问:
Evaluating the integral of $e^{\sin x}$ from $0 \to \frac{\pi}{2}$
$$
\xi = \int_0^{\frac{\pi}{2}} e^{\sin(x)}dx
$$
We can take advantage of the fact that $e^x$ has a simple McLaurin series:
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
$$
Substituting $\sin(x)$
$$
\xi = \int_0^{\frac{\pi}{2}} \sum_{n=0}^\infty \frac{\sin^nx}{n!} dx
$$
Switching the bounds of integration:
$$
\xi =\sum_{n=0}^\infty \frac{1}{n!} \int_0^{\frac{\pi}{2}}\sin^nx \, dx
$$

We let
$$
I_n = \int_0^{\frac{\pi}{2}}\sin^nx \, dx = \int_0^{\frac{\pi}{2}}\sin^{n-1}x \sin x\, dx
$$
Using integration by parts:

$$
I_n = \left[-\sin^{n-1}x \cos x \right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\left( \sin^{n-2}x - \sin^{n}x\right)
$$

$$
I_n = \frac{n-1}{n}I_{n-2}
$$

We can see that we get different results when $n$ is odd or even. We know that $I_0 = \frac{\pi}{2}$ and $I_1 =1$. So for any $n$ we have:
$$
\begin{cases}
n \,\textrm{even}\, , \quad I_{2n} = \frac{\pi}{2} \prod_{k=1}^n \frac{2k-1}{2k}\\
n \,\textrm{odd}\, , \quad I_{2n+1} =  \prod_{k=1}^n \frac{2k}{2k+1}
\end{cases}
$$
We can now use this for $\int_0^{\frac{\pi}{2}}\sin^nx$ in $\xi$:
$$
\xi = \sum_{n=0}^\infty \left[\frac{\pi}{2(2n)!}\prod_{k=1}^n \frac{2k-1}{2k} \right] + \sum_{n=0}^\infty \left[\frac{1}{(2n+1)!}\prod_{k=1}^n \frac{2k}{2k+1} \right]
$$
Merging:
$$
\int_0^{\frac{\pi}{2}} e^{\sin(x)}dx = \sum_{n=0}^\infty \left[\frac{\pi}{2(2n)!}\prod_{k=1}^n \frac{2k-1}{2k} + \frac{1}{(2n+1)!}\prod_{k=1}^n \frac{2k}{2k+1} \right]
$$


问:
Find generator of principal ideal
The ideal $(9, 2 + 2\sqrt{10})$ of $\mathbb{Z}[\sqrt{10}]$ is a principal ideal; it is generated by $1+\sqrt{10}$.

This is easy enough to check once it's been found, but can anyone tell me some way to arrive at this (or some other) generator? That could be done with pencil & paper?

Ordinarily I would use the Euclidean algorithm, but we don't have that here...

答: We have the norm (even though it is not always positive) obtained by multiplying $a+b\sqrt {10}$ with its conjugate $a-b\sqrt {10}$, that is $N(a+b\sqrt{10})=a^2-10b^2\in\mathbb Z$. So we have $N(9)=81$ and $N(2+2\sqrt{10})=-36$. We can restrict our search to elements of norm dividing both $81$ and $36$, that is we must have $N(a+b\sqrt{10})\in\{\pm1,\pm3,\pm9\}$. Looking for small solutions of this you will stumble upon $1+\sqrt {10}$, which is both a divisor of $9$ and $2+2\sqrt{10}$ and a linear combination of these, namely $(1+\sqrt {10})\cdot 9-4\cdot (2+2\sqrt{10})$.</p>

ContentLicense: CC BY-SA 3.0

Title: Find generator of principal ideal


问:
Partial sum of Geometric sequence
Can someone explain to me how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = \frac{n(1-(1-p)^r)}{p}$.

I have used the formula for a Geometric series. My common ratio is: $(1-p)$. I get the following:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n\cdot\frac{1-(1-p^r)}{1-(1-p)}$.

From here, I simply don't know how to get to $\frac{n(1-(1-p)^r)}{p}$

ContentLicense: CC BY-SA 4.0

Title: Partial sum of Geometric sequence

Body:
\begin{align*}
\sum_{i=1}^{r}n(1-p)^{i-1}
&
=n\sum_{i=0}^{r-1}(1-p)^i
=[\text{\(r\) terms}]
=n\cdot\frac{1-(1-p)^r}{1-(1-p)}
=\frac{n\bigl(1-(1-p)^r\bigr)}{p}
\end{align*}


问:
Confusion in resultant velocity and relative velocity?
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is.

I drew the diagram like this (blue one)and got answer as north east or east north direction.

My questions are

It is normal right if relative velocity and resultant velocity have same direction and drawn like this.

(This is the red diagram ).Also , I saw online that they drew velocity of train towards left without even making it$ -V_t$. I didn’t understand why they did that.

答:
Resultant vector is the sum of two vectors in the parallelogram diagram.

Difference vector is the relative vector.

Note that there are two possibilities of relative vector depending on the sense of the vectors considered. Shown as $A-B$ and $B-A$ in the diagram. The relative vector is obtained by changing the sense (direction) of one of the given vectors and then finding the resultant, found in the arrow tip to head placed the usual way.


问:
About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime
The number $30$ has a curious property:

All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!

I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.

Are there any such numbers $\gt 30$?

答:
Let $\, p_i\,$ be the $i$'th prime. Many proofs (e.g. below) use $\, p_{n+1}\! < 2 p_n\, $ (by Bertrand) to prove that $\,p_{k+1}^2\! < p_1 p_2 \cdots p_k.\:$  It is worth remarking that this follows from  a much weaker hypothesis than Bertrand. Indeed, it needs only $\ \color{#c00}{p_{k+1}\! < p_k^{3/2}}.\ $ Then the inductive step is
$$\begin{eqnarray} && \qquad\ p_k^2 &<\,& p_1 p_2 \cdots p_{k-1} \\
&& p_{k+1}^2/p_k^2 &<\,& p_k\ \ \  {\rm  by}\ \ \ \color{#c00}{p_{k+1} < p_k^{3/2}},\ \ \ \text{by hypothesis}\\
\Rightarrow && \qquad p_{k+1}^2 &<\,& p_1 p_2 \cdots p_k\ \ \ \text{by multiplying the above}
\end{eqnarray}$$

What is the simplest proof of this weaker hypothesis: $\ p_{k+1}\! < p_k^{3/2}\,$ for $\, k > 1$?

Below is a typical proof of said form, from PlanetMath.


问:
How to solve this? I am having difficulty in the very last step of the problem.
The general solution of $\vert\sin x\vert = \cos x$ is -

(A) $2n\pi+{\pi\over4}$, $n\in I$

(B) $2n\pi±{\pi\over4}$, $n\in I$

(C) $n\pi+{\pi\over4}$, $n\in I$

(D) None of these

So what I did was - I made a case for when sinx is greater than 0 and equated it to $\cos x$ to get $\tan x = 1$ which implies x = $\pi\over 4$. The other case was when $\cos x = -\sin x$. Here, x = $3\pi\over 4$. I don't understand how to proceed from here.

答:
$$\cos x=|\sin x|\ge0$$

If $\sin x<0, \cos x=-\sin x\iff\tan x=-1$

As $x$ will lie in the $4$th quadrant, $ x=2n\pi-\dfrac\pi4$

What if $\sin x\ge0?$


问:
Prove that if $Σ$ is consistent and $Σ \vdash A$ , then $Σ ∪ \{A\}$ is consistent.
Maybe I am wrong, but isn't $Σ$ being consistent equivalent or very similar to $Σ$ being satisfiable?

Since $Σ$ is consistent it is not a contradiction. Since $Σ \vdash  A$, we know that $Σ$ is tautologically equivalent to $A$. Therefore $Σ ∩ \{A\}$ is consistent because it is satisfiable and not a contradiction.

Is this the right line of thinking here?

How would one go about answering this question properly?

