问:
Estimate amount spent
I am reading the following simple problem on rounding.

10 people ordered calculators.

The least expensive was \$19.95 and the
most expensive \$39.95 Half ordered a \$29.95 calculator.

Select the
best estimate of the amount spent:

a) \$240

b) \$310

c) \$345

d) \$355

My approach:

The rest of the 5 could have ordered either \$20 or \$40 or \$10

So we have (on top of the $5$ that the problem mentions they bought the \$30 calculator)

$150 + 5 * 20 = 250$

or

$150 + 5 * 40 = 350$

or

$150 + 5 * 30 = 300$

So we get an average of 250 + 350 + 300 = \$900 which divided by 3 gives \$300

That is not part of the options though and the solution states that the correct answer is b

What am I messing up here?

答:
To round things out, let $$A=20\quad B=30\quad C=40$$

Assuming some kind of bell-curve in purchasing, we have

$$2A+5B+2C+??=40+150+ 80+??=270+??$$

Given that this is already above the $240$-estimate where $??=-30$, we must assume that $??=40$ and the sum $$270+??=310$$

Title: Estimate amount spent


问:
How do I get that $11^{35} \equiv 6 \hspace{0.1cm} \text{mod} (13)$ from $11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$
I have no clue how to do this, I manage to get I get that $11^{36} \equiv 1  \hspace{0.1cm} \text{mod} (13)$ but I can't get anywhere from there.

答:
The exercise here is to calculate the multiplicative inverse of $11$, written $11^{-1}$, in $\bmod 13$ arithmetic. Then we can get that since $11^{36}\equiv 1$, we will have $11^{35}\equiv 11^{36}\cdot 11^{-1}\equiv 11^{-1}\bmod 13$

Now $13$ and $11$ are coprime (of course, since $13$ is prime) so it's guaranteed that there is a value $a$ such that $11a \equiv 1 \bmod 13$. For a small number like $13$, it's not too hard to find this by guessing, but the extended Euclidean algorithm will find the value rapidly for less straightforward cases.

To show the logic of this algorithm, we find progressively smaller results which we can make with linear combinations of $13$ & $11$ until we make $1$:
\begin{align}1\cdot 13 + 0\cdot 11  &= 13 \tag{setup: 1}\\
0\cdot 13 + 1\cdot 11 &= 11 \tag{setup: 2}\\
1\cdot 13 + -1\cdot 11 &= 2 \tag{(1)-(2): 3 }\\
-5\cdot 13 + 6\cdot 11 &= 1 \tag{(2)-5(3): result }\\\end{align}
gives us that $6\cdot 11 \equiv 1$ so $11^{-1}\equiv 6 \bmod 13$.

ContentLicense: CC BY-SA 4.0


问:
Two definite integrals in terms of one another
While practicing definite integration I stumbled upon following question:

Q.) If $\int_{0}^{π/2}\ln(\sin x)dx=k$ then, find $\int_{0}^{π/2}\frac{x²}{(\sin x)²}dx$, in terms of $k$

My attempt:
I have memorised the first integral answer as

$\int_{0}^{π/2}\ln(\sin x)dx$=$-\frac{π}{2}\ln 2$=$k$

For second integral I did following:

Let $$I=\int_{0}^{π/2}\frac{x²}{(\sin x)²}dx$$

$$I=\int_{0}^{π/2}\frac{(\frac{π}{2}-x)²}{(\cos x)²}dx$$

$$[\int_{b}^{a}f(x)dx=\int_{b}^{a}f(a+b-x)dx]$$

and then i did by parts, but things become only more complicated. So please help if you can solve the second integral easily. THANKS IN ADVANCE

'ANSWER IS $-2k$'

答:
Integrate by parts so that $$ \begin{aligned}\int_0^{\frac{\pi}{2}} \dfrac{x^2}{\sin^2 x} \mathrm{d}x
&=\underbrace{-x^2 \cot x \bigg|_0^{\frac{\pi}{2}}}_0+2\int_0^{\frac{\pi}{2}}x\cot x \mathrm{d}x \\&=2\left(\underbrace{x \ln \sin x \bigg|_0^{\frac{\pi}{2}}}_0-\int_0^{\frac{\pi}{2}}\ln \sin x \mathrm{d}x\right) \\
&=-2k \\
&=\pi\ln 2
\end{aligned}$$


问:
Determine if there's a linear map such that...
Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix with the inputs for $\phi$ written in rows in the left hand side of the matrix and outputs of $\phi$ in the right hand side, also in rows. Then the matrix has been row reduced until an identity appeared in the left hand side... it seemed to me like some black magic. I have no idea why or how it works. This whole method doesn't make any sense to me. Is there any better way of doing it or some explanation why something like this works?

Body:
The first three input vectors are linearly independent,

so we can evaluate $\phi$ on the standard basis vectors by linearity:

$\phi(1,0,0)=\phi(\frac12((4,2,1)-(2,2,1))=\frac12((4,1,0)-(2,3,3))=(1,-1,-\frac32);$

$\phi(0,1,0)=\phi((2,2,0)+(2,2,1)-(4,2,1))=(3,2,1)+(2,3,3)-(4,1,0)=(1,4,4)$;

and $\phi(0,0,1)=\phi((2,2,2)-(2,2,1))=(0,0,1)-(2,3,3)=(-2,-3,-2).$

So   $\phi$ corresponds to $\pmatrix{1&1&-2\\-1&4&-3\\-\frac32&4&-2}$.

Check that this matrix applied to the fourth input vector $(2,2,2)$ yields $(0,0,1)$.


问:
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$
I came across this integral

$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$

According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.

A promising way might be the following

\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
 &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3}   \right )^2 + \frac{8}{3}}} \\
 &= \cdots
\end{align*}

A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )

So, any ideas how to evaluate it?

P.S: Are there techniques available for these type of problems?

答:
Since the integrand involves the square root of a quartic without repeated roots, elliptic integrals are needed. However, computer algebra systems usually give suboptimal results for such integrals – this is why I put together Elliptic Integrals and Functions so the "nicest" result can be obtained.

The corresponding Byrd and Friedman entry in this case is 225.00:
$$\int_0^1\frac1{\sqrt{3x^4+2x^2+3}}\,dx=\frac1{\sqrt3}\int_0^1\frac1{\sqrt{(x^2+z^2)(x^2+\overline z^2)}}\,dx\qquad z=\sqrt{\frac23}+\frac1{\sqrt3}i$$
$$=\frac1{\sqrt3}\cdot\frac12F(2\tan^{-1}1,m=1/3)=\frac1{\sqrt{12}}K\left(\frac13\right)$$
The numerical value is most definitely not $\frac12$; it is $0.500538690228\dots$


问:
Using mathematical induction to prove an identity related to combinatorics
Using Mathematical induction on $k$, prove that for any integer $k\geq 1$,

$$(1-x)^{-k}=\sum_{n\geq 0}\binom{n+k-1}{k-1}x^n$$

How should I proceed? The tutorial teacher attempted this question and forgot halfway through... facepalm

答: The idea is that you prove it for $k=1$ (which is exactly the sum of a geometrical progression), then assuming it is true for $k$, differentiate both members of the equality and obtain the result for $k+1$. The computation must be quite elementary.</p>

ContentLicense: CC BY-SA 3.0


问:
Quick question about integrating a surface integral over a sphere.
I am reading lecture notes on PDEs and I came across this equality, where $z_0\in \mathbb{R}^2$ and $u\in H^{1}(B;\mathbb{R}^3)$ where $B=B_1(0)\subset \mathbb{R}^2$ and we assume $B_R(z_0)\subset B_1(0)$.

$$\int_{R/2}^{R} \frac{1}{2\pi S}\int_{\partial B_S(z_0)} u(z) do(z) \frac{4S}{3R^2} dS = \frac{4}{3\pi R^2}\int_{B_{R}(z_0)\setminus B_{R/2}(z_0)} u dx.$$

I think there is a factor of $2$ missing, the correct equation should be,

$$\int_{R/2}^{R} \frac{1}{2\pi S}\int_{\partial B_S(z_0)} u(z) do(z) \frac{4S}{3R^2} dS = \frac{2}{3\pi R^2}\int_{B_{R}(z_0)\setminus B_{R/2}(z_0)} u dx.$$

答:
If $do$ means arc length on the circle, i.e. $do = Sd\theta$, then the formula is just the change of variables $dxdy = r dr d\theta$, taking polar coordinates centered in the disk $B_R(z_0)$, and you are right
about the 2. There is no sphere or surface integral though, since this is all in the plane.


问:
For $x_{n+1}=x_n-x_n^3$, with $|x_1|>1$. What about the convergence?
I see Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$ that if $|x_1| \le 1$, then $x_n\to 0$. However, if $|x_1|>1$, what can we say about this sequence? It seems hard to find the tendency of it. For $x_1$ near $1$, it converges to $0$, but for $x_1$ large, it is not the case.

答:
We only care about when $|x|>1$. The condition for it to get further away from the origin occurs when,

$$|x-x^3|>|x|$$

We can divide by $|x|$ since it's nonzero, and the inequality is unchanged because it's positive.

$$|1-x^2|>1$$

Since $|x|>1$ we know $x^2>1$ and $x^2-1$ is positive, and so we can replace the absolute value with it.

$$x^2-1 > 1$$

$$x^2 > 2$$

This means when $x$ is outside $[-\sqrt{2}, \sqrt{2}]$ it is repelled further away from the origin.

Similarly, working through with the reverse inequality gives that it is attracted to the origin when $x$ is inside $(-\sqrt{2}, \sqrt{2})$ (knowing that when $|x|\le 1$ it goes to the origin as well as given in the original question.)


问:
Morphism of quasi affine varieties giving algebra homomorphism and back.
$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$
$\newcommand{\A}{\mathbb A}$

I am new to algebraic geometry and in the following I ask two related questions that have come to my mind given the material I have read so far.
I believe the notation and terminology I have used is standard.
However, I have supplied most of the relevant definitions and notations in the appendix.
Please feel free to correct and inaccuracies.

In what follows $k$ is an algebraically closed field.

Question 1

Let $X$ and $Y$ be affine varieties.
It is a fact that a morphism $\vp:X\to Y$ is an isomorphism if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras.

Question.
Can we extend this fact to the case when $X$ and $Y$ are quasi affine varieties, that is, is a morphism $\vp:X\to Y$ an isomorphism if and only if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras?

The problem I am facing in proving this is the following.
When $X$ and $Y$ are affine varieties in $\A^m$ and $\A^n$ respectively, then we have $\mr O(X) = k[x_1, \ldots, x_m]/\mr I(X)$ and $\mr O(Y) = k[y_1, \ldots, y_n]/\mr I(Y)$.
Now once we have a $k$-algebra homomorphism $\mr O(Y)\to \mr O(X)$, we get a morphism of varieties $X\to Y$ by using the induced map $\text{MaxSpec}(\mr O(X))\to \text{MaxSpec}(\mr O(Y))$ and then use the nullstellensatz to get a map $X\to Y$.

However, when $X$ and $Y$ are quasi affine varieties in $\A^m$ and $\A^n$ respectively, then one cannot think of $\mr O(X)$ or as a quotient of $k[x_1, \ldots, x_m]$ (and similarly for $\mr O(Y)$).
For instance, let $X=\A^1\setminus \set{0}$. Then $\mr O(X) = k[x]_{x}$ which is not a quotient of $k[x]$ but of $k[x, y]$.

So I am unable to identify $\text{MaxSpec}(\mr O(X))$ and $\text{MaxSpec}(\mr O(Y))$ as subsets of $\A^m$ and $\A^n$.

Question 2

Let $\mc V$ be the category of affine varieties and $\mc A$ be the category of finitely generated $k$-algebras that are integral domains.
We get an equivalence between two categories by associating to an affine variety its coordinate ring and to a morphism its corresponding map between the algebras, namely the pullback map.

Question.
Is there such a statement for the category of quasi affine varieties, that is, is there a nice algebraic category to which the category of quasi affine varieties equivalent?

Question.
Suppose we consider the category of subsets of $\A^n$, $n\geq 0$.
We can still define $\mr O(X)$ for any such subset and form a category with morphisms as maps which pullback regular functions to regular functions.
Is this category useful in any way? Is there a reason why it is not studied?

Appendix

Let $k$ be an algebraically closed field.
We write $\A^n$ to denote $k^n$ with the Zariski topology.
An affine variety is an irreducible closed subset of some $\A^n$.

Given an affine variety $X$ in $\A^n$, the coordinate ring $\mr A(X)$ of $X$ is defined as $k[x_1, \ldots, x_n]/\mr I(X)$, where $\mr I(X)$ is the set of all the polynomials in $k[x_1, \ldots, x_n]$ which vanish on each point of $X$.
It turns out that $\mr I(X)$ is a prime ideal.

A quasi affine variety is any open subset of an affine variety.
A regular function on a quasi affine variety $Y\subseteq \A^n$ is a map $\vp:Y\to k$ which can be locally written as a quotient of polynomials in $k[x_1, \ldots, x_n]$.
The set of all the regular functions on $Y$ is denoted by $\mr O(Y)$.
It turns out that $\mr O(Y)$ has a natural $k$-algebra structure.
It is a fact that if $X$ is an affine variety, then $\mr A(X)$ can be naturally identified with $\mr O(X)$.

A morphism of quasi affine varieties $X$ and $Y$ is a map $\vp:X\to Y$ which is continuous (in the Zariski topology) such that $\vp$ pulls back regular functions on $Y$ to regular functions on $X$.
Given a morphism $\vp:X\to Y$, we get the pullback map $\vp^*:\mr O(Y)\to \mr O(X)$, and it is easy to check that $\vp^*$ is a $k$-algebra homomorphism.

