问:
What are cohomology rings good for?
I am studying some concepts of algebraic topology myself, and I read lately a bit about cohomology rings (created by the direct sum of cohomology groups) but besides all definitions I could not find any information.

What are they good for?

How do we use them?

Could someone give me a very short descriptive example?

Thank you

答:
One reason that the cohomology (and homology, and homotopy group) functors are so important is that they give a means of testing how similar two spaces are. For instance, the $\pi_1$ functor suffices to distinguish $\mathbb{R}^1$ from $\mathbb{R}^2$, up to homeomorphism (by removing a point); more generally, the $k$th homotopy (or homology) group will distinguish $\mathbb{R}^k$ from $\mathbb{R}^m$, $m > k$.

Now cohomology, homology, and homotopy groups (and K-theory and other cohomology theories) are all functors from some category of spaces to some algebraic category. Presumably, the idea is that the algebraic category will be easier to think about -- one can compute explicitly, at least if the functor applied to nice spaces produces simple results (which often happens).

At the same time, if these are to be a powerful tool at distinguishing spaces, we want these functors to take values not in the simplest category (e.g. the category of sets) but a category with more structure: groups, or, better, rings. The cup-product shows that the cohomology functor takes its value in the category of rings.

This means in practice that if you have two spaces with similar looking cohomologies as abelian groups, you may still be able to tell that they are different if the ring structures are different. As an example, you can see that complex projective space $\mathbb{CP}^2$ and $S^2 \vee S^4$ are not homotopy-equivalent, even though their homology and cohomology groups are all the same. The reason is that the cup-product is not trivial in the first, while it is trivial in the second.

So the more algebraic structure you can put on a functor from spaces to sets, the finer this functor will be as a tool to distinguish spaces. Cohomology has this nice ring structure, but in fact it has more: it turns out that the cohomology ring (with $\mathbb{Z}/2$-coefficients) is a module over the Steenrod algebra.

Title: What are cohomology rings good for?


问:
difference of recursive equations
Lets have two recursive equations:

\begin{align}
f(0) &= 2  \\
f(n+1) &= 3 \cdot f(n) +  8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot  g(n) + 12
\end{align}

We want a explicit equation for f(x) - g (x).

I firstly tried to do in manually for first $n$ numbers

\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 &    4 \\ \hline
1 & 6  & 6 &    0 \\ \hline
2 & 26  & 30 &  -4 \\ \hline
3 & 94 & 102 &  -8 \\ \hline
4 & 306  & 318 & -12 \\ \hline
\end{array}

We can deduce that $f(n) - g(n) = 4 - 4n$

But now we have to prove it.

Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$

We can simply check this by induction but I will skip it, so the question won't be so long.

Now lets  put it together:

$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n  + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$

As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.

How to prove that:

$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$

All I could do is this:

\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}

And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$

What to do next? Did I go the wrong direction anywhere?

Thank you for your fast responses.

答:
There is no need for induction, use a straight proof.

The equation is

$$f(n+1)-g(n+1)=3(f(n)-g(n))+8n-12$$ with $$f(0)-g(0)=4$$ and it does verify the solution

$$f(n)-g(n)=4-4n$$ as is shown by substitution,

$$4-4(n+1)=3(4-4n)+8n-12,$$ equivalent to $$-4n=-4n.$$

Furthermore, $$4=4-4\cdot 0.$$

ContentLicense: CC BY-SA 4.0

Title: difference of recursive equations


问:
Expectation of cubic brownian motion: $E[B_t^3]$
At time t, $B_t \sim N(0,t)$. By symmetry, the $E[B_t^3]=0$.

Obviously, we know $E[B_t]=0$ from the question, but I am not sure how to reach $E[B_t^3]=0$ using symmetry of Brownian motion?

答:
You have that $B_t \stackrel{d}{=} -B_t$ so that
$$\mathbb{E}[B_t^3] = \mathbb{E}[-B_t^3]$$
which implies that $$\mathbb{E}[B_t^3] = 0.$$


问:
A periodic cohomology theory?
I'm studying Lurie's paper, A survey of elliptic cohomology. He defines, in the third page, a periodic cohomology to be a cohomology theory $A$ in which there is an element $\beta$ in $A^{-2}(*)$ so that $\beta$ is invertible in $A^*(*)$. He says that this means $\beta$ has an inverse in $A^2(*)$. I cannot understand why this inverse should be in $A^2(*)$. I'm aslo confused with the definition of periodicity here. Is there a relationship between this definition and other definitions of periodicity in other contexts? (for example, periodic functions ...)

答:
The inverse of $\beta$ lives in $A^2(*)$ purely for degree reasons. We know that is some $\gamma\in A^*(*)$ such that $\beta\gamma=1\in A^0(*)$, so $|\beta|+|\gamma|=0$ where $|\cdot|$ denotes the degree of a (homogeneous) element.

The reason such a theory deserves to be called periodic is that multiplication by $\beta$ is invertible, and so induces an isomorphism $A^k(*)\to A^{k+2}(*)$ for all indices $k$. In fact, for any space $X$, since $A^*(X)$ is a (graded) $A^*(*)$-module, multiplication by $\beta$ induces an isomorphism $A^k(X)\to A^{k+2}(X)$, so $A^*$ is periodic in the usual sense.

Title: A periodic cohomology theory?


问:
Tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, $1 \le i \lt j \le k$
To generalize this question I would like to count how many tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, for any $i,j$ such that $1 \le i \lt j \le k$, are there.

The case $k=2$ has been solved in the linked question and the result is $\frac{3^n-1}{2!}$.

The case $k=3$ can be stated as $\frac{7^n-1}{3!}-\frac{3^n-1}{2!}$ and a computer test confirms the result.

For case $k = 4$ I would have expected something like $\frac{15^n-1}{4!}-\frac{7^n-1}{3!}+\frac{3^n-1}{2!}$, but the first addendum is not an integer. The values computed (hopefully correctly) for $n=1,2,3,4,5,6$ are $0,1,67,1546,27550,445531$.

To complete with computing data, the case $k = 5$ gives $0,0,56,4144,180096,6480656$, for $k = 6$ we have $0,0,28,7896,866432,69656776$, for $k = 7$: $0,0,8,11408,3308736,601192496$.

Any hint? If possible, for the general case, but especially for $k = 4$. Thank you.

答:
I will prove the following formula, valid for all $m\in \mathbb N$:
$$
\#\text{ sets of $m$ distinct sets whose union is $[n]$}=\frac1{k!}\sum_{i=1}^k(-1)^{k-i}{k\brack i}(2^i-1)^n
$$

Let $X$ be the set of ordered $k$-tuplets of sets whose union is $[n]$. This is different than what we actually want in two ways; first of all, this includes $k$-tuplets with repeated sets, and secondly, this distinguishes the order of the $k$-tuples. To deal with the first problem, we will use an alternating summation where the "bad" tuples with two ore more repeated entries cancel out. Once these are gone, we fix the order issue by dividing by $k!$.

In more detail, let $S_k$ be the set of permutations of $[k]$. Any such permutation $\pi$ can be applied to an element $x$ of $X$ to permute the indices of $x$. Now, consider the set of ordered pairs
$$
(x,\pi)\quad\text{   such that   }\quad x \text{ is unchanged by $\pi$}
$$
I claim that
$$
\sum_{(x,\pi)}\text{sign}(\pi)=\#\text{ of ordered tuples in $X$ whose entries are distinct}
$$
This follows from the following sign-reversing involution. Given an ordered pair $(x,\pi)$ where $x$ is unchanged by $\pi$, and where $x$ has at least two identical entries, let $(i,j)$ be the lexicographically first pair of indices for which the entries of $x$ at those indices are equal. Then the ordered pair matched to $(x,\pi)$ is $(x,(i\;\;j)\pi)$. This changes the sign of $\pi$, so the two ordered pairs matched together will cancel each other out in the above summation. All that remains are ordered pairs $(x,\pi)$ where $x$ has all distinct entries. Therefore, we have canceled out the bad tuples, and can account for the ordering by dividing by $k!$:
\begin{align}
\#\text{ sets of $m$ distinct sets whose union is $[n]$}
  &=\frac1{k!}\sum_{(x,\pi)}\text{sign}(\pi)
\\&=\frac1{k!}\sum_{\pi\in S_k}\text{sign}(\pi)\cdot \#\{x:x\text{ is unchanged by $\pi$}\}
\end{align}
You finally conclude in two steps:

For a given $\pi$, being unchanged by $\pi$ means that you can freely choose one of the entries of $x$ for each cycle of $\pi$, so $\#\{x:x\text{ is unchanged by $\pi$}\}$ is just $(2^i-1)^n$, where $i$ is the number of cycles of $\pi$.

You then group all permutations together with the same number of cycles, leading to the Stirling numbers of the first kind appearing.


问:
An interesting Olympiad Geometry problem.
The point P lies in the interior of $\Delta ABC$. A line is drawn through P, parallel to each side of a triangle. The line divides BC into three parts of length $a , a', a''$ (in that order); AB into three parts, lengths  $c, c', c''$ (in that order); CA into three parts length $b, b', b''$ (in that order) . Prove that $abc = a'b'c' = a''b''c''$ .
This is one of those types of question which I have never encountered before.