答:
These two inference rules are logically equivalent in a classical meta-level with (D |/- B) = ~(D |- B). They are just contraposition of each other:

```<code>   G |- A     G, A |- C
----------------------- (Cut)
         G |- C

   G |- A     G |/- C
----------------------- (Contra-Cut)
       G, A |/- C
</code>```

So the OPs theorem emerges for C = f and when the system has (Cut). Would work for incomplete calculus as well, since OP asked for consistency and not satisfiability.

Body:
These two inference rules are logically equivalent in a classical meta-level with (D |/- B) = ~(D |- B). They are just contraposition of each other:

   G |- A     G, A |- C
----------------------- (Cut)
         G |- C

   G |- A     G |/- C
----------------------- (Contra-Cut)
       G, A |/- C

So the OPs theorem emerges for C = f and when the system has (Cut). Would work for incomplete calculus as well, since OP asked for consistency and not satisfiability.


问:
Subspace topology induced by Inductive Limit is coarser than original topology on subspace
Let $\{X,\{X_i,T_i\}_{i\in I}\}$ be an inductive system of TVS's and let $T$ be the inductive limit topology. I am trying to show that the relative topology on $X_i$ induced by $T$ is coarser than $T_i$. This is where I've gotten so far:

$W\in T|_{X_i}$ implies $W=V\cap X_i$ for some $V\in T$.

But for all $v\in V$ there exists $U_v$, a balanced convex subset of $X$ such that $U_v\cap X_j\in T_j$ for all $j\in J$, and $(U_v+v)\subset V$.

Therefore $V=\cup_{v\in V}(U_v+v)$ and so $W=\cup_{v\in V}[X_i\cap(U_v +v)]$.

I would like to show that each $X_i\cap(U_v +v)$ is in $T_i$, as then the proof is finished, but not sure if this is true as I can't seem to find a way. Is my approach wrong?

答:
I think this way works;

$B=\{U\in X : U\text{ convex, balanced and } U\cap X_i\in T_i \text{ for all } i\}$ defines a basis for the $(X,T)$ neighbourhood system at $0$.

Then $A=\{U\cap X_i : U\in B\}$ defines a basis for the $(X_i,T|_{X_i})$ neighbourhood system at $0$.

It's clear that $V\in A \implies V\in T_i$.

Since the neighbourhood system at $0$ defines the topology for a TVS, this is sufficient to show that $T|_{X_i}\subset T_i$.


答: $\cos x$ is positive in the first and fourth quadrant and in these, $|\tan x|$ is $1$ for $x=\pm\dfrac\pi4$. Add $2n\pi$.

Title: How to solve this? I am having difficulty in the very last step of the problem.

Body:
The general solution of $\vert\sin x\vert = \cos x$ is -

(A) $2n\pi+{\pi\over4}$, $n\in I$

(B) $2n\pi±{\pi\over4}$, $n\in I$

(C) $n\pi+{\pi\over4}$, $n\in I$

(D) None of these

So what I did was - I made a case for when sinx is greater than 0 and equated it to $\cos x$ to get $\tan x = 1$ which implies x = $\pi\over 4$. The other case was when $\cos x = -\sin x$. Here, x = $3\pi\over 4$. I don't understand how to proceed from here.


答:
We will solve $\sin^{2}(x) = \cos^{2}(x)$, which is obtained upon squaring the equation (so we do not have to deal with sign issues).

$$\sin^{2}(x) = \cos^{2}(x)$$

$$\cos^{2}(x) -\sin^{2}(x) = 0$$

$$\cos(2x) = 0$$

$$x = \frac{\pi}{4} + \pi n, \frac{3\pi}{4} + \pi n$$

$$x = \frac{\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n, \frac{3\pi}{4} + 2\pi n, \frac{7\pi}{4} + 2\pi n$$

Here, we split into cases of $2\pi n$ periodicity to take advantage of the properties of $\sin$ and $\cos$.

Checking these solutions in the original equation $\vert \sin(x)\vert = \cos (x)$, we find that only $\frac{\pi}{4} + 2\pi n$ and $\frac{7\pi}{4} + 2\pi n$ work. $\frac{7\pi}{4} + 2\pi n$ is equivalent to $-\frac{\pi}{4} + 2\pi n$, so the answer is $\boxed{(\text{B})\ \pm\frac{\pi}{4} + 2\pi n.}$

You can do a sanity check by drawing a quick graph and seeing if the functions intersect in the general region of the conjectured solution.


问:
Torsion in absolute Galois groups
Let $K$ be a field, $K^{sep}$ its separable closure, and $G := Gal(K^{sep}/K)$ its absolute Galois group. What do we know about torsion in such groups? In particular, I am wondering if there is a restriction on the kind of subgroups, or if all finite subgroups can be embedded inside $G$.

If $K$ is a finite field then $G \cong \hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$, in particular it is torsion-free. How about $K = \mathbb{Q}$? How about local fields like $K = \mathbb{Q}_p$?

答:
The Artin-Schreier theorem says that for any field $K$ the only possibilities for torsion in $\operatorname{Aut}(\overline K/K)$ are in fact 2-torsion elements.
See: https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf

The absolute Galois group of $\mathbf Q$ does contain 2-torsion elements though, as we can choose an embedding
$$\overline{\mathbf Q} \hookrightarrow \mathbf C$$
and act via complex conjugation to get an involution of $\overline{\mathbf Q}$ fixing $\mathbf Q$.

Title: Torsion in absolute Galois groups


Title: Evaluating the integral of $e^{\sin x}$ from $0 \to \frac{\pi}{2}$

Body:
$$
\xi = \int_0^{\frac{\pi}{2}} e^{\sin(x)}dx
$$
We can take advantage of the fact that $e^x$ has a simple McLaurin series:
$$
e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
$$
Substituting $\sin(x)$
$$
\xi = \int_0^{\frac{\pi}{2}} \sum_{n=0}^\infty \frac{\sin^nx}{n!} dx
$$
Switching the bounds of integration:
$$
\xi =\sum_{n=0}^\infty \frac{1}{n!} \int_0^{\frac{\pi}{2}}\sin^nx \, dx
$$

We let
$$
I_n = \int_0^{\frac{\pi}{2}}\sin^nx \, dx = \int_0^{\frac{\pi}{2}}\sin^{n-1}x \sin x\, dx
$$
Using integration by parts:

$$
I_n = \left[-\sin^{n-1}x \cos x \right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\left( \sin^{n-2}x - \sin^{n}x\right)
$$

$$
I_n = \frac{n-1}{n}I_{n-2}
$$

We can see that we get different results when $n$ is odd or even. We know that $I_0 = \frac{\pi}{2}$ and $I_1 =1$. So for any $n$ we have:
$$
\begin{cases}
n \,\textrm{even}\, , \quad I_{2n} = \frac{\pi}{2} \prod_{k=1}^n \frac{2k-1}{2k}\\
n \,\textrm{odd}\, , \quad I_{2n+1} =  \prod_{k=1}^n \frac{2k}{2k+1}
\end{cases}
$$
We can now use this for $\int_0^{\frac{\pi}{2}}\sin^nx$ in $\xi$:
$$
\xi = \sum_{n=0}^\infty \left[\frac{\pi}{2(2n)!}\prod_{k=1}^n \frac{2k-1}{2k} \right] + \sum_{n=0}^\infty \left[\frac{1}{(2n+1)!}\prod_{k=1}^n \frac{2k}{2k+1} \right]
$$
Merging:
$$
\int_0^{\frac{\pi}{2}} e^{\sin(x)}dx = \sum_{n=0}^\infty \left[\frac{\pi}{2(2n)!}\prod_{k=1}^n \frac{2k-1}{2k} + \frac{1}{(2n+1)!}\prod_{k=1}^n \frac{2k}{2k+1} \right]
$$

Body:
.. and
$$
\eqalign{
  & \prod\limits_{k = 1}^n {{{2k - 1} \over {2k}}}
  = {{\prod\limits_{j = 0}^{n - 1} {2j + 1} } \over {\prod\limits_{j = 0}^{n - 1} {2j + 2} }}
 = {{\prod\limits_{j = 0}^{n - 1} {j + 1/2} } \over {\prod\limits_{j = 0}^{n - 1} {j + 1} }}
 = {{\Gamma \left( {1/2 + n} \right)\Gamma \left( 1 \right)}
 \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} =   \cr
  &  = \left( \matrix{  n - 1/2 \cr   n \cr}  \right) = {1 \over {n\,{\rm B}(1/2,n)}}  \cr
  & \prod\limits_{k = 1}^n {{{2k} \over {2k + 1}}}
  = {{\prod\limits_{j = 0}^{n - 1} {2j + 2} } \over {\prod\limits_{j = 0}^{n - 1} {2j + 3} }}
 = {{\prod\limits_{j = 0}^{n - 1} {j + 1} } \over {\prod\limits_{j = 0}^{n - 1} {j + 3/2} }}
 = {{\Gamma \left( {1 + n} \right)\Gamma \left( {3/2} \right)}
 \over {\Gamma \left( 1 \right)\Gamma \left( {n + 3/2} \right)}} =   \cr
  &  = \left( \matrix{  n + 1/2 \cr   n \cr}  \right) = {n \over {{\rm B}(3/2,n)}} \cr}
$$


答:
Using Weierstrass substitution with $t=\tan\dfrac x2,$

$$1-t^2=2|t|$$

For real $t, t^2=|t|^2$  $$\implies |t|^2-2|t|-1=0$$

$$\implies|t|=1\pm\sqrt2$$

As $|t|\ge0, |t|=\sqrt2+1=\csc\dfrac\pi4+\cot\dfrac\pi4=\cdots=\cot\dfrac\pi8=\tan\left(\dfrac\pi2-\dfrac\pi8\right)$

$$\iff\tan^2\dfrac x2=\tan^2\left(\dfrac\pi2-\dfrac\pi8\right)$$

$$\implies\dfrac x2=n\pi\pm\left(\dfrac\pi2-\dfrac\pi8\right)$$


问:
Unsure of attempt to determine $\dim(W_1+W_2)$ and $\dim(W_1 \cap W_2)$
Let $V=\mathbb{R}^4$. $W_1$ is a subspace of $V$ spanned by vectors $a_1=(1, 2, 0, 1)$ and $a_2=(1,1,1,0)$. $W_2$ is a subspace of $V$ spanned by vectors $b_1=(1,0,1,0)$ and $b_2=(1,3,0,1)$. Determine $\dim(W_1+W_2)$ and $\dim(W_1 \cap W_2)$.

Attempt

The vectors $a_1$ and $a_2$ are linearly independent and span $W_1$, so they form a basis for $W_1$. Hence $\dim(W_1)=2$. The vectors $b_1$ and $b_2$ are linearly independent and span $W_2$, so they form a basis for $W_2$. Hence $\dim(W_2)=2$.

This is the step that I'm unsure of. I added vectors from $W_1$ and $W_2$ together, i.e. $a_1 + b_1, a_1 + b_2, a_2 + b_1, a_2 + b_2$ to form $$W_1+W_2=\left\{(2,2,1,1),(2,5,0,2),(2,1,2,0),(2,4,1,1) \right\}.$$

The first three vectors are linearly independent, but $$(2,4,1,1)= (2,5,0,2)+(2,1,2,0)-(2,2,1,1).$$ Hence there are only $3$ vectors that are linearly independent and span $W_1+W_2$, so $\dim(W_1+W_2)=3$.

We know that $\dim(W_1+W_2) = \dim(W_1) + \dim (W_2) - \dim(W_1 \cap W_2)$. Therefore $\dim(W_1 \cap W_2)=2+2-3=1$.

Question

Is my attempt correct?

Thank you for your time.

答:
Your attempt is correct, but note that your understanding of $W_1 + W_2$ might be wrong. By definition, $W_1 + W_2 = \{\vec{w}_1 + \vec{w}_2 | \vec{w}_i \in W_i, i=1, 2\}.$ However, This doesn't mean that the bases have to be added together. In other words, you can simply take $\{a_1,a_2,b_1,b_2\}$ as a spanning set for the space $W_1 + W_2$ and then toss out any linearly dependent vectors. The easiest way to do this is to row reduce the matrix whose columns are the corresponding basis vectors. For your example, the matrix will row reduce to (why? -- try it!)
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}.$$
This tells you that you have three linearly independent vectors and so the dimension of the $W_1 + W_2$ is 3.

Edit: another related question  that is useful to ask yourself: if $a_1, a_2, a_3, a_4$ are linearly independent, are $a_1 + a_2, a_1 - a_2, a_3 + a_4, a_3 - a_4$? How would you check? (hint: checking independence can always be reduced to row reducing a matrix)


问:
Estimate volume of Tubes around a smooth hypersurface
Suppose $\Sigma \subset \mathbb{R}^n$ is a smooth compact hypersurface and the boundary of some set $\Omega$.  Now set $\Sigma_r:=\{x \in \Omega^c: \inf_{y\in \Sigma} |x-y|\leq r\}$ to be a tube of radius $r$ around the hypersurface. I was wondering if it is possible to somehow estimate the volume of these tubes in terms of the area of $\Sigma$, like for example
$$
vol(\Sigma_r) \leq C_{\Sigma} \cdot r \cdot area(\Sigma).
$$
It should be possible for small $r$ due to this formula. There is of course a trivial bound like if I take some ball $B_R(y)$ such that $\Sigma \in B_R(y)$ then
$
vol(\Sigma_r) \leq vol(B_{R+r}(y)).
$
Is there something better?

答:
First of all, suppose that $\Omega$ is orientable and that $\nu$ is a unit normal vector field on $\Omega$ (in fact, a compact hypersurface in the euclidean spane is forced to be orientable). The map $(p,t)\in \Omega \times \mathbb{R} \to p+t\nu(p)\in\mathbb{R}^n$ is smooth and its differential at a point $(p,0)$ is invertible. By compactness assumption and by the inverse function theorem, there exists $r_0>0$ such that it is a diffeomorphism from $\Omega\times (-r_0,r_0)$ onto its image that is the open tube around $\Omega$ with radius $r_0$.

Thus, we have a chart $f : \Omega \times (-r_0,r_0) \to \mathbb{R}^n$. If $\Sigma_r$ is the tube of radius $r$ around $\Omega$, then
$$
vol(\Sigma_r) = \int_{\Sigma_r} 1 ~\mathrm{d}vol_{eucl} = \int _{f^{-1}(\Sigma_r)}1 \left(f^*\mathrm{d}vol_{eucl}\right) = \int_{\Omega\times (-r,r)} 1
\left(f^*\mathrm{d}vol_{eucl}\right)$$
Thus, it suffices to compute the voume form $\left(f^*\mathrm{d}vol_{eucl}\right)$ to have an estimate on the volume of $\Sigma_r$.

First of all, the volume form $\left(f^*\mathrm{d}vol_{eucl}\right)$ on $\Omega\times (-r_0,r_0)$ is proportionnal to the volume form $\mathrm{d}vol_{\Omega}\wedge \mathrm{d}t$ because the top bundle of a manifold is of rank $1$. Thus, there exists a positive function $\lambda(p,t)$ such that
$$
\left(f^*\mathrm{d}vol_{eucl}\right) = \lambda\cdot \mathrm{d}vol_{\Omega}\wedge \mathrm{d}t
$$
In every good Riemannian geometry book, you can find a proof of the following:
$$
\partial_t \lambda = \lambda \times \mathrm{trace}S
$$
where $S$ is the shape operator of the level hypersurface $\Omega_t = f\left( \Omega\times \{t\} \right)$. We are now reduced to evaluating the trace of $S$. But in every good Riemannian geometry book, there is this well known fact:
$$
\partial_t S = -S \circ S
$$
in the euclidean space (in general, one has to add a curvature term on the RHS): this is the Riccati equation for the shape operator. Taking the trace and using the fact that $\mathrm{trace}(S^2) \geqslant \frac{(\mathrm{trace}S)^2}{n-1}$, it follows that
$$
\partial_t\left(\frac{\mathrm{trace} S}{n-1}\right) + \left(\frac{\mathrm{trace}S}{n-1}\right)^2 \leqslant 0
$$
From this, one can recover some information on $\mathrm{trace}S$, then on $\lambda$ and finally, one can show an estimate on the volume of $\Sigma_r$.

Note: here are listed some good Riemannian geometry books :

Riemannian Geometry, Gallot, Hulin, Lafontaine

Riemannian Geometry, Petersen

Introduction to Riemannian manifolds, Lee

One can find in them references for the above claims in the sections dedicated to extrinsic geometry and/or comparison estimates.

In some editions of Petersen's book, the function $\mathrm{trace}S$ does not appear explicitly; what is used is $\Delta r$, the Laplacian of $r$, where $r = d(\cdot,\Omega)$. But in fact this function is equal to $\mathrm{trace}S$.


问:
Show that $\sigma(\xi)$ is a sigma field.
I have an old exam question that reads:

Let $\xi:\Omega\rightarrow \mathbb{R}$ be an arbitrary mapping. Denote
