问:
Relation of equivalence and equivalence class
(ZFC)Prove that the two conditions are equivalent:

S/~≤T & T≤S

∃f (f:S↠T & ∀s1∀s2(f(s1)=f(s2)⇒s1~s2)).

My attempt: I start with the first, trying to prove the second.

S/~ is defined as the union of all equivalent classes of the form [a]/~, such that a ∈ S.
Hence S/ is the number of proper classes of S/~, such that ∪([a]/~|a∈ S).

Since T≤S ⇒∃(f:S↠T) and f is not an injection.

Proof: Let x1 and x2 be two arbitrary elements from S.
Since the number of elements of S is bigger than those in T⇒ under some condition f(x1)=f(x2).

But from AC it follows that if S/~ consists of pairwise disjoint non-empty sets, there exists at least one set C, such that it contains exactly one element in common with each of the sets in S/~.
Hence ∃(g:S⤖C) and since T≤C.

But from def of proper classes, namely ∀s1∀s2(<s1,s2> ∈ S/~ ⇒ Comp~(s1,s2)⇔f(s1)=f(s2).

Not sure wether it works, so please help me.

答:
In its current format (i.e., without assuming that range of $f$ is $T$), the second bullet point is equivalent to $|S/{\sim}|\le |T|$.

Indeed, if a function $f:S\to T$ satisfies $f(s_1)=f(s_2)\implies s_1\sim s_2$, then, using axiom of choice, choose a representative of every equivalence class by a function $u:S/{\sim}\,\to S$ that satisfies $\pi\circ u={\rm id}$ where $\pi:S\to S/{\sim}$ is the natural map $\pi(s)=[s]_\sim$.

Then the condition on $f$ implies that $f\circ u:S/{\sim}\,\to T$ is injective.

For the other direction, if $g:S/{\sim}\,\to T$ is injective, then the composition $g\circ\pi:S\to T$ satisfies the condition.

If, moreover $f$ in second bullet point is assumed to be surjective, then it also implies $|T|\le |S|$ (by taking any right inverse of $f$, which exists by the axiom of choice).

Finally, assuming injections $g:S/{\sim}\,\to T$ and $h:T\to S$, we can define $f$ so that

$\quad f(s)=t\quad$ if $h(t)=s$ and $t\notin{\rm range}(g)$

$\quad f(s)=g([s]_\sim)\quad$ otherwise,

then its range is $T$ by construction, and verify that it satisfies $f(s_1)=f(s_2)\implies s_1\sim s_2$.


问:
Every uniformly continuous function $\mathbb Q\to \mathbb Z$ is constant.
First of all, is continuity not enough?

Secondly, my attempt is since it is uniform continuous.

For $\epsilon=1/2$, for every ball on $\mathbb Z$

i.e. $B(n,\epsilon)\subset \mathbb Z$, with $B(n,\epsilon)\cap \mathbb Z=\{n\}$

on the inverse image $f^{-1}(B(n,\epsilon))$, $f$ is constant.

Since I have balls on $\mathbb Q$ with finite length, I can cover $\mathbb Q$ with infinitely many of them (centered at every rationals) and since every ball (interval) intersects and if two ball intersects their value under $f$ must be the same so I can conclude that $f$ is constant on whole $\mathbb Q$.

I believe the proof is correct (of course some details should be added), is there more elegant/clever way to prove it? Such as in the case for $f:\mathbb R\to \mathbb Z$ case using connectedness?

答: Less formal, more intuitive: If $f$ is uniformly continuous, then there is a $\delta>0$ such that $f$ varies by less than $1$ on all intervals of length $\delta$. But since it's a function to $\mathbb Z$, it then has to be constant on all those intervals. And since $\mathbb Q$ can be covered by overlapping intervals of that length, $f$ has to be constant in its entirety.

ContentLicense: CC BY-SA 4.0


问:
Illustration of negation of limit at infinity

If we consider a function $f: R \to R$ and $L\in R$.

Then $L$ is said to be the limit of $f(x)$ as $x$ goes to infinity iff:

$\forall \epsilon \in R: [\epsilon \gt 0 \implies \exists m\in R: \forall x\in R: [m\lt x \implies |f(x)-L|\lt \epsilon]]$

And we can say that $L$ is NOT the limit of $f$ as $x$ goes to infinity
iff:

$\exists \epsilon_0 \in R: [\epsilon_0 \gt 0 \land \forall m\in R: \exists x_0\in R: [m\lt x_0 \land |f(x_0)-L| \ge \epsilon_0]]$

I am trying to illustrate the negation with an example:

$f: R \to R$ such that $f(x)=\sin (x)$

If I consider $\epsilon_0 =1/2$, then there always exists $x_0 \in(2n\pi +\pi / 6, 2n\pi + 5\pi / 6) \cup (2n\pi +7\pi / 6, 2n\pi + 11\pi / 6) $ such that $\forall m (\gt 0) \in R : [m\lt x_0 \land |f(x_0)| \ge \epsilon_0]$ holds and thus the limit is not $L = 0$ as $x$ goes to infinity.

Am I going the right way? Please suggest.

答:
All the ideas are there but you've made a small logical mistake, in that the quantity $n$ has not been specified.

After setting $\epsilon_0 > 1/2$, next you say:

Let $m > 0$.

Now you have to find a value of $x_0 > m$ such that $|f(x_0)| > 1/2$. To find that value of $x_0$ here's what you do:

Choose an integer $n$ such that $2n\pi+\pi/6 > m$. Now choose $x_0 \in (2 n \pi + \pi/6, 2 n \pi + 5 \pi / 6)$.

To be even more rigorous, you might want to replace that first sentence by this:

Applying the Archimedean principle, choose an integer $n$ such that
$$n > \frac{m - \pi/6}{2\pi}
$$
and therefore
$$2n\pi+\pi/6 > m
$$


问:
a group is not the union of two proper subgroups - how to internalize this into other categories?
A well-known fact from group theory is that a group cannot be the union of two proper subgroups. I wonder, does this statement internalize into other categories than the category of sets? That is, is there some corresponding statement for group objects in, say, cartesian categories? It is certainly true for topoi, interpreted in a suitable way, since the usual proof is intuistic (however, note that $G=G_1 \cup G_2 \Rightarrow G=G_1 \vee G=G_2$ fails for disconnected topoi).

What about the following: Let $G:=(|G|,m,i,e)$ be a group object in the cartesian category $\mathcal{C}$. Let $G_1 \to G$ and $G_2 \to G$ be homomorphisms of group objects, which are no isomorphisms. Assume that $|G_1| \to |G|$ and $|G_2| \to |G|$ are monomorphisms. Is it not possible that $|G| \to |G|$ is their union, i.e. their sup in the preorder of subobjects of $|G|$?

Well as I've said, it fails for disconnected topoi. What do we have to assume on $\mathcal{C}$ so that it becomes true?

答:
First, another easy counterexample. Recall that an internal group in $\mathbf{Grp}$ is just an abelian group. So consider $G = \mathbb{Z} / 6 \mathbb{Z}$, let $G_1$ be the unique subgroup of order 2, and let $G_2$ be the unique subgroup of order 3. Then the smallest subgroup of $G$ containing $G_1$ and $G_2$ is $G$ itself, so the claim fails in $\mathbf{Grp}$ as well.

Now, for some slightly more positive results. Let's recall the standard proof. Let $G_1$ and $G_2$ be subgroups of $G$. Suppose $G = G_1 \cup G_2$, $g_1 \notin G_2$, $g_2 \notin G_1$. Consider $g_1 g_2$. If $g_1 g_2 \in G_1$, then $g_2 \in G_1$, a contradiction; and if $g_1 g_2 \in G_2$, then $g_1 \in G_2$ , also a contradiction. Thus, $g_1 g_2 \notin G_1 \cup G_2$, so $G \ne G_1 \cup G_2$.

As you say, this is intuitionistically valid. In fact it can even be interpreted in any coherent category. The only problem is that it does not mean what you think it means! The right formulation is the following:

If $G_1$ and $G_2$ are subgroups of $G$ and we (simultaneously) have (generalised) elements of $G \setminus G_1$ and $G \setminus G_2$, then $G \ne G_1 \cup G_2$.

More formally,
$$g_1 \notin G_2, g_2 \notin G_1 \vdash_{g_1 : G, g_2 : G} \lnot (\forall g : G . g \in G_1 \lor g \in G_2)$$
but having $g_2 \notin G_1$ is generally a stronger condition than $G_1 \hookrightarrow G$ not being an isomorphism. In general, $G_1 \hookrightarrow G$ in an isomorphism (in a topos) if and only if the following formula holds in the internal logic:
$$\forall g : G . g \in G_1$$
Unfortunately, it is not possible to express internally that a formula does not hold! Although we may write
$$\lnot (\forall g : G . g \in G_1)$$
the above formula asserts that $G_1 \hookrightarrow G$ is not even an isomorphism locally: in a topos, the above formula is valid if and only if, for every $U$, the induced map $\mathrm{Hom}(U, G_1) \to \mathrm{Hom}(U, G)$ is not a surjection. Clearly, this is a stronger condition than $G_1 \hookrightarrow G$ not being an isomorphism.

Nonetheless, we may deduce that the claim is true for complemented proper subgroups $G_1$, $G_2$ in a topos in which the subobjects of $1$ form an irreducible locale: because under these hypotheses, the locus $U$ over which both $G \setminus G_1$ and $G \setminus G_2$ is inhabited is non-empty, so $G \ne G_1 \cup G_2$ in the topos over $U$. But this is enough to prove $G \ne G_1 \cup G_2$ globally as well – because passing to the topos over $U$ is a logical functor.

Postscript. A more sophisticated approach allows us to deduce another version of the claim. Suppose we have subgroups $G_1$ and $G_2$ of an internal group $G$ in a topos such that $G_1 \times G \cup G \times G_2$ is a proper subobject of $G \times G$. Noting that $G = G_1 \cup G_2$ if and only if this is true in all covers of the topos, by passing to a boolean cover if necessary, we may assume the law of excluded middle. But then in that case it is clear that
$$(G \setminus G_1) \times (G \setminus G_2) = (G \times G) \setminus (G_1 \times G \cup G \times G_2)$$
so the LHS is non-empty. Thus we have reduced the question to the well-known case.

ContentLicense: CC BY-SA 3.0


问:
Concern regarding a proof from brilliant
Here in the proof part they deduced that the arc length is $$\lim_{n \to \infty}\sum_{n=1}^{\infty}\sqrt{1+(f'(x_{i}*))^2}\Delta x$$
then said that it was equal to the definite integral. But I didn't understand why. Shouldn't it be $f(x_{i})$ for the limit to be an integral? Since the $x_{i}*$ are not continuous and evenly spaced as $x_i$. Please explain.

答:
I would point you to another link.
But the confusion arises from the fact that your definition of integral admits only
right-Riemann sum. Whereas, you may pick any point in $\Delta x$ and in the $ limit \rightarrow \infty $, it will work out fine. So you may assume it's $x_i$ and the derivation still works.


问:
Find the field that $\mathbb{Z}_7[x,y]/\langle y-2x^2, 4xy + y +1 \rangle$ is isomorphic to
See the title. Here the set $\mathbb{Z}_7 = \{0,1,2,3,4,5,6\}$ is considered a ring with the obvious operations $+$ and $\cdot$; and $I := \langle y-2x^2, 4xy + y + 1 \rangle$ is an ideal in $\mathbb{Z}_7[x,y]$.

My idea was to show in two steps that:
$$ \mathbb{Z}_7[x,y]/I \cong \mathbb{Z}_7[x] / J \cong \mathbb{F}_{7^3}, $$
where $J := \langle x^3+2x^2+1 \rangle$.

The first isomorphism is true (on an intuitive level) because $y + I = 2x^2 + I$, so that:
$$(4xy + y + 1) + I = (8x^3 + 2x^2 + 1) + I = (x^3 + 2x^2 + 1) + I $$
To see why the second isomorphism holds, we need a bit more work. We know that $\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle$ is a field, because $x^3 + 2x^2 + 1$ is irreducible over $\mathbb{Z}_7$ (since no $x \in \mathbb{Z}_7$ is a root and its degree is 3). Now, "division" by $\langle x^3 + 2x^2 + 1 \rangle$ effectively gets rid of every polynomial in $\mathbb{Z}_7[x]$ of degree 3 or higher. We can reduce a polynomial of degree $\geq$ 3 to degree $\leq$ 2 by iteratively using the relation $x^3 = -2x^2-1 = 5x^2 + 6$. Moreover, every polynomial in $\mathbb{Z}_7[x]$ of degree $\leq$ 2 has a distinct representation in $\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle$. This means that $|\mathbb{Z}_7[x]/\langle x^3 + 2x^2 + 1 \rangle| = 7^3$, and hence this field must be isomorphic to $\mathbb{F}_{7^3}$.

I would like some tips on how to approach the details. Especially for the first isomorphism: I wanted to use the first isomorphism theorem to prove this. To this end, I used the ringmorphism:
$$\varphi : \mathbb{Z}_7[x,y] \to \mathbb{Z}_7[x]/J : p(x,y) \mapsto p(x,2x^2) + J$$
Then, I showed that $\ker(\varphi) = I$ and $\varphi(\mathbb{Z}_7[x,y]) = \mathbb{Z}_7[x]/J$. The first isomorphism theorem then tells us that $\mathbb{Z}_7[x,y]/\ker(\varphi) \cong \varphi(\mathbb{Z}_7[x,y])$, which is the desired result. I worked it out and it seems to be OK, but going through the details of the proof was rather cumbersome and I figured there should be an easier way.

The second isomorphism I think is pretty much proved already with the rough sketch I made earlier, I just need to work it out more precisely.

As a final question: is there maybe a way to find an isomorphism in just one step?

答:
A very useful trick to remember is that a multivariate polynomial ring is (isomorphic to) an iterated construction of polynomial rings. Namely, in your example, we have

$$
\Bbb F_7[x,y] \simeq \Bbb F_7[x][y].
$$

This is useful because polyomial rings in one variable may be more tractable.