答:
"Question. Can we extend this fact to the case when $X$ and $Y$ are quasi affine varieties, that is, is a morphism $φ:X→Y$ an isomorphism if and only if the corresponding map $φ∗:O(Y)→O(X)$ is an isomorphism of k-algebras?"

Example 1. If $\phi: \mathbb{P}^1_k \rightarrow \mathbb{P}^d_k$ is the $d$-uple embedding with $d\geq 2$, it follows $\phi$ is not an isomorphism. But

$$\Gamma(\mathbb{P}^1_k, \mathcal{O}_{\mathbb{P}^1_k}) \cong \Gamma(\mathbb{P}^d_k, \mathcal{O}_{\mathbb{P}^d_k}) \cong k$$

Hence the global sections are isomorphic, but the varieties are not (a projective variety is quasi projective). Hence it seems the correspondence you mention can not be extended to quasi projective varieties. You find this construction in Hartshornes book, chapter I and II. The projective varieties $\mathbb{P}^1_k$ and $\mathbb{P}^d_k$ have different dimensions, but their structure sheaves $\mathcal{O}_{\mathbb{P}^1_k}$ and $\mathcal{O}_{\mathbb{P}^d_k}$ have the same number of global sections. In chapter I you find the definition of the $d$-uple embedding and in chapter II you find the definition of the structure sheaf.

Example 2. If $k$ is algebraically closed, $X:=Spec(k[x,y])$ and $U:=X-\{(0,0)\}$ it follows there is an isomorphism

$$i^{\#}: \Gamma(X, \mathcal{O}_X) \cong \Gamma(U, \mathcal{O}_U)\cong k[x,y]$$

(Hartshorne Exercise I.3.6) but $X$ and $U$ are not isomorphic. Hence the correspondence seems not to hold for quasi affine varieties either.
The map $i^{\#}$ is induced by the inclusion

$$ i: U \rightarrow X.$$

The scheme $U$ is not affine.

You may change the  notion "regular map" - this is discussed briefly here:

Regular rational functions over real projective lines and projective spaces.

Example 3. If $k:=\mathbb{R}$ is the field of real numbers and $f:=x^2+y^2$
it follows $Z:=V(f):=Spec(A)$ where $A:=k[x,y]/(f)$. $Z$ has $(0,0)$ as the only rational point: $x^2+y^2=0$ iff $(x,y)=(0,0)$. Why? Let

$$\phi: A \rightarrow k$$

be a map of $k$-algebras.It follows $\phi(x)=a, \phi(y)=b$ for $a,b\in k$.
Since the map $\phi$ is well defined it follows $\phi(x^2+y^2)=a^2+b^2=0$, and from this it follows $a=b=0$. Hence the only $k$-rational points of $Z$ is the point $(0,0)$. Hence we may claim that $Z=\{(0,0)\}$ is the "one point space"
with only rational point $(0,0)$. It is defined as closed subscheme of $X:=Spec(k[x,y])$ corresponding to the ideal $I:=(f)$, hence $Z$ is indeed a hypersurface in $X$.

Let $U:=X-Z:=X-\{(0,0)\}$. Since $U$ is the complement of the "hypersurface" $Z$, it follows $U\cong D(f)\cong Spec(B)$ where

$$B:=k[x,y,t]/(tf-1)$$

hence $U$ is affine. Hence over a non-algebraically closed field $k$ it follows $U:=\mathbb{A}^2_k-\{(0,0)\}$ may be an affine scheme.

Note: If $\mathfrak{m}:=(x,y)$ and $P:=Spec(k[x,y]/\mathfrak{m})$ we may realize $V(\mathfrak{m})$ as a closed subscheme of $Z$: There is a canonical surjective map

$$ A \rightarrow k[x,y]/\mathfrak{m}\cong  k.$$

Hence the hypersurface $Z$ is the rational point $(0,0)$ with a non-trivial scheme structure.

Example 4. This generalize as follows. Let $k$ be the field of real numbers and let $J:=(f_1,..,f_k)\subseteq A:=k[x_1,..,x_n]$ be an ideal generated by the polynomials $f_i$. Let

$$F(x_1,..,x_n):=f_1^2+\cdots +f_k^2 \in A.$$

It follows $F$ defines a hypersurface. Let $B_1:=A/J, B_2:=A/(F)$ and let $X_i:=Spec(B_i)$ for $i=1,2$.

Lemma: The two affine schemes $X_1,X_2$ have the same $k$-rational points.

Proof: Assume $\phi: B_1\rightarrow k$ is a map of $k$-algebras. It follows

$$ \phi(x_i)=a_i$$

for $i=1,..,n$ and since $\phi$ is well defined it follows $f_j(a_1,..,a_n)=0$
for all $j=1,..,k$. Hence the point $a:=(a_1,..,a_n)$ is a zero of the polynomials $f_j$. It follows

$$F(a_1,..,a_n):= \sum_j f_j(a_1,..,a_n)^2=0.$$

Conversely if $\phi: B_2\rightarrow k$ is a map of $k$-algebras with $\phi(x_i):=a_i\in k$ it follows

$$F(a_1,..,a_n)=0.$$

It follows

$$F(a_1,..,a_n):= \sum_j f_j(a_1,..,a_n)^2=0$$

and since $f_j(a_1,..,a_n)^2\geq 0$ it follows $f_j(a_1,..,a_n)=0$ for all $j=1,..,k$. This sets up a 1-1 correspondence between $k$-rational points of $X_1$ and $k$-rational points of $X_2$. QED.

Note: $X_1$ and $X_2$ have the same rational points but are not isomorphic as affine schemes in general. Over the field of real numbers, any "real affine algebraic variety" may set-theoretically be realized as the zero set of a single polynomial.

Since there is an inclusion of ideals $(F) \subseteq J$, we get a canonical map of affine schemes

$$\phi: X_2\rightarrow X_1$$

where $\phi$ induce a bijection $X_2(k)\cong X_1(k)$ at the level of $k$-rational points, but $\phi$ is not an isomorphism of schemes. Moreover the hypersurface $(F)$ depends on the choice of generators of the ideal $J$. If $X_i$ are "smooth algebraic varieties" we get an induced isomorphism of real smooth manifolds $(X_1)_k \cong (X_2)_k$.


答:
HINT: Your induction hypothesis is that

$$(1-x)^{-k}=\sum_{n\ge 0}\binom{n+k-1}{k-1}x^n\;.$$

For the induction step take a look at this calculation:

$$\begin{align*}
\sum_{n\ge 0}\binom{n+k}kx^n&=\sum_{n\ge 0}\left(\binom{n+k-1}{k-1}+\binom{n+k-1}k\right)x^n\\
&=(1-x)^{-k}+\sum_{n\ge 0}\binom{n+k-1}kx^n\\
&=(1-x)^{-k}+\sum_{n\ge 1}\binom{n+k-1}kx^n\\
&=(1-x)^{-k}+x\sum_{n\ge 1}\binom{n+k-1}kx^{n-1}\\
&=(1-x)^{-k}+x\sum_{n\ge 0}\binom{n+k}kx^n\;.
\end{align*}$$

Title: Using mathematical induction to prove an identity related to combinatorics

Body:
Using Mathematical induction on $k$, prove that for any integer $k\geq 1$,

$$(1-x)^{-k}=\sum_{n\geq 0}\binom{n+k-1}{k-1}x^n$$

How should I proceed? The tutorial teacher attempted this question and forgot halfway through... facepalm


问:
Given a sequence $a_{n}= \frac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\in (0, 1).$ Prove that $\sum a_{n}< \frac{2\,a_{1}}{1- a_{1}}.$

Given a sequence $a_{n}= \dfrac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\in (0, 1).$ Prove that $\sum a_{n}< \dfrac{2\,a_{1}}{1- a_{1}}.$

How can I seperate $n$ parts from $\dfrac{2\,a_{1}}{1- a_{1}}$ and each of them $u_{n}> a_{n}{\it ?}$ I observe
$$a_{n}- a_{n+ 1}= \dfrac{a_{n+ 1}\left ( 1- a_{n+ 1}^{2} \right )}{1+ a_{n+ 1}^{2}}= \dfrac{a_{n}\left ( 1- a_{n+ 1}^{2} \right )}{2}$$
So, what $u_{n}- u_{n+ 1}$ sould be ? I need to the help, thanks a real lot.

Edit: I found one relationship that is
$$\frac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\geq a_{n+ 1}\left ( 2- a_{n+ 1} \right )$$
Sorry I can't talk it clearly now, because it's too late here. I'll post the proving of the inequality as soon as well.

答:
Using the double-angle formula for the hyperbolic tangent one can obtain an explicit formula for $a_n$ and get the desired inequality:

We know that $a_n \in (0, 1)$, so that we can substitute $a_n = \tanh(x_n)$ with $x_n > 0$. Then
$$
 \tanh(x_n) = \frac{2 \tanh(x_{n+1})}{1+\tanh^2(x_{n+1})} = \tanh(2 x_{n+1})
$$
implies $x_n = 2x_{n+1}$. It follows that $x_n = x_1/2^{n-1}$ for all $n$. Now the sum can be estimated:
$$
 \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \tanh\left(\frac{x_1}{2^{n-1}}\right) < \sum_{n=1}^\infty \frac{x_1}{2^{n-1}} = 2 x_1 \, .
$$
On the other hand,
$$
\frac{2a_1}{1-a_1} = \frac{2\tanh(x_1)}{1-\tanh(x_1)} = e^{2x_1} - 1 > 2x_1 \, .
$$


答:
Another approach you may like is as follows:

Let the input vectors are $u_1,u_2,u_3,u_4$ and the corresponding output vectors are $v_1,v_2,v_3,v_4$ so that for a $3\times 3$ matrix $A=\left (x_{ij}\right )$, the system $Au_i=v_i, (i=1,2,3)$ gives you a set of $9$ equations involving $9$ variables $x_{ij}, (1\le i\le 3, 1\le j\le 3)$.

Solving the above you can obtain matrix $A$. Now just verify whether $Au_4=v_4$ holds or not. If it does then such linear transformation is possible.

Title: Determine if there's a linear map such that...

Body:
Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix with the inputs for $\phi$ written in rows in the left hand side of the matrix and outputs of $\phi$ in the right hand side, also in rows. Then the matrix has been row reduced until an identity appeared in the left hand side... it seemed to me like some black magic. I have no idea why or how it works. This whole method doesn't make any sense to me. Is there any better way of doing it or some explanation why something like this works?


问:
Degeneracy of the scalar product $\langle A,B \rangle = \mathrm{tr}(BA)$
I have to solve the following linear algebra problem:

Consider the vector space formed by the set $M_{n}(\mathbb{R})$ of all real square matrices of order $n$ with the usual operations. A scalar product $\langle\cdot,\cdot\rangle:M_{n}(\mathbb{R})\times M_{n}(\mathbb{R}) \to \mathbb{R}$ can be defined as $\langle A,B\rangle=\mathrm{tr}(BA)$. Is this scalar product degenerate?

I "know" this product is non-degenerate because I did solve the case $n=2$ and I guess that it's valid for all $n\in\mathbb{N}$. However, I have no clue of how to solve the general case. I did note that, given two matrices $A=(a_{ij})$ and $B=(b_{ij})$, it follows that $$ \langle A,B \rangle = \mathrm{tr}(BA) = \sum_{i=1}^n B_{i}\cdot A^{i},$$ where $B_{i}$ is the $i$-th row of the matrix $B$ and $A^{i}$ is the $i$-th column of the matrix $A$, so I thought that I could use the non-degeneracy of the dot product and extend it to the matrix scalar product, but so far I haven´t been able to do that. Any hint on how to solve this problem? (I don't want a solution, just a hint).

答: Hint. By definition, the scalar product is degenerate if there exists some nonzero $A$ such that $\langle A,B\rangle=0$ for every $B$. This, in particular, implies that $\langle A,B\rangle=0$ for every rank-one matrix $B$.


问:
Ratios of similar triangles
This is a basic question, and based on my thoughts I really think this is true, but I just want some verification. If I have similar triangles, are the ratios between two sides of one triangle equal to the ratio of the corresponding sides on the other triangle?</p>

答: The scaling operation that defines similarity scales all sides by the same factor. So when going from one triangle to the other, $\frac{\text{side A}}{\text{side B}}$ becomes$\frac{k×\text{side A}}{k×\text{side B}}$ and the two instances of $k$ cancel. You are right.

Title: Ratios of similar triangles


问:
Prove that semi-simple rings are Dedekind-finite
Just to be consistent with the terminology, let me define the words I'm using.

A module $M$ is simple if it has no proper non-zero submodules, and semi-simple if it can be written as a direct sum $M = \bigoplus_i A_i$ where the $A_i$'s are simple submodules.

A ring $R$ is semi-simple if the left $R$-module $_RR$ is semi-simple. This means that $R$ is a direct sum of "minimal" ideals. I want to prove that such rings are also Dedekind-finite, i.e. if $a,b\in R$ and $ab = 1$, then $ba = 1$. I want to do this without invoking Artin-Wedderburn. (EDIT: We have "semi-simple $\implies$ Noetherian $\implies$ Dedekind-finite", but I don't know how to prove the first implication without using composition series. It seems to me that my problem can be solved without using the fact that semi-simple rings are Noetherian.)

So far, I can show that a semi-simple ring $R$ is a direct sum of finitely many minimal ideals: if $R = \bigoplus_\lambda A_\lambda$ where each $A_\lambda$ is a minimal ideal of $R$, then write $1 = e_1 + \cdots + e_n$ where $e_i \in A_{\lambda_i}$. It follows that $r = re_1 + \cdots + re_n$ for any $r \in R$, so $R = \bigoplus_{i=1}^n A_{\lambda_i}$. Let's abbreviate the notation by setting $A_i := A_{\lambda_i}$.