I was not able to use Ceva's theorem as there were no Cevians, I could not understand how to apply a similarity argument here, as that is what I think might work.

Any solution is highly appreciated, Thanks!

答:
Hint: If the ternary plot of $P$ is $(x,y,z)$, with $x+y+z=1$, show that

$$xyz \times |AB| \times |BC| \times |CA| = abc = a'b'c' = a''b''c''.$$

The ternary plot of $P$ is given by:

Let $D, E, F$ be the foot of the perpendiculars from $A, B, C$ to the respective sides.

Let $P_A, P_B, P_B$ be the foot of the perpendiculars from $P$ to the respective sides.

Then, $ x = \frac{PP_A } { AD} , y = \frac{ PP_B} { BE}, z = \frac{ PP_C}{CF}$.

In addition, we can show that $ x+y+z = \frac{PP_A } { AD} + \frac{ PP_B} { BE} + \frac{ PP_C}{CF} = \frac{ [PBC] + [PCA] + [PAB ] } { [ABC ] } = 1$.

Now, show that

$a'' = y \times |AB|$

$ a = z \times |AB| $

$a' = x \times |AB|$

And similar equations. Hence, the hint follows.


答:
Outline :

The three small triangles are similar to original triangle with similarity ratio $l,m,n$ respectively.

If the original triangle has sides $A,B,C$, the small triangles have sides $lA,lB,lC$, $mA,mB,mC$ and $nA,nB,nC$.

Also identify three parallelograms. Equating opposite sides, one can find all of $a,b,c,\ldots,b'',c''$. Substituting should conclude.

Here is another problem on same setting : Determining the average area as the point $X$ varies

Title: An interesting Olympiad Geometry problem.

Body:
The point P lies in the interior of $\Delta ABC$. A line is drawn through P, parallel to each side of a triangle. The line divides BC into three parts of length $a , a', a''$ (in that order); AB into three parts, lengths  $c, c', c''$ (in that order); CA into three parts length $b, b', b''$ (in that order) . Prove that $abc = a'b'c' = a''b''c''$ .
This is one of those types of question which I have never encountered before.

I was not able to use Ceva's theorem as there were no Cevians, I could not understand how to apply a similarity argument here, as that is what I think might work.

Any solution is highly appreciated, Thanks!


问:
Does there exist an initial condition to make this sequence integer?
I am interested if the following is true:

Given a natural number $N$, and $\omega_1, \omega_2 >0$ real numbers,  does there exist a $k_0$such that $k_0,k_1,\dots, k_N$ defined by
$$ k_n = (k_0 - \tfrac12 ) \left( \dfrac{1+ \omega_2}{1+\omega_1} \right)^n + \frac12$$
are all integers?

I think this is a number theory related problem which is not my strong suit, but I would appreciate any thoughts on this.

答:
$$2k_n-1=(2k_0-1) \left(\frac{1+\omega_2}{1+\omega_1}\right)^n$$

Therefore $\frac{1+\omega_2}{1+\omega_1}$ is rational. Let $\frac{1+\omega_2}{1+\omega_1} = \frac ab$ where $\gcd(a,b)=1$, then $b^n | (2k_0-1), n\le N$ so $b$ must be odd, and we can have $k_0=(1+b^N)/2$. Now
$$2k_n-1=b^{N-n} a^n$$

so $a$ must be odd too, and if $a$ is odd then $k_n$ is an integer.


问:
Coherent sheaves vs sheaves locally of finite presentation
I think it is clear for most people that finitely generated and finitely presented modules are important and, in a certain sense, the "right" notions. However, the corresponding notions for sheaves are not clearly the right notions (at least to me).

For completeness, let me define some relevant notions. Let $X$ be a ringed space.

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be quasi-coherent if every point of $X$ has a neighborhood $U$ over which there is an exact sequence $$\mathscr{O}_X^{\oplus I}|_U\to \mathscr{O}_X^{\oplus J}|_U\to \mathscr{F}|_U\to 0.$$

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be locally of finite presentation if it is quasi-coherent and the indices $I$ and $J$ as above can be taken to be finite.

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be coherent if every point of $X$ has a neighborhood $U$ over which there is an exact sequence $ \mathscr{O}_X^{\oplus m}|_U\to \mathscr{F}|_U\to 0$ (that is, $\mathscr{F}$ is finitely generated) and for any open set $V\subset X$, any natural number $n$, and any morphism $\varphi:\mathscr{O}_X^{\oplus n}|_V\to \mathscr{F}|_V$, the kernel of $\varphi$ is finitely generated.

It is very clear to me that quasi-coherentness is a relevant notion. But I don't understand why coherent is a more usual notion than locally of finite presentation, which seems more natural to be.

In https://mathoverflow.net/questions/68125/qcoh-algebraic-stack and in Stalks of the sheaf $\mathscr{H}om$? A. Vistoli and M. Brandenburg seem to imply that the "correct" notion is indeed that of local finite presentation.

I would like to understand what are the differences between those notions. For example, over an arbitrary ringed space the category of coherent sheaves is abelian. Is the same true about locally of finite presentation? What are other advantages or problems of either notions?

答:
First of all, if $\mathscr{O}$ is coherent over itself, then a sheaf of $\mathscr{O}$-modules is coherent if and only if it is locally finitely presented. This is due to the fact, that cokernels of morphisms between coherent sheaves are always coherent themselves. Thus, for $\mathscr{O}$ the sheaf of regular functions on a noetherian scheme and for $\mathscr{O}$ the sheaf of holomorphic functions on a complex analytic variety both concepts agree.

Moreover, it is definitely not true in general, that the category of locally finitely presented $\mathscr{O}$-modules is abelian. Indeed, obviously $\mathscr{O}$ is always (locally) finitely presented itself, however the answer to this question provides an example of a morphism $\mathscr{O} \rightarrow \mathscr{O}$ (where $\mathscr{O}$ is the sheaf of smooth functions on $\mathbb{R}$) for which the kernel is not even of finite type.

Sadly, I do not really know anything more about cases in which $\mathscr{O}$ is not coherent over itself (I suppose you are rather interested in something algebraic).


问:
Conditional Expected Value with Min/Max
Let's say I was given
$X\sim U[0,5]$ and $Y\sim\max[2,X]$

Then

$E[Y]=2\cdot P(X\leq2)+\int^5_2 x\cdot \frac{1}{5}dx$

My question is, why is it NOT

$E[Y]=2\cdot P(X\leq2)+\int^5_2 x\cdot \frac{1}{5}dx\cdot P(X\geq2)$

答:
I believe the formula you are thinking of is the law of total expectation:
$$E[Y] = E[Y|X\leq 2]P(X\leq 2) + E[Y|X > 2]P(X> 2)$$
Now obviously $E[Y|X\leq 2] = 2$. To calculate $E[Y | X>2]$, consider the conditional CDF
\begin{align*} P(Y \leq y| X > 2)&=P(X \leq y | X>2)\\
 &= \frac{P(2 < X \leq y)}{P(X>2)}
\end{align*}
for any $y \in (2,5)$. Differentiating the conditional CDF with respect to $y$ we get the conditional density
$$f_{Y|X>2}(y) = \frac{1}{P(X>2)}\frac{1}{5},$$
and thus we get
$$E[Y|X>2] = \frac{1}{P(X>2)}\int_2^5 \frac{y}{5} \: dy,$$
and inserting this in the formula, we see that $P(X>2)$ "cancels out" and we are left with the correct expression.

Title: Conditional Expected Value with Min/Max


问:
Compute limit of the sequence given by $x_1=1$, $x_{n+1}^2=\frac{x_n+3}{2}$
Let $\left(x_n\right)$ be a real sequence such that $x_1=1,x_{n+1}^2=\dfrac{x_n+3}{2},\forall n\geq 1$.Compute $\lim_{n\to+\infty} 3^n\sqrt{\dfrac{9}{4}-x_n^2}$</p>

答:
You know that $x_n\to 3/2$ as $n\to\infty$.  You are doing fixed point (Picard) iteration for the function
$$f(x) = \sqrt{(x + 3)\over 2}$$
converging to $x = 3/2$ and you know that $$f'(3/2) = {1\over 6\sqrt{2}}.$$
Does this tell you anything?</p>

ContentLicense: CC BY-SA 3.0


问:
Eigenspaces of 2x2 Jacobian matrix with change of variable
Let

$$
A=
\begin{pmatrix}
v & \rho\\
v\left(v+2\rho\right) & \rho\left(2v+\rho\right)
\end{pmatrix}
$$
where $\rho>0$, $v=\alpha-\rho$ with $\alpha\in\mathbb{R}$.

Show that $A$ has two distinct eigenvalues $\lambda_1(\rho,v)<\lambda_2(\rho,v)$ and compute a basis of eigenvectors assoiated to these eigenvalues.

My work
We have
$$\det(\lambda I_2-A)=\lambda^2-(v+\rho(2v+\rho))\lambda+v\rho(v-\rho)$$
The discriminant is
$$\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2$$

I cannot seem to find a way to show that it is strictly positive. Even if I assume it is, I do not know how to find easily the eigenvectors (apart from resolving a system which will take a very long time). There has to be something that I am missing.

