sin^2 x)]$ and hence $e^{\int^x P}$ and the actual integrating factor $e^{\int^x P}/[2(1+\sin^2 x)]$, or just note $\frac{\mathrm{d}}{\mathrm{d}x}[2(1+\sin^2 x)]=2\sin 2x$ almost gets you there so you just need to multiply by an appropriate power of $(1+\sin^2 x)$.

Title: Find integrating factor for $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$

Body: There's differential equation $2(1+\sin^2(x))\frac{dy}{dx}+\sin(2x)y=x$. Then what is integrating factor. First I convert it to standard form $\frac{dy}{dx}+P(x)y=Q(x)$. And integrating factor is $e^{\int P(x)dx}$. But first to find $P(x)$ I should rearrange it.


答:
If you rearrange your equation nicely and bring it to the form $y' + P(x)y = Q(x)$ you'll notice that $$P(x) = \frac{\sin 2x}{2(1+\sin^2 x)} , Q(x) = \frac{x}{2(1+\sin^2 x)}$$
where $\displaystyle y' = \frac{dy}{dx}$. You want to find $\displaystyle e^{\int P(x)dx}$, so I'll tell how to find $\int P(x)dx$.
$$\int \frac{\sin 2x}{2(1+\sin^2 x)} dx = \int \frac{du}{2u} = \frac{\ln u}{2} + C = \frac{\ln (1+\sin^2x)}{2} + C$$
where $u = 1+\sin^2 x \implies du  = \sin2x dx$. Let $C = 0$ (think why we can do this), to get $\displaystyle e^{\int P(x)dx} = \sqrt{1 + \sin^2x }$, which is your required integrating factor!


答:
If you divide by the coefficient of $y'$, your equation is written in the standard form with
$$
P(x)=\frac{\sin 2x}{2(1+\sin^2x)}=\frac{\sin x\cos x}{1+\sin^2x}
$$
The integrating factor is
$$
\mu(x)=\exp (\int P(x)\,dx)=\exp [\ln(1+\sin^2x)/2]=\sqrt{1+\sin^2 x}
$$


问:
Why does the cumulative summation of cos(n^2) have a strange symmetry?
I was messing around in Desmos and typed this in: $\displaystyle{\sum_{n=1}^{floor(x)}\cos(n^2)}$

It produces this graph in the $XY$-plane (keep in mind the axis ratios):

<img src="https://i.stack.imgur.com/uUl5Y.png" alt="Cumulative summation in the $XY$-plane" />

It appears to produce a highly symmetrical, fractal-like graph. I tried replacing $n^2$ with $n$ to different powers, but found no other symmetry like this. Does anyone have any idea why this phenomenon occurs?

Edit: Doing some more investigation, I realized that the graph is symmetric about the lines $x=(2k+1)177.5$, for non-negative integers $k$. As for the significance of $177.5$, the only thing I have realized so far is that $177.5$ is very nearly $\frac{113\pi}{2}$, and thus $\cos(177.5)$ is very nearly $0$.

答: The near periodity of this part of the graph is due to the fact that $2\pi$, the period of $\cos x$, is very close to $\frac{710}{113}$. This means that when we plug in $n$ and $n+710$, the resulting squares are very close to each other modulo $2\pi$. However, note that this phenomenon does not last very long:


问:
Is a single element a chain or an anti-chain?
Is a single element in a Hasse Diagram a chion or an anti-chain or both?

As a Hasse Diagram contains reflexive elements, it is compararble to itself. So it is a chain.

But I read somwhere that a single element is also an anti-chain

How is it an anti-chain if the Hasse diagram does not contain any other element to which it might be incomparable?

答:
A subset $A$ of a partial order $\langle P,\le\rangle$ is an antichain if and only if there do not exist $a,b\in A$ such that $a\le b$ and $a\ne b$. In other words, what keeps a set $A\subseteq P$ from being an antichain is having two different elements that are related by the order $\le$. If $A=\{x\}$, then $A$ doesn’t have two different elements at all, so it certainly doesn’t have two different elements that are related by the order, and it must therefore be an antichain.

Perhaps you’ve seen the following equivalent definition of an antichain: $A\subseteq P$ is an antichain if and only if for all $a,b\in A$, $a\le b$ implies that $a=b$. If you use this definition, it is if anything even clearer that $\{x\}$ is an antichain for each $x\in P$: if $a,b\in\{x\}$, then $a=x$ and $b=x$, so $a=b$ whether or not $a\le b$ (though of course it’s true in this case that $a\le b$).


问:
Show the map between orthogonal complement and range is a bijection
Let $H,K$ be Hilbert spaces.

