问:
Proving $|P(\mathbb{N})| $ is equinumerous to these 2 sets.

$ \left\{A\in \:P\left(\mathbb{N}\right):\:\:\left|A\right|=\left|\mathbb{N}\right|\right\} $

All equivalence,homogenous relations over $ \mathbb{N} $

For the first set I managed to prove 1 sided function because $ A\subseteq P\left(\mathbb{N}\right)\:\Rightarrow \:\left|A\right|\le P\left(\mathbb{N}\right) $

But I can't seem to think of another function from $ f:P\left(\mathbb{N}\right)\:\longrightarrow A $ that is a one to one function to use the Cantor-Bernstein theorem.

For the 2nd set I don't even even have a clue how to make those functions, having trouble defining the equivalence relations.

Would appreciate any guidance, I translated the assignment from Hebrew so I apologize if it's not clear.

答:
For the first one you could use the fact that if $S$ is uncountable, and $C$ is a countable subset of $S$, then $|S\setminus C|=|S|$ (if you’ve proved it): $\Bbb N$ has only countably many finite subsets. If you want to use the Cantor-Schröder-Bernstein theorem, however, you can get the other injection as follows. Let $E=\{2n:n\in\Bbb N\}$, $O=\Bbb N\setminus E$, and

$$\varphi:\Bbb N\to E:n\mapsto 2n\,,$$

and take

$$h:\wp(\Bbb N)\to\{A\in\wp(\Bbb N):|A|=|\Bbb N|\}:S\mapsto\varphi(S)\cup O$$

as your injection.

For the second question I am assuming that a homogeneous equivalence relation is one for which all of the equivalence classes have the same cardinality. Recall that there is a bijection between equivalence relations on $\Bbb N$ and partitions of $\Bbb N$, so we’re looking at the set of all partitions of $\Bbb N$ whose parts are all of the same cardinality.

First note that $|\{A\in\wp(E):|A|=|\Bbb N|\}|=|\wp(E)|=|\wp(\Bbb N)|$. Then observe that for each $A\in\wp(E)$ such that $|A|=|\Bbb N|$, $\{A,\Bbb N\setminus A\}$ is a homogeneous partition of $\Bbb N$. Thus, there are at least $|\wp(\Bbb N)|$ homogeneous partitions of $\Bbb N$: this indirectly gives you an injection from $\wp(\Bbb N)$ to the set of homogeneous partitions of $\Bbb N$.

For the other inequality note that if $\mathscr{P}$ is a partition of $\Bbb N$, we can list the parts of $\mathscr{P}$ in increasing order of their smallest elements. This gives us a sequence $\langle P_k:k\in\Bbb N\rangle$ of subsets of $\Bbb N$: if $|\mathscr{P}|=n$, we let $P_k=\varnothing$ for $k\ge n$. (Note that my $\Bbb N$ includes $0$.) We now have an injection from the set of homogeneous partitions of $\Bbb N$ to $\wp(\Bbb N)^{\Bbb N}$. I expect that you already know that $|\wp(\Bbb N)|=\left|\{0,1\}^{\Bbb N}\right|$, so

$$\left|\wp(\Bbb N)^{\Bbb N}\right|=\left|\left(\{0,1\}^{\Bbb N}\right)^{\Bbb N}\right|=\left|\{0,1\}^{\Bbb N\times\Bbb N}\right|=\left|\{0,1\}^{\Bbb N}\right|=|\wp(\Bbb N)|\,.$$


问:
Exactness and obstruction in sheaf cohomology
Given a topological space $X$, sheaf cohomology 'measures' the lack of exactness of the global section functor $\Gamma(X, -) : \textbf{Sh}(X) \to \textbf{Ab}$.

From another viewpoint, sheaf cohomology should be measuring the 'obstruction to lifting local to global data'.

I understand the notion of a sheaf as a local assignment to a topological space of algebraic structures that compatibly 'restrict' and 'glue'; and the global sections functor that maps $\mathcal{F} \mapsto \mathcal{F}(X)$.

Yet, I don't understand the connection between the exactness of $\Gamma(X, -)$ and the capability to 'lift local data to global'.

How do these two viewpoints connect?

答:
Well of course a sheaf is by definition something that can lift local datas to global ones in the following sense : if $\{U_i\}$ forms a cover of $X$ and if $f_i\in \mathcal{F}(U_i)$ are sections such that $f_i=f_j\in\mathcal{F}(U_i\cap U_j)$ then the $f_i$'s glue together to form a section $f\in\mathcal{F}(X)$ such that $f|_{U_i}=f_i$.

So what might ask what do not glue inside a sheaf ?

Consider a sheaf $\mathcal{F}$ and sections $f_i\in\mathcal{F}(U_i)$ over a cover of $X$, but this time you don't ask for the $f_i$'s to agree on the intersection. Instead you ask for the $f_i$'s to agree up to a given subsheaf. More precisely, let $\mathcal{G}\subset\mathcal{F}$ be a subsheaf and assume that $f_i-f_j\in\mathcal{G}(U_i\cap U_j)$. Does the $f_i$'s "glue" to a section $f\in\mathcal{F}$ ? I put some quotation marks because, of course you can't expect to find $f$ such that $f|_{U_i}=f_i$ (this would implies that $f_i=f_j$). But you can hope to find $f$ such that $f|_{U_i}-f_i\in\mathcal{G}(U_i)$.

(I believe that this problem was first ask by Cousin with the so-called Cousin's problem : assume you have meromorphic functions $f_i$ defined on $U_i$ such that $f_i-f_j$ are holomorphic. Is there a meromorphic function $f$ such that $f|_{U_i}-f_i$ are holomorphic ?)

It turns out that the sheaf $\mathcal{G}$, or rather its $H^1$, contains the obstruction to lift the local data $(f_i)$ to the global one $f$. Indeed, note that if $g_{ij}=f_i-f_j\in\mathcal{G}(U_i\cap U_j)$, then $(g_{ij})$ forms a $1$-cocycle and this cocycle is a trivial in cohomology iff you can find the $f$.

So now, what is the connection with $\Gamma$ not being exact ? Start with the $f_i\in\mathcal{F}(U_i)$ as above, well obviously $f_i$ defines a section $\overline{f_i}\in(\mathcal{F/G})(U_i)$, and the $\overline{f_i}$'s do glue in $\mathcal{F/G}$, since $\overline{f_i}-\overline{f_j}=0\in(\mathcal{F/G})(U_i\cap U_j)$. So you have a well defined element $\overline{f}\in \Gamma(X,\mathcal{F/G})$. Now you can find the $f$ as above iff $\overline{f}$ is in the image of $\Gamma(X,\mathcal{F})\to\Gamma(X,\mathcal{F/G})$.

The short exact sequence $0\to\mathcal{G}\to\mathcal{F}\to\mathcal{F/G}\to 0$ gives rise to the exact sequence

$$0\to \Gamma(X,\mathcal{G})\to\Gamma(X,\mathcal{F})\to\Gamma(X,\mathcal{F/G})$$
the non right exactness of $\Gamma$ means that we cannot extends this exact sequence by $0$, instead the derived functor $H^1(X,\mathcal{G})$ appears :
$$0\to \Gamma(X,\mathcal{G})\to\Gamma(X,\mathcal{F})\to\Gamma(X,\mathcal{F/G})\to H^1(X,\mathcal{G})$$
And the cocycle $(g_{ij})$ constructed above is exactly the image of $\overline{f}$ under the boundary map $\Gamma(X,\mathcal{F/G})\to H^1(X,\mathcal{G})$. Putting together what we said before we see that the following are equivalent :

you can find the global $f$ which "lift" the local data $f_i$

the cocycle $(g_{ij})$ is trivial, in other words $\overline{f}\in\ker(\Gamma(X,\mathcal{F/G})$ (here we see that cohomology contains the obstruction)

$\overline{f}$ comes from an $f\in\Gamma(X,\mathcal{F})$,  in other words $\overline{f}\in\operatorname{im}(\Gamma(X,\mathcal{F})\to\Gamma(X,\mathcal{F/G}))$ (here we see that $\Gamma$ does not preserve surjectivity).

ContentLicense: CC BY-SA 4.0


问:
Confused about when to treat probability calculation as small interval and use $\lambda$ in Poisson process
I am doing MITx 6.431x and I am really confused about Poisson process, in particular about when to treat the calculation as a small interval and when not.

Suppose we are merging two Poisson processes, say Red and Green. Then P(the kth arrival is red)=$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$.

Why? This is as if we are looking at a small interval here. Precisely speaking, we are taught to use the following when $\tau$ as an interval is 'small'.

$P(k,\tau) \approx
\begin{cases}
\ 1-\lambda\tau,  & \text{if $k$ =0} \\[2ex]
\lambda\tau, & \text{if $k$ =1} \\[2ex]
\ 0, & \text{if $k$ >1}
\end{cases}$

$\frac{\lambda}{\lambda_{Red}+\lambda_{Green}}$ is different from $\lambda\tau$ obviously, but in the lecture the former is derived from the latter, which seems to suggest we are still assuming a small interval. Furthermore, why shouldn't we use, say, $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$ to calculate, where $p_{N_\tau}(k)$ means the probability of having $k$ arrivals in interval $\tau$?

In the lecture, $\lambda$ is used to calculate probability when it is explicitly said that we are only concerning ourselves with small intervals. But it seems really vague when exactly do we use $\lambda$, consider the following:

All ships travel at the same speed through a wide canal. Each ship
takes $t$ days to traverse the length of the canal. Eastbound ships
(i.e., ships traveling east) arrive as a Poisson process with an
arrival rate of $λ_E$ ships per day. Westbound ships (i.e., ships
traveling west) arrive as an independent Poisson process with an
arrival rate of $λ_W$ ships per day. A pointer at some location in the
canal is always pointing in the direction of travel of the most recent
ship to pass it.

What is the probability that the next ship to pass will be westbound?

The solution here tells me that P(next ship is westbound)=$\frac{\lambda_W}{\lambda_{E}+\lambda_{W}}$. But nothing in the question suggests that we are considering a 'small' interval. Why shouldn't we use $p_{N_\tau}(k)=P(k,\tau)=\frac{(\lambda \tau)^ke^{-\lambda \tau}}{k!}$?

答:
So, the small interval thing is actually not particularly relevant here. You can take the small interval thing to be somewhat of a heuristic derivation of how a Poisson process is derived. (If you sum up n i.i.d. Bernoulli random variables with parameter $1/p$, you get a random variable $X_n$ whose distribution converges to Poisson(1)). You can argue loads of properties through it heuristically (and rigorously too, though at that point, arguing the properties through actual definitions of Poisson processes is simply easier.)

Under the hood, in this problem, you are actually using the splitting property of Poisson processes, if you are interested, their proofs you can find in Durrett EOSP (https://services.math.duke.edu/~rtd/EOSP/EOSP2E.pdf; Theorem 2.11). I will elaborate further on your ship example to show you the power of splitting.

Intuitively, for the ship problem, consider a big ship company that sends ships according to a Poisson process with rate $\lambda_W + \lambda_E$. Now, you have a conductor that tells each ship independently to be westbound with probability $\frac{\lambda_W}{\lambda_W + \lambda_E}$ and eastbound otherwise. What is remarkable is that now, both your eastbound and westbound ships are also going to be independent, and they will be Poisson processes with rates $\lambda_E$ and $\lambda_W$ respectively. So, conversely, whenever you are given two independent Poisson processes, you can imagine that they are actually coupled somewhere in the background together, and this magical splitting makes two independent streams. Hopefully from this view, it is clear as to why the probability of the first ship being westbound after any time $t$ will be $\frac{\lambda_W}{\lambda_W + \lambda_E}$.

Let me address your actual question: Here, it is kind of hard to work with the Poisson distribution itself, since when we are looking at the first ship, we are kind of subtly conditioning on it, and thus, the distributions kind of change. There is no fixed $\tau$; it is more like, at the random time $T$ when the next ship passes, what is the probability that it is westbound?


问:
Holomorphic mapping on open unit disk
Assume that $g$ is a holomorphic function on a neighborhood of $\overline{B}(0,1)$ (the closed unit disc centered at $0$) such that $g(0)=0$. Let $s=sup_{|z|=1}Re(g(z))$. Consider the function $h(z)=\frac{g(z)}{2s-g(z)}$.

Show that the function $h$ maps ${B}(0,1)$ (the open unit disk centered at $0$) to ${B}(0,1)$ and that $h(0)=0$.

Any hint will be greatly appreciated. Thanks!

答:
Assuming $g$ not identically zero (as otherwise $h$ doesn't make sense), we have by maximum modulus that $\Re g(z) < s, |z|<1$ and of course $s>0$ since $s > \Re g(0)=0$

But then $|2s-g(z)|^2-|g(z)|^2=4s^2-4s\Re g(z)=4s(s-\Re g(z)) >0$ for all $|z|<1$ so $|g(z)| <|2s-g(z)|$ hence $|h(z)|<1, |z|<1$, while $h(0)=0$ by defintion so we are done!

Title: Holomorphic mapping on open unit disk


问:
Proof by induction for a recursive sum
I have a homework assignment that has made my head hurt for the past several days. I found no explanation on the materials provided by the Professor on this matter. I'm given:

$$
sum: \mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\\
\begin{align}
s1&:sum(0, y)=y\\
s2&:sum(x+1, y)=sum(x,y)+1
\end{align}
$$

And the Prof. wants me to prove that $\forall x,y\in\mathbb{N}:sum(x,y)=sum(y,x)$.