Alright, now suppose $a,b\in R$ are such that $ab = 1$. We want to show that $ba = 1$. I have attempted to solve the problem in the case that $a \in A_i$ for some $i$. First, we get $ab = 1 \implies bab = b \implies (ba - 1)b = 0$. Since $b\neq 0$, we'd be done if $R$ is a domain. Even if $R$ is not a domain, I think we can show that we must have $ba - 1 = 0$, although my approach doesn't work.

Define an $R$-module homomorphism $F: R \rightarrow R$ by $F(r):= rb$. Suppose that $a\in A_i$ for some $i$. Then $F$ is non-zero on $A_i$, since $F(a) = ab = 1 \neq 0$. (Here's where we use that $a \in A_i$.) If we restrict $F$ to $A_i$, the kernel of this restriction is proper (since $F$ is non-zero) and therefore the kernel must be $\{0\}$ (because $A_i$ is minimal). Therefore the restriction of $F$ to $A_i$ is one-to-one.

Note that $F(ba - 1) = (ba - 1)b = 0$ as mentioned above. Since $F$ is one-to-one, we must have $ba - 1 = 0$, as required.

Problem with this last argument: we only know that $F$ is one-to-one on $A_i$, but we don't know that $ba - 1 \in A_i$. In fact, if $ba - 1 \in A_i$, then $1 = ba + -(ba - 1) \in A_i$ since $a \in A_i$. Thus $A_i$ is not proper, so $_RR = {_R}A_i$ is a simple $R$-module, which means that $R$ is a division ring (it's easy to prove that "$_RR$ simple $\implies$ $R$ is a division ring"). Since division rings are clearly Dedekind-finite, the problem is trivial in this case.

Anyway, so my attempt using a homomorphism $F$ doesn't really work. But it might be possible to salvage the approach by modifying it slightly. Even then, this would only give the result when $a \in A_i$ for some $i$, but hopefully this special case can be used to prove the more general claim.

Reminder that I want to prove that semi-simple rings are Dedekind-finite without using the Artin-Wedderburn theorem.

Cheers!

答:
As $R$ is semi-simple, $R$ is noetherian. If $ab = 1$, the map $F \colon R \to R$, $F(r) = rb$ is onto. We will show that it is an isomorphism. Let $I_j = \ker F^j$. As $R$ is noetherian, there is an $j$ such that $I_{j+1} = I_j$. Now suppose $F(a) = 0$, then $a = F^j(a')$ for some $a' \in R$, as $R$ is onto, so
$$ 0 = F(a) = F^{j+1}(a') $$
hence $a' \in I_{j+1} = I_j$, so $a = F^j(a') = 0$. Hence $F$ is one-to-one. As you wrote $F(ba - 1) = 0$ now gives $ba = 1$ and $R$ is Dedekind-finite.</p>


问:
Solving an AAA triangle using circles
I was asked to find the angles of an isosceles triangle by circumscribing a circle around it, then drawing another circle with the tip of the triangle as its center, and a radius of half of the leg.

I can see that the ratio of AD to DE is the same as AB to BC, but how does this help me know the angles of the triangle? The lengths are 10 for the legs and 6 for the base. I know I can use other methods to find the angles, but how do I use this method - meaning, how can I specifically use the circles to find the angles? I know the radius of the upper circle to be $\frac{1}{2} AB$, and that $DE=\frac12 BC$.

答: Hint. In the upper circle you have the information you need to find the $\sin$ of half the central angle.

Title: Solving an AAA triangle using circles


问:
$\Gamma: \mathcal C \to \text{Fun}(\mathcal Z, \mathcal C)$ has a left adjoint iff all $F \in \text{Fun}(\mathcal Z, \mathcal C)$ has a colimit.
Let $\mathcal Z$ be a small category and $\mathcal C$ any category. We then consider the category $\text{Fun}(\mathcal Z, \mathcal C)$ the category of functors from $\mathcal Z$ to $\mathcal C$ and the functor $\Gamma: \mathcal C \to \text{Fun}(\mathcal Z, \mathcal C)$ such that
$$
\Gamma(C):
\begin{pmatrix}
\mathcal Z &\longrightarrow &\mathcal C\\
Z & \longmapsto & C\\
f & \longmapsto & 1_C
\end{pmatrix}
$$
where $f: Z \to Z'$ and $1_C$ is the identity morphism on $C$.

I am trying to show that $\Gamma$ has a left adjoint if and only if the colimit of every functor $F: \mathcal Z \to \mathcal C$. The "if" part is not too hard but I have some difficulties to show the "only if" part.

Here is my idea: Suppose that there is $\Omega : \text{Fun}(\mathcal Z, \mathcal C) \to \mathcal C$ such that
$$\theta_{F, C}: \text{Hom}_\mathcal C(\Omega(F), C)\cong \text{Nat}(F, \Gamma(C)).$$
For each $\alpha \in \text{Nat}(F, \Gamma(C))$ we can associate a cocone, i.e. for $f: Z \to Z'$ the following diagram commutes
$$
\begin{array}{ccc} F(Z)&\xrightarrow{F(f)} &F(Z')\\ \searrow&&\swarrow\\
&\Gamma(C)(Z) = C = \Gamma(C)(Z')& \end{array}
$$
where the arrows $F(Z) \to C$ and $F(Z') \to C$ are given by $\alpha_Z$ and $\alpha_{Z'}$ respectively. Because of the above isomorphism, we can associate to $\alpha$ a unique morphism $\theta^{-1}_{F, C}(\alpha) =u: \Omega(F) \to C$ so $\Omega(F)$ is a good candidate to be the colimit of $F$. We just have to find a family of morphism $\mu_Z: F(Z) \to \Omega(F)$ such that $\alpha_Z = u \circ \mu_Z$. My guess is that this family of morphism is given by the unit of the adjunction $\eta_F \in \text{Nat}(F, \Gamma(\Omega(F)))$ but I am not able to show that
$$\alpha_Z = u \circ (\eta_F)_Z = \theta^{-1}_{F, C}(\alpha) \circ (\theta_{F, \Omega(F)}(1_{\Omega(F)}))_Z$$
where the last equality comes from the expression of $\eta_F$ in term of $\theta_{F, \Omega(F)}$. Does it seem right ? Does anyone know how to conclude ?

答:
Good so far!

You haven't explicitly written down how $\Gamma$ acts on morphisms, but it is clear that for $g\in\hom_\mathcal{C}(C,D)$, $\Gamma(g): \Gamma(C)\to\Gamma(D)$ is the natural transformation with $\Gamma(g)_Z = g$ for all objects $Z\in\mathcal{Z}$.  In particular, $u = \Gamma(u)_Z$.  From here, $u\circ(\eta_F)_Z = \Gamma(u)_Z\circ(\eta_F)_Z = (\Gamma(u)\circ\eta_F)_Z$, so showing that $u\circ(\eta_F)_Z = \alpha_Z$ for every $Z\in\mathcal{Z}$ is the same as simply showing that $\Gamma(u)\circ\eta_F = \alpha$.  This statement is true by the adjunction identity $\Gamma(g)\circ\eta = \theta(g)$, true for all $g$ and $g = u = \theta^{-1}(\alpha)$ in particular.


问:
Determine the interval of the result of a definite integral
For the integral
$$
\int_1^2 \frac {3^x + 2}{3^{2x} + 3^x} dx
$$
choose the interval of its result: $(-\infty, 0], (0, \frac 12], (\frac 12, 1], (1, 3], (3, \infty)$. According to the author of this task you do not have to compute the integral itself to determine the interval. I have managed to compute the integral but I am unsure how to determine the interval without computing it. How can I do that?

答: Since the function $f(x)=\frac{x+2}{x^2+x}$ is decreasing for all  $x\ge 1$ we have $$g(x)=\frac{3^x+2}{3^{2x}+3^x} \implies g(x) \le g(1) =5/12 \space \forall x\in[1,3]$$ thus $$0\le \int_1^2\frac{3^x+2}{3^{2x}+3^x}\le \frac{5}{12}(2-1)< \frac{1}{2}$$


Body:
$$A=\left(
\begin{array}{ccc}
 a & b & c \\
 d & e & f \\
 g & h & i \\
\end{array}
\right)$$
$$A\cdot (2,1,0)=(3,2,1),A\cdot (4,2,1)=(4,1,0),A\cdot (2,2,1)=(2,3,3),A\cdot (2,2,2)=(0,0,1)$$
$$\begin{cases}
2 a + b=3\\
2 d + e=2\\
2 g + h=1\\
4 a + 2 b + c=4\\
4 d + 2 e + f=2\\
4 g + 2 h + i=1\\
2 a + 2 b + c=2\\
2 d + 2 e + f=3\\
2 g + 2 h + i=3\\
\end{cases}
$$
which gives
$$a= 1,b= 1,c= -2,d= -1,e= 4,f= -3,g= -\frac{3}{2},h= 4,i= -2$$
These results satisfy the fourth set of conditions
$$\begin{cases}
2 a + 2 b + 2 c=0\\
2 d + 2 e + 2 f=0\\
2 g + 2 h + 2 i=1\\
\end{cases}
$$

Therefore the matrix is
$$
A=\left(
\begin{array}{ccc}
 1 & 1 & -2 \\
 -1 & 4 & -3 \\
 -\frac{3}{2} & 4 & -2 \\
\end{array}
\right)
$$


问:
If $\left| f'(x) \right| \leq A |f(x)|^\beta $ then f is a constant function
Problem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \ge 0$ and $\beta \ge 1$ such that the inequality

$$\left| f'(x) \right| \leq A \left| f(x) \right|^\beta $$

holds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.

My attempted solution

We shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \neq 0$.

Put $S:= \left\lbrace x \right. $ such that $a \le x \le b $ and $f(x) \neq 0 \left. \right\rbrace$ and $c:=\inf S$ (the infimum value of set $S$). If $c=b$ then $ f(x) = 0$   $\forall  x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \leq c c$ and close enough to $c$ such that $|f(x)| \le 1$ and $$\tag{$\star$} A(x-c) \le \frac12\quad\text{for all }x\text{ with }c \le x \le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \le t \le d$ such that $|f(t)| = \max_{c \le x \le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\epsilon>0$, there is some $c<m<c+\epsilon$ satisfying $f(m) \neq 0$. This observation, together with the definition of $t$, gives us $$\tag{$\star\!\star$}f(t) \neq 0,$$ $t \neq c$ and so $c<t$. Applying the Lagrange theorem gives:

$$ |f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \leq|t-c|A|f(u)|^\beta $$ where $c<u<t\le d$. By $(\star)$:
$$ |t-c|A|f(u)|^\beta \le \frac12 |f(u)|^\beta \leq \frac12 |f(u)| $$

Combining the two inequalities and noting that $|f(t)| \ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\star\star)$.

Question

I hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.

I like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.

Thank you.

答:
Let $B = \{x\in [a,b]: |f(x)|=0\}$. By continuity, this set is closed. Since $f(0) = 0$, the set is non-empty. We show that it is also open.

Let $x\in B$, i.e. $f(x) =0$. By continuity, there exists $0<r<(2A)^{-1}$, such that $|f(y)|<\frac 12$ for all $y\in B_r(x)$. Fix such a $y\in B_r(x)$. We want to show that $f(y) =0$.

By the mean value inequality, we have $$|f(y)| = |f(y) - f(x)| \le |f'(\xi_0)| |y-x| \le |f'(\xi_0)|r$$
for some $\xi_0\in B_r(x)$ lying on the segment between $x$ and $y$. By assumption, we have $|f'(\xi_0)|\le A |f(\xi_0)|^\beta$, hence $$|f(y)|\le |f'(\xi_0)|r \le A|f(\xi_0)|^\beta (2A)^{-1} \le \frac 12|f(\xi_0)|^\beta.$$ By the same argument applied to $\xi_0\in B_r(x)$ instead of $y$, we see that there exists $\xi_1\in B_r(x)$, such that $|f(\xi_0)|\le \frac 12|f(\xi_1)|^\beta$. Iterating yields a sequence $\xi_n \in B_r(x)$, such that $|f(\xi_{n})|\le \frac 12|f(\xi_{n+1})|^\beta$ for all $n$. Note that by our choice of $r$, we have $|f(\xi_{n})|\le \frac 12$ for all $n$. We conclude that
$$|f(y)|\le \frac 12 |f(\xi_0)|^\beta \le \frac 1{2^2}|f(\xi_1)|^{\beta^2} \le \frac{1}{2^3}|f(\xi_2)|^{\beta^3}\le \dots \le \frac{1}{2^n}|f(\xi_{n-1})|^{\beta^n}\le \frac 1{2^{n+\beta^n}},$$
for all $n\in \mathbb N$. Letting $n\to \infty$ shows that $|f(y)| = 0$ (using also $\beta \ge 1$). This implies that $B_r(x)\subset B$, hence $B$ is open.

Since $[a,b]$ is connected, it follows that $B$ must be all of $[a,b]$, hence $f= 0$.


问:
Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$
Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:

Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
      &\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}

Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$

Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$

So $x = 13$ or $x = 11$.

Is it correct that way?

答:
At no point in your proof did you split the congruence into $\bmod5$ and $\bmod7$, so you missed two solutions as I will show below:
$$3x^2-2x+4=0\bmod5\implies x^2+x+3\equiv0\bmod5\implies x\equiv\{1,3\}\bmod5$$
$$3x^2-2x+2\equiv0\bmod7\implies x^2+4x+3\equiv0\bmod7\implies x\equiv\{-1,-3\}\bmod7$$
Combining with the Chinese remainder theorem this gives $x\equiv\{6,11,13,18\}\bmod35$.