PS: this comes from an hyperbolic PDE where I need to show the system hyperbolicity:
$$
\left\{
\begin{array}{l}
\partial_t \rho+\partial_x(\rho(\alpha-\rho))=0\\
\partial_t(\rho\alpha)+\partial_x(\rho\alpha(\alpha-\rho))=0
\end{array}
\right.
$$
and we chose the change of variables ${\bf u} =(\rho,\alpha\rho)\to {\bf v} = (\rho,v)$ with $v=\alpha-\rho$.

答:
Note that we can write
$$
\Delta=v^2+4\rho^2v^2+\rho^4+4\rho^3v+6v\rho^2\\
= (1+4\rho^2)v^2+(4\rho^3+6\rho^2)v + \rho^4
$$
Consider the polynomial $p(v) = \Delta$. We see that this is a quadratic function with positive coefficients, so it will hold that $p(v) \geq 0$ for all $v$ if and only if its discriminant is either zero or negative. We find the discriminant of $p$ to be
$$
\Delta_v = (4\rho^3 + 6\rho^2)^2 - 4\rho^4(1 + 4 \rho^2) = 48\rho^5 + 32\rho^4.
$$
For positive $\rho$, $\Delta_v$ is positive. So, there must exist a value of $v$ for which $\Delta$ is negative, which in turn means that there are complex eigenvalues, contrary to your supposition.


问:
$R$ is a fractional ideal itself?
In Algebraic number theory by cassels

Proving $I^{-1}=\{\ a\in K\ |\ aI\subset R\}$

and $R(I)=\{\ a\in K\ |\ aI\subset I\}$

are fractional ideals of $R$

using $J=\{\ a\in K\ |\ aI_{2}\subset I_{1}\}$ ($I_1$ and $I_2$ are fractional ideals) is a fractional ideal.

It means $R$ is a fractional ideal of $R$ itself.

Am I right?

答: The product of fractional ideals is a fractional ideal, and given any fractional ideal $I$ of a ring $R$ there exists a fractional ideal $J$ such that $IJ = R$, so yes, $R$ is a fractional ideal of itself.

Title: $R$ is a fractional ideal itself?


问:
Calculate integral $ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx $ using the Dirichlet integral
I try to calculate the integral:

$$ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx, $$

using the Dirichlet integral

$$ \int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha. $$

I integrate this integral in parts, but I can't substitute the limits of integration because the limit is infinity.

Body:
Integrating twice by parts gives
$$
\int_0^T {\frac{{x - \sin x}}{{x^3 }}dx}
 = \frac{1}{2}\frac{{\cos T}}{T} + \frac{1}{2}\frac{{\sin T}}{{T^2 }} - \frac{1}{T} + \frac{1}{2}\int_0^T {\frac{{\sin x}}{x}dx} .
$$
Thus
$$
\int_0^{ + \infty } {\frac{{x - \sin x}}{{x^3 }}dx}  = \frac{1}{2}\int_0^{ + \infty } {\frac{{\sin x}}{x}dx}  = \frac{\pi }{4}.
$$


答:
$$\int\frac{x-\sin (x)}{x^3}\,dx=\int\frac{1}{x^2}\,dx-\int\frac{\sin (x)}{x^3}\,dx=$$
$$=-\frac{1}{x}-\int\frac{\sin (x)}{x^3}\,dx$$
by parts
$$\int\frac{\sin (x)}{x^3}\,dx=-\frac{\sin (x)}{2 x^2}-\int -\frac{\cos (x)}{2 x^2} \, dx=$$
again by parts
$$=-\frac{\sin (x)}{2 x^2}-\left(\frac{\cos (x)}{2 x}-\int -\frac{\sin (x)}{2 x} \, dx\right)=$$
$$=-\frac{\sin (x)}{2 x^2}-\frac{\cos (x)}{2 x}-\int \frac{\sin (x)}{2 x} \, dx$$
And finally
$$\int\frac{x-\sin (x)}{x^3}\,dx=-\frac{1}{x}+\frac{\sin (x)}{2 x^2}+\frac{\cos (x)}{2 x}+\int \frac{\sin (x)}{2 x} \, dx$$
the improper integral becomes
$$\int_0^{\infty } \frac{x-\sin (x)}{x^3} \, dx=\left[-\frac{1}{x}+\frac{\sin (x)}{2 x^2}+\frac{\cos (x)}{2 x}\right]_0^{\infty}+\int_0^{\infty } \frac{\sin (x)}{2 x} \, dx=$$
$$=\underset{M\to \infty }{\text{lim}}\left(\frac{\sin (M)}{2 M^2}+\int_0^M \frac{\sin (x)}{2 x} \, dx-\frac{1}{M}+\frac{\cos (M)}{2 M}\right)-\underset{M\to 0}{\text{lim}}\left(\frac{\sin (M)}{2 M^2}+\int_0^M \frac{\sin (x)}{2 x} \, dx-\frac{1}{M}+\frac{\cos (M)}{2 M}\right)$$
first limit gives $\frac{\pi }{4}$.

Second limit is zero because obviously $\int_0^0 \frac{\sin (x)}{2 x} \, dx=0$,
furthermore
$$\underset{M\to 0}{\text{lim}}\frac{-2 M+\sin (M)+M \cos (M)}{2 M^2}=\underset{M\to 0}{\text{lim}}\frac{\frac{\sin (M)}{M}^*+\cos (M)-2}{2 M}=$$
$$=\underset{M\to 0}{\text{lim}}\frac{\cos (M)-1}{2 M}=^{**}\underset{M\to 0}{\text{lim}}-\frac{\sin (M)}{2}=0$$

$^*$ fundamental limit $\frac{\sin (M)}{M}\to 1$ as $M\to 0$

$^{**}$ L'Hopital rule

Title: Calculate integral $ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx $ using the Dirichlet integral

Body:
I try to calculate the integral:

$$ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx, $$

using the Dirichlet integral

$$ \int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha. $$

I integrate this integral in parts, but I can't substitute the limits of integration because the limit is infinity.


问:
Adjoint operator, a condition for closed range
Let $X$ and $Y$ be two Hilbert spaces and $A\in\mathcal{L}(X,Y)$. Suppose that there's $\beta > 0$ such that
$$\inf_{z\ \in\ \text{Ker}(A)}\|x-z\|\ \leq\ \beta\|A(x)\|,\quad \forall\ x\in X.$$
Show that, $\text{R}(A) = \text{Im}(A)$ is closed.

Please, somebody can help with this problem? I tried to prove that $\text{R}(A) = \text{Ker}(A^*)^{\perp}$ in order to show that $\text{R}(A)$ is closed, but I really don't know how I can use the fact of $\text{dist}(x, \text{Ker}(A)) \leq \beta\|A(x)\|$. Please, I need some hints.

Thanks in advance.

答:
Consider well defined linear operator
$$
\hat{A}: X/\operatorname{Ker}(A)\to Y:\hat{x}\mapsto A(x)
$$
It is bounded and have the same image as $A$. It is enough to prove that $\hat{A}$ have closed image.

Since $X$ is complete and $\operatorname{Ker}(A)$ is closed then $X/\operatorname{Ker}(A)$ is complete. Note that $Y$ is also complete.

Inequality in your question preciesly means that $\hat{A}$ is bounded below. Just recall the definition of the quotient norm.

Any bounded below operator between Banach spaces have closed range.

From $1$, $2$, and $3$ it follows that $\hat{A}$ have closed range, and so does $A$. Note that this proof valid for any Banach spaces. Hilbertability is redundant here.


问:
Range of the transformation $T$ such that $T(v_1)=w_2,T(v_2)=T(v_3)=w_1+w_3$

With bases $v_1,v_2,v_3$ and $w_1,w_2,w_3$, suppose $T(v_1)=w_2$ and $T(v_2)=T(v_3)=w_1+w_3$. $T$ is a linear transformation. Find a vector that is not in the column space of $A$. Find a combination of $w$'s that is not in the range of the transformation $T$.

The solution given is $(1,0,0)\notin C(A)$, which is fine, and $w_1\notin range(T)$

But how come we can say $w_1\notin range(T)$ ?
$$
T(v_1)=w_2\\
T(v_2)=w_1+w_3\\
T(v_3)=w_1+w_3\\
$$
The matrix for the given linear transformation is
$$
A=\begin{bmatrix}0&1&1\\1&0&0\\0&1&1\end{bmatrix}
$$
$(1,0,0)\notin C(A)$ which is clear. but how do we conclude that $w_1\notin range(T)$ ?

$\color{blue}{\text{Can we equate $(1,0,0)$ to $w_1$ ?}}$

What I know

In the standard basis $\{e_1,e_2,e_3\}$ for the given vector $\vec{x}=x_1e_1+x_2e_2+x_3e_3$
$$
T(\vec{x})=x_1T(e_1)+x_2T(e_2)+x_3T(e_3)=\begin{bmatrix}T(e_1)&T(e_2)&T(e_3)\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=A\vec{x}
$$

If the input basis is $\{v_1,v_2,v_3\}$ and output basis is $\{w_1,w_2,w_3\}$ then $\vec{x}=y_1v_1+y_2v_2+y_3v_3$
$$
T(\vec{x})=x_1[T(v_1)]_w+x_2[T(v_2)]_w+x_3[T(v_3)]_w=\begin{bmatrix}[T(v_1)]_w&[T(v_2)]_w&[T(v_3)]_w\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=A_{v\to w}\vec{x}
$$

答:
I would assume that $(1,0,0)=w_1$... what else could it be?