My induction basis is $P(0)=sum(0, y)=sum(y,0)$, which, by commutativity, is valid.

My induction hypothesis: $HI:sum(x,y)=sum(y,x)$.

I really am stuck. Any and all help is greatly appreciated! Not even the Isabelle proofer could help me figure this out...

答:
You can do induction within induction.

What I mean is, since the claim to be proved is $\forall x \ \forall y \ sum(x,y) = sum(y,x)$, you can set up the 'outer' inductive proof on variable $x$ as:

Base: Show that $\forall y \ sum(0,y) = sum(y,0)$

Step:  Let $n$ be some arbitrary number, and show that if $\forall y \ sum(n,y) = sum(y,n)$, then  $\forall y \ sum(n+1,y) = sum(y,n+1)$

Now, to show the Base $\forall y \ sum(0,y) = sum(y,0)$ of this outer induction, you can;t just say, as you do, 'by commutativity'.  In fact, some sort of commutativity is the very thing we are proving over and over here.  Anyway, to prove this, we can do induction on the variable $y$:

'Base's Base':  Show that $sum(0,0)=sum(0,0)$

'Base's Step': Let $m$ be some arbitrary number, and show that if $sum(0,m) = sum(m,0)$, then $sum(0,m+1) = sum(m+1,0)$

Likewise, to prove the 'outside' Step (if $\forall y \ sum(n,y) = sum(y,n)$, then  $\forall y \ sum(n+1,y) = sum(y,n+1)$) we can use induction. To be specific, once we have assumed that $\forall y \ sum(n,y) = sum(y,n)$, we can show that $\forall y \ sum(n+1,y) = sum(y,n+1)$ by induction on the $y$:

'Step's Base': Show that $sum(n+1,0) = sum(0,n+1)$

'Step's Step': Let $m$ be some arbitrary number, and show that if $sum(n+1,m) = sum(m,n+1)$, then $sum(n+1,m+1) = sum(m+1,n+1)$

Title: Proof by induction for a recursive sum


问:
A compact metric space is totally bounded
There are lots of questions on this issue in the site. The original statement reads

A metric space is compact if and only if it is both complete and totally bounded.

I am only interested in proving If a metric space is compact, then it is totally bounded, following my definition of totally bounded:

A metric space X is totally bounded if for each $r>0$ there exists a finite set $F \subseteq X$ such that $\operatorname{dist}(x,F)<r$ for each $x\in X$; i.e. $X = \cup \lbrace B_r(y): y\in F \rbrace$. (For $x\in X$, $\operatorname{dist}(x,A) = \operatorname{inf}\lbrace \rho(x,a): a\in A \rbrace$)

Now, if $X$ is compact, then every open cover of $X$ has a finite subcover, say $\mathcal{U}$. Then we can select $y\in U_y$, for $U_y\in\mathcal{U}$, such that $B_{r_y}(y) \subseteq U_y$. However, $X\neq \cup B_{r_y}(y)$, if for instance $B_{r_y}(y) \subset U_y$ (more concretely I am thinking here in $\mathbb{R}$ such that $U_y=(0,1)\cup (5,5.1)$, then the neighborhood of $y$ does not include some real numbers)

答:
You don't have a cover, yet. You're assuming that $X$ is compact, so you know that every open cover has a finite sub cover. So, what you want to do is take an appropriate open cover, then reduce it to a finite subcover using the compactness, and somehow, this should give you the finite set that you're looking for. This is how you use an assumption like compactness.

Hint: Since you want to show that $X$ is totally bounded, you need to start by fixing $r$. Then, cover your space with balls of radius $r$ centered at every point. Use compactness to extract a finite subcover.


问:
Decoding [23,11,7] quadratic residual code
I'm reading this paper about decoding the [23,11,7] Golay code (Also QR code) generated by the polynomial $g(x)=x^{11}+x^9+x^7+x^6+x^5+x+1$, it says that if $\alpha$ is root of $g(x)$ and the associated cyclotomic set $B=\{1,2,3,4,6,8,9,12,13,16,18\}$ shows that $g(x)$ has roots $\alpha,\alpha^3,\alpha^9$. Why these are roots of $g(x)$ and not others? Also why the associated cyclotomic set is the same as the nonzero quadratic residues modulo $23$?

And why in this paper the author takes only the odd syndromes?

Thank you all.

答:
Due to conjugacy constraints (since $g(x)$ has binary coefficients) if $z$ is a root so are the elements of $\{z^{2^i}: i\geq 0\}$ so you pick up all the elements of the cyclotomic coset as a root, if any of its elements is a root.

Quadratic residues property is also preserved by squaring. Clearly $\alpha^2$ is a QR and thus so are $\alpha^4,\alpha^8,\cdots.$

Expand your last question in the body of the post, I didn’t click on the link. That way others reading the question may benefit.


问:
the conditional expectation of truncated multivariate normal distribution
I want to calculate the conditional expectation of truncated multivariate normal distribution. Specifically,
$$ \mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2 \geq a), $$
where
$
\tilde{Y} =
\begin{pmatrix}
\tilde{y} \\
\tilde{x}_1 \\
\tilde{x}_2
\end{pmatrix} \sim N(\mu, \Sigma)
$
with $\mu =
\begin{pmatrix}
b \\
b \\
b
\end{pmatrix}$ and $
\Sigma = \begin{pmatrix}
\sigma^2_y & \sigma^2_y & \sigma^2_y \\
\sigma^2_y & \sigma^2_x + \sigma^2_y & \sigma^2_y \\
\sigma^2_y & \sigma^2_y & \sigma^2_x + \sigma^2_y
\end{pmatrix}.
$

My attempt is as follows:
$$ \mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2 \geq a) = \mathbb{E} \left[ \mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2) \mid \tilde{x}_2 \geq a \right] = \mathbb{E} \left[ \frac{b \sigma_x^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{ (x_1 + \tilde{x}_2) \sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} \mid \tilde{x}_2 \geq a \right]. $$
For the last term, it seems to me that since $x_1$ is a realization but $\tilde{x}_2$ is still a random variable, the term with $x_1$ can go outside of expectation term. Then, it will be
$$ \frac{b \sigma_x^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{ x_1 \sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{\sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} \mathbb{E}(\tilde{x}_2 \mid \tilde{x}_2 \geq a),$$
where the last expectation is nothing but $b + \sigma_x\frac{\phi(Z)}{1-\Phi(Z)}$ with $Z = \frac{a-b}{\sigma_x}.$

Do you think that the calculations are correct?

答:
The answer to your question is available here:

Conditional expectation of multivariate normal distribution with inequality condition

When following the steps, just replace the formula

$$ \mathrm{E}\left(X_{1} \mid X_{3}<x_{3}\right)=\mu_{1}-\frac{\sigma_{13}}{\sigma_{3}} \cdot \frac{\varphi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)}{\Phi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)} $$

with

$$ \mathrm{E}\left(X_{1} \mid X_{3}>x_{3}\right)=\mu_{1}+\frac{\sigma_{13}}{\sigma_{3}} \cdot \frac{\varphi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)}{1 - \Phi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)} $$

and you should be good to go.

The reason your approach is not quite right is that when you use the Law of Total Expectation, you must still condition on $\tilde{x}_1 = x_1$. Your last formula will depend on $\mathbb{E}[\tilde{x}_2 = x_2 | \tilde{x}_2 > a , \tilde{x}_1 = x_1]$ instead of $\mathbb{E}[\tilde{x}_2 = x_2 | \tilde{x}_2 > a ]$.


问:
$\frac{1}{2+\sin(x)}$: function is finite positive everywhere, but antiderivative is periodic. Where's the error?
While $\frac{1}{2+\sin(x)}$ is positive, finite, and well-defined everywhere in $\mathbb{R}$, the antiderivative I found following in the footsteps here, using one round of Weierstrass integration, and a second round of trigonometric integration, appears to be periodic:
$$\int \frac{1}{2+\sin(x)} dx = \frac{2}{\sqrt{3}}\arctan\left(\frac{2}{\sqrt{3}}\tan\frac{x}{2}+\frac{1}{\sqrt{3}}\right)+C$$

There clearly is an error -- somehow the $\arctan$ and $\tan$ should've canceled each other magically, but I can't quite put my finger on where the error is.

What is the rigorous way to do this, i.e. to obtain a non-periodic, monotonically increasing antiderivative?

答:
A continuous, strictly increasing antiderivative of $1/(2+\sin x)$ is
$$
\frac{2}{\sqrt{3}}\arctan\left(\frac{2}{\sqrt{3}}\tan\frac{x}{2}+\frac{1}{\sqrt{3}}\right)
+\frac{2\pi}{\sqrt3} \bigg\lfloor \frac{x-\pi}{2\pi} \bigg\rfloor
+C.
$$
This is certainlly reassembling it piecewise manually, and doesn't suggest how one might come up with this expression (or an equivalent one) more directly.


问:
Every left ideal is complemented on the regular representation module
In a proof they used the fact that if R is a ring (with unit) and $_RR$ is the R-module with the action given by multiplication in the ring, then I a left ideal of R is complemented by another left ideal J (complemented in $_RR$, $_RR=I\oplus J$).

The proof also had the hypothesis of $_RR$ being semisimple, but I don't know if that was even used for this statement. I can't seem to figure out why it would necessarily be true (perhaps we need another property on the ring?).

My attempt:

Given the fact that I want an ideal, unique writing and for the sum to be R, my guess would be to consider: $1-i=x_i\ \text{for some}\ i\in I$, $x_i$ clearly doesn't belong to I, otherwise $1\in I$ and then I is R, so I would just consider J={$0$} (WLOG we can also assume I is not the zero-ideal).

Then $J=(x_i)_L$ is a left ideal trivially (and the smallest one whose sum makes R, so as to reduce the chance of intersection). Now I need to see $I\cap J=\lbrace 0\rbrace$.

Let $x\in I\cap J\implies x =y=rx_i=r(1-i)$ for $y\in I$ and $r\in R$.
$$x=r(1-i)=r-ri=y\implies r=ri-y\in I$$
In order for x to be $0$, then r should be a left divisor of $1-i$ and be in I. Yet, I don't see how to choose i so that $I\cap J ={0}$ (or even if I can do so).

答:
This is certainly not true without any hypotheses on $R$. For instance, if $R$ is an integral domain and $I,J<R$ are non-zero ideals, then we have $IJ\neq\{0\}$ and so $I\cap J\geqslant IJ\neq\{0\}$. Thus $R$ cannot be decomposed into a direct sum of any two non-zero ideals.

However, if $_RR$ is semisimple, then the result does hold. In fact, we have the following:

Lemma: For any ring $R$ and any (left) $R$-module, if $M$ is semisimple then $M$ is "completely reducible": ie, for any $A\leqslant M$, there is $B\leqslant M$ such that $A\oplus B=M$.

Proof: Let $A\leqslant M$. Since $M$ is semisimple, we can write $M=\bigoplus_{i\in I}M_i$, where each $M_i\leqslant M$ is simple. Now, let $S=\{N\leqslant M:A\cap N=\{0\}\}$, partially ordered by inclusion. $S$ contains the trivial submodule, and thus is non-empty. Also, if $(N_j)_{j\in J}$ is a chain in $S$, then clearly $N:=\bigcup_{j\in J}N_j$ lies in $S$; indeed, if we have $n\in N\cap A$, then $n\in N_j$ for some $j$, whence $n\in N_j\cap A=\{0\}$ and so $n=0$. Thus we may apply Zorn's lemma to find $B\in S$ maximal with respect to inclusion.

We claim that $C:=A\oplus B=M$. Since $M=\bigoplus_{i\in I}M_i$, it suffices to show $M_i\leqslant C$ for every $i$, so let $i\in I$ and suppose for contradiction that $M_i\nleqslant C$. Since $M_i$ is simple, this means that $C\cap M_i=\{0\}$. In particular, $M_i\cap B=\{0\}$, so $B':=B\oplus M_i$ strictly contains $B$. We claim that $B'\cap A=0$, which will contradict maximality of $B$ in $S$. Indeed, suppose we have $x\in B'\cap A$. Then we have $x=b+m$ and $x=a$ for some $b\in B$, $m\in M_i$, and $a\in A$. This means $a-b=m\in C\cap M_i=\{0\}$, whence $a=b$. Since $A\cap B=\{0\}$ by construction, this means $a=b=0$, and so $x=0$, giving the desired contradiction. $\square$

In fact, the converse of this lemma holds as well; as an exercise, try to prove it! (Hint: first show that a module is semisimple if and only if it is a sum (not necessarily direct) of simple submodules, using Zorn's lemma.) This characterization of semisimple modules is so common that it is sometimes taken as definition! In any case, the (left) submodules of $R$ are precisely the left ideals, so your desired result then follows immediately from the lemma.


问:
How to develop patience in mathematics?
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?

答:
It's all part of the character building nature of the subject. But I find that it helps to have multiple questions on the go; once I can no longer stand one problem, I just put it aside, work on another problem, then return to it or yet another once I'm impatient with that one.