问:
Simplifying a trig relation
My physics teacher wanted to prove Young's equation (my problem is pure mathematics), and in the derivation, he got the following expression:
$$\tag{1}
\begin{align}
\left(\gamma_{s l}-\gamma_{s o}\right) \cdot\left(\sin ^{2} \theta+\cos \theta\left(-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)\right) \\[6pt]
\quad+\;\gamma \cdot\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)=0
\end{align}$$
Which he says leads to:
$$\tag{2}
\gamma_{s l}+\gamma \cos \theta=\gamma_{s o}$$
It is easy to see from eq. (1) and (2) that we must have:
$$\tag{3}
\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)\left(\sin ^{2}(\theta)-\cos \theta \frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)^{-1}=\cos \theta
$$
But I have a hard time showing this (my trig-skills might not be up to par). Will someone help me show this?

答:
The coefficient of $\gamma$ can be simplified as,

$\begin{align}(1-\cos\theta)\frac{2+2\cos\theta - 2- \cos\theta}{1+\cos\theta} = \boxed{\cos\theta\ \left[\frac{1-\cos\theta}{1+\cos\theta}\right]} \end{align}$

Now, the coefficient of $\gamma_{sl} - \gamma_{so}$ can be simplified as,

$\begin{align}\sin^2\theta -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) &= (1-\cos\theta)(1+\cos\theta) -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) \\
&= (1-\cos\theta)\left[\frac{(1+\cos^2\theta +2\cos\theta) - (2\cos\theta+\cos^2\theta)}{1+\cos\theta}\right] \\& = \boxed{\frac{1-\cos\theta}{1+\cos\theta}}\end{align}$

Now the coefficients differ by a factor $\cos\theta$.

Assuming the coefficient $\frac{1-\cos\theta}{1+\cos\theta}$ is non-zero,

$\gamma_{sl} - \gamma_{so} + \gamma \cos\theta = 0$

or

$$\boxed{\gamma_{so} = \gamma_{sl} + \gamma \cos\theta}$$

Title: Simplifying a trig relation


Title: Determine the interval of the result of a definite integral

Body:
For the integral
$$
\int_1^2 \frac {3^x + 2}{3^{2x} + 3^x} dx
$$
choose the interval of its result: $(-\infty, 0], (0, \frac 12], (\frac 12, 1], (1, 3], (3, \infty)$. According to the author of this task you do not have to compute the integral itself to determine the interval. I have managed to compute the integral but I am unsure how to determine the interval without computing it. How can I do that?

Body:
$$f(x) = \frac {3^x + 2}{3^{2x} + 3^x} $$
$$ f'(x) = -\dfrac{\ln\left(3\right)\left(3^{2x}+4{\cdot}3^x+2\right)}{3^x\left(3^x+1\right)^2} < 0 ~\forall  ~ x \in \mathbb{R}$$
Thus, $f(x)$ is a decreasing function.
$$ \implies f(1) >f(x) > f(2)$$
$$\implies \dfrac{5}{12} >f(x) > \dfrac{11}{90}$$
$$\implies \int_1^2 \dfrac{5}{12} dx > \int_1^2 f(x) dx > \int_1^2 \dfrac{11}{90} dx$$
$$\implies \dfrac{5}{12} > \int_1^2 f(x) dx > \dfrac{11}{90}$$
$$ \therefore I \in \left(0, \dfrac12\right] $$


问:
Why does averaging over orthogonal transformations mimic averaging over the sphere?

The group of orthogonal transformations carries an invariant probability measure. This means that we can average a function over the group in a natural way. In particular, if $f$ is a function on the sphere and $θ$ is some point on the sphere, the average over orthogonal $U$ of the value $f(Uθ)$ is just the average of $f$ on the sphere: averaging over $U$ mimics averaging over the sphere:
$$\text{avg}_U f(Uθ) = \int_{S^{n-1}}f (\phi) dσ(\phi)$$

See Pg. $22-23$ of these notes for context.

Is there an easy way to understand/prove this? I'm not able to figure out what it means!

My intuition:

Consider $\theta$ on the sphere $S^{n-1}$. Orthogonal transformations rotate this without changing the length. All orthogonal transformations send this point to someplace on the sphere, and for any two points on the sphere, we can always find an orthogonal transformation that relates them. Hence, averaging over all orthogonal transformations is the same as averaging over the sphere.

Is that correct?

答:
Let $\sigma$ be the usual rotation-invariant surface measure on the sphere $S^{n-1}$.  We know that in fact $\sigma$ is the only rotation-invariant measure on the sphere.  Now, if we let $m$ denote the rotation-invariant (Haar) measure on the group $SO(n)$, let us define a new measure $\nu$ on the sphere, defined by duality via
$$
\int_{S^{n-1}} f \,d\nu := \int_{SO(n)} f(U \theta) \,dm(U)
$$
for $f: S^{n-1} \to \mathbb{C}$ continuous, where $\theta \in S^{n-1}$ is a fixed arbitrary point.  Then the rotation invariance of $m$ on $SO(n)$ implies that $\nu$ is also a rotation-invariant measure on $S^{n-1}$:
$$
\int f \circ M \,d\nu = \int_{SO(n)} f(MU \theta) \,dm(U) = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\nu
$$
for any $M \in SO(n)$.

Therefore we conclude that $\nu$ and $\sigma$ are the same measure, i.e.
$$
\int_{S^{n-1}} f \,d\nu = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\sigma
$$


问:
Convergence of $\sum \frac{a_n}{(a_1+\ldots+a_n)^2}$
Assume that $0 < a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that
$$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} < \infty $$
?

I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon > 0$ this is obvious.

Any idea? Thanks.

答: Recall how do you prove $\sum \frac{1}{n^2}$ converges.</p>


答: In geometry, we say that two objects are similar if they have the same shape; more precisely, if one shape can be turned into the other by a combination of enlargement, rotation, and reflection. None of these transformations alter the ratios between the lengths of the sides of the triangles. Hence, two triangles can only be similar if they have the same ratios of lengths. And if these ratios are the same, then the two triangles are similar.

Title: Ratios of similar triangles

Body: This is a basic question, and based on my thoughts I really think this is true, but I just want some verification. If I have similar triangles, are the ratios between two sides of one triangle equal to the ratio of the corresponding sides on the other triangle?</p>


问:
Prove that a matrix doesn't exist with a given characteristic polynomial
I've been trying to understand this following issue:

Show that a matrix A, where $\rho(A) = 3$ doesn't exist,

Where the characteristic polynomial is: $p(x) = x^7 - x^5 + x^3$

I've disassembled the polynomial to be: $p(x) = x^3(x^2(x^2 -1) + 1)$. It seems like it cannot be disassembled over $\mathbb R$.

I'm not sure how does the information of the matrix being from rank 3 plays in, if there aren't any eigenvalues to the characteristic polynomial.

答: Hint: Note that the dimension of the kernel (AKA nullspace) of $A$ is given by $7 - \rho(A) = 4$. However, the nullspace of $A$ is also an eigenspace of $A$.


问:
When is a Markov of order 1 a Martingale?
I have two questions and I am very confused about the concepts

Can a Markov process of order one also be a a Martingale?
Is any Markov process of order one also a Martingale?
For 1. I would say yes, since a Markov of order one is also an AR(1) it would be a martingale if c=0 and phi=1. However, I am very confused about the second. My answer for 2 would be no, but I cannot explain it properly.

Can anyone maybe help me?

答:
A first-order Markov chain can be a martingale, but need not be.  In order for it to be a martingale, you need $\mathbb E[X_{n+1} | X_n] = X_n$.  So for example a simple unbiased random walk (equal probability of steps of $+1$ and $-1$) is a martingale.  If you want it to be a finite Markov chain on states $a, \ldots, b$, make $a$ and $b$ absorbing.
On the other hand, a biased random walk is not a martingale.


问:
Extreme values of $f(x, y)=\frac{(x+y)^{2}}{2}+\frac{(x-y)^{3}}{3}$ on $D=\left\{(x, y) \in \mathbb{R}^{2}:|y| \leq 1-|x|\right\}$

Extreme values of $f(x, y)=\frac{(x+y)^{2}}{2}+\frac{(x-y)^{3}}{3}$ on $D=\left\{(x, y) \in \mathbb{R}^{2}:|y| \leq 1-|x|\right\}$

I resolved by taking the partial derivatives and computing the Hessian matrix, so $(0,0)$ is a saddle point.

For the first edge I tried to parametrize $D$
$$\begin{cases}x=1-t\\y=t\end{cases}$$
and resolve for $f(x,y)$:
$$f(t) = \frac{1}{2} + \frac{(2t-1)^3}{3}$$
$$f'(t)=4t-2=0 \implies t=\frac{1}{2}$$
and follows that $\left(\frac{1}{2},\frac{1}{2}\right)$ could be and extreme.

It's clear that $(1,0)$ is a max value, but I can't find it.
Can someone help me out in this?

答: Substitute $u=x+y,v=x-y$ to transform the problem into finding the extreme points of $\frac{u^2}2+\frac{v^3}3$ on $u,v\in[-1,1]$. Then it is clear that the maxima are at $u=\pm1,v=1$ and the minimum is at $u=0,v=-1$, which when translated to the $xy$ plane gives the maxima at $x=0,y=-1$ and $x=1,y=0$ with $f(x,y)=\frac56$, and the minimum at $x=-\frac12,y=\frac12$ with $f(x,y)=-\frac13$.


答:
As other comments and solutions have noted, you need to consider that 35 is composite.  But nearly all of your work is correct and completing the square is a great strategy for arbitrary quadratic congruences.

$$(3x-4)(3x+2)\equiv0\pmod{35}$$
This is correct.  But then you need to go from there to
$$(3x-4)(3x+2)\equiv0\pmod 5 \qquad\text{ and }\qquad (3x-4)(3x+2)\equiv0\pmod 7$$

we'll tackle those one at a time.

$$3x-4\equiv0\pmod 5\qquad\text{ or }\qquad 3x+2\equiv0\pmod 5\\
3x\equiv4\pmod 5\qquad\text{ or }\qquad 3x\equiv3\pmod 5\\
x\equiv3\pmod 5\qquad\text{ or }\qquad x\equiv1\pmod 5\\
$$

$$3x-4\equiv0\pmod 7\qquad\text{ or }\qquad 3x+2\equiv0\pmod 7\\
3x\equiv4\pmod 7\qquad\text{ or }\qquad 3x\equiv5\pmod 7\\
x\equiv6\pmod 7\qquad\text{ or }\qquad x\equiv4\pmod 7\\
$$

Pairing those solutions together with the CRT (Chinese Remainder Theorem) gives you the four solutions others have mentioned.

Title: Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$

Body:
Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:

Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
      &\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}

Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$

Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$

$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$

So $x = 13$ or $x = 11$.

Is it correct that way?


问:
Does "triangle" in English exclude degenerate triangles?
Just for fun read few problems on the projeteuler.net website.

Number 276 found interesting:

Consider the triangles with integer sides a, b and c with a ≤ b ≤ c.
  An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1.
  How many primitive integer sided triangles exist with a perimeter not
  exceeding 10 000 000?

Question:

Does the definition "triangle" in English also enforces the following condition?

a + b > c

答:
A CW answer to remove this question from the Unanswered queue.

Yes, "triangle" can be taken to mean "non-degenerate triangle" unless explicitly stated otherwise. As far as I know, this convention is universal.


问:
Analyzing functions with absolute values
I have a question. I am having trouble analyzing functions with absolute values, for example the following:

$$\frac{\ln{|x|}}{x}-x+1$$

Now, what I usually would do if I got a function with absolute values is try to break it down, for example if I had simply $|x|$ I know it's $ x, x\ge0$ and $-x, x<0$. But since I have $\ln|x|$ I don't know how to proceed.

Firstly, I know that the function is defined for all real x except $0$. I know that the function is zero at $-1, 1$, but if I try to calculate the sign of the function (where it's above and below the $x$-asis) I have trouble because I don't know how to break $\ln{|x|}$ into smaller parts.

Can anyone help?

答:
Hint: $\log|-5|=\log|5|$, for instance. The graph is symmetrical:

We know that $\log x$ is negative if $0<x<1$. Hence, $\log |x|$ is negative if...


问:
Is the number of Lebesgue covering dimensions a topological property?
Is it correct to characterize the number of dimensions that a topological space has as a 'topological property'? By 'dimensions,' I have in mind Lebesgue covering dimensions.

If the answer is yes, then why? If the answer is no, then what kind of a property is the number of dimensions?

答: There are many different definitions of dimension in different contexts. Some are topological, most are not.  Look up: topological dimension.</p>


问:
Morphism of quasi affine varieties giving algebra homomorphism and back.
$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$
$\newcommand{\A}{\mathbb A}$

I am new to algebraic geometry and in the following I ask two related questions that have come to my mind given the material I have read so far.
I believe the notation and terminology I have used is standard.
However, I have supplied most of the relevant definitions and notations in the appendix.
Please feel free to correct and inaccuracies.

In what follows $k$ is an algebraically closed field.

Question 1

Let $X$ and $Y$ be affine varieties.
It is a fact that a morphism $\vp:X\to Y$ is an isomorphism if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras.

Question.
Can we extend this fact to the case when $X$ and $Y$ are quasi affine varieties, that is, is a morphism $\vp:X\to Y$ an isomorphism if and only if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras?