Assume $\mathbf{v}:=a\cdot v_1+b\cdot v_2+c\cdot v_3$ is in your first vector space and solve:

$T(\mathbf{v})=w_1$.

If you are unsure about why $w_1=(1,0,0)$, but accept the matrix representation for $T$ (which is with respect to $w_1=(1,0,0)$ by the way), I can ask you to solve

$$\left[\begin{array}{ccc}0 & 1 & 1
\\ 1 & 0 & 0
\\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c}a\\b\\c\end{array}\right]=\left[\begin{array}{c}1\\0\\0\end{array}\right],$$
and you will see that $(1,0,0)$ is not in the range of the matrix representation of $T$...

I don't really understand your confusion to be honest... if you choose the basis $\{w_1,w_2,w_3\}$ for your target vector space, then you can choose $w_1=(1,0,0)$ and if you do that you get the matrix representation for $T$ that you have... how else can you get that matrix representation without making such a choice?


问:
Minimum estimation
Hi could anyone explain me how one come sup with the term circled with red? The proof tries to show that for integer numbers the mimimum is somewhere else (a bit higher) than it would be if f would be continous ($k_{0}$ is where the minimum is btw). Thanks!

答:
The first inequality probably comes from the Taylor expansion of $\ln(f)$ about $k_0$, since you're told the second derivative.

Since $k_1$ is the integer closest to the right of $k_0$ the difference between them is less than $1$ so the square of the difference is less than $1$. Then omitting that factor strengthens the inequality.

Title: Minimum estimation


问:
Combinatorics summation problem - $\sum_{k=0}^r k.{n \choose k}. {m \choose {c-k}} = ?$
Is there any way to calculate this - $$\sum_{k=0}^r k.{n \choose k}. {m \choose {c-k}}$$ where $r, n, m$ and $c$ are constants and $k \leq c$.

I have no idea how to approach this.

Thanks!

EDIT - To be more specific I actually need to calculate this - $$\sum_{k=0}^{150} k.{1000 \choose k}. {14000 \choose {150-k}}$$

答:
In your case, you actually have $c = r$, which simplifies the problem considerably.

Basically, it allows you to use the fact that
$$ \binom{n+m}{x} = \sum_{k=0}^{x} \binom{n}{k}\binom{m}{x-k}.$$
The only problem is getting your original expression in this form in the first place.

To do so, note that
$$ k\binom{n}{k} = k\frac{n!}{k!(n-k)!} = n\frac{(n-1)!}{(k-1)!((n-1)-(k-1)!)} = n\binom{n-1}{k-1}. $$

Therefore,
\begin{align*}
\sum_{k=0}^{r} k \binom{n}{k}\binom{m}{r-k}
   &= \sum_{k=1}^{r} k \binom{n}{k}\binom{m}{r-k} \\
   &= \sum_{k=1}^{r} n\binom{n-1}{k-1}\binom{m}{(r-1)-(k-1)} \\
   &= \sum_{s=0}^{r-1} n \binom{n-1}{s}\binom{m}{(r-1)-s} \\
   &= n\binom{m+n-1}{r-1}.
\end{align*}

Also, for completeness I should point out that this is consistent with DerpShady's answer -- in terms of this notation, their answer is
$$ r \frac{n}{n + m} \binom{m+n}{r}. $$


问:
Is the set $M := \{z \in \mathbb{C}: |z| + \operatorname{Re}(z) \leq 1\}$ compact?
I want to investigate whether the set $M$ mentioned in the question is compact. I know that $M$ has to be closed and bounded in order to be compact.

$M$ is not bounded since for $x \in \mathbb{R}^{-}$ we have that $|z| + \operatorname{Re}(z) = x -x = 0 \leq 1$. Therefore $x \in M$. So we know that $M$ can't be compact. Still I am curious how I can investigate whether the set is closed or not. How can this be done here?

答:
If it isn't bounded, you have your answer, regardless of closedness. That being said, the function $f:\Bbb C\to\Bbb R$ given by
$$
f(z)=|z|+\operatorname{Re}(z)
$$
is continuous, and your $M$ is the inverse image of a closed set under this function.


问:
Is $(a_n)_n$ with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ a Cauchy sequence?
Let $0 < q < 1$ and $(a_n)_n$ be a squence with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ for all $n ∈ ℕ$.

I need to show that this is a Cauchy sequence. I'm not sure how to start this one, as we just had cauchy sequences in lecture.
Can I maybe prove this by using induction?
How do start on a problem like this and where do I need to get to?

答: The increments $|a_{n+1}-a_n|$ are dominated by a geometric progression and therefore have a finite sum.  Hence the sequence converges.</p>


问:
Video lectures on Partial Differential Equations
Would anyone happen to know any introductory video lectures / courses on partial differential equations? I have tried to find it without success (I found, however, on ODEs).

It does not have to be free material, but something not to expensive would be nice.

答:
The best PDE introduction that I have found is by Dr. Pavel Grinfeld on youtube. The link is:

https://www.youtube.com/playlist?list=PLlXfTHzgMRUK56vbQgzCVM9vxjKxc8DCr

These are really really excellent. He starts with ODEs and then moves to PDEs. He covers the introduction to elliptic, hyperbolic,and parabolic PDEs, heat equation, separation of variables, and Fourier series. Unfortunately he does not cover other common topics like Sturm-Liouville problems and Green's functions, etc. But this is probably the best place to start. After that you can find individual videos on Green's functions, etc.


问:
let $R$ be a Ring, show that p is a prime element if and only if (p) is a prime ideal in $R$.
let $R$ be a Ring, show that p is a prime element if and only if (p) is a prime ideal in $R$.

I am puzzled as to what the definition of a prime element would be in a general ring. I thought divisibility is only well defined in integral domains.

答:
This isn't true in general: $0$ isn't considered a prime element in $\mathbb Z$ (or any ring, for that matter), but it generates a prime ideal.

For a commutative ring and a nonzero element $p$, it is true that $(p)$ is a prime ideal iff $p$ is a prime element: one just has to check the definitions.

The choice to use commutativity is just an expediency to stick with the normal definition of "prime ideal" in commutative algebra. Prime ideals of noncommutative rings are slightly more complicated to explain in terms of elements.

I thought divisibility is only well defined in integral domains.

For a commutative ring, we usually say "$a$ divides $b$ if there exists a $c$ such that $ac=b$."  Where does being a domain come into it?  Asking for it to be a domain buys us other nice things, but it has nothing to do with the definition of divisibility. Divisibility is nothing more than the containment partial-order on the principal ideals, which is always there.

For noncommutative rings one can do something similar but you'd have to take side into consideration. That is not usually something authors take time to do in introductory texts.


问:
Same $\sigma$-field of different sets
I have a probability measure space with random variables $X,Y:\Omega \to \mathbb{R}$

Question is how can is show that the sets : $$ A =
\{\omega \in \Omega:X(\omega) =Y(\omega)\} $$,

$$ B =
\{\omega \in \Omega:X(\omega) >Y(\omega)\} $$

$$ \Gamma =
\{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} $$
belong in $\sigma$-field $\mathcal{F}$

My effort is the below but I am not sure about $\Gamma$

For every $t \in \mathbb{R}$ we have :
$$\bigcap_{n=1}^{\infty} \{ \omega: X(\omega)\leq t\}= \{ \omega:\sup X (\omega) \leq t\}
=\{\omega: X_{n} (\omega) \leq t \}$$  for every $$n \in \mathbb{N} = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}$$.

But $$X_{n},n \in \mathbb{N}$$ is a random variable and the class $\mathcal{F}$ is a $\sigma$ - field then :
$$ \bigcap_{n=1}^{\infty} \{ \omega : X_{n}(\omega) \leq t  \} \in \mathcal{F}$$

The same for $$ \{\omega:Y(\omega)>t \}= \{\omega:X_{n}(\omega)>t \} $$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) >t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq t \} \in \mathcal{F}$$ for every $t \in \mathbb{R}$

For $$ \{\omega:Y(\omega)= e^{t} \}= \{\omega:X_{n}(\omega)
=e^t \} $$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) \geq e^t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq e^t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq e^t \} \in \mathcal{F}$$

答:
I understand that you have a a probability measure space $(\Omega, \mathcal{F}, P)$ and random variables $X,Y:\Omega \to \mathbb{R}$.

You want to prove that the sets : $$ A =
\{\omega \in \Omega:X(\omega) =Y(\omega)\} $$
$$ B =
\{\omega \in \Omega:X(\omega) >Y(\omega)\} $$
$$ \Gamma =
\{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} $$
belong in $\sigma$-field $\mathcal{F}$.

There is a simple way prove those results. Let us see it.

Proof:  Note that $X-Y$ is a random variable, so $(X-Y)^{-1}(\{0\}) \in \mathcal{F}$. Since
$$ A =
\{\omega \in \Omega:X(\omega) =Y(\omega)\}  = (X-Y)^{-1}(\{0\})$$
we have that $A\in \mathcal{F}$.

