问:
Tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, $1 \le i \lt j \le k$
To generalize this question I would like to count how many tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, for any $i,j$ such that $1 \le i \lt j \le k$, are there.

The case $k=2$ has been solved in the linked question and the result is $\frac{3^n-1}{2!}$.

The case $k=3$ can be stated as $\frac{7^n-1}{3!}-\frac{3^n-1}{2!}$ and a computer test confirms the result.

For case $k = 4$ I would have expected something like $\frac{15^n-1}{4!}-\frac{7^n-1}{3!}+\frac{3^n-1}{2!}$, but the first addendum is not an integer. The values computed (hopefully correctly) for $n=1,2,3,4,5,6$ are $0,1,67,1546,27550,445531$.

To complete with computing data, the case $k = 5$ gives $0,0,56,4144,180096,6480656$, for $k = 6$ we have $0,0,28,7896,866432,69656776$, for $k = 7$: $0,0,8,11408,3308736,601192496$.

Any hint? If possible, for the general case, but especially for $k = 4$. Thank you.

答:
Another, non direct, way to answer the linked question is by using binomial inversion for the numerical part and using an inclusion-exclusion argument, so that if $$\binom{2^m}{2}=\sum_{n=0}^m {m \choose n}\frac{3^n-1}{2},$$ then $$\frac{3^m-1}{2}=(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{2}.$$ In this way, I think you are looking for
$$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}.$$
which agrees with your computation. You can check this combinatorially using inclusion-exclusion by considering which element is missing in your decomposition.

Edit:
Consider then
$$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}=\binom{2^{m}}{k}-\sum _{n=1}^{m}\binom{m}{n}(-1)^{n-1}\binom{2^{m-n}}{k},$$
this corresponds to doing $$\left |\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}\right |,$$ where $\mathcal{A}$ is all possible ways to get $k$ different subsets $\{A_i\}$ of $[n]$ and $\mathcal{A}_{\ell}$ is the number of ways to get $k$ different subsets of $[n]$ such that $\ell$ is in non of them.

 Show that
$|\mathcal{A}|=\binom{2^m}{k},$ and that if $X\subseteq [m],$ then  $$\left |\bigcap _{x\in X}\mathcal{A}_x\right |=\binom{2^{n-|X|}}{k}.$$
Notice that $$\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}$$ is exactly the objects you want. You eliminate every time their union was not exactly $[n].$


问:
Proof Verification of Exercise 2.9 Baby Rudin: Prove that the complement of $E^{o}$ is the closure of the complement of $E$.
I am stuck on part d of this exercise:

Let $E^{o}$ denote the set of all interior points of a set. Prove that the complement of $E^{o}$ is the closure of the complement of $E$.

I solved it using double inclusion:

Let $p$ $\in (E^{o})^{c}$. Since $p$ is not an interior point of $E$, every neighborhood of $p$ contains a point not in $E$ (or else we have a neighborhood of $p$ that is contained in $E$ and p is an interior point) so $p$ is a limit point of $E^{c}$ so $p$ $\in$ $\overline{E^{c}}$. Conversely, let $q$ $\in$ $\overline{E^{c}}$. Then $q$ belongs to $E^{c}$ and is not an interior point of E, or $q$ is a limit point of $E^{c}$ and no neighborhood of $q$ is contained in $E$, so $q$ is not an interior point of $E$ and so $q \in (E^{o})^{c})$

Is this proof correct? If not, where exactly does it break down? Thank you for your time.

答:
Another proof :

Let's prove that $E\backslash \overset{\circ}{A}\subset\overline{E\backslash A}$, which is equivalent to proving $E\backslash \overline{E\backslash A} \subset \overset{\circ}{A}$. Since $E\backslash \overline{E\backslash A}$ is an open set (the complementary of a closed set), and since it is included in $A$ (because $E\backslash A \subset \overline{E\backslash A}$), we get $E\backslash \overline{E\backslash A} \subset \overset{\circ}{A}$ because $\overset{\circ}{A}$ is the biggest open set included in $A$.

Let's prove that $\overline{E\backslash A}\subset E\backslash \overset{\circ}{A}$. Itr is true because $\overline{E\backslash A}$ is a closed set that contains $E\backslash A$ (because $\overset{\circ} A\subset A$) and $\overline{E\backslash A}$ is the smallest closed set that contains $A$.

ContentLicense: CC BY-SA 4.0


问:
am Prove about set $X$ in vector space such that if $x,y \in X$ then $(x-y)/2 \in X$
I trying to prove the following:

Let $V$ be a real vector space and $X \subset V$ such that for every $x,y \in X$ is $\frac{x-y}{2} \in X$, then given $0 < \mu < \lambda$ is $\mu X \subset \lambda X$

I know that $0 \in X$ and I know how to prove it if $X$ was convex, but I don't see how to prove it with this apparently weak subtraction condition.

Any ideas?

答: That is not true. Take $V=\Bbb R$, seen as a real vector space. Take $X=\Bbb Q$. Then you clearly have$$(\forall x,y\in X):\frac{x-y}2\in X.$$However, $2X\not\subset\pi X$.


问:
the domain for $\dfrac{1}{x}\leq\dfrac{1}{2}$
What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$

according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$

however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?

答:
The equivalence
$$A\leq B\iff \frac{1}{A}\geq\frac{1}{B}$$
only holds for numbers $A$ and $B$ that have the same sign (i.e., are both positive or both negative). Remember, when $c>0$, we have
$$A\leq B \iff c A\leq c B$$
and when $c<0$, we have
$$A\leq B\iff cA\geq cB.$$
To go from $A\leq B$ to
$$\frac{1}{A} \mathbin{\fbox{$\leq$ or $\geq$}} \frac{1}{B}$$
you'll multiply both sides by $c=\frac{1}{AB}$.  This $c$ is greater than $0$ when $AB>0$, which happens when $A$ and $B$ have the same sign, and it is less than $0$ when $AB<0$, which happens when $A$ and $B$ have opposite signs.

Thus,
$$\frac{1}{x}\leq\frac{1}{2}\iff\begin{cases} x\geq 2 &\text{ if }x\text{ has the same sign as 2},\\
x\leq 2 &\text{ if }x\text{ has the opposite sign as 2}.
\end{cases}$$

ContentLicense: CC BY-SA 3.0


问:
Show that a continuous function between topological vector spaces is uniformly continuous when the domain is compact
Let $(E_i,\tau_i)$ be a topological vector space and $\mathcal N_{\tau_i}(x)$ denote the set of $\tau_i$-neighborhoods of $x\in E_i$.

I want to show that if $f:E_1\to E_2$ is $(\tau_1,\tau_2)$-continuous and $(E_1,\tau_1)$ is compact, then $f$ is uniformly $(\tau_1,\tau_2)$-continuous, i.e. $$\forall N\in\mathcal N_{\tau_2}(0):\exists M\in\mathcal N_{\tau_1}(0):\forall x,y\in E_1:x-y\in M\Rightarrow f(x)-f(y)\in N\tag1.$$

The idea for the proof is clear to me, but some steps need rigorous justification:

Let $N\in\mathcal N_{\tau_2}(0)$.

Since $f$ is $(\tau_1,\tau_2)$-continuous, $$\forall x\in E_1:\exists M_x\in\tau_1\cap\mathcal N_{\tau_1}(0):f(x+M_x)\subseteq f(x)+N\tag2.$$

Since $(E_1,\tau_1)$ is compact and $\left(x+\frac{M_x}2\right)_{x\in E_1}$ is a $\tau_1$-open cover of $E_1$, there is a finite $F\subseteq E_1$ with $$\bigcup_{x\in F}\left(x+\frac{M_x}2\right)=E_1\tag3.$$

Let $$M:=\frac12\bigcap_{x\in F}M_x$$ and $x,y\in E_1$ with $x-y\in M$.

If we could show that $$f(x)-f(y)\in N\tag4,$$ we would be done.

By $(3)$, there is a $x_0\in F$ with $$x\in x_0+\frac{M_{x_0}}2\tag5.$$

Now $x-y\in M\subseteq\frac{M_{x_0}}2$.

We somehow need to show that $$y=x-(x-y)\in M_{x_0}\tag6$$ and proceed from there. Can we do this?

答:
In my answer, I will replace $E_{1}$ by a compact subset $K \subseteq E_{1}$ since the uniform continuity you want holds when you restrict $f$ to any compact subset.

To start with, it's not clear that $M_{x_{0}}$ is convex so there seems to be no guarantee that $\frac{1}{2} M_{x_{0}} + \frac{1}{2} M_{x_{0}} \subseteq M_{x_{0}}$.  A way around this is since a TVS is an Abelian topological group, each neighborhood $U$ of $0$ has a subneighborhood $V \subseteq U$ satisfying
\begin{equation*}
0 \in V, \quad V = -V, \quad V + V \subseteq U.
\end{equation*}
(See here.  Munkres's Topology also has a nice section on these groups where this property is covered.)

Thus, above, given the neighborhood $M_{x}$ of the origin associated with $x$, I would have defined a corresponding neighborhood $V_{x} \subseteq M_{x}$ as above (to replace $\frac{1}{2}M_{x}$), and then defined $V$ (in place of $M$) by
\begin{equation*}
V = \bigcap_{x \in F} V_{x}
\end{equation*}
with $F$ a finite subset of $K$ satisfying $K \subseteq \cup_{x \in F} (x +V_{x})$.

Now $x \in x_{0} + V_{x_{0}}$ and $y - x \in V_{x_{0}}$ implies $y, x \in x_{0} + M_{x_{0}}$.  Therefore, $f(x), f(y) \in f(x_{0}) + N$.  You wanted $f(x) - f(y) \in N$, which is not quite what we just proved.  To get what you want, again, choose a symmetric neighborhood $\tilde{N}$ of the origin in $E_{2}$ so that $\tilde{N} + \tilde{N} \subseteq N$.  If $f(x), f(y) \in f(x_{0}) + \tilde{N}$, then
\begin{equation*}
f(x) - f(y) \in (f(x_{0}) + \tilde{N}) - (f(x_{0}) - \tilde{N}) \subseteq N.
\end{equation*}
In other words, instead of choosing $M_{x_{0}}$ so that $f(x_{0} + M_{x_{0}}) \subseteq f(x_{0}) + N$, fix $\tilde{N}$ after choosing $N$ and then choose $M_{x_{0}}$ so that $f(x_{0} + M_{x_{0}}) \subseteq f(x_{0}) + \tilde{N}$.

Again, to the best of my knowledge, this kind of "restrict to a smaller, symmetric neighborhood whose double is contained in the original one" is a standard trick for working in Abelian topological groups.


问:
Alternative measures for the unit interval
Is there a way to construct a measure for the unit interval such that:

For the Borel sets, the measure is countably additive, translation invariant, and equal to the length of the set

For the other sets, the measure is still countably additive, but not necessarily translation invariant

答:
I will assume that you are asking about measures defined on every subset of $[0,1]$. As I pointed out in my comment, if the question is "Does there exist a sigma-algebra extending the Borel algebra for which (1) and (2) hold?", then an affirmative answer is witnessed by the Lebesgue measure.

Ulam showed that there is an extension of the Lebesgue measure to every subset of $[0,1]$ if and only if there exists an atomlessly-measurable cardinal, and that atomlessly-measurable cardinals are weakly inaccessible (this is 1D in the Fremlin paper).

Moreover, if ZFC is consistent, then so is ZFC + "there does not exist a weakly inaccessible cardinal". So ZFC cannot prove that there is an extension of the Lebesgue measure to every subset of $[0,1]$ (unless it is inconsistent).

So, in order to answer your question in the affirmative, we need axioms beyond ZFC or we need to assume that ZFC is inconsistent.


答:
Let $f$ be the minimum-cost flow. That here, there [probably] are negative cost cycles does put a twist into what $f$ may look like. In particular, here is no reason why the arcs $e$ for which $f(e)$ is positive have to form a connected graph; indeed $f$ could consist of a  path $P$ between $a_1$ and $a_{n+1}$ and then vertex-disjoint cycles [disjoint from $P$ as well] where each vertex participates in a cycle.

In fact you also need to require that the capacity of each vertex is also 1 and that the net flow into a vertex besides $a_1$, $a_{n+1}$ is 0; without this even edge-disjointness of the cycles suffices, and the minimum-cost flow could be less than $-n$.

Title: Using Bellman-Ford to find a Hamiltonian cycle? (NP-complete)

Body:
Let $G(V,E)$ be a directed graph, where $V=\{a_1,\ldots,a_n\}$ is a set of vertices and $E$ is a set of ordered pairs of $V$, with $|V|=n$.

Now, let be $G(W,F)$ be a graph where $W$ is a set of vertices, such that $W=\{a_1,\ldots,a_{2n}\}$ with $|W|=2n$, and $F$ is a set of ordered pairs of $W$ defined as follows:
$$\forall i, j \in \{1, \ldots, n\} : (a_i,a_{j+n}) \in F \text{ if } (a_i,a_{j}) \in E \\
\forall i \in \{1, \ldots, n\} : (a_{i+n},a_i) \in F$$

Now we define the capacities:
$$c(a,b)=1 \text{ if } (a,b) \in F$$

And we define a cost:
$$\forall i \in \{1, \ldots, n\} : p(a_{i+n},a_i)=-1 \text{ if } (a_{i+n},a_i) \in F \\
p=0 \text{ else}$$

The source as $a_{n+1}$ and the sink as $a_{n+1}$.