by $\sigma(\xi)$ the following collection of subsets of
$\Omega:\{\xi^{-1}(B):B\in\mathcal{B}\},$ where $B$ runs through all
Borel subsets of $\mathcal{B}$.

Show that $\sigma(\xi)$ is a $\sigma-$algebra.

Find $\sigma(\xi)$ for a mapping $\xi$ taking just two distinct values.

When does $\sigma(\xi)$ for such $\xi$ coincide with the set $2^\Omega$ for all subsets of $\Omega$?

For the first one, I need to show that the conditions in the definition of a $\sigma-$algebra are fulfilled. The definition in my book (Grimmett and Stirzaker, Probability and random processes, 3d edition, page 3) they give the following definition:

A collection of $\mathcal{F}$ of subsets of $\Omega$ is called a
$\sigma-$algebra if it satisfies the following conditions:

i) $\varnothing\in\mathcal{F}$

ii) if $A_1,A_2...\in\mathcal{F}$ then $\bigcup_{i=1}^\infty A_i\in\mathcal{F}.$

iii) if $A\in\mathcal{F}$ then $A^c\in\mathcal{F}$.

I find these abstract "show that" questions a bit tricky as I've not seen many solutions for these, and those I've seen are are extremely short with not much explanation between the lines. I'd appreciate if anyone could break down these questions.

Attempt to solve 1:

i)

Since $\xi$ is a mapping from $\Omega$ to $\mathbb{R}$ we know that $\Omega=\xi^{-1}(\mathbb{R})\in\sigma(\xi).$ Because of iii) we also know that $\Omega^c=\varnothing\in\sigma(\xi).$

ii)

Let $I$ be a countable set, then we have a family of elements of $\sigma(\xi)$ denoted by $\{\xi^{-1}(B_i)\}_{i\in I}$ and all their unions should be in $\sigma(\xi):$

$$\bigcup\limits_{i\in I}\xi^{-1}(B_i)=\xi^{-1}\left(\bigcup\limits_{i\in I}B_i\right),$$

since $\bigcup_{i\in I}B_i\in\mathcal{B}.$

iii)

For some $B\in\mathcal{B}$ we know that $\xi^{-1}(B)\in\sigma(\xi)$. We have that $(\xi^{-1}(B))^c=\xi^{-1}(B^c)\in\sigma(\xi)$ since $B\in\mathcal{B}$.

答:
Let's break down question 1.

Claim. $\varnothing \in \sigma(\xi)$

We have $\sigma(\xi) = \{\xi^{-1}(B): B \in \mathcal{B}\}$, where I assume $\mathcal{B}$ is some $\sigma$-algebra of subsets of $\Omega$.

As Burb commented, observe $\varnothing \in \mathcal{B}$, so $\xi^{-1}(\varnothing) = \varnothing \in \sigma(\xi)$.

Claim. $\sigma(\xi)$ is closed under countable unions.

Suppose we have $A_1, A_2, \dots \in \sigma(\xi)$. Our goal is to show that $\bigcup_{i=1}^{\infty}A_i \in \sigma(\xi)$.

To show that $\bigcup_{i=1}^{\infty}A_i \in \sigma(\xi)$, we must show that $\bigcup_{i=1}^{\infty}A_i = \xi^{-1}(B)$ for some $B \in \mathcal{B}$.

Going back to our assumptions, let's assess
$\bigcup_{i=1}^{\infty}A_i$. Since $A_1, A_2, \dots \in \sigma(\xi)$, there exist sets $B_1, B_2, \dots \in \mathcal{B}$ such that $\xi^{-1}(B_i) = A_i$ for each $i \geq 1$.

Thus
$$\bigcup_{i=1}^{\infty}\xi^{-1}(B_i) = \bigcup_{i=1}^{\infty}A_i$$
but there is a trick here: one must make use of the fact that
$$\bigcup_{i=1}^{\infty}\xi^{-1}(B_i) = \xi^{-1}\left(\bigcup_{i=1}^{\infty}B_i \right)$$
and since $B_1, B_2, \dots \in \mathcal{B}$, we know that $\bigcup_{i=1}^{\infty}B_i \in \mathcal{B}$. Thus the claim holds.

Claim. $\sigma(\xi)$ is closed under complements.

The idea is similar as above, but you must instead make use of the fact that $$A_i^c = \Omega \setminus A_i = \xi^{-1}(\mathbb{R}) \setminus \xi^{-1}(B_i) =  \xi^{-1}(\mathbb{R}\setminus B_i) = \xi^{-1}(B_i^c)\text{.}$$

I will leave this part to you.


问:
Proving $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
Let $A$, $B$ and $C$ be sets. Prove that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

I am not sure how to formally prove and in special write this intuitive simple question about sets.

Thank you

答: Hint: TO prove two sets are equal, show they are subsets of each other.  So assume something is a member of the left hand side, and show it must be a member of the right hand side.  Then do the same in reverse.</p>


问:
How to calculate a combinatorial sum
I have a combinatorial sum in hand which I suspect equals zero. But I do not know how to prove it. Can you guys help me? (I am even not sure if this is a hard question or not)

Is
$$
\sum_{k=0}^n (-1)^k k^{n-1} \left(\begin{array}{l}
n \\
k
\end{array}\right) = 0
$$
?

Title: How to calculate a combinatorial sum

Body:
We have

$$\sum_{k=0}^n (-1)^k k^{n-1} {n\choose k}
= (n-1)! [z^{n-1}] \sum_{k=0}^n (-1)^k \exp(kz) {n\choose k}
\\ = (n-1)! [z^{n-1}] (1-\exp(z))^n.$$

Now $$1-\exp(z) = -z - z^2/2 - \cdots$$ and hence $(1-\exp(z))^n = (-1)^n z^n + \cdots$ so that

$$(n-1)! [z^{n-1}] (1-\exp(z))^n = 0.$$


问:
Can any function be approximated by periodic functions?
I'm sure there's a certain generic term for this kind of behavior but since I'm a little hard of understanding I don't know what to search for. Generally I want to know exactly what I wrote in the heading.

With periodic functions I mean not only sine like in Fourier-Series but also rectangular functions.

答:
The answer depends on the topology.

Philosophically, in any "nice" topology, a function $f$ can be approximated via convolution with an approximate identity/mollifiers by uniformly continuous functions. Moreover, if $f$ is periodic, then so are the approximants.

Note that if the approximants are uniformly continuous and periodic, they are Bohr/Bochner almost periodic and hence they can be approximated by trigonometric polynomials in $\| \, \|_\infty$.

It follows that if the topology is "nice" and weaker than the topology given by $\| \, \|_\infty$ (which most topologies on functions are), a function can be approximated by periodic functions if and only if it can be approximated by trig polynomials (i.e. almost periodic).

Note that there are many concepts of almost periodicity that are well studied/understood for functions (and even measures): Bohr/Bochner, weak, Stepanov, Weyl, Besicovitch almost periodicity.


问:
Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$

Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$

Given $n\in \mathbb{Z}$ then $n\in \lbrace [0],[1],[2],[3],[4],[5],[6]\rbrace$, of the same way
$n^3\in \lbrace[0],[1],[6] \rbrace$.

Since $7\mid x^3+y^3-z^3$ then $x^3+y^3-z^3\in [0]$ there be two possibilities

$x^3,y^3,z^3\in [0]$ and then $x,y,z\in[0]$ and the result is true.

$x^3\in [6],y^3\in [1],z^3\in[0]$ and then $z\in [0]$ therefore the results follows.

$x^3\in [0], y^3 \in [1], z^3 \in [1]$ and then $x\in [0]$and the afirmation is too true

The other possibilities only switch the order of the variables with the classes.

答:
By taste, I generally prefer solutions based on structural properties given implicitly in  the hypotheses. Here the natural structure is that of the field $\mathbf F_7=\mathbf Z/7\mathbf Z$, more precisely of its multiplicative group $\mathbf F_7^*$. From the properties of finite fields, we know that  $\mathbf F_7^*$ is cyclic of order 6 , hence $\cong  C_2 \times C_3$, where $C_2$ is generated by $-1$ and $C_3$ by $\zeta$, a 3-rd root of unity. In the given cubic equation, we may suppose that no variable is a multiple of 7, otherwise there is nothing to prove. Reduction modulo 7 then gives $a^3 + b^3=1$ in $\mathbf F_7^*$. The previous decomposition of $\mathbf F_7^*$ shows that the components of $a$ are $\pm 1$ and a power of $\zeta$, hence $a=\pm 1$, and the same for $b$. The  final equation mod 7 thus reads $(\pm 1) + (\pm 1) =1$, a contradiction.

This argument may seem circuitous, but I don't think so because: - first it avoids case by case verification, hence there is no risk of a possible omission  - second it appeals only to internal structural properties (e.g. no ordering relation was used), so it could be generalized to other analogous situations.

Addendum : To my surprise, I have just realized that the above argument proves (a weakened version of) Sophie Germain's result on the so called "first case" of Fermat's last theorem: if $p$ and $q= 2p+1$ are both prime (then $p$ is called a Sophie Germain prime), then for all integer solutions of $x^p + y^p = z^p$, $p$ must divide $xyz$ (see e.g. Wikipedia). This certainly cannot be checked case by case.


问:
How to find roots of $x^n -a$ =0
This is a question which i came across while trying some problems in abstract algebra. This question was not asked directly but It is useful while attempting questions.

Consider then equation $x^n -a$ =0. Then How 1 can find the roots of this equation.

What comes to mind is that this equation has n roots by Fundamental Theorem of Algebra and if I had $x^n -1=0$ instead of $x^n-a = 0$ then I could straight away use De Moivre's theorem and roots will be given by
$\cos(\pi /n ) + i \sin(\pi /n)$  and n=0, 1,2,...,n-1.  So, here along the same lines I can assume that $x^n -((a)^n){^{1/n}}$=0 , but there is a difference that $((a)^n )^{1/n}$ =x has n roots and so $x^n -((a)^n)^{1/n}$ =0 .

So, Can you please in detail explanation tell me what are the roots of $x^n -a$=0 which clear reason behind them.

答:
Let $a=r e^{i\phi}$, ehere $r>0$ and $0\le\phi<2\pi$ are real.

Then $$x=r^{\frac1n}e^{i\frac{\phi+2k\pi}n},$$ where $k=0,\dots,n-1$

Title: How to find roots of $x^n -a$ =0


问:
Question from PRMO 2019.
An ant leaves the anthill for its morning exercise. It walks 4 feet east and then makes a 160° turn
to the right and walks 4 more feet. It then makes another 160° turn to the right and walks 4 more
feet. If the ant continues this pattern until it reaches the anthill again, what is the distance in feet it would have travelled?

Well I have the solution but I am not able to understand it.
Let
$A_0 (0,0)$,$A_1 (4\cos0^\circ, 4\sin0^\circ)$, $A_2 (4\cos0^\circ + 4\cos160^\circ, 4\sin 0^\circ + 4\sin160^\circ)$,......, $A_n = (0,0)$ which gives
$$4(\cos 0^\circ + \cos160^\circ +......+\cos(160(n-1))^\circ) = 0$$ and $$4(\sin 0^\circ +\sin 160^\circ +..... +\sin(160(n-1))^\circ) = 0$$ which in turn, gives
$$\sin((80n)^\circ) = 0$$

$$n = 9$$ Then distance covered =$4×9$ =$36\space \text{feet}$.

Please explain how do we get that $\cos 0$ and $\cos 60$ etc. in it.
This question has been asked in PRMO 2019 in India

答:
You need not go into applying the coordinate system and trigonometry here.

Much of PRMO is based on deduction and observations. By observation, you can see that the ant will form a star en route its path to the starting point with each internal angle of the star being $20^{\circ}$. Hence you can calculate the number of sides in the star to be $\frac{180^{\circ}}{20^\circ}=9$,

$\because\space$ The sum of all the internal acute angles of a regular star $=180^{\circ}$, which can be quite easily proven.

Hope this methodology clears your doubt!

Title: Question from PRMO 2019.


问:
To sum $1+2+3+\cdots$ to $-\frac1{12}$
$$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

OwnerDisplayName: perplexed

Title: To sum $1+2+3+\cdots$ to $-\frac1{12}$

Body:
What a great method!  I tried a few...
$$
\sum_{n = 1}^{\infty} \left(\frac{1}{(2 n)^{s}} - \frac{1}{(2 n - 1)^{s}}\right) = \zeta (s) \Bigl(2^{1 - s} - 1\Bigr)
$$
put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{1}{4}$

or
$$
\sum_{n = 1}^{\infty} \left(\frac{1}{(2 n + 1)^{s}} - \frac{1}{(2 n)^{s}}\right) = \left(1 - 2^{1-s}\right)\zeta(s) - 1
$$
put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{3}{4}$


问:
showing that the Euler's number is irrational
Our teacher wants us to do the following:
Suppose that e is rational i. e $e=\frac{a}{b}$ where $a,b\in\mathbb{N}$. Choose $n\in\mathbb{N}$ such that $n>b$ and $n>3$. Use the following inequality

$0<e-(1+\frac{1}{1!}+\frac{1}{2!}+....+\frac{1}{n!})<\frac{3}{(n+1)!}$ and replace e with $\frac{a}{b}$. Then multiply both sides of the inequality with $n!$ to see that the inequality leads to existence of an integer N that satisfies $0<N<\frac{3}{4}$ which leads to a contradiction.

I have tried to to it but I am not getting there...
I choose $n=4$ and $b=3$ and
$0<\frac{a}{b}-1-1-\frac{1}{2}-\frac{1}{6}-\frac{1}{24}<\frac{3}{5!}$

$0<8a-\frac{17}{24}<\frac{3}{5}$

$0<8a<\frac{157}{120}$

$0<a<\frac{157}{960}$.....

How should I do it???

答:
"how can we conclude that an!/b is an integer whenever n>b? Could you elaborate a bit more?"

If n>b then b is in n x (n-1) x (n-2) x ... x (n-n+1)
therefore an!/b = a(n x (n-1) x ... x (n-b+1)) which is an integer.

Sorry about the formating, new to the forum.


答:
Hint:

I see you know how to find roots for $x^n-1=0$. Now you have $x^n-a=0$.

If $a\ge 0$, then you can write
$$
z^n-1=0, \quad z=\frac{x}{\sqrt[n]{a}}.
$$

The case when $a<0$ can be reduced to $x^n+1=0$. Now you can focus on this one. To apply the De Moivre's formula, note that
$$
-1=\cos(\pi)+i\sin(\pi).
$$

Title: How to find roots of $x^n -a$ =0

Body:
This is a question which i came across while trying some problems in abstract algebra. This question was not asked directly but It is useful while attempting questions.

Consider then equation $x^n -a$ =0. Then How 1 can find the roots of this equation.

What comes to mind is that this equation has n roots by Fundamental Theorem of Algebra and if I had $x^n -1=0$ instead of $x^n-a = 0$ then I could straight away use De Moivre's theorem and roots will be given by
$\cos(\pi /n ) + i \sin(\pi /n)$  and n=0, 1,2,...,n-1.  So, here along the same lines I can assume that $x^n -((a)^n){^{1/n}}$=0 , but there is a difference that $((a)^n )^{1/n}$ =x has n roots and so $x^n -((a)^n)^{1/n}$ =0 .

So, Can you please in detail explanation tell me what are the roots of $x^n -a$=0 which clear reason behind them.


问:
Is there an ambiguity on $\sum\limits^{\infty}_{i=0}a_i$?
Sorry if this is basic but I saw the theorem that says if $\sum\limits^{\infty}_{i=0}a_i$ is conditionally convergent then this series can be any number up to rearrangement. Then does it imply that $\sum\limits^{\infty}_{i=0}a_i$ can be any number? How about $\sum\limits^{\infty}_{i=-\infty}a_i$