In a similar way,  since  $X-Y$ is a random variable, so $(X-Y)^{-1}((0,+\infty))\in \mathcal{F}$. Since
$$ B =
\{\omega \in \Omega:X(\omega) >Y(\omega)\} = (X-Y)^{-1}((0,+\infty)) $$
we have that $B\in \mathcal{F}$.

Finally, note that, since $X$ is a random variable, so is $e^X$. So just apply the result for $A$ to have that $\Gamma \in \mathcal{F}$.

Title: Same $\sigma$-field of different sets


问:
Would a circle overlap a parabola's bottom by more than just its vertex?
I mean, out of the condition that a circle actually crosses the parabola.
My question is when a circle is "inside" a parabola, would it touch part of the parabola other than just the parabola's vertex at the bottom?

So just the curiosity brought me crazy...but would this be possible?? (There was an image to show but my reputation is not enough since I'm a freshmeat :P)

Thanks xx

答:
We may assume the parabola is $y=x^2$, since by the similarity $x=kx',y=ky'$ this becomes $y'=kx'^2$ (all parabolas are similar), and the problem is the same if the plane is stretched by a magnification factor. Let the circle be centered at $(0,k)$ ($k>0$) and have radius $k$ so that it goes through the origin.

Only the lower half of the circle might intersect the parabola, and this lower half has equation
$$y=k-\sqrt{k^2-x^2}.$$
Now it is desired in the question that the circle lie "inside the parabola", which becomes that, for $-k \le x \le k$, we have
$$[1] \ \ k-\sqrt{k^2-x^2} \ge x^2.$$
This boils down to $x^2 \ge 2k-1$, where at one step we divided by $x^2$. This shows that we need exactly $k \le 1/2$ in order for the circle to lie above the parabola: If $k \le 1/2$ then $x^2 \ge 2k-1$ holds for all $x$, in particular on $[-k,k]$, while if $k>1/2$ there will be positive $x \in [-k,k]$ for which $x^2<2k-1$, against the circle lying above the parabola.

Now assume that indeed $k \le 1/2$. The condition that the lower semicircle meet the parabola at any point is [1] with equality, call that [1']. Only for $k=1/2$ is there a solution of $x^2=2k-1$ and it is at the origin, while if $k<1/2$ the equation $x^2=2k-1$ has no real solutions, and the circle only meets the parabola at the origin. (Recall the step in which division by $x^2$ made the $0$ drop out as a solution to [1'].)

So in conclusion if a circle lying above the parabola goes through the vertex, then it does not meet the parabola at any other points.


问:
How to solve the matrix equations: $\frac{1}{2}(\Omega Q - Q\Omega)=F$ for $\Omega$
How to solve the following matrix equation: $$\frac{1}{2}(\Omega Q + Q\Omega^T)=F$$

$Q$ is of rank $1$, $\text{tr}(Q)=1$, $Q\succeq0$ (positive semidefinite).  So we can werite $Q = qq^T$ for some $q$ with $\|q\|=1$.

$\Omega$ is skew-symmetric, i.e., $\Omega^{T} = -\Omega$. So diagonal entries are all $0$.

$\Omega$, $Q$, $F$ are $4\times 4$ square matrices. $\Omega$ is the variable I want to solve.

I cannot find any method to solve it. Can anyone suggest me how to solve $\Omega$ in terms of $Q$ and $F$?

Even though we cannot obtain the closed form, any approximation or trick can we use?

Thanks!

Note:  $\Omega$ should look like
$$\begin{bmatrix}0 & \omega_3 & -\omega_2 & \omega_1 \\
  -\omega_3 & 0 & \omega_1 & \omega_2 \\
   \omega_2 & -\omega_1 & 0 & \omega_3 \\
   -\omega_1 & -\omega_2 & -\omega_3 & 0 \end{bmatrix}$$

答:
As a "brute-force" approach, it suffices to solve a $32 \times 16$ linear system.

Since it is given that $\Omega^T = -\Omega$, we can rewrite the equation as
$$
\frac 12 (\Omega Q - Q\Omega) = F.
$$
Note that this is a Sylvester equation. As such, this equation can be rewritten in terms of the vectorization operator and Kronecker product
$$
\frac 12 (Q^T\otimes I_4- I_4 \otimes Q) \operatorname{vec}(\Omega) = \operatorname{vec}(F) \implies\\
(Q^T\otimes I_4- I_4 \otimes Q) \operatorname{vec}(\Omega) = 2\operatorname{vec}(F).
$$
Note that this equation will not have a unique solution (if it has any solution at all), so we must impose the constraint that $\Omega$ is skew-symmetric. This constraint can be expressed with the help of the commutation matrix as
$$
(I_4 + K^{(4,4)})\operatorname{vec}(\Omega) = 0.
$$
Putting everything together, we can express the system to be solved as the block-matrix equation
$$
\pmatrix{Q^T \otimes I_4 - I_4 \otimes Q\\ I_4 + K^{(4,4)}} \operatorname{vec}(\Omega) = \pmatrix{2\operatorname{vec}(F)\\0}.
$$


答:
The following lacks justification (I don't know much about swapping integrals). But it sounds like the gist of what you're asking for.

$$\int_{0}^{\beta} \frac{\sin\alpha x}{x}\, d\alpha = \frac{1-\cos \beta x }{x^2}$$
$$\int_{0}^{\gamma} \frac{1-\cos \beta x}{x^2}\, d\beta = \frac{\gamma x - \sin \gamma x}{x^3}$$

So compute

$$\int_{0}^{1}\int_{0}^{\beta}\frac{\pi}{2}\text{sgn}(\alpha)\, d\alpha\, d\beta = \int_{0}^{1}\frac{\pi}{2}\beta\, d\beta = \frac{\pi}{4}$$


问:
If $Y_t=\int_0^t X_sdW_s$ where $\int_0^t X_s^2ds<\infty $ a.s., do we have that $\mathbb EY_t=0$?
Let $Y_t=\int_0^t X_sdW_s$. I know that if $\mathbb E\int_0^tX_s^2ds<\infty $, then $(Y_t)$ is a martingale, and $\mathbb EY_t=0$ for all $t$. But what if we only have $\int_0^t X_s^2ds<\infty $ ?

I know that in this case, $(Y_t)$ is a local martingale only, but do we have that $\mathbb E Y_t=0$ or not ? What I know, is thata there is a sequence of stopping time $\tau_n\nearrow \infty $ s.t. $(Y_{\tau_n\wedge t})$ is a martingale. Therefore, if $\mathbb E Y_{t\wedge \tau_n}=0$ for all $t$ and all $n$. But I think that we can't switch $\mathbb E$ and limit, so I think that that $\mathbb EY_{t\wedge \tau_n}$ won't hold, but I can't find a concrete example. Any idea ? And if it doesn't work, is there a condition on $Y=(Y_t)$ to have that $\mathbb E Y_t=0$ ?

答:
There is the so-called Dudley representation theorem: for any distribution, there exists an adapted process $X$ such that $\int_0^t X_s^2 ds<\infty$ and $\int_0^t X_s dW_s$ has this distribution. So $\mathbb E Y_t$ can be anything (and can be even undefined, as another answer suggests).

A sufficient condition in terms of $Y$? I doubt it exists. Actually, it is possible that $\sup_{0\le s\le t} \mathbb E e^{|Y_s|}<\infty$ so that $Y$ is uniformly integrable in a very strong sense, but still $\mathbb E Y_t\neq 0$.


问:
An Olympiad Geometry problem with incenter configurations.
Let the inscribed circle of triangle ABC touches side BC at D ,side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that $\frac{FG}{EG} = \frac{BF}{CE}$.

So this problem is equivalent to proving similarity between $\Delta BFG$ and $\Delta CEG$.

I was able to prove that $\angle GFB$ = $\angle GEC$ but after that, I hit a dead end. I found the point $G$ pretty annoying as I couldn't apply any circle theorems to it.

Any solution is highly appreciated but I am not very good at inversion and complex numbers so please don't use them.

答:
Draw $DE$, $DF$, $IB$ and $IC$.

Observe that, $\angle GED=\angle FED=90-\frac{\angle B}{2}=\angle BID$ and $\angle EGD=\angle IDB=90^{\circ}$; Hence, $\triangle EGD\sim \triangle IDB$.  Similarly, one can yield,  $\triangle FGD\sim \triangle IDC$.

Thus, $\frac{FG}{GD}=\frac{ID}{DC}$ and $\frac{EG}{GD}=\frac{ID}{BD}$.

Dividing these ratios gives, $\frac{FG}{EG}=\frac{BD}{DC}=\frac{BF}{CE}$  [Since $BD=BF$ and $DC=CE$]


问:
$\int_0^{\infty} \frac{f(t)}{t}dt<\infty \implies \lim\limits_{t\to\infty} f(t) = 0?$?
I was wondering if a function $f>0$ is such that $\int_0^{\infty} \frac{f(t)}{t}dt<\infty$ then we have that $\lim\limits_{t\to 0}f(t) = 0$. I was thinking it is not the case. Intuitively if I take $f(t)=\sum\limits_{n\in\mathbb{N}}\chi_{(n-{1/n},n+1/n)}$ then $\frac{f(t)}{t}$ would be "more or less" $f(t)=\sum\limits_{n\in\mathbb{N}}\chi_{(n-{1/n^2},n+1/n^2)}$ for which the integral if finite.