Suppose $T\in B(H,K)$.

Show that the restriction map of $T$, $T': \mathcal{N}^{\perp}(T) \rightarrow \mathcal{R}(T) $ is a bijection.

My attempt:

Note that $ \mathcal{N}^{\perp}(T) = \{ h\in H: \langle h, k \rangle=0, \forall k\in \mathcal{N} \}$

Injectivity:

Suppose $y_1=y_2 \in \mathcal{R}(T)$. Then there exists $x_1,x_2 \in H$ such that $Tx_1=Tx_2$.

But what we want to show is that $x_1,x_2 \in  \mathcal{N}^{\perp}(T)$.

So we want show that $\langle x_1,k \rangle = \langle x_2,k \rangle=0 $ for all $k$.

And this is where I got stuck.

Surjectivity:

Let $y \in \mathcal{R}(T)$. So there exists $x\in H$ such that $Tx=y$.

But we want to show that $x\in \mathcal{N}^{\perp}(T)$.

Again I'm stuck at this point.

Any help will be appreciated.

答:
Injectivity: Suppose $x,y \in \mathcal N^{\perp} (T)$ and $T'x=T'y$. Then $Tx=Ty$ so $x-y \in \mathcal N (T)$. Hence, $ \langle (x-y), (x-y) \rangle =\langle x, (x-y) \rangle-\langle y, (x-y) \rangle=0-0=0$. This implies that $\|x-y\|=0$ so $x=y$,

Surjectivity: Let $y \in \mathcal R (T)$. Then there exists $x$ such that $y=Tx$. By a standard result in Hilbert space theory we can write $x$ as $x_1+x_2$ where $x_1\in \mathcal N(T)$ and $x_2 \in  \mathcal N^{\perp} (T)$. Now  $T'(x_2)=T(x_2)=0+x_2=T(x_1)+T(x_2)=Tx=y$.


问:
I wanted to know in the statement $5$ added to a number divided by $3$ whether we divide after adding $5$ or the number is already divided?
If we were to convert the the statement

$5$ added to a number divided by $3$

to equation.

Would it be $5+\frac x3$ or $\frac{5+x}3$?

答:
I personally vote for $5+\frac x3$, just based on my gut.

Ultimately, however, you will have to ask the author of the phrase (or read their mind). There is no way to be absolutely certain.

Or try both, keep calculating, see what you get, and check whether only one result makes sense in the end.


问:
Linear Diophantine equation modulo $n$
Is there a way to find all solutions $(x,y)$ to the equation: $a x + b y = c\pmod n$ for fixed integers $a,b,c$ and $n$?

Regards,
Leon

答:
So, $ax+by=c+d\cdot n$ where $d$ is any integer

Now, if $(a,b)=g$ does nor divide  $(c,d\cdot n),$ for any $d$ there will be no solution.

If $g$ divides $(c,d\cdot n)$  let $\frac aA=\frac bB=\frac cC=\frac {d\cdot n}D=g$

So, we $Ax+By=C+D$ where $(A,B)=1$

Using Bézout's Lemma, we can find integers $u,v$ such that $Au+Bv=1$

So, we have $Ax+By=(Au+Bv)(C+D)$

$\implies A\{x-u(C+D)\}=B\{v(C+D)-y\}$

$\implies \frac{B\{v(C+D)-y\}}A=x-u(C+D)$ which is an integer

$\implies  A$ divides $B\{v(C+D)-y\}$

$\implies  A$ divides $\{v(C+D)-y\}$ as $(A,B)=1$

$\implies  v(C+D)-y=A\cdot m$ where $m$ is any integer

$\iff y=v(C+D)-A\cdot m$

Title: Linear Diophantine equation modulo $n$


答:
Injectivity: suppose $T(x_1)=T(x_2)$, with $x_1,x_2\in\mathcal{N}^\perp$. Since $T(x_1-x_2)=T(x_1)-T(x_2)=0$, $x_1-x_2\in\mathcal{N}$, Therefore $x_1-x_2\in\mathcal{N}\cap\mathcal{N}^\perp$ and hence $x_1-x_2=0$.

Surjectivity. Take $y\in\mathcal{R}(T)$. Then $y=T(x)$ for some $x\in H$. Decompose $x=x_1+z_1$ where $z_1\in\mathcal{N}$ and $x_1\in\mathcal{N}^\perp$.Clearly, $T(x_1)=T(x)=y$.

Title: Show the map between orthogonal complement and range is a bijection

Body:
Let $H,K$ be Hilbert spaces.

Suppose $T\in B(H,K)$.