答:
$\sum\limits_{i=0}^{\infty} a_i$ means one thing.  It means the limit of the sequence $\lim\limits_{n \to \infty} S_n$, where

$$S_n = a_0 + a_1 + a_2 + \cdots + a_n.$$

However,
$$\sum\limits_{i=-\infty}^{\infty}a_i$$

is ambiguous when it is only conditionally convergent.  It is not obvious in what order the terms are meant to be added here, and as you point out adding the terms in a different ways produces different results.


问:
How to find the center of a spiral given only the physical object?
I have a spiral cam from a belt sander tension assembly that needs to have a hole bored in the exact center and I am unsure of how to find it mathematically. I have the object and can measure the diameter at various points but I am stuck at how to get it into a mathematical equation

答:
Playing around with GeoGebra (which I'd suggest you to try) I found this one as my best logarithmic spiral:

and this other as my best archimedean spiral:

The log spiral looks better fitting with the sketch you gave.

Hope this helps.


问:
Find the interval in which $a$ lies such that the roots of the equation $(a+1)x^2 - 3ax + 4a$, is greater than 1.
Question:
Find the interval in which $a$ lies such that the roots of the given function, $f(x)$, is greater than 1.
$$f(x) = (a+1)x^2 - 3ax + 4a$$

My Attempt:

Using the quadratic formula,
$$x = \frac{3a±\sqrt{9(a)^2-16(a)^2-16a}}{2(a+1)}>1$$

First of all, the radicand must be greater than or equal to $0$. From that, we can restrict $a$ to
$$a \in [-\frac{16}{7}, 0]$$

Here, we can make two cases:

Case I:
$a+1>0$

$$±\sqrt{9(a)^2-16(a)^2-16a}>2 - a$$

Now squaring both sides and simplifying it further,

$$-7(a)^2-16a >a^2 + 4 -4a$$
$$2(a)^2+3a+1 <0$$
Hence, $a \in (-1, -\frac{1}{2})$

Intersecting this with the restricted set, we get $$a \in (-1, -\frac{1}{2})$$

Case II:
$a+1 < 0$

Here, we will end up getting,
$$a \in [-\frac{16}{7}, -1)$$

But this is not the required answer. Using desmos, it can be clearly seen that for $a\in(-1, -\frac{1}{2})$ there is only one root greater than $1$ and not both.

I suspect that I messed it up while squaring the radical. By squaring, I was trying to find the interval for $a$ where at least one root has a value greater than $1$. Right?

How am I supposed to get around it?
Also, is there a better approach for this particular problem?

答: Note that solving $\frac{3a}{2(a+1)}<1$ gets you to (-1,2).  In this region the - solution cannot be bigger than 1 since you are subtracting a positive number, so you have to exclude that region


问:
Find generator of principal ideal
The ideal $(9, 2 + 2\sqrt{10})$ of $\mathbb{Z}[\sqrt{10}]$ is a principal ideal; it is generated by $1+\sqrt{10}$.

This is easy enough to check once it's been found, but can anyone tell me some way to arrive at this (or some other) generator? That could be done with pencil & paper?

Ordinarily I would use the Euclidean algorithm, but we don't have that here...

答:
You can do something similar to the Euclidean algorithm, here.

Define the function $f:\Bbb Z\left[\sqrt{10}\right]\to\Bbb Z_{\ge0}$ by $$f(a+b\sqrt{10})=|a^2-10b^2|.$$ You should be able to show that it is a multiplicative function--that is, $f(xy)=f(x)f(y)$--and that $u\in\Bbb Z\left[\sqrt{10}\right]$ is a unit if and only if $f(u)=1.$

You need to find the greatest common factor $x+y\sqrt{10}$ of $9,2+2\sqrt{10}$. Note then that $f(x+y\sqrt{10})$ must be the greatest common factor of $f(9)=81$ and $f(2+2\sqrt{10})=36.$ (Why?) That is, we need $|x^2-10y^2|=9$. In particular, though, note that we have $$36=f(2+2\sqrt{10})=f(2)f(1+\sqrt{10})=4f(1+\sqrt{10}),$$ so $1+\sqrt{10}$ is a candidate, which we can quickly see is in fact a factor of $9$.

Title: Find generator of principal ideal


问:
Is the set that contains the empty set {∅} also a subset of all sets?
I think it isn't; although $\emptyset$ is a subset of every set, $\{\emptyset\}$ is only a subset of a set $A$ if $A$ contains the element $\emptyset$. So if $A = \{\emptyset\}$, then both $\emptyset$ and $\{\emptyset\}$ are subsets of A, but if we have a set $B$ where $B = \{0, 1\}$ then only $\emptyset$ is a subset of it.

If my reasoning is true, then if there was a set $C$ where $C = \{\emptyset, 0, 1\}$, would $\{\emptyset\}$ then be a subset of $C$?

Moreover, is this a valid set (i.e. can you have the empty set as an element of a set that contains other elements)?

答: Not only can you have the empty set as a member of other sets, but the classical way of building the natural numbers out of set theory does nothing but nest empty sets inside other sets.  We do that by defining $0=\emptyset$,  then $1=\{0\}$,   and in general,  $n=\{0,1,2,...,n-1\}$.   Writing these out long form gets you tons of empty sets!


答:
Your reasoning is right. If $A=\{ \emptyset \}$ then both $\emptyset, \{ \emptyset \}$ will be subsets of $A$.

Applying the same reasoning, if $C=\{ \emptyset, 0,1 \}$ as @oliverjones states we can say $\{1\} \subset C$ because we have a $1$ inside $C$ in its definition (the same way there it’s an $\emptyset$ inside $A$ in its definition). With this in mind it’s easy to see that in fact $\{\emptyset\} \subset C$ and it’s a valid set.

Title: Is the set that contains the empty set {∅} also a subset of all sets?

Body:
I think it isn't; although $\emptyset$ is a subset of every set, $\{\emptyset\}$ is only a subset of a set $A$ if $A$ contains the element $\emptyset$. So if $A = \{\emptyset\}$, then both $\emptyset$ and $\{\emptyset\}$ are subsets of A, but if we have a set $B$ where $B = \{0, 1\}$ then only $\emptyset$ is a subset of it.

If my reasoning is true, then if there was a set $C$ where $C = \{\emptyset, 0, 1\}$, would $\{\emptyset\}$ then be a subset of $C$?

Moreover, is this a valid set (i.e. can you have the empty set as an element of a set that contains other elements)?


答:
No, it cannot be any number. If the limit $\lim_{n\to\infty}\sum_{i=0}^na_i$ exists and it is a real number $a$, then (and only then)$$\sum_{i=0}^\infty a_i=a.$$So, since any sequence of real numbers has (at most) one limit, the expression $\sum_{i=0}^\infty a_i$ can be at most one number.

What you were told about rearrangements is that if $\sum_{i=0}^\infty a_i$ is conditionally convergent, and if $a\in\Bbb R$, then there is some bijection $b\colon\Bbb Z^+\longrightarrow\Bbb Z^+$ such that$$\sum_{i=0}^\infty a_{b(i)}=a.$$

Title: Is there an ambiguity on $\sum\limits^{\infty}_{i=0}a_i$?

Body: Sorry if this is basic but I saw the theorem that says if $\sum\limits^{\infty}_{i=0}a_i$ is conditionally convergent then this series can be any number up to rearrangement. Then does it imply that $\sum\limits^{\infty}_{i=0}a_i$ can be any number? How about $\sum\limits^{\infty}_{i=-\infty}a_i$


答: I think, you have confusion between begin  element and begin a subset. Consider, the following example $$A=\{\emptyset, \{\emptyset\}\}$$ Now, $\emptyset\in A$ and $\{\emptyset\}\in A$. But, $\{\{\emptyset\}\}\subset A.$ I hope that helps.


答:
To answer the question in the title, simply use the definition of "subsets".

$A$ is a subset of $B$ if and only if "every element of $A$ is an element of $B$".

So your question is equivalent to

is $\emptyset$ an element of an arbitrary set?

Of course not: the empty set is a subset of any set, but it is not necessarily an element of a set. The set $\{1\}$ is such an example.


问:
How to solve these calculations faster?
$\frac{7}{\sqrt{2}}$ and $\frac{\sqrt{8}}{23}$ and .$\frac{\sqrt{45}}{\sqrt{7}}$ .I am preparing for JEE exam in which we cannot use calculators. So , I want to know how can I solve these questions faster ?. We have to solve each one them i.e by dividing. How can I do that faster

A healthy young man standing at a distance of 7m from a 11.8m high building see a kid slipping from the top floor. With speed (assumed uniform) should he run to catch the kid at the arms height (1.8m).
I got square root of 2 seconds as the time need for young man to go . Hence it is answer for 1st one
a) 4.7m/s
b)4.9m/s
c)7m/s
d) none of these

答:
Take $g=9.8 \frac{m}{s^2}$ , it will make your calculations much easier:

The kid falls $h=10m$ in $t$ seconds:
$$h = \frac12 gt^2 $$
$$10 = \frac12 9.8 t^2$$
$$t^2 = \frac{10}{4.9}$$
$$t = \sqrt{ \frac{10}{4.9}}$$
Now, it helps to see that 10 and 4.9 are just one tenth of 100 and 49, which are perfect squares!
$$t = \sqrt{ \frac{100}{49}} = \frac{10}{7}$$
The rest is easy for you: the man should run $x=7m$ in $t=\frac{10}{7}s$
$$v=\frac{x}{t} = \frac{7}{\frac{10}{7}}=4.9$$

Title: How to solve these calculations faster?


答:
To answer the second question:

Indeed, the issue with conditional convergence is more pronounced when trying to sum $\sum_{i=-\infty}^\infty a_i$, for example $F(M,N)=\sum_{i=-M}^N\frac{1}{2i+1}$ will tend to different limits or not converge depending on which path of $\lim_{(M,N)\to(\infty,\infty)}$ you're taking.

However the ambituity goes away when both directional partial sums' limits $\lim_{N\to\infty}\sum_{i=0}^N a_i$ and $\lim_{M\to\infty}\sum_{i=-M}^0 a_i$ exist, in which case
$$\sum_{i=-\infty}^\infty a_i=\lim_{M\to\infty}\sum_{i=-M}^0 a_i+\lim_{N\to\infty}\sum_{i=1}^N a_i=\lim_{(M,N)\to(\infty,\infty)}\sum_{i=-M}^N a_i$$
will not depend on how $(M,N)\to(\infty,\infty)$.

In some other cases when $a_i\to 0$ as $i\to\pm\infty$, you may see the series $\sum_{i=-\infty}^\infty a_i$ being implicitly summed as $a_0+\sum_{i=1}^\infty \left(a_i+a_{-i}\right)$, even when it's conditionally convergent. For example recall the pole exansion series
$$\pi\cot(\pi x)=\sum_{n=-\infty}^\infty\frac{1}{x+n}:=\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{x+n}$$


问:
Is log the only choice for measuring information?
When we quantify information, we use $I(x)=-\log{P(x)}$, where $P(x)$ is the probability of some event $x$. The explanation I always got, and was satisfied with up until now, is that for two independent events, to find the probability of them both we multiply, and we would intuitively want the information of each event to add together for the total information. So we have $I(x \cdot y) = I(x) + I(y)$. The class of logarithms $k \log(x)$ for some constant $k$ satisfy this identity, and we choose $k=-1$ to make information a positive measure.

But I'm wondering if logarithms are more than just a sensible choice. Are they the only choice? I can't immediately think of another class of functions that satisfy that basic identity. Even in Shannon's original paper on information theory, he doesn't say it's the only choice, he justifies his choice by saying logs fit what we expect and they're easy to work with. Is there more to it?

答:
$\log$ or $\ln$ are definitely not the only way of measuring information - it depends upon what we understand by information. But the way we have chosen to define information (see below) confines us to using $\log$ or $\ln$.

I've tried to explain it here on stats.stackexchange.com. I'm pasting it below for quick reference.

There is a profound reason why the logarithm comes into picture, and it is not randomly chosen. The relationship between $\log$ and information stems from this simple way of writing any number $m$ (the symbols don't have any meaning yet), and the discussion that follows.

$$ m = \frac{1}{p} = 2^{i} \tag{1}$$

The above tells us that if we use exactly $i$ letters to encode a string where each letter can have one of 2 values at a time, we'll get $m$ different strings. A 2-valued letter is nothing else but a bit. So writing any number $m$ in this way brings into picture a property of the number - $i$ which can be used to construct the number $m$ again (uniquely) - using bits.

Now, it is easy to see that for a given outcome that has a probability $p$, the number of other outcomes in the same event that have probabilities greater than $p$ will always be less than or equal to $\frac{1}{p}$. For detail on this, check here.

This means that, as per $(1), $ $i=\log_2(\frac{1}{p})$ bits can be safely used to represent this outcome in an event unless there are lower probability outcomes. But even if there are lower probability outcomes, it is easy to see that we can still encode this outcome with $i=\log_2(\frac{1}{p})$ bits, and use more bits to encode the lower-$p$ outcomes. Check here for a detailed proof. In summary, $i=\log_2(\frac{1}{p})$ bits can be safely used to represent this outcome in any event.

Now, the information about an outcome that goes from the sender to receiver is actually the codeword that represents the outcome. And we just saw how the length of the codeword is determined by $\log_2(\frac{1}{p})$. So, we choose to call this special length $i$ - the information of the event, and that's how $\log$ comes into picture naturally.

$ p=0.25 \Rightarrow i = 2 $ means that we need $ 2 $ bits for encoding this outcome in any event.

$ p=0.125 \Rightarrow i = 3 $ means that we need ( 3 ) bits for encoding this outcome in any event.

Finally, what would be the information content of any event in total, that is, the information of all the outcomes combined? In other words, what is the information content of a system that can have different states with different probabilities? The answer is that each outcome or state adds its information to the system but only in the ratio of how much of it is there - i.e. its probability. This is just verbiage for the Entropy equation:
$$\begin{align}
H = & \sum_i{p_i.i} \\[6pt]
= & - \sum_i{p_i \log_2({p_i}})
\end{align}$$

The above has been explained in more detail here.


问:
Computing $\int_{-1}^1(1-x^2)^n\ dx$
