问:
Equality of Completeness property (Supremum and Infimum property) and Cauchy criterion in Real numbers
I need to prove Completeness property and Cauchy criterion in Real numbers are equal, and I think its true. to prove this equality I think should use Bolzano-Weierstrass Theorem. I be so thankful if anyone could help me.</p>

答:
supremum property $\Rightarrow$ Cauchy criterion

The supremum property (which is what you call completeness) also implies the infimum property: every subset $S\subset\mathbb R$ which is bounded below has an infimum. For proof, just apply the supremum property to $-S:=\{-s~\vert~s\in S\}$. These two imply the existence of limit superior and limit inferior for all bounded sequences, since they are defined as

$$\limsup a_n:=\inf_{k\in\mathbb N}\sup_{n\geq k} a_n,\\
\liminf a_n:=\sup_{k\in\mathbb N}\inf_{n\geq k} a_n,$$

both of which exist for bounded $a_n$ due to the supremum and infimum properties. If limit inferior and superior of a sequence are equal, then the sequence converges. Now the limits superior and inferior of a Cauchy sequence can't be different if they exist (I leave the proof to you). And since Cauchy sequences are bounded (again, I leave the proof to you), the limits superior and inferior do in fact exist. They are then equal, and thus the Cauchy sequence has a limit.

Cauchy criterion $\Rightarrow$ supremum property

Let $\emptyset\neq S\subset\mathbb R$ be a set which is bounded above. Let $b_0$ be an upper bound of $S$ and let $a_0\in S$. Now define recursively that if $\frac{a_n+b_n}{2}$ is an upper bound of $S$, then $a_{n+1}:=a_n$ and $b_{n+1}:=\frac{a_n+b_n}{2}$, and otherwise $a_{n+1}:=\frac{a_n+b_n}{2},~~b_{n+1}:=b_n$.

With these definitions, both $a_n$ and $b_n$ are Cauchy sequences since $\vert a_m-a_n\vert\leq\frac{1}{2^{\vert m-n\vert}}\vert b_0-a_0\vert$, same for $\vert b_m-b_n\vert$. Because of the Cauchy criterion, both sequences converge. Even more, both converge to the same number, since the sequence $b_n-a_n=\frac{1}{2^n}(b_n-a_n)$ converges to $0$. The limit of these sequences must be an upper bound of $S$ because all the $b_n$ are upper bounds. There also can't be a lower upper bound because none of the $a_n$ are upper bounds (if $b$ is an upper bound, then all $c>b$ are upper bounds, too). So this limit is a least upper bound, in other words, a supremum.


问:
$V_{\lambda}$ as a $\mathfrak{sl}(2)$ -module
Let $V_{\lambda} = \bigoplus_{i \in \mathbb{Z}_{\geq 0}} \mathbb{C}v_i$, where $\lambda \in \mathbb{C}$. This is a representation of $\mathfrak{sl}(2)$ according to the following action:

$$hv_{i} = (\lambda - 2i)v_{i}, \qquad fv_{i}= v_{i+1}, \qquad ev_{i} = i(\lambda - i +1)v_{i -1}.$$

I want to find all values of $\lambda$ for which $V_{\lambda}$ is irreducible and all the submodules of $V_{\lambda}$, for the values of $\lambda$ for which $V_{\lambda}$ is not irreducible.

So far, I have only noticed that a single subspace $\mathbb{C}v_{i}$ cannot be a submodule of $V_{\lambda}$, since it is not stable under the action. So, if $W$ is a submodule of $V_{\lambda}$, then $W = \bigoplus_{i \in \mathbb{Z}_{\geq 0}} W \cap \mathbb{C}v_i $. However, I can’t seem to understand how the choice of $\lambda$ can determine irreducibility of $V_{\lambda}$. Any hints?

答: Hint: If $\lambda\in\Bbb N$, then $ev_{\lambda+1}=0$ and $\bigoplus_{i=\lambda+1}^{\infty}\Bbb Cv_{i}$ is a subrepresentation.

ContentLicense: CC BY-SA 4.0


问:
Given Equation of Inner product , how to determine that its positive for all $x,y\in \mathbb{R}$
I'm confused on how to configure if an equation is positive for all $x,y$.

for example the inner product : $<(x,y),(x,y)> =2x^2+y^2-2xy$

How can I know if this equation is positive or it might be negative for some $x,y$ ?

Thank you in advance.

答:
Hint:

Observe that $x^2\ge 0$ and
$
x^2+y^2-2xy=(x-y)^2\ge 0
$.

In general, what you need is this: https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia


问:
Finite equivalence class same cardinality
For an equivalence relation $\sim$, if each partition has a finite number of elements, and $X$ is an infinite set, then is it true that $|X/\sim|=|X|$?

I can prove injectivity one way by defining the map $f:X/\sim\rightarrow X$ by picking an element from the equivalence class, but not sure how I would go about the other way.

答:
The quotient set $X / {\sim}$ is a partition of $X$ into finite sets $[X]_i$ with $i < |X / {\sim}|$ (the axiom of choice here is implicit: $i$ is an ordinal less than the cardinal number $|X / {\sim}|$).

Map each $f_i : [X]_i \to \omega \times \{ i \}$ injectively since they are finite.

Take the union $\displaystyle \bigcup_{i < |X / {\sim}|} f_i : X \to \omega \times |X / {\sim}|$, which is injective, since the images of the component functions are disjoint.

So we have $$|X| \le |\omega \times X / {\sim}| \le |\omega \times X| = |X|$$ since $X$ is an infinite set.

$|\omega \times X / {\sim}| = | X / {\sim}|$ because $X / {\sim}$ is infinite (easily seen because otherwise, $X = \bigcup X / {\sim}$ is the finite union of finite sets).

ContentLicense: CC BY-SA 3.0


问:
How to determine shape of level curves of $V(y,z)= \ln{ \left( \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)} $?
Context:

I expect that a circle will be written as
$$R^2 = (x-a)^2 + (y-b)^2,$$
where $R$ is the radius, and $(a,b)$ is the center of the circle.

Question:

How to determine analytically that the level lines of
$$V(y,z)=  \ln{ \left(  \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)}$$
are circles?

答:
Isocontours of $V$ are represented by constant values of the function, i.e., $V=C$, which yields

$$C = \ln{ \left(  \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)} \implies
\mathrm{e}^C = \left(  \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)$$

Now rearrange to have

$$A \left[y^2 + (z-z_o)^2\right] = y^2 + (z+z_o)^2 $$

or, equivalently

$$ (A-1) y^2 + A(z-z_o)^2 - (z+z_o)^2 = 0,$$

with $A = \mathrm{e}^C $. Upon completing the square*, one has (if I didn't mess up)

$$ y^2 + (z-B)^2 = \frac{D}{A-1}, $$

where $B = z_o(A+1)/(A-1)$, and $D = 4Az_o^2/(A-1)$.

*Credit to this step goes to this answer.


问:
Inverse Fourier Sine Transform of $\frac{1}{p^2+a^2}$
What is the inverse Fourier sine transform of $\frac{1}{p^2+a^2}$? I don't need just the answer, I need an explanation for its solution. I don't know how to start.

It appeared in one of my past assignments. However, now the deadline is over.

答:
Start with this:

$$\begin{align*}\dfrac{1}{2\pi}\int_{-\infty}^\infty \dfrac{1}{p^2+a^2}\sin(px)\;dp &= \dfrac{1}{2}\left[\dfrac{1}{2\pi i}\int_{-\infty}^\infty \dfrac{e^{ixp}}{(p+ia)(p-ia)}\;dp \\ \quad \quad \quad -\dfrac{1}{2\pi i}\int_{-\infty}^\infty \dfrac{e^{-ixp}}{(p+ia)(p-ia)}\;dp\right]\end{align*}$$

and then perform contour integration to solve the individual integrals on the RHS.

Note that for each contour integration you must consider the two cases of $x >0$ and $x < 0$ and change the closed contour accordingly.  The final answer will ultimately be expressed in terms of $|x|$.


问:
$R((x))$ Set of formal Laurent Series is an ordered field which is Cauchy Complete.
Let $R$ be any ordered field. Define $R((x)):=\{\sum\limits_{k=-n}^\infty a_kx^k:a_k \in R \ , \ n \in\mathbb N\}$

Addition:
$\sum_\limits{k=-n}a_kx^k+\sum_\limits{k=-m}b_kx^k:=
\sum_\limits{k=-max\{n,m\}} (a_n+b_n)x^n$

Multiplication:
$\sum_\limits{k=-n}a_kx^k \times\sum_\limits{k=-m}b_kx^k:=
\sum_\limits{k=-(n+m)} \left(\sum_\limits{i+j=k}(a_ib_j)\right)x^k$

Order Relation: $\left(\sum_\limits{k=-n}^\infty a_kx^k\right) \geq 0$ iff the first nonzero coefficient is greater than or equal to $0$. Using this we can define $\alpha \leq \beta$ iff $0 \leq \beta-\alpha.$

Then $\left(R((x)),+,\times, \leq\right)$ is an ordered field.

Archimedean Property: By the order relation defined above, we can see that there exists $x^{-1} \in R((x))$ with $1<2<...<n<...<x^{-1}$ $\implies$ the field is non-archimedean.

Defining modulus function:
$\left|\sum_\limits{k=-n}a_kx^k\right|_{R((x))}:=\begin{cases}
                      \sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient postive} \\
                     -\sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient negetive}
                                      \end{cases}$

Cauchy Sequences: $(\alpha_n)$ is said to be cauchy, if for every $(0<)\varepsilon \in R((x))$ there exists $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.

Now I want to see if $R((x))$ is Cauchy Complete, (Do I need to assume that $R$ is Cauchy Complete?)

Let $(\alpha_n)$ be a Cauchy Sequence in $R((x))$ that is for any $(0<)\varepsilon \in R((x))$ there exists an $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.

Taking the first nonzero coefficient from each $\alpha_n$ we can obtain a Cauchy sequence $(a_n)$ in $R$.

If I assume $R$ is Cauchy Complete then $(a_n) \to a(\in R)$.

But I don't know what to do next.

I am not sure whether the above definitions are meaningful. Any insight will be of great help.

答:
Given $(a_n)$ Cauchy,

For all $K$ there is $N_K$ such that for $n\ge m\ge N_K$ either
$0\le a_n-a_m < x^K$ or $0\le a_m-a_n < x^K$ which means that $a_n-a_m\in x^K R[[x]]$.

Writing $a_n = \sum_{j\ge J_n} A_{n,j} x^j$ we get that $(A_{n,j})_{n\ge 1}$ is constant for $n$ large enough and
$$\lim_{n\to \infty} a_n = \sum_{j\ge \min(0,J_{N_1})} (\lim_{n\to \infty} A_{n,j}) x^j\in R((x))$$

(the topology is generated by the open sets $\{ f\in R((x)), b-\epsilon<f<b+\epsilon\}$)


问:
Calculate the line integral where $F(x,y,z) = (\sin(x), \cos(y), xz)$, $r(t) = (t^3, -t^2, t)$ and $t\in[0,1]$
Let
$$F(x,y,z) = (\sin(x), \cos(y), xz)\quad\text{and}\quad r(t) = (t^3, -t^2, t)$$
and $t$ from $0$ to $1$.

Then, how to find $ \int_{C} F\cdot\mathrm{d}r$ such that the curve $C$ is specified by the vector function $r$.

Thanks in advance.

Body:
$F(r(t))=(sin(t^3),cos(-t^2),t^4)$ , $r`(t)=(3t^2,-2t,1)dt \Rightarrow\int F \cdot dr=\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt$

$\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt=sin(-1)-cos(1)+2$


问:
Schur's Lemma in Group Theory
The analogue of well celebrated Schur's Lemma in group theory will be

"If $G$ is a finite simple group, and $\phi$ is a non-identity homomorphism from $G$ to $G$, then $\phi$ is an isomorphism".

The proof follows exactly same lines as it is in representation theory.

I would like to ask the following question, which is a converse of Schur's lemma in some sense.

Question 1: If $G$ is a finite group, $|G|>2$, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

The progress towards the solution I made is the following: if $G$ satisfies the hypothesis in the question and $G$ is non-abelian, then $G$ is not solvable: if $G$ is non-abelian, solvable, then it has non-identity abelian quotient $G^{ab}=G/[G,G]$. Let $p$ be a prime divisor of $|G^{ab}|$ (hence $p$ also divides $|G|$). Then there is a surjective homomorphism from $G^{ab}$ to $C_p$ (cyclic group of order $p$). We can embedd $C_p$ in $G$. Then, consider the composition of the homomorphisms:

$G\rightarrow G^{ab} \rightarrow C_p\rightarrow G$. It is a non-identity homomorphism from $G$ to $G$ which is clearly not an automorphism.

Note that if $G$ is abelian, which is not cyclic of prime order, then we can easily find non-identity homomorphisms $G\rightarrow G$ which are not automorphisms.

I would like to ask a general question also (remove the condition of finiteness in above question).

Question 2: If $G$ is an infinite group, such that every non-identity homomorphism $\phi\colon G\rightarrow G$ is an automorphism, should $G$ necessarily simple?

答:
Nice question! The answer is no. The smallest potential counterexample works.