For example: you can show that in general, if $T-\alpha \in A[T]$ is a degree $1$ monic polynomial in a polynomial ring $A$ and $P \in A[T]$ then there always exist unique $Q \in A[T]$ and $R \in A$ such that $P = (T-\alpha)Q+R$.

In particular, for any poynomial $p$ in $\Bbb F_7[x,y]$ there are unique $q \in \Bbb F_7[x,y]$ and $r \in \Bbb F_7[x]$ such that

$$
p(x,y) = (y-2x^2)q(x,y)+r(x).
$$

From this observation you can see that the map $\tau \colon \Bbb F_7[x,y] \to \Bbb F_7[x]$ sending $x \mapsto x, y \mapsto 2x^2$ has kernel precisely $(y-2x^2)$. Thus

$$
\Bbb F_7[x,y]/(y-2x^2) \simeq \Bbb F_7[x]
$$

and by the third isomorphism theorem we get

$$
\Bbb F_7[x,y]/I \simeq \Bbb F_7[x]/J.
$$

Now, you already observe that $J = (x^3+2x^2+1)$ is maximal, hence the latter quotient a field.

Since $\{[1],[x],[x^2]\}$ is a $\Bbb F_7$-basis for $\Bbb F_7[x]/J$ as a vector space, we have $\dim_{\Bbb F_7}\Bbb F_7[x]/J = 3$ and so $|\Bbb F_7[x]/J| = 7^3$. But then by uniqueness of finite fields, it has to be $\Bbb F_7[x]/J \simeq \Bbb F_{7^3}$.

Edit: here are some generalizations that use the exact same arguments: if $p = y-\alpha \in R[x][y]$ is of degree one, then $R[x,y]/(p) \simeq R[x]$ for any ring $R$, the isomorphism sends $x$ to $x$ and $y$ to $\alpha$.

Moreover, if $g \in R[x,y]$ then its image via this iso is $g(x,\alpha)$ and thus

$$
R[x,y]/(p,g) \simeq R[x]/(g(x,\alpha)).
$$

Fix $d = \deg g(x,\alpha(x))$. Then $\{[1],\ldots, [x^{d-1}]\}$ is a basis of $R[x]/(g(x,\alpha))$ as a free module.

In particular if $R = k$ is a field, then $k[X]/(g(x,\alpha))$ is a $d$-dimensional $k$-algebra, and if $g(x,\alpha)$ is irreducible then it is a field extension of dimension $d$. In that case, some computations follow:

If $k = \Bbb F_q$ is finite, then $\Bbb F_q[X]/(g(x,\alpha)) \simeq \Bbb F_{q^d}$.

If $d = 2 \neq \mathbf{char} (k)$, then $k[X]/(g(x,\alpha)) = k(\sqrt{d})$ with $d^2 \in k, d \not \in k$. Indeed, if $g(x,\alpha) = x^2+bc+c$ we have $k[X]/(g(x,\alpha)) \simeq k(\omega)$ with $\omega$ a root of $g(x,\alpha)$ in some algebraic closure, and we can apply the classification of quadratic extensions.

In general, if $R$ at least has the invariant basis property, you can (maybe) distinguish such quotients by looking at the total degree of a polynomial.


问:
Why isn't $U = \{A \in \mathbb{R}^{3 \times 3} \mid \det(A) \neq 0 \}$ a subspace of $\mathbb{R}^{3 \times 3}$
I'm asked to see whether $U = \{A \in \mathbb{R}^{3 \times 3}\mid \det(A) \neq 0 \} $ is a subspace of $\mathbb{R}^{3 \times 3}$. I have no idea where to start though other than the subspace criterion which states that $U$ is a subspace of another vector space if two conditions are met:

$U \neq \emptyset$

for any $\lambda_1,\lambda_2 \in \mathbb{R}$ and $u, v \in U$, it holds that $\lambda_1u + \lambda_2v \in U$

答:
it's not necessarily true that given matrices $A,B$, with $\det(A)\neq 0$ and $\det(B)\neq0$ that $\det(A+B)\neq 0$.

For example $\det(I_{3\times 3})=1\neq 0$ and $\det(-I_{3\times 3})=(-1)^3 =-1 \neq 0$ and $\det(I-I)=\det(0)=0$


问:
Let $\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a Brownian motion
Let $X$ be an $\mathcal{F}_{t},\; t\geq 0$, adapted a.s. continuous process with $X_{0}=0$ and    $M^{u,X}_{t}:=\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a $\mathcal{F}_{t}$-Brownian motion.

My idea:

We need to show that $\operatorname{cov}(X_{t},X_{s})=t \land s$ and that $X$ is a Gaussian process.

Let $0 \leq t_{0} < ... < t_{n} <\infty$. I know that for $T > 0, \; u \in \mathbb R$ the process $M_{t\land T}^{u,X}=\exp( i u X_{t\land T}+\frac{1}{2}u^{2}(t\land T))$ is a local martingale too. Furthermore, $\lvert M^{u,X}_{t\land T}\rvert\leq \exp(\frac{1}{2}u^{2}T)< \infty$.

Therefore, $(M_{t\land T}^{u,X})_{t\geq 0}$ is a martingale. By Doob's optional sampling Theorem, we obtain for $t < T$,  $$E[M_{t}^{u,X}]=E[M_{t\land T}^{u,X}]=E[M_{0\land T}^{u,X}]=E[M_{0}^{u,X}]=1$$

This would then imply that $$ E[\exp(iuX_{t})]=\exp(-\frac{u^{2}t}{2})$$ which is the characteristic function of a normal random variable $\mathcal{N}(0,t)$ evaluated at $u$. But I am struggling to do the same for the random vector $(X_{t_{0}},...,X_{t_{n}})$ and $0 \leq t_{0} < ... < t_{n}< \infty$. I think I need to use induction and the tower property to reduce it all to the random variable $X_{t_{0}}$, but that seems less efficient.

On the issue of showing $ \operatorname{cov}(X_{t},X_{s})= t\land s$, I am completely stuck. Maybe I can set $u:=\log(E[X_{t}X_{s}])$ and by assumption: $ M^{u,X}_{r}:=\exp( i \log(E[X_{t}X_{s}]) X_{r}+\frac{1}{2}\log(2E[X_{t}X_{s}])r)$ is a local martingale

We then have:

$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=\exp(-\frac{\log(E[X_{t}X_{s}])r}{2})$$

Then

$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=E[X_{t}X_{s}]^{-\frac{r}{2}}$$

I do not see how I get to $t\land s $ from here. Any ideas?

Or is something simply missing in the question's assumptions?

答:
(Out of laziness I'm just going to write $M^u$,  rather than $M^{u,X}$, etc.)

What I had in mind: for $0<s<t$,
$$
E[M^u_sM^v_t] =E[M^u_sM^v_s]
$$
by the martingale property of $M^v$ and the adaptedness of $M^u$. Writing this out in detail:
$$
E[\exp(iuX_s+ivX_t]\cdot\exp(su^2/2+tv^2/2) = E[\exp(i(u+v)X_s)]\cdot\exp(su^2/2+sv^2/2).
$$
or, simplifying,
$$
\eqalign{
E[\exp(iuX_s+ivX_t]
&= E[\exp(i(u+v)X_s)]\cdot\exp(-(t-s)v^2/2)\cr
&=\exp(-s(u+v)^2/2)\cdot\exp((t-s)v^2/2)\cr
&=\exp( -[su^2+2uvs+tv^2]/2).
}
$$
This joint characteristic function of $(X_s, X_t)$ is that of a bivariate normal distribution with zero means and covariance matrix
$$
\left[\matrix{s&s\cr s&t\cr}\right],
$$
as hoped for. The calculation for the joint law of $X_{t_1}$, $X_{t_2},\ldots, X_{t_n}$ will be similar but more tedious.


问:
Deriving what $\cos(A+2B)$ is
We know that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ , but I don't understand with the $2B$. Would it just become $\cos(A+2B) = \cos(A)\cos(2B) - \sin(A)\sin(2B)$?

Thank you for any help you can give me.

答:
Yes, exactly! Just think of $2b=c$ and then substitute it back.

EDIT: It was marked that this not an answer, a requirement for substitution in general is that it has to be bijective, and indeed $2b=c$ is bijective.

Now you can use the formula $\cos(A+C) = \cos A\cos C - \sin A\sin C$
where $C=2B$ now you have it. Note that $\cos(2a)=\cos^2a-\sin^2(a)$. YoOu could use this as well later on.

Title: Deriving what $\cos(A+2B)$ is


答:
Further expansion is possible with repeated use of the angle addition identities.
$$\begin{align}
\cos (a + 2b) &= \cos a \cos 2b - \sin a \sin 2b \\
&= \cos a (\cos b \cos b - \sin b \sin b) - \sin a (\sin b \cos b + \cos b \sin b) \\
&= \cos a (\cos^2 b - \sin^2 b) - \sin a (2 \sin b \cos b) \\
&= \cos a \cos^2 b - \cos a \sin^2 b - 2 \sin a \sin b \cos b.
\end{align}$$

Title: Deriving what $\cos(A+2B)$ is

Body:
We know that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ , but I don't understand with the $2B$. Would it just become $\cos(A+2B) = \cos(A)\cos(2B) - \sin(A)\sin(2B)$?

Thank you for any help you can give me.


问:
Jordan measure of two intersecting cylinders
In the following example I need to find Jordan
measure of the following set:$$A=\{(x,y,z)\mid z>0,1-x^2=z, y^2+z^2=z\}.$$ Here we have two cylinders, but I am not sure to find Jordan
measure, is there formula or something?

答:
To find the limits of integral,

Take $y = r\cos\theta, z = r\sin\theta$ given the orientation of cylinder.

Equation of cylinder is $r = \sin\theta, 0 \leq \theta \leq \pi$

$-\sqrt{1-r\sin\theta} \leq x \leq \sqrt{1-r\sin\theta}$

$0 \leq r \leq \sin\theta$

$0 \leq \theta \leq \pi$

Order of integral $dx, $ then $dr$ and then $d\theta$.

In cartesian,

${-\sqrt{1-z}} \leq x \leq {\sqrt{1-z}}$

$-\sqrt{z-z^2} \leq y \leq \sqrt{z-z^2}$

$0 \leq z \leq 1$


问:
Sums of powers being powers of the sum
I'm looking for literature on solving problems of the form
$$
n_1^\alpha+\cdots+n_k^\alpha=(n_1+\cdots+n_k)^\beta
$$
for positive integers $n_1,\ldots,n_k$ and fixed parameters $k$ and $\alpha\ne\beta.$ Any ideas?

This is perhaps similar to multigrade equations or "equal sums of like powers" but not quite similar enough to use those results (as far as I can see). But surely this is too simple not to have been studied?

答:
First, note that $\alpha>\beta$ (if we consider $\alpha,\beta \in \mathbb{N}$).

$2$ limitations for $(n_1,\ldots, n_k)$:

A. Using generalized mean (power mean), one can obtain:

$\dfrac{n_1+\ldots+n_k}{k} \leqslant \sqrt[\alpha]{\dfrac{n_1^{\alpha}+\ldots+n_k^{\alpha}}{k}}$,

$\left(\dfrac{n_1+\ldots+n_k}{k} \right)^\alpha \leqslant \dfrac{n_1^{\alpha}+\ldots+n_k^{\alpha}}{k} = \dfrac{(n_1 + \ldots + n_k)^\beta}{k}$,

$\left(n_1+\ldots+n_k \right)^{\alpha-\beta} \leqslant k^{\alpha-1}$.

So, limitation for sum:
\begin{array}{|c|} \hline
n_1+\ldots+n_k\leqslant k^{\frac{\alpha-1}{\alpha-\beta}}. \\ \hline \end{array}

If we'll consider the case $\beta = \alpha-1$, then    $n_1+\ldots+n_k\leqslant k^{\alpha-1}$.

B.
We can obtain other (more strong) limitation:

$\dfrac{n_1+\ldots+n_k}{k}\leqslant\sqrt[\beta]{\dfrac{n_1^\beta+\ldots+n_k^\beta}{k}} \leqslant \sqrt[\alpha]{\dfrac{n_1^\alpha+\ldots+n_k^\alpha}{k}}$,

$\dfrac{n_1^\alpha+\ldots+n_k^\alpha}{k^\beta} = \left(\dfrac{n_1+\ldots+n_k}{k}\right)^\beta \leqslant \dfrac{n_1^\beta+\ldots+n_k^\beta}{k} \leqslant \left(\dfrac{n_1^\alpha+\ldots+n_k^\alpha}{k}\right)^{\frac{\beta}{\alpha}}$,

$\left(n_1^\alpha+\ldots+n_k^\alpha\right)^{\frac{\alpha-\beta}{\alpha}} \leqslant k^{\beta - \frac{\beta}{\alpha}} = k^{\frac{(\alpha-1)\beta}{\alpha}}$.

So, limitation for power sum:
\begin{array}{|c|} \hline
n_1^\alpha+\ldots+n_k^\alpha\leqslant k^{\frac{(\alpha-1)\beta}{\alpha-\beta}}.
\\ \hline \end{array}

If we'll consider the case $\beta = \alpha-1$, then    $n_1^\alpha+\ldots+n_k^\alpha\leqslant k^{(\alpha-1)\beta}$.

A few examples (for different $n_1, n_2,\ldots,n_k$):

$\color{#003388}{(\alpha,\beta)=(3,2), \;\;k\in \mathbb{N}}$: a famous identity:

$\color{#CC3300}{1^3+2^3+\ldots+k^3 = (1+2+\ldots+k)^2}$.

