问:
$X^3 + 22Y^3 + 3Z^3 = 0$ has no rational solution
This is Exercise 1.4 in Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory.

Write $\alpha = \sqrt[3]{22}$ and $k = \mathbb{Q}(\alpha)$. I have shown that $k$ has class number 3 and the norm form is
$$ \mathrm{norm}(a + b\alpha + c\alpha^2) = a^3 + 22b^3 + 22^2c^3 - 66abc. $$

Assuming that $(x,y,z)$ is a primitive solution of $X^3+22Y^3+3Z^3=0$ with $z > 0$, we find
$$ N(x+y\alpha) = |x^3+22y^3| = 3z^3. $$
Since $3$ is totally ramified in $k$, i.e. $3\mathcal{O}_k = \mathfrak{p}_3^3$, we can write
$$ (x+y\alpha) = \mathfrak{p}_3 \mathfrak{a}, $$
for some $\mathfrak{a}$ with $N\mathfrak{a} = z^3$.
The exercise says to show that $\mathfrak{a}$ is a cube, which would lead to a contradiciton since $\mathfrak{p}_3$ is not principal, and this is where I am having trouble.

Writing $\mathfrak{a} = \prod \mathfrak{p}^{a_{\mathfrak{p}}}$, the primes with $a_{\mathfrak{p}} \neq 0$ are above the primes dividing $z$ and I want to show that $a_{\mathfrak{p}} \equiv 0 \bmod 3$ for every $\mathfrak{p}$.
If $N\mathfrak{p} = p^{f_{\mathfrak{p}}}$, the fact that the norm of $\mathfrak{a}$ is a cube says
$$ \sum_{\mathfrak{p} \mid p} a_{\mathfrak{p}} f_{\mathfrak{p}} \equiv 0 \bmod 3.$$

If $p = 2,3$ or $11$, then $p$ is totally ramified in $k$, i.e. $p\mathcal{O}_k = \mathfrak{p}_p^3$, so $p \mid z$ implies $\mathfrak{p}_p^3 \mid \mathfrak{a}$ and the result holds.

I now consider a prime $p \neq 2,3,11$ which divides $z$.
The equation gives
$$ x^3 + 22y^3 \equiv 0 \bmod p^3, $$
and because of coprimality, we find that $22$ is a cube modulo $p$ so we cannot have $f_{\mathfrak{p}} = 3$ for any $\mathfrak{p} \mid p$. This gives $a_{\mathfrak{p}} \equiv 0 \bmod 3$ when $\mathfrak{p}$ is the only prime above $p$ dividing $\mathfrak{a}$.

The problem is that I do not see how to reach that conclusion when there are more than one prime factor of $p\mathcal{O}_k$ appearing in the decomposition of $\mathfrak{a}$. For instance, if $p\mathcal{O}_k = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$, why can we not have $\mathfrak{a} = \mathfrak{p}_1 \mathfrak{p}_2^2$?

答:
for what it may be worth, we can also ask when the integer polynomial
$$  x^3 + 22 y^3 + 484 z^3 - 66xyz \equiv 0 \pmod p $$
has no solutions except $$x,y,z \equiv 0 \pmod p$$
which is a finite check.

That happens  for small primes

$$     13,     19,     31,     37,     61,    109, $$


问:
Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible

Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.

Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.

答:
Suppose $\langle x \rangle$ is maximal and $x = yz$.  Then $\langle x \rangle \subseteq \langle y \rangle$.  From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $x$ is irreducible.

Suppose instead $x$ is irreducible.  Then $\langle x \rangle \subseteq \langle y \rangle$ would imply $x = yz$ for some $z$.  From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $\langle x \rangle$ is maximal.

ContentLicense: CC BY-SA 3.0


问:
$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$?
How can you compute the following summation:

$$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$$

答:
It probably was
$$ \sum_{k\geq 1}\frac{\binom{2k}{k}}{(2k-1)4^k} = 1$$
or the equivalent (via $k\mapsto k+1$)
$$ \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(k+1)} = 2$$
which are given by the Maclaurin series of $\sqrt{1-x}$ or by creative telescoping. Since
$$ \int_{0}^{\pi/2}(\cos\theta)^{2k}\,d\theta = \frac{\pi}{2\cdot 4^k}\binom{2k}{k} $$
holds by integration by parts, we also have
$$ \sum_{k\geq 0}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k\geq 0}(-1)^k(\cos\theta)^{2k}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta} $$
and via $\theta=\arctan t$
$$ \sum_{k\geq 0}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{2+t^2} =\frac{1}{\sqrt{2}}.$$

ContentLicense: CC BY-SA 4.0


问:
What is the point of disjoint union?
Disjoint union is seemingly ubiquitous in mathematical literature but it is rarely ever used properly. It is almost always the case that one forms $\bigsqcup_{i\in I} X_i$ and ends ups identifying $x\in X_i$ with $(x,i)\in\bigsqcup_{i\in I} X_i$. (Open any topology textbook and you will see this abuse.) Why not just assume that the $X_i$ are all pairwise disjoint and instead use ordinary union? Why bother with $X_i\times\{i\}$ when we don't ever use the notation $(x,i)$?

答:
As Brian said, the formal definition of disjoint union is important because it tells us that such a construction ("assume that the $X_i$ are all pairwise disjoint") is possible. Similarly, it's important to have a formal definition of ordered pairs (usually $(a,b) = \{\{a\}, \{a,b\}\}$). If there was no way to formally construct these things, it would be problematic to use them as mathematical objects! And, beyond telling us that these constructions make sense mathematically, these definitions allow us to prove things – without a formal definition of "ordered pair" it is impossible to prove that $(a,b) = (c,d) \iff a = c \land b = d$.

But the formal definition of these constructions is often separated from our mental models of these objects. Personally, I don't think of ordered pairs as sets of the form $\{\{a\}, \{a,b\}\}$ (even though, for me, this is what they technically are). Instead, I usually think of an ordered pair abstractly, as some kind of data structure that encodes two objects in order, without worrying about exactly how this is defined set-theoretically. This is fine, because I do have a formal definition of "ordered pair" to fall back on if necessary, and my intuition about ordered pairs is very strong (it comes from facts I've proven (long ago) from the formal definition).

I am confident that I will never make an incorrect claim about ordered pairs, and that I will never confuse readers by not using the formal definition of ordered pairs, even though I never deal with the formal definition in practice! Indeed, I think it would be much more confusing to most readers if I did use the formal definition of ordered pairs all the time – the important thing about a particular ordered pair is not that it's a set, but that it contains the data of two mathematical objects, in order.

Likewise, it's important that there is a formal definition of disjoint union. We can use this formal definition to prove things about disjoint unions, and thus build up our intuitions about how they behave. But our mental model of a disjoint union should not always be a collection of tuples; instead we should think of the disjoint union of a bunch of sets in the way you described: take all the sets $X_i$, "make them disjoint", and then stick them together. As such, it's often easier for a reader to digest an argument when we refer to elements of $\coprod_i X_i$ by the same names that they had in the sets $X_i$.

That's not to say that it's never a good idea to write $(x,i)$. For example, if all the sets in the coproduct are equal, it's always necessary to disambiguate the index. That is, I would always write $(x,i)$ for an element of $\coprod_{i \in I} X$.

Title: What is the point of disjoint union?


问:
what is the meaning of this inequality $x\le -5$ in $\mathbb{R} $?
what is the  meaning of  this inequality  $x\le -5$  in $\mathbb{R} $ ?

I thinks it may look like  this  $-9,-8 -7,-6 \le -5$

i mean $-6 \le -5$ ,

$-7\le-5$.......so on

Is its true ?

答: It simply means $x$ can be -5 or a real number less than -5. Can't comment.


问:
Show that $\vec{F}(x,y,z) = y\vec{i}-x\vec{j}$ is not a gradient field.
By definition, for this to be a gradient field, it must satisfy the condition that for a function $\phi$ : $\Bbb{R}^3 \to \Bbb{R}$, $F_1(x,y,z) = \frac{d\phi}{dx}$ $F_2(x,y,z) = \frac{d\phi}{dy}$ and $F_3(x,y,z) = \frac{d\phi}{dz}$

Or equivalently $\vec{F}(x,y,z)=\operatorname{grad}(\phi)(x,y,z) = \frac{d\phi}{dx}(x,y,z)\vec{\mathrm{i}}+\frac{d\phi}{dy}(x,y,z)\vec{\mathrm{j}}+\frac{d\phi}{dz}(x,y,z)\vec{\mathrm{k}}$

For my function, $\vec{F}(x,y,z)$ = $y\vec{\mathrm{i}}-x\vec{\mathrm{j}} +0\vec{\mathrm{k}}$, these conditions are satisfied, since $\frac{d\phi}{dx}(x,y,z) = y$ and $\frac{d\phi}{dy}(x,y,z) = -x$ and $\frac{d\phi}{dz} =0$

So therefore it is a gradient field. But I'm supposed to show that it is not a gradient field. How can I do so, if the conditions for it being a gradient field are satisfied?

答:
From $\phi_x = y$ we get that $\phi(x,y) = xy+C(y)$ for some function $C$ so
$$-x = \phi_y = x+C'(y)$$
which is a contradiction. Hence there is no such $\phi$.


问:
Chi squared from sum of squares of normal
Suppose $X=(X_1,X_2,X_3)^T$ is a random vector with a Gaussian distribution having mean vector $(0,0,0)$ and variance-covariance matric

$$
    \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 2 & 1 \\
    0 & 1 & 2 \\
    \end{pmatrix}.
$$
By definition, the $\chi^2 $ is the sum of squares of independent standard normal distributions. In my question, I was given different options on which one has the $\chi^2 $ distributions:

a) $(X_2 + X_3)^2/6 +(X_2-X_3)^2/2$

b) $(X_1+X_2)^2/5$

c) $X_1^2 + X_2^2 + X_3^2$

d)$X_1^2 + X_2^2$

e)$X_3^2/2 + (2X_2 -X_3)^2/6$

f) $(X_2^2+X_3^2)/2.$

The correct answers are (a) and (e). This doesn't make sense to me since X2 and X3 are not independent or standard Gaussian?
Similarly, X1 and X2 have a covariance of 0, so I don't understand how B, the sum of the two squared over 5, or even D, the sum of the squares of $X_1$ and $X_2$, aren't $\chi^2 $.

Any thoughts would be appreciated.

答:
(a) is $\chi^2$. Note that although $X_2$ and $X_3$ aren't standard normal, the rvs $$A=\frac{X_2+X_3}{\sqrt{6}}\quad\text{and}\quad B=\frac{X_2-X_3}{\sqrt{2}}$$ are normal (as the vector $(X_1,X_2,X_3)$ is Gaussian). You can compute the means and variances of $A$, $B$ to show that they are in fact standard normal. They are independent, as you can compute the covariance
$$\operatorname{Cov}(A,B)=\frac{1}{\sqrt{12}}\left[\operatorname{Cov}(X_2,X_2)-\operatorname{Cov}(X_2,X_3)+\operatorname{Cov}(X_3,X_2)-\operatorname{Cov}(X_3,X_3)\right]=0.$$
So $A^2+B^2$ is the sum of the squares of two independent standard normals, hence $\chi^2$.

If you expand (e), you will get the same expression as (a), proving that (e) is also $\chi^2$.

(b) is not $\chi^2$. If you compute the mean, $$\mathbb E\left[(X_1+X_2)^2/5\right]=(1+0+2)/5=3/5.$$ So it doesn't even have the same mean as any $\chi^2$ rv.

(d) isn't $\chi^2$ either, as $X_2$ is not standard normal. However, $X_1^2+\frac{X_2^2}{2}$ is $\chi^2$.

For completeness: for similar reasons as (d), the rv in (c) isn't $\chi^2$. Finally (f) isn't $\chi^2$ as $X_2$, $X_3$ aren't independent.


问:
Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$
Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$

One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?

答:
Extract the first order terms from $\int_0^t \sqrt{1+\sin(x^2)} dx $.

Expanding binomially, $$\sqrt{1+\sin(x^2)} \approx 1+\frac 12 \sin (x^2)\\ \approx 1+\frac 12 x^2 \\ \approx 1$$ That means $$\int_0^t \sqrt{1+\sin(x^2)} dx \approx \int _0^t dx =t$$ and $$\lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx = \lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{t} dx  =1$$


答:
Fix $t\neq 0$. By the Mean Value Theorem for Integrals, there is some $c_t$ between  $0$ and $t$ such that
$$
\int_0^t\sqrt{1+\sin\left(x^2\right)}dx=t\sqrt{1+\sin\left(c_t^2\right)}
$$
Therefore,
$$\int_0^t\frac{\sqrt{1+\sin\left(x^2\right)}}{\sin(t)}dx=\frac{t}{\sin(t)}\sqrt{1+\sin\left(c_t^2\right)}$$
If $t\to 0$, the quotient tends to $1$, while the root tends to $1$, so the limit is $1$.

For a (marginally different) approach, let $f(t):=\int_0^t\sqrt{1+\sin\left(x^2\right)}dx$ and apply L'Hospital's Rule and the Fundamental Theorem of Calculus to

$$
\lim_{t\to 0} \frac{f(t)}{\sin(t)}
$$

Title: Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$

Body:
Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$

One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?


问:
Why is 5.93e-56 considered to be close to zero?
I am very poor in math but have recently been studying some math courses, unfortunately they do not explain things well.

I have heard multiple times now that values like 5.93e-56 is very close to 0.

I do not know how this is. Is this not a 5.something value? And thus is it not supposed to be higher than 0?

Can anyone explain? I am specifically talking about when this value shows up as F-value in linear regression results.

答:
Scientific notation is used to make very large values and very small values that are out of the scale of normal every day observance manageable.

If we measure everything in miles it's it's easy to talk about distance between the corner of College and Ashby and The campanile on the U.C. Berkeley campus, ($1.09$ miles), between San Francisco and Sacramento, Ca.  ($87.9$ miles), Between Chicago and Madrid ($4176$ miles) and even the distance between the earth and the sun ($91409000$ miles).