答:
Not only the limit may not be equal to zero... But $f$ can even be unbounded.

An example is the piecewise linear map defined for $n \in \mathbb N$ by

$$f(x)=
\left\{
  \begin{array}{ll}
    0 & \mbox{if } x=0\\
    0 & \mbox{if } x=n-\frac{1}{2n^3}\\
    n^2 & \mbox{if } x=n\\
    0 & \mbox{if } x=n+\frac{1}{2n^3}\\
  \end{array}
\right.$$


问:
extention of the 6n$\pm1$ prime rule for $30n$, $210n$ ...
I am currently creating a program to exhaustively search for prime numbers. I head from sombody that this rule can be extended by multipliting 6 by the next prime number, giving 30, and then continuing this sequence 210.

The offsets from this value would be the prime numbers less than it, apart from the ones multiplied together to make this value, and also 1.

This works for 30, meaning the values you will need to check would be $30n+1$, $30n+7$,  $30n+11$, $30n+13$, $30n+17$, $30n+19$, $30n+23$ and $30n+29$

However, there is something that is missing for the larger values, the person that I learnt rule from said that there was a way to make it work for $210n$ and upwards, however they forgot how.

With a base value of 210 you need to check 1, and all the prime numbers lower than 210, not including the prime factors of 210, as above. However you also need to check $210n+121$, $210n+143$, $210n+169$, $210n+187$ and $210n+209$.

Any help in figuring out what the rules for the other values to be checked are for values larger than $210n$ would be greatly appreciated, since I need to automatically work these out for huge values of $n$.

答:
Let's construct the period $30$ sequence in a more instructive way, starting from the period $6$ sequence.

The period $6$ one consists of $6n+1$ and $6n+5$ (the latter is equivalent to $6n-1$, but it is easier to use only positive offsets). This sequence avoids all multiples of $2$ and all multiples of $3$, and note that $6=2\cdot3$.

The next prime we would like to avoid multiples of is $5$. So take five successive periods of the $6$-sequence:
$$1, 5\\7, 11\\13, 17\\19, 23\\25, 29$$
Each column has one multiple of $5$ in it, in this case $5$ and $25$. After removing those, you are left with the period $30$ sequence $30n+k$ where $k\in\{1,7,11,13,17,19,23,29\}$. This sequence avoids all multiples of $2$, $3$, and $5$.

To extend this with the next prime, $7$, you can do the same thing. Take $7$ successive periods of this sequence, get rid of the multiples of $7$, to give you the sequence of period $210$ that avoids all multiples of $2, 3, 5$, and $7$.

$$1,7,11,13,17,19,23,29\\
31,37,41,43,47,49,53,59\\
61,67,71,73,77,79,83,89\\
91,97,101,103,107,109,113,119\\
121,127,131,133,137,139,143,149\\
151,157,161,163,167,169,173,179\\
181,187,191,193,197,199,203,209$$

Again, each column has exactly one multiple of $7$ in it ($7$, $49$, $77$, $91$, $119$, $133$, $161$, $203$). Eliminate those, and the remaining $49$ numbers form your period $210$ sequence. Note that these are not all prime numbers, but also include products of larger primes that happen to be below $210$ such as $11^2=121$ and $11\cdot 19=209$.

This method soon becomes unwieldy, since for each extra prime $p$ you want to avoid, the period is multiplied by $p$, and the number of cases to check within that period is multiplied by $p-1$.

In effect, you are generating the result of a few rounds of the Sieve of Eratosthenes, but generally you are better off using the sieve fully.


答:
(Fill in the gaps as needed. If you're stuck, show your work and what you've tried.)

Just direct angle chasing and trigo will help you express those ratios.

Show that $BF /CE  = BD / DC = (BD / DG) / (DC / DG) = \tan ( 90^\circ - \beta /2)  / \tan (90^\circ - \gamma / 2) = \tan \gamma/2 / \tan \beta/2 $

Show that $ \angle IFE = \angle IAE  = \alpha / 2 $.

Show that $ FG / GE = (FG / GD) / (GE / GD) = \tan \gamma/2 / \tan \beta/2 $ .

Title: An Olympiad Geometry problem with incenter configurations.

Body:
Let the inscribed circle of triangle ABC touches side BC at D ,side CA at E and side AB at F. Let G be the foot of the perpendicular from D to EF. Show that $\frac{FG}{EG} = \frac{BF}{CE}$.

So this problem is equivalent to proving similarity between $\Delta BFG$ and $\Delta CEG$.

I was able to prove that $\angle GFB$ = $\angle GEC$ but after that, I hit a dead end. I found the point $G$ pretty annoying as I couldn't apply any circle theorems to it.

Any solution is highly appreciated but I am not very good at inversion and complex numbers so please don't use them.


问:
Is the ring of global functions on an integral scheme integral?
Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)? It is necessarily a reduced ring (https://stacks.math.columbia.edu/tag/01OL) so the question is whether it is irreducible.</p>

答:
This is an alternative answer to your question.

Question:
"Q1. Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)?

Q2. It is necessarily a reduced ring?"

Answer: If you use the definition in Hartshorne (page 82) you define a scheme $(X,\mathcal{O}_X)$ to be "integral" iff for any open set $U\subseteq X$ it follows
$\mathcal{O}_X(U)$ is an integral domain. In particular, since $X$ is an open set, it follows $\Gamma(X,\mathcal{O}_X):=\mathcal{O}_X(X)$ is an integral domain. Hence it seems to me the answer to Q1 is "yes, this is true by definition". Since an integral domain is a reduced ring it follows $\mathcal{O}_X(X)$ is a reduced ring. Hence the answer to Q2 is also "yes".


问:
What's the $\lim _{x \to \frac{π}{4}} \frac{\ln (\cot x)}{{1-\tan x}}$?

$$
\lim_{x \to \pi/4} \frac{\ln (\cot x)}{{1-\tan x}}
$$

I'm trying to solve it without using L'Hopital or Taylor series.

答:
Let the given limit be $L$ and $y = \tan x$. Then observe that as $x \to \pi /4 $ we have $y \to 1$. Then
$$
L = \lim_{x \to \pi /4} \frac{ \ln \cot x } {1 -  \tan x} = \lim_{y \to 1} \frac{ \ln y}{y - 1}
$$

Can you finish this?


问:
How to evaluate these improper integrals (if they’re even possible)?
I’m trying to evaluate the following two integrals: $$\int\limits_{-\infty}^\infty\frac{e^{-x^2}\cos(x)}{1+x^2}\,dx$$
and
$$\int\limits_{-\infty}^\infty\frac{e^{x^2}\cos(x)}{1+x^2}dx.$$

The first one I know is convergent (from wolframalpha) but I cannot calculate its value.

The second one i suspect to be divergent since the integrand oscillates a lot, but maybe you can find some sort of symmetry in it where things cancel out each other.

I’ve tried looking at some complex contour integrals along for example a semicircle closed by the real axis to be able to make use of the residue theorem, but I always end up with the integral being divergent on some part of the curve. I think a different approach is needed.

答:
Reference... Item 3.945.2 in G&R:

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.

\begin{align}
&\int_{0}^\infty \frac{e^{-\beta x^2}\cos (ax)}{x^2+\gamma^2}\;dx
\\ &=
\frac{\pi e^{\beta\gamma^2}}{4\gamma}\left[2\cosh(a\gamma)+
e^{-a\gamma}\Phi\left(-\gamma\sqrt{\beta}+\frac{a}{2\sqrt{\beta}}\right)
-e^{a\gamma}\Phi\left(\gamma\sqrt{\beta}+\frac{a}{2\sqrt{\beta}}\right)
\right],
\\
&\qquad \text{Re }\beta > 0, \text{Re }\gamma > 0, a > 0 .
\end{align}
Note that $\int_{-\infty}^\infty = 2 \int_0^\infty$; plug in $\beta=1, \gamma=1, a=1$; recall the notation $\Phi = \text{erf}$.  We then get
$$
\int_{-\infty}^\infty \frac{e^{-x^2}\cos x}{x^2+1}\;dx =
-\frac{\pi}{2}\, \left( {{\rm e}^{2}}{\rm erf} \left(3/2\right)-{{\rm e}^{2
}}+{\rm erf} \left(1/2\right)-1 \right)
$$


问:
Non-linear optimization: Levenberg-Marquardt gives different results using different forms of the same function
Background:

I'm using non-linear least squares to solve indoor trilateration problem using BLE (bluetooth low energy) beacons. My starting point is this: https://nrr.mit.edu/sites/default/files/documents/Lab_11_Localization.html B part of the document.

(I've been using this JAVA lib for trilateration ( https://github.com/lemmingapex/Trilateration ) which works fine but needed to switch to c++ and I couldn't find a similar lib. So I experimented with google's ceres solver and eigen's levenberg-marquardt lib)

Problem/question:

$(x*,y*) = argmin \sum_{i=0}^n [ri^2-(x-xi)^2 - (y-yi)^2$

For these non-linear optimization libs, you have to define the nonsquared function.