Does the graph $G(W,F,c,p)$ have a minimum cost flow of $-n$ if and only if the graph $G(V,E)$ has a Hamiltonian cycle?

(Comment to an answer in the comments section)

Michael, for your matrix with capacities G(V,E) the matrix G(W,F) is:

```<code>cap = [[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0]]

cost = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0]]
</code>```


问:
Lift morphism along a functor
Let $(\mathcal{C}, F)$ be a pair consisting of a (locally small) category $\mathcal{C}$ and a functor $F \colon \mathcal{C} \to \mathbf{Set}$ satisfying the following properties:

F is faithful.

$\mathcal{C}$ is complete and $F$ is continuous.

$\mathcal{C}$ admits filtered colimits and $F$ commutes with them.

$F$ reflects isomorphism.

Let $A,B,C$ be objects in $\mathcal{C}$ an let $f \colon A \to B$ and $g \colon C \to B$ be two arrows such that $g$ is a monomorphism.

Now suppose there is a map $\tilde{h} \colon FA \to FC$ such that
$$ Ff = Fg \circ \tilde{h}.$$

Is there a morphism $h \colon A \to C$ such that $Fh = \tilde{h}$ and $f = gh$?

The reason for this question is lemma 6.15.4 (tag 007P) of the Stacks Project. Following the proof of the lemma I understand why $FA = F(A \times_B C)$. Then it states that $A = A \times_B C$ since $F$ reflects isomorphisms. But why is the canonical isomorphism in the image of $F$. If the answer above is yes, then this is clear.

答:
Consider the diagram of sets that results as the image of the pullback $A \overset{s}\leftarrow P \overset{r}\to C$ of $f,g$:
$$
\begin{array}{ccc}
FP &\overset{Fr}\to & FC \\
_{Fs}\downarrow & \nearrow & \downarrow_{Fg} \\
FA &\underset{Ff}\to & FB
\end{array}
$$
where the "lifting" is the $\tilde h$ of your post; from this it's easy to see that $Fs$ is invertible, and so is $s$ because $F$ reflects isomorphisms; the image under $F$ of this pair of isomorphisms is an isomorphism $FP\cong FA$. $s$ is unique with this property, and so is its inverse.

All in all, your $h$ is just $rs^{-1}$ (the fact that $F$ is faithful has been secretly used).

Title: Lift morphism along a functor


答:
When you convert $A \le B$ to $\cfrac 1 A \ge \cfrac 1 B$, you are in fact dividing by $AB$. This works if $AB$ is positive, but if $AB \lt 0$ you have to reverse the inequality.

The inequality in the question is true whenever $x \lt 0$ because the left hand side is negative and the right-hand side is positive.

Title: the domain for $\dfrac{1}{x}\leq\dfrac{1}{2}$

Body:
What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$

according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$

however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?


问:
Exponential map on matrices
I was thinking about different definition of exponential:

Case 1:

Let $z=a+ib$ be a complex number so we have the exponential map $\exp(z)=\sum \frac{z^n}{n!}$.

But one can also view $z$ as a matrix (indeed $z$ is a special automorphism of $\mathbb{R}^2$) of the form $z_r:=Z:=\begin{pmatrix}a & -b \\ b & a \end{pmatrix}$ so one has the exponential matrix map defined by $\exp(Z)=\sum \frac{Z^n}{n!}$ it is not difficult to show that $\exp(z_r)=\exp(z)_r$ (I can give more details for this if it worth, the idea is to use the polar expression of $z$ and that write $Z=a\operatorname{Id}+bI$ and use the Taylor expansion of $\sin$ and $\cos$).

Case 2: (The question)

What about the more general case of the exp function on a linear operator $\phi:\mathbb{C}^n\longrightarrow\mathbb{C}^n$?

Indeed, let $X$ be a complex matrix of size $n$, and let $X_r$ the block-matrix obtained by replacing to each $x_{ij}$ the $2\times2$ matrix ${x{_{ij}}}_r$ is it true that $\exp(X_r)=\exp(X)_r$?

Is it a general thing?

My approach:

I've tried with a calculation like the Case 1 but it seems to be more difficult to calculate. Another approach I have in mind is to use the fact that $\mathbb{C}^n$ is a real smooth variety with standard complex structure and than use the flow theorem. This approach gives me a characterization for $\exp(X_r)$ but my problem is to prove that $\exp(X)_r$ is also a solution for the flow.

(Maybe if this is not very clear I would clarify)

Questions:

Is it possible to prove that $\exp(X_r)=\exp(X)_r$ by straightforward calculation?

Is it possible to prove that $\exp(X_r)=\exp(X)_r$ by using the flow theorem?

Any hint or suggestion would be great!

答:
Let's dig a bit further into your statement "let $X_r$ the block-matrix obtained by replacing to each $x_{ij}$ the $2\times2$ matrix $(x{_{ij}})_r$". I'd like to reframe that as follows: let $\Phi:M_{n \times n}(\Bbb C) \to M_{2n \times 2n}(\Bbb R)$ denote the map defined by
$$
\Phi(A + Bi) = A \otimes I + B \otimes J,
$$
where $A,B$ are real $n\times n$ matrices, $\otimes$ denotes the Kronecker product, $I$ denotes the size $2$ identity matrix, and
$$
J = \pmatrix{0&-1\\1&0}.
$$
I claim that if $X = A + Bi$, then $X_r = \Phi(X)$. Moreover, I claim the following:

For real $p,q$ and $X,Y \in M_{n \times n}(\Bbb C)$, $\Phi(pX + qY) = p\Phi(X) + q\Phi(Y)$

For $X,Y \in M_{n \times n}(\Bbb C)$, $\Phi(XY) = \Phi(X)\Phi(Y)$

For a convergent sequence $X_n \to X$, we have $\Phi(X_n) \to \Phi(X)$. That is, $\Phi(\lim_{n \to \infty}X_n) = \lim_{n \to \infty}\Phi(X_n)$.

These results can be proved using the properties of the Kronecker product and the fact that $J^2 = -I$. Now for any $X \in M_{n \times n}(\Bbb C)$, denote
$$
\exp_N(X) = \sum_{k=0}^N \frac{X^k}{k!}.
$$
Using properties 1 and 2 above, verify that $\Phi(\exp_N(X)) = \exp_N(\Phi(X))$. From there, we can conclude that
$$
\Phi(\exp(X)) = \Phi\left(\lim_{N \to \infty} \exp_N(X)\right)
= \lim_{N \to \infty}\Phi(\exp_N(X)) = \lim_{N \to \infty} \exp_N(\Phi(X))
= \exp(\Phi(X)).
$$
That is, we indeed have $\Phi(\exp(X)) = \exp(\Phi(X))$, which is what we wanted.

Title: Exponential map on matrices


问:
if $z=\frac {dt}{dx} $ why $ \frac {dz}{dx}=z^3\frac {d^2x}{dt^2}$
while solving a differential equation i encounter this derivative : let  $$z=\frac {dt}{dx} $$ i don't understand how they make that $$ \frac {dz}{dx}=z^3\frac {d^2x}{dt^2}$$

答:
Equivalently we have the expression

$$\frac{dx}{dt}=\frac{1}{z}$$

then take $\frac{d}{dt}$ on both sides and apply chain rule

$$\frac{d^2x}{dt^2} = \frac{d}{dt}\frac{1}{z} = \frac{dx}{dt}\cdot\left(\frac{d}{dx}\frac{1}{z}\right) = \frac{dx}{dt}\cdot\left(-\frac{1}{z^2}\frac{dz}{dx}\right)$$

Putting this together with first equation gives us

$$\frac{dz}{dx} = -z^3\frac{d^2x}{dt^2}$$


问:
Do the angle addition identities only work for positive angles?
The derivations of the angle addition identities that I have seen depend on right triangles so do these identities still work if the angles are negative, as allowed for angles on the unit circle? The angle addition identities that I am talking about are here: https://derive-it.com/2020/12/12/derivation-10-angle-addition-identities/
For example, what if one of the angles is negative?</p>

答:
The Angle Addition Identities were derived using $\sin\theta \equiv \frac{opp}{hyp}$ and $\cos\theta \equiv \frac{adj}{hyp}$. In addition, $\theta$ was constrained to the interval of $[0, \frac{\pi}{2})$.

The question is, do the Angle Addition Identities work if the arguments of the sine function and cosine functions are generalized to include negative numbers? In the following, $\theta$ will still be constrained to the interval of $[0, \frac{\pi}{2})$. The generalization of choice is $ \displaystyle \sin(-\theta) = -\sin\theta$ and $ \displaystyle \cos(-\theta) = \cos\theta$ This generalization is applicable to the unit circle, specifically in the 1st and 4th quadrants.

The Angle Addition Identities are

$\displaystyle \sin(x+y) = \sin x \cos y + \sin y \cos x$

$\displaystyle \cos(x+y) = \cos x \cos y - \sin x \sin y$.

Next use the above generalizations of the sine and cosine functions. There are three possible cases in which at least one of the input angles is negative.

The first case to investigate is $y \rightarrow -y$ while maintaining $y \in [0,\frac{\pi}{2})$:

$\displaystyle \sin(x+(-y)) \stackrel{?}{=} \sin x \cos (-y) + \sin (-y) \cos x$

$\displaystyle \cos(x+(-y)) \stackrel{?}{=} \cos x \cos (-y) - \sin x \sin (-y)$.

The second case to investigate is $x \rightarrow -x$ while maintaining $x \in [0,\frac{\pi}{2})$:

$\displaystyle \sin((-x)+y) \stackrel{?}{=} \sin (-x) \cos y + \sin y \cos (-x) $

$\displaystyle \cos((-x)+y) \stackrel{?}{=} \cos (-x) \cos y - \sin (-x) \sin y $.

The third case and final case to investigate is $x \rightarrow -x$ while maintaining $x \in [0,\frac{\pi}{2})$ and $y \rightarrow -y$ while maintaining $y \in [0,\frac{\pi}{2})$:

$\displaystyle \sin((-x)+(-y)) \stackrel{?}{=} \sin (-x) \cos (-y) + \sin (-y) \cos (-x) $

$\displaystyle \cos((-x)+(-y)) \stackrel{?}{=} \cos (-x) \cos (-y) - \sin (-x) \sin (-y) $.

One realization to make is that the form of the previous six "equations" is no different from the original Angle Addition Identities. So the question that remains is whether these identities make geometrical sense. This question can be further investigated with geometry. Three new pictures would be needed along with six separate proofs.


问:
Problem finding limit value with squeeze theorem
I have been working on a limit:  $$\lim \limits_{n \to \infty} \sqrt[n]{\frac{1}{3} - \frac{1}{3^n}} $$
using squeeze theorem to try and put it between 2 limits. I found the bigger limit ($\lim \limits_{n \to \infty} \sqrt[n]{\frac{1}{3}} = 1 $), but can't really find the smaller one. For n equals 1 I get the limit:  $\lim \limits_{n \to \infty} \sqrt[n]{\frac{1}{3} - \frac{1}{3}} $ . Anything smaller is a negative value and anything bigger is bigger than the main limit. Any help is appriciated and sorry for messy post, this is my first time on the forum.

答:
Note that in order to apply the squeeze theorem your inequalities don't need to hold for all $n$, it is sufficient if they hold for all $n\geq n_0$ for some $n_0$ (as finitely many terms doesn't influence the convergence). In this case we can take $n_0=2$, then we have for all $n\geq2$:$$\frac{1}{3}-\frac{1}{3^n}\geq\frac{1}{3}-\frac{1}{3^2}$$
and therefore $$\sqrt[n]{\frac{1}{3}-\frac{1}{3^n}}\geq\sqrt[n]{\frac{2}{9}}\xrightarrow{n\to\infty}1$$


问:
Is the empty set a superset of itself?
I know the empty set is a subset of itself, so I'm assuming that it's also the case that it is a superset of itself?

∅ ⊇ ∅

答:
Yes.  $A \subseteq B \iff B\supseteq A$ so .... $\emptyset \subseteq \emptyset \iff \emptyset\supseteq \emptyset$

..... Furthermore

Every set is a subset and a superset of itself.  Or in other words $A \subseteq A$ and $A\supseteq A$ for all sets $A$. (This holds true even if $A$ is the empty set.)

The empty set is a subset of every set. Or in other words $\emptyset \subseteq A$ for all sets $A$. (This holds true even if $A$ is the empty set.)  And

Every set is a superset of the empty set. or in other words $A \supseteq \emptyset$ for all sets $A$. (This holds true ev.... well, you know the drill.)

Title: Is the empty set a superset of itself?


答:
First, it looks like you are off by a minus sign.  So it should be:

$$\frac{dz}{dx} = -z^3\frac{d^2x}{dt^2}$$

Then, it follows naturally from a more algebraic view of differentials.  The standard notation for the second derivative of $x$ with respect to $t$ is $\frac{d^2x}{dt^2}$.  The problem is that this is not algebraically manipulable.  If you find the second derivative by actually applying the quotient rule to the first derivative of $x$ with respect to $t$, the result winds up being $\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}$.

So, if $z$ is $\frac{dt}{dx}$ then what is actually being multiplied (taking into account my suspected negative sign) is:

$$(-1)\left(\frac{dt}{dx}\right)^3\left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right) \\
(-1)\frac{dt^3}{dx^3}\left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right) \\
(-1)\left(\frac{d^2x}{dx^2}\frac{dt}{dx} - \frac{d^2t}{dx^2}\right) \\
\frac{d^2t}{dx^2} - \frac{d^2x}{dx^2}\frac{dt}{dx}
$$
This is the second derivative of $t$ with respect to $x$ (using the algebraically manipulable form of the notation for the second derivative).  Since $z = \frac{t}{x}$ (the first derivative of $t$ with respect to $x$, then $\frac{dz}{dx}$ is the second derivative of $t$ with respect to $x$, which, as we showed, is equal to your formula.

Title: if $z=\frac {dt}{dx} $ why $ \frac {dz}{dx}=z^3\frac {d^2x}{dt^2}$

Body: while solving a differential equation i encounter this derivative : let  $$z=\frac {dt}{dx} $$ i don't understand how they make that $$ \frac {dz}{dx}=z^3\frac {d^2x}{dt^2}$$


问:
A ring of polynomials is isomorphic to a set of functions
Let $\mathbb{R}\left[ X \right]$ be a ring of polynomials over the set of real numbers. Every element of the set $f\in \mathbb{R}\left[ X \right]$ can be uniquely represented as a finite sum $f=\sum\limits_{k=0}^{m}{{{a}_{k}}{{X}^{k}}}$, where ${{a}_{k}}\in \mathbb{R}$.

Let's define a set $P$ as all the possible mappings $\left\{ g:\mathbb{R}\to \mathbb{R}\left| \exists n\in \mathbb{N}:g\left( x \right)=\sum\limits_{k=0}^{n}{{{a}_{k}}{{x}^{k}}} \right. \right\}$.
Let $\phi $ be a mapping $\phi :\mathbb{R}\left[ X \right]\to P:\sum\limits_{k=0}^{m}{{{a}_{k}}{{X}^{k}}}\to \sum\limits_{k=0}^{m}{{{a}_{k}}{{x}^{k}}}$.

In order to prove that $\phi $ is an isomorphism I need to prove that $\phi $ is injective.
I know this result is pretty straight forward.
I tried to show that if $f,h\in \mathbb{R}\left[ X \right]$ and $\sum\limits_{k=0}^{m}{{{a}_{k}}{{X}^{k}}}=f\ne h=\sum\limits_{k=0}^{m}{{{b}_{k}}{{X}^{k}}}$ then there is as least one $i$ such that ${{a}_{i}}\ne {{b}_{i}}$. We have $\phi \left( f \right)=\sum\limits_{k=0}^{m}{{{a}_{k}}{{x}^{k}}}$ and $\phi \left( h \right)=\sum\limits_{k=0}^{m}{{{b}_{k}}{{x}^{k}}}$. If $\phi \left( f \right)=\phi \left( h \right)$ then $\phi \left( f \right)-\phi \left( h \right)=0$ but we already know that at least one ${{a}_{i}}\ne {{b}_{i}}$, therefore, $\left( {{a}_{i}}-{{b}_{i}} \right){{x}^{i}}=0$. I intuitively feel that my proof is somewhat incorrect.

答:
The logic is wrong in your sketched proof of injectivity. The "direct" meaning of injectivity is that if $f(x) = h(x)$ for all $x \in \mathbb R$ then $a_k = b_k$ for all $k=0,1,2,...$ One could also restate it as a contrapositive: if $a_k \ne b_k$ for some $k=0,1,2,...$ then $f(x) \ne h(x)$ for some $x \in \mathbb R$. It looks like you have mixed together bits of the "direct" meaning and the "contrapositive".

There is a nice proof of injectivity using calculus, by proving that if $f(x) = \sum_{k=0}^m a_k x^k$ then each coefficient $a_k$ is completely determined by the function $f$ itself:
$$a_k = \frac{f^{(k)}(0)}{k!}
$$
So if $f=h$ then $a_k=b_k$ for all $k$.


问:
How to graph the equation: $y=\frac {x-2}{x+1}$?
the title says it all.
I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?

"Graph the equation $y=\frac {x-2}{x+1}$"

I know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.

Any help would be appreciated.
If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!

答:
Rewrite as $y=1-3/(x+1)$, or $(y-1)= -3/(x+1)$

This is simply the regular recriprocal equation, with the axies rescaled.  The standard $Y$ axis, representing $X=0$, is now marked $x=-1$.  That is, the x-numbers are moved one to the right.  The old X axis of 0,1,2,3 become -1, 0, 1, 2...

The old $X$ axis, or $Y=0$ becomes $y=1$, and we have to rescale by a factor of 3.  So the old Y coordinates 0, 1, 2, 3... become 1, -2, -5, -8...

And that's it.


答:
Here it is another way to approach it for the sake of curiosity.

Definitions

Let $(X,d_{X})$ be a metric space and $E\subseteq X$. We say that $x_{0}\in X$ is an interior point of $E$ iff there exists an open ball $B(x_{0},\delta)\subseteq E$. We also say that $x_{0}\in X$ is an exterior point of $E$ iff there exists an open ball s.t. $B(x_{0},\varepsilon)\cap E = \varnothing$. Finally, we say that $x_{0}$ is a boundary point of $E$ iff $x_{0}$ is neither an interior point nor an exterior point of $E$.

Based on such definitions, we can conclude that the interior, the exterior and the boundary of $E$ are pairwise disjoint subsets of $X$.

Hence we conclude that $\{\text{int}(E),\partial{E},\text{ext}(E)\}$ is a partition of $X$.

Proposition

Let $A\subseteq X$ and $B\subseteq X$. If $A\cap B = \varnothing$, then $A\subseteq B^{c}$.

Proof

Suppose otherwise that $A\not\subseteq B^{c}$. Then we can conclude there exists $a\in A$ such that $a\not\in B^{c}$, that is to say, $a\in A\cap B$, which contradicts our assumption. Consequently, the proposed claim is true.

Lemma

Given $E\subseteq X$, then $\text{ext}(E) = \text{int}(E^{c})$ and $\partial E = \partial E^{c}$.

Proof

Given a point $x_{0}\in X$, we say that it is an exterior point of $E$ iff there exists an open ball $B(x_{0},\delta)$ s.t.
\begin{align*}
B(x_{0},\delta)\cap E = \varnothing & \Longleftrightarrow B(x_{0},\delta)\subseteq E^{c}\\\\
& \Longleftrightarrow x_{0}\in\text{int}(E^{c})
\end{align*}

On the other hand, we say that $x_{0}\in\partial E$ iff $x_{0}$ is neither an interior point of $E$ nor an exterior point of $E$.

This means that, for every $\delta > 0$, $B(x_{0},\delta)\not\subseteq E$ and $B(x_{0},\delta)\cap E\neq \varnothing$.

This is the same as claiming that, for every $\delta > 0$, $B(x_{0},\delta)\cap E^{c}\neq\varnothing$ and $B(x_{0},\delta)\not\subseteq E^{c}$.

Thus $x_{0}\in X$ is a boundary point of $E$ iff $x_{0}$ is a boundary point of $E^{c}$.

Solution

In accordance to this answer as well as the previous results, we have that
\begin{align*}
X - \text{int}(E) & = \text{int}(E)\cup\partial{E}\cup\text{ext}(E) - \text{int}(E)\\\\
& = (\text{int}(E)\cap\text{int}(E)^{c})\cup(\partial E\cap\text{int}(E)^{c})\cup(\text{ext}(E)\cap\text{int}(E)^{c})\\\\
& = (\partial E\cap\text{int}(E)^{c})\cup(\text{ext}(E)\cap\text{int}(E)^{c})\\\\
& = \partial E\cup\text{ext}(E)\\\\
& = \partial E^{c}\cup\text{int}(E^{c})\\\\
& = \overline{E^{c}}
\end{align*}

Title: Proof Verification of Exercise 2.9 Baby Rudin: Prove that the complement of $E^{o}$ is the closure of the complement of $E$.

Body:
I am stuck on part d of this exercise:

Let $E^{o}$ denote the set of all interior points of a set. Prove that the complement of $E^{o}$ is the closure of the complement of $E$.

I solved it using double inclusion:

Let $p$ $\in (E^{o})^{c}$. Since $p$ is not an interior point of $E$, every neighborhood of $p$ contains a point not in $E$ (or else we have a neighborhood of $p$ that is contained in $E$ and p is an interior point) so $p$ is a limit point of $E^{c}$ so $p$ $\in$ $\overline{E^{c}}$. Conversely, let $q$ $\in$ $\overline{E^{c}}$. Then $q$ belongs to $E^{c}$ and is not an interior point of E, or $q$ is a limit point of $E^{c}$ and no neighborhood of $q$ is contained in $E$, so $q$ is not an interior point of $E$ and so $q \in (E^{o})^{c})$

Is this proof correct? If not, where exactly does it break down? Thank you for your time.


问:
How to prove a problem is not P but NP?
As I try to do some reading On P and NP, I keep finding confusing statements like the Salesman Problem cannot be solved in P, but it has solution in NP.

Is that entirely true, or is it more correct to say "There is no known solution in P"?

Furthermore, as I read papers, many times the authors prove that a problem in "NP-Hard" or "NP-Complete" for example.

Wouldn't that simply mean that they found "NP-Hard" algorithm, but there may still be a simpler solution?

Take sorting problem of a list for example. I can brute force every possible solution and check if the list is sorted, and I can claim this is NP, and show equivalency between this type of solution to some NP Class problems.

Am I thinking about this correctly?

答:
Your first point is correct, but the second isn't.

For all we know, it might be that P = NP. So when a problem is in NP, we shouldn't say about such a problem that it isn't in P. We can say it's "not known to be in P" (as you suggested) or in many cases that it's "not believed to be in P" or even (in the case of NP-hard problems) that it's "not in P unless P = NP."

Concerning the second part of your question, about NP-hardness, the first thing to notice is that there's no such ting as an "NP-hard algorithm". To say that some problem X is NP-hard doesn't say anything about actual algorithms for solving X. Rather, it means that all other NP problems Y can be reduced to X by polynomial-time reductions.  The theory of NP-completeness tells us that, in order to  verify NP-hardness of X, we don't actually have to give such reductions for all NP-problems Y; it suffices to give such a reduction for one NP-complete problem Y, for example 3-SAT. Once we've done that, we have proved that X is NP-hard, and we don't need to include any hedging like "known to be" or "believed to be" or "unless P = NP".

Title: How to prove a problem is not P but NP?


问:
Comparative statics with a system of equations
I am self-learning macroeconomics and have got stuck due to a mathematical difficulty. I have a set of equations like this-

$$f(y, r, p, \bar A, \bar M)=0$$
$$g(y, r, p, \bar A, \bar M)=0$$
$$h(y, r, p, \bar A, \bar M)=0$$

$\bar A$ and $\bar M$ are parameters and the rest of the variables are endogenous. For comparative statics, I would want to find, for example the expression for $\frac{\partial y}{\partial \bar A}$ in terms of the functions $f, g$ and $h$ and their various partial derivatives, such that all the three constraints are always/ identically satisfied. But I am completely stuck, as I do not know how to take into account all the equations simultaneously.

The reason I am asking the question in the Mathematics SE instead of Economics SE is because I want to learn more about the general theory behind such comparative statics analysis. That is, suppose there are $n$ identities

$$f_1(x, a)=0$$
$$f_2(x, a)=0$$
$$....$$
$$f_n(x,a)=0$$

$x$ is the vector of endogenous variables and $a$ is the vector of parameters. I want to know how to find out the derivative of $x$ with respect to $a$ given that the $n$ identities are always satisfied. How should I go about it in general? What conditions must be satisfied for an answer to exist? Also, what are some good sources to read more about such analysis?

答:
This will be more intuitive but less precise, but it may get you started. Start with implicit differentiation
$$f(x,y)=k$$ and start with the total differential
$$f_xdx+f_ydy=0$$
$$\frac{dy}{dx}=-\frac{f_x}{f_y}.$$
But this works with vectors and matrices as well, giving us something that looks like
$$\frac{dy}{dx}= f_x f_y^{-1}$$
See the wikipedia https://en.wikipedia.org/wiki/Implicit_function_theorem for more precision, or Spivak's Calculus on Manifolds for more on implicit differentiation with matrices.  It wouldn't hurt to look at at math econ book, such as Intrilligator's Mathematical Optimization and Economic Theory or Takayama's Mathematical Economics.


问:
There is concept of finite sets that can have only one "interpretation"?
In our mind we have a naive idea of what a set is, and in nature we can only observe something that behave like a finite set, ZFC (or set theories in general) tries to catch these properties in axioms. Like Peano axioms describe the nature of natural numbers, and all the "things" that behave in that way are "isomorphic to the naturals, that property seems is called categoricity (in my naive understanding is that all models of the naturals are isomorphic).

Well, I think that all agree on how sets behave, in the sense that in our experience of the word, we can see only a kind of finite set, or like a bag full of something (or empty) or as a property (comprension axiom), and all sets are finite.

My question is there is a concept of set (axioms), that have a "unique interpretation" and are exactly the naive concept of set?

答:
In what sense can we "see" a finite set? Sure we can see a bag, and see the marbles in the bag. But the bag isn't a set (we don't identify bags by their contents!), and the marbles aren't a set either (for the marbles are many and the set of marbles is one).

If, as you imply, we are allowed to specify a set as the set of things with a certain property (you mention the comprehension axiom), then to be sure there are infinite sets -- e.g. the set of natural numbers, the set of space-time points along some given line, etc.

But agreed, there are indeed problems about the "naive" notion of a set (I don't mean naive set theory, but the pre-theoretic concept of a set). Can we have an empty set on the naive conception of a set as a collection treated as a unity? You might find some of the early chapters of Michael Potter's Set Theory and its Philosophy helpful.

[Added] For an excellent treatmentof the relationship between PA and a canonical theory of finite sets (i.e. ZF with the axiom infinity negated) see this first-rate paper by Kaye and Wong.


问:
Galois group of the polynomial $f(X)=(X^4+1)(X^4-2)$
Here is my attempt:

Let $\alpha=e^{i\frac{\pi}{4}}$ and $\beta=\sqrt[4]{2}$ two roots of the polynomial. Then, the other ones are given by $\alpha\omega$, $\alpha\omega^2$, $\alpha\omega^3$ $\beta\omega$, $\beta\omega^2$ and $\beta\omega^3$, where $\omega=e^{i\frac{\pi}{2}}$. Hence, the splitting field of $f$ is $K_f=\mathbb{Q}(\alpha,\beta,\omega)$. The minimal polynomial of $\alpha$, $\beta$ and $\omega$ are $p_\alpha(X)=X^4+1$, $p_\beta(X)=X^4-2$ and $p_\omega(X)=X^2+1$, so the Galois group has order 32.

However, the solution given to this exercise says that the Galois group is $\mathcal{D}_4$. Can someone tell me where my mistake is?

Thanks in advance.

答:
$\omega = \alpha^2$, so $K_f = \mathbb{Q}(\alpha, \beta)$. Also be careful:

$$\deg(\mathbb{Q}(\alpha, \beta, \omega) : \mathbb{Q}) \neq \deg(\mathbb{Q}(\alpha) : \mathbb{Q}) \deg(\mathbb{Q}(\beta) : \mathbb{Q}) \deg(\mathbb{Q}(\omega) : \mathbb{Q}).$$

You have to compute this adding one element at a time:

$$\deg(\mathbb{Q}(\alpha, \beta, \omega) : \mathbb{Q}) = \deg(\mathbb{Q}(\alpha,\beta,\omega) : \mathbb{Q}(\beta,\omega)) \deg(\mathbb{Q}(\beta,\omega) : \mathbb{Q}(\omega)) \deg(\mathbb{Q}(\omega) : \mathbb{Q}).$$

So for instance

$$\deg(\mathbb{Q}(\alpha,\beta,\omega) : \mathbb{Q}(\beta,\omega)) = 2$$
since $\alpha$ is a root of $X^2 - \beta \in \mathbb{Q}(\beta,\omega)[X]$.


答: The roots of $x^4-2$ are $\sqrt[4]{2}, i \sqrt[4]{2}, - \sqrt[4]{2}, -i \sqrt[4]{2}$. Thus, the splitting field of $x^4-2$ already contains $i$. Therefore, adding $\sqrt{i}$ makes $x^4+1$ split as well.

Title: Galois group of the polynomial $f(X)=(X^4+1)(X^4-2)$

Body:
Here is my attempt:

Let $\alpha=e^{i\frac{\pi}{4}}$ and $\beta=\sqrt[4]{2}$ two roots of the polynomial. Then, the other ones are given by $\alpha\omega$, $\alpha\omega^2$, $\alpha\omega^3$ $\beta\omega$, $\beta\omega^2$ and $\beta\omega^3$, where $\omega=e^{i\frac{\pi}{2}}$. Hence, the splitting field of $f$ is $K_f=\mathbb{Q}(\alpha,\beta,\omega)$. The minimal polynomial of $\alpha$, $\beta$ and $\omega$ are $p_\alpha(X)=X^4+1$, $p_\beta(X)=X^4-2$ and $p_\omega(X)=X^2+1$, so the Galois group has order 32.

However, the solution given to this exercise says that the Galois group is $\mathcal{D}_4$. Can someone tell me where my mistake is?

Thanks in advance.


问:
Are there other kinds of trig functions other than the circular ones and hyperbolic ones?
Are sin, cos and tan defined for other kinds of geometry other than circular and hyperbolic?  Perhaps for multiple dimensions?  Sorry for the open-ended question, I'm just wondering.

答: One can define, for any real $c$, $\sin_c$ and $\cos_c$ as the solutions to the system $$\sin_c'=\cos_c\quad \cos_c' = -c\sin_c,$$subject to $\cos_c(0)=1$ and $\sin_c(0)=0$. Then we have that $$\cos_c^2(x)+c\sin_c^2(x) =1.$$In particular, $\sin_1=\sin$, $\cos_1=\cos$, as well as $\sin_{-1}=\sinh$ and $\cos_{-1}=\cosh$. We also have $\sin_0(x)=x$ for all $x$ and $\cos_0 = 1$. There are formulas such as $$\cos_c(x+y)=\cos_c(x)\cos_c(y)-c\sin_c(x)\sin_c(y)$$and$$\sin_c(x+y)=\sin_c(x)\cos_c(y)+\cos_c(x)\sin_c(y).$$These functions are used in the study of geometries of constant curvature $c$. If I remember right, Christian Bär's Elementary Differential Geometry covers some of this.


问:
Find the minimum values of $a,b,c,d,e,f$ that satisfy following equations
${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right)  }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right)  }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right)  }^{ 2 }={ f }^{ 2 }$

where $a,b,c,d,e,f$ positive integers. Moreover, prove that we can always find positive integer solutions when the number of equations goes to infinity, i.e.

${ a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c \right)  }^{ 2 }={ d }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d \right)  }^{ 2 }={ e }^{ 2 }\\ { a }^{ 2 }+{ \left( b+c+d+e \right)  }^{ 2 }={ f }^{ 2 }\\ \vdots $

Note: You probably see this problem for the first time here.

Edit: A small example:

${ 24 }^{ 2 }+{ 7 }^{ 2 }={ 25 }^{ 2 }\\ { 24 }^{ 2 }+{ \left( 25+7 \right)  }^{ 2 }={ 40 }^{ 2 }$

答:
I've used the Python interface of the Z3 solver to find solutions.

For the a, b, c case:

```<code>a, b, c = Ints('a b c')
s = Solver()
s.add(a > 0, b > 0, c > 0)
s.add(a * a + b * b == c * c)
#  additional constraint to look for "smaller" solutions
s.add(c < 6)
print s.check()
m = s.model()
print "a = %s" % m[a]
print "b = %s" % m[b]
print "c = %s" % m[c]

a = 3
b = 4
c = 5
</code>```

For the a, b, c, d case the solver returns "unknown". The Microsoft Solver Foundation solver is also unable to find a solution for a, b, c, d (value range 1..5000).

Body:
I've used the Python interface of the Z3 solver to find solutions.

For the a, b, c case:

a, b, c = Ints('a b c')
s = Solver()
s.add(a > 0, b > 0, c > 0)
s.add(a * a + b * b == c * c)
#  additional constraint to look for "smaller" solutions
s.add(c < 6)
print s.check()
m = s.model()
print "a = %s" % m[a]
print "b = %s" % m[b]
print "c = %s" % m[c]

a = 3
b = 4
c = 5

For the a, b, c, d case the solver returns "unknown". The Microsoft Solver Foundation solver is also unable to find a solution for a, b, c, d (value range 1..5000).


问:
How to prove the concurrence of three lines?

Let quadrilateral $ABCD$ inscribe in circle whose center is $O$ and let $AC$ and $BD$ intersect at $P$,while $Q$ be a point on the straight line $OP$.The circumcenters of $\Delta QAB$, $\Delta QBC$, $\Delta QCD$, $\Delta QDA$ are $O_1$, $O_2$, $O_3$, $O_4$ respectively.Then how to prove the concurrence of the straight lines $O_1 O_3$, $O_2 O_4$, $OP$?($O$ is not coincide with $P$)

I'm sorry that I can't provide any useful ideas.(I have tried to violently calculate through trigonometric functions,but it only made me crazy)

And I'm very sorry for my possible wrong grammars and strange expressions because I'm still an English learner and my first language is very different from the English system.

I'd appreciate it if someone could share his ideas about the question.

答:
Here's a slick inversive proof.

Let $(QAB)$ meet $(QCD)$ again at $X$, let $(QBC)$ meet $(QDA)$ again at $Y$. It suffices to show that the centre of $(QXY)$ (which is $O_1O_3\cap O_2O_4$) lies on $OP$.

First, note that $(QAC)$ and $(QBD)$ meet again on line $OP$. Indeed, let $(QAC)$ meet $OP$ again at $R$. Then $PQ\cdot PR=PA\cdot PC=PB\cdot PD$, which implies that $R$ lies on $(QBD)$ too.

Now invert centre $Q$ with arbitrary radius, and denote inverses with $'$. The line $OP$ is fixed, and $ABCD$ maps to another cyclic quadrilateral $A'B'C'D'$, whose center $J$ still lies on $OP$. Since $(QAC)$ and $(QBD)$ meet on $OP$, we know that $R'=A'C'\cap B'D'$ is on $OP$.

Circles $(QAB)$, $(QCD)$ map to lines $A'B'$, $C'D'$ respectively, so $X'=A'B'\cap C'D'$. Similarly, $Y'=B'C'\cap D'A'$.

But by Brokard's theorem, we know that $JR'\perp X'Y'$, i.e. $OP\perp X'Y'$. This implies that the centre of $(QXY)$ lies on $OP$, as desired.


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
My answers:

yes.  The real numbers can be "said to be" a subset of the complex numbers.  Even if it is not a genuine subset, we can "identify" $\mathbb R$ with a subset of $\mathbb C$; after we do that, we say $\mathbb R \subset \mathbb C$.

depends.  (a) If $\mathbb C$ is constructed together with its topology, then there is a unique continuous automorphism $\sigma$ of order $2$ of $\mathbb C$ and its fixed set $\{z \in \mathbb C : \sigma(z)=z\}$ can be identified with the reals. (b)  You could conceivably construct $\mathbb C$ in some other way.  If you do not know what are the closed sets, or what are the Borel sets, or some structure in addition to the field structure, then what?  There are many, many automorphisms of order two.  The fixed set of any one of them is isomorphic to $\mathbb R$.  But there is no way to single out one of these copies of $\mathbb R$ from the others.

In case (b) would we say we have actually constructed $\mathbb C$?  Or merely that we have constructed some field isomorphic to $\mathbb C$?  If it is isomorphic to $\mathbb C$ up to a non-unique isomorphism, can we say it "is" $\mathbb C$?

Contrariwise, a field can be isomorphic to $\mathbb R$ in at most one way, so if we construct any of these, then we do say it "is" $\mathbb R$.


答:
1.- As already stated in your question, no: formally we can't say the reals are even a subset of the complex numbers. Yet there are many way in which we can embed $\;\Bbb R\;$ into $\;\Bbb C\;$ in such a way that the basic characteristics of these two fiels are mantained, and one of the most usual isomorphisms (=embeddings respecting the algebraic structure) is precisely $\;\phi:\Bbb R\to\Bbb C\;,\;\;\phi(r):=r+0\cdot i\;$ , or if you prefer the other very usual definition, $\;\phi(r):=(r,0)\;$. In this manner the real numbers become not only a subset but, as said, a subfield of the field $\;\Bbb C\;$ .

2.- Yes, we can define $\;\Bbb C\;$ as the algebraic closure of $\;\Bbb R\;$ , getting a fields extension of degree two, and in which in a rather canonical way, the basis field $\;\Bbb R\;$ is embedded in a the extension field $\;\Bbb C\;$ . In the sense of  a theorem by Artin, this is the only possible algebraic extension of degree two of a real closed field.

Title: Are the reals genuinely a subset of the complex numbers?

Body:
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?


问:
Compute value of $\pi$ up to 8 digits
I am quite lost on how approximate the value of $\pi$ up to 8 digits with a confidence of 99% using Monte Carlo. I think this requires a large number of trials but how can I know how many trials?

I know that a 99% confidence interval is 3 standard deviations away from the mean in a normal distribution. From the central limit theorem the standard deviation of the sample mean (or standard error) is proportional to the standard deviation of the population $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}$

So I have something that relates the size of the sample (i.e. number of trials) with the standard deviation, but then I don't know how to proceed from here. How does the "8 digit precision" comes into play?

UPDATE

Ok I think I am close to understand it. From CLT we have $\displaystyle \sigma_{M} = \frac{\sigma}{\sqrt{N}}$ so in this case $\sigma = \sqrt{p(1-p)}$ therfore $\displaystyle \sigma_{M} = \frac{\sqrt{p(1-p)}}{\sqrt{N}}$

Then from the Bernoulli distribution, $\displaystyle \mu = p = \frac{\pi}{4}$ therefore

$$\sigma_{M}=\frac{\sqrt{\pi(4-\pi)}}{\sqrt{N}}$$ but what would be the value of $\sigma_{M}$? and then I have $\pi$ in the formula but is the thing I am trying to approximate so how does this work? and still missing the role of the 8 digit precision.