Take it in incremental steps: do five minutes more, then ten, twenty; an hour, two hours; an afternoon; a day, a week, a month, a year!

Note that some textbooks deliberately put in open problems as exercises.

Also, if you're planning on doing a research degree at all, you've got to learn which questions are fruitful and which to leave unanswered, bearing in mind that some questions can take decades to answer.

Title: How to develop patience in mathematics?


问:
How do I construct this isomorphism?
Let $X$ be a completely ordered set, with the following properties:

a)$X$ does not have neither a maximum nor minimum element.

b)Any bounded subset of $X$ is finite.

Then $X$ is isomorphic to the set of al integers.

I was thinking on using a partition for $\mathbb{Z}$ which consists of only finite sets, such as intervals. However I'm struggling to see how can the isomorphism be constructed from there.

答:
SKETCH: Let $x\in X$; $X$ has no maximum element, so there is a $y\in X$ such that $x<y$. Use the fact that the interval $[x,y]$ is finite to show that $x$ has an immediate successor. Then use a similar argument to show that $x$ has an immediate predecessor. For each $x\in X$ let $x^+$ be the immediate successor of $x$ and $x^-$ the immediate predecessor of $x$.

Now define a function $f:\Bbb Z\to X$ as follows. Let $f(0)$ be any element of $X$. Given $f(n)$ for some $n\ge 0$, let $f(n+1)=f(n)^+$. Given $f(n)$ for some $n\le 0$, let $f(n-1)=f(n)^-$. Show that $f$ is an order-isomorphism.

Title: How do I construct this isomorphism?


问:
Number of possible states on a $6\times 6$ grid
You have a 6x6 square grid. You have 18 black and 18 white pieces. How many ways can you fill the board (one piece per square) such that there are 3 black and 3 white pieces per column, and such that there are 3 black and 3 white pieces per row?

I've tried to do it purely combinatorally , just looking row by row, but it gets confusing very fast. Is there perhaps some way to do it utilizing symmetries, like how a solution translated horizontally or vertically is a solution, or how the 90 degree rotation around the center of a solution is a solution?

答:
There is an action of the group $G=S_6\times S_6$ on the set of all configurations satisfying these conditions. One copy of $S_6$ permutes the rows and the other permutes the columns. If you write down representatives for each of the orbits under this action, then you can apply the orbit-stabilizer theorem to determine the size of the orbit. I would do this by writing down representatives that have black boxes as close to the top of the grid and as far to the left as possible. So, the top row will always have the three left-most tiles filled in black because this can always be obtained by permuting columns. Also, the top three tiles in the first column will be black because you can obtain this by permuting rows. Then, you go from there placing a black tile as far up and to the left as possible without repeating a previously studied case. The key issue, then, is determining the size of the stabilizer subgroup of $G$.

Let's make the count in the case of a $4\times 4$ grid (which I'll typeset as tic-tac-toe boards). In this case, there is an action of $G=S_4\times S_4$ and $|G|=(24)^2$ (Convention: the first copy acts on rows and the second acts on columns). There are two orbits:
$$\begin{array}{|c|c|c|c|}\hline X&X&O&O\\\hline X&X&O&O\\\hline O&O&X&X\\\hline O&O&X&X\\\hline \end{array}\;\;\;\;\;\;\;\;(1)
$$
and
$$
\begin{array}{|c|c|c|c|}\hline X&X&O&O\\\hline X&O&X&O\\\hline O&X&O&X\\\hline O&O&X&X\\\hline \end{array}\;\;\;\;\;\;\;\;(2)
$$
The stabilizer of configuration (1) is
$$H=\langle A, (w_0,w_0)\rangle\leq S_4\times S_4$$
where $w_0=(14)(23)$ and
$$A=\langle((12),1),((34),1),(1,(12)),(1,(34))\rangle\cong (S_2\times S_2)\times(S_2\times S_2).$$
Since $\langle (w_0,w_0)\rangle$ normalizes $A$, $|A|=16$, $|w_0|=2$, and $\langle (w_0,w_0)\rangle\cap A=\{1\}$, this group has order $|H|=32$ (you can represent this subgroup as a semi-direct product). Therefore, the size of the orbit is
$$|G|/|H| = (24)^2/32 = 18.$$
The stabilizer of configuration (2) is
$$
K=\langle B,(w_0,w_0)\rangle,
$$
where $B=\langle ((1342),(1243))\rangle$ and $w_0$ is as above. Again, $\langle(w_0,w_0)\rangle$ normalizes $B$ and $B\cap\langle (w_0,w_0)\rangle=\{1\}$. This time $|B|=4$, so $|K|=8$. Therefore the size of the orbit is
$|G|/|K| = (24)^2/8 =72.$

This gives a count of $18+72 = 90$ configurations.


答:
Hints: Choose arbitrary $x_0\in X$ and send to $0$.

We can define $\varphi(x):=|[x_0,x)|$ if $x\ge x_0$ and $\varphi(x):=-|[x,x_0)|$ if $x<x_0$.

Title: How do I construct this isomorphism?

Body:
Let $X$ be a completely ordered set, with the following properties:

a)$X$ does not have neither a maximum nor minimum element.

b)Any bounded subset of $X$ is finite.

Then $X$ is isomorphic to the set of al integers.

I was thinking on using a partition for $\mathbb{Z}$ which consists of only finite sets, such as intervals. However I'm struggling to see how can the isomorphism be constructed from there.


问:
Proving $\lim_{n\to+\infty} n\int_{0}^{\alpha}x^{n}\phi(x)dx=0$
Let $\phi : [0,1]\rightarrow\mathbb{R}$ be a Riemann integrable function. How can we show that $\forall\alpha\in[0,1)$, we have :
$$
\lim_{n\to+\infty} n\int_{0}^{\alpha}x^{n}\phi(x)dx=0
$$
I need to find an upper bound to be able to find the limit but the problem lies in finding an upper bound for $\phi(x)$. Any hint would be much appreciated, a full answer is not necessary, and thank you!

答:
[Hint] : Notice that $$\displaystyle\left|n\int_{0}^{\alpha}x^{n}\phi(x)\;\text{d}x\right|\leq n\int_{0}^{\alpha}x^{n}\left|\phi(x)\right|\;\text{d}x$$
Now define $\Omega_{x}:=\underset{x\in[0,\alpha]}{\sup}|\phi(x)|$, can you take it from here?


答:
Reading solutions step by step might be a good habit to increase the patience. By that I mean, when you struggle with a question for $30$ minutes, if you have access to a solution to that problem, you don't have to read the whole solution. For example, first you can check the first idea and after understanding it, you can simply keep struggling through that idea for another $30$ minutes maybe. If you are still not able to solve it, you can go to a further step in the solution and take it from there for another $30$ minutes, not reading the whole solution.

However, there are few things to add here. There are some problems which are hard to understand but easy to solve. For those kind of problems, the solution might be a one line proof for instance so you may ask "how can I go through the solution step by step if the solution is too short?". In these kind of cases, I would suggest trying to have a better understanding on question. This could be achieved by reading more about the topic or turning back to the points where you could not grasp the intuition. Note that this kind of "struggling" also develop some kind of patience.

Finally, there might be some questions for which you don't have an access to solution. In this case, I suggest trying to find similar problems to the one you have. Sometimes, it is more satisfying (and even harder) to adapt an existed solution of another problem to your problem, than solving the problem itself.

Title: How to develop patience in mathematics?

Body:
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?


答:
There are some significant typos in Delanoy's answer, but I'll accept it nonetheless as it pointed me in the right direction. (Sorry for the delay.)

Let $a^k+b^k+c^k = d^k+e^k+f^k,\;\; k =2,4\tag{1}$

Theorem 1 (Birck-Sinha): "Fact 1" should be instead, given (1) then,

\begin{aligned}&(a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k + (2d)^k + (2e)^k + (2f)^k =\\
&(2a)^k + (2b)^k + (2c)^k  + (-d+e+f)^k +  (d-e+f)^k + (d+e-f)^k + (d+e+f)^k\end{aligned}

for $k = 2,4,6,8$.

Theorem 2 (Gloden-Sinha): "Fact 2" should be, let (1) have the constraint $a+b-c = 2(d+e-f)$, then,

\begin{aligned}&(2a-3h)^k + (2a-h)^k + (2b-3h)^k + (2b-h)^k + (2f+h)^k =\\ &(2c+h)^k + (2c+3h)^k + (2d-h)^k + (2e-h)^k + (3h)^k\end{aligned}

where $h = (d+e-f)/2$ for $k = 1,3,5,7$.

We can now reverse-engineer Letac's solution and find that he started with,

$$195^k+71^k+98^k = 21^k+190^k+127^k,\;\; k = 2,4\tag{2}$$

and using Theorem 2,

$$132^k+174^k+8^k+50^k+148^k = 119^k+161^k+0^k+169^k+63^k,\;\; k = 1,3,5,7$$

after removing a common factor. Thus, he gave (1) the second constraint $2d-h = 3d-e+f = 0$, though how he found (2) is a question in itself.

Unfortunately, it can be shown these two constraints reduce (1) to the problem of making two quartic polynomials (in one variable) into a square, hence there are only a finite number of rational solutions, and I wasn't able to find additional ones.

P.S. I'm still checking if Letac's first solution also uses Theorem 2.

ContentLicense: CC BY-SA 3.0

Title: How did Letac solve $x_1^k + x_2^k + \dots +x_9^k = 0$ for $k = 1, 3, 5, 7$ in 1942?

Body:
It's quite easy to find integer solutions to,

$$x_0^k + x_1^k + \dots +x_9^k = 0$$

for $k = 1, 3, 5, 7$. One I found is, if $x^2-10y^2 = 9$, then,

$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = (-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$

However, Letac found in 1942 two special cases with $x_0 = 0$ (the case $x_0 = x_1 = 0$ is impossible), namely,

$$\{\,0, 34, 58, 82, 98\,\} = \{\,13, 16, 69, 75, 99\,\}$$

$$\{\,0, 63, 119, 161, 169\,\} = \{\,8, 50, 132, 148, 174\,\}$$

How did he do it?

The reference is A. Letac, Gazeta Matematica, 48 (1942), p. 68-69, and I'm hoping someone with JSTOR can explicitly give the answer.


问:
A few questions related to $a^x=x^a$
I have a few related questions I would like some help with. The numbers involved can be assumed to be positive real numbers strictly greater than 1.

Given a number $a>1$:

What is the easiest way to show that:
$$\lim_{x\to\infty} \frac{x^a}{a^x}=0$$

What is the easiest way to show that, for any particular $a$, there exists a number $x_0$ such that if  $x>x_0$ then $a^x > x^a$

Is it possible to determine the smallest such $x_0$? I think I can show that for $a > e$ then $x_0 = a$ should work, though I'm not sure it's rigorous. And I don't really know what do to for $1<a<e$

Given two real numbers $1<a<b$, is it possible to show the existence of and find a minimal $x_0$ such that if $x>x_0$ then $a^x > x^b$ ?

In some of the above questions it may be easier or interesting to assume $x \in \mathbb{Z^+}$ instead of being real.

What I started with was showing that you need this condition:
$$\frac{x}{\ln{x}} > \frac{a}{\ln{a}}$$

Since I think I can show that $\frac{x}{\ln{x}}$ is increasing for $a>e$, then I think this directly implies that $x_0 = a$ as discussed above. But this doesn't help for $a<e$ (which we can see easily with the example of $a=2$). And I haven't made much progress on question 4 at all.

Does the Lambert W function come into play here somehow?

答:
Show that $\lim_{x\to \infty} (a\ln x - x\ln a) = -\infty$

Both $a^x$ and $x^a$ are continuous. At $x = 0$ which is higher? At $x = \infty$ which is higher? But at $x = a$, they cross.

Yes. There are finitely many solutions to $a^x = x^a$. $x_0$ is the highest. That uniquely defines the value $x_0$. There are many, many root-finding techniques that will allow you to identify the intersection to whatever precision you need.

Only small adjustments need to be made to (2) and (3) to answer this.

And finally, as Chrystomath has said, yes, you can use Lambert's function. But then, using Lambert's function to find the values is not any easier or faster or more accurate than finding it by root-finding techniques. Slapping a name on a function only makes it easier to locate what other people have discovered about it.

Title: A few questions related to $a^x=x^a$


答: Firstly, if you are asked to prove a statement, think a counterexample even though the question say only prove (not disprove or give counterexample.) Write your all examples and check them which properties are satisfied and not satisfied. This is my method and I am very happy with that. By the way, please be patient in general. Learning something always take time. The point reached without climbing never be top.</p>

Title: How to develop patience in mathematics?


问:
Showing the set $\{x: x = 2^{-k} \ k\in \mathbb{N}\ \text{or}\ x = 0 \}$ is open, closed, or neither - solution feedback
This question comes from Shifrin's  Multivariable Mathermatics - Sec 2.2 - 1(b), it asks:
Show that the set  $B = \{x: x = 2^{-k} \ k\in \mathbb{N}\ \text{or}\ x = 0 \}$ is open, closed, or neither and prove your answer.

I've come to the conclusion that the set is closed and I wanted some assistance on tightening up my solution.

Solution

The tools I have up to this point to show that a set is closed are:

i) show all convergent sequences in the set converge to a point in the set.

ii) Show that the complement of the set is open.