But for large numbers: The distance between Earth and the Alpha Centauri ($25672000000000$ miles) or to the Andromeda galaxy ($14914073000000000000$ miles) or very small numbers. the size of a water molecule ($0.00000000000016777$ miles) things get.... awkward.

But we can make them manageable by considering them in terms of powers of $10$.

So width of a water molecule is $1.6777 \times 10^{-13}$ miles and the distance from college and ashby to the campanile is $1.09 \times 10^0$ miles; between San Francisco and Sacremento $8.79 \times 10^1$; Chicago to Madrid $4.176\times 10^3$ miles; between Earth and Sun $9.1409\times 10^7$; between earth and alpha centauri $2.5672\times 10^{13}$ and earth and andromeda $1.4914073\times 10^{19}$.

Now we have a way to refer to numbers from a wide range of scales.

Scientific notation:  We write $a.bc e{number}$ to mean $a.bc \times 10^{number}$.

So $5.93e-56 = 5.93\times 10^{-56} = 0.00000000000000000000000000000000000000000000000000000000593$.

Which is according to most people "very close to zero".  Of course, a mathematician must ask "very close to zero in comparison to what".

(To which the answer "compared to $1$" will usually suffice.  Except a mathematician might then ask "close on what scale" in which case "within $\frac 1{100000000000000000000000000000000000000000000000000000000}$ of the original" is an acceptable answer.)

(I think most people will concede that "$\frac {593}{100000000000000000000000000000000000000000000000000000000}$ is a very small number and close to $0$" to not be an unacceptable statement.)

.........

"And thus is it not supposed to be higher than 0?"

I'm a bit confused.  A number can be higher than $0$ and still be close to $0$ can't it.  "very close to $0$" doesn't mean less than $0$, does it?


问:
Stereographic projection and complex polynomial. (Do Carmo 2-3-16)
$\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\suchthat}{\;\big\vert\;}$

Let $ \mathbb{R}^{2} = \set{(x,y,z)\in\mathbb{R}^{3} \suchthat z=-1} $ be identified with the complex plane $ \mathbb{C} $ by setting $(x,y,-1)=x+iy=\zeta\in\mathbb{C}$. Let $ P: \mathbb{C}\to\mathbb{C} $ be the complex polynomial
\begin{equation*}
P(\zeta) = a_{0}\zeta^{n} + a_{1}\zeta^{n-1} + \cdots + a_{n}, \quad a_{0}\neq 0,\, a_{i}\in\mathbb{C},\, i=0,\dots,n.
\end{equation*}
Denote by $ \pi_{N} $ the stereographic projection of $ S^{2}=\set{(x,y,z)\in\mathbb{R}^{3} \suchthat x^{2}+y^{2}+z^{2}=1 } $ from the North pole $ N=(0,0,1) $ onto $ \mathbb{R}^{2} $. Prove that the map $ F: S^{2}\to S^{2} $ given by
\begin{align*}
F(p) &= \pi_{N}^{-1}\circ P \circ \pi_{N}(p), \quad \text{if } p\in S^{2}-\set{N}, \\
F(N) &= N
\end{align*}
is differentiable.

My attempt

By using Do Carmo's hint, $ F $ is differentiable in $ S^{2}-\set{N} $ as a composition of differentiable maps. To prove that $ F $ is differentiable at $ N $, consider the stereographic projection $ \pi_{S} $ from the South pole $ S=(0,0,-1) $ and set $ Q=\pi_{S} \circ F \circ \pi_{S}^{-1}: U \subset \mathbb{C}\to\mathbb{C} $ (of course, we are identifying the plane $ z=1 $ with $ \mathbb{C} $)

Now, I will show that $ \pi_{N}\circ\pi_{S}^{-1}: \mathbb{C}-\set{0}\to\mathbb{C} $ is given by $ \pi_{N}\circ\pi_{S}^{-1}(\zeta) = 1/\overline{\zeta} $.

Let $ \mathbf{u}=(u,v,1) $ be a point in plane $ z=1 $ (i.e., $ \mathbb{C} $); $ \ell_{S}(t) $ be the line through $ S $ with director vector $ \mathbf{u}-S=(u,v,2) $, then the parametrization of $ \ell_{S} $ is
\begin{equation*}
\ell_{S}(t) = (0,0,-1) + t(u,v,2) = (tu,tv,2t-1)
\end{equation*}
and the intersection point $ (x,y,z) $ between $ \ell_{S} $ and $ S^{2} $ (distinct of $ S $, i.e., $ t \neq 0 $) is given by the value of $t$ that satisfies
\begin{equation*}
(tu)^{2} + (tv)^{2} + (2t-1)^{2}= 1 \quad \implies \quad t^{2}u^{2} + t^{2}v^{2} + 4t^{2} - 4t + 1 = 1 \quad \implies \quad tu^{2} + tv^{2} + 4t = 4
\end{equation*}
then
\begin{equation*}
t=\dfrac{4}{u^{2} + v^{2} + 4}
\end{equation*}
yields
\begin{equation*}
(x,y,z) = \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right)
\end{equation*}
therefore
\begin{equation}
\pi_{S}^{-1}(u,v) = \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right)
\tag{1}
\end{equation}
Now, to calculate $ \pi_{N}\circ\pi_{S}^{-1} $, take (1) as a point in $ S^{2} $ and let $ \ell_{N}(t) $ be the line trough $ N $ with director vector $ \pi_{S}^{-1}-N $, then
\begin{align*}
\ell_{N}(t) &= (0,0,1) + t\left[ \pi_{S}^{-1} - (0,0,1) \right] \\[2ex]
 &= (0,0,1) + t\left[ \left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{4-u^{2}-v^{2}}{u^{2} + v^{2} + 4} \right) - (0,0,1) \right] \\[2ex]
 &= (0,0,1) + t\left( \dfrac{4u}{u^{2} + v^{2} + 4}, \dfrac{4v}{u^{2} + v^{2} + 4}, \dfrac{-2(u^{2}+v^{2})}{u^{2} + v^{2} + 4} \right)
\end{align*}
And for this point to belong to the plane $ z=-1 $ (i.e., $ \mathbb{C} $) clearly we need that
\begin{equation*}
\dfrac{-2(u^{2}+v^{2})}{u^{2} + v^{2} + 4}t+1=-1 \quad \implies \quad t=\dfrac{u^{2} + v^{2} + 4}{u^{2} + v^{2}}
\end{equation*}
thus
\begin{equation*}
(x,y,-1) = \left( \dfrac{4u}{u^{2}+v^{2}}, \dfrac{4v}{u^{2}+v^{2}}, -1 \right)
\end{equation*}
therefore
\begin{equation}
\pi_{N}\circ\pi_{S}^{-1}(u,v) = \left( \dfrac{4u}{u^{2}+v^{2}}, \dfrac{4v}{u^{2}+v^{2}}, -1 \right)
\tag{2}
\end{equation}
where $ (u,v-1) = u + iv = \zeta $. \
But the inverse of conjugate of $ \zeta $ should be
\begin{equation}
1/\overline{\zeta} = \left( \dfrac{u}{u^{2}+v^{2}}, \dfrac{v}{u^{2}+v^{2}}, -1 \right)
\tag{3}
\end{equation}

My Question. Where is my mistake? Do Carmo's book says the equation should be like (3). I'm confused.

Next step is conclude that
\begin{equation*}
 Q(\zeta) = \dfrac{\zeta^{n}}{\overline{a_{0}} + \overline{a_{1}}\zeta + \cdots + \overline{a_{n}}\zeta^{n}}
\end{equation*}
We have the next relations (if $ F \neq N $)
\begin{align*}
Q(\zeta) &= \left( \pi_{S}\circ F \circ\pi_{S}^{-1} \right) (\zeta) = \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \circ\pi_{N} \circ\pi_{S}^{-1} \right) (\zeta) \\
&= \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \right) \circ \left( \pi_{N}\circ\pi_{S}^{-1}(\zeta) \right) \\
&= \left( \pi_{S}\circ \pi_{N}^{-1} \circ P \right) (1/\overline{\zeta}) \\
&= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \circ P(1/\overline{\zeta}) \\
&= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left( a_{0}\left(1/\overline{\zeta}^{n}\right) + a_{1}\left(1/\overline{\zeta}^{n-1}\right) + \cdots + a_{n} \right) \\
&= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left(  \left( 1/\overline{\zeta}^{n} \right)\left( a_{0}+a_{1}\overline{\zeta} + \cdots + a_{n}\overline{\zeta}^{n} \right) \right)
\end{align*}
Since $ \pi_{S}\circ \pi_{N}^{-1} = \left( \pi_{N}\circ\pi_{S}^{-1} \right)^{-1} $ should be $ \pi_{S}\circ \pi_{N}^{-1} = 1/\overline{\zeta} $ as well. Thus
\begin{align*}
Q(\zeta) &= \left( \pi_{S}\circ \pi_{N}^{-1} \right) \left(  \left( 1/\overline{\zeta}^{n} \right)\left( a_{0}+a_{1}\overline{\zeta} + \cdots + a_{n}\overline{\zeta}^{n} \right) \right) \\
 &= \dfrac{\zeta^{n}}{\overline{a_{0}} + \overline{a_{1}}\zeta + \cdots + \overline{a_{n}}\zeta^{n}}
\end{align*}
Hence, $ Q $ is differentiable at $ \zeta=0 $. Thus, $ F = \pi_{S}^{-1}\circ Q \circ\pi_{S} $ is differentiable at $ N $.

答: DoCarmo's intent is that you project onto the plane $z=0$, not the plane $z=-1$. (This is the standard construction of stereographic projection.)


问:
Deriving Generating function for centered trinomial coefficients
Let $c_n$ denote the $n$-th center trinomial coefficient (OEIS sequence here).

It appears they cannot be generated by a linear recurrence relation, so how should I go about to find the generating function $G(x)$ for the sequence?

$$G(x)=\sum_{n=0}^{∞}c_nx^n=1+x+3x^2+7x^3+19x^4+51x^5+141x^6+...$$

The geometric ratio appears to have a limit close to $$\lim_{n\to ∞}\frac{c_{n+1}}{c_n}=2.95...$$ (these are the successive ratios of the last two listed terms in the OEIS sequence).

Also, what is the interval of convergence (and divergence)? Based on the geometric limit, it seems that $G(1/3)$ will converge.

Edit: The generating function is $$G(x)=\frac{1}{\sqrt{1-2x-3x^2}}$$
Any idea as to how this answer is derived?

答:
Here is a variation based upon G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. We start with the central trinomial coefficients:
\begin{align*}
[x^n](1+x+x^2)^n\qquad\qquad n\geq 0
\end{align*}
We consider the function
\begin{align*}
f(x)=1+x+x^2\tag{1}
\end{align*}
and derive a function $y=y(x)$:
\begin{align*}
y(x)=\frac{x}{f(x)}=\frac{x}{1+x+x^2}\qquad\qquad y^{\prime}(x)=\frac{1-x^2}{(1+x+x^2)^2 }\tag{2}
\end{align*}

With $f(x)$ and $y(x)=\frac{x}{f(x)}$ we can now apply the substitution rule (Rule 5, one-dimensional case) from section 1.2.2 in G. P. Egorychev's book as follows:
\begin{align*}
\color{blue}{[x^n](f(x))^n=[y^n]\left.\left(\frac{1}{f(x)y^{\prime}(x)}\right)\right|_{x=g(y)}}\tag{3}
\end{align*}
with $g(y)$ the inverted function given by $y=y(x)$ in (2).

We obtain from (1) - (3):
\begin{align*}
\color{blue}{[x^n]}&\color{blue}{\left(1+x+x^2\right)^n}\\
&=[y^n]\left.\left(\frac{1}{\left(1+x+x^2\right)\frac{d}{dx}\left(\frac{x}{1+x+x^2}\right)}\right)\right|_{x=g(y)}\\
&=[y^n]\left.\frac{1+x+x^2}{1-x^2}\right|_{x=g(y)}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-2y-3y^2}}}\tag{4}
\end{align*}
and the claim follows.

In (4) we use the identity
\begin{align*}
2y=\frac{2x}{1+x+x^2}&=1-3\left(\frac{x}{1+x+x^2}\right)^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
&=1-3y^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
\frac{1+x+x^2}{1-x^2}&=\left(1-2y-3y^2\right)^{-\frac{1}{2}}
\end{align*}


问:
In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$

Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.

Is the triangle $KLM$ also an equilateral side?
In what relation are the area of triangle $KLM$ the area of triangle$ ABC$?

My attempt:

I can see that $KLM$ is an equilateral triangle.  But, why the fraction $\frac{1}{7}$ is the area ratio of the triangle $KLM$ to the triangle $ABC$?

答:
Establish the area ratios

$$\frac{[ACL]}{[ACA_1]} = \frac{[ACC_1]}{[ACA_1C_1]}
= \frac{\frac13[ABC]}{[ABC]-[BC_1A_1]}= \frac{\frac13[ABC]}{[ABC]-\frac13\frac23[ABC]}
=\frac37 $$
which leads to
$$[ACL] =\frac37[ACA_1]=\frac37\frac23[ABC]=\frac27[ABC]
$$
Likewise, $[BCK]={[ABM]}=\frac27[ABC]$, Then
$$\frac{[KLM]}{[ABC]} = \frac{[ABC]-[ACL]-[BCK]-[ABM]}{[ABC]} = 1-3\cdot\frac27=\frac17
$$
In general, $\frac{[KLM]}{[ABC]} =\frac{(r-1)^2}{r^2+r+1}$, where $1:r$ is the given ratio.


问:
How do I solve that matrix operation?

Let $x$ be a column vector in $\Bbb R^n$, and $x$ different from matrix zero. Let
$A = \frac{ x x^{T}}{x^{T} x}$, T denotes tranpose. Calculate $A²$.

I hope that I am not the only one who was astonished with this exercise. Someone here knows what it means $\frac{ x x^{T}}{x^{T} x}$? I can imagine that a division is simple take the inverse of denominator, but so the exercise become so trivial that I discard it. Anyway, someone here knows what it means?