Using this function for L-M algorithm:
$\sqrt{(x-xi)^2 + (y-yi)^2} - ri$

c++ code:

```<code>fvec[i] = sqrt(pow(matrix(i, 0) - b[0], 2) + pow(matrix(i, 1) - b[1], 2)) - matrix(i, 2);
</code>```

I get different results when using this:
$1 - \frac{\sqrt{(x-xi)^2 + (y-yi)^2}}{ri}$

c++ code:

```<code>fvec[i] = 1 - sqrt(pow(matrix(i, 0) - b[0], 2) + pow(matrix(i, 1) - b[1], 2)) / matrix(i, 2);
</code>```

The latter gives much better results. Shouldn't this function defined in different forms give the exact same result when using L-M algorithm to find minimum? What am I missing?

By the way for predefined problems (in the JAVA github lib tests) both give correct results, but the considerable difference happens when I'm testing with reallife data (many beacons).

Notes:

First defined function (fvec) returns RelativeReductionTooSmall and second (which gives good results) returns with RelativeErrorTooSmall. So I guess the stopping condition may not be set properly. But this shouldn't have an effect on my problem.

I'm new to optimization. I understand how gradient-descent works, but didn't have the time to dig deeper into L-M algorithm. I know it "balances"/switches between gradient descent and Newton-Gauss approach. But my gut says I'm missing something more basic.

I'm treating this function as an eqution (hence the above mentioned 2 forms):
$0 = 1 - \frac{\sqrt{(x-xi)^2 + (y-yi)^2}}{ri}$ but maybe I can't treat the left side just as 0? Why not?

答:
It seems you missed a square, since the minimization problem refers to the function
$$
\sum_{i=0}^n f_i(x,y)^2,
$$
with
$$
f_i(x,y) = r_i^2 - (x-x_i)^2 - (y-y_i)^2.
$$


问:
How to show that that every finitely generated variety is locally finite
Revised Question:

From what I read, an algebra is locally finite if every finitely generated sub-algebra is finite, and a variety is locally finite if every member is locally finite (where a variety is the class of algebras closed under homomorphisms, sub-algebras, and direct products (direct product is denoted as $\prod_{i \in I}\mathbb{A}_i$)).

A variety is finitely generated if it is generated by a finite set of finite algebras.

Now, I read that there's a result (from universal algebra) that every finitely generated variety is locally finite.

The result says that given a variety generated by a set of finite algebras $\{\mathbb{A}_1,\mathbb{A}_2,...,\mathbb{A}_k\}$, we have that for any algebra $\mathbb{B}$ in the variety, there exists a homomorphism $\mathbb{C} \to \mathbb{B}$, for some algebra $\mathbb{C}$ in the variety, where $\mathbb{C} \cong \prod_{i\in I} \mathbb{C}_i$.

The proof uses the fact that $\textbf{Hom}(\mathbb{C},\mathbb{A}_i)$ is supposed to be a finite set - which is then supposed to imply that there are only finitely many maps from $\mathbb{C} \rightarrow \prod_{i \in I} \mathbb{C}_i\rightarrow \mathbb{C}_i$. Given this, there will be a finite subset $J$ of $I$ such that $\mathbb{C}$ embeds in $\prod_{j \in J}\mathbb{C}_j$, where $\prod_{j \in J}\mathbb{C}_j$ is a finite product of finite algebras. From this embedding of $\mathbb{C}$ into the finite product of finite algebras $\prod_{j \in J}\mathbb{C}_j$, we have that $\mathbb{C}$ is finite, and in turn, $\mathbb{B}$ is finite as well

But the proof did not go into any details as to why (1) $\textbf{Hom}(\mathbb{C},\mathbb{A}_i)$ is finite and how (2) this fact implies that $\mathbb{C} \to \prod_{i \in I}\mathbb{C}_i \to \mathbb{C}_i$ ... Any help for (1) and (2) ?

答:
First, let me include the full proof from the book you're most likely using. It'll be easier than rewriting it.

The book is Bergman's "Universal Algebra".

First, you stated that $\mathbb{C}$ is isomorphic to $\prod_{i\in I}\mathbb{C}_i$. This is not true, it's only isomorphic to a sub-algebra of $\prod_{i\in I}\mathbb{C}_i$.

Since $X$ generates $\mathbb{C}$, $C = \bigcup_{n=1}^\infty X_n$ where $X_0 = X$, $X_{n+1} = X_n\cup \{ p(a_1, ..., a_k): a_i\in X_n, p\text{ is } k\text{-ary}\}$.

Now if two homomorphisms $f, g:\mathbb{C}\to\mathbb{A}_i$ are equal on $X$, one can prove (and this is easy) by induction that they are equal on each of $X_n$, so that they are equal on all of $C$, so $f=g$.

Therefore the map $f\mapsto f|_X, \text{Hom}(\mathbb{C}, \mathbb{A_i})\to A_i^X$ which sends a homomorphism to its restriction to $X$ is injective. So there can be at most $|A_i^X|<\infty$ of such homomorphisms.

I hope this clears it up, I invite you to a further discussion if it's not clear.


问:
How can I simulate poisson process?
how can I simulate a single realization of the Poisson counting process as described above. Such that the discrete-amplitude random process X(t) counts the number of packets arriving in the time interval [0,t). Take your simulation total time to be 10 hours, time sampling step size to be 1 sec.

Assume that the average number of packets arriving per minute in a certain cell equal 10

I tried to simulate the poisson but I can't understand how to simulate it and how to compute the interval times between the arrival of each two successive packets

答: The chance of an event in $1$ second is $\frac 1{6}$ because you have $10$ events per minute.  Throw a random number with that probability for whether an event happens.  You have an error because you can't have two events in one second, but the chance is rather small.  I would have taken a shorter interval to reduce that, but $1$ second is specified in the problem.

Title: How can I simulate poisson process?


问:
Explanation for $\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $
I don't see how / why one can rewrite the integral as following:

$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi}  \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi}  \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $$

I think this should be rather easy, but I don't see what I'm missing.

答:
HINT:

Note that $\int_{(j-1)\pi+\pi/2}^{j\pi+\pi/2}|\cos(x) |\,dx=2$.  And the denominator of the integrand is a constant with respect to the variable of integration.


问:
Is $(a_n)_n$ with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ a Cauchy sequence?
Let $0 < q < 1$ and $(a_n)_n$ be a squence with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ for all $n ∈ ℕ$.

I need to show that this is a Cauchy sequence. I'm not sure how to start this one, as we just had cauchy sequences in lecture.
Can I maybe prove this by using induction?
How do start on a problem like this and where do I need to get to?

答:
In order to prove that it is a Cauchy sequence you have to prove that
$$|a_{n+p}-a_n|\longrightarrow 0,\text{ when }n\to\infty,p\to\infty,n,p\in\mathbb N$$

In your case , I 'll give you a HINT:

$$|a_{n+p}-a_n|=|a_{n+p}-a_{n+p-1}+a_{n+p-1}-\dots-a_{n+1}+a_{n+1}-a_n|\leq$$
$$|a_{n+p}-a_{n+p-1}|+\dots +|a_{n+1}-a_n|$$

Can you now continue? Express each difference above in terms of $q,a_0,a_1$


问:
Calculating Odds/Probability with Overlapping Type
Calculating Odds/Probability with Overlapping Type

I'm making a game where cards can have 1-2 types. Think pokemon types, where a 1-type pokemon is "normal" type and a two type pokemon is "normal/flying".

I'm trying to calculate odds of achieving certain hands given specific types. Something like: what are the odds of pulling 1 normal type in a draw of 5 cards, with a deck size of 50 and 4 normal types? Is relatively easy.

Where I'm running into trouble is where the types overlap. I'm trying to calculate something like:

What are the odds of pulling 2 normal types and 2 flying types in a draw of 5 cards with a deck size of 50, given 4 normal types, 3 flying types, and 4 normal/flying types (where normal/flying can count as a draw toward EITHER, but not BOTH, in other words, 1 normal/flying = 1 normal OR 1 flying, not 1 normal AND 1 flying).

Does anyone know where I could look for formulas or calculators that take into account the above scenario? Thank you.

答:
I believe motherboard's answer to contain a slight error. They list 5 possible types of compliant hands, but they should list 9, since there are 3 ways each type can be satisfied (with 0, 1, or 2 pure cards).