Let $G$ be the binary icosahedral group. This is a perfect group of order $120$ (in fact the smallest nontrivial such group which is not simple). Its only nontrivial normal subgroup is its center $\pm 1$, hence its only nontrivial quotient is the icosahedral group $A_5$, which cannot occur as a subgroup of $G$, since any subgroup of $G$ of order $60$ is necessarily normal. Hence $G$ is not simple but every nonzero homomorphism $G \to G$ is an isomorphism.

Dan Shved gives an example for your second question in the comments.

Title: Schur's Lemma in Group Theory


问:
How to transform this periodic boundary value problem to an integral equation?
While reading this paper, I find this passage interesting, but I dont know how to prove it exactly(I know that It uses Green's function..)

"We study the existence of a solution of the following periodic system:
\begin{align}
& u^{\prime}+\lambda_{1} u-\lambda_{2} v=f(t, u)+g(t, v)+\lambda_{1} u-\lambda_{2} v \\
& v^{\prime}+\lambda_{1} v-\lambda_{2} u=f(t, v)+g(t, u)+\lambda_{1} v-\lambda_{2} u
\end{align}
together with the periodicity conditions,
$$
u(0)=u(T) \quad \text { and } \quad v(0)=v(T)
$$
This problem is equivalent to the integral equations:
$$
\begin{align}
u(t)=\int_0^T & G_1(t, s)\left[f(s, u)+g(s, v)+\lambda_1 u-\lambda_2 v\right] \\
& {}+G_2(t, s)\left[f(s, v)+g(s, u)+\lambda_1 v-\lambda_2 u\right] \mathrm{d} s \\
v(t)=\int_0^T & G_{1}(t, s)\left[f(s, v)+g(s, u)+\lambda_1 v-\lambda_2 u\right] \\
& {}+G_2(t, s)\left[f(s, u)+g(s, v)+\lambda_1 u-\lambda_2 v\right] \mathrm{d} s
\end{align}
$$
where $$G_{1}(t, s)=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{e^{\sigma_{1}(t-s)}}{1-e^{\sigma_{1} T}}+\frac{e^{\sigma_{2}(t-s)}}{1-e^{\sigma_{2} T}}\right] & 0 \leq s<t \leq T \\
\frac{1}{2}\left[\frac{e^{\sigma_{1}(t+T-s)}}{1-e^{\sigma_{1} T}} + \frac{e^{\sigma_{2}(t+T-s)}}{1-e^{\sigma_{2} T}}\right] & 0 \leq t<s \leq T
\end{array}\right.$$
and
$$G_{2}(t, s)=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{e^{\sigma_{2}(t-s)}}{1-e^{\sigma_{2} T}}-\frac{e^{\sigma_{1}(t-s)}}{1-e^{\sigma_{1} T}}\right] & 0 \leq s<t \leq T \\
\frac{1}{2}\left[\frac{e^{\sigma_{2}(t+T-s)}}{1-e^{\sigma_{2} T}}-\frac{e^{\sigma_{1}(t+T-s)}}{1-e^{\sigma_{1} T}}\right] & 0 \leq t<s \leq T
\end{array}\right.$$
Here, $ \sigma_{1}=-\left(\lambda_{1}+\lambda_{2}\right)$  and $ \sigma_{2}=\left(\lambda_{2}-\lambda_{1}\right) .$"

答:
$\def\d{\delta}
\def\l{\lambda}
\def\s{\sigma}$We wish to solve
\begin{align*}
u'+\l_1 u-\l_2 v &= f_1(t,u,v) \\
v'+\l_1 v -\l_2 u &= f_2(t,u,v).
\end{align*}
This is equivalent to solving
\begin{align*}
a'-\s_2 a &= f_1+f_2 \\
b'-\s_1 b &= f_1-f_2,
\end{align*}
where
\begin{align*}
a &= u+v \\
b &= u-v.
\end{align*}
Note that $a(T)=a(0)$ and $b(T)=b(0)$.
We have decoupled the original system and solve
\begin{align*}
G_a'-\s_2 G_a &= \d(t-s) \\
G_b'-\s_1 G_b &= \d(t-s)
\end{align*}
with the appropriate boundary conditions.
It is a straightforward exercise to show that
\begin{align*}
G_a(t,s) &= \begin{cases}
\displaystyle \frac{e^{\s_2(t-s)}}{1-e^{\s_2 T}},
    & 0\le s<t\le T \\
\displaystyle \frac{e^{\s_2(t+T-s)}}{1-e^{\s_2 T}},
    & 0\le t<s\le T.
\end{cases}
\end{align*}
A similar result holds for $G_b$.

Thus,
\begin{align*}
a &= \int(f_1+f_2)G_a ds \\
b &= \int(f_1-f_2)G_b ds
\end{align*}
and so
\begin{align*}
u &= \frac12(a+b) \\
&= \int\left[
\frac12 (f_1+f_2)G_a + \frac12(f_1-f_2)G_b\right]ds \\
&= \int\left[\frac12(G_a+G_b)f_1 + \frac12(G_a-G_b)f_2\right]ds.
\end{align*}
A similar result holds for $v$.
Lastly, one can verify that

\begin{align*}
G_1&=\frac12(G_a+G_b) \\
G_2&=\frac12(G_a-G_b).
\end{align*}


问:
Non-computable functions by summing computable functions
The problem is to show that you may get a non-computable (non-recursive) function if you sum up an infinite number of computable (recursive) functions.

$$g(x) = \sum_{i=0}^\infty f(x, i)$$

show that $g(x)$ is non recursive, where $f(X, i)$ is recursive.

Can I use

$$g(x) = \sum_{i=0}^\infty 2^f,$$

where $f(i)$ is the Busy Beaver function?

And why can't I just use something like this: let infinite number of $f(x, i)$ equal $1$, so the sum will be divergent, which means its not computable.

答: Let $A$ be a non-recursive set. Let $f(x,i)$ be $1$ if $x=i\in A$ and $0$ in all other cases. So, for each fixed $i$, $f(X,i)$ is the characteristic function of the singleton $\{i\}$ if $i\in A$ and the identically zero function if $i\notin A$. In either case, it's recursive. But the sum over all $i$ is the characteristic function of $A$, which isn't recursive.


问:
Basic question on local fields
Maybe my question is very easy and typical, but I have a serious problem with it. Actually it is not a single question, and it is a mix of probably simple questions. Let $K/\mathbb{Q}$ be a number field. Let $p$ be a rational prime number and suppose that $p\mathcal{O}_K=\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_g^{e_g}$. I know how to construct $K_{\mathfrak{p}_i}$. Are these pairwise nonisomorphic? If yes, then there a correspondence between the $\mathfrak{p}_i$'s, primes lying over $p$, and the local fields $K_{\mathfrak{p}_i}$'s. Do we have this factorization $p=\mathfrak{p}_1^{e_1}$ in $K_{\mathfrak{p}_i}/\mathbb{Q}_p$?

What is the best way to think about the corresponding valuations?

I prefer to consider the $v_{\mathfrak{p}_i}$ as a map to $\dfrac{1}{e_i}\mathbb{Z}$, sending the uniformizers to $\dfrac{1}{e_i}$. If we consider that, then I think we have this relation $v_{\mathfrak{p}_i\big|\mathbb{Q}}(x)= v_p(x)$ for any $x \in \mathbb{Q}$.

Also, we can consider the $v_{\mathfrak{p}_i}$ as a map to $\mathbb{Z}$, sending the uniformizers to $1$. If we consider that, then I think we have this relation $v_{\mathfrak{p}_i\big|\mathbb{Q}}(x)=e_i v_p(x)$ for any $x \in \mathbb{Q}$.

Which one is more standard and compatible? Which one do you suggest me to consider?

Edit: Also I have another question. Maybe I have to ask it as an independent question. Let $K/\mathbb{Q}_p$ be an extension of local fields. Then there is a unique prime $\mathfrak{p}$ in $K$. Corresponding to this prime we have the inertia degree $f$ and the ramification index $e$, and we have $p\mathcal{O}_{K}=\mathfrak{p}^e$ and $ef=[K:\mathbb{Q}_p]$. How can we find a global number field $L$, such that $L_v=K$ for some prime $v$ of $\mathcal{O}_L$?

答:
Different completions may or may not be isomorphic.

In $\mathbf Q(i)$ we have $5=(1+2i)(1-2i)$ and both completions $\mathbf Q(i)_{(1+2i)}$ and $\mathbf Q(i)_{(1-2i)}$ are isomorphic to $\mathbf Q_5$.

In $\mathbf Q(\sqrt[3]{2})$ we have $(5) = \mathfrak p \mathfrak q$ where $\mathfrak p$ has norm 5 and $\mathfrak q$ has norm 25, so the completions of $\mathbf Q(\sqrt[3]{2})$ at these two primes do not have the same residue fields over $\mathbf F_5$ or in fact have the same degrees over $\mathbf Q_5$.

Concerning the different ways to normalize the discrete valuations, the two ways that you describe are the most commonly used. Neither is intrinsically better than the other, but if you want to treat $K$ as a field on its own (not as an extension of $\mathbf Q$) then it is common to normalize the different discrete valuations on $K$ to each have value group $\mathbf Z$.

The distinction between the two conventions you describe is related to whether you want to think of absolute values on an extension of local fields as an extension of functions, where the absolute value on the larger field restricts on the smaller field to the original absolute value used on the smaller field. If you want to get an absolute value on the algebraic closure of $\mathbf Q_p$ then you need to use absolute values on larger fields that are compatible with the ones already used on smaller fields.

Often an important condition to use that ties together all the absolute values on a number field is the product formula: $\prod_v |x|_v = 1$ for all $x$ in $K^\times$.

Title: Basic question on local fields


问:
Showing atomic $H^{1,p}$ is a Banach Space
Consider $1<p\leq\infty$ and let $H^{1,p}=H^{1,p}(\mathbb{R}^n)$ be an atomic Hardy space, that is $H^{1,p}\subset L^1(\mathbb{R}^n)$, and $f\in H^{1,p}$ if there exists a sequence of $p$-atoms $\{a_i\}_{i\in \mathbb{N}}$ and a sequence $(\lambda_i)\in \ell^1(\mathbb{N})$ such that

$f=\sum_{i=0}^\infty \lambda_i a_i $

Recall a $p$-atom $a$ satifies:

$supp(a)\subset Q$ for some cube $Q\subset\mathbb{R}^n$

$\int_Q a =0$

$||a||_p\leq \frac{1}{|Q|^{1-\frac{1}{p}}}$

The norm on $H^{1,p}$ is given by

$||f||_{H^{1,p}}=\inf\{\sum_i |\lambda_i| : f=\sum_i\lambda_i a_i\}$

I am trying to prove that $H^{1,p}$ is a Banach space. To start with, it is not hard to see that for any $f\in H^{1,p}$:

$||f||_{L^1}\leq||f||_{H^{1,p}}$

One can see this by considering

$\int |f|=\int |\sum_i \lambda_i a_i| \leq \sum_i |\lambda_i|\int|a_i|\leq \sum_i|\lambda_i|$

Where we have used dominated convergence and the property that for any $p$-atom $||a||_1 \leq 1$.

Using this, if we have $f_n$ a Cauchy sequence in $H^{1,p}$ then we have a Cauchy sequence in $L^1$, so it converges to some $f\in L^1$. I am now stuck on how to prove that $f\in H^{1,p}$.

I have attempted to construct a representation of $f$ in terms of $\lambda_i$ and $a_i$ where $\lambda_i=\lim_{n\rightarrow \infty} \lambda_i^n$ for representatives $f_n=\sum_i \lambda_i^n a_i^n$ (similarly for $a_i$). But this hasn't succeeded.

Thanks for any help

答:
I think I figured it out, for anyone who is interested: In general to prove a normed vector space $X$ is a banach space it is enough to prove:

Given a sequence $(x_n)$ such that $\sum_i ||x_i||<\infty$ then $\sum_i x_i \in X$.

This is my technique for proving $H^{1,p}$ is complete:

Suppose that we have $(f_n)_{n\in\mathbb{N}}$ such that $\sum_{n=0}^\infty||f_n||_{H^{1,p}}<\infty$ then using the estimate shown in the OP, we have $\sum_{n=0}^\infty||f_n||_{L^1}<\infty$, so in fact

$f:=\sum_{n=0}^\infty f_n \in L^1$ using the completeness of $L^1$. We want to show that $f\in H^{1,p}$ and $\sum_{n=0}^N f_n\rightarrow f$ as $N\rightarrow\infty$ with $H^{1,p}$ convergence.