The problem I am facing in proving this is the following.
When $X$ and $Y$ are affine varieties in $\A^m$ and $\A^n$ respectively, then we have $\mr O(X) = k[x_1, \ldots, x_m]/\mr I(X)$ and $\mr O(Y) = k[y_1, \ldots, y_n]/\mr I(Y)$.
Now once we have a $k$-algebra homomorphism $\mr O(Y)\to \mr O(X)$, we get a morphism of varieties $X\to Y$ by using the induced map $\text{MaxSpec}(\mr O(X))\to \text{MaxSpec}(\mr O(Y))$ and then use the nullstellensatz to get a map $X\to Y$.

However, when $X$ and $Y$ are quasi affine varieties in $\A^m$ and $\A^n$ respectively, then one cannot think of $\mr O(X)$ or as a quotient of $k[x_1, \ldots, x_m]$ (and similarly for $\mr O(Y)$).
For instance, let $X=\A^1\setminus \set{0}$. Then $\mr O(X) = k[x]_{x}$ which is not a quotient of $k[x]$ but of $k[x, y]$.

So I am unable to identify $\text{MaxSpec}(\mr O(X))$ and $\text{MaxSpec}(\mr O(Y))$ as subsets of $\A^m$ and $\A^n$.

Question 2

Let $\mc V$ be the category of affine varieties and $\mc A$ be the category of finitely generated $k$-algebras that are integral domains.
We get an equivalence between two categories by associating to an affine variety its coordinate ring and to a morphism its corresponding map between the algebras, namely the pullback map.

Question.
Is there such a statement for the category of quasi affine varieties, that is, is there a nice algebraic category to which the category of quasi affine varieties equivalent?

Question.
Suppose we consider the category of subsets of $\A^n$, $n\geq 0$.
We can still define $\mr O(X)$ for any such subset and form a category with morphisms as maps which pullback regular functions to regular functions.
Is this category useful in any way? Is there a reason why it is not studied?

Appendix

Let $k$ be an algebraically closed field.
We write $\A^n$ to denote $k^n$ with the Zariski topology.
An affine variety is an irreducible closed subset of some $\A^n$.

Given an affine variety $X$ in $\A^n$, the coordinate ring $\mr A(X)$ of $X$ is defined as $k[x_1, \ldots, x_n]/\mr I(X)$, where $\mr I(X)$ is the set of all the polynomials in $k[x_1, \ldots, x_n]$ which vanish on each point of $X$.
It turns out that $\mr I(X)$ is a prime ideal.

A quasi affine variety is any open subset of an affine variety.
A regular function on a quasi affine variety $Y\subseteq \A^n$ is a map $\vp:Y\to k$ which can be locally written as a quotient of polynomials in $k[x_1, \ldots, x_n]$.
The set of all the regular functions on $Y$ is denoted by $\mr O(Y)$.
It turns out that $\mr O(Y)$ has a natural $k$-algebra structure.
It is a fact that if $X$ is an affine variety, then $\mr A(X)$ can be naturally identified with $\mr O(X)$.

A morphism of quasi affine varieties $X$ and $Y$ is a map $\vp:X\to Y$ which is continuous (in the Zariski topology) such that $\vp$ pulls back regular functions on $Y$ to regular functions on $X$.
Given a morphism $\vp:X\to Y$, we get the pullback map $\vp^*:\mr O(Y)\to \mr O(X)$, and it is easy to check that $\vp^*$ is a $k$-algebra homomorphism.

答:
The problem with quasi-affine varieties is that, well, they are not affine. You should think of an affine variety as one that is ``determined by its global regular functions'' and this is not the case for arbitrary quasi-affine varieties.

Here is a classic example: let $X = \mathbb{A}^2 \setminus \{ (0,0) \}$. Then I claim $\mathcal{O}(X) \cong k[x,y]$ where the map $\mathcal{O}(\mathbb{A}^2) \to \mathcal{O}(X)$ induced by the inclusion $X \subset \mathbb{A}^2$ is an isomorphism. I highly recommend you try to prove this if you are not familiar with the example. It is often thought of as an algebraic version of the Harthogs's phenomenon.

This example shows that the answer to Question 1 is no. In Question 2, I don't know of any purely algebraic description and the category you suggest is useful it just does not give information about the quasi-affine varieties $X$ rather it tells you about the affine varieties $\mathrm{Spec}{(\mathcal{O}(X))}$.

Note: when you say subsets, I assume you mean locally closed subsets for the Zariski topology since an arbitrary subset of $\mathbb{A}^n$ need not have the structure of an algebraic variety.


问:
Improvement of $W^{1,p}$ regularity of a elliptic equation solution.
I'm looking for some reference for results like
$$ \mbox{div}(A(x) \nabla u) = 0, \ \ u \in H^1=W^{1,2} \Rightarrow u \in W^{1,p}, p>2 $$
where $A(x)$ is elliptic, this is, $Id\lambda \le A(x) \le \Lambda Id $ with $0< \lambda<\Lambda$  are constant matrices and $Id$ is the identity matrix.
I do not know if this is exactly true. Then if you know something like this, as adding hypothesis, please tell me or send me. I will be very grateful.</p>

答:
For the right hand side data is zero, the regularity of the solution depends on: the smoothness of the domain's boundary which is related to the smoothness of the boundary condition. Also certain properties of $A(x)$ (smoothness, monotonicity, and/or being in VMO/BMO).

Rewriting the equation as $-\mathrm{div}(A(x)\nabla u) = \mathrm{div}\mathbf{f}$ for some $\mathbf{f}$, we wanna check if there is such estimate:
$$
\|\nabla u\|_{L^p(\Omega)} \leq c\|\mathbf{f} \|_{L^p(\Omega)}, \tag{1}
$$
for some/all $p>2$, or some other similar estimates that involve the boundary data.

Caffarelli's paper is THE guide, if $A$ is sufficiently "close" to identity, then $u\in W^{1,p}$ for $p$ related to how close $A$ is to the identity.

If $A(x)$ is discontinuous, only some $p$ suffices for estimate (1), please see Theorem 1 to 3 in An $L^p$-estimate for the gradient of solutions of second order elliptic divergence  equations.

For an example of discontinuous and not monotone $A(x)$ would result very singular solution, please refer to Kellogg's paper on elliptic equations with intersecting interfaces, see (3.2) for that example, we can manipulate the coefficients such that $u\in H^{1+\epsilon}$, and $\epsilon$ is very small, there is much space left for you to trade the differentiability for integrability.


问:
How many subgroup has $\mathbb{Z_p}$ with $p$ prime
My try is how $\mathbb{Z_p} $ is a group of order $p$ prime by theorem we know thata group of prime order has only two subgroups: itself and ${e}$

is this right?

And how can i draw de hasse diagram of all subgroups of $\mathbb{Z_p}$ I think its just $Z_p$ and $e$ united by a line

答: Hint : How many elements of order $p$ are in $\mathbb{Z}_p$ ? And how many will create the same subgroup ?


答:
$p(x)=x^3(x^4-x^2+1)$ which has a triple root at $0$ and four simple roots at $$\pm\sqrt{\frac{1+\pm i\sqrt3}2}$$ (The second factor is a quadratic in $x^2$.)

Since eigenvectors corresponding to distinct eigenvalues are linearly independent, $A$, considered as a complex matrix, has rank $4$.  But the rank of a matrix doesn't change if we make the ground field bigger, so the rank of A is $4$ over the reals also.

Title: Prove that a matrix doesn't exist with a given characteristic polynomial

Body:
I've been trying to understand this following issue:

Show that a matrix A, where $\rho(A) = 3$ doesn't exist,

Where the characteristic polynomial is: $p(x) = x^7 - x^5 + x^3$

I've disassembled the polynomial to be: $p(x) = x^3(x^2(x^2 -1) + 1)$. It seems like it cannot be disassembled over $\mathbb R$.

I'm not sure how does the information of the matrix being from rank 3 plays in, if there aren't any eigenvalues to the characteristic polynomial.


问:
Lagrange multipliers method: this a maximum point or a minimum point?
Consider the function
$$f(x, y)=x^2 y -2xy^2 +4.$$
Find the absolute maxima and minima of $f$ constrained to $2x-4y+1=0$ with $x\in [-1, 1]$.

My solution: the Weierstrass theorem guarantees that the function has absolute minima and maxima. The Lagrange multipliers method has allowed me to find only the  point $(-1/4, 1/8)$.
My question is: how to understand if it is a maximum or a minimum? When I have two or more points, I evaluate the function in those points and the greatest value corrispond to the maximum point and the lowe to the minimum point.

In this case, how to proceed? Thank you in advance!

EDIT: Plotted function.

答:
Here's the surface and the constraint line:

Although the curvature on the constraint line is low, there is a single maximum, as is to be expected in a quadratic.  The minima occur at the boundaries of the region... and Lagrangian methods do not find those.

If you're given a bounded region, you may have to check boundaries to find maxima and minima.  After all, consider the trivial problem of finding the maxima and minima of $f(x)=x$ in the range $x \in [0,1]$.  Even basic calculus will not yield the extrema.  You must check boundaries.  The same sometimes occurs with Lagrangian methods.


问:
Problem with the definition of the degree of a morphism between curves
I need your help because there is something that I don't understand in the definition of the degree of a morphism of curves. Here it is:
"Let $\phi: C_1 \longrightarrow C_2$ be a morphism between two (smooth) projective curves. We define the degree of $\phi$ to be 0 if it is constant and $deg\phi=[k(C_1):\phi^*k(C_2])$ otherwise."

I don't understand why the curves can be singular. The point is, the degree is only defined by the fraction fields of the curves, but why does the smoothness have no "impact" on them?

Thanks.

答:
Here I assume that the ground field is algebraically closed for simplicity but this is not essential. When a morphism of curves $\phi : C_1 \to C_2$ is finite and flat then you can show that ``the degree of each fiber'' equals the degree defined as $\deg{\phi} = [k(C_1) : k(C_2)]$. Explicitly, this means for each $Q \in C_2$,
$$ \sum_{P \in f^{-1}(Q)} e_P = \deg{\phi} $$
where $e_P$ is the ramification degree at $P$. You can think of $[k(C_1) : k(C_2)]$  as the degree of the generic fiber.

Now, any nonconstant morphism of projective curves is automatically finite and if $C_2$ is smooth it is also automatically flat. However, when $C_2$ is not smooth it does not need to be flat and accordingly the above formula may not hold.

For example, let $X$ be the nodal cubic $y^2 = x^2(x - 1)$ in $\mathbb{P}^2$. Then the normalization is a map $\nu : \mathbb{P}^1 \to X$ defined by $t \mapsto (t^2 + 1, t(t^2 + 1))$. You can show this morphism is an isomorphism away from the node at $(0,0)$ so $\deg{\nu} = 1$. However, $\pm i$ both map to $(0,0)$ so the degree over the node is $2$.


问:
Topological entropy is finite?
Let $f:X\rightarrow X$ be a continuous function on a compact metric space $X$ with distance $d$. For each $n\in \mathbb{N}$ we can introduce the distance on $X$ ,$d_n(x,y)=\max_{k=0,...,n-1}d(f^k(x),f^k(y))$, and then we define the topological entropy of $f$ to be $h(f)=\lim_{\epsilon\rightarrow 0}\lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)$, where $N(n,\epsilon)$ is the smallest number of balls of radius $\epsilon$ needed to cover $X$. Now it's clear that $N(n,\epsilon)$ is always finite since $X$ is compact. It also makes sense to me that the function $\epsilon\rightarrow \lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)$ is non-increasing , what I don't get is why this gives us that the topological entropy acually exists , couldn't I have something like $\lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)=\frac{1}{\epsilon}$?  Or can we have the topological entropy to be $\infty$?

Any help is appreciated.

答:
Let me give a partial answer for now. Maybe I can think of a complete answer soonish. (see edit below)

Actually, I do not recall anyone claiming that the topological entropy is always finite. It is certainly well-defined as a number in $\mathbb{R} \cup \{\infty\}$ since the limit in $\epsilon$ is monotone as you pointed out, but it may very well be infinite.

Unfortunately, I cannot think of an example with infinite entropy right now, but I am quite sure there must be one. (see edit below)

Instead, let me mention that there are some hypothesis that can guarantee finiteness of the entropy.
For instance, it suffices if your system is Lipschitz continuous and your space $X$ is somewhat nice.
Here, "nice" in the sense that the "ball dimension"
$$ BD(X) = \limsup_{\epsilon \rightarrow} \frac{\log(N(1,\epsilon))}{|\log(\epsilon)|} $$
is finite. If your space is, e.g., a manifold, then the ball dimension is exactly the dimension of the manifold.
Now, for a Lipschitz dynamical system with Lipschitz constant $L \geq 1$, we have $N(n,\epsilon) \leq N(1,\epsilon L^{-n})$ and, hence,
$$ \frac{1}{n} \log(N(n,\epsilon)) \leq \underbrace{\frac{|\log(\epsilon L^{-n})|}{n}}_{\rightarrow \log(L)} \underbrace{\frac{\log(N(1,\epsilon L^{-n}))}{|\log(\epsilon L^{-n})|}}_{\rightarrow BD(X)} \xrightarrow{n \rightarrow \infty}{ \log(L) BD(X)}. $$
Thus, the entropy is bounded by $\log(L)BD(X)$.

Another possible hypothesis is positive expansivitiy: if there exists a constant $\delta > 0$ such that $d_n(x,y) < \delta$ for all $n \geq 0$ implies that $x=y$, then the entropy is finite.
I will note give the details, but, given $0 < \gamma < \epsilon < \delta/2$, positive expansivity yields the existence of a constant $C(\gamma,\epsilon)$ such that for all $n \geq 0$
$$ N(n,2\gamma) \leq C(\gamma,\epsilon) N(n,\epsilon).$$
From this it follows that $\limsup_{n \rightarrow \infty} \frac{1}{n}\log(N(n,\epsilon))$ is independent of $\epsilon$ for $\epsilon < \delta/2$ and, hence, $h(f)$ is finite.