答: If $x$ is the column vector$$\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix},$$then $x^T=\begin{bmatrix}a_1&a_2&\cdots&a_n\end{bmatrix}$ and so$$x^Tx=\begin{bmatrix}a_1&a_2&\cdots&a_n\end{bmatrix}\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}=a_1^{\,2}+a_2^{\,2}+\cdots+a_n^{\,2}.$$It's just a number.

Title: How do I solve that matrix operation?


问:
$x^2\ast y = y = y \ast x^2$ implies commutativity?
I would appreciate some help with this task:

Let be $S$ a non-empty set equipped with an operation $\ast :S\times S\to S$ such that $\ast$ is associative and has the property $x^2\ast y = y = y \ast x^2$ for every $x,y\in S$ (here, $x^2 = x\ast x)$. I have to show that $(S,\ast)$ is an abelian group. In other words, I have to show that there is an identity element $e$ such that $x\ast e = e\ast x = x$ for all $x$ and the commutativity $x\ast y = y \ast x$, for all $x,y \in S$

Note that, assuming there exists such element $e$, we have by virtue of the property  $x^2\ast y = y = y \ast x^2$ :
$$ x^2 = x^2\ast e = e  \quad \forall x\in S$$
This means that, in case an identity element exists, every element of $S$ is involutive. But how can I assert the existence of $e$ and, even more, the commutativity?.

Any hint is welcome. Thanks in advance.

答:
I'll drop the $\ast$'s.

First, if $x^2y=y=yx^2$ for all $x$ and $y$, then every $x^2$ works as an identity, and moreover $\{x^2:x\in S\}$ consists of single element, call it $e$. In particular, every element is its own inverse, $S$ is a group.

Finally, let $x,y\in S$. Then

$$
e=(xy)^2=xyxy\implies xy=y^{-1}x^{-1}=yx
$$


问:
Cantor set invariant under the dilation of ratio 3
We may use the explicit formulas of Cantor sets
$$
\mathcal{C} = [0,1] \setminus\bigcup_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right)
$$
to show that $C$ is invariant under the dilation
Let $T:x \longmapsto T_3(x)=3x \quad \text{mod} \, \, 1$.
That is
Prove that $T(C) = C.$

We use the fact that for any set $A$ and a map $T:E \longrightarrow F$ we have $f(A^c)=f(A)^c$, where $A^c$ is the complement of $A$ in $E$.
Let
\begin{align*}
A=& \bigcup_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right)\\
=& \bigcup_{n=0}^{\infty} \bigcup_{k=0}^{3^{n}-1} \left(\left(\frac{3k+1}{3^{n+1}}\,,\, \frac{3k+2}{3^{n+1}}\right)\right) \quad (n=m-1)\\
=&  \bigcup_{k=0}^{3^{0}-1} \left(\left(\frac{3k+1}{3^{1}}\,,\, \frac{3k+2}{3^{1}}\right)\right) \\
& = (\frac{1}{3}, \frac{2}{3})   \bigcup_{n=1}^{\infty} \bigcup_{k=0}^{3^{n}-1} \left(\left(\frac{3k+1}{3^{n+1}}\,,\, \frac{3k+2}{3^{n+1}}\right)\right)\\
\end{align*}
So
$$
C=\bigcap_{m=1}^{\infty} \bigcup_{k=0}^{3^{m-1}-1} \left(\left(\frac{3k+1}{3^{m}}\,,\, \frac{3k+2}{3^{m}}\right)\right)
$$
Which gives  $T_3(A)= (1/9;2/9) \cup A$, and by the Morgan law, we get
\begin{align*}
T_3(A^c)=(T_3(A))^c=(1/9,2/9)\cup A^c= A^c
\end{align*}
Question: I can't seem to find the power $n-1$ in the explicit formula of $C$?

Any help very much appreciated!

答:
Perhaps not an answer to your specific question. Let $C_0 = [0,1], C_1 = [0,1/3] \cup [2/3,1],$ and so on, i.e removing the middle third of each interval at each iteration. Write

$$C = \bigcap_{k=1}^\infty C_k.$$

To show that $C$ is invariant under $T(x) = 3x \mod 1$ one observes that $T(C_k) = C_{k-1}$ for each $k = 1,2,...$ . We then have

$$T(C) = T\bigg(\bigcap_{k=1}^\infty C_k\bigg) = \bigcap_{k=1}^\infty T(C_k) = \bigcap_{k=1}^\infty C_{k-1} = [0,1] \cap \bigcap_{k=1}^\infty C_k = C.$$

You can of course use the construction used in your question but the one above is the most natural and intuitive one in my opinion.

Exercise: Show that the Sierpinski Carpet and the Sierpinski triangle are invariant under some $kx \mod 1.$


答:
The binom formula gives that for all $\alpha \in \mathbb{C}$, $$\sum\limits_{k\geqslant 0} \binom{\alpha}{k}x^k = (1+x)^\alpha$$

By taking $\alpha  = -1/2$, we get $$\sum\limits_{k\geqslant 0} \binom{-1/2}{k} x^k = (1+x)^{-1/2}$$

Then we remark that $\binom{-1/2}{k}=(-1)^k \binom{k+1/2}{k}$.

So we have $$\sum\limits_{k\geqslant 0} \binom{k+1/2}{k}(-1)^k x^k = \frac{1}{\sqrt{1+x}}$$.

And as $\binom{k+1/2}{k}=\binom{2k}{k}\cdot \frac{1}{4^k}$, we get the answer $$\sum\limits_{k\geqslant 0} \binom{2k}{k}(-1)^k \frac{x^k}{4^k} = \frac{1}{\sqrt{1+x}}$$

By replacing with $x=1$, we get $\frac{1}{\sqrt{2}}$.

Title: $\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$?

Body:
How can you compute the following summation:

$$\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}=\frac{1}{\sqrt{2}}$$


问:
Deciding whether given infinite series converges or diverges
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am having some trouble determining the convergence or divergence of the below infinite series. I have some initial thoughts which I have put down. Any inputs would be really helpful.

How do I build more skill at finding out if a given infinite series is convergent/divergent?

Exercise 2.7.2

(c) $1 - \frac{3}{4} + \frac{4}{6} - \frac{5}{8} + \frac{6}{10} - \frac{7}{12} + \ldots$

(d) $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + \ldots$

(e) $1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} + \frac{1}{7} - \frac{1}{8^2} + \ldots$

Proof.

(c) Let $(s_n)$ be the sequence of partial sums of the infinite series $\sum_{n=1}^{\infty}a_n$.

The general expression for the infinite series is,

\begin{align*}
1 + \sum_{n=1}^{\infty}(-1)^{n}\cdot\frac{n+2}{2n}
\end{align*}

I cannot apply the alternate series test, because
\begin{align*}
\lim_{n\to\infty}a_n = \lim_{n\to\infty}\frac{n+2}{2n} = \frac{1}{2} \ne 0
\end{align*}

I can't think of a known convergent or divergent series to compare with, in order to use the Alternate Series test.

(d) One observation is, this is not a rearrangement of the alternating harmonic series.

答:
(c) Your series is $\sum_{n=1}^\infty(-1)^{n+1}\frac{n+1}{2n}$ and, since you don't have $\lim_{n\to\infty}(-1)^{n+1}\frac{n+1}{2n}=0$, the series diverges.

(d) The sum of the first $3$ terms is greater than $1$. The sum of the first $6$ terms is greater than $1+\frac14$. The sum of the first $9$ terms is greater than $1+\frac14+\frac17$. So, your series diverges.

(e) The sum of the positive terms diverges, whereas the sum of the negative terms converges, and therefore this series diverges.


问:
What would be the value of $a$ and $b$ in following rational expression?
If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,

How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?

答:
What was written was
$$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3b},$$
but I'm going to boldly assume that's a typo and what was really intended was
$$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3}b.$$

Multiply the numerator and denominator by the conjugate of $7+\sqrt{3}$, which is $7-\sqrt{3}$:
\begin{align}
& \frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = \frac{(5 + 2\sqrt{3})(7 - \sqrt{3})}{(7 + \sqrt{3})(7 - \sqrt{3})} = \frac{29 + 9\sqrt{3}}{7^2-3} = \frac{29 + 9\sqrt{3}}{46} \\[10pt]
& = \frac{29}{46} + \frac{9}{46}\sqrt{3} = \frac{29}{46} - \frac{-9}{46}\sqrt{3}.
\end{align}
So $a=\dfrac{29}{46}$ and $b=\dfrac{-9}{46}$.


答:
The components of $A$ are

$$ A_{ij}=\frac{x_i x_j}{|x|^2}$$

Where $|x|^2$ is the magnitude squared of $x$. Thus the components of $B=A^2$ are

$$ B_{ik} = A_{ij} A_{jk} =\frac{x_i x_j}{|x|^2} \frac{x_j x_k}{|x|^2}=\frac{x_i x_k}{|x|^2}=A_{ik}$$

Which are identical to the components of $A$. Thus $A^2=A$.

Title: How do I solve that matrix operation?

Body:
Let $x$ be a column vector in $\Bbb R^n$, and $x$ different from matrix zero. Let
$A = \frac{ x x^{T}}{x^{T} x}$, T denotes tranpose. Calculate $A²$.

I hope that I am not the only one who was astonished with this exercise. Someone here knows what it means $\frac{ x x^{T}}{x^{T} x}$? I can imagine that a division is simple take the inverse of denominator, but so the exercise become so trivial that I discard it. Anyway, someone here knows what it means?


问:
Proving $\left|1-e^{2ik\pi / n}\right|=2\sin \frac{k\pi}{n}$

Prove the  identity
$$\left|1-e^{2ik\pi / n}\right|=2\sin \frac{k\pi}{n}$$
for $k=1,\ldots,n−1$.

Body:
$$|1-e^{2ik\pi / n}|=|e^{ik\pi / n}(e^{-ik\pi / n} - e^{ik\pi / n})|$$

$$= |e^{ik\pi / n} \cdot 2i\sin \frac{k\pi}{n}| = 2\sin \frac{k\pi}{n}$$


问:
How to get Euler angles with respect to initial Euler angle
I have a sensor which gives me Euler angles (roll,pitch,yaw). There is a baseline value of Euler angle (assume it is $5,10,15$) at the beginning.I want to calibrate from this baseline values all subsequent value. How can I get those values? Is it just subtract $5,10,15$ from all values Or is there any rotational matrix for doing so? As an example if any time the value is $5,10,15$ then it should show $0,0,0$ and on the same way show other angle values with respect to the baseline values.I don't know how to do this.

Please advise.

答:
If the sensor is reporting the orientation with respect to a fixed frame (for example the absolute orientation w.r.t. an earth-fixed frame), and you are only interested in the change in the Euler angles with respect to the initial orientation of the same sensor, you can actually substract them, and the results will be the same as using relative rotation matrices/quaternions. It is simple to construct an artificial example in MATLAB and test this.

Also, in @Tpofofn's answer, the order of multiplication for obtaining the relative rotation matrix is incorrect.


问:
What is the functional inverse (with respect to $h$ (!)) of $f^{\circ h}(x)={F(h) +x F(h-1) \over F(1+h) +x F(h) }$?
I've considered the fractional iteration of $f(x) = {1 \over 1+x} $ for which the general expression depending on the iteration-height parameter $h$ might be assumed by the formula
$$ f^{\circ h}(x) = f(x,h)={F(h) +x \cdot F(h-1) \over F(1+h) +x \cdot F(h)  }$$  where

$ \displaystyle \qquad \small F(h)=\operatorname{fibonacci}(h) = {\varphi^h - (1-\varphi)^h \over \sqrt5 }$ and

$ \displaystyle \qquad \small \varphi = {1+\sqrt5 \over 2} \sim 1.618... $

(Note, that for fractional iterates we need $x \in \mathbb C$)

Today I tried to find a formulation for the functional inverse with respect to h, but don't find a good starting point so:

Q: How would look a function $ h = \operatorname{hgh}(x_0,x_h) $ which would indicate the required iteration-height $h$ given $x_0$ and $x_h = f(x_0,h)$ ? Is there even a closed form for it (I'd call it still closed-form  if it possibly includes e.g. the Lambert-W function)

[update]: for more background see the older exercise of mine

答:
Fractional linear transformation $f$ can be represented by the $2\times 2$ matrix $\displaystyle \left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)$, which is diagonalized as
$$\left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)=
P\left(\begin{array}{cc} \varphi & 0 \\ 0 & 1-\varphi\end{array}\right)P^{-1},\qquad
P=\left(\begin{array}{cc} \varphi-1 & -\varphi \\ 1 & 1\end{array}\right).$$
Therefore the matrix corresponding to $h$th iteration is
$$P\left(\begin{array}{cc} \varphi^h & 0 \\ 0 & (1-\varphi)^h\end{array}\right)P^{-1}=
\frac{1}{\sqrt{5}}\left(\begin{array}{cc} (\varphi-1)\cdot\varphi^h +\varphi\cdot(1-\varphi)^h & \varphi^h -(1-\varphi)^h \\ \varphi^h -(1-\varphi)^h &
\varphi\cdot\varphi^h +(\varphi-1)\cdot(1-\varphi)^h \end{array}\right)$$
and we obtain
$$\left(\frac{1-\varphi}{\varphi}\right)^{h}=\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}.$$
Taking the logarithm, one finds the expression
$$\mathrm{hgh}(x_h,x_0)=\ln\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}\Bigl/\Bigr.\ln\frac{1-\varphi}{\varphi}.$$
Note, however, that the expressions in the logarithms can become negative or even complex (in the denominator logarithm it is always negative), so some care should be taken to choose correct branches.</p>


问:
What's the normal line when partial derivative with respect to $x = 0$?
So I know the equation of a normal line is: $$\frac{x-a}{f_1(a,b)}=\frac{y-b}{f_2(a,b)}=\frac{z-f(a,b)}{-1}$$. But in my textbook there is a litle note that says "with suitable modifications if either $f_1(a,b)=0$ or $f_2(a,b)=0$". Funny that they forget what that modification would look like. Now I have a problem where $f_1=0$, and don't know what to do. I searched for this topic, but frankly the given formula we use is apparantly not so common. Anyhow If someone would know the modification. It would be very helpful.

Thanks in advance.

答:
We are talking of the normal line, not the "tangent" line. Talking of surfaces we have tangent planes not tangent lines.