To do this choose an "efficient" representative of $f_n$, that is choose sequence's and $p$-atoms so that $f_n=\sum_{i=0}^\infty \lambda_i^n a_i^n $ and

$\sum_{i=0}^\infty |\lambda_i^n| \leq ||f_n||_{H^{1,p}} + \epsilon 2^{-n} $

Such expressions can always be chosen because the norm is defined in terms of an infimum. It is clear that $f$ has an atomic decomposition of the form

$f=\sum_{n=0}^\infty f_n = \sum_{n=0}^\infty \sum_{i=0}^\infty \lambda_i^na_i^n$

We need to check that our sequence $(\lambda_i^n)$ is in $\ell^1(\mathbb{N})$:

$\sum_{n=0}^\infty \sum_{i=0}^\infty |\lambda_i^n|\leq \sum_{n=0}^\infty (||f_n||_{H^{1,p}} +\epsilon 2^{-n}) = \sum_{n=0}^\infty ||f_n||_{H^{1,p}} +\epsilon <\infty$

Lastly, to check $H^{1,p}$ convergence $f-\sum^N_{n=0}f_n$ has the representation $f-\sum^N_{n=0}f_n=\sum^\infty_{n=N+1}f_n = \sum^\infty_{n=N+1}\sum_{i=0}^\infty \lambda_i^na_i^n$. Hence we have:

$||f-\sum^N_{n=0}f_n||_{H^{1,p}}\leq \sum^\infty_{n=N+1}\sum_{i=0}^\infty |\lambda_i^n|\rightarrow 0$ as $N\rightarrow \infty$

Because we have an "efficient" cover, meaning $\sum_{n=0}^\infty \sum_{i=0}^\infty |\lambda_i^n|<\infty$.


答:
We use the mother of all inequalities about the exponential function:
$$ e^t\ge 1+t\qquad\text{for all }t\in\Bbb R$$
(with equality iff $t=0$).

Given $x>0$, we plug in $t=-\frac 1{1+x}\ne 0$ to arrive at
$$ e^{-\frac1{1+x}}> 1-\frac1{1+x}=\frac x{1+x}.$$
As both sides are positive, we take reciprocals and arrive at
$$ e^{\frac1{1+x}}<\frac{1+x}x=1+\frac1x.$$
Taking logarithms,
$$ \frac1{1+x}<\ln\left(1+\frac1x\right).$$

Title: How $\frac{1}{x+1}-\log{(1+\frac{1}{x})}<0$

Body:
To prove that $\frac{1}{x+1}-\log{(1+\frac{1}{x})}<0$ $\forall x>0$:
If I consider $f(x)=\frac{1}{x+1}-\log{(1+\frac{1}{x})}=\frac{1}{x+1}-\log{(x+1)}-\log{x}$ then $f'(x)=\frac{1}{(x+1)^2}+\frac{1}{x+1}+\frac{1}{x}>0 \forall x>0$ then $f(x)>f(0)=0$.
to do: check if my idea is right


问:
$n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Prove that there exist distinct, pairwise coprime natural numbers $n_1,n_2,...,n_{11}$ such that $n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Let $N = n_1n_2...n_{11}$ My approach was to firstly notice that none of the $n_1n_2...n_{11}$ is an even integer, otherwise $n_1n_2...n_{11}-1$ would be odd. Then I thought about using $CRT$ \begin{align} N&=1\mod x, \\ N&=1 \mod x+1,\end{align} $x,x+1$ to conssecutive natural numbers. By $CRT$ we know that this system of congruences has a solution. Now I want to prove that the sol. is $0$ $\mod x(x+1)$ and deduce that $n_1n_2...n_{11}-1 = x(x+1)$ is true. How to proceed? Any help appreciated.

答:
The following is an answer to the old version of the question, where $n_{i}$ are not necessarily distinct:

As lulu comments, you may take the the example $n_{1},...,n_{10} = 1$ and $n_{11} = 7$.

Note that in the same fashion, it is possible to prove in general that there are infinitely many.

Take $M=k(k+1)+1$, for any $k$, with $n_{1},...n_{m-1}=1$, and $n_{m}=M$.


问:
Evaluate $\int_{-1}^{3}\int_{0}^{\sqrt{\pi }}\int_{0}^{\sqrt{\pi-y^2}} \sin(x^2 + y^2 ) \,dxdydz$

Evaluate $$\int_{-1}^{3}\int_{0}^{\sqrt{\pi }}\int_{0}^{\sqrt{\pi-y^2}}  \sin(x^2 + y^2 )  \,dxdydz$$

I am stuck trying to express the  sine  as series. What is the best move?

答:
What about cylindrical coordinates?:

$$0\le r\le \sqrt\pi\;,\;\;0\le\theta\le\frac\pi2\;,\;\;-1\le z\le 3$$

So your integral becomes

$$\frac1\pi\int_0^{\sqrt\pi}dr\int_0^{\pi/2}d\theta\int_{-1}^3 r\,\sin r^2\,dz =2$$


问:
There exists a differentiable function for which $f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$
Prove or disprove: There exists a differentiable function for which $$f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$$

My way:

False. Assume towards a contradiction that there exists such function.

Then we have:

$$f'\left(0\right)=1$$ $$f'\left(2\pi\right)=0$$

By Darboux's theorem, there exists $$0<c<2\pi$$ such that:
$$f'\left(c\right)=\frac{1}{2}$$
Thus:
$$f'\left(c\right)=\frac{1}{2}=\frac{\cos c-1}{c^{2}}\Rightarrow$$ $$\cos c=1+\frac{c^{2}}{2}>1$$
Contradiction!

.

My question is - would it been easier to disprove by integrating and trying to find the original function?

Even if not, could I see the shortest method on this case please?

Thank you.

答: That's a clever argument. I had a different application of Darboux's Theorem in mind. By Taylor polynomial considerations (or, if you insist, l'Hôpital's rule), you see that $\lim\limits_{x\to 0} f'(x)=-1/2$. But Darboux's Theorem tells you that $f'$ cannot have a jump discontinuity, so no such $f$ exists.


问:
Prove or disprove continuous functions
If f is differentiable on an interval J in R then |f| is differentiable on J.

my answer is false.
i took a counter example. x-1 is differentiable on 1, but |x-1| is not differentiable on 1. is this correct?

答:
Yes, Your answer seems right.

You can also take $f(x)=x$ which is differentiable, but $|f(x)|=|x|$ is not differentiable at $x_0=0$, since $\lim_{x \to 0^+}\frac{|x|}{x} \neq \lim_{x \to 0^-}\frac{|x|}{x}$

Title: Prove or disprove continuous functions


问:
Converting problem with norm bound to SOCP problem
I am currently researching some robust control problems, and I ended up with the following optimization problem:

\begin{align}
\max_{x,w} ~&~ f(x,w) \\ {\rm s.t.} ~&~ g(x,w) \le 0 \\ ~&~ \|w\|\le \|x\| \\ ~&~ x,w\in\mathbb{R}^n
\end{align}

The functions $f,g$ are linear, e.g. $f(x,w) = c^\top x + d^\top w$ and $g(x,w) = Ax + Bw - g$ for matrices $A,B$ and vectors $c,d,g$.

If $n=1$, the problem can be recast as an SOCP as $\|w\| \le \|x\|$  is equivalent to $w^\top w \le x^2$, which can be recast as the pair of contstraints $w^\top w \le xv$ and $x=v$, and the first can be rewritten as $\left\|\begin{bmatrix} w \\ x-v\end{bmatrix}\right\| \le x+v$.

Can this program be recasted as an SOCP problem for $n\ge 2$?

Edit:
The problem might be recasted as a convex program even if the constraint itself is non-convex. For example, if $g = g(x)$ is independent of $w$, and $f(x,w)=c^\top x + d^\top w$, then for any fixed $x$, we have:

$$
\max_{w: \|w\|\le \|x\|} f(x,w) = c^\top x + \|d\|\|x\|
$$

Indeed, the inequality $\le$ follows from Hoelder's inequality, and $\ge$ follows from choosing $w = \frac{\|x\|}{\|d\|}d$. Thus, the problem is equivalent to:
\begin{align}
\max_{x} ~&~ c^\top x + \|d\|\|x\| \\ {\rm s.t.} ~&~ g(x) \le 0 \\ ~&~ x\in\mathbb{R}^n
\end{align}
which is convex. I wonder if "tricks" like that can be used for the general problem.

答:
You can omit $x$ from the objective since you can always perform a reformulation via the epigraph. For fixed $w$, the problem is equivalent to deciding if there is a $x$ such that $||x|| \geq ||w||$ and $Ax \leq g - Bw$.

This is NP-hard since it a generalization of maximizing a quadratic function over the unit box $\{ x : ||x||_1 \leq 1\}$, which is known to be NP-complete (see Some optimal inapproximability results by Håstad).


Body:
Note

$$\log{(1+\frac{1}{x})}=\frac{1}{x+1}+\int_x^\infty\frac{dt}{t(1+t)^2}> \frac{1}{x+1}
$$


问:
Unbounded linear map between finite dimensional spaces
I'm trying to spot the error in the following argument, but without success. Can you help me out?

Let $Y$ be an infinite dimensional normed space on, say, $\mathbb{R}$, and let $f: Y \rightarrow \mathbb{R}$ be an unbounded linear map, which always exists (by using the existence of an algebraic basis of $Y$ and the axiom of choice).

By the homomorphism theorem we get $\bar{f}: Y/\ker f \rightarrow\mathbb{R}$ linear, bijective and such that $\pi\circ \bar{f}=f$, $\pi$ being the natural projection to the quotient. Therefore, since $f$ is unbounded and $\pi$ is boundend, then $\bar{f}$ is unbounded.

But $\bar{f}$ is a linear bijective map, so $Y/\ker f$ has dimension $1$ and I found a linear unbounded map between two finite dimensional spaces. What is wrong?

答: As $f$ is unbounded, $\ker(f)$ is not closed. That means that $Y / \ker(f)$ is not a normed space and you can't apply the homomorphism theorem.


问:
Show that $f_n$ converges uniformly to $f=0$
To any integer $n \geq 0$, we associate the function $f_n : [0,1] \to \mathbb R$ defined by $$f_n(x)=x^n(1-x)$$
Show that $f_n$ converges uniformly.
I already have a proof involving $\sup\{f_n(x)-f(x)\}$ but I am looking for an alternative proof using something else.

答:
Here is an alternative approach:

If $0 \leq x \leq 1$, we have $x^n \geq x^{n+1}$ and thus $f_n(x) \ge f_{n+1}(x)$ for all $x \in [0,1]$. Clearly we also have $\lim_n f_n(x) = 0$ for all $x \in [0,1]$, so $(f_n)_n$ is a non-increasing sequence that converges pointwise to a continuous function. By  Dini's theorem, the convergence is also uniform.


问:
The relationship between positive operators and positive matrices and forms
I'm a little confused about the definitions of positivity. I'm following Linear Algebra by Hoffmann and Kunze, and a positive operator is defined to be any operator such that $T$ is self adjoint and $<Tv,v> $ is positive for a non-zero vector $v$. The positive matrix on the other hand is a Hermitian  matrix which has the property $X^{*}AX>0$ for any complex $nx1$ matrices.

Can we say any operator is positive if and only if its matrix wrt an orthonormal basis is positive?

Similarly we have defined a positive sesqui linear form $f$ which has property $f(v,v)>0$ for any non- zero $v$, and proved $f$ is positive if and only if its matrix is positive, but we also have showed that $f$ is one to one correspondence with the set of operators on $V$, so again can we link the positivity on these two things?

答: Yes, you can link the positivity of operator and sesqui linear form in the way that w.r.t. orthonormal basis matrix of sesqui linear form with respect to that basis is same as matrix of the operator $T$ which determines that form. If $T$ is positive then, form will be positive and correspondingly, matrices are positive.


问:
Finding the intersection of 3 sets when given all information (except the intersection)

Info:

104 students were asked if they like math, science or humanities.

35 don't like either

21 students like math only

17 students like science only

4 students like humanities only

15 students like math and humanities

13 students like math and science

17 students like humanities and science

Question:

how many students like all three subjects?

Work:

I'm not even sure what the formula is for finding n(A∩B∩C) is. But when I used the following formula, this is what I got:

formula: n(A∩B∩C) = n(A) + n(B) + n(C) - n(A∪B∪C)

established facts:

where n(A) = 49 (21 + 13 + 15)

where n(B) = 47 (13 + 17 + 17)

where n(C) = 36 (15 + 17 + 4)

where n(A∪B∪C) = 69 (104 [total] - 35 [neither])

substitution:

n(A∩B∩C) = 49 + 47 + 36 - 69
n(A∩B∩C) = 132 - 69
n(A∩B∩C) = 63?!

Could someone please let me know what I am doing wrong?? Or if there is a formula that is applicable to this particular situation (missing intersection) could someone please let me know of that??
Because the intersection cannot be 63...

答:
In general using the Inclusion-Exclusion Principle,

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C|-|B \cap C| + |A \cap B \cap C| \implies$

$|A \cap B \cap C| = |A \cup B \cup C|  -(|A| + |B| + |C|) + |A \cap B| + |A \cap C|+|B \cap C|$.

In this case, your conditions are a bit different as we have the strict conditions of "only" for $A,B,C$ so the subtraction becomes easier for those terms. Then using a Venn Diagram one can see that $A \cap B,A \cap C,B \cap C \subseteq A \cap B \cap C$, and are oversubstracted. Hence now we know how to substract so that we have the following:

$-2|A \cap B \cap C| = (104-35)-(21+17+4)-(15+13+17)=-18 \implies$

$|A \cap B \cap C| = 9$.

What was done was this: Cancel out easily the elements (students) strictly in singleton sets, then substract the $2$-intersections so that we are left with $69 - (STUFF)= |A \cup B \cup C| - (STUFF) = -3|A \cap B \cap C|.$


答:
Your proof looks perfectly fine, great job!