$\bf{EDIT:}$ The dynamical system on the circle $S^1 \rightarrow S^1$, $x \mapsto qx ~\mathrm{ mod }~ 1$ has entropy $\log(q)$.
Now consider the dynamical system $f \colon \mathbb{D} \rightarrow \mathbb{D}$ on the unit disc in the complex plane given by $f(0) = 0$ and
$$f(x) = |x| e^{\mathrm{arg}(x) * i/|x|} \qquad \text{for }x \ne 0.$$
Then restricted to each circle $\{|x| = r\}$, $r \in (0,1]$, the restricted system is conjugate to $S^1 \rightarrow S^1$, $x \mapsto x/r ~\mathrm{ mod }~ 1$.
Since the entropy of a system is greater than or equal to the entropy of the system restricted to any closed invariant subset, we conclude that $h(f) \geq \log(1/r)$ for any $r < 1$, i.e. $h(f) = \infty$.

Title: Topological entropy is finite?


问:
For which $p\in\mathbb{R}$ does $\int_{\mathbb{R}\times [0,1]}\frac{1}{\left(1+x^2+y^2\right)^p}d\mathcal{L}^2(x,y)$ exist?
For which $p\in\mathbb{R}$ and if so, why does $\int_{\mathbb{R}\times [0,1]}\frac{1}{\left(1+x^2+y^2\right)^p}d\mathcal{L}^2(x,y)$ exist?

Considering the integrand, the natural approach here would be to use polar coordinates but the integration domain doesn't seem to allow that. I have therefore tried using Fubini but I don't know how to compute $\int_0^1 1/(1+x^2+y^2)^p dy$.

How do I continue?

答:
As the integrand is non-negative then from Tonelli's theorem we have that

$$
\int_{\mathbb{R}\times [0,1]}(1+x^2+y^2)^{-p}\mathop{}\!d (x,y)=\int_{[0,1]}\int_{\mathbb{R}}((1+y^2)+x^2)^{-p}\mathop{}\!d x\mathop{}\!d y\tag1
$$
and
$$
\int_{\mathbb{R}}(C+x^2)^{-p}\mathop{}\!d x=C^{-p}\int_{\mathbb{R}}\left(1+\left(\frac{x}{\sqrt C}\right)^2\right)^{-p}\mathop{}\!d x=C^{1/2-p}\int_{\mathbb{R}}(1+z^2)^{-p}\mathop{}\!d z\tag2
$$

for any $C>0$. Now note that $(1+z^2)^{-p}\sim_\infty z^{-2p}$, therefore

$$
\int_{\mathbb{R}}(1+z^2)^{-p}\mathop{}\!d z<\infty \iff \int_{\mathbb{R}\setminus [-1,1]}z^{-2p}\mathop{}\!d z<\infty \tag3
$$
Hence we can see that (2) is finite if and only if $p>1/2$. Then setting $K_p:=\int_{\mathbb{R}}(1+z^2)^{-p}\mathop{}\!d z$ from (1) and (2) we have

$$
\int_{\mathbb{R}\times [0,1]}(1+x^2+y^2)^{-p}\mathop{}\!d (x,y)=K_p\int_{[0,1]}(1+y^2)^{1/2-p}\mathop{}\!d y\tag4
$$

Finally note that as $f_p:[0,1]\to \mathbb{R},\, y\mapsto (1+y^2)^{1/2-p}$ is continuous for any chosen $p\in \mathbb{R}$ we find that (1) is finite if and only if $p>1/2$.∎


问:
Converting a point in a finite field to its real (x, y) coordinate
Let curve $A = y^2 = x^3 + 3$ and curve $B = y^2 \equiv x^3 + 3 \pmod{19}$

Let $G$ be the positive y-valued point in the curve where $x = 2$

Let $r$ be a random scalar integer, for example, $r = 5$

Compute the point $G*r$ in both curves $A$ and $B$

Now, assume I give you the point $G*r$ in curve $B$, can you find what my point $G*r$ in curve $A$ is? You don't know the value for $r$, but you do know all other parameters. You can't bruteforce the curve.

答:
In the question (as in rev. 3), it can be shown by enumeration that there are 12 solutions to $y^2 \equiv x^3 + 3\pmod{19}$ :
$$\begin{array}{}
(1,2),&(2,7),&(3,7),&(7,2),&(11,2),&(14,7)\\
(1,17),&(2,12),&(3,12),&(7,17),&(11,17),&(14,12)
\end{array}$$
Thus with the neutral element $\infty$ (aka point at infinity), the order of curve $B$ is $12+1=13$. The point $B$ has the same $x=2$ coordinate on curves $A$ and $B$. We get $y=\sqrt{11}$ on $A$, and¹ $y=7$ on $B$.

Point addition $R\gets P+Q$ can go by the same formulas for $A$ and $B$, with addition, multiplication, division², and equality in the field $\mathbb R$ for $A$, the field $\mathbb F_{19}$ for $B$:

If $P=\infty$, $R\gets Q$.

Otherwise, if $Q=\infty$, $R\gets P$.

Otherwise, if $x_P\ne x_Q$
$$\begin{align}
\lambda&\gets\frac{y_Q-y_P}{x_Q-x_P}\\
x_R&\gets\lambda^2-x_P-x_Q\\
y_R&\gets\lambda(x_P-x_R)-y_R
\end{align}$$

Otherwise, if $P=Q$ (that is $x_P=x_Q$ and $y_P=y_Q$)
$$\begin{align}
\lambda&\gets\frac{3{x_P}^2}{2y_P}\\
x_R&\gets\lambda^2-2x_P\\
y_R&\gets\lambda(x_P-x_R)-y_R
\end{align}$$

Otherwise (that is $x_P=x_Q$ and $y_P\ne y_Q$), $R\gets\infty$.

In order to compute $G*5$ on each curve, we can compute $G_2\gets G+G$, $G_4\gets G_2+G_2$, $G*5\gets G_4+G$ per these formulas. Try it online!. We get
$$\begin{align}
G*r&=\left(\frac{60503882}{151321},-\frac{141898736429 \sqrt{11}}{58863869}\right)&\text{ on curve }A\\
&\approx(399.838,-7995.14)\\
\\
G*r&=(7,2)&\text{ on curve }B
\end{align}$$

In general, we don't know how to compute $G*r$ on curve $A$ from $G*r$ on curve $B$ other than by finding $r$ first. If we could, that would³ allow finding $r$, which would solve the discrete logarithm problem on a (non-singular) elliptic curve on $\mathbb F_p$ in time polynomial w.r.t. $\log p$. That feat is widely conjectured impossible with a classical Turing machine, and the basis of the conjectured security of Elliptic Curve cryptography, e.g. on secp256k1.

¹ Finding a modular square root of $11$ and taking "positive" as $\in[0,9]$. Square roots in the field $\mathbb F_p$ can be efficiently computed even when prime $p$ is large, see the Tonelli–Shanks algorithm.

² Division $\frac uv$ in field $\mathbb F_{19}$ can be computed as $v^{-1}\,u$ where the modular inverse $v^{-1}$ can be found using the extended Euclidean algorithm.

³ If we get the coordinates of $G*r$ accurately enough.

Body:
In the question (as in rev. 3), it can be shown by enumeration that there are 12 solutions to $y^2 \equiv x^3 + 3\pmod{19}$ :
$$\begin{array}{}
(1,2),&(2,7),&(3,7),&(7,2),&(11,2),&(14,7)\\
(1,17),&(2,12),&(3,12),&(7,17),&(11,17),&(14,12)
\end{array}$$
Thus with the neutral element $\infty$ (aka point at infinity), the order of curve $B$ is $12+1=13$. The point $B$ has the same $x=2$ coordinate on curves $A$ and $B$. We get $y=\sqrt{11}$ on $A$, and¹ $y=7$ on $B$.

Point addition $R\gets P+Q$ can go by the same formulas for $A$ and $B$, with addition, multiplication, division², and equality in the field $\mathbb R$ for $A$, the field $\mathbb F_{19}$ for $B$:

If $P=\infty$, $R\gets Q$.

Otherwise, if $Q=\infty$, $R\gets P$.

Otherwise, if $x_P\ne x_Q$
$$\begin{align}
\lambda&\gets\frac{y_Q-y_P}{x_Q-x_P}\\
x_R&\gets\lambda^2-x_P-x_Q\\
y_R&\gets\lambda(x_P-x_R)-y_R
\end{align}$$

Otherwise, if $P=Q$ (that is $x_P=x_Q$ and $y_P=y_Q$)
$$\begin{align}
\lambda&\gets\frac{3{x_P}^2}{2y_P}\\
x_R&\gets\lambda^2-2x_P\\
y_R&\gets\lambda(x_P-x_R)-y_R
\end{align}$$

Otherwise (that is $x_P=x_Q$ and $y_P\ne y_Q$), $R\gets\infty$.

In order to compute $G*5$ on each curve, we can compute $G_2\gets G+G$, $G_4\gets G_2+G_2$, $G*5\gets G_4+G$ per these formulas. <a href="https://tio.run/##dVRdb5swFH3nV3g8Jalpm0R7WDNPClNXdVI3u3lEBEEgLYIAI7DiIH57dm0IOOkqgYTv1zn33It3bvEa7Nwi3LjHo3b/naZhUtzdlXv3JSB6d7Yq7NnIK8PY3yMXZcKG0gTlgRujTZn/DZCH@HpGqvX8yrtGQ1rt4az5X@ou9cvYzbvsPlfYUXatL7S@hgNFHJw5jX1HzDjdRFZ9IDR9C/Kn1AeIOc7sK6/Bj1vrp7tJvXDFd14aWwewf73Fj8k2TMKC47oNz2zcJx/w9GYGlsa2zwA9R8WSxGT9w1m9Cq/@5IV1aNMhf@n7inJwsuqKYk4biGWYs0ao6PoghJRhjyhyEx8xoUenomxcZPYwzMGCDevs1BkYCDvVeiQHoByJBV9MfF3qVhGpATOAFujA5ZEzg4uj7LD6RG5xGzbq3EjR2piKvMwer2fGqKJXFRvDaQFSyFAwGdVYKcfJZTnxziu6FrKfCiNrBilq7Zkk@FHdrn@7ndpH3csh/ijjeBXusjjccqtXoVWAE9m62veI35BKEBCATOAP2GpD0Ixs4mYEzMcd43fhA892P57CRNkPOFlUbEQeFGWe7JFBLxdBhCjTJqfvk6@GXeXKoInccUMuOW8uo0AQUoNTeISrjFU2ZWxF@JxPNHnPCMKcx6QIXoL8fBHPuIlifRjtfiiQL/o2xd0UGGnjzLBYvYbb4jl8eS2AwtTG3aZLNvBI4X/7PrOiwceEj9m28EXwW7bFjAgPwopHeMVoqToNTRtN9PvKhbUIULkP9MlYy8nnhbYk84VGc3EL6A9LhIiOH5bkdDHM8FIMsvdPcvA/d23kEHnmR0jm/7KebayjkZtleVqFcNMGMR/rEGeSeo6nX5qhokkQAkRTQTRVRFMi9nim9B6P/wA" rel="nofollow noreferrer">Try it online!. We get
$$\begin{align}
G*r&=\left(\frac{60503882}{151321},-\frac{141898736429 \sqrt{11}}{58863869}\right)&\text{ on curve }A\\
&\approx(399.838,-7995.14)\\
\\
G*r&=(7,2)&\text{ on curve }B
\end{align}$$

In general, we don't know how to compute $G*r$ on curve $A$ from $G*r$ on curve $B$ other than by finding $r$ first. If we could, that would³ allow finding $r$, which would solve the discrete logarithm problem on a (non-singular) elliptic curve on $\mathbb F_p$ in time polynomial w.r.t. $\log p$. That feat is widely conjectured impossible with a classical Turing machine, and the basis of the conjectured security of Elliptic Curve cryptography, e.g. on secp256k1.

¹ Finding a modular square root of $11$ and taking "positive" as $\in[0,9]$. Square roots in the field $\mathbb F_p$ can be efficiently computed even when prime $p$ is large, see the Tonelli–Shanks algorithm.

² Division $\frac uv$ in field $\mathbb F_{19}$ can be computed as $v^{-1}\,u$ where the modular inverse $v^{-1}$ can be found using the extended Euclidean algorithm.

³ If we get the coordinates of $G*r$ accurately enough.


问:
Every regular space curve with polynomial coordinates of degree less or equal to 2 is a plane curve
I'm having difficulty on the following problem of differential geometry:

$a)$ Verify that the curve $\alpha\colon (0,\infty)\to\mathbb R^3$ given by $\alpha(t)=\left(t,\dfrac{1+t}{t},\dfrac{1-t^2}{t}\right)$ is a plane curve.

$b)$ Verify that every regular curve of $\mathbb R^3$, whose coordinate functions are polynomials of degree less than or equal to $2$, is a plane curve.

So, for the first item I tried the following:

$$
\begin{array}{cc}
\left\{
\begin{array}{ccc}
x &=& t \\
y &=& \dfrac{1+t}{t} \\
z &=& \dfrac{1-t^2}{t}
\end{array}
\right.
&
\Rightarrow
\left\{
\begin{array}{ccc}
y = & \dfrac{1+x}{x} & & \\
z = & \dfrac{1-x^2}{x} & = &  \dfrac{(1-x)(1+x)}{x}
\end{array}
\right.
\end{array}
$$

Hence,
$$
z=(1-x)y=y-xy=y-(1+x)=y-1-x
$$

Thus, $\alpha$ lies in the plane $-x+y-z=1$.