答:
I may completely off the mark, but I guess that Monte Carlo here refers to approximating some integral by a random process, and that the integral must be chosen in a way that knowing its value allows us to calculate $\pi$.

Let's guess that the integral we are interested in is
$$
\frac\pi4=\int_0^1\sqrt{1-x^2}\,dx
$$
giving the probability that a random point in the square $[0,1]\times[0,1]$ (so $x$ and $y$ independent and uniformly distributed in $[0,1]$) is also in the unit disk.

Assume that we generate $N$ such points $(x,y)$ and record the number of successes (points in the unit disk) $M$. We know from crude estimates for $\pi$, say $3<\pi<4$, that the success rate $p$ of an individual point is between $3/4$ and $1$. Therefore the standard deviation of $M$ is bounded from above by
$$\sigma(M)=\sqrt{Np(1-p)}<\frac{\sqrt{3N}}4.$$

We approximate
$$
\pi\approx\frac{4M}N.
$$
This has SD $\sigma(\pi)<4\sigma(M)/N=\sqrt{3/N}$.

For 99% confidence we want $3\sigma(\pi)<10^{-8}$ or equivalently $\sqrt{N}>3\sqrt{3}\cdot 10^8$. This suggests that generating $N\approx 27\cdot10^{16}$ random points on the unit square would do the trick :-)

After this gedankenexperiment I quite appreciate Machin's formula. Even the alternating series for $\arctan 1$ beats this.

So this leaves open the possibility that something completely different was wanted?

Title: Compute value of $\pi$ up to 8 digits


问:
How to solve trigonometric equations in 2 variables?
Find out both $x$ and $y$, $$\cos(x)=-\cos(x+y)$$ I come up with this equation when I was finding out maxima and minima of a two variable function $$f(x,y)=\sin x+\sin y+\sin (x+y);$$ however I get a solution by hit and trial approach, that is $$x=y={\pi\over 3}$$ will satisfy this equation, but how to solve it, as I have many other problems of the same kind and this hit and trial is time consuming.

答:
I would try $x=a+b$, $x+y=a-b$. Then
$$\begin{align}\cos x&=\cos(a+b)=\cos a\cos b-\sin a\sin b\\
&=-\cos(x+y)=-\cos(a-b)=-\cos a\cos b-\sin a\sin b\end{align}$$
So that brings us to $2\cos a\cos b=0$ so either $\cos a = 0$ in which case $a=\left(n+\frac12\right)\pi$, $y=(a-b)-(a+b)=-2b$, so $x=\left(n+\frac12\right)\pi-\frac12y$ or $\cos b=0$ so $b=\left(n+\frac12\right)\pi$, $y=-(2n+1)\pi$, and $x=a-(2n+1)\pi$.

Summarizing, we could have $y=\text{anything}$ but then $x=\left(n+\frac12\right)\pi-\frac12y$ or $x=\text{anything}$ but $y=(2m+1)\pi$.

EDIT: I guess another way of expressing the solution would be to say that $x=\text{anything}$ and $y=(2n+1)\pi$ or $y=(2n+1)\pi-2x$.


问:
Liu's "Algebraic Geometry and Arithmetic Curves" - Proposition 4.4. (Ch. 7) and Exercises 5.1.29 and 5.1.30
In Liu's book, Chapter 7, Proposition 4.4, the question is about closed embeddings (or: closed immersions) coming from very ample divisors. The theorems are quite well-known when the ground field $k$ is algebraically closed (e.g., Hartshorne's Proposition 7.3 and Sec. 7, Chap. 2 there, in general). But Liu does not assume $k$ to be algebraically closed because of the Exercises 5.1.29 and 5.1.30. and not assuming $k$ to be algebraically closed is also a major difference to various other texbooks!

Here it is said that the statements in the exercises could be incorrect. Can one prove them (at least) in special cases?

Ex. 5.1.29: Let $\pi: X'\rightarrow X$ be a faithfully flat morphism of schemes, $F$ quasi-coherent sheaf on $X$.

(a) There is a morphism $\phi: \mathcal{O}_{X}^{(I)} \rightarrow F$ such that $\phi(X)$ is surjective. Further, $F$ is generated by global sections iff $\phi$ is surjective.

(b) $F$ generated by global sections iff $\pi^{\ast}F$ generated by global sections.

(c) Now $X,X'$ quasi-compact, $L$ invertible sheaf on $X$. If $\pi^{\ast}L$ is ample, then $L$ is ample.

Ex. 5.1.30 (light version): Let $X$ be aproper scheme over $\mathbb{R}$, $L$ invertible sheaf on $X$, $\pi: X_{\mathbb{C}} \rightarrow X$ projection (to base change $\mathbb{R}$ to $\mathbb{C}$) and let $\pi^{\ast}L$ be very ample.

(a) Show that $L$ is ample and generated by some global sections $s_0, \ldots , s_n$. Let $f: X \rightarrow Y:=\mathbb{P}^{n}_\mathbb{R}$ be the morphism associated to $L$ and the $s_i$. Show that $f_\mathbb{C}: X_\mathbb{C} \rightarrow Y_\mathbb{C} $ is a closed immersion.

(b) Let $y \in Y$ and let $y' \in Y_\mathbb{C}$ be a point lying over $y$. Show that the fiber of $X_\mathbb{C}$ over $y'$ is isomorphic to $X_y \times_{Spec(k(y))} Spec(k(y')) $. Deduce from this that $f^{-1}(y)$ consists of at most one point, and that $f$ is a topological closed immersion.

(c) Show that $\mathcal{O}_Y \rightarrow f_{\ast}\mathcal{O}_X$ is surjective and that $f$ is a closed immersion.

答:
$\newcommand\msO{\mathscr O}\newcommand\pull[1]{#1^*}\newcommand\C{\mathbb C}\newcommand\by\times$
There are a lot of parts, so these are more hints/sketches than full arguments.

For 5.1.29(a), use that a morphism $\msO_X\to F$ is the same thing as a choice of global section $s\in F(X)$.

For 5.1.29(b), you want to show $\msO_X^I\to F$ is surjective its pullback $\msO^I_{X'}\to\pi^*F$ is. Surjectivity can be checked on stalks. Use that $\msO_{X,x}\to\msO_{X',\pi(x)}$ is a faithfully flat ring extension so a map $A\to B$ of $\msO_{X,x}$-modules is surjective iff the map $A\otimes_{\msO_{X,x}}\msO_{X',\pi(x)}\to B\otimes_{\msO_{X,x}}\msO_{X',\pi(x)}$ is surjective.

For 5.1.29(c), use part (b). If $F$ is a f.g. qcoh sheaf on $X$, then $\pull\pi(F\otimes L^n)=\pull\pi F\otimes(\pull\pi L)^n$ is generated by global sections for $n\gg0$ since $\pull\pi L$ ample, so $F\otimes L^n$ is generated by global sections for $n\gg0$, so $L$ is ample.

For 5.1.30(a), faithful flatness is preserved by base change, so $X_\C\to X$ is faithfully flat, so $L$ is ample by 5.1.29(c) and globally generated by 5.1.29(b). Finally, $f_\C$ is a closed immersion since it is the morphism associated to the very ample line bundle $\pull\pi L$ (gives immersion) and since $X_\C$ is proper (gives closed).

For 5.1.30(b), check/convince yourself that the square
$$\require{AMScd}\begin{CD}
X_\C @>>> X\\
@VVV @VVV\\
Y_\C @>>> Y
\end{CD}$$
is Cartesian. Then it is a formal fact that the fiber above some $y'\in Y_\C$ is the same as the fiber above the corresponding $y\in Y$. That $\#f^{-1}(y)\le1$ follows from $X_\C\to Y_\C$ being topologically injective. This gives that $X\to Y$ is a continuous injection with closed image. I don't know an easy way off the top of my head to see that it is an open map as well (so that $f$ will be a topological closed embedding). One possible method is to appeal to the fact that if $S'\to S$ is faithfully flat and quasi-compact, then the topology on $S$ coincides with the quotient topology induced by this map. Since $Y_\C\to Y,X_\C\to X$ are faithfully flat and quasi-compact, this would let you deduce openness of $X\to Y$ from openness of $X_\C\to Y_\C$.

For 5.1.30(c), use the same strategy as in 5.1.29(b). The point is that surjectivity can be checked after faithful flat base change where we already know it to hold. Knowing that $\msO_Y\to f_*\msO_X$ is surjective and that $f$ is a topological closed embedding says that $f$ is a closed immersion.

Edit: As KReiser points out in the comments, my suggestion for 5.1.29(b) is incomplete. Hence also are my arguments for 5.1.29(c) and 5.1.30(a). KReiser also explains how to recover a correct argument for 5.1.30 in their posted answer, so anyone looking at this question should read their answer as well.


问:
Question about Root locus and Nyquist analysis
Currently i m preparing for my control systems exam, and i saw a question about nyquist which asks 'For which K value transfor function is critically stable?' there is function and nyquist plot. So i decided to solve question with root locus too for double check and K i found on root locus plot is different than Nyquist plot and i checked with mathlab too.Did i miss something i thought root locus plot and Nyquist plot suppose to give me same answer.

Thank you for reading..

答: Do you have an expression of the open-loop transfer function? If so, just try to substitute both $K$ values and chack the closed-loop poles.


答:
My answer:

Yes and No. If a the Naturals can be said to be a subset of the Integers, or if the Integers can be said to be a subset of the Rationals, then by analogy we can say the Reals are subsets of the Complex. That is because the contruction of the Complex from the Reals is the same as the construction of Integers from the Naturals or the construction of the Rationals from the Integers.

But No. I think there is a better way to express the relation between the two sets. One way to say it is: The set of all complex numbers with zero imaginary part is isomorphic to the set of Reals. Isomorphic in the sense that any two corresponding elements can be used in the place of the other in any arithmetic/algebraic calculations.

(1), explains it.


问:
help calculating $\sum_{k=0}^n k!{n \choose k} ^2= \text{?}$
I am trying to find the number of relations on $[n] = \{1,\ldots,n\}$ s.t. for all $x,y,z\in R.xRz \wedge yRz\to x=y$. One idea I had was choosing $k$ elements from $n$ then choosing another $k$ elements from $n,$ there are $k!$ bijections between these two sets so we just have to sum over $k$ and here I got stuck, tried finding another approach but came up with nothing.

答:
Let $R$ be such a relation. For each $z\in[n]$ there is at most one $x\in[n]$ such that $x\mathrel{R}z$. The relation $R$ therefore defines a function

$$f_R:[n]\to[n+1]:z\mapsto\begin{cases}
x,&\text{if }R^{-1}[\{z\}]=\{x\}\\
n+1,&\text{if }R^{-1}[\{z\}=\varnothing\,.
\end{cases}$$

In other words, $f_R(z)$ is the unique $x\in[n]$ such that $x\mathrel{R}z$ if there is one and is $n+1$ if no such $x\in[n]$ exists.

Conversely, if $g$ is any function from $[n]$ to $[n+1]$, the relation

$$R_g=\{\langle f(k),k\rangle:k\in[n]\}\cap\big([n]\times[n]\big)$$

has the desired property, and $f_{R_g}=g$. Thus, the map $R\mapsto f_R$ is a bijection from the set of such relations to the set of functions from $[n]$ to $[n+1]$. There are $(n+1)^n$ functions from $[n]$ to $[n+1]$ and therefore $(n+1)^n$ such relations.


问:
Quadratic Question
Problem: Find all values of $b$ for which the equations $1988x^2 + bx + 8891 = 0$ and $8891x^2 + bx + 1988 = 0$ have a common root.

What I have done so far: Let the roots of the first quadratic be $r_1$ and $r_2$, and the roots of the second quadratic be $s_1$ and $s_2$. We have the following equations.

$r_1+r_2 = \frac{-b}{1988}$

$r_1r_2=\frac{8891}{1988}$

$s_1+s_2 = \frac{-b}{8891}$

$s_1s_2 = \frac{1988}{8891}$

Without loss of generality, we can let $r_1=s_1$, but I am not sure how to solve these equations after that. Tips?

答:
Tip. Euclid's division algorithm (if before the remainder of the last division you have a polynomial of degree $\ge$ 1 then the 2 polynomials share a common root)

Tip. A polynomial of degree n (in the complex field) has exactly n roots (in the complex field). Since the solutions are finite the first polynomial will have 2 roots and the second polynomial will have 2 roots and none of the roots of the first polynomial have to be a root for the second polynomial.

Tip. The solutions for which those 2 polynomials are the same must verify that $x=x^{-1}$, that means that either x=1 or x=-1, in which case b=-10879 or b=10879 (respectively).

Title: Quadratic Question


问:
Proving that exists $k'$ such that $3k'+2$ is prime and $3k'+2\mid 3k+2$
In an exercise I'm asked to prove the following:

Let $3k + 2$ be a natural number. Prove that, $3k + 2$ has a prime factor of the same form.

So I was able to prove that, if $k$ is an even number, then $2=3\cdot0+2\mid 3k+2$ but I'm not being able to conclude anything if $k$ is an odd number. How can I prove that, if $k$ is an odd number then exists $k'$ such that $3k'+2$ is prime and $3k'+2\mid 3k+2$?

答:
Note that a prime $p$ dividing $3k+2$ cannot be $0\pmod{3}$. Then, every prime dividing $3k+2$ is either $1\pmod{3}$ or $2\pmod{3}$.