$\color{#003388}{(\alpha,\beta)=(4,3), \;\;k=5}$:

$6^4+7^4+12^4+15^4+19^4=(6+7+12+15+19)^3$;

$\color{#003388}{(\alpha,\beta)=(4,3), \;\;k=6}$:

$2^4+9^4+14^4+21^4+22^4+23^4=(2+9+14+21+22+23)^3$;

$4^4+5^4+6^4+11^4+13^4+20^4=(4+5+6+11+13+20)^3$;

$\color{#003388}{(\alpha,\beta)=(4,3), \;\;k=7}$:

$(1,2,11,14,17,20,26)$;

$(1,8,12,22,23,28,29)$;

$(1,22,23,26,29,33,39)$;

$(3,11,18,21,26,27,35)$;

$(5,9,16,20,26,30,33)$;

$(5,10,11,19,23,24,33)$;

$(6,7,18,21,25,31,33)$;

$\ldots$

$\color{#003388}{(\alpha,\beta)=(5,4), \;\;k=6}$:

$1^5+3^5+14^5+21^5+22^5+39^5 = (1+3+14+21+22+39)^4$;

$3^5+57^5+81^5+96^5+99^5+114^5 = (3+57+81+96+99+114)^4$;

$4^5+45^5+57^5+90^5+91^5+98^5 = (4+45+57+90+91+98)^4$;

$\ldots$

$\color{#003388}{(\alpha,\beta)=(5,4), \;\;k=7}$:

$1^5+2^5+27^5+38^5+54^5+69^5+80^5 = (1+2+27+38+54+69+80)^4$;

$1^5+5^5+17^5+26^5+34^5+59^5+63^5 = (1+5+17+26+34+59+63)^4$;

$2^5+20^5+23^5+36^5+38^5+43^5+79^5 = (2+20+23+36+38+43+79)^4$;

$3^5+4^5+12^5+16^5+17^5+45^5+48^5 = (3+4+12+16+17+45+48)^4$;

$3^5+18^5+19^5+27^5+35^5+37^5+71^5 = (3+18+19+27+35+37+71)^4$;

$\ldots$

Title: Sums of powers being powers of the sum


问:
Types of convergence of a Martingale
I'm studying an exercise in which we must determine what kinds of convergence holds for the following martingale
$$
X_n = \sum_{i=1}^n Y_i
$$
where $Y_i$ are i.i.d standard normal distributed r.v.'s.

First, we check convergence in $L^2$, and we can see that
$$
E[X_n^2] = \sum_{i=1}^n E[Y_i^2] = n
$$
From which we conclude that $\sup_n E[X_n^2] = \infty$. By the $L^2$ martingale convergence theorem, we do not have convergence in $L^2$.

Then, we check convergence in $L^1$
$$
E[|X_n|] = \sqrt{n}\sqrt{\frac{2}{\pi}}
$$
from which we again conclude that $\sup_n E[|X_n|] = \infty$.

1) How does $\sup_n E[|X_n|] = \infty$ directly imply that $X_n$ is not uniformly integrable and so we do not have convergence in $L^1$?

Finally, we check almost sure convergence. The exercise hints that we should show that
$$
P(\liminf X_n = -\infty) = 1, P(\limsup X_n = \infty) = 1
$$
using Kolmogorov's 0-1 law to show that $X_n$ does not converge a.s.

2) How does one show this using Kolmogorov's 0-1 law?

I know you're not supposed to say this in posts, but... Many thanks for your help.

答:
If $(X_n)$ were uniformly integrable then it would be $L^1$ bounded, in the sense that  $\sup_n\|X_n\|_1<\infty$.

Can you show that $P(\limsup_nX_n =+\infty)>0$? (As a start, $P(\sup_{k\ge n}X_k>C)\ge P(X_n>C)$ for each $n$ and each $C>0$. That last probability tends to $1/2$ as $n\to\infty$.)

Title: Types of convergence of a Martingale


问:
Representation-Dependent Properties of $\mathrm{C}^*$-algebras
Consider a $\mathrm{C}^*$-algebra. By the Gelfand-Naimark Theorem it has a faithful representation $A\subset B(\mathsf{H})$ of bounded operators on a Hilbert space. Let $\pi_1:A\rightarrow B(\mathsf{H}_1)$ and $\pi_2:A\rightarrow B(\mathsf{H}_2)$ be two faithful representations. These representations are called unitarily equivalent if there exists a unitary intertwiner $U:\mathsf{H}_1\rightarrow \mathsf{H}_2$, such that $U\pi_1=\pi_2U$.

Let us quote nLab:

From the physical viewpoint unitarily equivalent representations describe the same system, so that the classification of not unitarily equivalent representations is an important topic.

QUESTION 1

What properties of a $\mathrm{C}^*$-algebras are representation dependent? Do these properties coincide for unitarily equivalent faithful representations?

QUESTION 2

What is a good reference to read about unitary equivalence and such matters?

I will give an example of the kind of thing I am looking to hear about.

In this question we see that two projections can have trivial range intersection in one faithful representation and non-trivial in another. Note that this means that it is possible that if we look at $\pi_1(p),\,\pi_1(q)\in B(\mathsf{H}_1)$, and $\pi_2(p),\,\pi_2(q)\in B(\mathsf{H}_2)$, by von Neumann's theorem of alternating projections, both $\pi_1(pq)^n$ and $\pi_2(pq)^n$ converge strongly, but one to zero and the other to something non-zero. $\mathrm{C}^*$-algebras are not strongly closed, so this might be more about 'full' algebras bounded operators on Hilbert spaces... but the below I think gives something slightly more 'affiliated' with the $\mathrm{C}^*$-algebra

Consider the reduced group $\mathrm{C}^*$-algebra of the infinite dihedral group $D_\infty\cong \mathbb{Z}_2\star \mathbb{Z}_2$, denoted $\mathrm{C}^*_{\text{red}}(D_\infty)$. Faithfully represented on $\ell^2(D_\infty)$, where $u$ and $v$ are the free order two generators, the projections $p=(U_e+U_u)/2$ and $q=(U_e+U_v)/2$... I believe these have empty intersection (there is another faithful representation of this $\mathrm{C}^*$-algebra of Pederson in which I am more sure of this empty-range-intersection but harder to describe), but the counit, as $D_\infty$ is amenable, $\varepsilon(U_g)=1$ is a bounded linear functional, and if the GNS space of the counit is constructed, and included in a direct sum type faithful representation, then the cyclic vector of the counit is in the intersection of the ranges of $p$ and $q$. I do not know if these representations are unitarily equivalent (hence Question 2).

答:
This question is undoubtedly a very broad one so let me instead concentrate on a small contribution to the part related
to the intersection of the ranges of idempotent elements, which seems to be relevant for the OP.

So let us suppose throughout that $A$ is a C*-algebra and that  $p$ and $q$ are  projections in $A$.  Assuming that
there is a
representation $\pi $ such that the ranges of $\pi (p)$ and $\pi (q)$  have a nontrivial intersection then, picking a nonzero
vector $\xi $ in that intersection, we have that
$\pi (pq)\xi =\xi $,
which implies that $\Vert \pi (pq)\Vert =1$, and hence also  $\Vert pq\Vert =1$.

Conversely, if $\|pq\|=1$, then also
$$
  \|pqp\| = \|pq(pq)^*\| = \|pq\|^2 = 1.
  $$
Observing that $pqp$ is a positive element, we deduce that $1\in \sigma (pqp)$, so there exists a character $\varphi $ defined on the
C*-algebra generated by $pqp$, such that $\varphi (pqp)=1$.   Extending $\varphi $ to a state on $A$, and letting $\pi $ be the
associated  GNS representation, one has that
$$
  1 = \varphi (pqp) = \langle \pi (pqp)\xi ,\xi \rangle \leq \|\pi (pqp)\xi \Vert \Vert \xi \Vert  \leq 1.
  $$
From this one easily proves that $\pi (pqp)\xi =\xi $, whence $\xi $ lies in the range of $\pi (p)$.  Consequently
$$
  \|\xi \| = \|\pi (pqp)\xi \|=\|\pi (pq)\xi \|\leq \|\pi (q)\xi \|,
  $$
and we then deduce that $\xi $ lies in the range of $\pi (q)$ as well.

We have then proved:

Theorem.  Given two projections $p$ and $q$ in a C*-algebra $A$, there exists a representation $\pi $ of $A$ such that
the
ranges of $\pi (p)$ and $\pi (q)$  have a nontrivial intersection iff $\Vert pq\Vert =1$.

Speaking about the regular representation of ${\mathbb Z}_2\star{\mathbb Z}_2$, it is correct to say that the projections $p$ and $q$ defined in
the question have trivially intersecting ranges.  To see this, let me denote the images of the two standard generators
simply by $u$ and $v$,  so that
$$
  p = \frac{1+u}2\quad\text{and}\quad q = \frac{1+v}2.
  $$
If $\xi $ is a vector lying in the intersection of the ranges of $p$ and $q$,  then
$$
  \xi =p\xi  = \frac{\xi +u\xi }2,
  $$
from where one deduces that $u\xi =\xi $, and similarly that $v\xi =\xi $.  This would imply that $\xi $ is fixed for the whole of
${\mathbb Z}_2\star{\mathbb Z}_2$.

However  it is easy to see that the regular representation of an infinite group cannot have a fixed point
except for the zero vector, so we conclude that $p$ and $q$ indeed have trivially intersecting ranges.

On the other hand,  as observed by the OP,  the co-unit maps $pq$ to 1,  so $\|pq\|=1$,  and hence the above Theorem
could be used to find another representation (as the one constructed by the OP) in which the ranges have a non-trivial intersection.


问:
How do we take care of the case illustrated in figure 3.8?

Consider figure 3.8, if we pick $k$ to be the lowest point between $f(x_1)$ and $f(x_2)$ as in the illustration, then it is not the case that $f(x_1) - k < 0$ and $f(x_2) - k > 0$. Why does the proof still work in this case?

答:
We are picking $k$ between the $y$-values $f(x_1)$ and $f(x_2)$.  That is not the same as picking an arbitrary $x$ between $x_1$ and $x_2$ and setting $k = f(x)$.  Not every $x\in [x_1,x_2]$ necessarily corresponds to a value of $f(x)$ such that $f(x) \in [f(x_1),f(x_2)]$.

See the image below: the "lowest point" is not a valid choice of $k$, so there is no issue with the proof.


答:
The curve $L$ is the intersection of the two surfaces (see the picture below)
$$\begin{cases}
y^2+z^2=z\\
z=1-x^2\\
\end{cases}
$$
Its parametric equations are

$$L=\left(\cos t,\frac{1}{2} \sin 2 t,\sin ^2t\right);\;t\in[0,2\pi]$$

$$\frac{dL}{dt}=\left(-\sin t,\cos 2t,\sin 2t\right)$$

Length of $L$ is given by
$$L=\int_{0}^{2\pi}\sqrt{\left\lVert \frac{dL}{dt}\right\rVert^2}\,dt$$
$$L=\int_{0}^{2\pi} \sqrt{1+\sin ^2t}\,dt\approx 7.64$$

$\int \sqrt{1+\sin ^2t}\,dt$ is an elliptic integral and the value that can be found is only an approximated value.

Hope this helps

$$...$$

Title: Jordan measure of two intersecting cylinders

Body:
In the following example I need to find Jordan
measure of the following set:$$A=\{(x,y,z)\mid z>0,1-x^2=z, y^2+z^2=z\}.$$ Here we have two cylinders, but I am not sure to find Jordan
measure, is there formula or something?


问:
Proving prefix/suffix exists for finite automata
My task is to show that if a language $L \subseteq \{a, b\}^*$ is recognised by a finite automaton then there exists a finite automaton that recognise all the prefixes and suffixes of the word in $L$.

$$\text{Pref}(L) = \big\{x \in \{a,b\}^* \big| xy \in L\text{ for some }y \in \{a,b\}^*\big\}$$

and

$$\text{Suff}(L) = \big\{y \in \{a,b\}^* \big| xy \in L\text{ for some }x \in \{a,b\}^*\big\}$$

I've come so far to understand that:

A word $u$ is a prefix of a word $v$ if there is a word $w$ such that $v=uw$.
And intuitively I somehow believe that for a word $w$ a prefix/suffix can be all the partitions(?) in order for the word $w$, for example if the word is $abab$ then the possible prefixes/suffixes are

$$\big\{\emptyset, \{a\}, \{ab\}, \{aba\}, \{abab\}\big\}$$

However, I'm having a hard time actually proving this in a mathematical way. What I assume is that for a given finite automata, we'd have to turn all the states on a path to an accepting state into accepting states? Anyone care to help me prove these two?

答:
SKETCH: You have the right idea for half of the problem. Suppose that $M$ is a DFA for $L$. You can modify $M$ to get a DFA $M_P$ for $\operatorname{Pref}(L)$ by letting $S_0$ be the set of all states of $M$ from which an acceptor state of $M$ can be reached, including the acceptor states of $M$ themselves, and making $S_0$ the set of acceptor states of $M_P$. No other changes are needed.

You can modify $M$ to get an NFA $M_S$ for $\operatorname{Suff}(L)$ in either of two ways: if $S_1$ is the set of states of $M$ that are accessible from the initial state of $M$, you can either make all of the states in $S_1$ initial states of $M_S$, or add $\epsilon$-transitions from the initial state of $M$ to each state in $S_1$.

Now use a standard construction to combine $M_P$ and $M_S$ to get an NFA for $\operatorname{Pref}(L)\cup\operatorname{Suff}(L)$.


问:
Left inverse and nullspace
I am doing some independent studying, and came across one note from a professor, which states that if $A$ is an $m \times n$ matrix with zero nullspace, $m \geq n$. I simply have no idea on how to prove it. Hopefully someone can help me out. Thank you. </p>

答: Hint: this follows from the rank-nullity theorem, using the fact that the rank of a matrix with $m$ rows is no greater than $m$.</p>

Title: Left inverse and nullspace


问:
Homomorphisms from $\mathbb{Z}[x] \to R$

Problem: Let $R$ be a ring with identity. Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R.$

I can't seem to find any approach for this. There are no generators of $\mathbb{Z}[x]$. Can we try something else? Any hint is highly appreciated.

Thank you.

答:
Let $\phi:\mathbb{Z}[X]\to R$ be a ring homomorphism. Since $1\in\mathbb{Z}[X]$ must always be mapped to $1\in R$, we have no choice for what $\phi$ does with $\mathbb{Z}\subset\mathbb{Z}[X]$.

This leaves only what $\phi$ is going to do with $X\in\mathbb{Z}[X]$. What could $\phi(X)$ be? The answer is "anything you want in $R$". And once we know what $\phi(X)$ is, by the rule of ring homomorphisms, we know what $\phi$ does to every element of $\mathbb{Z}[X]$, thus determining $\phi$. There are exactly $|R|$ choices for $\phi(X)$, hence the result.