The function given for $f'(x)$ has a removable discontinuity at $x=0$. There's a theorem in analysis stating that the derivative of a differentiable function can't have removable or jump discontinuities. But if you look at a typical proof of this theorem, it's pretty much the same argument as your solution here — such a proof would either refer to Darboux's Theorem or would reproduce the Intermediate Value Theorem argument that lies behind Darboux's Theorem.

So short of stating that "there's a theorem for this", your solution is the best approach, imho.

Title: There exists a differentiable function for which $f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$

Body:
Prove or disprove: There exists a differentiable function for which $$f'\left(x\right)=\begin{cases} \frac{\cos x-1}{x^{2}} & x\ne0\\ 1 & x=0 \end{cases}$$

My way:

False. Assume towards a contradiction that there exists such function.

Then we have:

$$f'\left(0\right)=1$$ $$f'\left(2\pi\right)=0$$

By Darboux's theorem, there exists $$0<c<2\pi$$ such that:
$$f'\left(c\right)=\frac{1}{2}$$
Thus:
$$f'\left(c\right)=\frac{1}{2}=\frac{\cos c-1}{c^{2}}\Rightarrow$$ $$\cos c=1+\frac{c^{2}}{2}>1$$
Contradiction!

.

My question is - would it been easier to disprove by integrating and trying to find the original function?

Even if not, could I see the shortest method on this case please?

Thank you.


问:
Proof that $\mathbb{R}$ is not countable
I know that this proof may sound ridiculous, but I'm really curious to find out if it's logically correct(and whether there are some circularities). Since $|[0,1]|=|\mathbb{R}|$, we have  to simply show that $[0,1]$ is not countable. This theorem holds:

$f:[a,b]\to \mathbb{R} \\ \text{Let } D_f \text{ be the set of discontinuities of }f \\ |D_f|\leq\aleph_0 \Rightarrow f \text{ is Riemann-integrable in }[a,b] $

Now I can define the Dirichlet function in $[0,1]$:
$\chi(x)=\begin{cases} 1 \ \ \text{if }x\in\mathbb{Q} \\ 0 \ \ \text{otherwise} \end{cases}$

Clearly $D_\chi=[0,1]$. By contradiction, if $|[0,1]|=\aleph_0$, then it would mean that $\chi$ is Riemann-integrable in $[0,1]$, but it's not: this is a contadiction!

So $|[0,1]|>\aleph_0$.

答:
It is correct and as far as I can see this does not contain circularities. Perhaps it is a little cleaner to use contraposition instead of contradiction though.

FYI, a more general statement than the one you described holds. Let $f: [a,b]\to \mathbb{R}$ be a bounded function. Then $f$ is Riemann-integrable if and only if $D_f$ has Lebesgue measure $0$. Of course, if $D_f$ is countable then it has Lebesgue measure $0$ since singeltons don't have a length.


问:
Compute $\iint_D (x^4-y^4) \,dx\,dy$
How should I proceed? the question is :
$$\iint_D (x^4-y^4) \,dx\,dy$$
$$D= \left\{(x,y):1<x^2-y^2<4, \quad \sqrt{17}<x^2+y^2<5,\quad x<0,\ \  y>0\right\}$$
I've tried to solve it with change of variable:
$$u=x^2-y^2, \quad v=x^2+y^2$$ $$|J|=\frac{1}{8xy}, \quad \frac{1}{8}\iint uv\frac{1}{xy} \,du\,dv$$

How should I proceed?

Any suggestion would be great, thanks

答:
If you do

$$u + v = 2x^2 \text{ and } v - u = 2y^2$$

you can solve for $x$ and $y$ in terms of $u$ and $v$. Just keep in mind the conditions $x < 0$ and $y > 0$ when you take the square roots. Also note that $u + v$ and $v - u$ are both positive from the hypotheses.

Therefore,

$$x = -\sqrt{\frac{u + v}{2}}, \quad y = \sqrt{\frac{v-u}2}, \quad \text{and } xy = -\frac12\sqrt{v^2 - u^2}.$$

And before I forget: you want the absolute value of the Jacobian, so

$$|J| = \frac{1}{8|xy|} = \frac{1}{4\sqrt{v^2 - u^2}}.$$

Title: Compute $\iint_D (x^4-y^4) \,dx\,dy$


问:
GCSE( how to find the volume of a pencil)
In the figure below, the dimensions of an unsharpened pencil are given. The unsharpened pencil have graphite in a cylindrical shape, surrounded by wood. The diameter of the graphite is 2mm. The density of graphite is $640\text{ kg/m}^3$. The density of wood is $420\text{ kg/m}^3$. Find the mass of 500 pencils.

My attempt was the following:
I found the volume of the wood: $$3.14\times (0.004)^2\times 0.15= 0.000007536$$
The volume of graphite:
$$3.14\times (0.001)^2\times 0.15= 0.000000471$$
I found the mass of 1 pencil: $$0.000007536\times  420 + 0.000000471\times 640= 0.0034\\
0.0034\times 500= 1.7 \text{ kg}$$

I think I messed up the solution!

All help/solutions appreciated.

Thanks

答: You forgot to take out the volume of the graphite when calculating the volume of the wood. In fact the volume of the wood is $$[\pi (0.004)^2(0.15)-\pi (0.001)^2(0.15)]\text{ m}^3$$


问:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$?
The question:

How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?

My attempt:

To find the slope of the tangent line to the curve, we need to get the first derivative of $y = \frac{x}{x+1}$, which is $y' = \frac{1}{(x+1)^2}$.

So, the line equation of the tangent line is $y = m(x-1) + 2$, where $m = \frac{1}{(x+1)^2}$ for the given $x$.

We also know that the point where the tangent line intersects the curve must be both on the tangent line and the curve, so using that,

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

$$\frac{x-1}{(x+1)^2} + 2 - \frac{x}{x+1} = 0$$

$$\frac{x - 1 + 2x^2 + 4x + 2 - x^2}{(x+1)^2} = 0$$

$$\frac{x^2 + 5x + 1}{(x+1)^2} = 0$$

And I'm stuck. I'd appreciate if you can point out where I'm wrong and how to better approach this problem.

答:
For a fraction $\frac{a}{b}$ to be equal to 0, we must have $a=0$. In this case we have $$x^2+4x+1=0$$ and the result follows.

EDIT:
Note that $$\frac{x-1}{(x+1)^2}+2-\frac{x}{x+1}=\frac{x^2+4x+1}{(x+1)^2}$$
You might wanna check your steps.


答:
For a finite Galois extension of $\Bbb Q$, its completions at different primes over $p$ are isomorphic to each other and are Galois over $\Bbb Q_p$, so they are all literally the same field inside an algebraic closure of $\Bbb Q_p$.
In terms of polynomials, if $K$ is the splitting field over $\Bbb{Q}$ of a polynomial in $\Bbb{Q}[x]$ then all the completions of $K$ at primes above $p$ are (isomorphic to) the splitting field of that same polynomial over $\Bbb{Q}_p$. In terms of decomposition groups, if $\mathfrak p$ is a prime over $p$ in a finite Galois extension  $K/\Bbb Q$ then ${\rm Gal}(K_\mathfrak p/\Bbb Q_p) \cong D(\mathfrak p|(p))$: there is a natural embedding $D(\mathfrak p|(p)) \to {\rm Gal}(K_\mathfrak p|\Bbb Q_p)$ that is surjective since both groups have the same size.

If $K/\Bbb{Q}$ is not Galois then it is more complicated to describe which primes over $p$ in $K$ have isomorphic completions. Let $L$ be the normal closure of $K/\Bbb{Q}$ and $L_p$ be "the" $p$-adic completion (the common completion of $L$ at an arbitrary prime over $p$ in $L$) and $f$ a choice of embedding $L\to L_p$.  The different subfields of $L$ that are isomorphic to $K$ are $\sigma(K)$ as $\sigma$ runs over ${\rm Gal}(L/\Bbb Q)$.

The completions of $K$ at the different primes over $p$ in $\mathcal O_K$ are the fields $\overline{f(\sigma(K))}$ in $L_p$ as $\sigma$ runs over ${\rm Gal}(L/\Bbb{Q})$, where the overline means the closure in $L_p$, and the corresponding embedding $K\to \overline{f(\sigma(K))}$ induces a discrete valuation on $K$ and thus a (nonzero) prime of $\mathcal O_K$.

Let a completion $\overline{f(\sigma(K))}$ come from a prime $\mathfrak p$ over $p$ in $\mathcal O_K$:  $\overline{f(\sigma(K))} = K_\mathfrak p$. Let $\mathfrak P$ be a prime over $\mathfrak p$ in $\mathcal O_L$, so $\mathfrak p =\mathfrak P \cap \mathcal O_K$.

For each $g\in {\rm Gal}(L_p/K_{\mathfrak p})=D(\mathfrak P|\mathfrak p)$ we have $\overline{f(\sigma(K))}=\overline{f(g\sigma(K))}$.

If $g\in {\rm Gal}(L_p/\Bbb{Q}_p)=D(\mathfrak P|(p))$ then $\overline{f(\sigma(K))} \cong g (\overline{f(\sigma(K))}) =\overline{g(f(\sigma(K)))}=\overline{f(g\sigma(K))}$.

Let $H=\{h\in {\rm Gal}(L/\Bbb{Q}), h(K)=K\}$. If $h\in H$ then $\overline{f(\sigma(K))} = \overline{f(\sigma h(K))}$.

My guess is that

$$\overline{f(\sigma(K))} \cong \overline{f(\sigma'(K))} \ \iff\ \sigma'\in D(\mathfrak P|(p)) \sigma H$$

Is it true ??

Title: Basic question on local fields

Body:
Maybe my question is very easy and typical, but I have a serious problem with it. Actually it is not a single question, and it is a mix of probably simple questions. Let $K/\mathbb{Q}$ be a number field. Let $p$ be a rational prime number and suppose that $p\mathcal{O}_K=\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_g^{e_g}$. I know how to construct $K_{\mathfrak{p}_i}$. Are these pairwise nonisomorphic? If yes, then there a correspondence between the $\mathfrak{p}_i$'s, primes lying over $p$, and the local fields $K_{\mathfrak{p}_i}$'s. Do we have this factorization $p=\mathfrak{p}_1^{e_1}$ in $K_{\mathfrak{p}_i}/\mathbb{Q}_p$?

What is the best way to think about the corresponding valuations?

I prefer to consider the $v_{\mathfrak{p}_i}$ as a map to $\dfrac{1}{e_i}\mathbb{Z}$, sending the uniformizers to $\dfrac{1}{e_i}$. If we consider that, then I think we have this relation $v_{\mathfrak{p}_i\big|\mathbb{Q}}(x)= v_p(x)$ for any $x \in \mathbb{Q}$.

Also, we can consider the $v_{\mathfrak{p}_i}$ as a map to $\mathbb{Z}$, sending the uniformizers to $1$. If we consider that, then I think we have this relation $v_{\mathfrak{p}_i\big|\mathbb{Q}}(x)=e_i v_p(x)$ for any $x \in \mathbb{Q}$.

Which one is more standard and compatible? Which one do you suggest me to consider?

Edit: Also I have another question. Maybe I have to ask it as an independent question. Let $K/\mathbb{Q}_p$ be an extension of local fields. Then there is a unique prime $\mathfrak{p}$ in $K$. Corresponding to this prime we have the inertia degree $f$ and the ramification index $e$, and we have $p\mathcal{O}_{K}=\mathfrak{p}^e$ and $ef=[K:\mathbb{Q}_p]$. How can we find a global number field $L$, such that $L_v=K$ for some prime $v$ of $\mathcal{O}_L$?


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
What we have here should really be interpreted as multiple statements:

(1.) If $ \lim_{h \to 0} \frac{2hx + h^2}{h} $ exists then
$ \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$ exists and is equal to $\lim_{h \to 0} \frac{2hx + h^2}{h} $.

(2.) If $ \lim_{h \to 0} [2x + h] $ exists then
$ \lim_{h \to 0} \frac{2hx + h^2}{h}$ exists and is equal to  $\lim_{h \to 0} [2x + h]$.

(3.) If $ \lim_{h \to 0} 2x$ exists then $ \lim_{h \to 0} [2x + h]$ exists and is equal to $ \lim_{h \to 0} 2x$.

(4.) $ \lim_{h \to 0} 2x$ exists and is equal to $ 2x $.

Note that once we have (4.) the "if" (conditional) part of (3.) is satisfied and so on all the way up to (1.). You can see that assuming that the limit exists in statements 1 to 3 is not a problem because you haven't used used that assumption to prove that it actually does. That would be circular logic and no good.

Your log example is different to this in the way that you don't have a statement that takes the role of statement (4.) above, which would allow you to escape the conditional. You have only proven that $\log(0) = 0$ IF $\log(0)$ exists, not that $\log(0)$ exists! This in itself is not an incorrect conclusion.


问:
When are two elements of a group equivalent in terms of their action on an element of an orbit of X?
I have strong intution that if $g_1$ and $g_2$ are two members of a group $G$ acting on $X$; then,
if $g_1(x)= g_2(x)$ for some $x$ in a $G$-orbit, $O$  of $X$; then $g_1(x)=g_2(x)$ for all $x$ in $O$. But I can't prove this.

I am using this as a preliminary result to derive that each orbit corresponds to a subgroup of G, so only want to use the definition of orbits and actions to prove this result.