If all primes dividing $3k+2$ were $1 \pmod{3}$ then their product (regardless of their power) would be $\equiv 1 \pmod{3}$ which gives a contradiction.


答:
Note that, $$p(x)=1988x^2+bx+8891\implies q(x)=x^2p\left(\frac 1x\right)=8891x^2+bx+1981$$

Since $p(0)\neq 0$ we want $b$ such that $p(x),p\left(\frac 1x\right)$ have a common root.

It is easy to see that this implies that the common root, $r$, satisfies $r=\frac 1r$ so that $r\pm 1$.

And from there, the problem is straight forward.

Title: Quadratic Question

Body:
Problem: Find all values of $b$ for which the equations $1988x^2 + bx + 8891 = 0$ and $8891x^2 + bx + 1988 = 0$ have a common root.

What I have done so far: Let the roots of the first quadratic be $r_1$ and $r_2$, and the roots of the second quadratic be $s_1$ and $s_2$. We have the following equations.

$r_1+r_2 = \frac{-b}{1988}$

$r_1r_2=\frac{8891}{1988}$

$s_1+s_2 = \frac{-b}{8891}$

$s_1s_2 = \frac{1988}{8891}$

Without loss of generality, we can let $r_1=s_1$, but I am not sure how to solve these equations after that. Tips?


问:
Maximal ideal in $\mathbb{Z}[x]$ containing a prime $p$ is $(p,f(x))$
Algebra qualifying exam question here. Suppose that $I$ is a maximal ideal in $\mathbb{Z}[x]$ and it contains a prime number $p$. Show that $I=(p,f(x))$, for some $f(x)$ monic in $\mathbb{Z}[x]$ and irreducible in $(\mathbb{Z}/p\mathbb{Z})[x]$.

I'm very stuck on this. A friend told me to try Bezout's identity on $(\mathbb{Z}/p\mathbb{Z})[x]$, but I don't see how I can take a gcd downstairs in this ring and pull information back upstairs to $\mathbb{Z}[x]$. I could also be misinterpreting what they meant by "use Bezout's identity". Any hints would be tremendous. Thank you!

答: The quotient map $p:\mathbb{Z}[X]\rightarrow \mathbb{Z}[X]/I$ factor by a map $q:\mathbb{Z}/p[X]\rightarrow\mathbb{Z}[X]/I$. The kernel of this map is a maximal ideal $J$ since $\mathbb{Z}[X]/I$ is a field. Since $\mathbb{Z}/p$ is a field, $\mathbb{Z}/p[X]$ is a principal domain. This implies that there exists a irreducible polynomial $f$ of $\mathbb{Z}/p[X]$ which generates $J$. Up to a coefficient, you can suppose that $f$ is monic since $\mathbb{Z}/p$ is a field.


答:
hint

Observe that zero is not a root and  if $ r\ne 0 $ is a root of the first,
$ \frac 1r $ will be a root of the second.

So, we should have $$r=\frac 1r$$

or $ r=\pm1 $.

thus

$$1988\pm b+8891=0$$

Title: Quadratic Question


问:
Properties of a hyperbola slice of a cone
I'm attempting to locate some points (in a 3D coordinate system) on the surface of a cone by slicing the cone with a plane, and using the resultant ellipse, parabola or hyperbola to calculate the points.  I've had success with the ellipse & parabola solutions, but am seeking help with the hyperbola case.

Known data:  The slope of the cone, the slope of the plane, and distance from the vertex of the cone to the intersection of the plane and cone's longitudinal axis.  Given these knowns, I'm looking for "a" & "b".

From reading, I have the formulas x²/a² - y²/b² = 1 & c² = a² + b².  I understand distance "a" to be half the distance between the two vertices.  Question 1 is, is "a" always half the distance between the two vertices, even when the plane is sloped?  If not always half, how is "a" calculated when the plane is sloped?  Question 2 is how to determine "b"?  Question 3 is how to determine the angle of the asymptotes? ...is that the angle of the cone, or if the plane is sloped, some other angle?

I'm concerned only with the upper nappe of the cone.

As I massage the above formula, attempting to solve for an "x" & "y" & "b", as the plane angle gets shallower, "a" gets longer.  And when "a" gets longer than an "x" value that I am using, my solutions fail due to an attempt to take the square root of a negative number.  So, I know I'm messing up somewhere.  Any help is appreciated.

This is my first post here so I don't know my way around.  I'm going to attempt to post a sketch.

Thanks in advance.

答:
I did not see a diagram for the Dandelin Spheres with a hyperbola, so I drew one. This page has a hyperbola

This is a side view, a cross section. Each circle means a sphere, both spheres are tangent to the cone, which is shown by the two black lines. The red/purple line is
(two generating lines of) the cone.


答:
Hint:

All primes $\ne 3$ are congruent to $1$ or $-1\mod 3$, and $n\equiv -1 \mod 2$.

Title: Proving that exists $k'$ such that $3k'+2$ is prime and $3k'+2\mid 3k+2$

Body:
In an exercise I'm asked to prove the following:

Let $3k + 2$ be a natural number. Prove that, $3k + 2$ has a prime factor of the same form.

So I was able to prove that, if $k$ is an even number, then $2=3\cdot0+2\mid 3k+2$ but I'm not being able to conclude anything if $k$ is an odd number. How can I prove that, if $k$ is an odd number then exists $k'$ such that $3k'+2$ is prime and $3k'+2\mid 3k+2$?


问:
Negative value for area
I have a problem with the following example:Given the surfaces:$$x^2+y^2+z^2=5, x^2+z^2\ge (y-1)^2$$ find the area bounded with them. First problem appears when I try to sketch this in 3D. For the first one is sphere and that is OK, and the second one should be infinite cone right? (but not sure what part of it left or right, since it is on y axis).

Then for intersection I get $x^2+z^2=1$ and after cilindrical coordinates it is easy to see that $0\le r\le 1$ and $0\le \theta \le 2\pi$.
After that, I am calculating the area of cone and sphere seperatly. So for sphere i get $$\sqrt{5}\int\limits_{0}^{2\pi}d\theta \int\limits_{0}^{1}\frac{rdr}{\sqrt{5-r^2}}$$. But this integral has a negative value, so I am obviously mistaking somewhere but not sure where.

BTW i used that $y=\sqrt{5-x^2-z^2}$ from the equation of sphere and than plugged that into formula.

答:
$x^2+y^2+z^2=5$ and $ x^2+z^2= (y-1)^2$ intersect at

$y^2+(y-1)^2=5\to y_1=-1;\;y_2$

The surface is in the picture below and its area is $S=2\pi rh$.

where $r=\sqrt 5$ is the radius of the sphere, $h$ is the height $h=|y_2-y_1|=|2-(-1)|=3$

Therefore $$S=6\sqrt 5\pi$$
which is $67\%$ of the area of the sphere $20\pi$

Title: Negative value for area


问:
How do I read this proportion?
Geneticists say that the color of Italian squash should follow the $12:3:1$ ratio. A group of researchers collects the following information: White 155, yellow 40 and green 10. Are these data consistent with the hypothesis of the geneticists?

I have to perform a chi-square test, but I need the expected frequencies which are obtained with the following formula

$e_i=nπ_i  $ where $n$ is the sample size and $π_i$ is the assumed proportions

What I don't understand is how to occupy $12:3:1$, is 12% or $\frac{2}{3}$ or $\frac{\frac{2}{3}}{1}$ ?

答: The total number of observations is $n = 155 + 40 + 10 = 205$.  The theoretical or hypothesized proportions are $\pi_1 = 12/16$ white, $\pi_2 = 3/16$ yellow, and $\pi_3 = 1/16$ green; hence the expected frequency of white is $205(12/16) = 153.75$, and so forth.

Title: How do I read this proportion?


问:
Directional Derivatives on a Manifold.
Let $M$ be a manifold (if its easier then say a Riemannian Manifold).

My aim is to understand directional derivatives of some function $\Phi: M\to \mathbb{R}$.

$\underline{\textit{My Understanding So Far :}}$

For a manifold $M$ the tangent space at a point $p$ (denoted $T_p M$) can be seen as the set of equivalence classes of curves on $M$ passing through $p$. More specifically : https://en.wikipedia.org/wiki/Tangent_space#Definition_via_tangent_curves .

A Riemannian Manifold is a manifold $M$ equipped with an inner product on $T_p M \times T_p M$ (which will let us talk about angles and lengths of curves on $M$).

The classical directional derivative (in direction $v$ and at point $p$) of a function acting on $\mathbb{R}^d$ tells you how much the function changes at point $p$ and in direction $v$.

$\underline{\textit{Help : }}$

Im finding it hard to go from the above to the definition of directional derivatives given here https://en.wikipedia.org/wiki/Tangent_space#Tangent_vectors_as_directional_derivatives . Any help ? :)

答:
I think it's helpful to think about how we compute the directional derivative in $\mathbb{R}^n$, this is done as $$D_vf(x_0)=\lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}{h}$$

If we have some differentiable curve $\gamma: I\to \mathbb{R}^n$ such that $\gamma(t_0)=x_0$ then $\frac{d}{dt}f(\gamma(t))$ at $t_0$ is given by the directional derivative of $f$ in the direction of $\dot{\gamma}=v$, i.e. $$\frac{d}{dt}f(\gamma(t))=\nabla{f}\cdot v=v[f]_{x_0}.$$ This implies that we can think of $\dot{\gamma}$ as the directional derivative operator satisfying $$\frac{d}{dt} f(\gamma(t)):=\dot{\gamma}[f]_{x_0}$$ Since there are many such curves that give the same result we can take an equivalence class of them. In the end we don't need the Riemannian structure on $M$ in order to talk about these vectors, we only need a differentiable structure on $M$ in order to get consistent answers for our derivatives. When going to a manifold, we'll have the map $\gamma: I\to M$ the map $f: M\to \mathbb{R}$ meaning that $f\circ\gamma: I\to \mathbb{R}$ where we can do normal calculus and get coordinate independent results.

Title: Directional Derivatives on a Manifold.


答:
PROOF BY CONTRADICTION:

Hint: the product of any two integers of the form $(3m+1)$, is $(3m+1)$

Solution:
if $p_1$, $p_2$, ... are prime factors of the number, and if $p_1 \equiv 1 \pmod{3}$, $p_2 \equiv 1 \pmod{3}$ ..., then $p_1 \cdot p_2 \cdot ... \equiv 1 \pmod{3}$


问:
Finding a four-digit number $\overline{abcd}$ such that $(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $
Consider a four digit number $\overline{abcd}$ such that $$(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $$
$\overline{ab}$, $\overline{bc}$ and $\overline{cd}$ are two digit numbers

What is this number ?

What I have tried is

$$(10a+b)^2+(10b+c)^2+(10c+d)^2=\overline{abcd} $$ then I used $(a+b)^2$ formula but this method didn't help.

I need mathematical solution

答:
It's not a pretty answer, but here is a brute-force solution in Python</p>
```<code>>>> from itertools import product
>>> digits = '0123456789'
>>> for a,b,c,d in product(digits, digits, digits, digits):
...     if int(a+b)**2 + int(b+c)**2 + int(c+d)**2 == int(a+b+c+d):
...         print(a + b + c + d)
...
0000
0001
3334
</code>```

Body:
It's not a pretty answer, but here is a brute-force solution in Python

>>> from itertools import product
>>> digits = '0123456789'
>>> for a,b,c,d in product(digits, digits, digits, digits):
...     if int(a+b)**2 + int(b+c)**2 + int(c+d)**2 == int(a+b+c+d):
...         print(a + b + c + d)
...
0000
0001
3334


问:
binary search tree rotations
if I have two binary search trees with the same keys(n keys) but not identical,
I can get from the first tree to the other by maximum n tree rotations.
how can I prove it?
for example I can get from tree 1 to tree 2 by one rotation:

答:
This is probably the same thing as the famous problem posed by Daniel
Sleator, Robert Tarjan, and William Thurston asking for the diameter of the associahedra or the diameter of the rotation graph on plane binary trees.

The correct diameter is rather $2n - 2$.

This has been solved by Pournin in 2012 in the language of flips of triangulations, see his article.

Title: binary search tree rotations


答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did a quick search where I look for values that can be written as $1000\text{a}+100\text{b}+10\text{c}+\text{d}$ where $\left\{\text{a},...,\text{d}\right\}\in\left\{1,...,9\right\}$

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c +
      1*d == (a*b)^2 + (b*c)^2 + (c*d)^2],
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={{{{1284}}}}
</code>```

Because:

$$\left(1\cdot2\right)^2+\left(2\cdot8\right)^2+\left(8\cdot4\right)^2=1284\tag1$$

That is the only four digit number with that property.

If you mean that two digit number thing, I got:

```<code>In[2]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c +
      1*d == (a*10 + b)^2 + (b*10 + c)^2 + (c*10 + d)^2],
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing

Out[2]={{{{3334}}}}
</code>```

Because:

$$\left(10\cdot3+3\right)^2+\left(10\cdot3+3\right)^2+\left(10\cdot3+4\right)^2=3334\tag2$$

That is the only four digit number with that property.

Title: Finding a four-digit number $\overline{abcd}$ such that $(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $

Body:
Consider a four digit number $\overline{abcd}$ such that $$(\overline{ab})^2+(\overline{bc})^2+(\overline{cd})^2=\overline{abcd} $$
$\overline{ab}$, $\overline{bc}$ and $\overline{cd}$ are two digit numbers

What is this number ?