I chose to approach this problem using (ii), but I do have a question about how I could use (i) I'll ask at the end.

So first I have to figure out what it means to be in $C = \mathbb{R} - B$. To me this meant $\mathbb{R} - B = \{x: x \neq 2^{-k}\ \text{and} x \neq 0\}$. To build off of this idea I drew a picture of the number line with the points belonging to $B$ and observed how the points of $C$ would have to be to satisfy my argument. So after some fiddling around I defined the points of $C$ as follows:

$$C = \mathbb{R} - B = \{y: y = x\pm \delta\ \text{depending on which side of an $x$ from $B$ that the point $y$ is in relation to}\}$$.

So how do I choose my $\delta$? So keep in mind that each $x \in B$ is actually a function of a particular $k$, so with this in mind I defined $\delta = \min(|2^{-k}-y|, |2^{-(k+1)}-y|)$. In this set up it means I'm taking a $\delta$ that will take into consideration the possibility of having an open ball over a value $x \in B$ which we don't want.

So with this setup I now need to show $C$ is open. Define a point $z \in C$ as $z = x \pm \frac{\delta}{2}$.

By this definition we see that $z \in B(y, \delta)$. It's here where I'm having trouble completing the proof. So to show the ball is open is dependent on where my $y$ value is, so for a concrete argument let's suppose that $y < 2^{-k}$. To show  $B(y, \delta) \subset C$ means I have to show for all $z \in B(y, \delta)$ that:

$$|z - y| < |2^{-k} - y| \\
\Rightarrow\ |z - y| < 2^{-k} - y\ \text{(since assuming $y < 2^{-k}$ case)}$$

I'm having trouble arriving at the conclusion...What I envision happening after some algebra is:

$$2^{-(k+1)} < z < 2^{-k}$$

Which would then give me my open ball.

So I have two questions:

My idea seems to be leading me in the right direction, but needs some fine-tuning. What do I need to do to tighten it up?

With regards to the other way of showing closed. In this set it seems like the only sequence is the set itself since the $x = 2^{-k}$ and that will converge to $0$, but can I actually say that this is true? Couldn't there be some rare sequences that I have not found that are in the set, but since I have not explicitly stated all convergent sequences then I can't use the definition of closed explicitly?

答: You have the right idea, by either of the two techniques you described you could arrive to the conclusion (see that $2^{-k}$ aproaches $0$ when $k$ goes to infnity) but for a more topological aproach I would sugest you try to write the compliment as the union of disjoint open intervals.


问:
How to fit a square, a rectangle into an arbitrary triangle?
I need to check if a square or rectangle with given sides fits into a given arbitrary triangle. Shapes can be inscribed tangentially or not.

Does anyone know how to do that?
Thanks in advance.

答:
Possible strategy, perhaps an answer, too long for a comment.

You can check each of six possible cases (either rectangle orientation on each side) with simple trigonometry. That would be a straightforward efficient computer program.

That assumes the optimal placement is in fact with one edge of te rectangle coinciding with an edge of the triangle. I think that's true but needs proof.


答:
Some ideas:

By the given terms, we're talking of a 2$\pi\,-$ periodic function ( since, for example, the Fourier series is given in terms of $\;\cos nx=\cos\frac{n\pi x}L\implies L=\pi,\,T=2\pi\;$, and the same fo the sines terms ), and thus we can translate the interval from $\;[13\pi,\,15\pi]\;$ by $\;2\pi\cdot (-7=)-14\pi\;$ to the much more nice interval $\;[-\pi,\,\pi]\;$

As the Fourier coefficients fulfill

$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=0\;{}{}(\text{Riemann-Lebesgue Lemma})$$

but not

$$\lim_{n\to\infty} na_n=\lim_{n\to\infty} nb_n=0$$

we have here a function $\;f:[-\pi,\pi]\to\Bbb R\;$ which is $\;2\pi\,-$ periodic, but not continuous on its whole periodic continuation, or with a derivative which is not piecewise continuous. As the given function which expresses $\;a_n, b_n\;$ as integrals is $\;x^2\;$ (and thus undoubtedly with a piecewise continuous derivative over the basic interval $\;[-\pi, \pi)\;$), we can safely deduce the function is not continuous on its periodic continuation, meaning $\;f(-\pi)\neq f(\pi)\;$ ...

Finally, by what we're being told the Fourier coefficients are given from, we have that

$$f(x)=\begin{cases}x^2,&0<x<\frac\pi2\\{}\\
0,&-\pi<x\le0\,,\,\,\frac\pi2\le x<\pi\end{cases}$$

With the above function, we get exactly the definition of $\;a_n,\,b_n\;$ you were given, and for example

$$a_0=\frac1\pi\int_{-\pi}^\pi f(x)\,dx=\frac1\pi\int_0^{\pi/2} x^2\,dx=\left.\frac1{3\pi}x^3\right|_0^{\pi/2}=\frac1{3\pi}\frac{\pi^3}8=\frac{\pi^2}{24}\implies \frac{a_0}2=\frac{\pi^2}{48}$$

and we indeed get the first Fourier coefficient as given...

By Dirichlet Theorem, the Fourier series converge pointwise to the function at every continuity point of it, and to the average value of the function where it is discontinue. In symbols:

$$\frac{f(x^+)+f(x^-)}2= \text{ Fourier Series}$$

Because we don't have a continue function over its periodic continuation, the above convergence cannot be uniform.

Title: Finding the sum of the Fourier series

Body:
How to find the sum of the Fourier series defined by:

$$a_0 = \frac{\pi^2}{24}$$

$$a_n = \frac{1}{4\pi n^3}\left((\pi^2n^2-8)\sin(n\pi/2)+4\pi n\cos(n\pi/2)\right)$$
$$b_n = \frac{1}{4\pi n^3}\left((8-\pi^2n^2)\cos(n\pi/2)+4\pi n\sin(n\pi/2)-8\right),$$
for $x\in [13\pi,15\pi]$. This terms were obtained from
$$a_n = \frac{1}{\pi}\int_0^{\pi/2}x^2\cos(nx)dx,~b_n = \frac{1}{\pi}\int_0^{\pi/2}x^2\sin(n x)dx. $$

I started by observing that if $n=2k+1,$ then,
$$a_n = \frac{1}{4\pi n^3}(\pi^2n^2-8)(-1)^{2k+1}$$
$$b_n = \frac{1}{4\pi n^3}(4\pi n (-1)^{2k+1} -8)$$
for $k = 0,1,2,\ldots.$

Moreover, if $n = 2k$, then,
$$a_n = \frac{(-1)^{k}}{n^2}$$
$$b_n = \frac{8-\pi^2n^2}{4\pi n^3}(-1)^{k}$$

So the series is reduced to
$$S = \frac{\pi^2}{48} + \sum_{k=0}^{\infty}\left(\frac{(-1)^k}{4k^2}\cos(2kx) + \frac{8-\pi^24k^2}{32\pi k^3}\sin(2kx)\right) + \sum_{k=0}^{\infty}\left(\frac{(\pi^2(2k+1)^2-8)(-1)^{2k+1}}{4\pi (2k+1)^3}\cos((2k+1)x) + \frac{4\pi (2k+1)(-1)^{2k+1}-8}{4\pi(2k+1)^3}\sin((2k+1)\pi x)\right).$$

From here I don't know how to proceed.


问:
Proving inequality using Chebyshev's inequality
I got in trouble with this:

suppose that $X_1,\dotsc,X_n$ are independent random variables. Show that for any $a>0$, $P(\max|S_k|>3a) < 3\max P(|S_k|>a)$ where $S_k=X_1+\dotsb+X_k$.

I tried to solve it using Chebyshev's inequality, but nothing gains.
I've been try to solve this problem about 2 days, but I have no idea what the clue is!

Please give me some advice.

答:
Hint: Let $a>0$, for $k=1,2,\ldots, n$ let's do $$A_{k}=\left\{\max_{1\leq j \leq k-1}|S_{j}|\leq 3a, |S_{k}|>3a \right\}$$
so, we can see that $$\left\{ \max_{1\leq j \leq k} |S_{j}|>3a\right\}=\bigcup_{k=1}^{n}A_{k}, \quad A_{i}\cap A_{j} =\emptyset $$
it's to say the sets $\{A_{k}\}_{k=1}^{n}$ are disjoint.
Then use Chebyshev's inequality.


答:
$4$ years ago I started my journey. Initially , used just can't sit for more than half an hour with simple real analysis stuff. I found that reading multiple books has given me insight how basically theory develops and I look at similar problem. For better insight. Even if I have solved to get a better understanding(but I do this in semester breaks else my grades suffer a lot). I just push myself to do assignment by setting alarm for let's say 10 or 20 or 50 minutes. For me initially I thought that doing assignment without proper understanding is not good but my here in Indian pattern my professor said I am a tourist guide who will show you various places but to know your places and how long you have to stay depends on number of days you are with me.

So, you should choose wisely. If you are stuck ask help. Some problems might take a month or so just to show existence but after that it's cake walk.

And I asked my professor same question that you just asked he says it's part of planning.

Your planning improves with practice. He says even I plan how to cover course and make my lecture interactive over years still I was not able to do so($15$ years of experience) .
Keep practice - know your constraints - optimize output.

Title: How to develop patience in mathematics?


答:
Using the result in https://en.wikipedia.org/wiki/Fourier_series, then $P=2\pi$ and your function is
$$f(x)=\left\{\begin{array}{ll}x^2 \text{ if }x\in(0,\pi/2), \\
0 \text{ if }x\in(\pi/2,2\pi).
\end{array}\right.$$
and $f(x)=0$ if $x\in(\pi/2,2\pi)$. Extend $f(x)$ periodically and you will get $f(x)$ in $(-\infty,\infty)$. So in $[13\pi,15\pi]$,
$$f(x)=\left\{\begin{array}{ll}x^2 \text{ if }x\in(13\pi,13.5\pi), \\
0 \text{ if }x\in(13.5\pi,15\pi). \end{array}\right.$$
For $x=13\pi, 13.5\pi, 15\pi$, use
$$ f(x)=\frac12(f(x^+)+f(x^-). $$

Title: Finding the sum of the Fourier series


问:
Matrices that satisfy $A^2,B^2,C^2$ and commute
Is there a set of matrices that satisfy all of the following constraints?

1) $A^2=0, B^2=0, C^2=0...$ where $A,B,C,D..$ are different matrices.

2) All of them commute.

Edit: 3) $AB \neq 0, AC \neq 0, BC \neq 0, ...$.

答:
Let $$E=\left(\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right), \ \ \ \ I=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right).$$

Then set

$$
A = E\otimes I\otimes I\otimes I,\\
B = I\otimes E\otimes I\otimes I,\\
C = I\otimes I\otimes E\otimes I,\\
D = I\otimes I\otimes I\otimes E.
$$

Where $\otimes$ is the Kronecker product. These were constructed following the procedure outline by @HenningMakholm here:

More generally if only you have an $A$ and $B$ with $A^2=B^2$, $AB=BA=0$, you can combine $n$ copies of this structure using Kronecker products. That automatically makes matrices from different pairs commute, and ensures that cross-level products are nonzero as long as the generators are.


问:
Graph Theory - Can two vertices have two distinct edges?
My question is simple:

Can two vertices in a graph have two distinct edges connecting them? Why or why not?

答: It just depends on what you define a graph to be. Typically, graphs with at most one edge between vertices (and no loops) are referred to as simple graphs.</p>


问:
Dropping the "Lowest Grade" Problem
Let $A=\{a_1, a_2, ... , a_n\}$ be a set of non-negative real numbers and $B=\{b_1, b_2, ..., b_n\}$ be sets of positive real numbers.

Let $s = \dfrac{ \sum_A a}{\sum_B b} =  \dfrac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}$.

What I want to do find $k$ so that

$s_1 = \dfrac{\sum_{A-\{a_k\}} a }{\sum_{B-\{b_k\}} b}$ is the largest.

And more generally, find

$s_p = \dfrac{\sum_{A-\{a_{k_1},a_{k_2},\dots,a_{k_p}\}} a }{\sum_{B-\{b_{k_1},b_{k_2},\dots,b_{k_p}\}} b}$

I'm not exactly sure what this is called, and I tried doing a google search but I wasn't able to find what I am looking for.

And of course, I can do this by exhaustion, but I feel that there should be a smart way to figure this out...

答:
You can solve the problem via mixed integer linear programming as follows.  Let binary decision variable $x_j$ indicate whether you keep exam $j$ or not.  The problem is to maximize
$$\frac{\sum_{j=1}^n a_j x_j}{\sum_{j=1}^n b_j x_j} \tag1$$
subject to $$\sum_{j=1}^n x_j = n - p \tag2$$
You can linearize the objective $(1)$ by introducing nonnegative decision variables $z$ and $y_j$ and the following linear constraints:
\begin{align}
\sum_{j=1}^n a_j x_j &= \sum_{j=1}^n b_j y_j \tag3 \\
y_j &\le x_j &&\text{for all $j$} \tag4 \\
y_j &\le z &&\text{for all $j$} \tag5 \\
y_j &\ge z - (1-x_j) &&\text{for all $j$} \tag6 \\
\end{align}
The idea is that $y_j = z\cdot x_j$.
Now maximize $z$ subject to $(2)$-$(6)$.