Any help is appreciated.

答:
Note that, by action properties:

\begin{alignat}{1}
&g_1\cdot x=g_2\cdot x \iff \\
&(g_2^{-1}g_1)\cdot x=x \iff \\
&g_2^{-1}g_1\in G_x \\
\end{alignat}

where $G_x$ is the stabilizer of $x\in X$. Since the stabilizers of the points of one same orbit are conjugate, a sufficient condition for your claim to hold is that $G_x\unlhd G$, as in this case $G_{x'}=G_x$ for every $x'\in O(x)$.


问:
Prove that the determinant of the following $n\times n$ symmetric matrix that consists can be found by the following formula
A is an $n\times n$ symmetrical matrix that has a diagonal of zeros, first row and column are ones with the rest of the entries equal to a real number d

\begin{bmatrix}0&1&1&.&.&.&1\\1&0&d&.&.&.&d\\1&d&0&\
.&.&.&d\\.&.&.&.&.\\.&.&.&.&.\\.&.&.&.&.&.\\1&d&d&.&.&.&0\end{bmatrix}

prove that $$|A|= ((-1)^{(n-1)}\cdot(d^{(n-2)})\cdot(n-1)$$

I have tried finding the determinant the usual way but that turned out to be so complicated.

答:
Hint:

replacing one column at a time of the matrix with a combination (to be specific, a linear combination where any coefficients are either $1$ or $0$ ) of the other columns e.g. for the 2nd column of $\mathbf{A}$ denoted as $\mathbf{A}^2$, we could replace it thusly $$\mathbf{A}^2 \to \mathbf{A}^1 + \mathbf{A}^2 + \mathbf{A}^3 ...$$
And the determinant will not change (see Row replacement operation not changing the determinant - this works with row as well as column replacements). See you if you can use this fact to simplify the matrix.


问:
How to read $\sqrt[\frac{1}{x}]{n}$
How do I read $$\sqrt[\frac{1}{x}]{n}$$

I was curious if this was equal to $n^x$. This may sound a weirdly obvious question and instinctively I thought the answer was yes (since a root is a fractional exponent and we are just using the reciprocal here) but I wanted to be sure, and on the usual math sites I never see this used.

答:
$$\sqrt[a]{n}=n^{1/a}$$
is a general rule for the root notation and so:
$$\sqrt[\frac1x]{n}=n^x$$

However one thing to note is that often the root notation means to take the primary root only, so for complex numbers where there are many roots only the primary is taken

Title: How to read $\sqrt[\frac{1}{x}]{n}$


答:
hint

The equation of the tangent line to the curve at the point $ (a,\frac{a}{a+1}) $ is

$$y=\frac{1}{(a+1)^2}(x-a)+\frac{a}{a+1}$$

it passes through the point $ (1,2) $ if

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

So,
$$2(a+1)^2=(1-a)+a(a+1)$$

or

$$a^2+4a+1=0$$
thus
$$a=-2\pm \sqrt{3}$$

Body:
The question:

How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?

My attempt:

To find the slope of the tangent line to the curve, we need to get the first derivative of $y = \frac{x}{x+1}$, which is $y' = \frac{1}{(x+1)^2}$.

So, the line equation of the tangent line is $y = m(x-1) + 2$, where $m = \frac{1}{(x+1)^2}$ for the given $x$.

We also know that the point where the tangent line intersects the curve must be both on the tangent line and the curve, so using that,

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

$$\frac{x-1}{(x+1)^2} + 2 - \frac{x}{x+1} = 0$$

$$\frac{x - 1 + 2x^2 + 4x + 2 - x^2}{(x+1)^2} = 0$$

$$\frac{x^2 + 5x + 1}{(x+1)^2} = 0$$

And I'm stuck. I'd appreciate if you can point out where I'm wrong and how to better approach this problem.


问:
Question about isomorphism of Hilbert Spaces
Following is a question from Rudin :

Given Hilbert Spaces $H_1$ and $H_2$ show that one of them is isomorphic to a subspace of the other.

My thought process : wlog say $H_1$ has algebraic dim $\leq$ algebraic dim of $H_2$.
I took a Hilbert Basis of $H_1$ and mapped it injectively to a subset of a basis of $H_2$

How do I extend this to a Hilbert Space isomorphism i.e preserve inner products?

答:
Hint: Let $\{e_s: s \in S\}$ be an orthonormal base for $H_1$ and $\{f_t: t \in T\}$ be an orthonormal basis for $H_2$. Without loss of generality, assume $|S| \leq |T|$. I.e. there is an injection $\iota: S \hookrightarrow T$. Show that there is a linear isometry
$$\psi: H_1 \to H_2$$
such that $\psi(e_s) = f_{\iota(s)}$ for all $s \in S$.


问:
$n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Prove that there exist distinct, pairwise coprime natural numbers $n_1,n_2,...,n_{11}$ such that $n_1n_2...n_{11}-1$ is a product of two consecutive natural numbers.

Let $N = n_1n_2...n_{11}$ My approach was to firstly notice that none of the $n_1n_2...n_{11}$ is an even integer, otherwise $n_1n_2...n_{11}-1$ would be odd. Then I thought about using $CRT$ \begin{align} N&=1\mod x, \\ N&=1 \mod x+1,\end{align} $x,x+1$ to conssecutive natural numbers. By $CRT$ we know that this system of congruences has a solution. Now I want to prove that the sol. is $0$ $\mod x(x+1)$ and deduce that $n_1n_2...n_{11}-1 = x(x+1)$ is true. How to proceed? Any help appreciated.

答:
I assume you want to avoid the trivial solutions where some of the $n_i$ are $=1$.

Claim. Let $S$ be the set of all primes $p$ for which  $x^2+x+1\equiv 0\pmod{p}$ has a solution. Then $S$ is infinite.

Proof. Assume otherwise. Then  $S=\{p_1,\ldots, p_n\}$ is a finite set of primes. Let $y=p_1\cdots p_n$ and $z=y(y+1)+1$. Then  $z\ge 1\cdot 2+1=3>1$. Let $p$ be a prime divisor of $z$. Then $p\in S$ as witnessed by $x=y$. But then $z\equiv 1\pmod p$, contradicting $p\mid z$. $\square$

Pick $11$ primes $p_1,\ldots , p_{11}\in S$. For each of them, find $x_i$ with $x_i(x_i+1)\equiv -1\pmod {p_i}$ (which is possible because $p_i\in S$). Use CRT to find a natural number $x$ such that $x\equiv x_i\pmod {p_i}$.
Then $x(x+1)\equiv x_i(x_i+1)\equiv -1\pmod{p_i}$ for $1\le i\le 11$. So the number $N:=x(x+1)+1$ has at least $11$ different prime divisors. Thus $N$ can clearly be factored into $11$ pairwise co-prime numbers $>1$ (which are a fortiori pairwise distinct). Of course, $11$ plays no special role and can be replaced by any natural number.

Remark: In my first write-up, I had a wrong expression in $x$ instead of $x^3+x+1$. With the correct expression, $x^2+x+1$, it is of course clear from $x^3-1=(x-1)(x^2+x+1)$ that $x^2+x+1\equiv 0\pmod p$ implies $x^3\equiv 1\pmod p$, and conversely any non-trivial solution of the latter is a solution of the former. From this we read off that $S$ consists precisely of $3$ and the primes $p\equiv 1\pmod 3$. (So essentially the proof of the claim above turns into a proof of the fact that there are infinitely many such primes). It comes to no surprise that the solution by @Oleg567 involves the first such primes (only $1117$ is larger).


问:
Center of Group and Conjugacy Classes
I am trying to prove that the center of a group $G$ is the union of the trivial conjugacy classes of $G$.

So far what I have:

We know the center $Z(G)$ of group $G$ is defined by $\{ b \in G\mid ba= ab\,\forall a \in G\}$. We also know if there is an element $ x \in G $ such that $ b = x^{-1}ax$, then $a$ is conjugate to $b$.

Since we know the elements $b$ of $G$ commute with every element of $G$ in the center, we may say every element in $Z(G)$ has a conjugate. Therefore, the center is the union of all these conjugacy classes.

Is this correct, or where did I run off course?

答:
As @zibadawa timmy pointed out, this is equivalent to saying that $b$ is in the center if and only if every conjugate of $b$ is $b$.

In the forward direction, for all $a \in G$, there is $x \in G$ such that:

$$
\begin{equation}
\begin{split}
a \sim b & \Rightarrow ax = xb \\
 & \Rightarrow  ax = bx &\ \text{(because $b \in Z(G))$} \\
 & \Rightarrow a = b
\end{split}
\end{equation}
$$

In the backward direction, consider the conjugacy class of $b$. It contains all elements of the form $xbx^{-1}$. By the premise, the conjugacy class of $b$ contains only itself: $[b] = \{b\}$. So, for any $x \in G$, $b = xbx^{-1}$ and, therefore, $bx = xb$. In other words, $b$ commutes with all elements of $G$ and, as such, $b \in Z(G)$.

Title: Center of Group and Conjugacy Classes


问:
Is the closure of a subgroup of a quasitopological group a subgroup?
The closure of a subgroup in a topological group is a subgroup. Is the same true for a quasitopological group?</p>

答:
Just to have it as an answer, my comment from MO:

Yes: if $h_n \to g$ and $h'_m \to g'$, then $h_n h'_m \to g h'_m$ for each $m$, so each $g h'_m$ lies in the closure; and then $g h'_m \to g g'$, so $g g'$ lies in the closure.  (Identity and inverses are obvious.)


问:
Computing the euler Characteristic of $ S^2$
After defining the euler charateristic of a manifold $M$ as the intersection number of $M$ with it's tangent bundle , $\#(M,M,TM)$, Hirsch gives an example of computing for the special case where $M=S^n$.  So let $P$ denote the norh pole and $Q=-P$ the south pole. Let $\sigma: S^n-P\rightarrow \mathbb{R}^n$ and $\tau : S^n-Q\rightarrow \mathbb{R}^n$ be stereographic projections. Let $f$ be the vector field whose representation via $\sigma$ is the identity , I think what he means by this is that $T\sigma \circ f\circ \sigma^{-1}=Id:\mathbb{R}^n \rightarrow \mathbb{R}^n$. He claims that $f(x)\rightarrow 0$ as $x\rightarrow P$, and this I am not seeing why it's true. So let's suppose that $x\rightarrow P$ then we get that $\sigma(x)\rightarrow \infty$ and so we get that $T\sigma\circ f\circ \sigma^{-1} \rightarrow \infty$ but I don't see how we get that $f(x)\rightarrow 0$. I am not interaly sure of the affect of $T\sigma $ but I would say that $x\rightarrow P$ then $T_x\sigma (v)\rightarrow \infty$.

Then he defines the extension of $f$ to $S^n$ to be $f(P)=0$ and $f$ otherwise. Then in $\tau$ coordinates $f$ corresponds to $x\rightarrow -x$. And then he claims that $Ind_pf =1$ and $Ind_Qf=(-1)^n$. Shouldn't it be the other way around ? It doens't change the end result but I wanna make sure this makes sense or not.

Thanks in advance.

答:
In the $S^2$ case, you can gain intuition about the norm of $f(x)$ by considering circles centered around the origin of the plane of projection, which correspond to latitudinal circles on the sphere. In the plane, the circumference of such a circle is proportional to the norm of the identity vector field along it (under the usual metric on $\mathbb{R}^2$), but the corresponding latitudinal circle gets smaller as you go near the north pole and eventually reaches zero. This shows that the pullback metric from the sphere should decrease to zero faster than the identity vector field grows. This argument doesn't generalize very neatly to $S^n$, but a bit of computation can prove the point more concretely.

You can calculate the limit of $|f(x)|$ as $x\to P$ either in stereographic coordinates, or by pushing the identity vector field forward. Going the former route, the metric is (taken from Lee's textbook on Riemannian manifolds)
$$
\frac{4}{(1+ \sum u_i^2 )^2} \sum (d u_i)^2\ ,
$$
so the norm of $f(x)$ is $2r/(1+r^2)$ (where $r^2 = \sum u_i^2$), which approaches $0$ as $r\to\infty$.

In the $S^2$ case, the pushforward is not very hard either, where $f$ turns out to be $(-x z, -y z, 1-z^2)$, which approaches $0$ as $(x,y,z)\to (0,0,1)$.

Also I agree that the indices of $f$ at the poles should be the other way around.


问:
How to prove being conservative vector field of $R^3$
How to prove that $\mathbb{R}^3$ a conservative vector field $\mathbb{F}$ = (P, Q, R) satisfies
these 3 conditions:

$P_y = Q_x$

$P_z = R_x$

$Q_z = R_y$

答: If $F = \nabla \varphi$, then $P = \varphi_x$, $Q = \varphi_y$ and $R=\varphi_z$. So $$P_y = (\varphi_x)_y=\varphi_{xy}=\varphi_{yx}=(\varphi_y)_x=Q_x.$$Same for the rest.


答:
$104-35=69$ students like at least one subject.

Then, $69-(21+17+4)=27$ students like at least two subjects.

But all of the students that like all three (call this $n$) also like at least two, so we have counted those students three times by adding $15+13+17 = 45$ (once for each of the three groups). We only want to count them once, so we subtract out $2n$:

$45 - 2n = 27 \implies n = 9.$

Title: Finding the intersection of 3 sets when given all information (except the intersection)

Body:
Info:

104 students were asked if they like math, science or humanities.