What I have tried is

$$(10a+b)^2+(10b+c)^2+(10c+d)^2=\overline{abcd} $$ then I used $(a+b)^2$ formula but this method didn't help.

I need mathematical solution

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did a quick search where I look for values that can be written as $1000\text{a}+100\text{b}+10\text{c}+\text{d}$ where $\left\{\text{a},...,\text{d}\right\}\in\left\{1,...,9\right\}$

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c +
      1*d == (a*b)^2 + (b*c)^2 + (c*d)^2],
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{1284}}}}

Because:

$$\left(1\cdot2\right)^2+\left(2\cdot8\right)^2+\left(8\cdot4\right)^2=1284\tag1$$

That is the only four digit number with that property.

If you mean that two digit number thing, I got:

In[2]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1000*a + 100*b + 10*c +
      1*d == (a*10 + b)^2 + (b*10 + c)^2 + (c*10 + d)^2],
   1000*a + 100*b + 10*c + 1*d, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1,
    9}, {d, 1, 9}] //. {} -> Nothing

Out[2]={{{{3334}}}}

Because:

$$\left(10\cdot3+3\right)^2+\left(10\cdot3+3\right)^2+\left(10\cdot3+4\right)^2=3334\tag2$$

That is the only four digit number with that property.


问:
Sign of the eigenvalues of the sum of two matrices with positive eigenvalues
Let $A,B \in \mathcal M_n(\mathbb R^n)$ be two square real matrices of dimension $n$. Suppose that the eigenvalues of $A,B$ are real and strictly positive. Can $A+B$ have a strictly negative eigenvalue?

I know that if $A,B$ commute and are diagonalizable, then they are diagonalizable in a shared basis and therefore the answer to the question is negative.

However, I don’t see how to tackle the general case.

答: Take $A=\pmatrix{1&0\cr n&1}$ and $B=\pmatrix{1&n\cr 0&1}$, $A+B=\pmatrix{2&n\cr n&2}$, the polynomial characteristic of $A+B$ is $(2-X)^2-n^2=X^2-4X-n^2+4$. If $n$ is enough big, it has a negative root.


问:
How do I apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$?
I want to apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$. I'm not able to do it in the standard way, because one term has $z^{-1}$ term and the other has $z$. What is the approach in problems like this?</p>

答:
I think this is about Z-Transformation.

In order to make the overall transfer function causal, we rewrite the given transfer function as

$$ \dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})}. $$

Next, do:

$$ \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})} = \dfrac{A}{a+bz^{-1}} + \dfrac{B}{y+xz^{-1}} $$

It is easy to see that:

$$
aA + yB = 0 \\
xA + bB = K
$$

Solution of this equation set gives:

$$
A = \dfrac{Ky}{xy - ab} \\
B = \dfrac{Ka}{ab - xy}
$$

Therefore, the partial fraction expression is (in the standard first order Z-Transform format):

$$
\dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{\dfrac{Ky}{xy - ab}}{a+bz^{-1}} + \dfrac{\dfrac{Ka}{ab - xy}}{y+xz^{-1}} = \dfrac{\dfrac{Ky}{a(xy - ab)}}{1+\dfrac{b}{a}z^{-1}} + \dfrac{\dfrac{Ka}{y(ab - xy)}}{1+\dfrac{x}{y}z^{-1}}
$$

And the corresponding time domain expression of the given discrete time transfer function is:

$$
f[n] = \left[\dfrac{Ky}{a(xy - ab)} e^{-\frac{b}{a}n} + \dfrac{Ka}{y(ab - xy)} e^{-\frac{x}{y}n}\right] u[n]
$$


问:
What pairing function coincides with the Gödel pairing on the natural numbers?
Gödel's pairing function on ordinals is defined in this MO post. We may compare it with Cantor's pairing function
$$
p(m,n)=\frac{(m+n)(m+n+1)}{2}+n
$$
But these two pairing functions don't coincide on the natural numbers. For example, $p(1,1)=4$, while the ordertype of the initial segment determined by $(1,1)$ in the canonical ordering of ordinals is $3$: $(0,0)\prec(1,0)\prec(0,1)\prec(1,1)$.

So here is the question: is there a pairing function on the natural numbers which agrees with Gödel's pairing function (restricted to the naturals)?

答:
The inverse of the Cantor pairing function follows this path, where $(0,0)$ is the top left point:

In contrast, the inverse of Gödel's function does this:

Either Cantor invented many paths for reasons unknown, or your source just assumed they were the same without checking.

Noah Schweber is correct (of course). Gödel's pairing function is a pairing function on the natural numbers. It just also extends beyond them. Once you get the basic idea of these functions, it is easy to come up with new ones. Just pick some path through all the lattice points and invert it. For example, this path provides another pairing function:


问:
Prove that:$u_n$ and $v_n$ adjacent
WE have two sequence
$V_n=(1+\frac{1} {n}) (u_n) $

And

$u_n=\prod\limits_{1}^{n}(1+\frac{1} {k^2})$

I know that : $\lim_{n\to+\infty} v_n-u_n=0$

I want to prove $u_n$ is incremental and $v_n$ is diminishing for say : $u_n$ and $v_n $ adjacent

For $u_n$:

$ln(u_n) =\sum_{1}^{n}\frac{k^2 +1}{k^2} =2+\frac{5}{4}+\frac{10} {9}+.... +\frac{n^2 +1}{n^2} $

$ln(u_(n+1)) =\sum_{1}^{n+1}\frac{k^2 +1}{k^2} =2+\frac{5}{4}+\frac{10} {9}+.... +\frac{(n+1)^2 +1}{(n+1)^2} $

$\Rightarrow$$\frac{u_(n+1)}{u_n}=e^{\frac{1+(n+1)^2} {(n+1)^2}} >1 $

$\Rightarrow $$u_(n+1)>u_n$
$u_n$ is incremental sequence

For $V_n:$
$V_n=(1+\frac{1} {n}) (u_n) $

I am proved that : $v_(n+1)-v_(n)=\frac{(u_(n+1)-u_n)(n^2 +2n)-u_n}{n(n+1)}$

And I US other ideas but I don't know how we can show $v_n$ is diminishing

Finally:

I want to prove $v_n$ is diminishing

Title: Prove that:$u_n$ and $v_n$ adjacent

Body:
hint

You want to prove that
$$\ln(V_{n+1})\le \ln(V_n)$$

or equivalently

$$\ln(1+\frac{1}{n+1})+\ln(U_{n+1})\le$$
$$\ln(1+\frac 1n)+\ln(U_n)$$

or

$$\ln(\frac{n(n+2)}{(n+1)^2})\le \ln(\frac{U_n}{U_{n+1}})$$

and

$$\frac{n(n+2)}{(n+1)^2}\le \frac{(n+1)^2}{(n+1)^2+1}$$

This last inequality is always true.
use for this
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$


问:
Question in one of the proofs of Terry Tao's Analysis I on chapter on Integration
We have this inequality:
\begin{equation}
0 \leq \overline{\int_I} \max(f,g) - \underline{\int_I} \max(f,g) \leq 4\epsilon
\end{equation}
for every $\epsilon$. The next line says since $\overline{\int_I} \max(f,g) - \underline{\int_I} \max(f,g)$ does not depend on $\epsilon$, we thus see that $\overline{\int_I} \max(f,g) - \underline{\int_I} \max(f,g)=0$ .

I don't understand the argument that if the difference does not depend on $\epsilon$ it is equal to zero. why?

答:
This is about a certain definite number $a\in{\mathbb R}$, maybe defined in a complicated way. In addition you know that for every $\epsilon>0$ the inequality
$$0\leq a\leq\epsilon$$
holds. This is only possible when in fact $a=0$. Because, when $a>0$ we could choose $\epsilon:={a\over2}>0$ and then would have
$$0<a\leq{a\over2}\ ,$$
which is impossible.


问:
Find all $f(X)\in\mathbb{Z}[X]$ such that $\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$ [Legendre symbol and $p$ is a fixed prime]
Question: Let $f(X)\in\Bbb{Z}[X]$ be a non-constant polynomial such that $$\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$$ where $\left(\frac{\cdot}{p}\right)$ is the Legendre symbol and $p$ is a fixed odd prime. Find all such polynomials $f$.

Basic Observations: We know that $f(n+p)\equiv f(n)\pmod{p}$ for all $n\in\Bbb{Z}$. Then $$\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$$ if and only if $$\left(\frac{f(j)}{p}\right)=\left(\frac{j}{p}\right)\tag{1}$$ for all $j\in\{0,1,\ldots,p-1\}$. By Euler's criterion for quadratic residues, we get that $$\boxed{\left(\frac{f(j)}{p}\right)\equiv f(j)^{\frac{p-1}{2}}\pmod{p}}\qquad\boxed{\left(\frac{j}{p}\right)\equiv j^{\frac{p-1}{2}}\pmod{p}}$$ Therefore by condition $(1)$, we get that $$f(j)^{\frac{p-1}{2}}-j^{\frac{p-1}{2}}\equiv0\pmod{p}\;(\,\forall\,0\leq j\leq p-1)$$ Therefore we get that the polynomial $f(X)^{\frac{p-1}{2}}-X^{\frac{p-1}{2}}$ is in the ideal $(X^p-X)$ of $\Bbb{Z}/p\Bbb{Z}[X]$. In fact, this is an iff condition. Any $f$ satisfies the condition of the question if and only if $f(X)^{\frac{p-1}{2}}-X^{\frac{p-1}{2}}$ is in the ideal $(X^p-X)$ of $\Bbb{Z}/p\Bbb{Z}[X]$. I don't know how to proceed further to characterize all such polynomials. It is obvious that any polynomial of the form $aX^{2k+1}$ works, where $a$ is a quadratic residue modulo $p$. But what are all such polynomials? Any help would be highly appreciated. Thanks in advance!

答:
It makes more sense to first work in $\mathbb F_p[x]$. Observe that for $f\in \mathbb F_p[x]$, we have $\left(\frac{f(n)}{p}\right) = \left(\frac{n}{p}\right)$ for all $n\in \mathbb F_p$ if and only if $f(n)=0$ and $\left(\frac{nf(n)}{p}\right)=1$ for all $n\in \mathbb F_p^{\times}$.

We claim that $g\in \mathbb F_p[x]$ satisfies $g(0)=0$ and $\left(\frac{g(n)}{p}\right)=1$ for all $n\in \mathbb F_p^{\times}$ if and only if $g\equiv h^2\mod (x^p-x)$ for some $h\in \mathbb F_p[x]$ with $h(0)=0$ and no zeroes in $\mathbb F_p^\times$. One direction is obvious. For the other, suppose $g$ satisfies $g(0)=0$ and $g(n)$ is a nonzero square for all $n\in \mathbb F_p^{\times}$. Let $\tau : \mathbb F_p^{\times}\to \mathbb F_p^{\times}$ be a function (not necessarily polynomial yet) with $g(n)=\tau(n)^2$. Via Lagrange interpolation, we can construct a polynomial $h$ with $h(0)=0$ and $h(n)=\tau(n)$ for all $n$. Then $g - h^2$ has zeroes everywhere in $\mathbb F_p$, so $g\equiv h^2\mod (x^p-x)$

Going back to $f$, this means that we need $f(0)=0$ and $xf(x)\equiv h(x)^2\mod (x^p-x)$ for some polynomial $h(x)\in \mathbb F_p[x]$ whose only root is $0$. This implies that $x\mid h(x)$, so we can write $h(x) = x r(x)$ for some $r$ which has no roots in $\mathbb F_p^\times$. Then $xf(x)\equiv x^2r(x)^2\mod (x^p-x)$. Since $f(0)=0$, this implies $f(x)\equiv xr(x)^2\mod (x^p-x)$. This classifies all possible $f\in \mathbb F_p[x]$.

So, for polynomials in $\mathbb Z[x]$, the answer is that all possible $f$ are those of the form $$f(x) = xr(x)^2 + p r_1(x) + (x^p-x)r_2(x)$$ for polynomials $r,r_1,r_2$, with the additional restriction that $p\mid r(n)\implies p\mid n$.

The example you gave of $ax^{2k+1}$ is when $r(x) = bx^k$, where $b^2\equiv a\mod p$, and $r_1$ is some suitable constant, and $r_2=0$.


问:
Fourier Series of mod sinx
The definition given on wikipedia for Fourier Coefficients is as follows:

$a_n=\frac{2}{P}\int_{P}^{} s(x)cos(\frac{2\pi xn}{P})dx$

where $P$= Period

Now, then while solving for Fourier series of $|sinx|$ I equated it to:

$\frac {a_o}{2} + \sum cos(\frac{2\pi xn}{P})dx$

Here, I got confused whether to take P as $\pi$ or $2\pi$ even when I know that the period of $|sinx|$ is $\pi$. This is is because most of the solution online equate the function to the series:

$\frac {a_o}{2} + \sum cos(nx)dx$ and then proceed while it should be $cos(2nx)$ according to me since period is $\pi$.

Also, in other places I have found the definiton as:

$a_n=\frac{2}{P}\int_{P}^{} s(x)cos(\frac{\pi xn}{P})dx$ which differs from wikipedia which has a 2 in the argument of $cos$.