Title: Dropping the "Lowest Grade" Problem


问:
show that $\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2 \left(\frac{a\pi}{2}\right) $

Show that $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2\left(\frac{a\pi}{2}\right) $$

I think we can solve it by contour integration, but how? and it would be better if there were two ways, the first with real analysis and the second with complex analysis.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax}
\over \sinh\pars{x}}\,\dd x =
{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}}:\
{\Large ?}.}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x\cos\pars{ax}
\over \sinh\pars{x}}\,\dd x} =
\totald{}{a}\int_{0}^{\infty}{\sin\pars{ax}
\over \sinh\pars{x}}\,\dd x
\\[5mm] = &\
2\,\totald{}{a}\Im\int_{0}^{\infty}
{\expo{\ic ax} - 1 \over \expo{x} - \expo{-x}}\,\dd x
\\[5mm] = &\
2\,\totald{}{a}\Im\int_{0}^{\infty}
{\expo{-\pars{1 - \ic a}x} - \expo{-x} \over 1 - \expo{-2x}}
\,\dd x
\\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,&
\totald{}{a}\Im\int_{0}^{\infty}
{\expo{-\pars{1/2 - \ic a/2}x}\,\,\, - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x
\\[5mm] = &\
\totald{}{a}\Im\left[\int_{0}^{\infty}
{\expo{-x} - \expo{-x/2} \over 1 - \expo{-x}}\,\dd x
\right.
\\[2mm] &\
\phantom{\totald{}{a}\Im\left[\right.}
\left.-\int_{0}^{\infty}
{\expo{-x} \expo{-\pars{1/2 - \ic a/2}x} \over 1 - \expo{-x}}
\,\dd x\right]
\\[5mm] = &\
\totald{}{a}\Im\bracks{\Psi\pars{1 \over 2} -
\Psi\pars{{1 \over 2} - {a \over 2}\,\ic}}
\\[5mm] = &\
\totald{}{a}\bracks{-\Psi\pars{1/2 - a\ic/2} +
\Psi\pars{1/2 + a\ic/2} \over 2\ic}
\\[5mm] = &\
-\,{\ic \over 2}
\totald{}{a}\bracks{\pi\cot\pars{\pi\bracks{{1 \over 2} - {a \over 2}\,\ic}}}
\\[5mm] = &\
-\,{\pi\,\ic \over 2}
\totald{}{a}\bracks{\tan\pars{{\pi a \over 2}\,\ic}}
=
{\pi \over 2}
\totald{}{a}\bracks{\tanh\pars{\pi a \over 2}}
\\[5mm] = &\
\bbx{{\pi^{2} \over 4}\on{sech}^{2}\pars{\pi a \over 2}} \\ &
\end{align}


答:
For a graph, or a simple graph, no, because a simple graph is a pair of sets, in which one is the set of vertices and the other is the set of edges. And elements of the set of edges are identified with endpoints, so, in this sense, there cannot be edges with same endpoints (if there are, then they are identified as the same one).

But there is also something called "multigraph," in which edges with the same endpoints are permitted. Check https://en.wikipedia.org/wiki/Multigraph. You may also want to look up "pseudograph."

Title: Graph Theory - Can two vertices have two distinct edges?

Body:
My question is simple:

Can two vertices in a graph have two distinct edges connecting them? Why or why not?


问:
Question on step of existing proof regarding area of disk
The following proof is from Apostol Calculus V1 Page 91.  I'm unsure how near the end of the proof:

$2r \int^1_{-1} \sqrt{r^2 - (rx)^2} dx = 2r^2 \int^1_{-1} \sqrt{1 - x^2} dx$.

How is $r * r$ on the LHS?

I'm using an image because I'm concerned the OCR may have caused on error.

答:
you have:
$$2r\sqrt{r^2-(rx)^2}=2r\sqrt{r^2(1-x^2)}=2r|r|\sqrt{1-x^2}$$
and if $r\in\mathbb{R}^+$ this simplifies to:
$$2r^2\sqrt{1-x^2}$$


问:
Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$
I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4}  }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4}  } |<ϵ.$$

I have this:

$\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot  \sqrt[3]{4}} \cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} $

= $\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac{\sqrt[3]{2}}{6} $

=$ \dfrac{6\cdot \sqrt[3]{2+h}-6\cdot \sqrt [3]{2}-\sqrt[3]{2}\cdot h}{6h} $

=$ \left |\dfrac{1}{(\sqrt[3]{2+h})^2 + \sqrt[3]{2+h}\sqrt[3]{2}+(\sqrt[3]{2})^2 } -  \dfrac {\sqrt[3]{2}} {6}\right | $

=$   $

I don't know how continue for use the fact $|h|<\delta$

The original exercise says: $f(x)=\sqrt[3]{x}$. Found $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$

答: Have you tried multiplying top and bottom of $\frac{\sqrt[3]{2+h}-\sqrt[3]{2}}{h}$ by $(2+h)^{\frac{2}{3}}+\sqrt[3]{2}\cdot\sqrt[3]{2+h}+2^{\frac{2}{3}}$?


问:
Edge length of hyperbolic tesselations
If I have a general uniform tesselation in hyperbolic plane (same configuration of regular polygons at every vertex, but multiple types of polygons allowed), how can I find the edge length and/or inner angles of the polygons so they'd fit?

For example: a truncated {3,7} tesselation has two hexagon and one heptagon at every vertex. What is its edge length and what are the inner angles of the two types of polygons?

答:
The angle $\theta$ at a vertex of a regular hyperbolic $n$-gon with side length $s$ is given by

$$\sin\tfrac12\theta=\frac{\cos\tfrac12\tfrac{2\pi}n}{\cosh\tfrac12s}.$$

This is monotonically decreasing in $s$. As the polygon shrinks we have $\lim_{s\to0}\sin\tfrac12\theta=\cos\tfrac12\tfrac{2\pi}n$, or $\lim_{s\to0}\theta=\pi-\frac{2\pi}n$, the Euclidean $n$-gon angle. As the polygon grows we have $\lim_{s\to\infty}\theta=0$.

If your hyperbolic tiling has a vertex type $(n_1.n_2.n_3.\cdots)$, for the polygons to fit exactly around the vertex we need

$$2\pi=\sum_k\theta_k=\sum_k2\arcsin\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}.$$

This sum is also monotonically decreasing, so there is at most one solution $s$. There is a solution if and only if the sum in the Euclidean case $s\to0$ is greater than $2\pi$ (using the Intermediate Value Theorem). I don't think we can find the solution in closed form, in general.

Multiplying that equation by $\tfrac12i$ and exponentiating gives us this product of unit complex numbers:

$$-1=\prod_ke^{i\theta_k/2}=\prod_k\left(\sqrt{1-\left(\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}\right)^2}+i\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}\right);$$

since $\cos\tfrac\pi n$ is always an algebraic number, this shows that $\cosh\tfrac12s$ is also an algebraic number.

Title: Edge length of hyperbolic tesselations


问:
Sub-Gaussian norm is proper norm
I am trying to show that the sub-Gaussian norm is a proper norm:

$$\|X\|_{\psi_2}=\text{inf}\bigg\{t>0:\mathbb{E}\bigg[\text{exp}\bigg(\frac{X^2}{t^2}\bigg)\bigg]\bigg\}$$

I was able to show the triangle inequality, and just want to make sure my logic on showing the homogeneity property is correct:

$$\|cX\|_{\psi_2}=\text{inf}\bigg\{t>0:\mathbb{E}\bigg[\text{exp}\bigg(\frac{c^2X^2}{t^2}\bigg)\bigg]\bigg\}$$

Def $k=\frac{t}{|c|}$ then

$$\|cX\|_{\psi_2}=\text{inf}\bigg\{|c|k>0:\mathbb{E}\bigg[\text{exp}\bigg(\frac{X^2}{k^2}\bigg)\bigg]\bigg\}=|c|\text{inf}\bigg\{k>0:\mathbb{E}\bigg[\text{exp}\bigg(\frac{X^2}{k^2}\bigg)\bigg]\bigg\}=|c|\|X\|_{\psi_2}$$

Is this  correct?

答: You proof is correct, and basically uses the fact that $\inf_{a\in A} \lvert c\rvert a=\lvert c\rvert \inf_{a\in A} a$.

Title: Sub-Gaussian norm is proper norm


问:
Convergence almost surely and SLLN
I'm studying probability theory, especially about limit theorem.
And I got some trouble in moving forward.

The problem is:

Let $(X_n)$ be i.i.d random variables with $E(X_1)=0$ and $E(|X_1|^p)<1$ for some $1<p<2$. Show that $\displaystyle \frac{S_n}{n^{1/p}}$ converges to $0$ almost surely, where $S_k=X_1+...+X_k$.

At first, I tried with Strong Laws of Large Numbers, but it doesn't help. I think the condition that $E(|X_1|^p)<1$ for some $1<p<2$ is crucial, but I don't figure out what it means.

Please give some advice!

答:
First note that $$\frac{S_{n}}{n^{\frac{1}{p}}}=\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}$$
Now, since that $(X_{n})$ are identically distributed, and since that $$\mathbb{E}[|X_{1}-0|^{p}]<1<\infty \implies \text{absolutely integrable:} \quad  \mathbb{E}[X]<\infty$$
So, by The Strong Law of Large Number (SLLN), we can conclude that $$\frac{S_{n}}{n}\overset{a.s}{\to}\mathbb{E}[X]=0$$
So, $$\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}\overset{a.s}{\to} 0\cdot c =0$$

Title: Convergence almost surely and SLLN


问:
Find the limit as $n \rightarrow \infty$ of $\sum_{k=1}^{n} \frac{k^3}{n^4}$
I would appreciate suggestions or hints for this homework question.

I'm asked to find the limit as $n \rightarrow \infty$ of the above series. The question is in a chapter on Riemann integration, so I assume that I need to formulate it as a Riemann sum (although I could be wrong about that).

So far, I have expanded the sum out to the partial, $S_{N} = \frac{1}{N^4} (1^3 + 2^3 + ... + (N-1)^3 + N^3)$.

But I'm not sure where to go from here.

答:
Use Stolz-Cesaro (look it up).

In this case it says
$\lim_{n \to \infty} \dfrac{\sum_{k=1}^n k^3}{n^4}
=\lim_{n \to \infty} \dfrac{n^3}{n^4-(n-1)^4}
=\lim_{n \to \infty} \dfrac{n^3}{4n^3-6n^2+4n-1}
=\dfrac14
$.


问:
The closure of $X = [0,1) \cup (1,2] \cup \{3\}$ and the closure of its complement
For $X = [0,1) \cup (1,2] \cup \{3\}$. I can say that $\overline{X} = [0,3]$ and the $\overline{X^c} = 1$?

答:
No.

$\overline X = [0,2] \cup \{3\}$

$X^c = (-\infty,0) \cup \{1\} \cup (2,3) \cup (3,\infty)$

$\overline{X^c} = (-\infty,0) \cup \{1\} \cup (2,\infty)$

A good intuition for whether some element $s$ is in the closure $\overline S$ of a subset $S$ of a metric space is whether you can find a sequence $\{s_n\}_{n \in \Bbb N}$ of elements of $S$ such that $s_n \to s$. If such a sequence $\{s_n\}_{n \in \Bbb N} \subseteq S$ exists such that $s_n \to s$, then $s \in \overline S$.

(This follows readily from the definition of $\overline S$ as $S \cup S'$, where $S'$ denotes the set of limit points. Apply the definition of sequence convergence in a metric space. Note in particular that this means $S \subseteq \overline S$: just take constant sequences in $S$ to justify this or use the original definition.)

So to see why your answers are wrong:

Can you find a sequence in $X$ converging to anything in $(2,3)$? Say, to $2.5$?

Knowing what $X^c$ is, it should be not difficult to see why you have much more than just $\{1\}$ in the closure - and, again, in particular, $X^c \subseteq \overline{X^c}$. But you can also trivially get a sequence converging to $3$, say the sequence
$$s_n = 2 + 9 \sum_{k=1}^n \left( \frac{1}{10} \right)^k = 2.\underbrace{999\ldots 999}_{n \; \text{dots}}$$
which lets us include that too.


问:
Asymptotic expansion of the inverse of $x\mapsto x+x^{\small\sqrt2}+x^2$ near zero
This is a follow-up to my previous question "Asymptotic expansion of the inverse of $x\mapsto x+x^\phi$ near zero".

Consider a continuous real-valued monotone increasing function $f:\mathbb R^+\to\mathbb R^+$ satisfying $f\big(x+x^{\small\sqrt2}+x^2\big)=x.$

I am interested in an asymptotic expansion of $f(z)$ for $z\to0^+$ in terms of powers of $z$. I was able to find a few initial terms by manually balancing coefficients:
$$f(z)=z-z^{\small\sqrt2}+\sqrt2\;z^{\small\unicode{x202f}2\unicode{x202f}\sqrt2-1}-z^2+\mathcal O\big(z^{\small\unicode{x202f}3\unicode{x202f}\sqrt2-2}\big), \quad z\to0^+.\tag{$\diamond$}$$
Computing next terms in an ad hoc fashion quickly becomes tedious, so I am looking for a more systematic approach that would allow to obtain a general formula for the terms of this series. I expect it to be a mix of integer powers of $z$ and irrational powers involving $\sqrt2$. Also, I would like to know the radius of convergence of that series.