问:
$I(\oplus_AM_\alpha)=\oplus_AIM_\alpha$ and $\oplus_AM_\alpha/I(\oplus_AM_\alpha)\cong\oplus_AM_\alpha/IM_\alpha$
Let $(M_\alpha )_{\alpha \in A}$ be an indexed set of left R-modules and let I be a left ideal of R. Prove that:

$I(\oplus_AM_\alpha)=\oplus_AIM_\alpha$ and $\oplus_AM_\alpha/I(\oplus_AM_\alpha)\cong\oplus_AM_\alpha/IM_\alpha$

If I can prove the first part then second part is the consequence of it. But I couldnt prove the first part. I will be very appreciate for any help.

答:
Hint: each $IM_\alpha$ is an $R$-submodule of $M_\alpha$. Hence we have a map
$$
\jmath = \oplus \iota_\alpha \ \colon \bigoplus_\alpha IM_\alpha\to \bigoplus_{\alpha}M_\alpha,
$$

which is by construction injective. Compute its image.

For this, pick $(x_\alpha)$ in the domain with finitely of them being non-zero, of course. Then $x_\alpha = \sum_{i=1}^{n_\alpha}r_{i,\alpha} y_{i,\alpha}$ with $r_{i,\alpha} \in I, y_{i,\alpha} \in M_\alpha$. The image via $\jmath$ is the same tuple.

 Suppose that the non-zero indices are $\alpha_1, \ldots, \alpha_j$, then $(x_\alpha) = \sum_{1 \leq k\leq j, 1 \leq i \leq n_{\alpha_k}}r_{i,\alpha}y_{i,\alpha_k}$. The latter is a sum of elements of $\oplus M_\alpha$ with coefficients in $I$, hence belongs to $I\oplus_\alpha M_\alpha$.

 Reciprocally, if $x = \sum_{i=1}^n r_i x_i$ with $r_i \in I, x_i \in \oplus_{\alpha} M_\alpha$, then each $x_i$ is a finite sum of elements of each submodule. Adding zero elements if needed, we can assume that there exist $\alpha_1, \ldots, \alpha_k$ such that $x_i = \sum_{j}y_{ij}$ and $y_{ij} \in M_{\alpha_j}$. But then $x$ is a sum of elements $r_i y_{ij} = \jmath(r_i y_{ij})$ that belong to each $I M_{\alpha_j}$.


问:
Union of two sets in $\mathbb{R}^2$
Let
\begin{align}
X &= \{ (x,y) : 1 < x,y < 3\} \\
Y &= \{ (x,y) : 2 < x,y < 4 \} \\
&\textrm{and} \, , \, \, X \, , \, Y \, \subseteq \mathbb{R}^2
\end{align}
I thought that
$$
X \cup Y =\{ (x,y) : 1 < x,y < 4\}
$$
but when I draw both sets I obtain
<img src="https://i.stack.imgur.com/WYbRb.jpg" alt="Union of sets $X$ and $Y$" />

How can I obtain this via rules of set theory and is there a way to represent this new set?

答:
The drawing is correct. No rules needed, simply think about it logically:

$$
X \cup Y = \{(x,y)\in \mathbb{R}^2 \colon \text{EITHER } (x \text{ AND } y \in (1,3)) \text{ OR } (x \text{ AND } y \in (2,4))\}.
$$

Title: Union of two sets in $\mathbb{R}^2$


问:
Check whether the following polynomial is irreducible over $\mathbf Q$
I was trying this problem from my Abstract Algebra book exercise that says:

Show that the polynomial $x^2+\frac 13x-\frac 25$ is irreducible in $\mathbf Q[x]$.

What I tried:  $x^2+\frac 13x-\frac 25 \equiv 15x^2+5x-6=f(x)$,say.

Now I compute ,$f(x+1)=15x^2+35x+14$. Now, $ 7 \mid 35, 7 \mid 14; 7\nmid 15,7^2 \nmid 14$. Hence using Eisenstein's criterion ,$f(x+1)$ is irreducible  over $\mathbf Q$. Hence $f(x)$ is irreducible over $\mathbf Q$.

Am I right?

I want to add another problem similar to the above :

Show that the polynomial $2x^3-x^2+4x-2$ is not irreducible in $\mathbf Z[x]$.

Here,$f(x+1)=2x^3+5x^2+8x+3$ where $f(x)=2x^3-x^2+4x-2$. Here ,I can not apply Eisenstein's criterion for $f(x+1)$. I can not also factorize $f(x)$.Then how Can I prove it?

Can someone point me in the right direction? Thanks in advance for your time.

Body: $\ \begin{eqnarray}{\rm Hint\!:}\quad\  &2x^3-x^2+4x-2 \\ = &\,x^2(2x-1)+ 2(2x-1)\end{eqnarray}$


问:
Calculus: Meaning of the differentiate sign $\frac{d}{dx}$, Why is $\frac{d}{dx}(\sin y)$ applied with chain rule but $\frac{d}{dx}(\sin x) =\cos(x)$?

I'm getting confused with the different signs.

I understand that $\frac{dy}{dx}$ reads "$y$ in respect to the derivative of $x$" & $\frac{d}{dx}$ is to differentiate a certain equation. But I don't think I really understand what it means when another function is mixed in apart of $x$.

A more concrete example of what I don't understand is:

Why is $\frac{d}{dx}(\sin y)$ applied with chain rule but $\frac{d}{dx}(\sin x) =\cos(x)$ applies the derivative-of-sine "rule"

$\frac{d}{dx}(y^2)$ applies the chain rule, but $\frac{d}{dx}(x^2)$ applies the power rule

sub-question: There's so many rules in derivatives of calculus alone, do I have to remember them all, will formula usually be provided? (ie. power rule, chain rule I get. But derivatives of $\sin(x)$, $\cos(x)$, $e^x$, $\ln(x)$, $\tan(x)$, $\cot(x)$, $\sec(x)$, $\csc(x)$, $\arcsin(x)$..?)

If I have to remember them, is there a few main ones to remember then derive them after.

答:
The way I remember it is to separate the actions of "differentiate" and "divide by $dx$".  I use $d()$ as the differential operator, and $\frac{d}{dx}\left(\right)$ combines the differential operator and "divide by $dx$".

So, for instance, $d(x^2) = 2x\,dx$.  However, for $\frac{d}{dx}(x^2)$, you divide by $dx$ afterwards, giving $2x$.

Now, let's look at a similar one.  If we have $d(y^2)$ we have $2y\,dy$.  However, for $\frac{d}{dx}(y^2)$ you divide by $dx$ afterwards, giving $2y\,\frac{dy}{dx}$.

The way I always do it is to differentiate first, then just solve for whichever derivative I'm looking for (i.e., solve for the ratio of differentials that I'm looking for).


问:
What the probabilty of selecting different congruences of numbers?
I was reading through an article the other day regarding the probability of selecting relatively prime integers, which is known to be $\frac{6}{\pi^2}$. How would you compute the probability of randomly selecting 2 integers, call them x and y such that x is congruent to 0 modulo 4, whereas, y is congruent to 1 modulo 4.

答:
I think this is much easier than the probability of relatively prime that you mention, although there's a little ambiguity in the problem description.

Choosing from all integers, the probabilities for $x$ and $y$ are independent and the probability for each remainder modulo 4 is 1/4.  So
$$ P(x \equiv 0 \bmod 4 \; \& \; y \equiv 1 \bmod 4) = P(x \equiv 0 \bmod 4) \cdot P(y \equiv 1 \bmod 4) = \left(\frac{1}{4}\right)^2 = \frac{1}{16}.$$

If you mean, though, the probability of selecting two numbers where one (call it $x$) is congruent to 0 modulo 4 and the other (call it $y$) is congruent to 1 modulo 4, then the probability is doubled to $1/8$ since the two numbers could be chosen in either order.  (Compare the probability of rolling a 3 with two 6-sided dice.)


问:
Is $\{(x,y) \in \Bbb R^2: x \cdot y=1\}$ a submanifold?
Meanwhile it is well know that
$$M:=\{(x,y) \in \Bbb R^2: x \cdot y=0\}$$
not a submanifold of $\Bbb R^2$ is. The next question is now, is for example
$$L:=\{(x,y) \in \Bbb R^2: x \cdot y=1\}$$
Is now $L$ a submanifold of $\Bbb R^2$?

答: You can use Submersion theorem to show this. If you don't want to use this theorem, then you must explicitly construct a smooth atlas of this submanifold which should be easy.


问:
Proof that there is a shortest path between two specific nodes in a weighted, connected digraph which has no negative cycles
Let $G=(V,E,W)$ is a directed graph, where $V$ and $E$ are the sets of nodes and edges, respectively, while $W: E \rightarrow \mathbb{R}$ is an arbitrary weight function. How to prove that there exists a shortest path from an $s \in V$ to a $t \in V$ if and only if $G$ does not contain negative-cost cycles.
The shortest path from $s$ to $t$ is the path $e_1,e_2,...,e_n$ for which $\sum W(e_i)$ is minimal.

I am sure that this is a very elementary proof, but I did not found it directly.

答:
Some thoughts that may be too long for a comment: assuming the length of a path $e_1 \ldots e_n$ to be $W(e_1) + \cdots + W(e_n)$, fix $s,t \in V$ and consider $\mathcal P$ the set of (finite) and $\mathcal P_{s,t}$ the subset of paths from $s$ to $t$. We have a length function

$$
\ell \colon \mathcal P \to \Bbb R, \quad \ell(e_1\ldots e_n) = W(e_1) + \cdots + W(e_n)
$$

which is compatible with path concatenation, that is $\ell(\omega\omega') = \ell(\omega) + \ell(\omega')$.

A shortest path, thus, would be a path $\omega \in \mathcal P_{s,t}$ that minimizes $\ell$ when restricted to $\mathcal P_{s,t}$.

If we have a cycle $c$ connected to both $s$ and $t$ via paths $\omega_s,\omega_t$, then $\omega_s c^k \omega_t$ is also a path from $s$ to $t$ that goes trough $\omega_s$ then loops around $c$ for $k$ times, and then goes through $\omega_t$. By direct computation,

$$
\ell(\omega_s c^k \omega_t) = \ell(\omega_s) +k\ell(c)+\ell(\omega_s).
$$

Hence if the cycle $c$ is negative, by the previous formula the function $\ell$ takes arbitrarily large negative values in $\mathcal P_{s,t}$ and thus can't attain a minimum value.

Now, your task is to show the converse. Also, note that to be a problem we needed the aforementioned cycle to be connected to $s$ and $t$. If we have negative cycles in, say, a disjoint connected component from the one of $s$ and $t$, it has no impact on the cost of paths $s \to t$.

Edit: okay, some more thoughts for the converse. What I've come up with may be a bit cumbersome, I hope someone posts a clever proof soon. But anyway, here's my attempt:

Fix $\omega$ a path for $s$ to $t$. If $\omega = \omega' c \omega ''$ with $c$ a positive (or rather non-negative) cycle, then $\ell(\omega) \geq \ell(\omega' \omega'')$. Thus, to check wether $\ell$ attains a minimum value, we can restrict ourselves to paths without non-negative cycles.

But then, by hypothesis, the paths considered have no cycles. If your graph is finite, there are a finite amount of vertices and so without making a cycle there are finite options for a path from $s$ to $t$, i.e. the subset $S \subset \mathcal P_{s,t}$ of acyclic paths is finite. Hence $\ell$ attains a minimum in $S$, and by the previous observation this is a minimum in $\mathcal P_{s,t}$.


问:
Check proof that two functions are linearly independent.
I am not very experienced with math and need help with a proof. I need to show that the pair of functions $t$ and $1/t$ defined for $t>0$ are linearly independent. Is the following a sufficient proof? If so, is there a simpler way?

Assume $at+b\frac{1}{t}=0$ is correct for any $t>0$. Substituting $t=1$ yields $a+b=0$ (Eq. 1). Substituting $t=2$ yields $2a+\frac{b}{2}=0$ (Eq. 2). Subtracting Eq. 1 from Eq. 2 yields $a-\frac{b}{2}=0$, so $a=\frac{b}{2}$. Substituting $a=\frac{b}{2}$ into Eq. 1 yields $\frac{3}{2}b=0$, therefore $b=0$, which implies $a=0$.

Since the only scalars $a, b$ such that $at+b\frac{1}{t}=0$ are $a=b=0$ then $t$ and $1/t$ are linearly independent.

答:
Your proof is fine and straightforward.

Maybe little shorter proof would be to see that $\lim_{t\rightarrow\infty} at+b\frac{1}t$ can be zero if and only if $a=0$. It follows immediately that $b=0$ too.


问:
Prove that $\sup\limits_{\alpha \in D} \|x_{\alpha}\| \lt \infty.$

Let $X$ be a complex normed linear space. Let $D$ be a directed set and $\{x_{\alpha} \}_{\alpha \in D}$ be a net in $X$ converging to $x \in X$ with respect to the weak topology i.e. for all $\varphi \in X^*$ we have $\varphi (x_{\alpha}) \to \varphi (x).$ If $\sup\limits_{\alpha \in D} \left \lvert \varphi (x_{\alpha} ) \right \rvert \lt \infty,$ for any $\varphi \in X^*$ then prove that $\sup\limits_{\alpha \in D} \|x_{\alpha} \| \lt \infty.$

Attempt $:$ Consider the collection $\{J_{\alpha} \}_{\alpha \in D}$ in $X^{**}$ defined by $J_{\alpha} (\varphi) = \varphi (x_{\alpha}),$ $\varphi \in X^*.$ Then it is easy to see that $\|J_{\alpha}\| \leq \|x_{\alpha}\|.$ Also By Hahn-Banach theorem there exists $\varphi_{\alpha} \in X^*$ for each $\alpha \in D$ with $\|\varphi_{\alpha}\| = 1$ such that $\varphi_{\alpha} (x_{\alpha}) = \|x_{\alpha}\|.$ This shows that $\|J_{\alpha}\| \geq |J_{\alpha} (\varphi_{\alpha})| = \|x_{\alpha}\|,$ proving that $\|J_{\alpha} \| = \|x_{\alpha} \|.$ Now by the given hypothesis it follows that for any $\varphi \in X^*$ we have $\sup\limits_{\alpha \in D} J_{\alpha} (\varphi) \lt \infty.$ Hence by uniform boundedness principle it follows that $$\sup\limits_{\alpha \in D} \|x_{\alpha}\| = \sup\limits_{\alpha \in D} \|J_{\alpha}\| \lt \infty$$ as required.