What do I don't understand here?

答:
In general, and for a periodic function $\;f(x+2a)=f(x)\;$ defined on $\;[-a,a]\,,\,\,\,a\in (0,\infty)\;$, we get

$$f(x)\sim \frac{a_0}2+\sum_{n=1}^\infty a_n\cos\frac{n\pi x}L+b_n\sin\frac{n\pi x}L\;,\;\;L=a=\frac12\left|[-a,a]\right|\;$$

and

$$a_n=\frac1L\int_{-a}^a f(x)\cos\frac{n\pi x}L\,dx\;,\;\;a_n=\frac1L\int_{-a}^a f(x)\sin\frac{n\pi x}L\,dx$$

Title: Fourier Series of mod sinx


问:
Prove that $e^{\frac{x+y}{2}} \le \frac{e^x + e^y}{2}$
To prove:  $e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$

We observe:

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le 1/2\sum_{n=0}^\infty \dfrac{x^n}{n!} + 1/2\sum_{n=0}^\infty \dfrac{y^n}{n!}$

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty 1/2\dfrac{x^n}{n!} + \sum_{n=0}^\infty 1/2 \dfrac{y^n}{n!} $

$\Leftrightarrow \sum_{n=0}^\infty  \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty \dfrac{x^n + y^n}{2\cdot n!}$

In essence, now we need to show that $ \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \dfrac{x^n + y^n}{2\cdot n!}$

I'm looking for an idea or a hint to help me solve this. I'm thinking about using something from the binomial theorem. Maybe I made a mistake as well.

答:
It is sufficient to notice that :

$e^{\frac{x+y}{2}}= \sqrt{e^xe^y}$

The Inequality of arithmetic and geometric(AM-GM) means is:

$ \sqrt{ab}\le \frac{a+b}{2}$

Let $a=e^x, b=e^y$. Then we are done!


答:
It could be done in a simpler way using  Mathematica

```<code>FindInstance[(10 a+b)^2+(10 b+c)^2+(10 c+d)^2==1000 a+100 b+10c+d
 &&0<a<10&&0<b<10&&0<c<10&&0<d<10,{a,b,c,d},Integers,2]
</code>```

and get
$$\{\{a\to 3,b\to 3,c\to 3,d\to 4\}\}$$
the number is $3334$.

Body:
It could be done in a simpler way using  Mathematica

FindInstance[(10 a+b)^2+(10 b+c)^2+(10 c+d)^2==1000 a+100 b+10c+d
 &&0<a<10&&0<b<10&&0<c<10&&0<d<10,{a,b,c,d},Integers,2]

and get
$$\{\{a\to 3,b\to 3,c\to 3,d\to 4\}\}$$
the number is $3334$.


问:
How to find Isomorphism between groups of order 8
Let $A=\{1,2,3\}$ and $2^A$ is all the subsets of A. Then $(2^A,\triangle)$ is a group. How can I show that the group $\big\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\},\emptyset\big\}$
with operation $\triangle$, where $\triangle$ is the symmetric diffrence $C△B=(C \cup B) \backslash (C \cap B)$
is isomorphic to the group

$S= \{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),(1,1,1)\}= \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ with operation $+$.

I did not manage to find a bijection, but both groups have order $8$ and the order of the elements correspond.

I did do the cayley tables for both groups but I still did not manage.

答:
HINT: In general, if $X$ is a finite set, $P(X)$ is the power set of $X$ and $A \in P(X)$, what is the identity element of $(P(X), \triangle)$? What do you get when you take symmetric difference $A \triangle A$? What does it tell you about the order of $A$ in $(P(X), \triangle)$? Is $\triangle$ a commutative operation?

If you are trying to find a bijection, you can consider the map $\phi: P(X) \to S$ such that $\phi(A) = (a_1,a_2,a_3)$ where
$$a_i =
\begin{cases}
1,  & \text{if $i \in A$} \\[2ex]
0, & \text{if $i \notin A$}
\end{cases}$$

where addition operation is under modulo $2$.

Then, you can show that $\phi$ is an isomorphism by taking arbitrary two sets $A,B \in P(A)$ and showing that $\phi(A\triangle B) = \phi(A)+\phi(B)$ by also considering the meaning of symmetric difference operation (for instance if we take $\{1,2\}$ and $\{1\}$, what is their symmetric difference and how is it related to addition modulo $2$ of $(1,1,0)$ and $(1,0,0)$? As an intuition, symmetric difference can be also thought as $A \triangle B = (A \cup B) \backslash (A \cap B)$. So, elements in the intersection do not belong to the resulting set).


问:
Pencil of cubics with infinitely many exceptional curves
In Beauville's Complex Algebraic Surface, exercise V.21(5), we must find a surface with infinitely many exceptional curves.

He gives the following hint (I'm paraphrasing):

let $P$ be a pencil of cubic in $\Bbb{P}^2$ such that every cubic in $P$ is irreducible and let $S$ be the blowup of the basepoints of $P$. Show that every divisor $D$ with $D^2=D\cdot K_X=-1$ is equivalent to an exceptional curve (*). Let $L\subset S$ be the strict transform of a general line in $\Bbb{P}^2$ and $E_i$ the exceptional divisors. Consider the following:
\begin{align*}
\delta&:=L-E_1-E_2-E_3\\\\
\phi:\text{Pic}(S)&\to\text{Pic}(S)\\
D &\mapsto D+(\delta\cdot D)\delta
\end{align*}

I've checked that $\delta^2=-2$, $K_S\cdot\delta=0$, $\phi(D)^2=D^2$ and $K_S\cdot\phi(D)=K_S\cdot D$. That also allows us to check that $\phi\circ\phi=\text{id}$ and that $\phi$ is an automorphism.

Assuming the claim (*), we can also show that for every exceptional curve $C$ we can get a new exceptional curve by applying $\phi$.

Now I'm stuck because 1) I don't know how to prove (*) and 2) I don't know how the information I have about $\phi$ will show that there are infinitely many exceptional curves.

答:
For (1), according to Tabes' comment, the only issue is to show smoothness. Actually, there is a adjunction formula for effective divisors: $$2p_a(D)-2=D\cdot(D+K_S),$$
where $p_a(D)$ is the arithmetic genus of an effective divisor $D$ (see Hartshorne, Chp. V, Ex. 1.3). Now the assumptions implies that $p_a(D)=0$. Now, if $D$ is irreducible, it must be a smooth rational curve. The irreducibility should be guaranteed by choosing 9 blowup points to be general (Also see the reference given in this MO answer).

For (2), you can iterate the $\phi$ with respect to different exceptional divisors. For instance, for each $\delta_{ijk}=L-E_i-E_j-E_k$ with $i<j<k$, the map

$$\phi_{ijk}:D\mapsto D+(D\cdot \delta_{ijk})\delta_{ijk}$$

defines an automorphism on $\text{Pic}(S)$. Of course, it is an involution, so in order to generate an infinite sequence of $(-1)$-curves, one should consider compositions of $\phi_{ijk}$'s with varying indexes.

Apply $\phi_{ijk}$ to a divisor $C=aL-\sum_nb_nE_n$, we have  $$\phi_{ijk}(C)=a'L-\sum_nb_n'E_n,~ \text{with}~a'=2a-b_i-b_j-b_k. $$

If $C$ is a $(-1)$-curve, then the condition $C\cdot K_S=-1$ implies that $3a-\sum_nb_n=1.$ So if $b_i,b_j,b_k$ are the minimal three integers among $\{b_1,...,b_9\}$, $$a-b_i-b_j-b_k\ge a-\frac13\sum_nb_n>0,$$ therefore $a'>a$. Namely, for each $(-1)$-class, we can always apply suitable $\phi_{ijk}$ such that the leading coefficient is strictly incresing. So by iterating this process, we can find a infinite sequence of $(-1)$-curves.


问:
The harmonic conjugate of $\Im e^{z^2}$?
It is obvious that $e^{z^2}$ is analytic, right? So the harmonic conjugate of $\Im e^{z^2}$ is $\Re e^{z^2}$, isnt' it?

However, the solutions manual I'm consulting gives the answer as $\Im (-ie^{z^2})$, which is not the same function, and I don't understand.

答:
For real-valued functions $u(x,y)$ and $v(x,y)$, we say that $v$ is a harmonic conjugate of $u$ if $u+iv$ is analytic. This is not a symmetric relation between $u$ and $v$. For example, $e^x\sin y$ is a harmonic conjugate of $e^x\cos y$ because $e^x\cos y+ie^x\sin y=e^z$ is analytic, but $e^x\cos y$ is not a harmonic conjugate of $e^x\sin y$ because $e^x\sin y+ie^x\cos y$ is not analytic (check the Cauchy-Riemann equations).

If $v$ is a harmonic conjugate of $u$, meaning that the function $f(z)=u+iv$ is analytic, then $-if(z)=v-iu$, being a constant multiple of $f(z)$, is also analytic; this shows that $-u$ is a harmonic conjugate of $v$, in other words, $-\Re f(z)$ is a harmonic conjugate of $\Im f(z)$.

To repeat one more time: although $\Im f(z)$ is a harmonic conjugate of $\Re f(z)$ (assuming $f(z)$ is analytic), it's $-\Re f(z)$, not $\Re f(z)$, that is a harmonic conjugate of $\Im f(z)$. This is true for any analytic function $f(z)$, including your problem function $f(z)=e^{z^2}$.


答: One more proof: define $f_y(x):=e^x+e^y-2e^{(x+y)/2}$ at fixed $y$ so $f_y(y)=0$ while $f_y^\prime(x)=e^x-e^{(x+y)/2}$. Since $f_y^\prime(x)$ has the same sign as $x-y$, $x=y$ is the global minimum.

Title: Prove that $e^{\frac{x+y}{2}} \le \frac{e^x + e^y}{2}$

Body:
To prove:  $e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$

We observe:

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le 1/2\sum_{n=0}^\infty \dfrac{x^n}{n!} + 1/2\sum_{n=0}^\infty \dfrac{y^n}{n!}$

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty 1/2\dfrac{x^n}{n!} + \sum_{n=0}^\infty 1/2 \dfrac{y^n}{n!} $

$\Leftrightarrow \sum_{n=0}^\infty  \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty \dfrac{x^n + y^n}{2\cdot n!}$

In essence, now we need to show that $ \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \dfrac{x^n + y^n}{2\cdot n!}$

I'm looking for an idea or a hint to help me solve this. I'm thinking about using something from the binomial theorem. Maybe I made a mistake as well.


答:
Note that a group morphism $\psi:  (\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2,+)\to (2^A ,\triangle)$ is fully determined by the values
$\psi(1,0,0), \psi(0,1,0)$ and $\psi(0,0,1)$. A reasonable guess for our isomorphism would be
$$\psi(1,0,0) = \{1\}, \ \psi(0,1,0) = \{2\}, \ \psi(0,0,1) = \{3\}.$$
I leave the explicit verification that this works as an exercise for you.

Alternatively, you can proceed as follows, once you observe that $\triangle$ is a commutative operation:

By the fundamental theorem of finite abelian groups, you know that $(2^A, \triangle)$ must either be isomorphic to $\mathbb{Z}_8, \mathbb{Z}_2 \times \mathbb{Z}_4$ or $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. The first two options are impossible because $(2^A, \triangle)$ does not contain elements of order $4$ and $8$.

Title: How to find Isomorphism between groups of order 8

Body:
Let $A=\{1,2,3\}$ and $2^A$ is all the subsets of A. Then $(2^A,\triangle)$ is a group. How can I show that the group $\big\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\},\emptyset\big\}$
with operation $\triangle$, where $\triangle$ is the symmetric diffrence $C△B=(C \cup B) \backslash (C \cap B)$
is isomorphic to the group

$S= \{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),(1,1,1)\}= \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ with operation $+$.

I did not manage to find a bijection, but both groups have order $8$ and the order of the elements correspond.

I did do the cayley tables for both groups but I still did not manage.


问:
What is the probability of getting 3 heads in 4 coin tosses, given you get at least 2 heads?
I came across this question while doing conditional probability questions and while I can get the answer by counting outcomes, I'm getting a different answers using Bayes' theorem. $\ P(2T)\ $ is the probability of getting at least 2 tails. $\ P(3T)\ $ is the probability of exactly 3 tails. We flip a fair coin 4 times and are looking for the probability of getting 3 heads given we get at least 2 heads.
$$P(3T|2T) = \frac{P(2T|3T)P(3T)}{P(2T)}$$
$$P(2T|3T) = 1$$
$$P(2T) = \frac{1}{4}$$
$$P(3T) = \frac{3}{8}$$
$$P(3T|2T) = \frac{2}{3} $$

The answer is supposed to be $P(3T|2T) = \frac{4}{11}$ which is easy enough to see when the outcomes are counted but there's clearly a problem with the above method I'm using.

答:
It is unclear from your question whether your asking for the probability of exactly three heads, but that's what I'll assume.

Given that there are at least two heads, the number of cases are as follows:

Exactly $2$ heads:  ${4 \choose 2} = 6$ cases

Exactly $3$ heads:  ${4 \choose 3} = 4$ cases

Exactly $4$ heads:  ${4 \choose 4} = 1$ case

Of these $11$ equally likely cases, four have exactly $3$ heads.  Hence the probability is $P(3|2) = \frac{4}{11}$.