Show that the restriction map of $T$, $T': \mathcal{N}^{\perp}(T) \rightarrow \mathcal{R}(T) $ is a bijection.

My attempt:

Note that $ \mathcal{N}^{\perp}(T) = \{ h\in H: \langle h, k \rangle=0, \forall k\in \mathcal{N} \}$

Injectivity:

Suppose $y_1=y_2 \in \mathcal{R}(T)$. Then there exists $x_1,x_2 \in H$ such that $Tx_1=Tx_2$.

But what we want to show is that $x_1,x_2 \in  \mathcal{N}^{\perp}(T)$.

So we want show that $\langle x_1,k \rangle = \langle x_2,k \rangle=0 $ for all $k$.

And this is where I got stuck.

Surjectivity:

Let $y \in \mathcal{R}(T)$. So there exists $x\in H$ such that $Tx=y$.

But we want to show that $x\in \mathcal{N}^{\perp}(T)$.

Again I'm stuck at this point.

Any help will be appreciated.


问:
Is a Cauchy sequence - preserving (continuous) function is (uniformly) continuous?
Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $f:X\to Y$ be a function and suppose for any Cauchy sequence $(a_n)$ in $X$, $(f(a_n))$ is a Cauchy sequence in $Y$.

Is $f$ continuous?

Let $f$ be continuous, is it uniformly continuous?

答: Consider $f:(0,1)\to \mathbb R$, defined by $f(x)=1/x$. $f$ is continuous. Sequence $1/n$ is Cauchy in $(0,1)$, while $f(1/n)=n$ is not a Cauchy sequence.


问:
Two problems on Autonomus ODE properties of unknow solutions
I have to explain a student two exercises but I don't know how continue at the first and how start at the second one.

Let $f(x)$ continously differentiable and such that $f(0)=0$. Show that the solutions of $x'=f(x)h(t)$ cannot change sign.

My attemp:
As $f$ is cont. diff. we know that $f$ is continous and also $\frac{\partial{f}}{\partial{x}}=f'(x)$, then for the existence and unicity theorem we have a unique solution $x(t)$. Since $f(0)=0$ we have that $t=0$ is an extreme value for $x(t)$. On a graph, I can see that $x(t)$ cannot change sign independent of the extreme value, and I know that we have to use the product $f(x)h(t)$ and law of signs but I can see how.

Find the limits as $t\rightarrow \pm \infty$, of the solution $\phi(t)$ of the ivp $x'=(x+2)(1-x^4)$, with $x(0)=0$.

I'm lost at this one

The first idea that came out to me are using algebra we can put: $x'=(x+2)(1+x)(1-x)(1+x^2)$ via partial fractions we can get an antiderivative and then estimate that limits, but this appears a lot of  unnecessary work. There's an elegant way to get this excersice?

答:
For the second one, simply looking at the sign of $x'$ in different intervals gives you the phase portrait, where you should be able to easily read off where you end up if you start at $x=0$:
$$
\longrightarrow
\quad
-2
\quad
\longleftarrow
\quad
-1
\quad
\longrightarrow
\quad
1
\quad
\longleftarrow
$$


答:
It's ambiguous. You could parse it as "(5 added to a number) divided by 3" or "5 added to (a number divided by 3)".

When speaking the phrase, you can sort of indicate the "parenthesis" with a pause.

Title: I wanted to know in the statement $5$ added to a number divided by $3$ whether we divide after adding $5$ or the number is already divided?

Body:
If we were to convert the the statement

$5$ added to a number divided by $3$

to equation.

Would it be $5+\frac x3$ or $\frac{5+x}3$?


问:
Finding the general equation to the two-body problem
So, just finished reading the Three Body Problem by Cixin Liu (very good novel), and was curious about the two-body problem. Of course, the two-body scenario is much simpler and more predictable than the two-body problem, so I was hoping to find a general solution to the two-body problem but haven't found one online, just solutions of specific cases. I decided to try to go at it on my own for the 2-dimensional system, but I got stuck.