In the above proof I didn't use anywhere the weak convergence of the net $\{x_{\alpha} \}_{\alpha \in D}$ and also didn't see any special role of the directed set $D$ as well. Are they of no importance here? Any suggestion regarding this will be highly appreciated.

Thanks in advance.

答:
Without loss of generality, we may assume that $X$ is a Banach space.
For, let $\bar{X}$ be the completion of $X$. Recall that every bounded
linear function $\varphi\in X^{\ast}$ has a unique continuous extension
to $\bar{\varphi}\in\bar{X}^{\ast}$. On the other hand, given $\varphi\in\bar{X}^{\ast}$,
we have that $\overline{\varphi\vert_{X}}=\varphi$, i.e., the continuous
extension of the restriction of $\varphi$ on $X$ is itself. Hence,
there is a one-one correspondance between $X^{\ast}$ and $\bar{X}^{\ast}$.
Now, it is clear that for each $\varphi\in\bar{X}^{\ast}$, we have
that $\sup_{\alpha\in D}|\varphi(x_{\alpha})|<\infty.$

Assume that $X$ is complete. Consider the canonical embedding $X\rightarrow X^{\ast\ast},$
$x\mapsto\hat{x}$, where $\hat{x}(\varphi)=\varphi(x)$, $\varphi\in X^{\ast}$.
Recall that the canonical embedding is an isometric homomorphism,
and hence $||x||=||\hat{x}||$. The given condition implies that:
$\{\hat{x}_{\alpha}\mid\alpha\in D\}$ is a family of bounded linear
functionals and for each $\varphi\in X^{\ast}$, $\sup_{\alpha\in D}|\hat{x}_{\alpha}(\varphi)|<\infty$.
By Uniform Boundedness Principle, we have $\sup_{\alpha\in D}||\hat{x}_{\alpha}||<\infty$.
Note that weak convergence is irrelevant.


问:
Need help with algebra portion of calculus finding slope of secant line
The example problem is: Given f(x) = $x^2$,find and simplify the slope of the secant line for a = 1 and h = any non-zero number.

The answer is as follows:

For a = 1 and h any non-zero number, the secant line goes through  (1, f(1)) = (1,1) and (1+h,f$(1+h))$ = $(1+h,(1+h)^2)$ and its slope is

(1)$$\frac {f((1+h) - f(1)} {h} =$$

(2) $$\frac{(1+h)^2 - 1^2} {h} =$$

(3) $$\frac{1 + 2h + h^2 -1} {h} = $$

(4) $$\frac {h(2+h)} {h} = $$

(5) $$2 + h$$  where h != 0.

My question here is the algebra.  I do not understand how they got from step (2) to step (3) to step (4).  In the book step (2) is annotated as "square the binomial".  Step (3) is annotated as "combine like terms and and factor the numerator".  (Step (4) is annotated as "cancel" - that much I understand.

So can anyone walk me through steps 3 to 5?  Thanks!

答:
From steps 2 to 3 they used:
$$(a+b)^2=a^2+2ab+b^2$$

From steps 3 to 4 they used $$1+2h+h^2-1=2h+h^2$$
then $$2h+h^2=h(2+h)$$


答:
Well they are actually not linear independent. Take $t=1$ and $a=1;b=-1$

and You have $1+(-1*1/1$)

@billiam Thanks for your answer, if it is only positive, then it is quite trivial, here is my previous answer. Well, I don't know whether it may help, but since $t>0;1/t>0$ therefore $t+1/t$ can only be zero if the coefficients are zero.

But i could prove that $1/t$ is at least poaitive. first lets show that if $0<x |*x \Leftrightarrow0<x^2$ so the sqaure is always positive. Now $0<t|*(1/t)^2\Leftrightarrow0* (1/t)^2<t*(1/t)^2 \Leftrightarrow0<1/t$ hope that helps

For any positive scalars the functions as it takes only positive values is positive again.

Title: Check proof that two functions are linearly independent.

Body:
I am not very experienced with math and need help with a proof. I need to show that the pair of functions $t$ and $1/t$ defined for $t>0$ are linearly independent. Is the following a sufficient proof? If so, is there a simpler way?

Assume $at+b\frac{1}{t}=0$ is correct for any $t>0$. Substituting $t=1$ yields $a+b=0$ (Eq. 1). Substituting $t=2$ yields $2a+\frac{b}{2}=0$ (Eq. 2). Subtracting Eq. 1 from Eq. 2 yields $a-\frac{b}{2}=0$, so $a=\frac{b}{2}$. Substituting $a=\frac{b}{2}$ into Eq. 1 yields $\frac{3}{2}b=0$, therefore $b=0$, which implies $a=0$.

Since the only scalars $a, b$ such that $at+b\frac{1}{t}=0$ are $a=b=0$ then $t$ and $1/t$ are linearly independent.


问:
How to set up Zeno's paradox as a summation
Let's suppose that a projectile travels 1 m/s and I want to find out how long it takes to travel 2m -- it should be two seconds, of course. But how could this be set up with a summation? For example, I would think that it would be something along the lines of:

$$
\sum_{t=0}^{?}
delta_t *rate = 2m
$$

Or, how could this be shown with a summation? I'm not very good at latex but perhaps something along the lines of (just halving the time each time until it gets to 2...but this of course assumes the answer is already 2...):