The general equation of the normal is
$$n:(x_0,y_0,z_0)+t (f'_{x_0},f'_{y_0},f'_{z_0})$$


问:
How do I approach optimizing this multivariable function? What should I learn to be able to do that?

Function to maximize

$R(P) = \sum\limits_{i=1}^4 (-\frac{1}{2}p_i^2 + 1000p_i)$

Parameters

$P = (p_1, p_2, p_3, p_4)$

Constraints

$p_1, p_2, p_3, p_4 > 0$

$\sum\limits_{i=1}^4 (-\frac{1}{2}p_i + 1000) \leq 2000$

答:
This is the first time I have answered a question, so please bear with me.

Maximizing $R(P)=\sum_1^4(−1/2p^2_i+1000p_i)$ is the same as minimizing its negative. Doubling it and adding 1000000 to each summand doesn't change the minimum and we get
$R(P)=\sum_1^4(p^2_i-2000p_i+1000000)$
$    =\sum_1^4(p_i-1000)^2$.
Setting each $p_1=1000$ satisfies the constraints and makes $R(P)=0$ and, since it is a sum of squares this must be the minimum.


问:
why $f(x)$is divergent?
Given the improper integral
$$f(x)=\int_0^1 x^m \sin x\, dx$$ where  $ m\in \mathbb{R}$

Is $f(x)$ is divergent at $m=-4/3?$

My try :  we have the inequality  $\sin x \leq x$

$f(x)  = \int_0^1 x^m \sin x \, dx$

so $$f(x) \le \int_{0}^1 x^{1+m}\,dx \le \frac{1}{2+m} <\infty$$

clearly we can see if $-2<m<0$ then $f(x)$ is convergent

and $m= -4/3 \in (-2,0)$

so $f(x)$ is convergent

But answer key show that $f(x)$ is divergent I don't understand why $f(x)$ is divergent.

答:
When $m=-4/3$, compare it with the second integral as follows:$$\int_0^1 \frac{1}{x^{4/3}} \sin x dx \le \int_0^1 \frac{1}{x^{4/3}}x dx = \int_0^1 \frac{1}{x^{1/3}} dx$$
The second integral is convergent since $1/3 < 1$.

So, your reasoning is right.

Title: why $f(x)$is divergent?


答:
Neat $ 0 $ bound, we have

$$x^m\sin(x)\sim x^{m+1} \;\;(x\to 0^+)$$

or

$$x^m\sin(x)\sim \frac{1}{x^{-m-1}}$$

but

$$\int_0^1\frac{dx}{x^{-m-1}}\;converges$$
$$\;\iff -m-1<1$$
$$\iff m>-2$$

So, if $ m=\frac{-4}{3} $, the integral is Convergent.

Title: why $f(x)$is divergent?

Body:
Given the improper integral
$$f(x)=\int_0^1 x^m \sin x\, dx$$ where  $ m\in \mathbb{R}$

Is $f(x)$ is divergent at $m=-4/3?$

My try :  we have the inequality  $\sin x \leq x$

$f(x)  = \int_0^1 x^m \sin x \, dx$

so $$f(x) \le \int_{0}^1 x^{1+m}\,dx \le \frac{1}{2+m} <\infty$$

clearly we can see if $-2<m<0$ then $f(x)$ is convergent

and $m= -4/3 \in (-2,0)$

so $f(x)$ is convergent

But answer key show that $f(x)$ is divergent I don't understand why $f(x)$ is divergent.


问:
Check if $u_y+u^2 u_x = 0$ with rectangular initial data has a shock

I have the p.d.e. $$u_y+u^2 u_x =0$$ $x \in \mathbb{R}$, $t>0$
with $$u(x,0)=h(x)= \begin{cases} 0 & x<0 \\ 1 & 0<x<1 \\ 0 & x>1 \end{cases}$$
Check if a shock forms and solve the p.d.e.

My attempt, the characteristic curves are $(s,0,h(s))$, so I have $$\frac{dy}{dt}=1 \implies y=t$$
$$\frac{du}{dt}=0 \implies u=h(s)$$
$$\frac{dx}{dt}=u^2 \implies x=h^2(s)t+s$$

So the solution is $$u=h(x-u^2y)$$ $$u= \begin{cases} 0 & x<0 \\ 1 & t <x <t+1 \\ 0 & x>1 \end{cases}$$
Now this is where I am stuck, how do I check if a shock forms?
And how do I proceed to solve the problem?

答:
The present problem is very similar to the case of (...) Burgers' equation with rectangular data. However, the present PDE rewrites in conservation form as $u_y + (\frac13 u^3)_x = 0$, hence the flux $\frac13 u^3$ is nonconvex. Following this post, the solution for small times reads
$$
u(x,y) = \left\lbrace\begin{aligned}
&0 &&x\leq 0\\
&\sqrt{x/y} &&0\leq x\leq  y\\
&1 &&y\leq x< 1+y/3\\
&0 && 1+y/3< x
\end{aligned} \right.
$$
which is valid up to $y = 3/2$. One notes that the initial discontinuity located at $(x,y)=(0,0)$ leads to the formation of a rarefaction wave, while the initial discontinuity located at $(x,y)=(1,0)$ leads to a shock wave (these facts follow from the entropy condition -- or from the intersection of characteristic curves, if you prefer). Then, the interaction of the rarefaction and shock waves can be computed in a similar way to the case of the Burgers' equation.


问:
The unity of a subdomain
Let $D$ be an integral domain and $R$ a subdomain , I'm trying to show that both $R$ and $D$ have the same unity. I did it using the fact that $1$ and $0$ are the only idempotent elements inside any integral domain. I'm trying to do it in another way which is the following:

Since $R$ is a subdomain it has unity call it $c$, let $a\in R$ then $ac=a$ since we can cancel we get that $c$ which is in $R$ is equal to $1$ then $1\in R$.

I'm not really confident this right. If this is wrong what other way is there that does not use the fact that there are two idempotent elements in an integral domain.

答:
Let $ \ c \ $ be the unity of the subdomain $ \ R \ $ and $ \ e \ $ the unity of the domain $ \ D \ $ itself. For any $ \ a \in R \ $, $ \ ea = ca \ $ and thus $ \ (e-c)a=0 \ $.

Hope this is right.

Title: The unity of a subdomain


问:
Symmetries in a dihedral group
How many symmetries does $D_{2n}$ have exactly? I imagine it has rotations written in the form $\frac{x\pi}{n}$ such that $x \in \{0, 1, \dots, n-1\}$, and can also be flipped.

But what about rotations reflections across lines from vertices to midpoints of opposite sides? I know this differs depending on whether $n$ is even or odd, but I still don't grasp this concept...

答:
I assume that $D_{2n}$ refers to the symmetries of the $n$-gon (note that sometimes this group is referred to as $D_n$). For the purposes of geometric description, I will suppose that the $n$-gon is oriented with a vertex on the $x$-axis.

As you correctly note, we have rotational symmetries; there are $n$ of these. Each is a rotation by angle $\frac{2\pi}{n} \cdot x$ for $x = 0,1,\dots,n-1$.

On the other hand, we have $n$ reflection symmetries. In particular, we have the reflections across the line with angle $\frac {\pi}{n}\cdot x$ clockwise from the horizontal for $x = 0,1,\dots,n-1$. When $n$ is odd, each of these is a reflection across a line connecting a vertex to the midpoint of its opposite edge. When $n$ is even, the axis of reflection connects opposing vertices when $x$ is even and opposing edge midpoints when $x$ is odd.

Title: Symmetries in a dihedral group


问:
Calculate the limit of a sequence $\lim_{n\to \infty} \frac{I_{n}}{n^{n}}$

Calculate $$\lim_{n\to \infty} \frac{1}{n^{n}}\int_{0}^{n}x^n\arctan(\sqrt[n]{x})dx$$

My approach: I was trying to solve first the integral, so for example for $n=1$, we have that
$$\int x\arctan(x)dx=\frac{1}{2}x^{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)+c$$
and for $n=2$, we can see that $$\int x\arctan(\sqrt{x})dx=\frac{1}{6}(3(x^{2}-1)\arctan(\sqrt{x})-(x-3)\sqrt{x}+c$$
and for $n=3$, we have
$$\int x\arctan(\sqrt[3]{x})dx=\frac{1}{30}(-3x^{5/3}+15(x^{2}+1)\arctan(\sqrt[3]{x})+5x-15\sqrt[3]{x})+c$$

I was trying to see if by induction I could give a closed formula for the integral but the calculation of the integral does not show any pattern. How can I solve this limit?

答:
Hint :

$u = \frac{x}{n}$ gives you a simpler formula :
$$\lim_{n\to \infty} \int_{0}^{1}n u^n\arctan(\sqrt[n]{nu})du$$


问:
Stationary shock for Burgers' equation with Murman-Roe scheme

Consider Burgers' equation $u_t + f(u)_x = 0$ where $f(v)=\frac{v^2}{2}$, with initial condition $$u_0(x)=\begin{cases} -1 & x<0 \\ 1& x>0 .\end{cases} $$ It's clear that $u_l=-1<u_r=1$, so the discontinuity is not admissible and the entropy solution is a rarefaction wave.
Consider the following discrete initial condition $$v_{j}^{0}= \begin{cases} -1 & j\le -1 \\ 1& j\ge 0 \end{cases} $$
Show that using the Murman-Roe scheme that the numerical solution is a stationary shock and is given by $$v_{j}^{n}=v_{j}^{0}= \begin{cases} -1 & j\le -1 \\ 1& j\ge 0 \end{cases} $$

Can you give me some hint pls, or can you solve it.Because I know the Roe Scheme in case of linear function f

What is the solution using Godunov Scheme? Is it by using the Riemann Solver $W_R(\frac{x}{t},u_l,u_r)$ ,where $\frac{x}{t}$ is non-zero?

答:
Roe-type approximate Riemann solvers are known to fail at providing the entropy-satisfying solution in the particular case of transonic rarefaction waves (i.e., rarefaction waves that cross the zero-velocity state $u=0$). The present example illustrates this feature. First, we need to write the Murman-Roe scheme. To do so, consider the mesh size $\Delta x$ and the time step $\Delta t$, such that the coordinates $(x,t)$ are discretized as $x_j = j\Delta x$ and $t^n = n\Delta t$. The scheme reads
$$
v_j^{n+1} = v_j^n - \frac{\Delta t}{\Delta x} \left(g(v_{j}^n, v_{j+1}^n) - g(v_{j-1}^n, v_{j}^n)\right)
$$
with numerical flux
$$
g(u,v) = \tfrac12\left(f(u) + f(v) - |a(u,v)|(v-u)\right)
$$ $$
a(u,v) = \left\lbrace\begin{aligned}
&\tfrac{f(u) - f(v)}{u-v} &&\text{if}\quad u\neq v\\
&f'(u) &&\text{if}\quad u= v .
\end{aligned}\right.
$$
Then, we need to show that the unphysical stationary shock solution is obtained by iterating this scheme -- instead of the physical rarefaction solution. Note that this bad feature doesn't show up for the Godunov scheme (i.e., the exact Riemann solver), which numerical solution will be an approximation of the correct rarefaction wave.


问:
Why is $d(T \times N)/ds$ orthogonal to the unit tangent vector $T$?
Consider a curve in $\mathbb{R}^3$ and its Frenet-Serret frame $(T,N,B)$. A calculation gives $$\frac d{ds} (T \times N)= T \times \frac {dN}{ds}.$$

Question:  I read that this implies that $\frac d{ds} (T \times N)$ is orthogonal to $T.$ Why is it so?

答:
Let $\times$ the vector cross product in $\mathbb{R}^3$ and $b,c\in \mathbb{R}^3$, then we have that $a:=b\times c$ is an orthogonal vector to the plane defined by the span of $b$ and $c$, therefore $T\times N'$ is orthogonal to both $T$ and $N':=\frac{\partial}{\partial s}N$.

An elementary proof is given by: let $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ be vectors in $\mathbb{R}^3$ who coordinates are given by the standard orthonormal basis, then

$$
\langle x\times y, x \rangle=\langle (x_2y_3-x_3y_2,x_3y_1-y_3x_1,x_1y_2-y_1x_2),(x_1,x_2,x_3) \rangle\\
=\color{red}{x_1x_2y_3}\color{blue}{-x_1x_3y_2}+x_2x_3y_1\color{red}{-x_2x_1y_3}\color{blue}{+x_3x_1y_2}-x_3x_2y_1=0
$$
where $\langle \cdot ,\cdot  \rangle$ is the standard dot product in $\mathbb{R}^3$.∎


问:
Is it true that: $f(x,y)=constant \implies \nabla f(x,y)= \pmb{0}$?
Let $f: \mathbb{R^2 \rightarrow R}$, be a differential function. Is it true that:
$$f(x,y)=constant \implies \nabla f(x,y)= \pmb{0}$$

I think this should be true since differentiability $\implies\exists\nabla$ and the partial derivative of a constant function is $0$. The answer to the problem is that it is false though, according to my book.