35 don't like either

21 students like math only

17 students like science only

4 students like humanities only

15 students like math and humanities

13 students like math and science

17 students like humanities and science

Question:

how many students like all three subjects?

Work:

I'm not even sure what the formula is for finding n(A∩B∩C) is. But when I used the following formula, this is what I got:

formula: n(A∩B∩C) = n(A) + n(B) + n(C) - n(A∪B∪C)

established facts:

where n(A) = 49 (21 + 13 + 15)

where n(B) = 47 (13 + 17 + 17)

where n(C) = 36 (15 + 17 + 4)

where n(A∪B∪C) = 69 (104 [total] - 35 [neither])

substitution:

n(A∩B∩C) = 49 + 47 + 36 - 69
n(A∩B∩C) = 132 - 69
n(A∩B∩C) = 63?!

Could someone please let me know what I am doing wrong?? Or if there is a formula that is applicable to this particular situation (missing intersection) could someone please let me know of that??
Because the intersection cannot be 63...


问:
definition of an array
I'm teaching a programming class in Python, and I'd like to start with the mathematical definition of an array before discussing how arrays/lists work in Python.

Can someone give me a definition?

答: I'd go with comparing an array to a matrix. That way when you introduce arrays of arrays, the mental leap will be easier for your students to make.</p>

Title: definition of an array


问:
Calculus IVT question
Let $f$ be a function that is continuous on $[a,b]$ for some $a<b$. Suppose that f(a)=f(b)=0, and let $M=\sup${$f(x) : x \in [a,b]$}, while $M>0$, and let $0 \leq r < M$.

Prove that the set {$x \in [a,b] : f(x)=r$} contains at least two distinct points.

--

So, because $M>0$ and $M=\sup${$f(x) : x \in [a,b]$}, there exists $x_{0} \in [a,b]$ such that $f(x_{0})=M>0$, so we get $f(a) \leq r < f(x_{0})$.

$f$ is continuous on $[a,b]$ for $a<b$, so from IVT, there exists $c,r \in R$, $f(a)\leq r<f(x_{0})$ such that $f(c)=r$.

From here I am a bit confused on what to do next.

Any tip will be very appreciated!

Thanks a lot!

答:
hint

Assume $ r>0$.

$ f $ is continuous at the compact $ [a,b] \implies $

$$(\exists c\in(a,b) )\;:\;f(c)=M$$

let
$$g(x)=f(x)-r$$
$ g $ is continuous at $ [a,c] $ and $ [c,b]$.
$$g(a)g(c)=-r(M-r)<0$$

and

$$g(c)g(b)=-r(M-r)<0$$

conclude by IVT.

Title: Calculus IVT question


答:
You are almost there! Here I present the reasoning in its entirety.

Since $f$ is continuous defined on the compact interval $[a,b]$, it attains its maximum $f(x_{\max}) = M > 0$ for some $x_{\max}\in(a,b)$.

Consequently, once we have that $f(a) \leq r < f(x_{\max})$ and $f(b) \leq r < f(x_{\max})$, we conclude through the IVT there exists $x_{1}\in[a,x_{\max}]$ as well as $x_{2}\in[x_{\max},b]$ such that $f(x_{1}) = f(x_{2}) = r$, and we are done.

Hopefully this helps!

Title: Calculus IVT question

Body:
Let $f$ be a function that is continuous on $[a,b]$ for some $a<b$. Suppose that f(a)=f(b)=0, and let $M=\sup${$f(x) : x \in [a,b]$}, while $M>0$, and let $0 \leq r < M$.

Prove that the set {$x \in [a,b] : f(x)=r$} contains at least two distinct points.

--

So, because $M>0$ and $M=\sup${$f(x) : x \in [a,b]$}, there exists $x_{0} \in [a,b]$ such that $f(x_{0})=M>0$, so we get $f(a) \leq r < f(x_{0})$.

$f$ is continuous on $[a,b]$ for $a<b$, so from IVT, there exists $c,r \in R$, $f(a)\leq r<f(x_{0})$ such that $f(c)=r$.

From here I am a bit confused on what to do next.

Any tip will be very appreciated!

Thanks a lot!


问:
Calculating Density Function in Probability?
Given that:
$Y \sim \text{Exp}(j)$ and $X \sim \text{Uni}(a,b)$ independent from each other.
How may I calculate the density of $Z=Y-X$ knowing that:
$$
f_{X+Y}(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\mathrm dx.
$$

with your help I have calculated the density function for $-x$ which is

$$f_{-X}(-x) = \begin{cases} \dfrac{1}{b-a} &-x \in [a,b] \\ 0 & \text{otherwise}. \end{cases}$$

And I got too: $$\int_{-\infty}^\infty f_Y(y)f_{-X}(-x)dy$$

But how to continue from here? Plus how may I seperate the fields in which this is zero?

I know that:
$$f_{Y}(y) = \begin{cases} \lambda e^{-\lambda y} &y >=0 \\ 0 & \text{otherwise}. \end{cases}$$

Update: I got

$f_Z(z)$= 0 if $y<0$ or $x<-b$ or $x>-a $

else $1/{b-a}$

My my book has a totally different Answer:

if $z<-b$ then $0$

if $z>-a$ then $\dfrac {e^{- \lambda(a+z)}-e^{- \lambda(b+z)}}{b-a}$

else $\dfrac {1- e^{- \lambda(b+z)}}{b-a}$

答:
HINT: you have that
$$
\begin{align*}
\Pr [X-Y\leqslant c]&=\int_{\{(x,y)\in\mathbb{R}^2:x-y\leqslant c\}}f_X(x)f_Y(y)\mathop{}\!d (x,y)\\
&=\int_{\mathbb{R}}f_Y(y)\left(\int_{(-\infty ,c+y]}f_X(x)\mathop{}\!dx\right)\mathop{}\!d y\\
&=\int_{\mathbb{R}}f_Y(y)F_X(c+y)\mathop{}\!d y
\end{align*}
$$

Can you follow from here?


问:
Terms of $\sum_{q=0}^{\infty} \frac{\eta^{q}e^{q}}{q^{q(1-\beta)}}$ tend to 0 more rapidly than terms of any geometric progression?
Could anyone explain to me this reasoning in a part of a proof?

$$\sum_{q=0}^{\infty} \frac{\eta^{q}e^{q}}{q^{q(1-\beta)}}$$

"Since the terms of the series tend to zero more rapidly than the terms of any geometric progression, then starting with some number the $q$-th term of the series becomes less than $\frac{1}{2^{q}}$."

How do I see that they tend more rapidly to zero than terms of a geometric progression?

EDIT: I forgot to mention that $\beta <1 $

答:
The reasoning explained: For any fixed positive number $A$ and any $\beta < 1$, there is a $q'(A)$ s.t. the inequality

$$q^{(1-\beta)q} > A^q$$

holds for all $q  \ge q'(A)$; namely all $q \ge A^{\frac{1}{1-\beta}}+1 = q'(A)$. [Indeed, taking natural log on both sides of the inequality $q^{(1-\beta)q} > A^q$ and dividing by $1-\beta$ gives $$q \ln q > \frac{q \ln A }{1-\beta}.$$ Dividing by $q$ on each side of this gives $$\ln q > \frac{\ln A}{1-\beta}.$$ Then exponentiating both sides of this gives $$q > A^{\frac{1}{1-\beta}}.$$ Note that one can reverse the steps here to show that the strict inequality $q^{(1-\beta)q}  > A^q$ is satisfied if the strict inequality $q > A^{\frac{1}{1-\beta}}$ is satisfied,  so $A^{\frac{1}{1-\beta}}+1 = q'(A)$ will do.]

So to show that there is a $q'$ such that $\frac{\eta^q e^q}{q^{(1-\beta)q}} < 2^{-q}$ for all $q' \ge q$, take $A = 2 \eta e$; then the inequality $q^{(1-\beta)q} < A^q = 2^q \eta^q e^q$ holds for all $q \ge q'(A)$:

$$\frac{\eta^q e^q}{q^{(1-\beta)q}} = \frac{2^{-q}A^q}{q^{(1-\beta)q}} = 2^{-q}\left(\frac{A^q}{q^{(1-\beta)q}}\right) < 2^{-q}.$$


问:
Compute $\iint_D (x^4-y^4) \,dx\,dy$
How should I proceed? the question is :
$$\iint_D (x^4-y^4) \,dx\,dy$$
$$D= \left\{(x,y):1<x^2-y^2<4, \quad \sqrt{17}<x^2+y^2<5,\quad x<0,\ \  y>0\right\}$$
I've tried to solve it with change of variable:
$$u=x^2-y^2, \quad v=x^2+y^2$$ $$|J|=\frac{1}{8xy}, \quad \frac{1}{8}\iint uv\frac{1}{xy} \,du\,dv$$

How should I proceed?

Any suggestion would be great, thanks

答:
Oddly enough polar coordinates will do the trick and does it well as an alternative. The region is one integral if done angular first (which is easy to see because the region is between two circular arcs)

$$1<x^2-y^2<4 \implies \pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)<\theta<\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)$$

in the second quadrant. Which makes the integral

$$\int_{\sqrt[4]{17}}^{\sqrt{5}}\int_{\pi-\frac{1}{2}\cos^{-1}\left(\frac{1}{r^2}\right)}^{\pi-\frac{1}{2}\cos^{-1}\left(\frac{4}{r^2}\right)}r^5\cos2\theta\:d\theta dr = \int_{\sqrt[4]{17}}^{\sqrt{5}} \frac{r^3}{2}\left(\sqrt{r^2-1}-\sqrt{r^2-4}\right)dr$$

$$ = \int_{\sqrt{17}}^{5} \frac{t}{4} \left(\sqrt{t-1}-\sqrt{t-4}\right)dt$$

Integration by parts takes care of the rest.

$$ = \boxed{\frac{113}{15}-\frac{\sqrt{17}}{6}\left[(\sqrt{17}-1)^{\frac{3}{2}}-(\sqrt{17}-4)^{\frac{3}{2}}\right]+\frac{1}{15}\left[(\sqrt{17}-1)^{\frac{5}{2}}-(\sqrt{17}-4)^{\frac{5}{2}}\right]}$$

As a side note, this answer would have come out a lot more cleanly had the lower bound been $4$ instead of $\sqrt{17}$ giving

$$\frac{113}{15}-\frac{7\sqrt{3}}{5}$$

Title: Compute $\iint_D (x^4-y^4) \,dx\,dy$


问:
Why the differential of a function is a linear transformation?
I try to find some applications of the differential, that weird thing that is a linear transformation, what is the reason of linearity?
And how this linearity works, I mean, there's something fundamental about that linearity? And where we can see the importance of the differential, like in physics or another field of application.</p>

答:
Yes, there is something absolutely fundamental about linearity.  The whole idea of calculus is to approximate a complicated function by a linear one.  You have probably heard the slogan that the tangent line to a function at a point is the best linear fit at that point; this is exactly the connection between the derivative and linearity.

What may trouble you is that what we usually think of as the derivative in Calculus I is not linear.  For example, if $f : \mathbb R \to \mathbb R$ is given by $f(x) = x^3$ for all $x \in \mathbb R$, then $f' : \mathbb R \to \mathbb R$ is given by $f'(x) = 3x^2$ for all $x \in \mathbb R$, which is certainly not linear.

The point of multi-variable calculus is that we shift our view:  instead of thinking about the derivative as a single function, we think of it as a family of functions, one per point.  We'll often write something like ${\operatorname d}f_x$ to emphasise that we are thinking of the derivative function at $x$.  So, with $f$ as above, we would say that ${\operatorname d}f_7$ is, not the number $f'(7) = 147$, but the linear function ${\operatorname d}f_7 : \mathbb R \to \mathbb R$ given by ${\operatorname d}f_7(\Delta x) = 147\Delta x$ for all $\Delta x \in \mathbb R$.  That is, the derivative in the usual Calculus I sense should really just be thought of as the slope of the derivative in the more general sense.

In Calculus III, you learn to speak of the Jacobian matrix of a multi-variable function; for example, for the change-of-coördinate function $g : \mathbb R^2 \to \mathbb R^2$ given by $g(r, \theta) = (r\cos(\theta), r\sin(\theta))$ for all $(r, \theta) \in \mathbb R^2$, we have the Jacobian matrix $\operatorname{Jac} g(r, \theta) = \begin{pmatrix}
\cos(\theta) & \sin(\theta) \\
-r\sin(\theta) & r\cos(\theta)
\end{pmatrix}$ for all $(r, \theta) \in \mathbb R^2$.  This is just another name for the derivative in the sense above, where we think of specifying a matrix by its linear transformation.  For example, we would write ${\operatorname d}g_{(2, \pi/6)}$ for the linear transformation with matrix $\begin{pmatrix}
\sqrt3/2 & 1/2 \\
-1 & \sqrt3
\end{pmatrix}$ in the standard basis, which is to say the linear transformation given by ${\operatorname d}g_{(2, \pi/6)}(\Delta r, \Delta\theta) = (\sqrt3\Delta r/2 + \Delta\theta/2, -\Delta r + \sqrt 3\Delta\theta)$ for all $(\Delta r, \Delta\theta) \in \mathbb R^2$.