$$
\sum_{i=1}^{∞}
time * rate =
\sum_{i=1}^{∞}
1/2^{n-1} * rate = 2m
$$

I know this isn't a programming site, but here is the C code I used to emulate the last summation:

```<code>int n=1, max_n;
double rate=1.0, distance=2.0;

printf("Enter the number of iterations you want: \n");
if (scanf("%d", &max_n)  != 1) printf("Error! Items parsed != 1\n");

for (double t=0.0, distance=0.0, dt; n <=max_n ; n++) {
    dt =  1.0 / pow(2, n-1);
    t += dt;
    distance += dt * rate;
    printf("%2d. Time: %8.4f | Distance: %.4f \n", n, t, distance);
}
</code>```

Enter the number of iterations you want: 12

1. Time:   1.0000 | Distance: 1.0000

2. Time:   1.5000 | Distance: 1.5000

3. Time:   1.7500 | Distance: 1.7500

...

12. Time:   1.9995 | Distance: 1.9995

答:
Per OP's request:

First see my comment.  I derived the expression in my comment simply by noting that for $|x| < 1, ~\sum_{t=0}^{\infty} x^t = \frac{1}{1 - x}.$

Since I was looking for a value of $x$ so that $\frac{1}{1-x} = 2$, I set $x = \frac{1}{2},$ and then noted that $\left(\frac{1}{2}\right)^t = 2^{(-t)}.$


问:
Is this algebraic expression a polynomial?
I'm having a hard time finding the rules to identify monomials I know a monomial cannot have variables below the square root sign, I try to find in Google a monomial rule where it cannot have a variable as an index in the square root but not found any, I'm testing my self in Khan Academy about identifying polynomials but the example below I get it wrong, could you verify why is this not a polynomial?

$$\sqrt[k]{23} + k$$

答:
A polynomial is an expression of the form

$$
a_0  + a_1 T + \ldots + a_n T^n \tag{1}
$$

where $n \geq 0$ is some natural number and each $a_i$ are constants in your favourite "set of numbers", say, each $a_i$ can be a real number.

In your example, you have something of the form

$$
a_0^T + a_1T
$$

where $a_1 = 1, a_0 = \sqrt{23}$ which is not of the aforementioned form.

Moreover, even if polynomials can be thought of as functions $\Bbb R \to \Bbb R$ in some cases, one should not in general treat $T$ as a number; $T$ is a "placeholder", an indeterminate. There is (a priori) no meaning for doing $a_0$ to the power of $T$.


问:
Lattices: What does it mean that every pair of elements has an infimum and a supremum?
what does it mean that every relation of two (or more?*) elements has an infimum or maximum?

Could you please give me some examples?

Thanks
von Spotz

* while, so far as i know, the operations of both the order-relation interpretation as well as the algebraic-structure interpretation of a lattice are binary only.

答:
Let $P$ be a poset ordered by $\leq$. For $a, b \in P$ we say that $c$ is a supremum of $a$ and $b$ if $a \leq c$ and $b \leq c$ and furthermore for every $d \in P$ such that $a \leq d$ and $b \leq d$ we have $c \leq d$. In words this means that a supremum of two elements is greater than both elements, so it is an upper bound, and it the least upper bound.

The notion of infimum is dual: $c$ is the infimum of $a$ and $b$ if $c \leq a$ and $c \leq b$ while for every $d \in P$ with $d \leq a$ and $d \leq b$ we have $d \leq c$.

Exercise. The supremum of $a$ and $b$, if it exists, is unique. That is, if $c$ and $c'$ are both the supremum of $a$ and $b$ then $c = c'$. Likewise the infimum is unique.

We call a poset $P$ a lattice if any two elements have both a supremum and infimum. For the supremum of $a$ and $b$ we write $a \vee b$ (by the exercise this makes sense because there is a unique element with this role). For the infimum of $a$ and $b$ we write $a \wedge b$.

We can generalise things to finite subsets. For example: for a finite subsets $\{a_1, \ldots, a_n\} \subseteq P$ we say that $c$ is a supremum of $\{a_1, \ldots, a_n\}$ if $a_i \leq c$ for every $1 \leq i \leq n$ and for any $d$ with $a_i \leq c$ for all $1 \leq i \leq n$ we have $c \leq d$. Similarly for infimum.

Exercise. A poset $P$ is a lattice if and only if every nonempty finite subset has a supremum and an infimum. The right to left direction should be trivial. For the left to right direction you can use the hint that given $\{a_1, \ldots, a_n\}$ we can consider $(((a_1 \vee a_2) \vee a_3) \vee \ldots) \vee a_n$ (to be precise: use induction).


问:
Continuity and the existence of the limits
This question is due to a clarification asked by an engineer friend. Supposing that I have this function

$$f(x)=\sqrt{1-x^2} \tag 1$$

that it is simply a semicircle in the first and second quadrant with the domain $[-1;1]$ and $f([-1;1])=\operatorname{im} f=[0;1]$. One of his students of an high school thinks that at point $-1$ and $1$ the function is not continuous, because

$$\bbox[pink,5px,border:2px solid lightgray]{\lim_{x\to (-1)^-}f(x)=\nexists, \quad \lim_{x\to (-1)^+}f(x)=0}$$

($ (-1)^-$ it is out of the domain). I think instead that it is true that $ (-1)^-$ it is out from the domain but are true the two conditions:

\begin{cases}
-1 \in \mathrm{dom}\, f(x), \\
\lim_{x\to (-1)^-}f(x)=\lim_{x\to (-1)^+}f(x)=f(-1)=0
\end{cases}
The same procedure for $1$.

Is it correct or are there specific other observations?

答:
Function $f\left(x\right)=\sqrt{1-x^2}$ is continuous as the composition of continuous functions.

Remember that function $f$ is continuous at the point $x_0\in\mathrm{dom}(f)$ if the limit of $f\left(x\right)$, as $x$ approaches $x_0$ through the domain of $f$, exists and is equal to $f\left(x_0\right)$. As you can see, definition (or characterization) says that variable $x$ must approach through the domain.


问:
In $\mathbb{F}_p[X]$ with $p$ prime and $p = 1 \text{ mod } 3$, prove that there exists an element of order 3.
Let $p$ be prime, and $p = 1 \text{ mod } 3$. Consider $\mathbb{F}_p[X]$.

Prove that in $\mathbb{F}_p$ there exists an element of order 3. Use that to conclude that $X^2 + X + 1$ has a root in $F_p[X]$.

I am preparing for an exam and this is one of the example questions. I've thought about it for a bit, but to be honest I don't really know where to start. Can somebody give me a hint, so that I know where to start looking?

答:
Hint: if $k$ is a field and $G \leq k^\times$ a finite subgroup, then $G$ is cyclic.

More hints below (don't read them without thinking about this for a while):

 by Cauchy's theorem, you only need $3$ to divide the order of a group $G$ to get that it has an order $3$ element.

 We know that $\Bbb F_p^\times$ has $p-1$ elements and by hypotehsis $3 \mod p-1$, so there exists some $\omega$ such that $\omega^3 = 1$. Plug $\omega$ in $x^2+x+1$, play around with this.

 We have $(x-1)(x^2+x+1) = x^3-1$. Evaluate in $\omega$, use that $\omega \neq 1$ to conclude that it is a root of $x^2+x+1$.


答: Hint: $\mathbb{F}_p^\times$ is cyclic.

Title: In $\mathbb{F}_p[X]$ with $p$ prime and $p = 1 \text{ mod } 3$, prove that there exists an element of order 3.

Body:
Let $p$ be prime, and $p = 1 \text{ mod } 3$. Consider $\mathbb{F}_p[X]$.

Prove that in $\mathbb{F}_p$ there exists an element of order 3. Use that to conclude that $X^2 + X + 1$ has a root in $F_p[X]$.

I am preparing for an exam and this is one of the example questions. I've thought about it for a bit, but to be honest I don't really know where to start. Can somebody give me a hint, so that I know where to start looking?

Body: Hint: $\mathbb{F}_p^\times$ is cyclic.


问:
Need help finding an eigenvector
I'm trying to compute the eigenvector of the following matrix:

$$
   A  =   \begin{bmatrix}
        0.3889 & 0.3456  \\
        0.3456 & 0.4044  \\
        \end{bmatrix}
$$

where one of the eigenvalues is $\lambda=0.7423$. This example was taken from my course and I only have the solution:

$$\vec{v}=\begin{bmatrix}
        0.6991\\
        0.7150\\
        \end{bmatrix}$$

The solution is correct since $A\vec{v}=\lambda\vec{v}$, but I don't understand how to find it. The system of equations I'd have to solve would be:

$$-0.3534v_{1}+0.3456v_{2}=0\\0.3456v_{1}-0.3379v_{2}=0$$

The solution is clearly $0$ for both components. What am I missing here?

答:
I think you're  missing that there are solutions other than $v_1=v_2=0$

If $-.3534v_1+.3456v_2=0$, then $.3456v_2=.3534v_1$, or $v_2=1.022569v_1$. So $\left(\begin{matrix}1\\1.022569\end{matrix}\right)$ is a solution, as well as $\left(\begin{matrix}1*.6991\\1.022569*.6991\end{matrix}\right)=\left(\begin{matrix}.6991\\.7149\end{matrix}\right)$

Title: Need help finding an eigenvector


问:
How Many 4-Digit Positive Even Numbers Use Only Digits 0 Through 4, With No Digits Repeating?
I am currently working on this problem:

How many 4-digit multiples of 2 use only digits 0 through 4, with no digits repeating?

I understand/know the basic idea of how to solve this kind of problem:

You multiply the number of possible numbers that could be in each place, for example, for How many 4 digit numbers use only digits 0 through 3, with no digits repeating?, the answer would be 3 * 3 * 2 * 1, or 18.

In my problem, the last digit of any even number is either 0, 2, or 4. That gives us this current multiplication equation:

_ * _ * _ * 3

The reason I am stuck on this problem is because, in the first space, there could either be 4 possibilities(1, 2, 3, 4) or 3 possibilities(1, 3, 4 OR 1, 2, 3),  depending on if 2 or 4 is used in the ones column.
And depending on the 1st column, the following numbers of possibilities will be one minus the last number.

Now, because there can be two possible number of possibilities in the left-most/4th space, there can be two multiplication problems, which makes this problem confusing for me:

4 * 3 * 2 * 3

OR

3 * 2 * 1 * 3

So my question is: How would you solve this problem?

答:
The answer can be written as

$$1\cdot4\cdot3\cdot2+2\cdot3\cdot3\cdot2$$

The idea is to split the count into two cases, according to whether the ones digit (which makes the number even) is $0$ or not.  There's only one way to choose $0$, after which the remaining digits can be chosen in $4\cdot3\cdot2$ ways. There are two ways to choose nonzero ones digit (either $2$ or $4$), but that means there are only three choices for the lead digit, since you have to avoid the $0$. But after that, the two interior digits can be chosen in $3\cdot2$ ways.

Added later: The heart of this solution is the calculation of $1\cdot4+2\cdot3$, which amounts to counting the number of two-digit even numbers out of the digits $0$ to $4$; the remaining $3\cdot2$ is routine. I was bemused by the observation that $1\cdot4+2\cdot3=10={5\choose2}$, where the latter counts the number of (unordered) choices of $2$ of the $5$ digits, and wondered if this could be attributed to anything more than numerical coincidence (aka, Richard Guy's Strong Law of Small Numbers). It turns out it can:

Note that each odd digit can be paired with a distinct nonzero even digit, e.g., $1$ with $2$ and $3$ with $4$. Now pick any pair of digits, which can be done in $5\choose2$ ways. If one of the chosen digits is $0$, put it in the ones digit and use the other as the lead digit.  If both are even but neither is $0$, put the smaller in the ones digit and use the larger as the lead digit. If both are odd, replace each one with the even digit it's been paired with and then put the larger of these in the ones digit and use the smaller as the lead digit.

This works as long as each odd digit can be paired with a nonzero even digit, i.e., if the set of available digits consists of $0$, $k$ odd digits, and $k$ nonzero even digits, giving a count of $2k+1\choose2$ two-digit even numbers. (It also works without the $0$, the count being simply $2k\choose2$ in that case.)


答:
For instance, $A+(B \cap C) = (A+B) \cap (A+C)$ only when $B$ or $C$ contains $A$, in which case the identity can be written as $$A+(B \cap C) = \begin{cases} B \cap (A+C) & \textrm{if } A \subseteq B, \\ (A+B) \cap C & \textrm{if } A \subseteq C. \end{cases}$$
We say that the lattice of submodules of a module over a ring forms a modular lattice. Can you take it from here? Also, what role plays $L$ and $L'$ in this problem?

Body:
Let $M=K\oplus K'=L\oplus L'$. Prove $K\subset H\leq M$ implies $H=K\oplus (H\cap K')$

My attempt:

If $K\subset H\leq M$ then $H=K+ (H\cap K')=(K+H)\cap (K+K')$

Then$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=H\cap(K+K')$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=H\cap M=H$

But I think something is wrong.


问:
Unique solution of $\dot{x} = f(x), x(0) = x_0$
I teach myself to solve IVP and could use some help regarding this exercise I found:

Let $f:[0,+\infty) \to [0,+\infty)$ be continuous such that $0$ is the only zero of $f$. Consider the differential equation $\dot{x} = f(x)$ with initial value $x(0) = x_0 \in [0, +\infty)$.

I am asked to show that $\int_{0}^{1} \frac{1}{f(x)}\ \mathrm{d}x = +\infty$ implies that the IVP has a unique solution for $x_0 = 0$ . It's obvious that $x(t) = 0$ is a solution for $x_0 = 0$ but I have no idea how to proof the statement.

Are there any important theorems that might help here?

答:
Assume there is a solution with $x(t_1)=1$. Let $t_0=\sup\{t>0:x(t)=0\}$. Then for the time difference one gets
$$
\frac{dx}{dt}=f(x)\implies t_1-t_0=\lim_{\varepsilon\to0}\int_ε^1\frac{dx}{f(x)}.
$$
This of course only works if the limit of the integral has a finite value. If the integral diverges, then no such solution exists, the zero function is the unique solution.

Remarks: To properly apply existence theorems, one could extend $f$ as an odd function $f(-x)=-f(x)$, so that $f$ is defined and continuous on all of $\Bbb R$, and $(t_0,x_0)=(0,0)$ is an inner point of the domain.

For any $ε>0$, $f$ has a positive minimum on the interval $[ε,1]$, so that any solution that deviates from the zero solution will be strictly increasing after leaving the zero axis.

As a consequence, if a solution with $x(t_1)=1$ exists, then for every $ε>0$ there is a unique $δ>0$ so that $x(t_0+δ)=ε$. Now by the substitution rule one can integrate
$$
t_1-(t_0+δ)=\int_{t_0+δ}^{t_1}dt=\int_{t_0+δ}^{t_1}\frac{\dot x(t)}{f(x(t))}dt=\int_{ε}^1\frac{dx}{f(x)}.
$$
In view of the last two points, the emphasis for $t_0$ should perhaps be more that it is the start of the positive values, $t_0=\inf\{t>0:0<x(t)\}$.

If $f$ is Lipschitz on $[0,1]$, then any solution is unique. On the other side of the claim, $|f(x)|\le L|x|$ implies
$$
\int_{ε}^1\frac{dx}{f(x)}\ge\frac{-\ln(ε)}{L},
$$
which is unbounded for $ε\to0$.


问:
Prove that vectorspace is the result of a direct sum
So I'm given the vectorspace $(\mathbb{R}, \mathbb{R}^{n \times n}, +)$ and the subspace $U$ which contains all symmetric matrices and W which contains all skew matrices. I'm asked to prove that the given vectorspace $\mathbb{R}^{n \times n} = U \bigoplus W$. How would I even start on this? I suppose I'll try to prove that every matrix is the sum of some symmetric and skew matrix?

答:
Yes.

That is, you want to write arbitrary matrix $A$ as $A=B+C$ where $B=B^T$ and $C=-C^T$. What can you say about $A^T$= Then solve for $B$ and $C$.


答:
You can choose one's place in $3$ ways $(0, 2, 4)$.

Once you do that, you now have $4$ digits left for the second place (ten's), $3$ for the third (hundred) and $2$ for the $4$th (thousand).

That gives you $3 \times 4 \times 3 \times 2 = 72 \ $ numbers.

But this also counts numbers that have $0$ in thousand place. We need to remove them as they are not $4$ digit numbers really.

So if I fix $0$ in thousand place, I have $2$ choices for one's place ($2, 4$), $3$ remaining for the ten's and $2$ remaining for the hundred place, which is $12$ numbers in total (= $2 \times 3 \times 2$).

So the count of numbers you are interested in $ = 72 - 12 = 60$.

Title: How Many 4-Digit Positive Even Numbers Use Only Digits 0 Through 4, With No Digits Repeating?

Body:
I am currently working on this problem:

How many 4-digit multiples of 2 use only digits 0 through 4, with no digits repeating?

I understand/know the basic idea of how to solve this kind of problem:

You multiply the number of possible numbers that could be in each place, for example, for How many 4 digit numbers use only digits 0 through 3, with no digits repeating?, the answer would be 3 * 3 * 2 * 1, or 18.

In my problem, the last digit of any even number is either 0, 2, or 4. That gives us this current multiplication equation:

_ * _ * _ * 3

The reason I am stuck on this problem is because, in the first space, there could either be 4 possibilities(1, 2, 3, 4) or 3 possibilities(1, 3, 4 OR 1, 2, 3),  depending on if 2 or 4 is used in the ones column.
And depending on the 1st column, the following numbers of possibilities will be one minus the last number.

Now, because there can be two possible number of possibilities in the left-most/4th space, there can be two multiplication problems, which makes this problem confusing for me:

4 * 3 * 2 * 3

OR

3 * 2 * 1 * 3

So my question is: How would you solve this problem?


问:
Can a function be big Oh and big theta of another function?
I'm trying to understand if a function $f(n)$ can be both $O(g(n))$ and $\Theta(g(n))$.

答: For any function $g$ holds $g\in O(g), g\in \Theta(g),g\in \Omega(g)$.


问:
Finding the Area of a Piecewise Function
I'm learning applications of integration and this is my first experience with this sort of question. How would I get started on a problem like this?

答: Try to calculate the integral $\lim_{a\rightarrow1}\int_{0}^{a}2-x^2  dx+\int_{1}^{2} (x-2)^2dx$


答:
There are two initial conditions to Zeno's paradox:  The relative speed of the tortoise and the amount of head-start the tortoise is given.  Say the tortoise is given a 50cm head start and Achilles moves twice as fast.  Then Achille's movement becomes:
$$\frac{1}{2}+\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2^2}...$$
So set your starting conditions Headstart=H and speed differential=$\frac{1}{s}$.
Your summation is then:
$$\sum_{n=0}^{\infty} H\frac{1}{s^n}$$
For your example, pick the initial distance of the projectile - say H=1, and the relative size you want of each term - say $\frac{1}{2}$.  That will give you 2m.

More generally, we know
$$\sum_{n=0}^{\infty} \frac{1}{x^n} = \frac {1}{1-\frac{1}{x}} = \frac{x}{x-1}$$  So if you want an answer of 2m, you need $H.\frac{s}{s-1}=2$ or $H=2\frac {s-1}{s}$.  Interestingly, note that this answer is completely independent of the speed of the projectile, since the important criterion is that speed relative to another object. Are you comparing an ICBM to a bullet or a bullet to an arrow?

Title: How to set up Zeno's paradox as a summation

Body:
Let's suppose that a projectile travels 1 m/s and I want to find out how long it takes to travel 2m -- it should be two seconds, of course. But how could this be set up with a summation? For example, I would think that it would be something along the lines of:

$$
\sum_{t=0}^{?}
delta_t *rate = 2m
$$

Or, how could this be shown with a summation? I'm not very good at latex but perhaps something along the lines of (just halving the time each time until it gets to 2...but this of course assumes the answer is already 2...):

$$
\sum_{i=1}^{∞}
time * rate =
\sum_{i=1}^{∞}
1/2^{n-1} * rate = 2m
$$

I know this isn't a programming site, but here is the C code I used to emulate the last summation:

```<code>int n=1, max_n;
double rate=1.0, distance=2.0;

printf("Enter the number of iterations you want: \n");
if (scanf("%d", &max_n)  != 1) printf("Error! Items parsed != 1\n");

for (double t=0.0, distance=0.0, dt; n <=max_n ; n++) {
    dt =  1.0 / pow(2, n-1);
    t += dt;
    distance += dt * rate;
    printf("%2d. Time: %8.4f | Distance: %.4f \n", n, t, distance);
}
</code>```

Enter the number of iterations you want: 12

1. Time:   1.0000 | Distance: 1.0000

2. Time:   1.5000 | Distance: 1.5000

3. Time:   1.7500 | Distance: 1.7500

...

12. Time:   1.9995 | Distance: 1.9995


问:
Taylor's theorem and evaluating limits when x goes to infinity
In this problem I have to evaluate the following limit
$$\lim_{x \to \infty} \left(\left(x^3-x^2+\frac{x}{2}\right)e^{1/x} - \sqrt{x^6+1}\right)$$
using the Taylor's formula with the Peano form of the remainder.

In class I have seen examples of how to handle limits using Taylor series expansion when $x$ goes to $0$, but I am not sure how to solve something like this. I tried substituting $t=1/x$, which led to
$$\lim_{t \to 0^+}\frac{(2-2t+t^2)e^t-2\sqrt{1+t^6}}{2t^3}.$$
Now it looks a little bit better but I got stuck at this point. Is this approach correct? If so, how do I proceed?

Body:
$$e^t \left(t^2-2 t+2\right)=2+\frac{t^3}{3}+O\left(t^4\right)$$
$$2\sqrt{t^6+1}=2+O\left(t^4\right)$$
thus
$$\frac{e^t \left(t^2-2 t+2\right)-2 \sqrt{t^6+1}}{2 t^3}\sim \frac{2+\frac{t^3}{3}-2}{2 t^3},\text{ as }x\to 0^+$$
Therefore
$$\underset{t\to 0^+}{\text{lim}}\frac{e^t \left(t^2-2 t+2\right)-2 \sqrt{t^6+1}}{2 t^3}=\frac{1}{6}$$


问:
For any bounded set $A$ in a metric space $(X,d)$, $A$ is inside $B(x,\epsilon)$ for any $x\in X$ and for some $\epsilon>0$
The idea came to my mind from $\mathbb R^n$, if a set $A$ is bounded there, then you can find a ball centered at $\vec 0$ which consists $A$.