答:
It is clear that if

$$(\forall (x,y)\in \Bbb R^2)$$
$$\;f(x,y)=\;Constant\;=K$$

where $ K\in\Bbb R $, doesn't depend neither on $ x $, nor on $ y $,

we will have
$$(\forall (x,y)\in\Bbb R^2)$$
$$\frac{\partial f}{\partial x}(x,y)=0\;\;
and \;\;\frac{\partial f}{\partial y}(x,y)=0$$


答:
Substitute $x = n - u$ to rewrite

$$ \frac{1}{n^n} \int_{0}^{n} x^n \arctan(x^{1/n}) \, \mathrm{d}x
= \int_{0}^{\infty} \left(1 - \frac{u}{n}\right)^n \arctan((n-u)^{1/n}) \mathbf{1}_{[0,n]}(u) \, \mathrm{d}u. $$

Now by the dominated convergence theorem, this converges to

$$ \int_{0}^{\infty} \lim_{n\to\infty} \left(1 - \frac{u}{n}\right)^n \arctan((n-u)^{1/n}) \mathbf{1}_{[0,n]}(u) \, \mathrm{d}u
= \int_{0}^{\infty} e^{-u} \arctan(1) \, \mathrm{d}u
= \frac{\pi}{4}. $$

Title: Calculate the limit of a sequence $\lim_{n\to \infty} \frac{I_{n}}{n^{n}}$

Body:
Calculate $$\lim_{n\to \infty} \frac{1}{n^{n}}\int_{0}^{n}x^n\arctan(\sqrt[n]{x})dx$$

My approach: I was trying to solve first the integral, so for example for $n=1$, we have that
$$\int x\arctan(x)dx=\frac{1}{2}x^{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)+c$$
and for $n=2$, we can see that $$\int x\arctan(\sqrt{x})dx=\frac{1}{6}(3(x^{2}-1)\arctan(\sqrt{x})-(x-3)\sqrt{x}+c$$
and for $n=3$, we have
$$\int x\arctan(\sqrt[3]{x})dx=\frac{1}{30}(-3x^{5/3}+15(x^{2}+1)\arctan(\sqrt[3]{x})+5x-15\sqrt[3]{x})+c$$

I was trying to see if by induction I could give a closed formula for the integral but the calculation of the integral does not show any pattern. How can I solve this limit?


问:
Average number of strings with edit distance at most 3 (larger alphabet)
Consider a string of length $n \geq 3$ over an alphabet $\{1,\dots, \sigma\}$.   An edit operation is a single symbol insert, deletion or substitution.  The edit distance between two strings is the minimum number of edit operations needed to transform one string into the other one.  Given a string $S$ of length $n$ with $S_i \in \{1,\dots, \sigma\}$, my question relates to the number of distinct strings which are edit distance at most $3$ from $S$.

Let us write $g_{k, \sigma}(S)$ for the number of distinct strings over the alphabet $\{1,\dots, \sigma\}$ which are edit distance at most $k$ from $S$, i.e. $g_{k,\sigma}(S) = |\{S' : d(S', S) \leq k\}|$ where $d(-,-)$ is the edit distance.

Let $X_n$ be a random variable representing a random string over the alphabet $\{1,\dots, \sigma\}$ of length $n$, with the symbols chosen uniformly and independently.

This leads directly to my question:

Let $X_n$ be a random variable representing a random string of
length $n$, with the symbols chosen uniformly and independently.  What is:

$$\mathbb{E}(g_{3, \sigma}(X_n))\;?$$

For $\sigma=2$ we can get an explicit formula $(40+6n-4n^2)/2^n-83/2+(331/12)n-6n^2+(2/3)n^3$. So my question is, what does the dependency on the alphabet size $\sigma$ look like?

答:
Varying v. Unchanged String Length

If, as you initially indicated in response to my comment, the length of the transformed string can differ from the length of the original, then this problem becomes vastly more difficult because the set of distinct editing operations (operations that might potentially yield a distinct result) includes all 18 of the following:

length +3 = 3 insertions

length +2 = 2 insertions and 0 or 1 substitutions

length +1 = 1 insertion and 0, 1, or 2 substitutions

length unchanged = 0, 1, 2, or 3 substitutions; 1 deletion, 1 insertion, and 0 or 1 substitutions

length -1 = 1 deletion and 0, 1, or 2 substitutions

length -2 = 2 deletions and 0 or 1 substitutions

length -3 = 3 deletions

Whenever multiple insertions or multiple deletions are performed, moreover, counting becomes inordinately difficult.  If, on the other hand, we require that the length remain unchanged, we have only 6 editing combinations to consider and the problem becomes more tractable because none of those 6 combinations involves multiple insertions or multiple deletions.  Indeed, the counting for each of the six cases becomes relatively straightforward; the trickiest bit is discounting to avoid double-counting instances when two different editing operations will produce the same string--a problem solved in an answer to another question.

The Six Cases and the Danger of Overcounting

To get our bearings initially, we can generalize this logic:

The string must maintain $n$ symbols.

The expected number of groups of identical symbols is $\frac{n+1}{\sigma}$

The expected number of adjacent, identical symbol pairs is $\frac{n-1}{\sigma}$

The number of ends is 2.

A fine-grained consideration of the five possible types of single edits thus yields:

The number of possible substitutions is $n(\sigma-1)$

The expected number of shrinkages of a group of identical symbols is $\frac{n+1}{\sigma}$

The expected number of expansions of a group of identical symbols with the same symbol is $\frac{n+1}{\sigma}$

The expected number of insertions into a group of identical symbols with the same symbol is $\frac{n-1}{\sigma}$

The number of possible insertions of a different character at the beginning or end is $2(\sigma-1)$

We can now apply that basic logic to each of our six cases:

no edits

Performing no edits whatsoever yields only the original string, so 1 result for this case.

one substitution

There are $n$ different symbols and $\sigma-1$ ways each can be substituted into a different symbol, so $n(\sigma-1)$ results.

two substitutions

There are $\binom{n}{2}$ different pairs and $(\sigma-1)^2$ ways to modify each: $\binom{n}{2}(\sigma-1)^2$ results.

three substitutions

There are $\binom{n}{3}$ different trios and $(\sigma-1)^3$ ways to modify each: $\binom{n}{3}(\sigma-1)^3$.

one deletion, one insertion, no substitutions

For this case, we can generalize this solution for $\sigma=2$ to any $\sigma$, using the same logic to avoid double-counting those instances where two substitutions would yield the same result as one deletion and one insertion.

Let's count the cases where the insertion is to the left of the deletion and then multiply by 2. The combined effect of the insertion and the deletion is to shift all  bits between them to the right while replacing the first one and removing the last one. This result can also be achieved by at most  substitutions, so we need >2. Inserting  within a run of s has the same effect as inserting  at the end of the run. Thus we can count all insertions with different effects once by always inserting the bit complementary to the one to the right of the insertion. Similarly, a deletion within a run has the same effect as a deletion at the start of the run, so we should only count deletions that follow a change between 0 and 1.  That gives us an initial count of:

$2\cdot\frac12\sum_{k=3}^n(n+1-k)=\sum_{k=1}^{n-2}k=\frac{(n-1)(n-2)}2\;$

Because the tricky logic to prevent double-counting carries directly over, the only modification required is to substitute a variable $\sigma$ for the fixed $\sigma=2$:

$2\cdot\frac{1}{\sigma}\sum_{k=3}^n(n+1-k)=2\cdot\frac{1}{\sigma}\sum_{k=1}^{n-2}k=\frac{(n-1)(n-2)}{\sigma}\;$

The overcount of results that have already been tallied as two substitutions can be calculated as follows when $\sigma=2$:

If there are no further changes in the  shifted bits other than the
one preceding the deletion, then only the bits next to the insertion
and deletion change, and we can achieve that with 2 substitutions, so
we have to subtract

$\sum_{k=3}^n\left(\frac12\right)^{k-2}(n+1-k)=\sum_{k=1}^{n-2}\left(\frac12\right)^{n-k-1}k=n-3+2^{-(n-2)}\;$

Again, our only modification is to substitute $\sigma$ for 2:

$\sum_{k=3}^n\left(\frac1{\sigma}\right)^{k-2}(n+1-k)=\sum_{k=1}^{n-2}\left(\frac1{\sigma}\right)^{n-k-1}k=n-3+{\sigma}^{-(n-2)}\;$

Also, if the entire range of shifted bits consists of alternating
zeros and ones, then swapping the insertion and the deletion yields
the same effect, so in this case we were double-counting and need to
subtract

$\sum_{k=3}^n\left(\frac12\right)^{k-1}(n+1-k)\;$

Swapping in $\sigma$ a final time yields:

$\sum_{k=3}^n\left(\frac1{\sigma}\right)^{k-1}(n+1-k)\;$

These two overcounts (which, alas, cannot be combined as cleanly as when the symbols are binary) are then subtracted from the initial count of deletion/insertion operations to yield the overall results produced by this case, but not by case 3 above:

$\frac{(n-1)(n-2)}{\sigma}\ - \left(n-3+{\sigma}^{-(n-2)}\right) - \sum_{k=3}^n\left(\frac1{\sigma}\right)^{k-1}(n+1-k)\;$

one deletion, one insertion, one substitution

That same calculation carries over to the final case.  Here, however, each combination of one deletion and one insertion--likewise discounted to avoid double-counting the triple substitutions already tallied in case 4 above--is accompanied by a third edit: a substitution involving one of the $n-1$ original symbols remaining after the deletion.  Since each of these $(n-1)$ symbols admits $(\sigma-1)$ novel substitutions, the total count for the sixth and final case becomes:

$\left(\frac{(n-1)(n-2)}{\sigma}\ - \left(n-3+{\sigma}^{-(n-2)}\right) - \sum_{k=3}^n\left(\frac1{\sigma}\right)^{k-1}(n+1-k)\right)(n-1)(\sigma-1);$

Summing the (previously uncounted) results produced by each of these six cases should yield the expected count when the length of the string remains unchanged.  It's ugly (perhaps unnecessarily), but I hope correct.


答:
I assume the term "versor" $\equiv$ "vector".

Note that

$\dfrac{d(T \times N)}{ds} = \dot T \times N + T \times \dot N, \tag 1$

where

$\dot T = \dfrac{dT}{ds} \tag 2$

and so forth. We have the Frenet-Serret equations

$\dot T = \kappa N \tag 3$

and

$\dot N = -\kappa T + \tau B; \tag 4$

substituting (3) into (1) yields

$\dfrac{d(T \times N)}{ds} = \kappa N \times N + T \times \dot N = T \times \dot N, \tag 5$

the calculation mentioned in the question itself; we further substitute (4) into (5):

$\dfrac{d(T \times N)}{ds} = T \times(-\kappa T + \tau B)$
$= -\kappa T \times T + \tau T \times B = -\tau N, \tag 6$

and thus

$T \cdot \dfrac{d(T \times N)}{ds} = -\tau T \cdot N = 0. \tag 7$

In short, $d(T \times N)/ds$, being proportional to $N$, is orthogonal to $T$.  $OE\Delta$.

Title: Why is $d(T \times N)/ds$ orthogonal to the unit tangent vector $T$?

Body:
Consider a curve in $\mathbb{R}^3$ and its Frenet-Serret frame $(T,N,B)$. A calculation gives $$\frac d{ds} (T \times N)= T \times \frac {dN}{ds}.$$

Question:  I read that this implies that $\frac d{ds} (T \times N)$ is orthogonal to $T.$ Why is it so?


问:
Does Frechet differentiability of the first order partials imply twice Frechet differentiability?
Let $f:\mathbb{R}^n \to \mathbb{R}$. If all first order partial derivatives of $f$ are Frechet differentiable, must $f$ be twice Frechet differentiable?

This question is inspired by my comment on this answer: https://math.stackexchange.com/a/1181928/95800

答:
Yes, it's true; the proof is a simple matter of carefully organizing everything you know. We can prove the following general theorem:

Let $E_1,\dots, E_n, F$ be Banach spaces over the same field, let $E:= E_1\times \cdots E_n$ and $U\subset E$ be a non-empty open set. Suppose $f:U\to F$ is a function such that all its first order partials $\partial_if$ (which are maps $U\to L(E_i,F)$) are differentiable on $U$. Then, $f$ is twice differentiable on $U$.

It is a classic theorem that continuity of first order partials implies $f$ is $C^1$, so it makes sense to even talk about $f$ being twice differentiable. Next, recall that for $a\in U,h\in E$, we have
\begin{align}
Df_a(h)&= \sum_{i=1}^n(\partial_if)_a(h_i) \equiv\sum_{i=1}^n \partial_if_a(h_i)
\end{align}
In other words, we take the $i^{th}$ partial derivative at the point $a\in U$, $(\partial_if)_a$, and evaluate on $h_i\in E_i$. Now, introduce the projections $\pi_i:E\to E_i$. The above equation can be written more concisely as
\begin{align}
Df_a&=\sum_{i=1}^n\partial_if_a \circ \pi_i
\end{align}
(this is essentially the meaning of the equation $df=\sum_{i=1}^n\frac{\partial f}{\partial x^i}\, dx^i$ which you may have seen several times before). Now that we have related the first derivative to the partial derivatives, we can now compute: for any $a\in U$ and $h\in E$ with sufficiently small norm so that $a+h\in U$, we have
\begin{align}
Df_{a+h}&=\sum_{i=1}^n (\partial_if)_{a+h} \circ \pi_i\\
&=\sum_{i=1}^n \bigg(\partial_if_a + D(\partial_if)_a[h] + \phi_i(h)\bigg)\circ \pi_i
\end{align}
for some $\phi_i:U\to L(E_i,F)$ such that $\frac{\lVert \phi_i(h)\rVert}{\lVert h\rVert} \to 0$ as $h\to 0$ (hopefully it's clear which norm is which). This is simply due to the differentiability assumption on $\partial_if$ at the point $a\in U$. Now, we simplify some more:
\begin{align}
Df_{a+h}&=\sum_{i=1}^n (\partial_if)_a\circ \pi_i + \sum_{i=1}^n \left(D(\partial_if)_a[h]\right)\circ \pi_i + \sum_{i=1}^n \phi_i(h) \circ \pi_i \\
\end{align}
The first term is simply $Df_a$. The second term defines a bounded linear function of $h$, which we shall call $T(h)$. Finally, I claim the last term is $o(\lVert h\rVert)$ as $h\to 0$. To see this, we just calculate:
\begin{align}
\dfrac{1}{\lVert h\rVert}\left\lVert\sum_{i=1}^n \phi_i(h)\circ \pi_i \right\rVert &\leq \sum_{i=1}^n \frac{\lVert \phi_i(h)\circ \pi_i\rVert}{\lVert h\rVert}
\leq \sum_{i=1}^n\frac{\lVert \phi_i(h)\rVert}{\lVert h\rVert} \cdot \lVert \pi_i\rVert.
\end{align}
There are a bunch of different norms here, so you should make sure you know which is which. As $h\to 0$, each of the fractions approaches $0$, therefore, the entire RHS approaches $0$. In other words, we have shown that
\begin{align}
Df_{a+h}&= Df_a + T(h) + \rho(h)
\end{align}
for some bounded linear map $T:E \to L(E,F)$ and some function $\rho:U\to L(E,F)$ such that $\frac{\lVert\rho(h) \rVert}{\lVert h\rVert}\to 0$ as $h\to 0$. This is precisely what it means for $Df$ to be differentiable at $a$.