For the second item I took the curve

$$
\beta(t)=(a_0+a_1t+a_2t^2,b_0+b_1t+b_2t^2,c_0+c_1t+c_2t^2), a_i,b_i,c_i\in\mathbb R.
$$

And wlog I assumed $\beta$ parameterized by the arc length. I tried something over the relation $|\beta'(t)|=1$, but I did not get anything because of the lack of control of the coefficients $a_i,b_i,c_i$.

答: Every space curve with zero torsion is a plane curve.  If the (time) derivative of the acceleration is zero, then so is torsion.</p>


问:
Computing the probability that a function of random variables takes particular values
If one rolls two fair die, the sum of the two rolls is most likely to be $7$, and $2$ and $12$ are very improbable. Likewise, if one averages two random numbers in the interval $[0,1]$, the most likely result is $0.5$ and the most unlikely are $0$ and $1$.

More generally, I would like to compute the continuous probability density $P\left(A=\frac{X+Y}{2}, \,\, B=X-Y\right)$ given independent continuous probability distributions $P(X),P(Y)$, and cannot seem to remember how to do this. I thought it went as
$$
P\left(A=\frac{X+Y}{2}, \,\, B=X-Y\right) \\= \int_0^1 \int_0^1 P(X)P(Y) \delta\left(A-\frac{X+Y}{2}\right)\delta\left(B-(X-Y)\right) dX dY
$$
but this cannot be right. For uniform $P(X),P(Y)$ it just gives either 0 or 1.

What is the proper way to calculate this probability? For that matter, for the life of me, I can't even remember what the right distribution for uniform $P(X),P(Y)$ in the above example is called.

答:
Note that we have
$$\begin{pmatrix}A\\B\end{pmatrix}=M\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}1/2&1/2\\1&-1\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}.$$
Observe that $\det M=-1$ and $M^{-1}=\begin{pmatrix}1&1/2\\1&-1/2\end{pmatrix}$. Then the density of $(A,B)$ is given by
\begin{align*}
f_{A,B}(a,b)=\frac{1}{\lvert\det M\rvert}f_{X,Y}(M^{-1}y)&=f_{X,Y}(a+b/2,a-b/2)\\
&=f_X(a+b/2)\cdot f_Y(a-b/2).
\end{align*}
This is a special case of the general change of variables formula for integrals. Since our transformation is linear, the Jacobian is just $\lvert J\rvert=\det(M^{-1})$.

The marginal density of $A$ is
$$f_A(a)=\int_{-\infty}^\infty f_{A,B}(a,b)\mathop{}\!\mathrm{d}b=\int_{-\infty}^\infty f_X(a+b/2)f_Y(a-b/2)\mathop{}\!\mathrm{d}b.$$
When $X$ and $Y$ are both $\mathcal U[0,1]$, we have $f_X(x)=\mathbf1(0\leq x\leq1)$, so the integrand is positive iff $a\in[0,1]$ and $\max\{-a,a-1\}\leq b/2\leq\min\{1-a,a\}$. Then we have
$$f_A(a)=\int_{\max\{-a,a-1\}}^{\min\{1-a,a\}}1\mathop{}\!\mathrm{d}b=2\min\{1-a,a\},$$
which is clearly maximised for $a=1/2$.


问:
Prove $\tan:(-\pi/2,\pi/2) \to \mathbb{R}$ can be extended to a homeomorphism $\overline{\tan}:[-\pi/2,\pi/2] \to \overline{\mathbb{R}}$
Let $\overline{\mathbb{R}} = \{-\infty\} \cup \mathbb{R} \cup \{\infty\}$ with the topology given by the subbase composed by the rays $[-\infty,a)=\{-\infty\}\cup\{x;x<a\}$ and $(b,\infty]=\{x;x>b\} \cup \{\infty\}$, where $a,b$ vary in $\mathbb{R}$.

Prove the homeomorphism $\tan:(-\pi/2,\pi/2) \to \mathbb{R}$ can be extended to a homeomorphism $\overline{\tan}:[-\pi/2,\pi/2] \to \overline{\mathbb{R}}$.

I'm kinda lost on what do I need to do here, so I'd appreciate any guidance. I believe what is need to be done here is prove that $\overline{\tan}$ is an homeomorphism.

Here is my "attempt" on proving continuity of $\overline{\tan}$: Let $A \subseteq \overline{\mathbb{R}}$ an open set. Then $A = \bigcup_{\lambda \in \Lambda} A_{\lambda}$, where each $A_{\lambda}$ is of the form $[-\infty,a), (a,b)$ or $(b,\infty]$.

$\overline{\tan}^{-1}(A) = \overline{\tan}^{-1}\left(\bigcup_{\lambda \in \Lambda} A_{\lambda}\right)=\bigcup_{\lambda \in \Lambda} \overline{\tan}^{-1}(A_{\lambda})$

Now I need to show each $\overline{\tan}^{-1}(A_{\lambda})$ is open on $[-\pi/2,\pi/2]$, right? I'm not quite sure how to do that.

Any help would be appreciated.
Thanks.

答:
Show that

$\overline\tan^{-1}\big[(a,b)\big]=\left(\tan^{-1}a,\tan^{-1}b\right)$,

$\overline\tan^{-1}\big[[-\infty,a)\big]=\left[-\frac{\pi}2,\tan^{-1}a\right)$, and

$\overline\tan^{-1}\big[(b,\infty]\big]=\left(\tan^{-1}b,\frac{\pi}2\right]$.

That will show that $\tan$ is continuous. Showing that it is also open is very similar.


问:
Are polynomials dense in $C^k\left(\bar{B}\right)$?
Let $\bar{B}$ be the closed unit ball in $\mathbb{R}^n$, $C^k\left(\bar{B}\right)$ the Banach space of all real function defined on $\bar{B}$ with continuous derivatives up to order $k$, with norm
$$\Vert f \Vert = \sum_{h\le k} \Vert \partial_{i_1}\dots \partial_{i_h}f\Vert_\infty$$
Are polynomials dense in $C^k\left(\bar{B}\right)$?

Can you give me references about that subject?

Thanks a lot...

答: By Stone-Weierstrass, the polynomials are dense in $C(\overline{B})$.  Do the rest by induction: if polynomials are dense in $C^{k-1}$, write a function in $C^k$ in terms of integrals of its partial derivatives, and approximate those partial derivatives by polynomials...</p>


问:
if a subset in $C([0,1])$ is bounded and equicontinuous
Let $F= \{ f \in C([0,1]) :  \int_0^1 |f(x)| dx \leq 1 \}$ then does $F$ bounded and equicontuous?

I am not sure how can I relate the function values with the integral of the function.

答:
Consider
$$f_n(x)=(n+1)x^n$$

Notice that for $x \neq 1$,
$$
|f_n(1) - f_n(x)| = (n+1)(1 - x_0^n)
$$
and as $x^n \to 0$ for $x \neq 1$, we see that the difference $|f_n(1) - f_n(x)|$ will not be bounded as $n$ came large, therefore this family is not equicontinuous.


问:
The relation $a^n|b^n$ implies that $a|b$, where $a$ and $b$ are positive integers and $n\ge1$.
Let $gcd(a, b) = d$

Let $a = rd$ and $b = sd$, where $gcd(r, s) = 1$.

$\therefore gcd(r^n, s^n) = 1$

Now, how do I show that $r = 1$?

答:
You're on the right track:

$b^n = a^n t \implies s^n = r^n t \implies r \mid s^n \implies r = 1$ because $\gcd(r,s)=1$.

Another way

$a^n\mid b^n$ iff $v_p(a^n) \le v_p(b^n)$ for all primes $p$. Here, $v_p(x)$ is the exponent of $p$ in the prime factorization of $x$.

Now, $v_p(a^n) \le v_p(b^n)$ iff $n v_p(a) \le n v_p(b)$ iff $v_p(a) \le  v_p(b)$ iff $a \mid b$.


问:
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$
I came across this integral

$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$

According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.

A promising way might be the following

\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
 &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3}   \right )^2 + \frac{8}{3}}} \\
 &= \cdots
\end{align*}

A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )

So, any ideas how to evaluate it?

P.S: Are there techniques available for these type of problems?

答:
A trigonometric substitution does indeed help here: Consider $f \colon [0,1] \to \mathbb{R},$
\begin{align}
f (a) &= \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{1 + 2 a x^2 + x^4}} = \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{(1 + x^2)^2 - 2 (1-a) x^2}} \\
&\!\!\!\!\!\!\!\!\stackrel{x = \tan\left(\frac{t}{2}\right)}{=} \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - 2 (1-a) \sin^2 \left(\frac{t}{2}\right) \cos^2 \left(\frac{t}{2}\right)}} = \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - \frac{1-a}{2} \sin^2(t)}} \\
&= \frac{1}{2} \operatorname{K} \left(\sqrt{\frac{1-a}{2}}\right) \, .
\end{align}
Your integral is
$$ \mathcal{J} = \frac{1}{\sqrt{3}} f \left(\frac{1}{3}\right) = \frac{\operatorname{K} \left(\frac{1}{\sqrt{3}}\right)}{2 \sqrt{3}} > \frac{1}{2} \, .$$


问:
Finding ALL values of $h$ such that a given set of vectors spans $\mathbb{R}^3$
Find ALL value's of $h$ so that the vectors $$a_1= (-1,h,7)$$ $$a_2=(4,-2,5)$$ $$a_3=(1,-6,2)$$ span $\mathbb{R}^3$.

Previously, I posted asking how to find a single value of $h$ so that the given vectors span $\mathbb{R}^3$. I am now trying to figure out how to find ALL values of such $h$. A got a great answer: Trying to find the value of $h$ such that the set of vectors spans $\mathbb{R}^3$

Don Thousand restated the question as "for what values of $h$ will the vectors NOT span $\mathbb{R}^3$ and he demonstrated that the vectors don't span $\mathbb{R}^3$ $\iff$ $h=60$

His response was based on this particular problem, where he used that in this case, if one vector can be written as a linear combination of the others, than they all can; however, this will not hold in general. Does anyone have a systematic way to approach this problem? Thank you.

Also, I'm trying to solve this problem without resorting to arguments involving linear independence, because we haven't covered that in class yet. Thank you.

答:
Well, if $v=\lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3$ for some $\lambda_1, \lambda_2, \lambda_3 \in \mathbb R$, then $v=A\left(w\right)$ where $w=(\lambda_1, \lambda_2, \lambda_3)\in\mathbb R^3$ and $A$ is linear transformation with matrix $$\left[\begin{matrix} -1& 4 & 1 \\ h & -2 & 6 \\ 7 & 5 & 2\end{matrix}\right].$$
So, every $v\in \mathbb R^3$ can be written as linear combination of $a_1,a_2,a_3$ if and only if $A$ is surjective. And $A$ is surjective if and only if $\det A\ne 0$ (note that $A$ is injective iff it is surjective). This give us linear equation in terms of $h$. Zero of this (linear) polynomial is precisely value of $h$ for which $A$ is not surjective.

This is systematic way for solving this.

More elementary way for solving would be to write $v=Aw$ as system of linear equation and then with elementary transformations reduce this system to simpler one from which value of $h$ can be easily deduced.


问:
Finding values of $a$ which this system of linear equations has one, no solution, or infinite solutions
Given the following equation:

$x+ay -z = 1$

$x+y -az = a$

$2x-4y +2az = -2$

Determine the value of $a$ (if any) for which the system has no solutions.

Determine the value of $a$ (if any) for which the system has exactly one solution.

Determine the value of $a$ (if any) for which the system has infinitely many solutions

And in case the system has infinite solutions, find them

So I'd like to put the coefficients of $x,y,z$ into a matrix like the following:

$\begin{pmatrix} 1 & a & -1 \\ 1 & 1 & -a \\ 2  & 4 & 2a \end{pmatrix}$

And I want to do this by the method of checking when the determinant not equals to $0$.

So I calculate it and got to a place where $(a-1)(-4a-6)$ which means that $1$ or $-3/2$ will zero the determinant.

So I realize that if $a$ will be $0$ or $3/2$ then the system has no solutions,

but then I don't know how to continue,

How do I find out about the unique infinitely soulutions? I guess for unique solution $a$ needs to be not $0$ or $3/2$
but for some reason I can't put it all together..

答:
Calculate the determinant of the system's matrix:
$$
\begin{vmatrix}
1 & a & -1 \\
1 & 1 & -a \\
2 & 4 & 2a \\
\end{vmatrix} = -4a^2+6a-2
$$

You must know that, if that determinant is different from $0$, then the system has unique solution. When it's $0$, it can either have infinite or none. Let's calculate the values which make the determinant be $0$:

$$-4a^2+6a-2=0 \ \implies \ a=\frac{1}{2},\ a=1.$$

So now you can conclude that the system has unique solution when $a\in\mathbb{R}\setminus\{\frac{1}{2},1\}$. Now the only work you have left is to study what happends when $a=\frac{1}{2}$ and what when $a=1$ separately.


问:
Definition of proper morphism between schemes
The definition of proper morphism by separated, finite type, universally closed properties doesn’t look enlightening to me. I am trying to look for a simpler characterization for this concept.

As we know properness in algebraic geometry is the analog notion of compactness in topology. In topology the definition of the proper map is that the inverse image of every compact subspace is a compact subspace, I am wondering is there an analog definition for proper morphisms between schemes in algebraic geometry.