Since $X$ is arbitrary metric space. So there is no unique element as $0$ (by saying unique element, it is unique and special in our hearts). Ergo, I should be able to do it for arbitrary elements in $X$.

Attempt:

Fix any $x_0\in X$ and let $A$ be bounded in $X$. So there is a ball centered at, say $a_0\in A$, for some $\epsilon_0$

$$A\subset B(a_0,\epsilon_0)$$

Now define $m=\sup\limits_{x\in B(a_0,\epsilon_0)} d(x_0,x)$

So $$A\subset B(a_0,\epsilon_0)\subset B(x_0,m)$$

Is there any amusing mistake? If it is wrong can you give me hint for counter-example? Or directly a counter-example?

答:
What you did is correct, but a bit too complicated. If $A \subset X$ is bounded then $A \subset B(a_0, r)$ for some $a_0 \in X$ and some $r > 0$.

If $x \in X$ then for all $a \in A$, using the triangle inequality for the metric $d$,
$$
 d(x, a) \le d(x,a_0) + d(a_0, a) < d(x,a_0) + r
$$
so that $A \subset B(x, R)$ with $R = d(x,a_0)+ r$.


问:
How to identify the independent and dependent variables in a differential equation?
I had doubt regarding the equation $y^4 = x^5$. Which one is the dependant and the independent variable here? Thanks in advance.

答:
You can considerate the indipendent variable the classic $x$ and dependent the $y$. I remember that

$$y^4=x^5 \iff y=\sqrt[4]{x^5}=x\sqrt[4]{x} \quad  \wedge \quad y=-\sqrt[4]{x^5}=-x\sqrt[4]{x}$$

Plotting $y=\sqrt[4]{x^5}$ you have

and plotting $y=-\sqrt[4]{x^5}$ you will have:

Hence $$y^4=x^5$$ is:


问:
How to prove that $A\cap B\subseteq C$ and $A^c\cap B\subseteq C$ imply that $B\subseteq C$?
How do you solve this problem??

Suppose that $A\cap B\subseteq C$, and $A^c\cap B\subseteq C$. Prove that $B$ is a subset of $C$.

I don't know where even to begin

Can anyone help?

Thank you

答:
You could also think of it another way: What if $B \nsubseteq C$? Then you have several possibilities:

$(A \subseteq C) \land (A^C \nsubseteq C) \implies A^C\cap B\nsubseteq C \\
(A \nsubseteq C) \land (A^C \subseteq C) \implies A\cap B\nsubseteq C \\
(A \nsubseteq C) \land (A^C \nsubseteq C) \implies (A\cap B\nsubseteq C) \land (A^C\cap B\nsubseteq C) \\
(A \subseteq C) \land (A^C \subseteq C) \implies C = U \implies B \subseteq C
$

Regardless, all these cases would contradict the original (given) statements, or, as in the last case, the assumption that $B \nsubseteq C$.

EDIT: Last two cases added, and previous cases edited, after @Asaf Karagila pointed out the incompleteness/incorrectness of my answer.


问:
How does one calculate the temperature dissipation time of a metal in ambient air?
So, say I have some mild steel. The steel is 250F. I take it away from the heat and place it in ambient temperature, $T_a$, now I'd like to know how long it takes to get from 250F to some arbitrary temperature, $T_{a'}$. I think for mild steel the heat transmission coefficient is 7.9 $\frac{W}{m^2K}$... Not sure how to set this up. Not homework. Actual real world problem in my life. I'd like to see how you got the answer, as I'd like to use it over and over again for different temperatures.

Edit:

Ok, so my initial conditions are $T(0s)=102F, T(270s)=122F$, and I'm working with $\frac{dT}{dt}=-k(T-T_a)$. Where $T_a=250F$ is the ambient temperature. Now in my new test case I'm heating a block of steel from ambient to 250F (assuming homogeneous dissipation and constant temperature regulation...). So, one thing to another we get this:

$\frac{1}{T-T_a}\textbf{dT}=-k\textbf{dt}$

$\int\frac{1}{T-T_a}\textbf{dT}=-\int k\textbf{dt}$

$\ln{|T-T_a|}=-kt+c$

$T-T_a=e^{-kt}e^{c}$

${
T(t)=
\begin{cases}
      T_{>}(t)=Ce^{-kt}+T_a, T>T_a \\
      T_{<}(t)=T_a-Ce^{-kt}, T<T_a \checkmark
\end{cases}
}$

So, for $T_{<}(0)$ we have

$102=250-Ce^{-0k}$

$148=C$

Ok, so then $T_{<}(270)=122$

$122=250-148e^{-270k}$

$122=250-148e^{-270k}$

$\frac{32}{37}=e^{-270k}$

$-\frac{\ln\frac{32}{37}}{270}=k$???

So I guess that makes

$T_{<}(t)=250-148e^{\frac{\ln\frac{32}{37}}{270}t}$

But... If I wanted to see how long it takes to get to 250F I get no solution? What's happening here?

Note: $T(0)$ isn't the true zero. It's staggered from an arbitrary time after heating has actually begun. The steel block starts at ambient 50F, and picked a random time after it got to 102, called that 0, let 270 seconds pass and marked the second time at 122.

答:
This is the approach I took.
From Newtons Law of Heating :)
$$\frac{\mathrm{d}}{\mathrm{d}t}T(t)=k\left[T_a-T(t)\right]$$
$$\frac{1}{T_a-T(t)}\frac{\mathrm{d}}{\mathrm{d}t}T(t)=k$$
$$\ln\left|T_a-T(t)\right|=\ln\left(e^{kt+c}\right)=\ln\left(M{e^{kt}}\right)$$

SInce $T_a-T\gt 0$
$$\frac{1}{T_a-T(t)}=M{e^{kt}}$$

Using the values given $T_a=250$, $T(0)=102$:
$$\frac{1}{250-102}=M{e^{k0}}\Rightarrow M=\frac{1}{148}$$

Using the values given $T_a=250$, $T(270)=122$:
$$\frac{1}{250-122}=\frac{1}{128}=\frac{1}{148}{e^{k(270)}} \Rightarrow \frac{148}{128}=\frac{37}{32}=e^{k(270)}$$
Therefore:
$$k=\frac{1}{270}\ln\left(\frac{37}{32}\right)$$

Putting it all together:
$$250-T(t)=\frac{1}{e^{\frac{1}{270}\ln\left(\frac{37}{32}\right)t}}$$


问:
Showing $\mathbb{S}^1$ with a diameter attached is a deformation retract of plane minus two points
I'm trying to solve the following exercise:

I think I've understood the "union of two circles with a point in common" case but I'm lost on the other one. I tried to come up with examples of a retraction $r: \mathbb{R}^2 \setminus \{p, q\} \to \mathbb{S}^1 \cup \text{diameter} $ but I couldn't come up with any. The usual retraction $\mathbb{R}^2 \setminus \{p\}$ onto a circle with center $p$ doesn't work either, because in this case we have to attach a diameter. Can anyone help me? I'd appreciate it!

答:
Hint: such a homotopy may be not so fun to make explicit. But here's an idea: if you collapse the diameter to a point, continuously, you get the two circles glued in a point as before.

What may work is picking a "center" $x,y$ in each half circle to remove, and sending each point $z$ to the intersection of the line $\vec{xz}$ (or $\vec{yz}$, depending on which half of the circle lies $z$) with $S^1$ Then for the outside of the circle we do "the same", fix $o$ to be the center of the circle and send some point $z$ outside the circle to $S^1 \cap \vec{zo}$.

That should define a continuous retraction $r$; we're in the plane so for concrete $o,x,z$ we can cook up a concrete formula and all, but I don't know if the homotopy $ir \simeq 1$ is easy to describe. My intuition is that since our retraction was defined by assigning each point $p$ a curve $\gamma_p$ from $p$ to $r(p)$, the homotopy should look like $H(t,p) = \gamma_p(t)$. Again, this has to be made precise.


问:
Number of Paths on a Grid From S to M
I was reading about a the derivation of the formula for the number of paths from one corner to another corner of a H by W grid here and I wondered whether it is possible to apply the result: $\binom{(H-1)(W-1)}{H-1}$ to find the number of paths from a given square on the top row of the grid to another selected square in the bottom row.

For example the number of paths from B to J.
I thought of reducing the grid to just the columns B to D, counting the paths there and then adding the possible paths from every other square outside the reduced grid. However I had trouble in finding a formula for the possible paths from outside of the reduced grid.

In a path you cant repeat a square, and you can move to any adjacent square.

答:
In the original source, note that each step in the path was Right or Down.  With those rules, any path from B to J requires 2 Rights and 5 Downs, in any order.  There are $\binom{5+2}{2} = 21$ such paths.  But to go from B to G, say, you would need Left and Down moves.  If you allow Left and Right both, then the number of paths is infinite, as you could move Right Left Right Left... as long as you like.

Edit: With the restriction about not visiting a position more than once, the number of paths with every direction allowed is finite but large (compared to the board size).  For instance, the number of paths between opposite corners of a $2 \times n$ grid is $2^{n-1}$, already exponential.

Title: Number of Paths on a Grid From S to M


问:
Positive Totient function?
We know that:
$$\phi(n) = n(1- 1/p_1)(1-1/p_2)\ldots(1 - 1/p_k)$$
where $\phi$ is the Euler totient function and $p_i$ are the primes dividing $n$.
Is anything known about the similar function
$$\varphi(n)= n(1+1/p_1)(1+1/p_2)\ldots (1+1/p_k)?$$
I'm just looking reading material and references. Does it have a name?

答:
Next time you have a similar question, you can try to put the few starting values into the search bar on OEIS.org. In this case, you would've found your answer. (Sometimes it is useful to omit the first value due to different conventions, e.g. here the empty product is taken to be $1$.)

From wikipedia:

The Dedekind psi
function is the
multiplicative function on the positive integers defined by

$$\psi(n)=n \prod_{p \mid n}\left(1+\frac{1}{p}\right),$$

where the product is taken over all primes $p$ dividing $n$. (By
convention, $\psi (1)=1$.)

It starts with $\psi(n)=1, 3, 4, 6, 6, 12, 8, 12, 12, 18,\dots$, which is OEIS entry A001615.

The wikipedia article references:

Leonard Eugene Dickson "History of the Theory Of Numbers", Vol. 1, p. 123, Chelsea Publishing 1952.

Journal für die reine und angewandte Mathematik, vol. 83, 1877, p. 288. Cf. H. Weber, Elliptische Functionen, 1901, 244-5; ed. 2, 1008 (Algebra III), 234-5

Similarly, the Mathworld article references:

Cox, D. A. Primes of the Form x2+ny2: Fermat, Class Field Theory and Complex Multiplication. New York: Wiley, p. 228, 1997.

Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, p. 96, 1994.

Sloane, N. J. A. Sequence A001615/M2315 in "The On-Line Encyclopedia of Integer Sequences."

Title: Positive Totient function?


问:
Concrete Mathematics problem 1.9c proof by contradiction
I'm self-studying the Concrete Mathematics book and I'm trying to work through the following problem:

Consider the following statement:

$ P(n): x_1 ... x_n \leq \left( \frac{x_1 + ... + x_n}{n}\right)^n$ ,
if $ x_1, ..., x_n \geq 0 $

This is true when $n = 2$, since $(x_1 + x_2)^2 - 4x_1x_2 = (x_1 - x_2)^2 \geq 0$.

a) By setting $x_n = \frac{x_1 + ... + x_{n-1}}{n-1}$, prove that
$P(n) \implies P(n-1)$ whenever $n > 1$.

b) Show that $P(n)$ and $P(2)$ imply $P(2n)$

c) Explain why this implies the truth of $P(n)$ for all $n$.

I've already done the first two parts, but I'm trying to reason about why (c) is true. I guess it's because the question asks us to explain why this implies truth for all $n$ instead of prove it, but I'm not satisfied with the book's answer. I've attempted to come up with my own rigorous proof to convince myself of this assertion inspired by this post. I'm wondering, would this be a valid proof by contradiction to show that $P(n)$ is true for all $n$?

Proof by Contradiction

Assume there is a non-empty set $S = \lbrace s \in \mathbb{N} | P(s) \text{ is false } \rbrace $.

Due to the well-ordering principle, we know that $S$ must have a minimum if it is non-empty. Let's take the minimum to be $s_0 = \text{min}(S)$. Because we know the base case P(2) is true, $s_0 > 2$.

Since $s_0$ is the minimum of set $S$, $\forall x \in [2, n), P(x)$ is true.

Without loss of generality, assume that $s_0$ is odd as it can be easily proven if $s_0$ is even. $P \left( \frac{s_0+1}{2} \right)$ is true since $\frac{s_0+1}{2} < s_0$ for $s_0 > 1$ and we've established that $s_0 > 2$.

Since $ P(n) \land P(2) \implies P(2n)$, $P(\frac{s_0+1}{2}) \land P(2) \implies P(s_0 + 1)$. Thus, $P(s_0+1)$ must be true.

Because $ P(n) \implies P(n-1)$ when $n > 1$, $P(s_0+1) \implies P((s_0 + 1) - 1) = P(s_0)$. Thus, $P(s_0)$ must be true.

However, this contradicts our assumption that there is a minimum of $S$, $s_0 = \text{min}(S)$. Thus, if there is no minimum of $S$, $S = \emptyset$ and there are no $x \in \mathbb{N}$ such that $P(x)$ is false.

答:
Your argument is correct, if a bit over-complicated. And I would not say that you can assume without loss of generality that $s_0$ is odd: the argument in the even case is pretty trivial, but an argument is required.

Alternatively, one can observe that it follows from (b) by a trivial induction that $P(2^n)$ holds for all $n\in\Bbb Z^+$, and it follows from (a) by an equally easy (downward) induction that if $2\le m\le n$, and $P(n)$ holds, then $P(m)$ holds. Since for each $m\ge 2$ there is an $n\in\Bbb Z^+$ such that $m\le 2^n$, the result follows.