By the way, the above even gives a formula for $D^2f_a$, which is $T$. If you want, you can verify (after identifying with appropriate bilinear maps) that
\begin{align}
D^2f_a[h,\cdot] &= \sum_{i=1}^n D(\partial_if)_a(h) \circ \pi_i(\cdot)
=\sum_{i,j=1}^n (\partial_j\partial_if)_a[h_j, \pi_i(\cdot)].
\end{align}
Also, it's a classic theorem that $D^2f_a$ is a symmetric function of its two arguments. So, we can be sloppy with the order of indices, and this is essentially the meaning of the formula $D^2f = \sum_{i,j=1}^n \frac{\partial^2f}{\partial x^i \partial x^j}\,dx^i\, dx^j$ which you may have seen before.


问:
Is this proof for countability correct?
Conjecture: The set A = {(a1, a2, a3,..)} of infinite sequence of integers is  countably infinite.

I think that this is true (correct me if I'm wrong) but I'm not sure if my proof is correct. Can someone help?

Here's my proof:

Firstly, we agree to list the sequence in an increasing order, i.e. a1≤a2≤a3.... so on.
Then we can list the elements of A as follows:

Compare the first entry in the tuples(a1). Then arrange the tuples in an increasing order of them. Repeat the same for the second entry and so on...
By the end, we would have a list of the elements of A. Thus it is countably infinite.

I highly doubt the validity of this "sorting the tuples" proof. If it's wrong, please also supply a correct proof.
Thanks.

答:
As to where your proof fails:

Firstly, we agree to list the sequence in an increasing order, i.e. a1≤a2≤a3.... so on.

Many sequences do not have a least element to act as $a_1$, for example, the sequence $-1, -2, -3, \ldots$.

To avoid this, let's just talk about sequences of natural numbers instead. Then we have a least element always. But:

For any increasing sequence that is not eventually constant, there are infinitely many sequences which when reordered to be increasing will give that same sequence. So by making this reordering, you are treating all of those infinitely many sequences as if they were only one. For example, all of the following sequences reorder to be $1,2,3,\ldots$:

$2,1,4,3,6,5,\ldots$

$2,1,3,4,5,7,6,\ldots$

$2,1,3,5,4,7,9,11,6,\ldots$

$\ldots$

Alright, to sidestep that issue, let's just restrict our attention to increasing sequences right from the start. Then we can proceed.

Then we can list the elements of A as follows:

Compare the first entry in the tuples(a1). Then arrange the tuples in an increasing order of them.

There are infinitely many increasing sequences that start with $1$. For example

$1,2,3,4,5,\ldots$

$1,3,4,5,6,\ldots$

$1,4,5,6,7,\ldots$

$\ldots$

So you will have completely filled out your "countable" list before even reaching the sequences that start with $2$.

Title: Is this proof for countability correct?


问:
Functoriality of the maximal unramified extension
Let $K$ be a finite extension of $\mathbb{Q}_p$ and $L$ a finite Galois extension of $K$. We have $$\text{Gal}(K^{nr}/K)\cong\hat{\mathbb{Z}}\cong \text{Gal}(L^{nr}/L) $$ where $K^{nr},L^{nr}$ denotes the maximal unramified extension of $K$ and $L$ respectively. I would like to understand the map
$$\text{Gal}(L^{nr}/L)\to \text{Gal}(K^{nr}/K).$$
I think, well maybe hope, it is induced by a morphism
$$\mathbb{Z}\to \mathbb{Z}.$$
Would that make sense?

答:
$K^{nr} = \bigcup_{p\ \nmid\ n} K(\zeta_n)$

$L^{nr} = \bigcup_{p\ \nmid\ n} L(\zeta_n)$

$Gal(K^{nr}/K) \cong Gal(\overline{k}/k)$ where $k=O_K/(\pi_K),\overline{k}=O_{K^{nr}}/(\pi_K)$, the isomorphism is the reduction map $O_{K^{nr}}\to O_{K^{nr}}/(\pi_K)$.

The map $ Gal(\overline{k}/\ell)\to Gal(\overline{k}/k)$ is a bit obvious, right ?

$ Gal(\overline{k}/\ell)$ is topologically generated by the Frobenius $\phi_\ell(a) = a^{|\ell|}=\phi_k^f(a)$ where $f=[\ell:k]$ and $\phi_k(a)=a^{|k|}$.

The isomorphism  $ Gal(\overline{k}/k)\to \hat{\Bbb{Z}}$ sends $\phi_k$ to $1$.

So the map $\hat{\Bbb{Z}}\to \hat{\Bbb{Z}}$ corresponding to $ Gal(\overline{k}/\ell)\to Gal(\overline{k}/k)$ is just $c\to fc$.

No need that $L/K$ is Galois.

On the $L^{nr}$ side the Frobenius lifts uniquely to the automorphism defined by $\sigma_L(\zeta_n)=\zeta_n^{|\ell|}$.


问:
PDE $u_y+e^uu_x=0$ solution where initial data is discontinuous

Given a pde $$u_y+e^uu_x=0$$ $x \in \mathbb{R}$, $t>0$ with the initial conditions $$u(x,0)=f(x)=\begin{cases} 2 & x<0 \\ 1 & x>0 \end{cases}$$
Solve the pde.

My attempt so far, I find the characteristic curves for $(s,0,f(s))$
$$\frac{dy}{dt}=1$$
which gives $$y=t$$
$$\frac{du}{dt}=0$$ which gives $$u=f(s)$$
and $$\frac{dx}{dt}=e^u \implies $$ $$x=e^{f(s)}t+s$$
So
$$u(x,y)=f(x-e^uy)=\begin{cases} 2 & x< e^2 t \\ 1 & x>et \end{cases}$$

Now I know that to find the time the shock forms I must find $t$ such that $$(e^u,1) \cdot (0,-e^{f(s)}f'(s)-1)=0 \implies $$ which gives $$t=\frac{1}{e^{f(s)} f'(s)}$$

But the problem is that $f(s)$ is constant on each interval so the derivative vanishes, therefore the shock time becones infinite, this is what I don't understand, how can I find the time the shock forms?

答:
The initial data is discontinuous. Therefore, this is a Riemann problem for the PDE $$
u_y + (e^u)_x = 0
$$
rewritten in conservation form, whose flux function $u\mapsto e^u$ is convex. In the present case, the base characteristics intersect, and the solution is a single shock wave with speed $$
s = \frac{e^2 - e^1}{2 - 1} = e\, (e-1) \approx 4.67
$$
deduced from the Rankine-Hugoniot condition. The shock forms immediately, at $y=0^+$.


问:
Formula for range of a projectile landing at some distance above the ground
What will be the formula for range of the projectile if it is launched from the ground level but lands at some height about the ground?
I have searched everywhere for the answer, but everyone has discussed the case of "launching from some height" but not "landing at some height".
The picture of the scenario is attached below.

答:
In a physical framework you may also perform some simplification, assuming that $200\,m$ are negligible with respect to $h_m$, as the picture seems to suggest. Removing the last step you have a range equal to $\frac{u^2}{g}\sin(2\theta)$, and assuming the last part of the trajectory to be approximately linear you have that the answer to your question is fairly close to
$$ \frac{u^2}{g}\sin(2\theta)-f\cot\theta. $$
You can compare this with the exact expression found by Tavish:

$$\begin{eqnarray*} \frac{u^2 \sin2\theta + u\cos\theta\sqrt{4u^2\sin^2\theta -8gf}}{2g}&=&\frac{u^2\sin(2\theta)}{2g}\left(1+\sqrt{1-\frac{2gf}{u^2\sin^2\theta}}\right)\\&\approx&\frac{u^2\sin(2\theta)}{2g}\left(2-\frac{gf}{u^2\sin^2\theta}\right)\\&=&\frac{u^2}{g}\sin(2\theta)-f\cot\theta\end{eqnarray*}$$
and the approximation is accurate as soon as $gf\ll u^2\sin^2\theta$.


问:
How do I solve that matrix operation?

Let $x$ be a column vector in $\Bbb R^n$, and $x$ different from matrix zero. Let
$A = \frac{ x x^{T}}{x^{T} x}$, T denotes tranpose. Calculate $A²$.

I hope that I am not the only one who was astonished with this exercise. Someone here knows what it means $\frac{ x x^{T}}{x^{T} x}$? I can imagine that a division is simple take the inverse of denominator, but so the exercise become so trivial that I discard it. Anyway, someone here knows what it means?

答:
If $x\in\mathbb R^n$ is nonzero and $A=\dfrac{xx^T}{x^Tx}$, then by associativity of matrix multiplication and scalar multiplication we obtain

$$\begin{array}{rcl}A^2&=&\left(\dfrac{xx^T}{x^Tx}\right)\left(\dfrac{xx^T}{x^Tx}\right)\\
&=&\dfrac{x(x^Tx)x^T}{(x^Tx)^2}\\
&=&\dfrac{x^Tx}{(x^Tx)^2}\cdot xx^T\\
&=&\dfrac{1}{x^Tx}\cdot xx^T\\
&=&\dfrac{xx^T}{x^Tx}\\
&=&A.
\end{array}$$

Title: How do I solve that matrix operation?


问:
Rotman's Homological Algebra, Theorem 7.8, showing two definitions of Ext have the same values
I have a question regarding one detail in his proof of this theorem below.

Note: in the proof, he uses $\operatorname{Ext}^n(C,..)$ to denote $R^n\operatorname{Hom}(C,..)$ and $\operatorname{ext}^n(...,A)$ to denote $R^n \operatorname{Hom}(...,A)$.

Now, I understand everything except for the highlighted isomorphism near the end of his proof.  How did he conclude that $\text{Ext}^1(K_{n-2}, L^0) \cong \text{ext}^1(K_{n-1}, A) $?

At first, I thought it was because we prove $\text{Ext}^1 \cong \text{ext}^1$  and we also have this isomorphism $\text{Ext}^1(K_j, L^i) \cong \text{Ext}^1(K_{j-1}, L^{i+1})$ for $i,j \geq 0$, so that would imply
$$  \text{ext}^1(K_j, L^i) \cong \text{Ext}^1(K_j, L^i) \cong \text{Ext}^1(K_{j-1}, L^{i+1}) .$$  But that would means we would have
$$\text{Ext}^1(K_{n-2}, L^0) \cong \text{ext}^1(K_{n-1}, L^{-1}).  $$
But he never defined what $L^{-1}$ is and if we defined $L^{-1} = A$, then he would have to show that this isomorphism $\text{Ext}^1(K_j, L^i) \cong \text{Ext}^1(K_{j-1}, L^{i+1})$ is true for $i,j$ not necessarily greater than or equal to 0 as well.

Thank you very much.

Edit: Corollary 6.16 is below

答:
$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\ext{ext}$The string of isomorphisms at the end of the proof has a typo in the first step that carries through and makes for the senseless step that you have highlighted (the highlighted step is senseless because it is saying $\ext^1(K_{n-2},A)\cong \ext^1(K_{n-1},A)$, once we recall $\Ext^1\cong \ext^1$ and $L^0=A$). Here's what the first isomorphism should be:
$$\Ext^{n+1}(C,A) = \Ext^{n+1}(C,L^0) \cong \Ext^1(C,L^n)$$
with that correction in place, we get the desired isomorphism of $\Ext^{n+1}$ and $\ext^{n+1}$ (assuming I haven't introduced any new typos :)):
\begin{align*}
&\Ext^{n+1}(C,A)\cong\Ext^1(C,L^n)\cong \Ext^1(K_{n-1},L^0)\cong\ext^1(K_{n-1},L^0) \\
&= \ext^1(K_{n-1},A) \cong \ext^{n+1}(C,A)
\end{align*}


答:
Assuming that your definition of subdomain requires that the unity is non zero, then you have for this unity $c\in R$

$$
cc=ce
$$

where $e$ is the unity of $D$, because both, by definition, give $c$. Hence

$$
c(c-e)=0
$$

and, since $c\ne0$, $c=e$.

Of course, this is just using that $c$ is idempotent. Which it is, so why bother?

Title: The unity of a subdomain

Body:
Let $D$ be an integral domain and $R$ a subdomain , I'm trying to show that both $R$ and $D$ have the same unity. I did it using the fact that $1$ and $0$ are the only idempotent elements inside any integral domain. I'm trying to do it in another way which is the following:

Since $R$ is a subdomain it has unity call it $c$, let $a\in R$ then $ac=a$ since we can cancel we get that $c$ which is in $R$ is equal to $1$ then $1\in R$.

I'm not really confident this right. If this is wrong what other way is there that does not use the fact that there are two idempotent elements in an integral domain.


问:
Lines intersecting in a segment? Coloring the plane
I am reading about the problem of cutting/coloring a plane using 2 colors black/white. The problem states:

A number of straight lines are drawn across a piece of paper and each
line extends all the way from one border to another. As a result the
paper is divided into regions.   Show that it is possible to color
each region black or white so that no adjacent regions have the same
color

The proof is given using induction on the number of lines and inversion of colors of one of the segments created per new line and is understood.

What I don't understand is the following.

The proof states that (emphasis mine):

The requirement for the solution is that the lines are straight i.e.
intersect in a single point. If the lines are not straight and the
intersect in a segment of a line, inverting the colors of one of the
two regions does not guarantee that the coloring of adjacent regions
at the boundary of the left and right regions is satisfactory.

What is meant by lines intersecting in a segment of a line? Trying it out with 2 lines that are curved and intersecting in 2 points it seems the solution would still hold. So does it mean the lines are overlapping?

答:
The problem is if the line segments overlap in some configuration that is more than just a finite set of distinct points:

In this image, the blue and red curves overlap along the purple segment, forming 4 regions which cannot be colored in the desired manner.


问:
How to solve for $x$ : $ae^{bx}+x = c$
How to solve for $x$ :  $ae^{bx}+x = c$

I've tried to solve it with the Lambert W function, but, in all of the methods I tried, I had $x$  as the Lambert W function parameter. Can I solve it without graphs? Thank you for your help.