A natural statement one might wish to hold is the following:

Let S be a scheme. A morphism between S schemes is proper if and only if the inverse image of every proper S subscheme is a proper S subscheme.

Here we put the reduced induced subscheme structure on the inverse image of a closed subscheme. Is this statement true? If not, is it true in some special cases, say k schemes, where k is a field? Or are there some related statements that hold? Thank you!

答:
You are correct that if $f : X \to Y$ is proper over some base $S$ then it has the property you described. If $Z \subset Y$ is a closed subscheme proper over $S$ then $f^{-1}(Z) \subset X$ is a closed subscheme proper over $S$.

Because $f^{-1}(Z)$ has $f^{-1}(Z)$ has a natural subscheme structure as $f^{-1}(Z) = Z \times_Y X$ which is not always reduced, it is probably better to require that $f^{-1}(Z)$ be proper with this standard subscheme structure.

However, it's also true if we give $f^{-1}(Z)$ the reduced subscheme structure because this is a closed subscheme of $f^{-1}(Z)$ with its standard structure and closed subschemes of proper schemes are proper.

Now let me prove the claim. The map $f^{-1}(Z) \to Z$ is proper because it is the base change of $f : X \to Y$. Therefore the composition $f^{-1}(Z) \to Z \to S$ is proper showing that $f^{-1}(Z)$ is proper over $S$.

However, I do not believe this property is enough to characterize proper morphisms even for varieties over an algebraically closed field. The problem is ``there aren't enough proper closed subvarieties''. For example, $\mathbb{A}^1 \setminus \{ 0 \} \to \mathbb{A}^1$ is not proper but the only proper closed subvarities of $\mathbb{A}^1$ are points whose preimages are also points.

In fact, we can make examples where $f : X \to Y$ is a closed surjective map of varieties over $\mathbb{C}$ with your property but not proper. For example, let $Y$ be the affine nodal curve $y^2 = x^2(x-1)$ and $X = \mathbb{A}^1 \setminus \{ i \}$. The normalization map $\nu : \tilde{Y} \to Y$ where $\tilde{Y} = \mathbb{A}^1$ sends $t \mapsto (t^2 + 1, t(t^2 + 1))$ so take $X \to \tilde{Y} \to Y$ i.e. we remove one of the two preimages of the node. Then $f$ is closed and surjective (even bijective!) pulling back proper closed subvarities (points) to proper closed subvarities. However, $f$ is not proper. To see this, consider $\tilde{f} : X \times \tilde{Y} \to Y \times \tilde{Y}$ and the closed set
$$\Delta = \{ (x,x) \mid x \in X \} \subset X \times \tilde{Y} $$
however $\tilde{f}(\Delta) = \{ (f(x), x) \mid x \in X \} \subset Y \times \tilde{Y}$ is the graph of $f$ minus one point and thus not closed.

However, for smooth varieties over $\mathbb{C}$, we can view an algebraic map $f : X \to Y$ as a map of complex manifolds $f^{\mathrm{an}} : X^{\mathrm{an}} \to Y^{\mathrm{an}}$ viewing $X, Y$ as complex manifolds with the analytic topology. Then it is a comforting fact that $f$ is proper iff $f^{\mathrm{an}}$ is proper in the usual topology sense.


问:
Positive integers $(x,y)$ satisfying $x^2-y^2= 12345678$?
I saw this question in one of the previous year Olympiad questions.

Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ satisfying the condition : $x^2-y^2= 12345678$

Then what can be said about the number of elements in set $S$ ?

The solution of this question doesn't seem to be correct. Look at it :

Can someone provide me a reasonable approach ?

答: $x^2-y^2\equiv 12345678\equiv 2$ modulo $4$ which is not possible since squares are $0$ or $1$ modulo $4$. So the answer given is correct.


问:
Oscillating String attached to Spring
Here is the problem I am woking on:

You are given a string of length L that is elastic. The function $v(x,t)$ measures the displacement of the string from the horizontal axis. At the left endpoint of the string ($x=0$) the string is fixed and $v(0,t)=0$. At the right endpoint of the string ($x=L$) the string is attached to a spring. The spring is only able to move up and down and it has elasticity k, linear density $\rho$ and tension force T (all constants). Moreover, the end of the spring not attached to the string is fixed at point $(L,h)$. We are given that the string has resistance force proportional to the displacement from the x axis.
Overall the system is given by $\rho v_{tt} = T v_{xx} -r v$, with $v(0,t)=0$, and $T v_x(L,t)+k(v(L,t)-h)=0$.

How can we determine the equilibrium position of the string?

Secondly, if given the boundary condition $v(0,t)=sint$ how can we solve the BVP?

I'm entirely lost on where to start and what to do to solve this. From extensive google searching I think it may have something to do with the wave equation but I'm very lost on how to do any of this. Any suggestions, hints or explanations will be very appreciated.Even if you could just point me to some good reference material I would appreciate that.

答:
At $x=L$,  at rest, the forces acting on the movable hinge (positive downwards) will be

$T$, directed along the string;

$mg/2+k(v-h)$, directed along the vertical.

Normally, the end movement $v$ is taken to be much smaller than $L$.

Therefore $T$ is considered to be constant, so that its vertical projection would be approximated by
$$
T_v  = T{v \over L}
$$

So that at equilibrium we shall have
$$
mg/2 + k(v - h) + T_v  = mg/2 + k(v - h) + T{v \over L} = 0
$$
which means
$$
v(L,0) = {{ - mg/2 + kh} \over {k + {T \over L}}}
$$
and this answers your first question.

Now consider that a little vertical starting displacement of the end at $x=L$, if much smaller than $L$, can be neglected,
i.e. that $kh=mg/2$, which is to tell that the offset positioning of the spring is just to compensate the
(half) weight of the string.

In conclusion at $x=L$ you may consider to have just a reaction (on the side of the string) of $-kv$.

The second step is that an elastic clamp as the above can be "bartered" with a little increase $\Delta L$ of the length $L$
and clamp the string to a fixed hinge there.

That's because near  the end the displacement $v$ is even smaller than in the central part of string where it is already
taken to be much less than $L$.

That allows to put
$$
kv \approx T{{dv} \over {dx}} \approx T{v \over {\Delta L}}\quad
 \Rightarrow \quad \Delta L \approx {T \over k}
$$
if the spring is stiff enough to give $\Delta L << L$. This allows as well to neglect the effect of the dynamic reaction of the final
part of the spring.

If instead the spring is quite soft then we cannot adopt the approximation above: there will be reflections at the end .

Analog of the mechanical system above is the transmission along an electrical line of finite length, terminating on a capacitive load.
This goes under the name of Telegrapher's Equation and you can find
for that various sources for how to solve the problem.

--- clarifications  ---

As a very concise reply to your comments:

a) The modelling of the tansverse vibration of  a "normal" elastic string is derived as shown
in this article as (using your notation)
$$
\rho v_{\,t\,t}  = Tv_{\,x\,x}
$$
and this is analogous to the Telegrapher's equation for a lossless line.

b) The presence of the further term $-r v$ in the equation that you propose
$$
\rho v_{\,t\,t}  = Tv_{\,x\,x}  - rv
$$
corresponds to a string that lies on an elastic "bed" (with negligible mass and shear).

So the mechanical system is in any case lossless.

c) The $v(0,t) = \sin t$ is the exciting signal fed at $x=0$ in terms of an imposed displacement $v(t)$..

That means that the actual mechanical model underlying your mathematical model is the following

that is
$$
\left\{ \matrix{
  T_{\,n} (x) = T\,v_{\,x}  \hfill \cr
  T_{\,n} (x + dx) - T_{\,n} (x) - r\,v\,dx = \rho dx\,v_{\,t\,t} \quad  \Rightarrow  \hfill \cr
   \Rightarrow \quad \rho v_{\,t\,t}  = Tv_{\,x\,x}  - r\,v \hfill \cr
  v(0,t) = U(t)\sin t \hfill \cr
  0 =  - T_{\,n} (L) - k\,v(L) = T\,v_{\,x} (L) + k\,v(L) \hfill \cr}  \right.
$$
where $U(t)$ denotes the Unit Step function.

Clearly this physical model is lossless.

d) For finding the steady state solution we adopt the method
described in the Telegrapher's equation article and well known to electrical engineers,

and which essentially consists in taking the Fourier transform.
We put
$$
v(x,t) =\Re \left( {\hat V(x)\,e^{\,j\omega t} } \right) = V(x)e^{\,j\left( {\omega t + \alpha \left( x \right)} \right)}
$$
where $\hat V(x)$ is in general complex, and replace in the system above.

Now that the system is in the correct shape (if  I understood properly your post) can you
proceed with the math part ?

Title: Oscillating String attached to Spring


问:
Proof of Ptolemy's Theorem

I am trying to prove using similar triangles without considering an arbitrary point on the diagonals which is the most common proof what we see in text books.

Let $HI=a$, $IJ=b$, $JC=c$ ,$CH=d$, $HL=x$, $LJ=e-x$, $CL=f-y$ and $LI=y$

we have $$\Delta HLC \sim \Delta ILJ $$ so we have

$$\frac{d}{b}=\frac{f-y}{e-x}=\frac{x}{y} \tag{1}$$ Also

$$\Delta CLJ \sim \Delta HLI$$ so

$$\frac{a}{c}=\frac{e-x}{y}=\frac{f-y}{x} \tag{2}$$  from $(1)$ and $(2)$ is it possible to prove that $$ac+bd=ef$$

答:
As a practice to me learning geometry, I write an answer to this question :^) Note: $[ab]$ denotes the distance between point $a$ and point $b$

Introduction to geometric inversion:

For an origin centered circle C of radius R, geometric inversion map is defined in the following way:

$$ \tilde{z} = I_C (z) = \frac{R^2}{\overline{z}}$$

From this we get that a relation between conjugate and inverted point,

$$ \tilde{z} \cdot \overline{z} = R^2$$

Note that:

$$ | \tilde{z} | | \overline{z} | = | \tilde{z}| |z| = R^2$$
Also note that the transformation is involutive (self inverse)

Now let's see this transformation's effect on geometrical objects such as triangles:

Under inversion $ a \to \tilde{a}$ and $ b \to \tilde{b}$, note that:

$$ [qa][q \tilde{a} ] = R^2$$

Also,

$$ [qb][ q \tilde{b} ] =R^2$$

Hence,

$$ \frac{ [qa]}{[qb] } = \frac{ [q \tilde{b} ] }{ [ q \tilde{a} ]}$$

Hence with the common angle q, this implies

$$ \triangle qab \sim \triangle q\tilde{b} \tilde{a}$$

Hence,

$$ \frac{ [ab]}{ [\tilde{a} \tilde{b}] } =  \frac{[qa]}{[q \tilde{b}]}$$

Now,

$$ [ q \tilde{b} ] = \frac{R^2}{ [qb] }$$

This and rearrangement gives:

$$ [ \tilde{a} \tilde{b} ] = \frac{R^2}{ [ qa] [qb]}[ab] \tag{1}$$

How are circles passing through center of inverting circle mapped under inversion?

Consider a Line outside the circle, take two points $a$ and $b$ and consider the inverses $ \tilde{a}$ and $ \tilde{b}$, from the previous triangle arguements we find $ \angle q \tilde{b} \tilde{a} = \frac{\pi}{2}$, now notice that as we vary $b$ this angle condition remains same. From inscribed angle theorem, we can say that locus of $ \tilde{b}$ is a circle.(2)

Proof

Under inversion of a circle K centered at $a$, the circle containing the cyclic quadrilateral gets mapped abcd gets mapped to a line L (2)with the $ b \to \tilde{b}$ , $ c \to \tilde{c}$ , $ d \to \tilde{d}$

It is obvious that:

$$[ \tilde{b} \tilde{c} ] + [\tilde{c} \tilde{d}] = [\tilde{b} \tilde{d}]$$

Now,

$$ \frac{[bc]}{[ab][ac]} + \frac{[cd]}{[ac][ad]} = \frac{[bd]}{[ab][ad]}$$ (1)

$$ [bc][ad] + [cd][ab] = [bd][ac]$$

QED

All pictures from Tristan Needham's Visual Complex Analsysis

Title: Proof of Ptolemy's Theorem


问:
How to solve $\int_0^\infty J_0(x)\ \text{sinc}(\pi\,x)\ e^{-x}\,\mathrm dx$?
I need some help with solving this integral involving Bessel function:

$\hspace{2in}\displaystyle\int_0^\infty$$J_0(x)\ $$\text{sinc}(\pi\,x)\ $$e^{-x}\,\mathrm dx.$

答:
Here's an evaluation of the integral used by O.L..

$$\begin{align} \int_{0}^{\infty} \frac{1-e^{-px}}{x} J_{0}(x) \ dx &= \int_{0}^{\infty} \int_{0}^{\infty} (1-e^{-px})J_{0}(x) e^{-xt} \ dt \ dx  \\ &= \int_{0}^{\infty} \left( \int_{0}^{\infty} J_{0}(x) e^{-tx} \ dx \ - \int_{0}^{\infty} J_{0}(x) e^{-(p+t)x} \ dx \right) \ dt  \\ &= \int_{0}^{\infty} \Bigg( \frac{1}{\sqrt{1+t^{2}}} - \frac{1}{\sqrt{1+(p+t)^{2}}} \Bigg)\ dt \tag{1} \\ &= \text{arcsinh}(t) -\text{arcsinh}(p+t) \Bigg|^{\infty}_{0} \\ &= \text{arcsinh}(p) \\ &= \ln \left(p+\sqrt{p^{2}+1} \right) \end{align}$$

$(1)$ http://mathworld.wolfram.com/LaplaceTransform.html