More generally, I am looking for a uniform approach for inverting generalized polynomials of a single variable that may contain both rational and irrational powers.

答:
Too long for a comment.

Look at $$f(x)=x(1+x^{\sqrt2-1}+x)\in S=\Bbb{R}[[x,x^{\sqrt2-1}]]$$
The exponents are well-ordered.

Let $$\phi : x\to f(x), \ x^{\sqrt2-1} \to x^{\sqrt2-1} \sum_{k\ge 0} {\sqrt2-1 \choose k} (x^{\sqrt2-1}+x)^k$$
Then $\phi$ is an injective endomorphism of $S$, since $\phi(x^a)= x^a(1+O(x^{\sqrt2-1})+O(x))$ then $\phi$ is surjective. Whence $\phi(f^{-1}(x))=x$ for some $f^{-1}(x)\in S$.

It remains to find the coefficients one by one as you did. I don't see much reason to believe that $f^{-1}(x)$ converges for some $x>0$.


问:
Exactness and obstruction in sheaf cohomology
Given a topological space $X$, sheaf cohomology 'measures' the lack of exactness of the global section functor $\Gamma(X, -) : \textbf{Sh}(X) \to \textbf{Ab}$.

From another viewpoint, sheaf cohomology should be measuring the 'obstruction to lifting local to global data'.

I understand the notion of a sheaf as a local assignment to a topological space of algebraic structures that compatibly 'restrict' and 'glue'; and the global sections functor that maps $\mathcal{F} \mapsto \mathcal{F}(X)$.

Yet, I don't understand the connection between the exactness of $\Gamma(X, -)$ and the capability to 'lift local data to global'.

How do these two viewpoints connect?

答:
As another viewpoint, I would propose considering Čech complexes of the sheaf. These give rise to a more refined version of the "local-to-global" principle in sheaves. This became way longer than I wanted, but it was fun to write anyway.

Intro

In a lot of practical cases, if you have a sheaf $S$ over a space which is "sufficiently small" or "looks sufficiently like $\mathbb{R}^n$", then the sheaf cohomology of $S$ vanishes in nonzero degree: Think about the cohomology of the constant sheaf, or the sheaf of holomorphic functions. We often understand this as "topologically, this space isn't very interesting".

The cohomology of these sheaves certainly doesn't vanish over every space, but we can often cover our topological spaces by "good" covers, where every element (and every intersection of elements) in the cover is an open set over which the sheaf cohomology becomes trivial -- well, the quotation marks aren't even necessary, people really call these covers "good covers" :) Very often, these good covers will consist of open sets where every participant (all the sets themselves, and all their intersections) is homeo-/diffeo-/biholomorphic to an open disk.

Mathematics

Let us make this precise: Let $M$ be a topological space, $S$ a sheaf on $M$ with values in the category of abelian groups, and $\mathcal{U}$ an open cover of $M$ so that the sheaf cohomology of $S$ vanishes when restricted to any intersection of elements in $\mathcal{U}$, so
$$H^i(S|_{U_{i_0} \cap \dots \cap U_{i_n}}) = 0 \quad \forall U_i \in \mathcal U, i \geq 1.$$
Then $\mathcal{U}$ is a Leray cover for the sheaf $S$, and by the Leray theorem, the sheaf cohomology of $S$ is isomorphic to the cohomology of the Čech complex associated to the open cover $\mathcal{U} = \{U_i\}_{i \in I}$, which has the shape

$$ \prod_i S(U_i) \to \prod_{i,j} S(U_i \cap U_j) \to \prod_{i,j,k} S(U_i \cap U_j \cap U_k) \to \dots $$

The differentials in this complex are certain antisymmetrizations of the restriction maps of the sheaf (writing down the details is tedious, but they are easy to look up and the idea is easy to imagine, I'd argue).

Now we can deduce how the sheaf cohomology groups can be read as "local-to-global"-obstructions! The sheaf properties of $S$ immediately show that the above complex has zeroeth cohomology equal to $S(M)$: If an element in $\prod_i S(U_i)$ is in the kernel of the first map, that means its components are equal on intersections. By the gluing axiom this means they glue to an element in $S(M)$, and by the local identity axiom this element is unique.

We can try to do the same with elements $(x_I) \in \prod_{i_1,\dots,i_k} S(U_{i_1} \cap \dots \cap U_{i_k})$ in the higher groups of the complex: Here, the condition of being in the kernel of the differential will be some complicated equations on the restrictions of the elements $x_I$ to intersections of the $U_i$; whatever these look like, they are basically the higher-order-equivalent of "the components of $(x_I)$ agree on intersections". And, of course, these "agreeing-on-intersections"-conditions are necessary for $(x_I)$ to come from an element in the $(k-1)$-part of the complex, the higher-order-equivalent of gluing.

But now, a nontrivial sheaf cohomology group tells us: Even if the higher-order-"agreeing-on-intersections"-condition is fulfilled, the $(x_I)$ may not be higher-order-gluable! Now, this viewpoint does not directly read as a local-to-global condition, but it is a kind of "very-local-to-a-bit-less-local"-condition, in the sense that we try to glue data from more refined intersections to less refined intersections.


问:
Bezier curves algorithm complex yet simple
I've got a path that is made up of multiple points - i.e. 0,0  0,50  50,50  70,20

If I just draw this line on the screen it looks quite harsh as it puts a sharp angle at the joining of each point.

Hence I was wondering what the bezier curves algorithm/method would look like which I could call that automatically change the sharp angles to "tight" curves?

I wouldn't want the curve to be too big or generally effect the fall of main path, just soften the join out. If you have a look at the below here is a quick sample I have put together. The line on the left is what I have now, the one in the middle is the line I want.

The image on the right represents what I think I need the algorithm to do. Essentially, I add an additional point to each arc that makes up the join at a point which 10% away from the join, then I remove the join point and adjust the handles so that they are where the point was (not in the diagram they are slightly apart, this is just so that you can see). This is what I need to be able to do.

答: Well, Bézier curves are always guaranteed to lie within the convex hull of the control points; it is only a matter of taste/convenience for you to adjust the control points to get the shape you wish. Try finding a program that allows you to interactively move control points around, and experiment!</p>

ContentLicense: CC BY-SA 2.5


问:
Real Analysis Of Taylor Polynomial
I have a random function $f(x)$. I know nothing about the shape of the function, but I know

$f'(x)$ exists, $x \in \mathbb{R}$

$f(3)=1$, $f'(3)=5$, $f''(3)=7, f''''(3)=3$

If $x\in (3,6)$, then $f'(x) \in[1,5]$

$-3\leq f''(x)\leq 9,  \forall x \in \mathbb{R}$

$-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$

So having this information, how do I determine the second degree taylor polynomial, $T_{2, 3}$ for this function? I know the value of $f'(3)$ up to $f''''(3)$, which is perfect for this question. But the taylor polynomial theorem requires the function is differentiable everywhere in order for this to work right? Do I need to prove that first, and how should I prove it?

Also, how do we find the upper bound for the error using $T_{2,3}$ for f(5)? Using taylor's remainder theorem is fine but which points should I choose for $f'''(c)$?

答:
Second degree taylor polynomial: $1+5(x-3)+\frac{7}{2}(x-3)^2$

Upper bound on error for $f(5)$: $\frac{16}{3}$

We already know the function is differentiable everywhere, because $f'(x)$ exists $x \in R$

Because the function is differentiable, we can write out its Taylor series expansion (centered at 3 because that's where we know the derivatives, and up to a degree of 2 because we're only trying to find the second degree Taylor polynomial).

$f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2=1+5(x-3)+\frac{7}{2}(x-3)^2$

To find the upper bound on the error, we use Taylor's Remainder Formula:

if $\left|f^{(k+1)}(z)\right|\leq M$ for all z between x and a, then
$$\left|R_k(x)\right|\leq \frac{M}{(k+1)!}|x-a|^{k+1}$$
Where $R_x$ is the error in the approximation

So in our case $k=2$, $\left|f^{(3)}(z)\right|\leq M = 4$ (because $-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$), $x=5$, and $a=3$, so $$\left|R_2(x)\right|\leq \frac{4}{(2+1)!}|5-3|^{2+1}=\frac{16}{3}$$

Title: Real Analysis Of Taylor Polynomial


问:
$log(X)$ is normally distributed. how to find CDF of $X$?
i know that $y = ln(X)$ is normally distributed with a certain mean and standard deviation.

how do i find the CDF of $X$?

suppose my CDF of $y = F(y)$.

my initial idea is that i just calculate $e^{ln(X)}$ because this would give $X$. then i would take this and put it into the CDF for $y$, so $F(x) = F(e^{ln(X)}) = F(x)$. but this is not correct.

another idea that i just came up with is to put the CDF of $y$ into the exponent of $e$, so $e^{F(y)}$. when i do this and visualize it i observe that $e^{F(y)}$ is normally distributed.

is this second approach the correct one? or am i still wrong?

thank you in advance for any insight.

答:
CDF of $X$ is the probability $F_X(x) = \Pr(X\le x)$.

For $x \le 0$ the probability is zero, and otherwise it's equal to $\Pr(\log X \le \log x) = F_Y(\log x)$.


问:
Subspaces being homeomorphic implies complements being homeomorphic?
If $U$ and $V$ are subspaces of $X$ and $U$ and $V$ are homeomorphic, is it true that $X \setminus U$ and $X \setminus V$ are homeomorphic?

答: Not in general. $\{0,3\}$ and $\{1,2\}$ are homeomorphic subspaces of $[0,3]$, but $\{0,3\}\setminus\{0,3\}$ is connected, while $[0,3]\setminus\{1,2\}$ is not.


答: This structure is called a multigraph. Multigraphs are graphs except you can have an arbitrary number of edges between any pair of vertices and you can have edges where both ends are the same (this is called a loop).</p>


问:
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
In the below solved problem, every thing is okay, but if we have $4$ consonants then why we are giving $5!$? and is this a combination problem? how to distinguish?

Question:
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Answer:
The word 'OPTICAL' contains $7$ different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, $5$ letters can be arranged in $5! = 120$ ways.
The vowels (OIA) can be arranged among themselves in $3! = 6$ ways.
Required number of ways $= (120*6) = 720$.

答:
For this problem I use the multinomial coefficient.
We have 7 letters in total 3 of which are vowels and 4 are consonants

The number of vowels combinations are 3! and the number of consonants combinations are 4! (I've used the multinomial coefficient)
So by the end we've 3!*4! ways


答:
In general, no. Take $X=\mathbb{R}$, $U=(0,1)\cup (1,2)$, and $V=(0,1)\cup(2,3)$. Then $U$ and $V$ are homeomorphic, but $X\setminus U$ is not homeomorphic to $X\setminus V$: in the latter, in each of the connected components one can remove a single point and obtain four connected components in the result, but this is not possible in the former.

But what you want to prove does not require such a strong (yet false) claim. There is a linear automorphism of $\mathbb{R}^4$ that sends $\{(0,0,x,y)\mid x,y\in\mathbb{R}\}$ to $\{(x,y,x,y)\mid x,y\in\mathbb{R}\}$, and hence complements to complements. Being a linear map, it is continuous; being a linear automorphism, it has a continuous inverse.

Title: Subspaces being homeomorphic implies complements being homeomorphic?

Body: If $U$ and $V$ are subspaces of $X$ and $U$ and $V$ are homeomorphic, is it true that $X \setminus U$ and $X \setminus V$ are homeomorphic?


问:
How to prove |– (U → V ) → ((¬U → V ) → V)? Or how to deduce V from (U → V ), (¬U→V) using Deduction theorem?
This is what I need to ultimately prove: |– (U → V ) → ((¬U → V ) → V)

We have these axioms and modus ponens:

Axiom 1 is: P→(Q→P)

Axiom 2 is: (P→(Q→R))→((P→Q)→(P→R))

Axiom 3 is: (¬Q→¬P)→ (P→Q)

Modus ponens is: from P and P→Q infer Q

A proof in this manner means a sequence of lines where each line is an axiom, or a formula from your "base" set ({α,¬α} below), or an inference (using modus ponens), with the last line being the formula being proved.

I used the deduction theorem to say that a finding a proof from the empty set of that formula is equivalent to finding a proof:

(U → V ), (¬U→V) |– V

Using these axioms and modus penons I want to infer V from (U → V ), (¬U→V)

I'm a bit stuck here. Any help please?

答:
The proof that I know of is fairly tricky, and consists on bootstrapping up several results from axiom 3 in turn, using the Deduction Theorem.

Lemma schema 1: $\vdash \lnot A\rightarrow(A \rightarrow B)$.

Proof: By the Deduction Theorem, it suffices to show $\lnot A \vdash A \rightarrow B$.  Then, using axiom 3 and a modus ponens, we reduce to showing that $\lnot A \vdash \lnot B \rightarrow \lnot A$.  This is now easy. $\square$

Lemma schema 2: $\vdash (\lnot A\rightarrow A)\rightarrow A$.