I'd like to calculate $\int_{-1}^1(1-x^2)^n\ dx$, but I have no idea how. After searching through this website, I found out that we can transform the integral into Beta function, by using substitution. However, I'm having trouble finding the right substitution. I tried using $x = 1 - 2t$. Then the integral becomes
$$-\frac{4^n}{2}\int_{3}^{-1}t^n(1-t)^n\ dt.$$
What's inside the integral is similar to Beta function. But the boundaries are not. It should have been
$$\int_0^1t^n(1-t)^n\ dt.$$
Any idea what substitution should I choose?

答: Your substitution leads to the correct bounds. It seems like you used $x = 1 - 2t$ and substituted $t = -1, 1$ when it should have been the other way around: Use $t = \frac{1}{2}(1-x)$ and substitute $x = -1, 1$.

Title: Computing $\int_{-1}^1(1-x^2)^n\ dx$


问:
Intuition behind Quadratic Order of Convergence
I am trying to understand Order of Convergence and specifically quadratic, cubic and so on.

The definition I came across was : $$\lim_{n \to \infty }  \frac{|x_{n+1}-L|}{|x_n -L|^q}=M$$

From there two cases where defined were :

$q=1$ and $M \in [0,1]$

$q\geq 2$ and $ 0< M < \infty$

I understand case $(1)$ as by dividing the current error with the previous error you can see how they compare.

However I am struggling to understand case $(2)$ and I got some questions :

Why do we need to square or even cube the previous error?

Why are there no limitations for $M$ in case $(2)$? Would a sequence still converge if $M$ was really large number?

What would happen if say in Newton's method where convergence is quadratic we compared the errors like in case $(1)$?

答:
Why are there no limitations for  in case (2)? Would a sequence still converge if  was really large number?

The convergence rate only makes sense if the sequence converges... When $q=1$, if $M\ge 1$ there would be no convergence.

Why do we need to square or even cube the previous error?

The convergence rate $q$ is a measure of how fast $x_n$ is converging to $L$. Imagine that $M=1$ and $|x_n-L|\leq 10^{-2}$. If $q=2$, we would have $|x_{n+1}-L| \leq 10^{-4}$ while if $q=3$ we would have $|x_{n+1}-L| \leq 10^{-6}$.

We are not "squaring" or "cubing", we are just trying to see if the inequality holds for $q=2$ or $q=3$.


问:
Markov chain where you can either hop right or stay
Can someone help me with this question? I have searched everywhere but cannot find how to answer this.

Consider a random walk in discrete time, $t=0,1,2, \ldots,$ and on a set of discrete sites $i=0,1,2, \ldots$ The walker starts from $i=0$. In each time step, the walker hops to the next site $(i \rightarrow i+1)$ with probability $q ;$ with probability $1-q$ the walker stays at its current location. The walker cannot move backward.
State the probability that the walker is at site $i$ at time $t .$ What is the expected position $\langle i\rangle$ at time $t ?$ (A detailed calculation is not required.)

答: Think of it like flipping a (biased) coin $t$ times. Suppose you are interested in $t=2$. If you flip two coins, then you could either get HH, HT, TH, TT. Heads heads corresponds to going forward by $2$, HT and TH by $1$, etc. The probabilities of each are $P(i =2) = q^2$, $P(i = 1) = 2 q(1-q)$ (two ways to get 1 head), and $P(i = 0) = (1-q)^2$. Then extend this argument to more flips - you may recall the binomial distribution.


问:
Number of grid points outside a rook circuit
When I draw a circuit on a 8x8 chess board that passes through all squares, no matter how many turns I draw, I always end up with 18 grid points outside my circuit (and 31 grid point inside). It looks to me that I get also constant values for any $n$ x $m$ board. Has anybody ever seen a formula of how many grid points are inside/outside for an $n$ x $n$ board? It looks obvious that there are (n/2 - 1) * (n - 2) points outside if $n$ is even, but I am searching for a proof.

If you could point me to some literature, that would be great!

答:
The main idea here is to create a bijection from the enclosed grid points, to the area of the polygon, to the angle sum on the polygon. We will set this up in reverse.

Because the angle sum is a constant, hence the number of enclosed grid points is a constant, which we can then determine. From there, the number of exterior grid points is a constant.

Let the square be of size $ 2k \times 2k$. The question is for $k = 4$.

(It has to be of even size in order to form a circuit. The standard coloring argument applies.)

Show that

Let the circuit connect through the mid points of the grid squares. This gives us a polygon whose (interior) corner angles are $90^\circ$ or $ 270^\circ$.

For each grid square, there are exactly 2 lines that connect to the mid point. This splits the square into 2, one which is on the inside of the curve, and one which is on the outside of the curve. (This should be obvious, else appeal to high-power Jordan curve theorem.)

For each mid point of a grid square, associate it to the angle that lies on the inside of the curve. This angle is either $ 90 ^ \circ, 270^\circ$ (if the mid point is also a corner of the polygon, then it's equal to the interior angle), or $180^\circ$ (if the mid point lies within an edge of the polygon).

For each mid point of a grid square, the associated angle is also equal to "$180 ^ \circ -$ exterior angle of the polygon". Note that I'm taking  "exterior angle = $180^\circ - $interior angle" as a signed angle, which could be negative. (Verify this for mid points that are on the edges and corners of the polygon.).

There are $4k^2$ such mid points of the grid squares.

The sum of all associated angles is thus

$\sum_{\text{mid point}}(180 ^ \circ -$ exterior angle) = $180^\circ \times 4k^2 - \sum $ exterior angles$ = 180^\circ \times 4k^2 - 360^\circ$   . (Using that summation of signed exterior angles for a non-self intersecting polygon is $360^\circ$.)

Show that there is a bijection between each $90^\circ$ angle and the (obvious) $\frac{1}{2} \times \frac{1}{2}$ square that is contained within the curve. (There's some work here to establish the seemingly obvious bijection.)

Hence, the area enclosed by the curve is $ \frac{1}{4} \frac{1}{90^\circ} (180^\circ \times 4k^2 - 360^\circ) = 2k^2 - 1 $.

Show that there is a bijection between an enclosed grid point and the $1 \times 1$ square around it (which intersects 4 grid squares) that lies within the polygon. (There's some work here to establish the seemingly obvious bijection.)

Each enclosed grid point contributes exactly 1 to the area, hence there are $ 2k^2  -1 $ enclosed grid points.

Since there are $(2k-1)^2$ grid points, there are $ (2k-1)^2 - (2k^2 -1) =  2(k-1)^2$ excluded grid points.

For the general $ n \times m$ grid where $nm$ is even, you can follow the above proof to conclude that

area of enclosed polygon is $nm/2-1$, hence so is the number of enclosed grid points

$(n-1)(m-1)$ grid points,

$(n-1)(m-1) - ( nm/2 - 1) = \frac{1}{2} (n-2)(m-2)$ excluded grid points.


问:
Finding optimal x given symmetric matrix A
Trying to find $$\text{min} \{x | xI-A\succeq 0\}$$ where a $A$ is a symmetric matrix.

This seems pretty simple but I'm not quite sure where to start. I was taking the derivative of $xI-A$ but that doesn't seem to help.

答:
Let $A = U \Lambda U^{\mathsf{T}}$ be the spectral decomposition of $A$. Then we can rewrite:
$$
x I - A = x UU^{\mathsf{T}} - UDU^{\mathsf{T}} = U \left(x I - D \right) U^{\mathsf{T}}
$$

Because $U \in \mathbf{R}^{n \times n}$ is an orthogonal matrix, requiring $U(xI - D) U^{\mathsf{T}} \succeq 0$ is equivalent to requiring $xI - D \succeq 0$. The smallest $x$ that satisfies this is equal to $\max_{i} D_{ii} = \lambda_{\max}(A)$.


问:
What is a more rigorous method of solving modulus function inequalities?
I have the inequality

$$\vert x-3\vert < 3x-4$$

To solve this I first break the modulus containing function into the two cases using the critical value

$$\vert x-3 \vert  \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.  $$

For the case where $x>3$:

$$x-3<3x-4$$
$$x>\frac{1}{2}$$

For the case where $x<3$:

$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.

The value does however seem to hold for the function $f(x)=x-3$

My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?

答:
There are no false answers.

The case you have taken is $x > 3$. So you cannot take $x > \frac12$ without taking $x > 3$. You have to take the intersection of both the inequalities so it will be $x > 3$ for the first case and $3 > x > \frac74$ for the second case.

This is the most rigorous method.


问:
Proof of Alternating Binomial Coefficient Identity relating
I am looking for a proof of the following binomial identity. I encountered it in an article on Euler’s derivation of the gamma function. Euler begins by evaluating the integral:

$$\int_0^1 x^a(1-x)^n\,dx$$

He performs a binomial expansion on the integrand and makes use of the following identity involving alternating binomial coefficients:

$$\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{1}{a+k+1} = n! \prod_{k=0}^{n}\left( \frac{1}{a+k+1} \right).$$

It is not obvious to me on inspection that this formula is true, but I have found it cited elsewhere in the mathematical literature, without proof. I have verified it by hand for the cases of n=2 and n=3, but I am unable to prove the identity in general. I am looking for a good combinatorial or inductive proof of this very interesting fact.

I am aware that this formula relates to the beta function, but I am trying to derive this relationship without reference to this topic.

Thank you for enlightening me.

答:
In trying to evaluate

$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{r+a+1}$$

where $a$ is not in $\{-1, -2, \ldots, -n-1\}$
we introduce

$$f(z) = \frac{(-1)^n \times n!}{z+a+1}
\prod_{q=0}^n \frac{1}{z-q}$$

which has the property that for $0\le r\le n$

$$\mathrm{Res}_{z=r} f(z) =
\frac{(-1)^n \times n!}{r+a+1}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = \frac{(-1)^n \times n!}{r+a+1}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= (-1)^r {n\choose r} \frac{1}{r+a+1}.$$

It follows that

$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$

Now residues sum to zero and the residue at infinity of $f(z)$ is zero
by inspection, therefore

$$S_n = - \mathrm{Res}_{z=-1-a} f(z) =
(-1)^{n+1} \times n! \times
\prod_{q=0}^n \frac{1}{-1-a-q}
\\ = n! \prod_{q=0}^n \frac{1}{a+q+1}.$$

This is the claim.


问:
Continued radical of powers of 4 equals 3
Can someone explain to me why
$$\sqrt{4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \dots}}}} = 3???$$
I need an answer

答:
Let $f_0(x)=0$ and recursively $$ f_{n+1}(x)=\sqrt{x+f_n(4x)}.$$
Next, let $$ f(x)=\lim_{n\to\infty}f_n(x).$$
Our goal is to find $f(4)$.

Claim 1. For $x\ge 0$ and $n\in\Bbb N$,
$$\tag1 f_{n}(x)\le f_{n+1}(x).$$
Proof.
This is trivial for $n=0$. Assume $(1)$ holds for $n$ and all $x\ge0$.
Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le \sqrt{x+f_{n+1}(4x)} =f_{n+2}(x)$$
for all $x\ge 0$. Hence the claim follows by induction. $\square$

Claim 2. For all $n\in\Bbb N_0$ and all $x\ge 0$,
$$\tag2 f_n(x)\le\sqrt x+1.$$

Proof. The claim is certainly true for $n=0$. Assume $(2)$ holds for all $x\ge0$ and for some $n\in\Bbb N$. Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le\sqrt{x+2\sqrt{x}+1} =\sqrt{(\sqrt x+1)^2}=\sqrt x+1,
$$
so $(2)$ holds for all $x\ge0$ also for $n+1$. By induction, the claim follows. $\square$

Combining claims 1 and 2, we see that $f_n(x)$ converges point-wise (so $f(x)$ is in fact defined for all $x\ge0$) and that
$$\tag3 \sqrt x\le f(x)\le \sqrt x+1. $$
In particular, $f(4)\le 3.$
We also note that
$$f(x)=\sqrt{x+f(4x)}. $$

Let $g(x)=\frac{f(x)}{\sqrt x+1}\in[0,1]$.
Then
$$ g(x)=\frac{\sqrt{x+g(4x)(\sqrt{4x}+1)}}{\sqrt x+1}=\sqrt{\frac{x+g(4x)(2\sqrt x+1)}{x+2\sqrt x+1}}$$
i.e.,
$ g(x)^2$ is a convex combination of $g(4x)$ and $1$ and hence is not further from $1$ than $g(4x)$ is. Then also $|g(x)-1|\le |g(4x)-1|$. By $(3)$, $\lim_{x\to\infty}g(x)=1$, so that we must have $g(x)=1$ for all $x>0$. We conclude

$$ f(x)=\begin{cases}0&x=0\\\sqrt x+1&x>0\end{cases}$$


问:
Describing a distance vector in space
I have two points in cartesian coordinates $P_1 = (1,0,2)$ and $P_2 =(0,2,1)$ and need the distance vector from $P_1$ to point $P_2$. The vector from origin to $P_1$ is defined as $\textbf{R}$ while the vector from origin to point $P_2$ is defined as $\textbf{R'}$. The vector I am looking for is then $\textbf{R-R'}$ between the first and the second one in space. I have forgotten linear algebra a bit so I am not sure of the proper way of doing it. Could you please help me out?

Is this right?

$$ \textbf{R} = (1,0,2) \\ \textbf{R'} = (0,2,1) \\ \textbf{R-R'} = (1,0,2) - (0,2,1) = (1,-2,1) $$

答: What you wrote is the opposite: the vector from $P_2$ to $P_1$. The vector from $P_1$ to $P_2$ will be $$\mathbf{R’} -\mathbf R =(-1,2,-1)$$

Title: Describing a distance vector in space


问:
(Wikipedia Bug?) Equilibrium in natural ordering of Auction prices
According to wikipedia:

Natural ordering

Order the buyers in decreasing order of their bid: b1≥b2≥...≥bn.

Order the sellers in increasing order of their bid: s1≤s2≤...≤sn.

Let k be the largest index such that bk≥sk (the "breakeven index"). Every price in the range [max(sk,bk+1),min(bk,sk+1)] is an equilibrium
price, since both demand and supply are k. It is easier to see this by
considering the range of equilibrium prices in each of the 4 possible
cases (note that by definition of k, bk+1 < sk+1):

I've found a counterexample where the price range formula doesn't work, and I wanted to check if I'm on the right track.

Let:

```<code>Asks = [2.1, 2.2, 2.3, 2.4, 2.5]
Bids = [2.4, 2.3, 2.2, 2.1, 1.0]
              ^
</code>```

In this case, k = 2 (indexing from 1).

According to the price range formula, the equilibrium would lie between [2.2, 2.3]. But if it was 2.2, then 2 sellers would accept it and*3* buyers would accept it. This isn't a state of equilibrium since the number of accepted buyers and sellers are unequal.

答:
Strictly between $2.2$ and $2.3$ you are in equilibrium at a quantity of $2$.  If the price exactly matches a bid or ask that person doesn't care if the transaction happens or not.  This makes the quantity unclear at that point and equilibrium is within the range of possibility.  If someone does a careful presentation they should make clear what happens when the transaction price exactly equals a bid or ask.  The Wikipedia article does say that the transaction will happen if the price matches the bid or ask but it is not clear that the analysis really reflects that as your example shows.  I think using the indifference of buyers/sellers will cover that, but have not worked it carefully.

The proof that this works is easy if you assume all the prices are distinct.  Say you have $m$ bids and $n$ asks.  At a price of $0$ you have demand of $m$ and supply of $0$.  Define the imbalance as the demand minus the supply.  At a price of $0$ the imbalance is $m$.  At a very high price you have demand of $0$ and supply of $n$ for an imbalance of $-n$.  As you raise the price from $0$ the imbalance is reduced by $1$ for each price, bid or ask, that you cross.  Somewhere you have to go through $0$ and the equilibrium will hold for an interval between two prices.  If some prices are exactly the same, the imbalance can change more than $1$ and you might miss $0$.  It is reasonable to say that if the price is exactly at a bid or ask the person doesn't care if they transact or not.  In that case you can have prices that match and the equilibrium can be found at one of the prices.  Some of the people who have quoted that price will transact and some will not.


问:
Finding Sources or Citations for Theorems
I just started writing math papers, and I'm finding I often use existing theorems or statements in mathematics without knowing who to cite. Often times the theorems I'm using don't have a clear source, and figuring out who developed them is quite challenging. Is there something I'm missing? Is there some trick or streamlined method of getting this information that I'm unaware of?</p>

答:
Many (most) results aren't named, so often you can just point to any one source that has them - maybe a text that you learned it from or another paper you've read. If it's used but not proven in another paper, then hopefully those authors will point towards an original source. If you don't know where it comes from, try looking in papers that proved similar results to yours, or used a similar technique.

Also, sometimes a result is considered sufficiently trivial in the discipline/sub-discipline that its basically taken for granted and authors assume everyone knows. I have found it difficult to tell whether I should prove something by hand or take it for granted, especially in areas which are not my primary interest. In that case I tend to err on the side of caution try to find where the result has been used, and failing that include a brief proof (often in an appendix).


问:
Prove that if $\varphi$ is injective then $G \cong \varphi(G)$
Let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism  and  the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
Prove that if $\varphi$ is injective then  $G   \cong \varphi(G)$

My attempt:

If $\varphi$ is injective, then for every $a \in \varphi(G)$,there exist $b \in G$ such that $\varphi(b)=a$.

After that I'm not able to proceed any further.

答:
Hint:

 $$G/\ker(\phi)\cong\phi(G)$$

Body:
Hint:

 $$G/\ker(\phi)\cong\phi(G)$$


答: Use the first isomorphism theorem and the fact that $\varphi$ is injective iff $\ker\varphi$ is trivial.

Body:
Let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism  and  the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
Prove that if $\varphi$ is injective then  $G   \cong \varphi(G)$

My attempt:

If $\varphi$ is injective, then for every $a \in \varphi(G)$,there exist $b \in G$ such that $\varphi(b)=a$.

After that I'm not able to proceed any further.


问:
Is "lines remain lines" sufficient to characterise linearity in $\mathbb R^2$?
Note: There are other questions on this site similar to this one, but they are either unanswered or have a different definition of "lines  remain lines".

3Blue1Brown uses the following intuitive definition of linearity in his video series on linear algebra:

A linear transformation is a transformation which fixes the origin and keeps lines straight.

I mainly care about this as an intuition for the plane, so I do not really want to generalise it to $\mathbb R^n$. So my question is, if $f\colon\mathbb R^2\to\mathbb R^2$ is a map such that

$f(\boldsymbol 0) = \boldsymbol 0$, and

for all $\boldsymbol a,\boldsymbol b\in\mathbb R^2$, there exist $\boldsymbol u, \boldsymbol v\in\mathbb R^2$ such that $f(\boldsymbol a+\mathbb R\boldsymbol b) = \boldsymbol u+\mathbb R\boldsymbol v$,

can I show that $f$ is a linear map on $\mathbb R^2$?
(here $\boldsymbol a+\mathbb R\boldsymbol b$ denotes the obvious coset $\{\boldsymbol a+t\boldsymbol b:t\in\mathbb R\}$).

I've been playing around with these two properties a lot, but 2 doesn't seem to be strong enough to allow me to go from statements about lines to statements about individual vectors in an obvious way.

答:
Let $k:\def\R{\mathbb R}\R\rightarrow\R$ be any non-linear bijection such that $k(0)=0$ (for instance, $k(x)=x^3$, or even something non-continuous). Define $f:\R^2\rightarrow\R^2$ by $a\boldsymbol x+b\boldsymbol y\mapsto k(a)\boldsymbol x$. Then every line (except vertical ones, which is mapped to a point) is mapped to the $x$-axis, but this map is obviously not linear.

UPDATE: If we assume bijectivity in addition (and do not assume $f(\boldsymbol0)=0$), the conditions give rise to a collineation, which I found in this linked question thanks to pregunton. According to Wikipedia, in the case of $\R^n(n\geq2)$:

In general, some collineations are not homographies, but the fundamental theorem of projective geometry asserts that is not so in the case of real projective spaces of dimension at least two.

So we arrive at the notion of a homography.


问:
A question regarding etale morphisms of affine varieties
Suppose that $p:Y\rightarrow X$ is a surjective etale morphism of affine varieties (both over an algebraically closed field of characteristic zero), and let $x \in X$.

Then is there a subvariety $V$ of $Y$ containing $y \in p^{-1}(x)$, such that $V$ isomorphic to $X$?

答:
Let $C \subset \mathbb{A}^2$ be the elliptic curve $\{ y^2 = x^3 - x \}$, let $Y = C \setminus \{ (-1, 0), (0, 0), (1, 0) \}$, and let $X = \mathbb{A}^1 \setminus \{ -1, 0, 1 \}$. It is not hard to see that $X$ and $Y$ are affine varieties: in fact, $X$ corresponds to the $k$-algebra $A = k [t, (t-1)^{-1}, t^{-1}, (t+1)^{-1}]$, and $Y$ corresponds to the $k$-algebra $B = k [x, y, y^{-1}] / (y^2 - x^3 + x)$. Consider the morphism $p : Y \to X$ defined by $(x, y) \mapsto x$. This corresponds to the unique $k$-algebra homomorphism $p^* : A \to B$ where:
\begin{align}
p^* (t) & = x &
p^* ((t-1)^{-1}) & = \frac{x^2 + x}{y^2} \\
p^* (t^{-1}) & = \frac{x^2 - 1}{y^2} &
p^* ((t+1)^{-1}) & = \frac{x^2 - x}{y^2}
\end{align}
It is also easy to check that $p : Y \to X$ is surjective, so $p^* : A \to B$ is injective. By considering what is happening in the projective closure, we see that $p : Y \to X$ is a finite étale morphism. But there is no open subset of $Y$ that is isomorphic to $X$, because an elliptic curve has genus 1 and $\mathbb{P}^1$ has genus 0, and the genus of a smooth projective curve can be computed from any non-empty open subset (via the function field).</p>


问:
Dimensionality theorem irreps finite dimensional algebra
For representations of finite groups it is known that the regular representation of a finite group is the direct sum of irreps, each with dimension equal to the multiplicity with which they appear. I assume the same does not hold for the algebra case since we can't really talk about conjugacy classes and the left action of an algebra on itself does not necessarily permute the basis elements of the algebra.

I do wonder whether there exists some (probably weaker) statement about the decomposition of the regular representation of irreps of a finite dimensional complex algebra?

答:
If $A$ is any ring (and in particular any finite-dimensional complex algebra) then the regular representation of $A$ on itself by left multiplication has submodules consisting of exactly the left ideals, more or less by definition. It is completely reducible iff $A$ is semisimple, again more or less by definition, and for $A$ finite-dimensional over a field (or more generally for $A$ Artinian) this occurs iff $A$ has vanishing Jacobson radical.

If $A$ is Artinian but does not have vanishing Jacobson radical then $A/J(A)$ is semisimple and is the maximal quotient with this property, and by the Artin-Wedderburn theorem it is a finite direct product of matrix rings $M_n(D)$ over division rings. In particular if $A$ is finite-dimensional over a field $k$ then $A/J(A)$ is a finite direct product of matrix rings over division $k$-algebras, so if $k$ is algebraically closed $A/J(A)$ is a finite direct product of matrix rings $M_n(k)$ over $k$.

In the case of semisimple algebras over an algebraically closed field $k$, each $d$-dimensional simple module $S$ corresponds to a matrix ring factor $M_d(k) = \text{End}_k(S)$ which, in terms of the left action of $A$ on itself consists of $d$ copies of $S$. A conceptual way to see this is that when $A$ is a semisimple module then the multiplicity of $S$ in $A$ is $\dim_k \text{Hom}_A(A, S)$, and we have $\text{Hom}_A(A, S) \cong S$ since $A$ is the free $A$-module. More generally, if $A$ is a semisimple ring and $S$ a simple module with endomorphism ring $\text{End}_A(S) \cong D$, corresponding to a matrix ring factor $M_d(D)$ in $A$, then the multiplicity of $S$ in $A$ is

$$\dim_D \text{Hom}_A(A, S) = \dim_D S = d.$$

So $S$ still occurs with multiplicity equal to its dimension but its dimension needs to be understood as a module over its endomorphism ring $D$ (which is a division ring by Schur's lemma).


问:
For a symmetric matrix, are eigenvectors of an eigenvalue with a multiplicity $> 1$ orthogonal to each other?
I have learned so far that the eigenvectors to distinct eigenvalues are orthogonal to each other, given a symmetric matrix.

Does the same apply for eigenvectors of an eigenvalue with multiplicity > 1?

Let's suppose I have an eigenvalue with a multiplicity of 2 and found 2 eigenvectors. Are those orthogonal given a symmetric matrix?

答:
No.

If $\lambda$ is a $k$-fold eigenvalue, then the eigenvectors with eigenvalue $\lambda$ form a $k$-dimensional subspace. Of course, not every pair of independent vectors in such a subspace is orthogonal.

Then again, you can pick an orthogonal basis of said eigenspace, thus obtaining an orthogonal basis consisting only of eigenvectors for the whole space.


问:
System of linear equations and column space of matrix:
If I have a system of equations:
$$a_1x+b_1y+c_1z = d_1$$
$$a_2x+b_2y+c_2z = d_2$$
$$a_3x+b_3y+c_3z = d_3$$
where the coefficients $a_i, b_i, c_i$ and constants $d_i$ are real, then I know that a zero determinant on the coefficient matrix tells us that we have an inconsistent system, and there are either infinitely many solutions or zero solutions. I read that if the column vector $(d_1,d_2,d_3)^T$ lies outside of the "column space" of our coefficient matrix then we have zero solutions (makes sense as the matrix cannot map to this vector). My question is, e.g., for the matrix
$$\begin{bmatrix}5 & 2&3\\1 & 2&1\\3&2&2\end{bmatrix}$$
how can we check if it maps to e.g. $$\begin{bmatrix}4\\2\\1\end{bmatrix}$$
using this column space property? (i.e. what are the actual steps involved?)
Also, can anyone clarify what exactly is meant by the column space?

答:
You seem to already understand what you are asking.