You may notice that I have regarded my derivatives as functions of "the same" variables as the original function, annotated with $\Delta$s—hopefully this reminds you of the idea of the differential approximation, according to which we write $f(x + \Delta x) \approx f(x) + f'(x)\Delta x$—but this is really a bit deceptive.  In the utmost generality, we have a function $h : M \to N$ between smooth manifolds, and get, for each $m \in M$, a differential function ${\operatorname d}h_m : \operatorname T_mM \to \operatorname T_{h(m)}N$ that is a linear map of tangent spaces.  It just happens that, when our manifolds are vector spaces $V$, there is a natural identification $V \cong \operatorname T_vV$ at each point.

For a very slightly more complicated example, we have maps $h_1 : \mathbb R \to \operatorname S^1$ and $h_2 : \mathbb R \to \mathbb C$ given by $h_1(t) = e^{i t}$ and $h_2(t) = h_1(t)$ for all $t \in \mathbb R$.  These are the same maps except for the codomain, but their derivatives are quite different.

Namely, for all $t \in \mathbb R$, we have canonical identifications $\mathbb R \cong \operatorname T_t\mathbb R$ and $\mathbb C \cong \operatorname T_{h_2(t)}\mathbb C$, with respect to which ${\operatorname d}h_{2, t}$ is multiplication by $i h_2(t)$.  On the other hand, we have to choose identifications of the tangent spaces to $\operatorname S^1$ with $\mathbb R$—one very natural choice is to identify, for each $z \in \operatorname S^1$, $\Delta t \in \mathbb R$ with the tangent vector at $t = 0$ to the curve $t \mapsto z e^{i\Delta t\cdot t}$—and then ${\operatorname d}h_{1, t}$ is the identity map for all $t \in \mathbb R$.


问:
Can a function be uniformly continuous on an open interval?
I am learning analysis and all the uniformly continuous functions I have seen are over a closed interval. So, can a uniformly continuous function be defined on an open interval?</p>

答:
Sure, take a function which is continuous on a finite closed interval, and remove the endpoints.

And there are many functions defined on the whole real line that are uniformly continuous.


问:
Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$

My attempt:

Consider the following series:

$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$

Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.

Simplfying S,

$$ S = - \ln \left(  \cos x \cdot \cos \frac{x}{2} ... \cos \frac{x}{2^n} \right)$$

or,

$$ S = - \ln \left(  \frac{ \sin(2x)}{2^{n+1} \sin (\frac{x}{2^n})} \right)= \ln(2^{n+1}) + \ln( \sin \frac{x}{2^n}) - \ln( \sin(2x) )$$

Now,

$$ \frac{dS}{dx}  = \frac{1}{2^n} \cot \frac{x}{2^n} - 2 \cot(2x)$$

The problem I'm having is proving that

$$ \lim_{ n \to \infty} \frac{1}{2^n} \cot \frac{x}{2^n} = \frac{1}{x}$$

答: For a calculus-free solution, prove by induction $\sum_{i=0}^n2^{-i}\tan\tfrac{\theta}{2^i}=2^{-n}\cot(2^{-n}\theta)-2\cot2\theta$ for $n\ge0$. Since $\tan x=\cot x-2\cot2x$, the left-hand side is a telescoping series.


问:
Sorting an array using reverse
I ran into an Olympiad question recently, and one challenging question surprised me:

We have an array $A$ with $n$ elements. $\operatorname{Rev}(i, j)$ for $1 \leq i < j \leq n$ reverses subarray $A[i\ldots j]$, i.e., swaps $A[j-k]$ and $A[i+k]$ for all $0 \leq k \leq j-i$.

How many times must we use $\operatorname{Rev}(i, j)$ to sort $A$?

Answer is: $\operatorname O(n)$!

How can I sort $A$, using $\operatorname{Rev}(i, j)$ $\operatorname O(n)$ times?

答:
Although it seems clear to me that this is not an answer in the spirit in which the question was meant, I post it by request:

I think the parameters of the question need to be more clearly stated. Obviously the specific $\operatorname{Rev}$'s used need to depend on $A$, but how are we allowed to implement that dependence? Here's an algorithm that only uses $n - 1$ $\operatorname{Rev}$'s:

```<code>def sortfrom(i, A):
{
    if (i < n) {
        m = A[i];
        mi = i;
        for j in (i..n) {
            if ( A[j] <= m ) {
                mi = j;
                m = A[mi]
            }
        }
        Rev(i, mi, A);
        sortfrom(i + 1, A)
    }
    return A;
}
sortfrom(1, A)
</code>```

I used to use repl.it as my go-to one-shot compiler, but now it appears that you can't run code there without logging in.  Here is the above pseudoccode translated to Perl (with your array hardcoded), which you can run on OnlineGDB:

```<code>use v5.10;

sub sortfrom {
    my (<span class="math-container">$i, @a) = @_;
    my $</span>n = <span class="math-container">$#a;
    if ($</span>i < <span class="math-container">$n) {
        my $m = $a[$i];
        my $mi = $i;
        for my $j ( $i .. $n ) {
            if ( $a[$j] < $m ) {
                $m = $a[$j];
                $mi = $j;
            }
        }
        @a[$i .. $mi] = reverse @a[$i .. $mi];
        say "(", (join ", ", @a), ")";
        sortfrom($</span>i + 1, @a);
    }
    return @a;
}

sortfrom(0, (5, 9, 0, 2, 1, 3, 4, 6));
</code>```

Edit from  @Joffan as a way of posting my comment as requested, since I don't feel this merits a separate answer, being effectively a description of the code posed by LSpice above:

One obvious way to guarantee a maximum of $(n{-}1)$ uses of a $\text{Rev}$ function would be: find the $i$th smallest element at location $k$, use $\text{Rev}(i,k)$ to put that in place, with $i$ running from $1$ to $n{−}1$ (after which the last & greatest element must already be at the end). And $n{−}1$ is of course in $O(n)$.

From another comment, illustrating one step in the process:

Suppose we are at step $i=4$, and the array looks like this $$(\color{blue}{3,5,6},20, 15, 11, 18, \color{red}7, 10)$$ where the first three steps have put the 3 smallest elements at the start. The $4$th smallest element is $7$ (as shown in red) and is found at position $8$ so we undertake $\text{Rev}(4,8)$ which produces  $$(\color{blue}{3,5,6},\color{red}7, \color{green}{18, 11, 15, 20}, 10)$$

Title: Sorting an array using reverse

Body:
Although it seems clear to me that this is not an answer in the spirit in which the question was meant, I post it by request:

I think the parameters of the question need to be more clearly stated. Obviously the specific $\operatorname{Rev}$'s used need to depend on $A$, but how are we allowed to implement that dependence? Here's an algorithm that only uses $n - 1$ $\operatorname{Rev}$'s:

def sortfrom(i, A):
{
    if (i < n) {
        m = A[i];
        mi = i;
        for j in (i..n) {
            if ( A[j] <= m ) {
                mi = j;
                m = A[mi]
            }
        }
        Rev(i, mi, A);
        sortfrom(i + 1, A)
    }
    return A;
}
sortfrom(1, A)

I used to use repl.it as my go-to one-shot compiler, but now it appears that you can't run code there without logging in.  Here is the above pseudoccode translated to Perl (with your array hardcoded), which you can run on OnlineGDB:

use v5.10;

sub sortfrom {
    my ($i, @a) = @_;
    my $n = $#a;
    if ($i < $n) {
        my $m = $a[$i];
        my $mi = $i;
        for my $j ( $i .. $n ) {
            if ( $a[$j] < $m ) {
                $m = $a[$j];
                $mi = $j;
            }
        }
        @a[$i .. $mi] = reverse @a[$i .. $mi];
        say "(", (join ", ", @a), ")";
        sortfrom($i + 1, @a);
    }
    return @a;
}

sortfrom(0, (5, 9, 0, 2, 1, 3, 4, 6));

Edit from  @Joffan as a way of posting my comment as requested, since I don't feel this merits a separate answer, being effectively a description of the code posed by LSpice above:

One obvious way to guarantee a maximum of $(n{-}1)$ uses of a $\text{Rev}$ function would be: find the $i$th smallest element at location $k$, use $\text{Rev}(i,k)$ to put that in place, with $i$ running from $1$ to $n{−}1$ (after which the last & greatest element must already be at the end). And $n{−}1$ is of course in $O(n)$.

From another comment, illustrating one step in the process:

Suppose we are at step $i=4$, and the array looks like this $$(\color{blue}{3,5,6},20, 15, 11, 18, \color{red}7, 10)$$ where the first three steps have put the 3 smallest elements at the start. The $4$th smallest element is $7$ (as shown in red) and is found at position $8$ so we undertake $\text{Rev}(4,8)$ which produces  $$(\color{blue}{3,5,6},\color{red}7, \color{green}{18, 11, 15, 20}, 10)$$


问:
Strange statistic
Let $N$ be a random integer and let $p_N(n)$ be its discrete density.
The statistic
\begin{equation}S(N)\triangleq\sum_{n=2}^\infty p_N(n)\,n\,(n-1)\end{equation}
has some specific meaning?
It is clear that in general it is not the variance of $N$, but it is instead
\begin{equation}S(N)=\mathbb{E}[N^2]-\mathbb{E}[N]\end{equation}
where $\mathbb{E}[N]$ denotes the expected value of $N$. In fact
\begin{equation}\begin{aligned}S(N)&=\sum_{n=2}^\infty p_N(n)\,n^2 -\sum_{n=2}^\infty p_N(n)\,n\\
&=\left(\sum_{n=0}^\infty p_N(n)\,n^2-p_N(1)\right)-\left(\sum_{n=0}^\infty p_N(n)\,n-p_N(1)\right)\\
&=\sum_{n=0}^\infty p_N(n)\,n^2-\sum_{n=0}^\infty p_N(n)\,n\\
&=\mathbb{E}[N^2]-\mathbb{E}[N]\end{aligned}\end{equation}
The statistic $S(N)$ is the variance only if $N$ has expected value equal to zero.

答:
\begin{align}
& \text{The $k$th} \textbf{ factorial moment } \text{of the} \\
& \text{distribution of the random variable $X$} \\[10pt]
= {} & \operatorname E\big( \,\underbrace{X(X-1)(X-2)\cdots(X-k+1)}_\text{$k$ factors} \, \big)
\end{align}

The only reason that I know of (but I am ignorant of many things) why this is of interest is that it can be neatly computed in cases where $X$ has a Poisson distribution, and then the moments of the Poisson distribution may be computed by applying some algebra after that.

\begin{align}
& \text{Suppose } X \sim\operatorname{Poisson}(\lambda). \text{ Then} \\[10pt]
& \operatorname E\big( X(X-1)(X-2)\cdots(X-k+1) \big) \\[8pt]
= {} & \sum_{x\,=\,0}^\infty x(x-1)(x-2)\cdots(x-k+1) \Pr(X=x) \tag 1 \\[8pt]
= {} & \sum_{x\,=\,k}^\infty x(x-1)(x-2)\cdots(x-k+1) \Pr(X=x) \\
& \text{because the sum of the first $k$ terms on line $(1)$ above is 0} \\[8pt]
= {} & \sum_{x\,=\,k}^\infty x(x-1)(x-2)\cdots(x-k+1)\cdot\frac{\lambda^x e^{-\lambda}}{x!} \\[8pt]
= {} & \sum_{x\,=\,k}^\infty \frac{\lambda^x e^{-\lambda}}{(x-k)!} = \sum_{y\,=\,0}^\infty \frac{\lambda^{y+k} e^{-\lambda}}{y!} \quad \text{where } y = x-k \\[8pt]
= {} & \lambda^k \sum_{y\,=\,0}^\infty \frac{\lambda^y e^{-\lambda}}{y!} \qquad = \lambda^k.
\end{align}

Proposition: The first $n$ moments of the distribution of the number of fixed points of a uniformly distributed random permutation of the set $\{1,\ldots,n\}$ are the same as the first $n$ moments of the Poisson distribution with expected value $1.$

Proof: The first $n$ factorial moments are the same; therefore the first $n$ moments are the same. $\qquad\blacksquare$

Title: Strange statistic


问:
For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$
For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$

I immediately thought of saying that from symmetry we have that $x\le y \le z$.

Also, $y^2+z^2 \equiv 0 mod x$, $x^2+z^2\equiv 0mody$ and $x^2+y^2\equiv 0modz$.

Moreover through trial and error I worked out that the solutions for $k$ must be $k=1$ or $k=3$ but I have not managed to prove it. I attempted to use inequalities, but that didn't work out either. Could you please explain to me how to solve this question?

答:
this is called a CW answer; recommend beginning with

Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.

Here are a number of posts  about
Vieta Jumping/ Hurwitz/Markov         Grundlösung

Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$.