Title: How to solve for $x$ : $ae^{bx}+x = c$

Body:
$$ae^{bx}+x=c$$

$$ae^{bx}=c-x$$

$$ae^{bc-b(c-x)}=c-x$$

$$abe^{bc}=b(c-x)e^{b(c-x)}$$

$$b(c-x)=W(abe^{bc}).$$

$$x=c-\frac1bW(abe^{bc}).$$

(By hand.)


问:
Combinatorics- 6 digits decreasing code
I need to find out how many combinations can you use to make a code sequence like this :
All digits are 0-9 and they can't increase
For example: 887665 is a valid combination.

I tried putting  a monotonic hard increasing sequence and then subtract it from $ 10^6 $ but I can't figure out how many combinations there are for increasing sequences, only that if we start at 4 there is only 1 combination.
$ \begin{pmatrix}0&1&2&3&4&5\\ 1&2&3&4&5&6\\ 2&3&4&5&6&7\\ 3&4&5&6&7&8\\ 4&5&6&7&8&9\end{pmatrix} $

答:
You are asking for the number of tuples $(x_1,\dots,x_6)$ such that $$9\geq x_1\geq\dots\geq x_6\geq 0.$$
We can see this as the number of ways to split $6$ balls into $10$ bins, which is $\binom{6+10-1}{6}=5005$.

Alternatively, add $6$ to the first digit, $5$ to the second, and so on down to $1$ for the last. This transforms your number into a six-element subset of $\{1,\dots,15\}$. Further, given any subset of $\{1,\dots,15\}$ with size $6$, we can reverse this process to get a $6$ digit number with non-increasing digits. So the answer is just $\binom{15}{6}$.

A third method: count the number of paths from $(0,9)$ to $(6,0)$, where you are only allowed to move one step right or down at a time. What is the connection to the original problem?

Title: Combinatorics- 6 digits decreasing code


问:
$\mathrm{Tor}_1^R(R/I,R/J)=(I\cap J)/(IJ)$
I am proving that $\mathrm{Tor}_1^R(R/I,R/J)=(I\cap J)/(IJ)$. For that, I consider the exact sequence
$0\rightarrow I \rightarrow R\rightarrow R/I\rightarrow 0$, and then I have an long exact sequence

$\begin{align}\DeclareMathOperator{\Tor}{Tor}
\dots\rightarrow0=\Tor^R_1(R,R/J)\rightarrow\Tor^R_1(R/I,R/J)\rightarrow I\otimes_R R/J \rightarrow R/J\rightarrow R/I\otimes_R R/J \rightarrow 0
\end{align}$

By that, to conclude I only need to prove that the map
$
\Tor^R_1(R/I,R/J)\rightarrow I\otimes_R R/J $ is surjective, because in that case by the exactness of the sequence I have the result.

¿Is that map necessary surjective?

答:
By definition of exact sequence, $\operatorname{Tor}^R_1(R/I,R/J)$ is the kernel of the map $I\otimes_RR/J\to R/J$.

On the other hand, you also have the exact sequence $0\to J\to R\to R/J\to0$ and tensoring with $I$ you get
$$
I\otimes_RJ\to I\otimes_RR\to I\otimes_R R/J\to 0
$$
so $I\otimes_R R/J\cong I/K$ and it's not difficult to show that $K=IJ$. Then check that the map $I\otimes_RR/J\to R/J$ translates to the map
$$
I/IJ\to R/J, \quad a+IJ\mapsto a+J
$$
whose kernel is $(I\cap J)/IJ$.


问:
Splitting multivariate normal into individual (correlated) components
I have a multivariate normal variable $X$ with mean zero and variance $\Sigma$. I would like to write every component $X_i$ of $X$ as:
$$
X_i = \phi_i Z_i
$$
where $\phi_i$ is a scalar and $Z_i$ is a scalar standard normal random variable (for every $i$). Question: what are the values of $\phi_i$ (given $\Sigma$), and what is the correlation matrix of the random variables $Z$?</p>

答:
Using the standard properties of variance and covariance, we have that $\phi_i = \sqrt{\Sigma_{i,i}}$ and
$$\displaystyle \operatorname{cov}(Z_i,Z_j)=\operatorname{cov}(\phi_i^{-1}X_i,\phi_j^{-1}X_j)= \frac{\Sigma_{i,j}}{\phi_i\phi_j}=\frac{\Sigma_{i,j}}{\sqrt{\Sigma_{i,i}}\sqrt{\Sigma_{j,j}}} = \rho_{i,j}$$ where $\rho_{i,j}$ is
the correlation coefficient of $X_i$ and $X_j$ (and also of $Z_i$ and $Z_j$).


问:
Explicit description of nilpotent cones
Given a Lie algebra $\mathfrak{g}$, one can define its nilpotent cone $\mathcal{N}$ to be the subspace that consists of element $x \in \mathfrak{g}$ such that $ad(x)$ is a nilpotent operator.

In the case that $\mathfrak{g} = \mathfrak{sl_2}$, each $ad(x)$ is an endomorphism of the vector space $\mathfrak{sl_2}$. By choosing a basis, this gives a $3 \times 3$ matrix. Explicitly, for the standard basis $\beta=\{h, e, f\}$ and

$$x =  \begin{pmatrix}
a & b \\
c & -a
\end{pmatrix},$$

the matrix $[ad(x)]_\beta$ is

$$
\begin{pmatrix}
0 & -c & b \\
-2b & 2a & 0 \\
2c & 0 & -2a
\end{pmatrix}.
$$

It is then a linear algebra exercise to show that the subset of nilpotent operators is exactly

$$ \mathcal{N} = \{ x \in \mathfrak{g} \,|\, \det(x)=0 \}.$$

The computation is elementary but long.. However, the result is quite elegant. I fail to spot any insight why the defining equation is given by the determinant.

Question

Is this a coincidence? In general, or at least for those semisimple Lie algebras, what are their nilpotent cones in terms of explicit equations?

答:
There's two different things at work here:

A rather deep theorem that says that for a semisimple Lie algebra written as matrices, an element $x$ is $ad$-nilpotent if and only if it is nilpotent as matrix.

A more coincidental thing, namely, that a matrix of the form $\pmatrix{a &b\\c&-a}$ is nilpotent if and only if its determinant is $0$. For bigger matrices, only one direction holds, and one cannot conclude anything from the determinant being zero. Example: $\pmatrix{1&0&0\\0&-1&0\\0&0&0} \in \mathfrak{sl}_3$, in spite of having zero determinant, is not nilpotent in any reasonable sense of the word.

Title: Explicit description of nilpotent cones


问:
Calculating percent reduction in variance after deductible

The amount of a claim that a car insurance company pays out follows an exponential distribution. By imposing a deductible of $d$, the insurance company reduced the expected claim payment by $10\%$. Calculate the percentage reduction on the variance of the claim payment.

This question was previously discussed here. Based on this discussion, I came up with this solution:

Let $X$ be the random variable for the claim amount before the deductible, d,  is applied and $Y$ be the random variable for the claim amount after the deductible, d,  is applied. Then, $Y = X-d$ if $d \le X <\infty$ and $Y = 0$ otherwise.

Now, $E[Y] = E[X] - 0.1 E[X] \implies E[Y] = 0.9E[X] = E[0.9X] \implies Y = 0.9X$.

So, $Var[Y] = Var[0.9X] = 0.9^2Var[X] = 0.81 Var[X]$

which means that the variance has been reduced by approximately $19\%$ but the correct answer is $1\%$. What have I done incorrectly?

答:
Note that
$$
\mathbb{E}[Y] \overset{\star}= \mathbb{E}[0.9 X]
$$
does not in general imply $Y = 0.9X$. And what about the deductible? As you already noted, we actually have the following relationship between $X$ and $Y$:
\begin{align*}
Y = \boldsymbol{1}_{\{X\geq d\}} (X-d).
\end{align*}
The idea is to use $\star$ to determine $d$, and then calculate $\text{Var}[Y]$ more or less directly.

Proof

Note that $X \sim \text{exp}(\lambda)$, so that in particular $\mathbb{E}[X] = \lambda$. To simplify matters, let us first note that conditionally on $(X \geq d)$, the random variable $Y$ also follows an $\text{exp}(\lambda)$-distribution. This may be seen from the calculation
\begin{align*}
\mathbb{P}(Y \leq y \, | \, X \geq d)
&=
\mathbb{P}(X - d \leq y \, | \, X \geq d) \\
&=
\frac{\mathbb{P}(d \leq X \leq y + d)}{\mathbb{P}(d \leq X)} \\
&=
\frac{e^{-\frac{d}{\lambda}}-e^{-\frac{y+d}{\lambda}}}{e^{-\frac{d}{\lambda}}} \\
&= 1 - e^{-\frac{y}{\lambda}}
\end{align*}
for $y > 0$. It is actually a direct consequence of the so-called memoryless property of the exponential distribution.

For $k\in\mathbb{N}$, we obtain from the law of total expectation that
\begin{align*}
\mathbb{E}[Y^k]
&=
\mathbb{E}[Y^k \, | \, X \geq d]\cdot\mathbb{P}(X \geq d) + \mathbb{E}[Y^k \, | \, X < d]\cdot\mathbb{P}(X < d) \\
&=
\mathbb{E}[Y^k \, | \, X \geq d]\cdot\mathbb{P}(X \geq d) \\
&=
k! \lambda^k e^{-\frac{d}{\lambda}},
\end{align*}
since the $n$'th moment of an exponential distribution with mean $\lambda$ is $n!\lambda^n$.

In particular,
\begin{align*}
\mathbb{E}[Y]
&=
\lambda e^{-\frac{d}{\lambda}}.
\end{align*}
Thus imposing $\star$, the deductible $d$ is given as the solution to
\begin{align*}
\lambda e^{-\frac{d}{\lambda}} = 0.9\mathbb{E}[X] = 0.9\lambda,
\end{align*}
which yields $d = -\log(0.9)\lambda \approx 0.11 \lambda$. (You may think about why $d>0.1 \lambda$ is reasonable.)

Using this value of $d$, which exactly yields $e^{-\frac{d}{\lambda}} = 0.9$, we find that
\begin{align*}
\text{Var}[Y]
&=
\mathbb{E}[Y^2] - \mathbb{E}[Y]^2 \\
&=
2\lambda^2 \cdot 0.9 - (0.9 \lambda)^2 \\
&=
0.99 \lambda^2 \\
&=
0.99 \text{Var}[X],
\end{align*}
as desired.


问:
The Hasse principle for bivariate cubic forms
In a talk he gave at the MSRI in 2011, Manjul Bhargava sketches a proof of why the Hasse principle holds for all bivariate cubic forms. It goes as follows:

Suppose the bivariate cubic forms $f$ has no root in $\mathbb{P}_{\mathbb{Q}}^1$.

Then $f$ factors over a cubic field.

By the Chebotarev density theorem, this implies that $f$ is irreducible mod $p$ for some prime $p$.

Thus $f$ has no root in $\mathbb{P}_{\mathbb{Q}_p}^1$ for some prime $p$.

I regret to say that I am having trouble following this proof. My main difficulties lie in understanding how the Chebotarev density theorem comes into it, seeing as I thought that to be a statement about the density of primes in non-Abelian extensions.

Also, step (2): Why does the non-existence of a root in $\mathbb{P}_{\mathbb{Q}}^1$ imply that $f$ must factor over some cubic field extension?

If a textbook proof of the above statement exists, I would much appreciate it if someone could point it out.

Lastly: He also mentions that the reason the Hasse principle holds for bivariate cubic forms is "basically the same as for bivariate quadratic forms". But surely, the Hasse principle for bivariate quadratic forms simply follows from the fact that any element $a$ in an algebraic number field $K$ is a square in $K$ if and only if it is a square in every completion of $K$, which is a much simpler argument, is it not?

Many thanks.

答:
$\sum_{j=0}^3 a_j x^{3-j}y^j=0$ has no non-zero solution in $\Bbb{Q}^2$ implies that $a_0\ne 0$ and $f(x)=\sum_{j=0}^3 a_j x^{3-j}$ has no solution in $\Bbb{Q}$ so it is irreducible.

Replacing $f$ by $a_0^{-3} f(a_0 x)$ We may assume that $f$ is monic.

With $a$ a root of $f$ the splitting field is  $L=\Bbb{Q}(a,\sqrt{Disc(f)})$.

$G=Gal(L/\Bbb{Q})$.

Any $\sigma\in G$ non-trivial on $\Bbb{Q}(a)$ has order $3$.

Chebotarev says that the density of $p$ such that there is a prime $P\subset O_L, p\in P$ with $\forall c\in O_K,\sigma(c)=c^p\bmod P$ is $|H|/|G|$ where $H = \{ g\sigma g^{-1},g\in G\}$.

For such a $p$ not dividing $Disc(f)$ then $f$ is irreducible over $\Bbb{F}_p[x]$ so $y^3f(x/y)=\sum_{j=0}^3 a_j x^{3-j}y^j=0$ has no non-zero solution in $\Bbb{F}_p^2$.

From a solution in $\Bbb{Q}_p^2$ we get a solution in $\Bbb{F}_p^2$.


问:
What's the normal line when partial derivative with respect to $x = 0$?
So I know the equation of a normal line is: $$\frac{x-a}{f_1(a,b)}=\frac{y-b}{f_2(a,b)}=\frac{z-f(a,b)}{-1}$$. But in my textbook there is a litle note that says "with suitable modifications if either $f_1(a,b)=0$ or $f_2(a,b)=0$". Funny that they forget what that modification would look like. Now I have a problem where $f_1=0$, and don't know what to do. I searched for this topic, but frankly the given formula we use is apparantly not so common. Anyhow If someone would know the modification. It would be very helpful.

Thanks in advance.

答:
You have a surface defined by $z=f(x,y)$ and a point $P(a,b, f(a,b))$ in that surface.