The column space of a matrix is just the set of all linear combinations of the columns of $A$.  It's the subspace spanned by those columns.

So, if $d^T$ is contained in this subspace, then the system is consistent for that specific $d^T$, otherwise it is not.

All of this is just another way of asking if row reducing the augmented matrix $[a^T b^T c^T d^T]$ is consistent.  It sounds like maybe you have not got to subspaces yet, but they are not that difficult of a concept if you already understand spans.


问:
Understanding the second condition in Fritz John's Theorem

Fritz John's Theorem: Each convex body $K$ contains a unique ellipsoid of maximal volume. This ellipsoid is $B^n_2$ iff: $B^n_2 \subset K$ and (for some $m$), there are Euclidean unit vectors $(u_i)_1^m$ on the boundary of $K$ and positive numbers $(c_i)_1^m$ satisfying
\begin{equation}
    \sum_{i=1}^m c_i u_i = 0
\end{equation}
and
\begin{equation}
    \sum_{i=1}^m c_i \langle x,u_i\rangle^2 = \|x\|^2 \text{ for each }x\in\mathbb{R}^n
\end{equation}

This is an excerpt from Keith Ball's notes. He goes on to say that the second condition (involving $\|x\|^2$) is equivalent to $$x = \sum c_i\langle x,u_i\rangle u_i$$ for all $x\in\mathbb{R}^n$.
This is easily seen so I've skipped the proof.

"...The $(u_i)$ behave rather like an orthonormal basis in that we can resolve the Euclidean norm as a weighted sum of squares of inner products..."

I think this is because for two vectors $x,y$ we can write $$\langle x,y\rangle = \sum_i c_i\langle x,u_i\rangle \langle y,u_i\rangle$$

"This guarantees that the $(u_i)$ do not all lie close to a proper subspace of $\mathbb{R}^n$. If they did, we could shrink the ball a little in this subspace and expand it in an orthogonal direction, to obtain a larger ellipsoid inside $K$."

What does it mean to lie close to a proper subspace of $\mathbb{R}^n$?

Thanks for reading, I would appreciate any help! I hope to better understand this theorem and its consequences.

Edit:

Later in the notes, the author says to interpret the second condition as a rigidity condition:

One may get a bit more of a feel for the second condition in John’s Theorem by interpreting it as a rigidity condition. A sequence of unit vectors $(u_i)$ satisfying the condition (for some sequence $(c_i)$ has the property that if $T$ is a linear map of determinant $1$, not all the images $Tu_i$ can have Euclidean norm less than $1$.

How do we prove this? I am unable to come up with a proof by contradiction (as follows).
Assume $\|Tu_i\| < 1$ for all $1\le i\le m$. Now I'm trying to put $x = u_i$ in the second condition but haven't gotten anything fruitful yet. Thanks!

Update:
I tried putting $x = Tu_j$ in condition $2$ and got something redundant. If you're interested, you may read ahead:
$$\|Tu_j\|^2 = \sum_i c_i \langle Tu_j,u_i\rangle^2 \le \sum_i c_i \|Tu_j\|^2 \implies \|Tu_j\|^2 (1 - \sum_i c_i) \le 0$$ which is nothing new since we already have $\sum_i c_i = n$. Nothing special about $Tu_j$ here, we would have gotten this from any $x$.

Update 2:
I'm trying to use Hadamard's inequality but haven't gotten anywhere yet. The bounty is about to end and I hope someone gives it a shot?

答:
we can resolve the Euclidean norm as a weighted sum of squares of inner products

I think this is literally the second condition in the theorem: $c_i$ are the weights, the $\langle x, u_i\rangle^2$ terms are the squares of inner products. This is in analogy to the usual Euclidean norm $\|x\|^2 = \sum_i \langle x, e_i \rangle^2$ for some orthonormal basis $\{e_i\}$.

$(u_i)$  do not all lie close to a proper subspace of $\mathbb{R}^n$.

The text points to Figure 14, which has a picture of an ellipsoid (solid circle) in $\mathbb{R}^2$ whose contact points (the $u_i$) lie close to a vertical line (a one-dimensional proper subspace of $\mathbb{R}^2$). By "squashing" this ellipsoid vertically (along the subspace) and elongating it horizontally (the orthogonal direction) we get a larger ellipsoid (dashed ellipsoid in the figure).


问:
Finding the following definite and indefinite integrals
I want to calculate the integral
$$\int_0^{\frac{\pi}{2}} e ^{ \sin t}\, dt.$$
Can we find a primitive function for $f(t) = e ^{\sin t}$?</p>

答:
We recall the following Jacobi-Anger expansion (DLMF 10.35.3):

$$e^{z\sin t}=I_0(z) + 2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\sin((2k+1)t)+2\sum_{k=1}^\infty (-1)^k I_{2k}(z)\cos(2kt)$$

Integrating both sides from $t=0$ to $\pi/2$:

\begin{align}
\int_0^{\pi/2}e^{z\sin t}\,dt
&= \frac{\pi}{2}I_0(z)+2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\frac{1+\sin k\pi}{2k+1}+2\sum_{k=1}^\infty (-1)^k I_{2k}(z)\frac{\sin \pi k}{2k}\\
&= \frac{\pi}{2}I_0(z)+2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\frac{1}{2k+1}
\end{align}

There's a few ways we could deal with this infinite sum. First, we can expand the modified Bessel functions in powers of $z$ and confirm agreement with the infinite sums found previously. Second, if we're satisfied with numerical results, we may note that when $z=1$ this alternating sum converges rapidly: To five decimals, the first five partial sums are
$$1.98873,\;3.11905,\;3.10427,\;3.10438 , \; 3.10438 $$
which already has converged to the desired precision.


问:
Prove that the dimension of the tangent space $T_x(X)$ of a k-dimensional manifold is k
In section 2, page 9 of Guillemin and Pollack's book $\textit{Differential Topology}$, he gave a proof that the dimension of the tangent space $T_x(X)$ is equal to the dimension of the manifold $X$. However I read the proof at least 5 times and had no idea what they were talking about.

"The dimension of the vector space $T_x(X)$ is, as you expect, the dimension $k$ of $X$. To prove this, we use the smoothness of he inverse $\phi^{-1}$. Choose an open set $W$ in $\mathbf{R}^N$ and a smooth map $\Phi': \mathbf{R}^N \rightarrow \mathbf{R}^k$ that extends $\phi^{-1}$. Then $\Phi'\circ\phi$ is the identity map of U, so the chain rule implies that the sequence of linear transformations

$\mathbf{R}^k\xrightarrow{d\phi_0}T_x(X)\xrightarrow{d\Phi_x'}\mathbf{R}^k$

is the identity map of $\mathbf{R}^k$. It follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism, so $dim T_x(X)=k$."

Just for everyone's information, in GP's book, the convention is that $\phi$ is a diffeomorphism from $\mathbf{R}^k$ to a k-dimensional manifold $X\subset \mathbf{R}^N$ ($\phi :\mathbf{R}^k\rightarrow X$).

I have two questions here:

why do we need to choose an open set $W$ in $\mathbf{R}^N$ and define $\Phi'$ which extends $\phi^{-1}$? Naively I would just take $\phi^{-1}$. Then $d\phi^{-1}$ can also map $T_x(X)$ to $\mathbf{R}^k$ without problems. The map $\Phi'$ seems a bit artificial and redundant here. In fact it seems really redundant to me as when you extend $\phi^{-1}$, you actually don't care the rest of the function except for the $\phi^{-1}$ part.

The proof seems only to serve as a proof that dim $T_x(X)=k$. However we know the tangent space itself is a manifold(actually a hyperplane). Why don't we find a diffeomorphism between $T_x(X)$ and $\mathbf{R}^k$? I haven't proved but I think we can use $d\phi_0$ as the diffeomorphism.

Thanks a lot for everyone's help!

Regards,
Evariste

答:
The strategy of the proof is to convert the statement about smooth maps between manifolds into a statement about linear maps between vector spaces. At this early point in the book, the authors have not yet presented a way to define $d(\phi^{-1})$, because $\phi^{-1}: V \rightarrow U$ has a domain $V$ that is an open subset of a manifold, whereas they have only defined differentials in terms of difference quotients, which require that the domain be a vector space, so that addition is defined; see the limit definition given on page 8 of your text.

Therefore, it is necessary to "straighten" $V$ by first extending $\phi^{-1}$ to a map between Euclidean spaces, and then defining the differential of the extension using difference quotients.

To address your second question, this proof shows precisely that $d\phi_0$ is a linear isomorphism between $\mathbb{R}^k$ and the tangent space $T_x(X)$. A linear map is infinitely differentiable, so it is also a diffeomorphism.


问:
complex rational integral
Let $z_1$ $z_2$ and $z_3$ non iqual complex numbers and be $\Omega$ an open set such that the 3 points are in the same connected component in $\mathbb{C}-\Omega$, then $$f(z)=\frac{z}{(z-z_1)(z-z_2)(z-z_3)}$$
admits a primitive

We use residue theorem in $\Omega\cup\lbrace z_1,z_2,z_3\rbrace$ and we think the problem is reduced to show that a closed path round the three points the same number of times.

答: There are complex numbers $\omega_1$, $\omega_2$, and $\omega_3$ such that$$f(z)=\frac{\omega_1}{z-z_1}+\frac{\omega_2}{z-z_2}+\frac{\omega_3}{z-z_3},$$which are$$\omega_1=\frac{z_1}{(z_1-z_2)(z_1-z_3)},\ \omega_2=\frac{z_2}{(z_2-z_1)(z_2-z_3)}\text{, and }\omega_3=\frac{z_3}{(z_3-z_1)(z_3-z_2)}.$$So, if $\gamma\colon[a,b]\longrightarrow\Omega$ is a closed path,$$\oint_\gamma f(z)\,\mathrm dz=\omega_1\operatorname{ind}_\gamma(z_1)+\omega_2\operatorname{ind}_\gamma(z_2)+\omega_3\operatorname{ind}_\gamma(z_3).\tag1$$Since $\Omega$ is connected, the $3$ numbers $\operatorname{ind}_\gamma(z_1)$, $\operatorname{ind}_\gamma(z_2)$, and $\operatorname{ind}_\gamma(z_3)$ are equal; let $j$ be their common value. Then, it follows from $(1)$ that$$\oint_\gamma f(z)\,\mathrm dz=(\omega_1+\omega_2+\omega_3)j=0,$$since $\omega_1+\omega_2+\omega_3=0$. Since this take place for every closed path, $f$ has a primitive.

Title: complex rational integral


问:
Are associates unit multiples in a commutative ring with $1$?
Recall the following relevant definitions. We say that

$b$ is divisible by $a$ in $R$, or $a\mid b$ in $R$, if $b = r a$ for some $r\in R$.

$a$ and $b$ are associates  in $R$ if $a\mid b$ and $b\mid a$, (or, equivalently, if $aR = bR$).

$u\in R$ is a unit if it has a multiplicative inverse (a $v\in R$ such that $uv=vu=1$).

$a$ and $b$ are unit multiples in $R$  if $a = ub$ for some unit $u\in R$.

Given these definitions, my question is,

If $R$ is a commutative ring with unity and $a,b\in R$ are associates in $R$, are $a$ and $b$ unit multiples in $R$?

I was told that this not always true. But I encountered some difficulties in finding a counterexample.

答:
For a counterexample, I will use the ring $R := \mathbb{Z}[a, b, r, s] / \langle b - ra, a - sb \rangle$.  In this ring, we certainly have $b = ra$ and $a = sb$.  It remains to show there is no unit $u \in R^*$ such that $b = ua$.

Thus, suppose we did have $b = ua$.  Then $a(u-r) = 0$.  On the other hand (and here is where the argument gets somewhat technically involved), it can be shown that the kernel of the $R$-module homomorphism $a\cdot : R \to R$ is generated by $rs - 1$.  Therefore, there exists $p \in R$ such that $u = r + p(rs - 1)$.  Also, if $u$ is a unit, then the image $\bar u$ of $u$ in $R / \langle a, b \rangle \simeq \mathbb{Z}[a,b,r,s] / \langle b - ar, a - sb, a, b \rangle = \mathbb{Z}[a,b,r,s] / \langle a, b \rangle \simeq \mathbb{Z}[r,s]$ would also have to be a unit.  But the only units of $\mathbb{Z}[r,s]$ are the constant polynomials $\pm 1$, whereas with $\bar u = r + \bar p (rs - 1)$ we would need to have $\bar u(1, 1) = 1 \ne -1 = \bar u(-1, -1)$ if we consider $\bar u$ as inducing a function $\mathbb{Z}^2 \to \mathbb{Z}$.  This gives a contradiction, so the assumption that $u$ is a unit must have been false.

The basic idea here is that we are in some sense taking the "universal example" of a ring with two associate elements $a,b$ (though with the caveat that the elements $u,v$ making the elements associate are also distinguished, rather than simply assumed to exist somewhere in the ring; and also that the actual universal example would be $\mathbb{Z}[a,b,r,s] / \langle b-ra, a-sb \rangle$ whereas I used $\mathbb{Q}$ for a slightly easier time with the technical details I omitted).  If we had found a unit within $\mathbb{Q}[a,b,r,s] / \langle b-ra, a-sb \rangle$ making $a$ and $b$ associates, then that expression would have worked in particular for any $\mathbb{Q}$-algebra.  On the other hand, once we instead found that there is no unit in this universal example, we already had our counterexample that we needed.