I started by saying the first planet, Planet $x$, had a 2-dimensional position vector $\vec{x}$ and a mass $m_x$, and the second (Planet $y$) had a position vector of $\vec{y}$ and mass $m_y$. With the generalization that the two bodies are point masses with the only force being gravity, and assuming that both the initial positions/velocities and the masses of the two bodies are known, I came up with these two second-order differential equations:

$$\frac{d^2\vec{x}}{dt^2}=\frac{Gm_y}{||\vec{x}-\vec{y}||^3}(\vec{y}-\vec{x}) \\
\frac{d^2\vec{y}}{dt^2}=\frac{Gm_x}{||\vec{x}-\vec{y}||^3}(\vec{x}-\vec{y}) \tag{1}$$

I got these from these classical physics equations:
$$\underbrace{F_{y \rightarrow x}}_\text{force of gravity on x} = \underbrace{\frac{G m_y m_x}{||\vec{x}-\vec{y}||^2}}_\text{gravitational force}=\underbrace{m_x\frac{d^2\vec{x}}{dt^2}}_\text{mass and acceleration}$$

Then, I saw that there was a lot similar in the terms $\frac{d^2\vec{x}}{dt^2}$ and $\frac{d^2\vec{y}}{dt^2}$, so I used substitution to get this relationship:

$$m_x\frac{d^2\vec{x}}{dt^2} + m_y\frac{d^2\vec{y}}{dt^2} = 0$$

And that makes sense with the equal and opposite forces. Then I integrated twice:

$$m_x\vec{x} + m_y\vec{y} = \vec{c}t + \vec{d} \tag{2}$$

... where $\vec{c}$ and $\vec{d}$ are vector integration constants (I realized that they'd need to be vectors and not just constants). Using initial conditions, this means:

$$\vec{c} = m_x\vec{x}'(0) + m_y\vec{y}'(0) \\
\vec{d} = m_x\vec{x}(0) + m_y\vec{y}(0) \tag{3}$$

I solved $(3)$ for $\vec{y}$ (and later for $\vec{x}$) to substitute into $\frac{d^2\vec{x}}{dt^2}$ and $\frac{d^2\vec{y}}{dt^2}$ and uncouple the two differential equations in $(1)$:

$$\frac{d^2\vec{x}}{dt^2}=\frac{-Gm_y^3}{||(m_x+m_y)\vec{x}-\vec{c}t-\vec{d}||^3}((m_x+m_y)\vec{x}-\vec{c}t-\vec{d}) \\
\frac{d^2\vec{y}}{dt^2}=\frac{-Gm_x^3}{||(m_x+m_y)\vec{y}-\vec{c}t-\vec{d}||^3}((m_x+m_y)\vec{y}-\vec{c}t-\vec{d}) \tag{4}$$

But this is where I got stuck. I coded something in Matlab that uses $(4)$ in the ode45 function to see how things move at least (code below), but I definitely have not been able to find any methods to solve either equation in $(4)$. Does such a closed form even exist? Of course that would mean that $\vec{x}$ and $\vec{y}$ aren't always functions, but I was wondering if maybe there's a parameterized solution or something like that. Thanks!