问:
Proof of Gallai Theorem for factor critical graphs
I am trying to find the proof of the following theorem:

Definition 1.2. A vertex $v$ is essential if every maximum matching of $G$ covers $v$ (or $\nu(G-v) = \nu(G)-1$). It is avoidable if some maximum matching of $G$ exposes $v$ (or $\nu(G-v) = \nu(G)$). A graph $G$ is factor-critical if $G-v$ has a perfect matching for any $v \in V(G)$.

Theorem 1.3. Let $G$ be a connected graph. Then $G$ is factor-critical if and only if every vertex is avoidable.

I don't think this is something I can prove on my own. However, I can't seem to find the proof of it online either.

It makes sense intuitively. If for every vertex $v$, in graph $G$, there exists a perfect matching that doesn't include $v$, then removing $v$ and its incident edges would result in a perfect matching because one existed that didn't include $v$ in the first place.

Could someone provide a more formal proof for this theorem, or provide a link to where I might find the proof?

Thank You

答:
A remark: if $M$ and $N$ are matchings of maximal size, the components of the subgraph formed by the edges in $M\cup N$ are even cycles and paths of even length.

The following is due to de Caen.

Lemma: Suppose $u$ and $v$ are vertices in the graph $X$ such that any maximum matching covers at least one of them. If $M_u$ is a maximum matching in $X$ that misses $u$ and $M_v$ is a maximum matching which misses $v$, there is a path of even length in $M_u\oplus M_v$ with $u$ and $v$ as its end-vertices.

Proof:
Let $H$ be the subgraph formed by the edges in $M_u\oplus M_v$.  The components of $H$ are even cycles and paths and, as both $M_u$ and $M_v$ have maximum size, each path must have even length.

Our hypothesis implies that $u$ is covered by $M_v$ and $v$ is covered by $M_u$.  Hence $u$ and $v$ have valency one in $H$ and therefore they are end-vertices of paths.

Suppose first that they lie on distinct paths and let $P$ be the path containing $u$.  Then $P$ is an $M_v$-alternating path with even length and $M_v\oplus E(P)$ is a maximum matching which misses both $u$ and $v$.

But no maximum matching misses both $u$ and $v$, therefore $u$ and $v$must be end-vertices of the same path from $H$.  As all paths in  $H$ have even length, we are finished.

Lemma:
Let $u$, $v$ and $x$ be distinct avoidable vertices in $G$.

If no maximum matching in $G$ misses both $u$ and $x$, and no maximum matching misses both $x$ and $v$, then no maximal matching misses both $u$ and $v$.

Proof:
Let $M_x$ be a maximum matching that misses $x$ and suppose that $N$ is a maximum matching that misses both $u$ and $v$. By de Caen's lemma applied to $u$ and $x$, there is an alternating path in $M_x\cup N$ joining $u$ to $x$.  By the same argument applied to $x$ and $v$, there is an alternating path from $x$ to $v$ in $M_x\cup N$.  Since
there can be only one alternating path in $M_x\cup N$ starting at $x$, this implies that $u=v$, a contradiction. Therefore there is no maximum matching in $G$ that misses $u$ and $v$.

Lemma (Gallai):
If $X$ is connected and each vertex is missed by a maximum matching, then for each vertex $v$ the graph $X\setminus v$ has a perfect matching.

Proof:
As $X$ is connected and all vertices are avoidable, from the previous lemma and a simple induction argument, we see that no two vertices of $G$ are missed by a maximum matching.  Therefore each maximum matching
of $G$ misses exactly one vertex.


问:
Is it possible to make integers a field?
Is it possible to define addition and/or multiplication on the set of

a) natural numbers (including $0$: $0,1,2,3,...$)

b) integers $(..., -2, -1, 0, 1, 2, ...)$

in such way that they will become fields?

Thanks in advance.

答: Apart from choosing arbitrary bijections to countable fields, there is actually a quite natural field structure on the set $\mathbb{N}$ which is motivated by game theory, namely nim addition and nim multiplication. See the wikipedia article on nimbers, or better directly consult Conway's On numbers and games. This field is the union (or rather colimit) of the finite fields $\mathbb{F}_{2^{2^n}}$


问:
Hölder inequality and interpolation
I was seeking the other day for intuition about the Hölder inequality and more precisely the geometric intuition behind the relation $\frac{1}{p} + \frac{1}{q} = 1$.

I found this interpretation of the Hölder inequality.

It says that we are trying to find an estimate of :

$$ \int f^a g^b$$ with the knowledge of $\int f^p$ and $\int g^q$.

For $(a, b) = (1,1)$, we need the relationship $\frac{1}{p} + \frac{1}{q} = 1$ so that the points : $(p, 0), (1,1), (0, q)$ lie on the same line. We can know use interpolation to have the Hölder inequality which give an estimate at the point $(1, 1)$.

But I feel like an argument is missing here. Why the estimate is using the geometric mean of the  function $f$ and $g$ ?  Why we don't have something like (using arithmetic mean) :
$$
\int \mid fg \mid \leq \frac{1}{p} \int f^p + \frac{1}{q}\int g^q \;\;?
$$
Moreover the article is saying we can generalize this approach to approximate $\int f^ag^b$ but in this case we need three points such that $(a,b)$ is in the convex hull of these three points. Why for $(1, 1)$ two points are needed while for general $(a, b)$ we need three points ?

答:
But you do have something like that: Young's inequality says that for $u,v\ge 0$ and conjugate exponents $p$ and $q$,
$$
uv\le p^{-1}u^p+q^{-1}v^q.
$$
(This can be proved with a picture once you notice that $u\mapsto u^{p-1}$ and $v\mapsto v^{q-1}$ (the derivatives of $u\mapsto p^{-1}u^p$ and $v\mapsto q^{-1}v^q$) are inverse functions.) Take $u=|f(x)|$ and $v = |g(x)|$ and integrate.

Title: Hölder inequality and interpolation


问:
Continuity and the existence of the limits
This question is due to a clarification asked by an engineer friend. Supposing that I have this function

$$f(x)=\sqrt{1-x^2} \tag 1$$

that it is simply a semicircle in the first and second quadrant with the domain $[-1;1]$ and $f([-1;1])=\operatorname{im} f=[0;1]$. One of his students of an high school thinks that at point $-1$ and $1$ the function is not continuous, because

$$\bbox[pink,5px,border:2px solid lightgray]{\lim_{x\to (-1)^-}f(x)=\nexists, \quad \lim_{x\to (-1)^+}f(x)=0}$$

($ (-1)^-$ it is out of the domain). I think instead that it is true that $ (-1)^-$ it is out from the domain but are true the two conditions:

\begin{cases}
-1 \in \mathrm{dom}\, f(x), \\
\lim_{x\to (-1)^-}f(x)=\lim_{x\to (-1)^+}f(x)=f(-1)=0
\end{cases}
The same procedure for $1$.

Is it correct or are there specific other observations?

答:
See here.

The limits the student is talking about simply makes no sense. Left limit at $x=-1$ or right limit at $x=1$ are nonsensical, like, for instance, limit at $x=1000$.

This is expressed by the sentence

"essa [la definizione] ha senso solo se $p$ è un punto di accumulazione per il dominio di $f$"

My daughter's math teacher says the same nonsense, like $\frac{1}{x}$ is not continuous because doesn't exist at $x=0$. This is WRONG.

Hope this can help


问:
Finding gradient of $\det(\mathbf{\Sigma})^{\frac{-1}{2}}$ with respect to $\mathbf{\Sigma}$ (a Covariance matrix)
In the book Mathematics for Machine Learning (chapter on Gaussian Mixture Models), I got stuck trying to figure out how they came up with

$$\frac{\partial \det(\mathbf{\Sigma})^{\frac{-1}{2}}}{\partial \mathbf{\Sigma}} = \frac{-1}{2}\det(\mathbf{\Sigma})^\frac{-1}{2} \mathbf{\Sigma}^{-1}$$

where $\mathbf{\Sigma}$ is the covariance matrix. The book mentions that the following identity was used:
$$\frac{\partial \det(f(\mathbf{X}))}{\partial \mathbf{X}} = \det(f(\mathbf{X}))Trace\left( f(\mathbf{X})^{-1} \frac{\partial f(\mathbf{X})}{\partial \mathbf{X}} \right)$$

But the question is, how did they use it to arrive at the first equation ? I can't seem to figure it out. Been stuck for quite a while.

Thanks!

答:
It's probably easier to work with the logarithmic form of the Jacobi formula
$$\eqalign{
\frac{\partial \log\det X}{\partial X} &= X^{-T}
 &\quad\big({\rm assuming\;}\det X > 0\big) \\
}$$
and to use $S=\Sigma\;$ for ease of typing.
$$\eqalign{
c &= (\det S)^{-1/2}
 &\qquad\qquad&\big({\rm Cost\;function}\big) \\
\log(c) &= -\tfrac 12\log(\det S)\\
\frac{1}{c}\frac{\partial c}{\partial S} &= -\tfrac 12S^{-T}
 &\qquad&\big({\rm Logarthmic\;derivative=Jacobi}\big) \\
\frac{\partial c}{\partial S} &= -\tfrac 12cS^{-1}
 &\qquad&\big(S\,{\rm \,is\,symmetric}\big) \\
}$$


问:
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$
$a$ and $b$ are both positive Real Numbers .

I saw this in a math olympiad, and within two days i couldn't solve it.

There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.

答:
I would start with noticing that $a^{2} + 1 \geq 2\sqrt{a^{2}} = 2a$.

Similarly, we have that $b^{2} + 1 \geq 2b$.

Then we have that
\begin{align*}
\frac{a^{2} + 1}{b} + \frac{b^{2} + 1}{a} \geq 2\left(\frac{a}{b} + \frac{b}{a}\right) \geq 4\sqrt{\frac{a}{b}\times\frac{b}{a}} = 4
\end{align*}

Hopefully this helps!


问:
General Addition Rule for Variances (Perfect Positive Correlation)
Confused on the general addition rule for variances.

Why is it that when two random variables, X and Y, have a perfect positive correlation (p=1) their standard deviations add. But when they are uncorrelated (p=0) their variances add?

答:
Further illumination on Robert Israel's answer:

Starting with the formula
$$
\begin{align}
\mathrm{Var}(X)
&=\mathrm{E}\left((X-\mathrm{E}(X))^2\right)\\
&=\mathrm{E}(X^2)-\mathrm{E}(X)^2\\
\end{align}
$$
we get
$$
\begin{align}
\mathrm{Var}(X+Y)
&=\mathrm{E}\left(X^2+2XY+Y^2\right)-\left(\mathrm{E}(X)^2+2\mathrm{E}(X)\mathrm{E}(Y)+\mathrm{E}(y)^2\right)\\
&=\left(\mathrm{E}(X^2)-\mathrm{E}(X)^2\right)+\left(\mathrm{E}(Y^2)-\mathrm{E}(Y)^2\right)+2\Big(\mathrm{E}(XY)-\mathrm{E}(X)\mathrm{E}(Y)\Big)\\[5pt]
&=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y)
\end{align}
$$
Furthermore, using Hölder,
$$
\begin{align}
\left|\mathrm{Cov}(X,Y)\right|^2
&=\left|\mathrm{E}\Big((X-\mathrm{E}(X))(Y-\mathrm{E}(Y))\Big)\right|^2\\
&\le\mathrm{E}\left((X-\mathrm{E}(X))^2\right)\mathrm{E}\left((Y-\mathrm{E}(Y))^2\right)\\
&=\mathrm{Var}(X)\mathrm{Var}(Y)
\end{align}
$$
For perfect positive correlation, we have $\mathrm{Cov}(X,Y)=\mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}$; therefore,
$$
\begin{align}
\mathrm{Var}(X+Y)
&=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Var}(X)^{1/2}\mathrm{Var}(Y)^{1/2}\\
\mathrm{Std}(X+Y)^2
&=\mathrm{Std}(X)^2+\mathrm{Std}(Y)^2+2\mathrm{Std}(X)\mathrm{Std}(Y)\\
\mathrm{Std}(X+Y)
&=\mathrm{Std}(X)+\mathrm{Std}(Y)
\end{align}
$$

For independence, we have $\mathrm{Cov}(X,Y)=0$; therefore,
$$
\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)
$$


答: I assume that $a,b$ are positive real numbers. Then we may assume $a\geq b$, hence $a^2+1\geq b^2+1$ and $\frac{1}{a}\leq\frac{1}{b}$, so the rearrangement inequality gives $$\frac{a^2+1}{b}+\frac{b^2+1}{a}\geq\frac{a^2+1}{a}+\frac{b^2+1}{b}\geq4$$ where the last inequality holds by what you already showed.

Title: Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$

Body:
$a$ and $b$ are both positive Real Numbers .

I saw this in a math olympiad, and within two days i couldn't solve it.

There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.


问:
Dummit and Foote $p$-Primary Components of a Module over a Principal Ideal Domain
Dummit and Foote problem 12.1.10 reads:

For $p$ a prime in the P.I.D. $R$ and $N$ an $R$-module prove that the $p$-primary component of $N$ is a submodule of $N$ and prove that $N$ is the direct sum of its $p$-primary components (there need not be finitely many of them).

I think there might be a typo here, or I might be misunderstanding the definition they give of $p$-primary components. I actually think this exercise is the same as 10.3.22 in D&F which is the chapter about modules in general. In that problem they define the $p$-primary component of an $R$-module $M$ (where $R$ is a P.I.D.) as the set of all elements of $M$ that are annihilated by some power of $p$. However, in this problem they are assuming that $M$ is a torsion $R$-module, and I think that's the crux of the proof. Otherwise we have $\mathbb{Z}$ as a module over itself. Every $p$-primary component of $\mathbb{Z}$ is just $\{0\}$, so certainly these don't generate $\mathbb{Z}$ via direct sum or anything.

My question is, does anyone have a definition of $p$-primary component which does make this statement true without assuming $M$ is a torsion module somehow? (I don't think it is true without the assumption but want to cover my bases.)

答: The definition that Dummit and Foote give of a $p$-primary component refers back to their Theorem 7, in which it is assumed that $M$ is a torsion module. So $M$ being a torsion module is part of the definition, but they don't make that obvious.