The plane tangent to the surface at point P is given by:

$z= f(a,b)+ \frac{\delta f}{\delta x}(a,b)(x-a)+ \frac{\delta f}{\delta y}(a,b)(y-a)$

Let's write it like

$f(a,b)+ \frac{\delta f}{\delta x}(a,b)(x-a)+ \frac{\delta f}{\delta y}(a,b)(y-a) -z =0$

Now compare with a generic plane $Ax+By+Cz+D=0$

We see that $A=\frac{\delta f}{\delta x}(a,b)\;\;$ and $\;\;B=\frac{\delta f}{\delta y}(a,b)\;\;$ and $\;\;C=-1$

We know that $(A,B,C)$ are the components of a vector perpendicular to the plane. So it's the normal to the surface in that point P.

Defining a line in 3D is better done in parametrics. So our normal is:

$x=a+At$

$y=b+Bt$

$z=f(a,b)-t$

If you eliminate "t" in these parametrics you arrive to your original expression

$\frac{x-a}{A}=\frac{y-b}{B}=\frac{z-f(a,b)}{-1}$

Notice that you can avoid division by zero:
$B(x-a)=A(y-b)$

Now, if $A=\frac{\delta f}{\delta x}(a,b)=0$ then the first parametric is just $x=a$, but you can still use the other expression $\frac{y-b}{B}=\frac{z-f(a,b)}{-1}\;\;$ or without denominators $\;\;y-b=B(f(a,b)-z)$


问:
Let $G$ be f.g. with $H\le G$ s.t. $[G:H]<\infty$. Then $\exists K\le H$ with $K\sqsubseteq G$ and $[G:K]<\infty$.
This is Exercise 4.29 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

Definition: A subgroup $H\le G$ is characteristic in $G$, written $H\sqsubseteq G$,  when, for all $\sigma\in {\rm Aut}(G)$, we have $\sigma(H)=H$ (or, equivalently, $\sigma(H)\le H$).

The Question:

Let $G$ be a finitely-generated group with $H\le G$ such that $[G:H]<\infty$. Then there exists $K\le H$ characteristic in $G$ with $[G:K]<\infty$.

Thoughts:

It is well-known that any finite index subgroup of a finitely-generated group is itself finitely-generated. Hence $H$ is finitely-generated. However, I don't recall this result in the book so far.

It might help to note that

$$[G:K]=[G:H][H:K]$$

as cardinal numbers, once we have found a candidate for $K$. This is proven early on in the book.

A special case is when $K=H$. All we need to prove there is that $K$ is characteristic in $G$. There's nothing to say, though, that equality is always possible.

I have asked a question here on characteristic subgroups before: An abelian, characteristically simple group is divisible (supposedly).

I have a few years of experience with combinatorial group theory, so, given that we're talking about finitely-generated groups, I feel as if I should be able to answer this; I guess my main difficulty is the property of being characteristic.

Please help :)

答:
Given a subgroup $H$ of a group $G$ there is a unique subgroup $K$ of $H$ which is maximal with respect to the property of being characteristic in $G$: namely, it's the subgroup given by the intersection

$$K = \bigcap_{\varphi \in \text{Aut}(G)} \varphi(H)$$

of the images of $H$ under all automorphisms of $G$. So there exists a finite index characteristic subgroup iff $K$ is finite index.

As a warmup, here's an easier problem: it's a classic exercise to show that if $H$ is finite index in $G$ then there exists a subgroup $H'$ of $H$ which is normal and finite index in $G$. (We don't need $G$ to be finitely generated here.)  The proof is very short; the maximal such subgroup is the intersection $\bigcap_{g \in G} gHg^{-1}$ of the conjugates of $H$, and this subgroup is exactly the kernel of the action of $G$ on the cosets $G/H$, which is finite. We get the more precise statement that if $H$ has index $n$ then $H'$ can be chosen to have index dividing $n!$, and setting $G = S_n, H = S_{n-1}$ shows that this bound is tight.

We can try to imitate this construction. Instead of just considering the action of $G$ on $G/H$ let's consider the action of $G$ on $G/\varphi(H)$ for every automorphism $\varphi$. We are done if we can show that this construction produces finitely many isomorphism classes of $G$-sets. Here is where we need the hypothesis that $G$ is finitely generated: we can actually show more, namely

Proposition: A finitely generated group $G$ has only finitely many isomorphism classes of actions on a finite set of size $n$.

Proof. There are finitely many homomorphisms $G \to S_n$. $\Box$

Translated back into subgroups this implies that $G$ has finitely many subgroups of a fixed finite index $n$, and since the subgroups $\varphi(H)$ all have the same index, in particular there are finitely many of these. So $K$ is the kernel of the action of $G$ on

$$\bigsqcup_{\varphi \in \text{Aut}(G)} G/\varphi(H)$$

which is finite, and we are done. More explicitly, if $G$ is generated by $r$ generators and $H$ has index $n$ then there are at most $n!^r$ homomorphisms $G \to S_n$, so the above set has size at most $n \cdot n!^r$, and $K$ has index dividing $(n \cdot n!^r)!$. Probably not best possible, but it works.


问:
Find null space of wide matrix with repeated columns and a $0$ row
The null space of
$$ R =
\begin{pmatrix}
1&0&x&x&x&0&x&\\
0&1&x&x&x&0&x&\\
0&0&0&0&0&1&x&\\
0&0&0&0&0&0&0&\\
\end{pmatrix}
$$

is equal to the span of these four column vectors (written transposed)

\begin{bmatrix} x&x&1&0&0&0&0\end{bmatrix}
\begin{bmatrix} x&x&0&1&0&0&0\end{bmatrix}
\begin{bmatrix} x&x&0&0&1&0&0\end{bmatrix}
\begin{bmatrix} x&x&x&0&0&1&0\end{bmatrix}

Is this correct? I just read off which combinations make sense off the matrix.

答:
You have 3 linearly independent columns: 1st, 2nd, 6th. Which means your Null Space should contain $7-3=4$ vectors. You need to express dependent columns via independent ones. But in your answer the last (7th) column isn't expressed by the Null Space - which is a hint that you did something wrong.

As mentioned in the comments - your vectors ($v$) in Null Space should $Rv=0$. But your answer gives e.g.:

$$
\begin{bmatrix}
1&0&x&x&x&0&x&\\
0&1&x&x&x&0&x&\\
0&0&0&0&0&1&x&\\
0&0&0&0&0&0&0&\\
\end{bmatrix}
\begin{bmatrix} x\\x\\1\\0\\0\\0\\0\end{bmatrix} =
\begin{bmatrix} x + x\\x+x\\0\\0\end{bmatrix}
$$

To get $0$ you need to subtract vectors, not add them:
$$
\begin{bmatrix}
1&0&x&x&x&0&x&\\
0&1&x&x&x&0&x&\\
0&0&0&0&0&1&x&\\
0&0&0&0&0&0&0&\\
\end{bmatrix}
\begin{bmatrix} x\\x\\-1\\0\\0\\0\\0\end{bmatrix} =
\begin{bmatrix} 0\\0\\0\\0\end{bmatrix}
$$
You need to do similar thing with all 4 vectors in Null Space. E.g. the last one would be:
$$
\begin{bmatrix} x\\x\\0\\0\\0\\x\\-1\end{bmatrix}
$$
I think you can take it from here and figure out the other 2.


问:
How can I get a longer calculation?
This is the example that my teacher gave in my economic class. I will write the formula and equation with the given answer below. Actually my teacher already calculate the example table of (modified internal rate of return - mirr). The only thing that I don't know is how to use the formula or get the answer even though the answer is already given.

Example

Formula for Compound Amount

S = P (1+i)^n,

i = 12%

Answer

Product F

16,431.7 = 9000(1+i)^4
= 16.24%

MY PROBLEM

The problem is can you show me the exact longer way to get those '16.24%'? I know that I'm stupid but please help me. I'm really can't remember how to calculate it.

*what should I do with that '16431.7'?
*how to get that '16.24%'?

Title: How can I get a longer calculation?

Body:
Given $P=9000$, $i=12\,\%$, $n=4$, we first write $12\,\%$ as $\frac {12}{100}$ and then compute
$$ \begin{align}S&=P\cdot (1+i)^n\\&=9000\cdot\left(1+\frac{12}{100}\right)^4\\&=9000\cdot 1.12^4 \\&= 9000\cdot 1.57351936 \\&= 14161.67424\end{align}$$
which I do not see in your post.

Perhaps, your actual task rather was: Given $S=16431.7$, $P=9000$, $n=4$, find $i$?
For this, transform the orignal equation:
$$P\cdot (1+i)^n=S$$
$$ (1+i)^n= \frac SP$$
$$ 1+i = \sqrt[n]{\frac SP}$$
$$ i = \sqrt[n]{\frac SP}-1$$
so
$$ \begin{align}i&=\sqrt[4]\frac{16431.7}{9000}-1 \\&\approx\sqrt[4]{\strut1.82574}-1\\&\approx 1.1624-1\\&=0.1624\\&=16.24\,\%  \end{align}$$


问:
Can $\frac{y - x}{\ln y-\ln x}$ be rational, for positive reals $x$ and $y$?
Consider two positive real numbers $x$ and $y$ such that $y > x \geq 0$. Do there exist pairs $\left\{x,y\right\}$ such that

$$ \frac{y - x}{\ln\left(\frac{y}{x}\right)} = \frac{y-x}{\ln(y)-\ln(x)} \in \mathbb{Q}_+, $$

where $\mathbb{Q}_+$ is the set of positive rational numbers?

答:
The map

$$f_x(y) = \frac{y-x}{\ln(y)-\ln(x)}$$ defined on $(x, \infty)$ is continuous and positive. As it is not constant, it takes rational values according to the intermediate value theorem as $\mathbb Q$ is dense in $\mathbb R$.


问:
For linear operator $T(t)$, what is the difference between "$T(t)$ is continuous", "$t\to T(t)$ is continuous", and "$t\mapsto T(t)$"?
Let $T(t)$ be a family of linear operators defined on a vector space $X$, $t\geq 0$. What is the difference between the following statements:

$T(t)$ is continuous

$t\rightarrow T(t)$ is continuous

$t\mapsto T(t)$

答:
Since $T(t)$ is a family of linear operators parameterized by $t$, we can re-phrase mathematically as $T:\mathbb{R}_+\rightarrow\mathcal{L}(X)$ (this is what $(3)$ is saying, actually).

$(1)$ means that, for a given $t$, $T(t)\in \mathcal{L}(X)$ is a continuous linear operator. $(2)$ is bad notation and is likely supposed to be $\mapsto$ instead of $\rightarrow$. The latter notation is the notation used to define functions between given sets, not an assignment of a parameter to an output value (which is what the former does). $(3)$ means that $T$ is a function from $\mathbb{R}_+$ to $\mathcal{L}(X).$ That is, you input $t\in\mathbb{R}_+$ and get back a linear operator $T(t)$. So, it's actually the same thing as saying that $T(t)$ is a family of linear operators. If we fix the notation in $(2)$, then $(2)$ means that $T$ as a map from $\mathbb{R}_+$ to $\mathcal{L}(X)$ is continuous. That is, if we fix the notation, then $(2)$ is the same as $(3)$ with the  additional stipulation of continuity of the parameterization.


问:
How to transform a Cauchy-Euler equation into a constant coefficients equation?
I have doubts with a problem that goes like this:

Let's consider the second order linear differential equation

$ y''(x)+R(x)·y'+S(x)·y=Q(x) $

a) How must the change of variable $z = z(x)$ be so that the equation
is transformed into one of constant coefficients?

b) Apply the above result to solve the equation:

$ y''(x)+(1+4e^{x})·y'+3e^{2x}·y=e^{2(x+e^{x})} $

I understand what the problem ask I don't know at all how to do it. I even wonder if the statement is right because the condition I get it's a bit abstract.

It's a Cauchy-Euler differential equation, so that:

$$ \frac {dy}{dx} = \frac {dy}{dz}· \frac {dz}{dx} = \dot yz'$$

$$ \frac {d^2 y}{dx^2} = ... = \ddot{y}z'^{2}+ \dot yz'' $$

And the differential equation turns to:

$$ \ddot{y}z'^{2}+ \dot yz'' + R(x)·\dot y z' + S(x)y = Q(x) $$

$$ \ddot{y}z'^{2}+ \dot y (z'' + R(x)z') + S(x)y = Q(x) $$

If the equation should transformed into a constant coefficients equation, I think it means:

$$ z'' + R(x)z' = K $$

But I don'tknow what to do from here..

I supposse (in case this result is fine) I can't make K=0 (because it's just a special case) but this is the only value that let me develop a little but more this condition.

Please, anybody know whether this result is right or it has some errors?
How could I solve the problem please? (Not solve it completely, of course. Just a tip to let me continue doing it).

Thanks a lot.

答:
$$\ddot{y}z'^{2}+ \dot y (z'' + R(x)z') + S(x)y =0$$
For this DE to be a constant DE you need:
$$ \dfrac {S(x)}{z'^2}=c_1$$
And:
$$\dfrac {z'' + R(x)z'}{z'^2}=c_2$$
Too many equations. So there must be some relation between $R(x)$ and $S(x)$.

The differential equation;
$$y''(x)+R(x)·y'+S(x)·y=0$$
has no known solution. We can only solve it for some particular functions $R,S$.


问:
Comparing pixels for color difference - taking Alpha into account
Say I want to compare pixels to see how similar in color they are. This wikipedia article describes how:

https://en.wikipedia.org/wiki/Color_difference

it says:

My question is, there not just a R,G,B component, but there is one more dimension - Alpha, which has to do with opacity. Is there a way to take the Alpha component into account?

Is it just one more independent variable as another dimension? If it's an independent variable, then it seems like we can just do:

$$distance = \sqrt{(R_2-R_1)^2 + (G_2-G_1)^2 + (B_2-B_1)^2 + (A_2-A_1)^2}$$

is that right or does it not make sense? I would prefer the using the squareroot instead of omitting it, to save a smaller number in a database.

答: You should initially ask what the alpha channel means in your specific context. May be you mean the transparency and the color as two distinct concepts, so it would be pointless to compare them mixed up as a single RGBA value. However it is worth to try the four elements version of the formula against your environment, as it may perform quite well on checking whether two pixels differs by a very small RGBA distance.</p>