Diophantine quartic equation in four variables

Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?  !!!!!!!!!  +++++++

http://math.stackexchange.com/questions/1411049/if-a-b-are-positive-integers-and-ab-1-mid-a%C2%B2-b%C2%B2-then-prove-that-q

Proving that all terms in sequence are positive integers (Recursive)

Diophantine equation $(x+y)(x+y+1) - kxy = 0$

Find the postive integers such $xy+x+y\mid x^2+y^2-2$

Find the integer values of c

Find all solutions to the diophantine equation $(x+2)(y+2)(z+2)=(x+y+z+2)^2$

Characterize the integers $a,b$ satisfying: $ab-1|a^2+b^2$

Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.     Follows Hurwitz very closely!!  (x+y+z)^2 = nxyz

Showing that $m^2-n^2+1$ is a square  LEMMA +

Showing that $m^2-n^2+1$ is a square   LEMMA -

How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$

Find all possible value of c

What are the solutions of the equation $3np+3n+2=n^2+p^2$, with n and p positive integers?

Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$

Integer points on a hyperbola two spiral arms

$m+n+p-1=2\sqrt{mnp}$ in positive integers

https://artofproblemsolving.com/community/c6h1726220_if_fraction_integer_then_equal_to_5   Poland 1991 training


答:
The usefulness of differentials  comes from the fact that a linear function is very simple to calculate and that, by definition, the differential of a function $f$ defined on a vector space $V$ at a point $x\in V$, among all possible approximations of $f(x+h)$ by a linear function, is the best possible approximation, in the sense that
$$f(x+h)=f(x)+\mathrm df_x(h)+o\bigl(\|h\|\bigr).$$

Title: Why the differential of a function is a linear transformation?

Body:
I try to find some applications of the differential, that weird thing that is a linear transformation, what is the reason of linearity?
And how this linearity works, I mean, there's something fundamental about that linearity? And where we can see the importance of the differential, like in physics or another field of application.</p>


问:
GCSE( how to find the volume of a pencil)
In the figure below, the dimensions of an unsharpened pencil are given. The unsharpened pencil have graphite in a cylindrical shape, surrounded by wood. The diameter of the graphite is 2mm. The density of graphite is $640\text{ kg/m}^3$. The density of wood is $420\text{ kg/m}^3$. Find the mass of 500 pencils.

My attempt was the following:
I found the volume of the wood: $$3.14\times (0.004)^2\times 0.15= 0.000007536$$
The volume of graphite:
$$3.14\times (0.001)^2\times 0.15= 0.000000471$$
I found the mass of 1 pencil: $$0.000007536\times  420 + 0.000000471\times 640= 0.0034\\
0.0034\times 500= 1.7 \text{ kg}$$

I think I messed up the solution!

All help/solutions appreciated.

Thanks

答:
first of all, I'd like to say it would be much easier for you to calculate with, say, $4.71×10^7$ instead of $0.000000471$.
The mistake you're making here is counting the graphite's volume twice. When calculating the volume of the wooden part, you need to subtract the graphite's volume:

$7.536×10^{-6}-0.471×10^{-7}=10^{-7}×(75.36-0.471)=74.889×10^{-7}=7.4889×10^{-6} m^3$

Now, you have the density of both materials. That is to say, you know how many $kg$ one $m^3$ of wood/graphite weighs. From that info, you can calculate the mass of wood/graphite in one pencil:

$m_{wood}=7.4889×10^{-8}×420=3145.338×10^{-8}=3.145338×10^{-5}kg$

$m_{graphite}=0.471×10^{-7}×640=301.44×10^{-7}=3.0144×10^{-5}kg$

$m_{pencil}=m_{wood}+m_{graphite}=10^{-5}×(3.0144+3.145338)=6.159738×10^{-5}kg$

Now getting the mass of 500 pencils is simple! The answer is

$6.159738×500×10^{-5}=30.79869×10^{-3}kg \approx 30.8 gr$

Hopefully this solves your problem :)


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
The derivative does not exist unless the limit of the difference quotient exists.

The "limit law" that says the limit of a sum of two functions is equal to the sum of the two separate limits is not applicable unless the two separate limits exist. Notice that

There are no cases where the two separate limits exist and the limit of the sum does not. If the two separate limits exist, then so does the limit of the sum.

However, there are cases in which the two separate limits do not exist and the limit of the sum does. A similar situation applying to products rather than sums arose in something I posted here recently (I can't find it right now). For one of the two factors the limit did not exist, but the function was bounded and therefore the limit of the product could be found by squeezing.


问:
How to show a derivative is a maximum, not a minimum
I am working through a maths text book. I study as a hobby. I have solved most of this problem except for the very last part.

The points A and B are on the same horizontal level and at a distance of $b$ apart. A particle P falls vertically at rest from B, so that at time $t$ its depth below B is $kt^2$, where $k$ is constant. At this time, the angle of depression of P from A is $\theta$.

Prove that $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$. Show that $\frac{d\theta}{dt}$ is greatest (and not least) when $\theta = \frac{1}{6}\pi$.

I proved the first part and to then I differentiated $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$ to get the second derivative:

$\frac{d^2\theta}{dt^2} = \{(b^2 + k^2t^4)(2bk) - (2bkt)(4k^3t^4)\}/(b^2 + k^2t^4)^2$

$= 0$ when $2b^3k +2bk^3t^4 - 8bk^3t^4 = 0$

$\implies b^2k = 3k^2t^4$

Now when $\theta = \frac{1}{6}\pi$, $\tan \theta = \frac{1}{\sqrt3}$ and

$\frac {kt^2}{b} = \frac{1}{\sqrt3}$, meaning $b^2 = 3k^2t^4$

So $\theta = \frac{1}{6}\pi$ is a maximum or a minimum.

But how do I show which? Taking a third derivative seems very complicated.

答:
To find a stationary point in general, we let the first derivative be equal to $0$. As you've said, this by itself gives us no information about whether the stationary point is a minimum or maximum. To determine which type it is, we use the second derivative of the function.

For a given curve, $y=f(x)$ which has a stationary point at $(a, f(a))$:

If $$f''(a)<0$$
then the stationary pint is a (local) maximum. This is because the second derivative is the rate of change of the gradient (the first derivative). If the gradient is decreasing at the stationary point then this means that the curve is about to slope downwards ; ie the stationary point is a maximum.

If $$f''(a)>0$$
then the stationary pint is a (local) minimum. This is because the second derivative is the rate of change of the gradient (the first derivative). If the gradient is increasing at the stationary point then this means that the curve is about to slope upwards ; ie the stationary point is a minimum.

However, if $f''(a)=0$ then we either have a point of inflection, or a point which isn't a point of inflection (No...really? ;) ). If it isn't a point of inflection then it is more tedious to determine which type of stationary point we have. I can elaborate about this if you like.


问:
Question about dual hypersurfaces
Leet $F: \mathbb{R}^n \to \mathbb{R}$ be a smooth function which defines a hypersurface $X = \{F=0\}$. Then one can define the dual hypersurface by $F^* = 0$ where
$$
F^*(y) = y \cdot (\nabla F)^{-1}(y) - (F \circ (\nabla F)^{-1})(y)
$$
at points where $(\nabla F)^{-1}$ is well defined. The only notion of dual hypersurface $X^*$ I knew before running into this notion is the one in Wikipedia https://en.wikipedia.org/wiki/Dual_curve

I'm struggling to understand if these two notions the same or not? Also Im interested in finding a reference for the first notion of dual hypersurface. Thank you for your help!

ps This notion came up here https://arxiv.org/pdf/1711.01390.pdf page 16

答:
The dual hypersurface is really a natural projective notion. In $\Bbb P^n$ (working over whatever field you want), you start with a hypersurface $X$ given by $F=0$ (with $F$ homogeneous). The dual hypersurface $X^*$ lives in $\Bbb P^{n*}$, which is the space of hyperplanes in $\Bbb P^n$, and is the locus of the tangent hyperplanes to $X$.

For any $[p]\in X$, Euler's Theorem gives us the equation
$$\sum x_i\frac{\partial F}{\partial x_i}(p) = 0$$
for the tangent hyperplane to $X$ at $[p]\in\Bbb P^n$, so $[\nabla F(p)]$ is the point on $X^*$ corresponding to $p$. (Check that everything is projectively well-defined here because, of course, $F$ is homogeneous.)

Now let's look at your formula for $F^*$. Set $y=\nabla F(x)$ (which should be a point on $X^*$). Then
$$F^*(y) = F^*(\nabla F(x)) = \nabla F(x)\cdot x - F(x).$$
There's no reason on earth this should be $0$. If $F$ is homogeneous of degree $k$ and the original hypersurface is given by $F=0$, then Euler tells us that $\nabla F(x)\cdot x = kF(x)$, so $F^*(y)$ with those hypotheses does simplify to  $(k-1)F(x)=0$.

Let's try a concrete, well-understood example. I'm still working projectively with a homogeneous $F$. Start with a quadratic hypersurface $F(x) = \frac12\sum c_i x_i^2 = 0$. Then $\nabla F = (c_1x_1,\dots,c_nx_n)$, so $y=\nabla F$ lies on $X^*$ if and only if $$G(y)=\sum \frac1{c_i}y_i^2 = 0.$$ This is all very classical. Now let's see if your formula gives this. With your formula, noting that $\nabla F$ here is the linear map with diagonal matrix $(c_i)$,
$$F^*(y) = y\cdot (\nabla F)^{-1}y - F((\nabla\ F)^{-1}(y)) = \sum \frac1{c_i}y_i^2 - \frac12 c_i(y_i/c_i)^2 = \frac12 G(y).$$

At any rate, your definition is actually is the Legendre transform of $F$. You might read the section on Geometric interpretation in this. Although I no longer have it to look at, in one of my joint papers we gave a reference to V. I. Arnold's Singularities of systems of rays, Russian Math Surveys 38 (1983), pp. 87-176, for a discussion of the Legendre transform. See in particular pp. 108-110.

Title: Question about dual hypersurfaces


问:
Number of elements of order $p$ in a (possibly infinite) group
If a group is finite, then it's easy to see by group action that $n_p\equiv-1\pmod p$.

However, is the result true for an infinite group? My conjecture is that if $n_p$ is non-zero and finite, then $n_p\equiv-1\pmod p$ of infinite groups. I proved the result for $n_2$, but don't know whether it is true for all primes.

答:
We know that the result is true for finite groups (by the group-action argument you're familiar with).

But if there's only finitely many elements of order $p$, we can look at $H$ the group generated by those elements. It is finite, and contains all the elements of order $p$. So the result holds for $H$, and since $H$ has all the elements of order $p$, the result holds for $G$.

I hope this helps ^_^

Added by Derek Holt: proof that $H$ is finite. Since $H$ is generated by the elements of order $p$ and there are only finitely many, they have only finitely many conjugates, so their centralizers have finite index in $H$. So the intersection of their centralizers, which is the centre $Z(H)$ of $H$ has finite index in $H$.

Now, by a result of Schur, the derived group $[H,H]$ is finite. Since $H/[H,H]$ is an abelian group generated by finitely many elements of order $p$, it is also finite, and hence $H$ is finite.


问:
Approximately 8.33% of men are colorblind. You survey 8 men from the population of a large city and count the number who are colorblind.
This question shows up in the binomial distribution chapter in my book.

g. How many men would you have to survey in order to be at least 95%
sure you would find at least $1$ who is colorblind?

Now I know that $P(X \geq 1) = 1 - P(X=0) = 1 - {8\choose0}.0833^0.9167^8 = 1 - .4986752922 = .5013247078$

But this is for a sample of 8 men. This question is asking me to find $n$ with $95$% confidence,

But this seems like a confidence interval question which I'm a little confused about since my book is about probability and not about statistics. Nothing like this shows up in my book, so I'm stuck. I feel like this is a "challenge" question.

Can someone give me a clue on what formula to use to deal with a question like this? Is this even a confidence interval question?

Thank You

Body:
You would have to solve
$1 - {n\choose0}(0.0833^0)(0.9167^n)\ge0.95$.


问:
Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$
The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.

The easy way to solve this is by taking the last digit of each number:

$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.

$4^{2n+1}$ ends in $4$ as $2n + 1$ is odd.

So the sum will end in $4+1=5$ and because $81^{3^{n}}+4^{2n+1}>5$, it is $5|81^{3^{n}}+4^{2n+1}$, thus the number is not prime.

Are/is there any other proof(s) which do not reuse the same idea?

答: Yes, there are other proofs. Instead of showing $81^{3^{n}}+4^{2n+1}\equiv 0\bmod 5$ you can also show that $81^{3^{n}}+4^{2n+1}\equiv 0 \bmod 17$ for all $n\in \Bbb N$.


问:
Determining a limit of a function doesnt match up
Consider this function

$$\lim_{x \to 2} \frac {x^3 - 8} {x-2}$$

Now what I've done is I've tried to alter the fraction to avoid division by zero. For the numerator, I've put it as this

$$ \frac {(x - 2)^3} {x-2} $$

After this I've simply gotten rid of the cube and the lower part of the fraction leaving me with this

$$(x-2)^2$$

And here I've applied the formula $$a^2 -2ab + b^2$$ And I than I put in 2 and I get as a result $$ x^2 +4x + 4$$
If i put in 2 now I get 16. Now the solutions say 12, but the solutions are notorius to be wrong more often than you would think.Is my solution correct? If not what am I doing wrong.

Thanks!

答:
$x^3-a^3\ne (x-a)^3$, except in characteristic $3$.

It can be much simpler to determine: $\frac{x^3-2^3}{x-2}$ is the rate of variation of the function $x^3$ from $x=2$, so the limit is its derivative $\:3x^2\bigr|_{x=2}=12$.