Proof: By the Deduction Theorem, it suffices to show that $\lnot A\rightarrow A\vdash A$.  Now, in general, for any propositional formula $P$, we have that $\lnot A\rightarrow A, \lnot A \vdash P$ through concluding $A$ as well, and then applying lemma schema 1.  But then, by the Deduction Theorem, we conclude that $\lnot A\rightarrow A \vdash \lnot A \rightarrow P$.  In particular, $\lnot A\rightarrow A \vdash \lnot A \rightarrow \lnot(\lnot A \rightarrow A)$; and now, combining this fact with axiom 3 and an application of modus ponens, we then see that $\lnot A\rightarrow A \vdash (\lnot A\rightarrow A)\rightarrow A$.  Since $\lnot A\rightarrow A \vdash \lnot A\rightarrow A$ also, a modus ponens gives the required conclusion $\lnot A \rightarrow A \vdash A$. $\square$

Lemma schema 3: $\vdash \lnot\lnot A \rightarrow A$.

Proof: By the Deduction Theorem, it suffices to show that $\lnot\lnot A \vdash A$.  Now, $\lnot\lnot A \vdash \lnot A\rightarrow A$ using lemma schema 1 and a modus ponens.  Now, from lemma schema 2 (and extension of context) we also have $\lnot\lnot A \vdash (\lnot A\rightarrow A)\rightarrow A$, and a modus ponens gives the desired conclusino $\lnot\lnot A \vdash A$. $\square$

Lemma schema 4: $\vdash A \rightarrow \lnot\lnot A$.

Proof: Using axiom 3 and a modus ponens, it suffices to show $\vdash \lnot\lnot\lnot A \rightarrow \lnot A$.  This is then a specialization of lemma schema 3. $\square$

Lemma schema 5: $\vdash (A\rightarrow B)\rightarrow(\lnot B\rightarrow \lnot A)$.

Proof: By the Deduction Theorem, it suffices to show $A\rightarrow B \vdash \lnot B\rightarrow \lnot A$.  Using axiom 3 and a modus ponens, we reduce to showing $A \rightarrow B \vdash \lnot\lnot A \rightarrow \lnot\lnot B$.  By another application of the Deduction Theorem, it suffices to show $A\rightarrow B, \lnot\lnot A \vdash \lnot\lnot B$.  But then, lemma schema 3 and a modus ponens gives $A \rightarrow B, \lnot\lnot A \vdash A$; a modus ponens gives $A \rightarrow B, \lnot \lnot A \vdash B$; and finally, lemma schema 4 and a modus ponens gives from this that $A\rightarrow B, \lnot\lnot A \vdash \lnot\lnot B$. $\square$

Final result: $\vdash (U\rightarrow V) \rightarrow [(\lnot U\rightarrow V)\rightarrow V]$.

Proof: By two applications of the Deduction Theorem, it suffices to show $U\rightarrow V, \lnot U\rightarrow V \vdash V$.  But by lemma schema 5, $U\rightarrow V, \lnot U\rightarrow V \vdash \lnot V\rightarrow \lnot U$; and combining with the assumption $\lnot U \rightarrow V$, an easy argument gives $\lnot V\rightarrow V$.  But now by lemma schema 2, $U\rightarrow V, \lnot U\rightarrow V \vdash (\lnot V\rightarrow V)\rightarrow V$ also; and a final modus ponens gives what we wanted. $\square$


问:
An easy computation of the laplacian of $\lvert{f(z)\rvert}^2$ for a holomorphic function $f$.
I'm reading Narasimhan's book "Complex Analysis in one complex variable" ($2$nd ed).
After proving the weak form of the Maximum Principle (p. $25$, Theorem $3$), he applies the result to the function $u(z)=\lvert{f(z)\rvert}^2=f(z)\overline{f(z)}$, and gets the equality (p. $26$, $2$nd row):
$$\frac{\partial^2 \lvert{f\rvert}^2}{\partial z \partial\overline{z}}=\frac{\partial f}{\partial z}\cdot \frac{\partial \overline{f}}{\partial \overline{z}}.$$

Recall that $\frac{\partial}{\partial z}=\frac{1}{2}(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}(\frac{\partial}{\partial x}+i \frac{\partial}{\partial y})$ by definition.

After many long computations (and a lot of time spent on it), I hope someone can help me to prove this apparently easily derived equality.

Please, tell me if I need to elaborate further my question.

答:
$$\frac{\partial }{\partial z}\left(\frac{\partial }{\partial \overline{z}}f\overline{f}\right)=\frac{\partial}{\partial z}\left(f\frac{\partial \overline{f}}{\partial z}\right)=\frac{\partial f}{\partial z}\cdot\frac{\partial \overline{f}}{\partial \overline{z}},$$
because $\frac{\partial f}{\partial \overline{z}}=\frac{\partial \overline{f}}{\partial z}=0$ from Cauchy-Riemann equation.


问:
What does this probability tree tells about the problem?
I have this image of probability tree but I am unsure of what problem it solves.

From the given data I am assuming that it is telling that there's a six-sided die and it is thrown three times to get a desired number e.g. '6'.

So,

$=\Large\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$

Would this be the probability needed for getting '6' on a third trial? Otherwise, what does it represent? Assuming a die is involved and we are finding probability of getting a number.

答: At each stem, the number represents the probability of that event happening. So $$\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$$ represents the probability of getting your desired number $6$ at your third throw.


问:
A question about covering number and maximal separated subset.
I am   thinking about the following question of covering number and maximal separated subset:

Let $X$ be a Banach space and let $B_X(y,r)$ denote the ball of $X$ whose center is at $y$ with radius $r$.

(Definition of covering number)

Denote $N_{\epsilon}(A,X):=\min\{n: \exists  y^1, y^2, ... ,y^n, where\ \ each \ \ y^j\in X: A\subset \cup_{j=1}^n B_X(y^j,\epsilon)\}$.

(Definition of  maximal separated subset)

Denote $F\subset A$ be the maximal $2\epsilon$-separated subset of $A$ in the metric of $X$, i.e. $||x-y
 ||\geq2\epsilon$ for any  two distinct  $x,y\in F$, and $A\subset \cup_{x\in F}B_X(x,2\epsilon)$.

Prove that:
$$|F|\leq N_{\epsilon}(A,X) $$ where $|F|$ is  the cardinality of  the finite set $F$.

My idea is that for the maximal $2\epsilon$-separated subset of $A$, the set $A$ is covered by the balls with radius at least $2\epsilon$. While for the cover number,  we use a set of balls with a smaller radius $\epsilon$ to cover $A$.  My intuition is that if the radius becomes smaller, the number of balls we used could be larger. But I don't know how to give a good proof since the cover number is considering the minimum of the cover.

Any suggestions would be welcome and thank you in advance!!

答:
We can assume w.l.g that $N_\epsilon(A,X)<\infty$ (otherwise the inequality is obvious). Note that for a non totally bounded set $A$ then $N_\epsilon(A,X)=\infty$ it will happen for some $\epsilon's$. First we claim that if $N_\epsilon(A,X)<\infty$ then $2\epsilon-$separated subsets of $A$ exist. Indeed, pick an $\epsilon-$cover of $A$, say $\{y^1,...,y^n\}$ (we can assume, probably by adding more points that $\{y^1,...,y^n\}\subseteq A)$ such that
$$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,\epsilon)$$
If the above balls are disjoint, then $||y^j-y^i||\geq \epsilon$ for every $i\neq j$. Since,
$$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,2\epsilon)$$
Then, $\{y^1,...,y^n\}$ is an $2\epsilon-$separated subset of $A$. If the above balls are not disjoint, then let
$$A_1=\{y^j:\, B_X(y^j,\epsilon)\cap B_X(y^1,\epsilon)\neq \emptyset,\ j=1,...,n\}$$
If $A_1$ contains only $y^1$ we keep the ball $B_X(y^1,\epsilon)$ which is disjoint from all the others and we continuing with the point $y^2$ and the set $A_2$ defined in similar fashion as with $A_1$. So, suppose that $A_1$ contains also other points than $y^1$. We claim that
$$\tag{1}\bigcup_{y\in A_1}B_X(y,\epsilon)\subseteq B_X(y^1,2\epsilon)$$
Indeed, if $z\in B_X(y,\epsilon)$ for some $y\in A_1$ then, since $B_X(y^1,\epsilon)\cap B_X(y,\epsilon)\neq \emptyset$ we can pick an elememnt $w$ in the intersection of the two balls. Then, by triangle inequality,
$$|z-y^1|\leq |z-w|+|w-y^1|<2\epsilon$$
Hence, $z\in B_X(y^1,2\epsilon)$ and $(1)$ is justified. Now, from the set $A_1$ we keep only $y^1$ and we throw all the other points. Now, by considering the set $\{y^1,...,y^n\}\setminus A_1$ if this is empty we will have $A\subseteq B_X(y^1,2\epsilon)$ and $\{y^1\}$ is an $2\epsilon-$separated subset of $A$. Otherwise, we pick a point $y\in \{y^1,...,y^n\}\setminus A_1$ and we define the set
$$A_2=\{y^j:\, B_X(y^j,\epsilon)\cap B_X(y,\epsilon)\neq \emptyset\}$$
and apply the same procedure as before. Continuing this way, with a recursive argument we end up with a subset $F'\subseteq \{y^1,...,y^n\}$ such that
$$A\subseteq \bigcup_{y\in F'}B_X(y,2\epsilon)$$
with the above union being disjoint, meaning that $|y-y'|\geq \epsilon$ for every $y\neq y'$ in $F'$. Hence, $F'$ is an $2\epsilon-$separated subset of $A$.

Now, that we justified that $2\epsilon-$separated subsets of $A$ exists, we can pick a random $F'=\{x^1,...,x^m\}\subseteq A,\ 2\epsilon-$separated subset of $A$, and a random $\{y^1,...,y^n\},\ \epsilon-$cover of $A$ (this time we dont assume that $\{y^1,...,y^n\}\subseteq A$ otherwise this restriction will violate the generality of the argument). We claim that $m\leq n$. If we prove this, we would have prove that $|F|\leq N_\epsilon(A,X)$. Now,
$$A\subseteq \bigcup_{j=1}^{n}B_X(y^j,\epsilon)$$
Now, since every $x^k$ belongs to $A$ there will exists $y^k$ such that $x^k\in B_X(y^k,\epsilon)$. Define the function $\phi:\{x^1,...,x^m\}\to \{y^1,...,y^n\}$ by $\phi(x^k)=y^k$ for every $k=1,...,m$. Our claim is that $\phi$ is one to one. Indeed, if $\phi(x^k)=\phi(x^l)$ then, $x_k,x_l\in B_X(y_k,\epsilon)$. This implies that
$$|x^k-x^l| \leq |x^k-y^k|+|x^l-y^k|<2\epsilon$$
and now using the fact that $|x^k-x^l|\geq 2\epsilon$ for every $k\neq l$ we end up with $x_k=x_l$. Hence, $\phi$ is one to one which means that $m\leq n$ as desired.


问:
An expression for the multilinear joint XOR polynomial over $\{1,-1\}^n$
Preliminary: If $g:\{0,1\}^n\rightarrow \{0,1\}$ then there is a unique multilinear polynomial $f:\{1,-1\}^n\rightarrow \{1,-1\}$ such that $f((-1)^{a_1},\ldots,(-1)^{a_n})=(-1)^{g(a)}$ of all $a\in \{0,1\}^n$. Indeed, one only needs to consider the indicator polynomial $$f(x_1,\ldots,x_n)=\sum_{a\in \{0,1\}^n}(-1)^{g(a)}\prod_{i=1}^n\left(\frac{1+(-1)^{a_i}x_i}{2}\right)$$ and recall that the monomials $x^S:=\prod_{j\in S}x_j$ for $S\subseteq [n]$ form a basis for the vector space of such functions. Hence, we can express $$f(x_1,\ldots,x_n)=\sum_{S\subset [n]} \hat{f_S} x^S,$$ where $\hat{f_S}$ are the Fourier coefficients (basis coefficients for $f$).

Question: Let $g$ be the XOR function i.e. $g_n(a)=1$ if $a\in \{0,1\}^n$ contains a single $1$ otherwise $g_n(a)=0$. Can we find a nice expression for the corresponding $f(x_1,\ldots,x_n)$ of the joint XOR function $g_n$?

eg. If $n=2$ we have $g_2(a_1,a_2)=a_1\oplus a_2$ and we can see that $f_2(x_1,x_2)=x_1x_2$.

So far I have tried to obtain a simplified formula by starting directly with the indicator polynomial for the XOR function but I haven't had any luck... even a formula for $n=3$ would be appreciated.

Body:
Let $g_n$ be the $n$-th XOR function and $f_n$ a corresponding polynomial. Take $x = (x_1, ..., x_n)$ and $y$. I provide a recursive formula.

\begin{align*} f_{n+1}(x, y) =& \sum_{a\in \{0,1\}^n}(-1)^{g_{n+1}(a, 0)}\prod_{i=1}^n\left(\frac{1+(-1)^{a_i}x_i}{2}\right)\left(\frac{1+y}{2}\right)\\& + \sum_{a\in \{0,1\}^n}(-1)^{g_{n+1}(a, 1)}\prod_{i=1}^n\left(\frac{1+(-1)^{a_i}x_i}{2}\right)\left(\frac{1-y}{2}\right)\\ =&f_n(x)\left(\frac{1+y}{2}\right)-2\prod_{i=1}^n\left(\frac{1+x_i}{2} \right)\left(\frac{1-y}{2}\right)+\\& \sum_{a\in \{0,1\}^n}\prod_{i=1}^n\left(\frac{1+(-1)^{a_i}x_i}{2}\right)\left(\frac{1-y}{2}\right)\\ =& f_n(x)\left(\frac{y+1}{2}\right)+\left(\frac{1}{2^{n-1}}\prod_{i=1}^n\left(1+x_i \right)-1\right)\left(\frac{y-1}{2}\right)\end{align*}

For $n = 1$, $g_1 \equiv \text{Id}$ and $f_1(x_1) = x_1$.