(On the other hand, this shows that trying to work directly with the universal example might not always be the best or easiest approach.  In this example, we had some mildly hairy calculations, ultimately based on Groebner basis theory, that were needed.  However, in the counterexample based on $R = C[0,3]$ in another answer, it was more straightforward to check the details.)

Addendum on the technical detail omitted above, verifying the fact in Macaulay2:

```<code>Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : R = ZZ[a,b,r,s] / (b - r*a, a - s*b)

o1 = R

o1 : QuotientRing

i2 : mult_a = map(R^1, R^1, matrix {{ a }})

o2 = | a |

             1       1
o2 : Matrix R  <--- R

i3 : kernel(mult_a)

o3 = image | rs-1 |

                             1
o3 : R-module, submodule of R
</code>```

Body:
For a counterexample, I will use the ring $R := \mathbb{Z}[a, b, r, s] / \langle b - ra, a - sb \rangle$.  In this ring, we certainly have $b = ra$ and $a = sb$.  It remains to show there is no unit $u \in R^*$ such that $b = ua$.

Thus, suppose we did have $b = ua$.  Then $a(u-r) = 0$.  On the other hand (and here is where the argument gets somewhat technically involved), it can be shown that the kernel of the $R$-module homomorphism $a\cdot : R \to R$ is generated by $rs - 1$.  Therefore, there exists $p \in R$ such that $u = r + p(rs - 1)$.  Also, if $u$ is a unit, then the image $\bar u$ of $u$ in $R / \langle a, b \rangle \simeq \mathbb{Z}[a,b,r,s] / \langle b - ar, a - sb, a, b \rangle = \mathbb{Z}[a,b,r,s] / \langle a, b \rangle \simeq \mathbb{Z}[r,s]$ would also have to be a unit.  But the only units of $\mathbb{Z}[r,s]$ are the constant polynomials $\pm 1$, whereas with $\bar u = r + \bar p (rs - 1)$ we would need to have $\bar u(1, 1) = 1 \ne -1 = \bar u(-1, -1)$ if we consider $\bar u$ as inducing a function $\mathbb{Z}^2 \to \mathbb{Z}$.  This gives a contradiction, so the assumption that $u$ is a unit must have been false.

The basic idea here is that we are in some sense taking the "universal example" of a ring with two associate elements $a,b$ (though with the caveat that the elements $u,v$ making the elements associate are also distinguished, rather than simply assumed to exist somewhere in the ring; and also that the actual universal example would be $\mathbb{Z}[a,b,r,s] / \langle b-ra, a-sb \rangle$ whereas I used $\mathbb{Q}$ for a slightly easier time with the technical details I omitted).  If we had found a unit within $\mathbb{Q}[a,b,r,s] / \langle b-ra, a-sb \rangle$ making $a$ and $b$ associates, then that expression would have worked in particular for any $\mathbb{Q}$-algebra.  On the other hand, once we instead found that there is no unit in this universal example, we already had our counterexample that we needed.

(On the other hand, this shows that trying to work directly with the universal example might not always be the best or easiest approach.  In this example, we had some mildly hairy calculations, ultimately based on Groebner basis theory, that were needed.  However, in the counterexample based on $R = C[0,3]$ in another answer, it was more straightforward to check the details.)

Addendum on the technical detail omitted above, verifying the fact in Macaulay2:

Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : R = ZZ[a,b,r,s] / (b - r*a, a - s*b)

o1 = R

o1 : QuotientRing

i2 : mult_a = map(R^1, R^1, matrix {{ a }})

o2 = | a |

             1       1
o2 : Matrix R  <--- R

i3 : kernel(mult_a)

o3 = image | rs-1 |

                             1
o3 : R-module, submodule of R


问:
Finite extensions of $\mathbb{Q}_p$ as completions of number fields with some valuation over a prime ideal
In my first post, I asked if $\mathbb{Q}_p$ is the completion of $\mathbb{Q}$ over some valuation of an prime ideal (and it seems to be true, according to an answer). Now I am asking if this can be generalized for finite extensions of $\mathbb{Q}_p$.

To put it in other words: Let $K$ be a finite extension of $\mathbb{Q}_p$. Question: Is there a valuation $v$ of some prime ideal in $\mathcal{O}_k$ on a global field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that the completion $k_v$ of $k$ over $v$ is equal to $K$?

答:
Yes.  Let $K$ be a finite extension of $\mathbb Q_p$.  We want to show that there is a finite extension $F$ of $\mathbb Q$, and a prime ideal $\mathfrak p$ in the ring of integers $\mathcal O_F$ of $F$, such that $K$ is the completion of $F$ with respect to the $\mathfrak p$-adic valuation on $F$.

Krasner's lemma tells us that there is an irreducible polynomial $g(t)$ (irreducible in the ring $\mathbb Q_p[t]$, that is) with rational coefficients, and a root $\beta$ of $g(t)$, such that $K = \mathbb Q_p(\beta)$.  Then $g(t)$ is also irreducible in the ring $\mathbb Q[t]$, so if we set $F = \mathbb Q(\beta)$, then $[F : \mathbb Q] = [K : \mathbb Q_p] = \operatorname{deg} g$.

Now let $| - |$ be the unique extension of the $p$-adic absolute value to $K$.  Restrict $|-|$ to an absolute value on $F$.  A version of Ostrowski's theorem for number fields states that this absolute value is either archimedean, or arises from a prime ideal $\mathfrak p$ in the ring of integers $\mathcal O_F$ of $F$.  Specifically, there is a real number $0 < \rho < 1$ and a prime ideal $\mathfrak p$ such that

$$|x| = \rho^{\operatorname{ord}_{\mathfrak p}(x)}.$$

Obviously this absolute value is not archimedean on $F$ (its restriction to the $p$-adic absolute value on $\mathbb Q$ isn't, after all).
Anyway, the completion of $F$ with respect to $\mathfrak p$ is the same as the metric space completion of $F$ with respect to $|-|$.  Finally, just as the metric space completion of $\mathbb Q$ with respect to $|-|$ gets you $\mathbb Q_p$, so does the metric space completion of $F = \mathbb Q(\beta)$ with respect to $|-|$ get you $\mathbb Q_p(\beta) = K$.


问:
Integral inequality related to increasing functions
Assume $f:[a,b] \to \mathbb R$ is an increasing function,if $P_n$ is a partition of $[a,b]$ into $n$ equal subintervals,then show that $$0\le U\left(P_{n},f\right)-\int_{a}^{b}f\left(x\right)dx\le\frac{b-a}{n}\left(f\left(b\right)-f\left(a\right)\right)$$

We have that $$ U\left(P_{n},f\right)=\sum_{i=0}^{n-1}M_{i}\Delta x_i=\frac{b-a}{n}\sum_{i=0}^{n-1}f(x_{i+1})$$

Where $\Delta x_i=x_{i+1}-x_i=(b-a)/n$ and $M_{i}=\sup \{f(x):x \in [x_{i},x_{i+1}]\}$
and $P_n=(a=x_0,x_1,...,x_n=b)$

On the other hand:

$$\int_{a}^{b}f\left(x\right)dx=\inf \{ U\left(P,f\right):P \in \mathcal P_{[a,b]} \}\le \inf \{ U\left(P_n,f\right):n \in \mathbb N^+\} \le U\left(P_n,f\right)$$

Where $\mathcal P_{[a,b]}$ is the set of all partitions of $[a,b]$
which shows that $$ 0\le U\left(P_n,f\right) -\int_{a}^{b}f\left(x\right)dx$$

But how to get the other inequality?

答: Use the Sup of $L(P_n,f)$'s for the definition of the integral of $f$, then for the same $P_n$:  $ \displaystyle \int_{a}^b fdx \ge L(P_n,f)= \dfrac{b-a}{n}\cdot \displaystyle \sum_{i=0}^{n-1} f(x_i)\implies U(P_n,f) - \displaystyle \int_{a}^b fdx \le \dfrac{b-a}{n}\left(\displaystyle \sum_{i=1}^n f(x_i) - \displaystyle \sum_{i=0}^{n-1} f(x_i)\right)= \dfrac{b-a}{n}\cdot (f(x_n) - f(x_0)) = \dfrac{b-a}{n}\cdot (f(b) - f(a))$.


问:
Markov Chain and Random Walks
I am having difficulty understanding the question here. It says that $\{X_n\mid n=0,1,2,...\}$ is a simple random walk on the integers $S=\{\dots,-2,1,0,1,2,\dots\}$,starting at $X_0=3$. The probabilities of jumping left and right are $1/2$.

Let $T=\min{n\mid X_n=-3\text{ or }Xn=7}$.

(a) Find $E[X(T-2)]$.

(b) Find $E[X(T+2)]$.

I was wondering if I should put $T=4$ here since the minimum no of transitions to reach $-3$ or $7$ from $3$ is $4$. But then it would be straightforward as $E[X(2)]$ and $E[X(4)]$ which I believe should be $3$ in both cases. I think I am missing something [I was thinking whether I should try $\sum_xP(X>x)$]. Any help here would be appreciated.

答:
It is a very nice exercise to show that $\mathbb P(X_T=7)=\frac{3}{5}$ and $\mathbb P(X_T=-3)=\frac{2}{5}$. As a hint:

 Let $h_n$ be the probability of hitting $7$ before $-3$ when you start
 out at $X_0=n$, and set up a system for $h_n$, which you can solve
 with the boundary conditions $h_7=1$ and $h_{-3}=0$.

As a consequence, we have
$$\mathbb E[X_T]=\frac{3}{5}\times7+\frac{2}{5}\times(-3)=3.$$
Now note that if $S_1, S_2$ are the two steps you take after $X_T$, then
$$\mathbb E[X_{T+2}]=\mathbb E[X_T+S_1+S_2]=\mathbb E[X_T]+\mathbb E[S_1]+\mathbb E[S_2]=\mathbb E[X_T]+0+0=\mathbb E[X_T]=3.$$
The case for $T-2$ is a bit trickier -- we can't immediately apply a similar argument as above, because, conditional on the fact that we hit $-3$ or $7$ two steps later, the increments may no longer distributed uniformly over $\pm1$.

We do have the following work around though. Note that $X_{T-2}$ is either $-1$ or $5$. Further, $X_{T-2}=-1$ iff $X_T=-3$ and $X_{T-2}=5$ iff $X_T=7$. So
$$\mathbb E[X_{T-2}]=\frac35\times5+\frac25\times(-1)=2.6.$$

What's going on under the hood here is that the simple random walk $X_n$ is an example of what is called a martingale, that is $X_n$ has the property
$$\mathbb E[X_{n+1}\mid X_n,\dots,X_1]=X_n.$$
For the random walk, this is obvious -- at each step, we are equally likely to move one space left or right. Martingales also have the property that
$$\mathbb E[X_n]=\mathbb E[X_{n-1}]=\dots=\mathbb E[X_0],$$
which again is especially obvious in the case of a simple random walk. For general martingales, you can prove this via the law of total expectation.

In general, if $\tau$ is a stopping time, and $M_n$ is a martingale, then the optional stopping theorem tells us that, so long as $\tau$ and $M_n$ satisfy some certain "finiteness" properties, then it is in fact true that $\mathbb E[M_\tau]=\mathbb E[M_0]$. This applies in the case of the simple random walk $X_n$ and our stopping time $T$, so we instantly know that $\mathbb E[X_T]=\mathbb E[X_0]$.

The fact that $\mathbb E[X_{T-2}]\neq\mathbb E[X_0]$ is because, as noted in the comments, $T-2$ is not a valid stopping time.

Title: Markov Chain and Random Walks


问:
How to find the center of a spiral given only the physical object?
I have a spiral cam from a belt sander tension assembly that needs to have a hole bored in the exact center and I am unsure of how to find it mathematically. I have the object and can measure the diameter at various points but I am stuck at how to get it into a mathematical equation

答:
I have assumed that the shape of the cam is an Archimedean spiral, as it is commonly used for converting uniform radial motion into uniform speed translation.

The polar equation of an Archimedean spiral is:

$$r=k\theta \ \ \ \text{with} \ \ \ k:=2a/\pi$$

meaning that each time the cam undergoes a $\pi/2$ ($ = 90°$) rotation , we have a new added length $a$, then $2a$, $3a$, $4a$, $a+4a=5a$, etc., this regularly added length explaining the regular translation.

See the theoretical curve in red with a rather good agreement with the shape you have given

Having just seen the proposal posted by Intelligenci Pauca, I see that our solutions should not differ that much.

Remark: the respective width ($14 a$ with an estimated extension of the cam) and height $12a$ of the "bounding box" of the cam should help to obtain an accurate positioning of the looked for center.

Edit: If the figure is at the right scale, as I have measured a 19.5 units in the horizontal direction and 17 in the vertical direction, ratio $19.5/17$ is in a very good agreement with ratio $14/12$.


问:
Prove every integer exists in this sequence?
Please could someone give me a hint on this sequences question? The question is to prove that every integer appears infinitely many times in the following sequence:

$$ \pm 1^{2} , \pm 1^{2} \pm 2^{2} , \pm 1^{2} \pm 2^{2} \pm 3^{2} , \pm 1^{2} \pm 2^{2} \pm 3^{2} \pm 4^{2} , ... $$

Any help would be much appreciated, thank you (:

答:
Hint: Note that $n^2 -(n+1)^2-(n+2)^2+(n+3)^2=4$. So by choosing signs of four consecutives appropriately, we can get $4$ or $-4$.

Any integer is of the form $4k+r$, where $r$ is one of $0$, $1$, $2$, or $3$. Now it is just a matter of showing we can get all of $1$, $2$, and $3$. And we don't even need $3$. By the way, $2$ has a length $4$ representation.