Here's the code for my demonstration, it's kinda janky:

```<code>% %%%%%%%%% Set stuff up %%%%%%%%%

% planet x
x_init = [6 9]';
dx_init = [0 -0.001]';
mx = 6900;

% planet y
y_init = [-6 -9]';
dy_init = [0 0]';
my = 420000;

% other things
tlength = 100000; % diffeq soln time length (bigger means longer duration)
scale = 300; % scale of velocity vectors
speed = 5; % speed of animation (bigger means faster)

% %%%%%%%%% Animate Motion %%%%%%%%%
plotPlanets([x_init dx_init y_init dy_init],[mx my tlength],scale,speed)

%% Functions
function [T,D] = getPositions(vecs, cons)
    % vecs := matrix containing info for the planet in this format:
    %    [[x0] [dx0] [y0] [dy0]]
    % cons := vector with relevant constants in this format:
    %    [[mass of x] [mass of y] [time length]]
    syms x1(t) x2(t) dx1(t) dx2(t) y1(t) y2(t) dy1(t) dy2(t)

    funcx = formula([x1 x2 dx1 dx2]);
    funcy = formula([y1 y2 dy1 dy2]);

    % Unpack
    x_init = vecs(:,1);
    dx_init = vecs(:,2);
    y_init = vecs(:,3);
    dy_init = vecs(:,4);

    mx = cons(1);
    my = cons(2);
    tlength = cons(3);
    c = mx*dx_init + my*dy_init;
    d = mx*x_init + my*y_init;

    eqnsx = getFormulas([c d],[mx my],[funcx t]);
    eqnsy = getFormulas([c d],[my mx],[funcy t]);
    eqns = [eqnsx eqnsy];
    vars = [funcx funcy];
    init = [x_init' dx_init' y_init' dy_init'];

    F = odeFunction(eqns,vars);
    [T, D] = ode45(F,[0 tlength],init);
end

function eqns = getFormulas(vecs, mass, funcs)
    % vecs := matrix containing info for the planet in this format:
    %    [[c] [d]]
    % mass := vector with relevant constants in this format:
    %    [[own mass] [other mass]]
    % funcs := symbolic functions in the format [z1 z2 dz1 dz2 t].

    % Set up
    G = 6.67408*10^(-11); % Gravitational constant

    % Unpack
    c = vecs(:,1);
    d = vecs(:,2);
    mz = mass(1);
    mw = mass(2);

    % fromz is vector between planets
    fromz = formula((mz+mw)*[funcs(1);funcs(2)] - c*funcs(5) - d);
    eqn1 = -G*mw^3/norm(fromz)^3*fromz(1);
    eqn2 = -G*mw^3/norm(fromz)^3*fromz(2);
    eqns = formula([funcs(3:4) eqn1 eqn2]);
end

function plotPlanets(vecs, cons, scale, speed)
    [T,D] = getPositions(vecs,cons);
    Dx = D(:,1:4);
    Dy = D(:,5:8);
    xmin = min([min(Dx(:,1)) min(Dy(:,1))]);
    xmax = max([max(Dx(:,1)) max(Dy(:,1))]);
    ymin = min([min(Dx(:,2)) min(Dy(:,2))]);
    ymax = max([max(Dx(:,2)) max(Dy(:,2))]);
    tlast = T(1);

    figure(1)
    hold on
    axis(sizeUp(xmin,xmax,ymin,ymax))
    pathx = plot(1);
    px = plot(Dx(1,1),Dx(1,2),"bo");
    vx = quiver(Dx(1,1),Dx(1,2),scale*Dx(1,3),...
        scale*Dx(1,4),"Color",[0 0 0]);
    pathy = plot(1);
    py = plot(Dy(1,1),Dy(1,2),"ro");
    vy = quiver(Dy(1,1),Dy(1,2),scale*Dy(1,3),...
        scale*Dy(1,4),"Color",[0 0 0]);
    pause(0.5)

    for i = 2:size(D,1)
        if ~any(findobj(allchild(0), 'flat', 'type', 'figure') == 1)
            break
        end
        delete(pathx);
        delete(px);
        delete(vx);
        delete(pathy);
        delete(py);
        delete(vy);

        pathx = plot(Dx(1:i,1),Dx(1:i,2),"b");
        px = plot(Dx(i,1),Dx(i,2),"bo");
        vx = quiver(Dx(i,1),Dx(i,2),scale*Dx(i,3),...
            scale*Dx(i,4),"Color",[0 0 0]);
        pathy = plot(Dy(1:i,1),Dy(1:i,2),"r");
        py = plot(Dy(i,1),Dy(i,2),"ro");
        vy = quiver(Dy(i,1),Dy(i,2),scale*Dy(i,3),...
            scale*Dy(i,4),"Color",[0 0 0]);
        pause((T(i)-tlast)/speed/1000)
        tlast = T(i);
    end
    if any(findobj(allchild(0), 'flat', 'type', 'figure') == 1)
        hold off
    end
end

function sizes = sizeUp(xmin, xmax, ymin, ymax)
    if xmax - xmin > ymax - ymin
        len = (xmax - xmin)/2;
        ymid = sum([ymax ymin])/2;
        ymin = ymid - len;
        ymax = ymid + len;
    elseif xmax - xmin < ymax - ymin
        len = (ymax - ymin)/2;
        xmid = sum([xmax xmin])/2;
        xmin = xmid - len;
        xmax = xmid + len;
    end
    pad = abs(len*0.1);
    sizes = [xmin-pad, xmax+pad, ymin-pad, ymax+pad];
end
</code>```

答:
The easiest way to solve the two-body problem is to reduce it into two independent one-body problems by introducing the center of mass.

This effectively decouples the motion of the two bodies into a (1) center of mass motion (= 1st one-body problem) and a (2) displacement vector motion (= 2nd one-body problem).

For detailed info, you may wish to refer to this Wikipedia link:
https://en.wikipedia.org/wiki/Two-body_problem#Center_of_mass_motion_(1st_one-body_problem)


问:
Marginal Distribution Calculation for use in Binomial Distribution.
[1

In the question I know how to calculate marginal distribution for X1 and X2.

Like for X1 the marginal distribution for each column is the sum of each joint probability mass function in that column. For example the marginal distribution for 0 column of X1 = 0.343 But I'm not able to understand how to use marginal Probability of X1 and X2 in the binomial distribution as asked in the question in the image!

答:
Just summing the probability per row and column you get  your distribution

For example, as $X_1$ is concerned you get

$$P(X_1=3)=0.027=p^3$$

That is $p=\sqrt[3]{0.027}=0.3$

Which means $X_1\sim Bi(3;0.3)$

Similar reasoning for $X_2$