For $n = 2$, $f_2(x_1, x_2) = x_1\left(\frac{x_2+1}{2}\right)+x_1\left(\frac{x_2-1}{2}\right) = x_1x_2$.

For $n = 3$, $$f_3(x_1, x_2) = x_1x_2\left(\frac{x_3+1}{2}\right)+\left(\frac{1}{2}x_1x_2+\frac{1}{2}x_1+\frac{1}{2}x_2-\frac{1}{2}\right)\left(\frac{x_3-1}{2}\right) = \frac{1}{4}[3x_1x_2x_3+x_1x_2+x_1x_3+x_2x_3-x_3-x_1-x_2+1] = \frac{1}{4}[3p_1+p_2-p_3+1]$$ where $(x+x_1)(x+x_2)(x+x_3) = x^3+p_3x^2+p_2x+p_1$ (elementary symmetric polynomials).


答:
The "success" labels on the tree suggest that the problem may be to calculate the probability that you get at least one of your desired outcomes in the three tries.

If that's the right interpretation then you fail with probability
$(5/6)^3 = 125/216$ and succeed with the complementary probability $91/216$.

The path outlined in green is the case in which you succeed  on the third trial. You have correctly computed that probability.

Note that you need not be looking for the same number each time. Success might mean seeing $1$ on the first roll or $2$ on the second or $3$ on the third.

Edit to respond to this comment:

Just a bit of confusion. If trials are  $T1$, $T2$ and $T3$. I noticed that
probability of $P(T2|T1)=P(T2)$. So, events are both independent and
conditional?

I think that's not quite right.
"Both independent and conditional" doesn't make sense.

You have to think carefully about the sample space. If it's the full tree of possibilities with three dice rolls looking for a $1$ (say) then there are $2^3 = 8$ possible states, corresponding to success and failure on each roll. Then the probability of success on any particular roll is independent of what happens on any other. No roll gives information about another. So
$$
P(\text{success on roll } 2 \, | \, \text{success on roll } 1)= P(\text{success on roll } 2).
$$
But the picture suggests that you stop the experiment as soon as you see one success. Then the sample space has $4$ elements, corresponding to the $4$ leaves of that tree. If roll $1$ succeeds you don't roll a second time, so
$$
P(\text{success on roll } 2 \, | \, \text{success on roll }  1)
$$
doesn't make sense. You could set its probability to $0$. The probability of the subset of the sample space that corresponds to the existence of a second roll is not independent of what happens on the first roll. It's $5/6$ before that roll and either $1$ or $0$ after that roll. So
$$
P(\text{success on roll } 2 \, | \, \text{failure on roll }  1) = 1/6
$$
while
$$
P(\text{success on roll } 2)= \frac{5}{6}\frac{1}{6}.
$$

Title: What does this probability tree tells about the problem?

Body:
I have this image of probability tree but I am unsure of what problem it solves.

From the given data I am assuming that it is telling that there's a six-sided die and it is thrown three times to get a desired number e.g. '6'.

So,

$=\Large\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$

Would this be the probability needed for getting '6' on a third trial? Otherwise, what does it represent? Assuming a die is involved and we are finding probability of getting a number.


问:
Square-Trangular Numbers Checking Answer
Problem: The first 2 numbers that are both squares and triangles are 1 and 36. Find the next one and if possible, the one after that.

Answer: 1225, 41616

Problem: Can you figure out an efficient way to find triangular-square numbers?

Answer: $s^2 = t(t+1)/2$, where $s,t \epsilon{} \textbf{Z}$.

Problem: Do you think that there are infinitely many?

Answer: Yes. (Can someone give me an justification for my answer?)

答:
Your answer is correct: the third and fourth both-triangular-and-square numbers are 1225 and 41616. There are infinitely many such numbers, all of them being solutions to the equation $s^2 = \frac{1}{2}t(t+1)$ in integers. What you didn't provide is an efficient method for finding further solutions to this equation.

The equation can be rewritten as $(2t+1)^2 - 8s^2 = 1$, which is just a slightly disguised Pell's equation. There are a few general methods for solving such equations. Applying one of them to your case yields a simple recurrence for values of $s$: $s_0=0$, $s_1=1$, $s_{k+2} = 6s_{k+1} - s_k$.


问:
Why does elements in the parabolic subgroups of $SL(n,k)$ take the form of upper block triangular matrices?
Let $G$ be either $SL(n,k)$ (or I guess any linear subgroup of $GL(n,k)$) for a field $k$. And $P$ be a parabolic subgroup of $G$, I have seen the fact that any $A\in P$ looks like the block diagonal matrix:

$$ A = \left( \begin{matrix} A_1 & * & * \\ 0 & \ddots & * \\ 0 & 0 & A_m \end{matrix} \right) $$

But how to prove this fact? Any solution or reference will be appreciated!

答:
Do you know how to show that the set of  upper triangular matrices $B$ of $G= \operatorname{SL}_n(k)$ forms a Borel subgroup of $G$?  This follows from the Lie-Kolchin theorem.

A parabolic subgroup of $G$ is by definition a subgroup $P$ of $G$ such that $gPg^{-1}$ contains $B$ for some $g \in G$.  So what you want to show is that all subgroups of $G$ containing $B$ are of the block diagonal form.

If you let $T$ be the group of diagonal matrices in $G$, and $N$ the normalizer of $T$ in $G$, this follows by showing that $(G,B,N)$ form a Tits system.  In any Tits system, the subgroups of $G$ containing $B$ are completely classified in terms of the Weyl group $N/(B \cap N)$.  If you write out what these are, you will see they all take the form you want.


问:
Division of a polynomial in three variables
I would like to demonstrate the following

Let $f(x,y,z) \in K[x,y,z]$ a polynomial such that $f(a^n,b^n,ab)=0$; then f is divisible by $xy-z^n$.

Hint: use a polynomial division algorithm

EDIT
At this link Showing that $\mathbb{C}[x,y]^{\mu_n}$ and $\mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic as rings
I found the same problem. The answer suggest:
$$ $$
If $f(x,y,z)=\sum_{i,j,k} a_{i,j,k}x^iy^jz^k$ so
$$0=f(u^n,v^n,uv)=\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$$
The equation $\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$ give some dependence relation amongst the $a_{i,j,k}$ (since the exponent of $u$ and $v$ are the same for some $i,j,k$ and since the element $u^av^b$ are indipendent over $K[x,y]$.

Can anyone tell me what are these relations between coefficients
$a_{i,j,k}$ they are talking of? How do they obtain? How do these
allow me to prove that $f(x,y,z)=g(x,y,z)(xy-z^n)$? TO PLEASE SOMEONE
CAN HELP ME?

答:
Here's how the polynomial division approach would work.

Lemma: Any polynomial $f\in k[x,y,z]$ can be written uniquely as $q(x,y,z)(xy-z^n) + \sum_{i=0}^{n-1} p_i(x,y)z^i$.

Proof: Induction on the highest power of $z$ appearing in $f$. Clearly any polynomial of degree $n-1$ or less in $z$ satisfies the statement, so suppose $f$ has degree $d\geq n$ in $z$. Write $f=\sum_{i=0}^d p_i(x,y)z^i$: then $$f= p_d(x,y)z^d + \sum_{i=0}^{d-1} p_i(x,y)z^i =$$ $$= p_d(x,y)xyz^{n-d}-p_d(x,y)z^{d-n}(xy-z^n)+\sum_{i=0}^{d-1} p_i(x,y)z^i$$ and so $f=-p_d(x,y)z^{d-n}(xy-z^n) + f'$, where $f'=p_d(x,y)xyz^{n-d}+\sum_{i=0}^{d-1} p_i(x,y)z^i$ has degree at most $d-1$ in $z$, so by the induction hypothesis we are done. $\blacksquare$

Now suppose $f$ is in the kernel of $\varphi:k[x,y,z]\to k[a,b]$ by $\varphi(x)=a^n$, $\varphi(y)=b^n$, $\varphi(z)=ab$. Writing it as above, we see that $\varphi(q(x,y,z)(xy-z^n))=\varphi(q(x,y,z))(a^nb^n-(ab)^n)=0$, so $f$ is in the kernel iff $$\varphi\left(\sum_{i=0}^{n-1} p_i(x,y)z^i\right)=0.$$

Distributing $\varphi$, we see that we should have $\sum_{i=0}^{n-1} p_i(a^n,b^n)(ab)^i=0$. Now I claim that no monomial $a^rb^s$ appears with nonzero coefficient in $p_i(a^n,b^n)(ab)^i$ and $p_j(a^n,b^n)(ab)^j$ for $i\neq j$: look at the residue class of the exponent of $a$ modulo $n$. If $a^rb^s$ appeared in both of those expressions with nonzero coefficient, we should have $i\equiv r \mod n$ and $j\equiv r \mod n$, or $i\equiv j\mod n$. But $0\leq i,j<n$, so $i\equiv j\mod n$ implies $i=j$. Hence the coefficient of $a^rb^s$ in $\sum_{i=0}^{n-1} p_i(a^n,b^n)(ab)^i$ comes from only one $p_i(a^n,b^n)(ab)^i$, so if all the coefficients of $a^rb^s$ are zero, then all the coefficients of $p_i$ are zero for all $i$.

But this exactly means that $f=q(x,y,z)(xy-z^n)$ and we're done.


问:
How is CS/programming used by mathematicians in industry?
I am currently writing about the relationship and I've read a lot about the relationship going the other way. However, it's usually about how mathematics taught in CS and nothing about the relationship in for example research or by mathematicians in white-collar jobs.

I would also appreciate any articles about the subject for the proper reference.

答:
Computers can be used to form conjectures such as the Birch and Swinnerton-Dyer Conjecture. You calculate a large number of values and look for patterns to form conjectures.

Another use can be to find counter-examples. One such case was Euler's Sum of Powers conjecture which was disproven by a brute force computer search.

A third use can be to prove theorems by checking a large number, but finite number of cases. One such case is the Four Color Theorem, which has only proven using a computer verified proof which could not be accomplished by humans. Software can be used to prove theorems directly as well.

But I think we should address one other point which is that the Curry-Howard
Correspondence shows that all proofs are programs and all programs are proofs. This means the idea that math being used for computer science and vice-versa might not be a meaningful question in the first place. They're both the same thing, so you're always using both of them. They're entangled with each other inherently.


问:
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$
I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$

I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2}  \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?

Body: Note that\begin{align}\int_1^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=-\int_1^0\frac{1+y^{-2}}{1+y^{-4}}\left(-\frac1{y^2}\right)\,\mathrm dy\\&=\int_0^1\frac{1+y^2}{1+y^4}\,\mathrm dy\end{align}and that therefore\begin{align}\int_0^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=\int_0^1\frac{1+x^2}{1+x^4}\,\mathrm dx+\int_1^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx\\&=2\int_0^1\frac{1+x^2}{1+x^4}\,\mathrm dx.\end{align}


问:
Majority function and Russo's Formula
I am studying Boolean Functions in Garban and Steif's book (Noise sensitivity of Boolean
functions and percolation). Eercise III.5 asks us to

Show that among all monotone Boolean functions on $\Omega_n$, $MAJ_n$ is the function with largest total influence (at $p = 1/2$).
Hint: use the Margulis-Russo formula.

Here $MAJ_n$ is the simple majority function on $n$ variables.

I have seen proofs of this facts using Fourier analysis, but the book hasn't introducted this subjec yet. I could'nt prove this using only Russo's Formula. Thank you for all the help.

答:
By abuse of notation, let $MAJ$ denote the set of sequences who have 1 as their majority element.

Now, at $p=1/2$, you can show that if $\omega \in MAJ, \frac{dP_p(\omega)}{dp} > 0$, and conversely, if it is not in it, then the derivative will be less than $0$. (Intuitively, this is because if a sequence has more 0s than 1s, as p exceeds 1/2, it will definitely be less likely. You can also compute the derivatives or plot the functions). So, by linearity of differentiation, the best $A$ here will be $MAJ$.

Title: Majority function and Russo's Formula


问:
Constraint qualification for linear constraints
How to formulate conditions under which the LI CQ is fulfilled at arbitrary feasible point of the set of feasible solutions given by linear inequalities $S=(x\in\mathbb{R}^{n}:Ax\le b, x\ge 0)$? I know I need to show that gradients of constraints are linear independent, however, it depends on the number of active inequalities. I do not know how to count with this fact. Any idea, thank you?

答:
LICQ is quite hard to show in this case since in order to show that LICQ is satisfied at every point, you need to show that for every possible active subset of the linear inequality constraints $Ax \le b$ and nonnegativity constraints $x \ge 0$, the gradients of the active constraints are linearly independent.

There are other constraint qualifications besides LICQ that you might use that could be much easier to establish.  For example, you could use linear programming to determine whether Slater's constraint qualification was satisfied.