I reckon the following categories and quantities (following motherboard's notation):

$\{PN,PN,PF,PF\}: {4 \text{ PN } \choose 2} \times {3 \text{ PF } \choose 2} \times {4 \text{ M } \choose 0} = 18$

$\{PN,PN,PF,MF\}: {4 \text{ PN } \choose 2} \times {3 \text{ PF } \choose 1} \times {4 \text{ M } \choose 1} = 72$

$\{PN,PN,MF,MF\}: {4 \text{ PN } \choose 2} \times {3 \text{ PF } \choose 0} \times {4 \text{ M } \choose 2} = 36$

$\{PN,MN,PF,PF\}: {4 \text{ PN } \choose 1} \times {3 \text{ PF } \choose 2} \times {4 \text{ M } \choose 1} = 48$

$\{PN,MN,PF,MF\}: {4 \text{ PN } \choose 1} \times {3 \text{ PF } \choose 1} \times {4 \text{ M } \choose 2} = 72$

$\{PN,MN,MF,MF\}: {4 \text{ PN } \choose 1} \times {3 \text{ PF } \choose 0} \times {4 \text{ M } \choose 3} = 16$

$\{MN,MN,PF,PF\}: {4 \text{ PN } \choose 0} \times {3 \text{ PF } \choose 2} \times {4 \text{ M } \choose 2} = 18$

$\{MN,MN,PF,MF\}: {4 \text{ PN } \choose 0} \times {3 \text{ PF } \choose 1} \times {4 \text{ M } \choose 3} = 12$

$\{MN,MN,MF,MF\}: {4 \text{ PN } \choose 0} \times {3 \text{ PF } \choose 0} \times {4 \text{ M } \choose 4} = 1$

This would yield a total of 293 possible 4-card combinations which satisify the (2 normal, 2 flying) constraint and a resultant probability that a randomly selected, 5-card hand will have at least 2 normal and at least two flying of $$\frac{293\times 46}{{50 \choose 5}}=\frac{293}{46060}\simeq0.0064.$$

For exactly 2 normal and exactly 2 flying, you would need to multiply by 39 instead of 46 (thus excluding the extra normal and flying cards from the fifth slot). This would yield a probability of $$\frac{293\times 39}{{50 \choose 5}}=\frac{11427}{2118760}\simeq0.0054.$$

In general, when you are looking to determine how many ways a set of $n$ elements can be partitioned into $k$ groups of $n_k$ size, here is the formula: $$P_{n_{1},...,n_{k}}=\frac{n!}{(\Pi_{i=1}^k n_i!)(n-\Sigma_{i=1}^k n_i)!}$$

Further reading here.


答:
Because

$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi}  |\cos(t)| \,dt = 2 $$ for any $j \in \mathbb Z$.

Title: Explanation for $\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $

Body:
I don't see how / why one can rewrite the integral as following:

$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi}  \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi}  \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $$

I think this should be rather easy, but I don't see what I'm missing.


答:
Get the denominator outside the integral, as it's a constant.

You are integrating $|\cos(x)|$, which is $\pi$-periodic, on an interval on length $\pi$. The result is the same as

$$\int_{-\pi/2}^{\pi/2}\cos(x)dx=[\sin x]_{-\pi/2}^{\pi/2}=2$$


问:
Question regarding the simple proof involving dual space and annihilator

If $\eta$ is a suspace in a finite-dimensional vector spce $v$, then
$\eta^{00}=\eta$

Proof: Observe that we use here the convention, established at the end
of 16, that identifies $v$ and $v''$. By definition, $\eta^{00}$ is
the set of all vectors $x$ such that $[x,y]=0$ for all $y$ in
$\eta^{0}\dots$

16 explains that in convention, we can identify $v$ with its double dual space $v''$, because they are isomorphic. But I'm not sure how the second sentence in the above proof follows.
From my understanding,
by definition, $\eta^{00}$ is the set of all vectors $x$ in $v''$ such that $[y, x]=0$ for every $y\in\eta^{0}$. I'm not sure how we can switch the position of $x$ and $y$ from the isomorphism of $v$ and $v''$.

答:
I believe it is probably this. The Convention you are talking about is to identify $x\in v$ with a functional $\overline{x}:v'\to k$ such that $\overline{x}(y)=y(x)$ for every $y\in v'$. That means that

$$[y,\overline{x}]=\overline{x}(y)=y(x)=[x,y]$$

Now, $\eta^{00}$ is the set of all vectors $\overline{y}\in v''$ such that $[x,\overline{y}]=0$ for all $x\in\eta^0$. (Note we can always see those vectors as being of the form $\overline{y}$ for some $y\in v$, because the correspondence $v\mapsto v''$ happens to be "onto".)

Swap the role of the letters $x$ and $y$, and you get that $\eta^{00}$ is the set of all vectors $\overline{x}\in v''$ such that $[y,\overline{x}]=0$ for all $y\in\eta^0$.

Now, identify $\overline{x}$ with $x$ and you get that $\eta^{00}$ is the set of all vectors $x\in v$ such that $[x,y]=0$ for all $y\in\eta^0$.


问:
Course-of-values induction, according to Kleene IM (1952)
I'm having troubles with exercise *162a in "Introduction to Metamathematics", by S.C. Kleene, 1952.

The ask is to prove:

(1) $ \vdash A(0) \& \forall x [\forall y (y \leq x \Rightarrow A(y)) \Rightarrow A(x')] \Rightarrow A(x)$

where x and y are variables distinct from each other, y is free for x in A(x) and does not occur free in A(x).

Kleene's hint is to proof:

(2) $ \vdash A(0) \& \forall x [\forall y (y \leq x \Rightarrow A(y)) \Rightarrow A(x')] \vdash \forall y (y \leq x \Rightarrow A(y))$

and

(3) $ \vdash \forall y (y \leq x \Rightarrow A(y)) \vdash A(x)$

from which (1) follows. He also suggests that (2) can be proofed using the induction rule, which requires the proof of:

(4) $\vdash A(0) \& \forall x [\forall y (y \leq x \Rightarrow A(y)) \Rightarrow A(x')] \vdash \forall y (y \leq 0 \Rightarrow A(y))$

and of

(5) $\vdash A(0) \& \forall x [\forall y (y \leq x \Rightarrow A(y)) \Rightarrow A(x')], \forall y (y \leq x \Rightarrow A(y)) \vdash \forall y (y \leq x' \Rightarrow A(y))$

I managed to prove (3), but I'm struggling with (4) and (5). Can you help? Thanks

答:
Hint for (4): by induction with $P(y) := (y \le 0 \to A(y))$.

$A(0)$ by assumption; thus, using $P \to (Q \to P)$ and MP, we get $(0 \le 0) \to A(0)$.

Assume $P(y)$ and prove $P(y+1)$. But $P(y+1)$ is $(y+1 \le 0 \to A(y+1))$ and $\lnot (y+1 \le 0)$, because $y+1$ is a successor and we know that no successor can be equal to $0$.

Thus, using tautology $\lnot P \to (P \to Q)$ and MP we get $P(y+1)$.

Thus: $P(y) \to (P(y+1)$; $y$ is not free in the assumption and we can generalize it to get: $\forall y (P(y) \to (P(y+1))$. With $P(0)$, by Induction: $\forall y ((y \le 0) \to A(0))$.

Hint for (5)

Using the second part of the fist premise instantiating the leading quantifier we have: $∀y(y≤x \to A(y)) \to A(x')$ and using 2nd premise and MP we get: $A(x')$ and thus $y=x' \to A(y)$. From which: $y=x' \vdash (y \le x' \to A(y))$.

If $y≤x'$ we have $y \le x \lor y=x'$. Thus, assuming $y \le x$ from 2nd premise we have $A(y)$ and so $y \le x \to A(y)$ and also $y \le x \vdash (y \le x' \to A(y))$.

Thus, from $y \le x \lor y=x'$ by cases, we get: $y \le x' \to A(y)$


问:
Isometries of $\mathbb{R}^n$
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be such that $\left\| f(x)-f(y)\right\| =\left\| x-y\right\|$.  Is $f$ necessarily surjective?

If this is so, you can prove (Mazur-Ulam Theorem) that $f$ is affine, and hence you could classify all isometries of $\mathbb{R}^n$.  However, at the moment, I can't think of any good ideas to prove that $f$ is surjective.  For that matter, is it even the case that $f$ must be surjective?

Any ideas would be most welcomed.

Thanks much!

答:
Let $ f $ be an isometry of $ \mathbb{R}^n $ ( i.e. a map $ f : \mathbb{R}^n \rightarrow \mathbb{R}^n $ such that $ || f(x) - f(y) || = || x - y || $ for all $ x,y $ ). We can write $ f(x) = f(0) + ( f(x) - f(0) ) $ and now $ g(x) := f(x) - f(0) $ is an isometry fixing $ 0 $ ( especially $ g $ preserves norm, i.e. $ || g(x) || = || x || $ for all $ x $ ).  We'll now focus on $ g $.

Notice $ g $ preserves inner product i.e. $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ for all $ x, y $ ( since expanding $ \langle g(x) - g(y), g(x) - g(y) \rangle = \langle x - y, x - y \rangle $ gives $ || g(x) ||^2 - 2 \langle g(x), g(y) \rangle + || g(y) ||^2 = || x ||^2 + || y ||^2 - 2 \langle x, y \rangle $ and so $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ ).

Turns out $ g $ is linear too : Expanding $ || g(x+y) - g(x) - g(y) ||^2 $ as $  \langle g(x+y) - g(x) - g(y), g(x+y) - g(x) - g(y) \rangle $ gives \begin{align*}
&|| g(x+y) ||^2 + || g(x) ||^2 + || g(y) ||^2 - 2 \langle g(x+y), g(x) \rangle - 2 \langle g(x+y), g(y) \rangle + 2 \langle g(x), g(y) \rangle \\
&= || x + y ||^2 + ||x||^2 + || y ||^2 - 2 \langle x + y, x \rangle - 2 \langle x + y, y \rangle + 2 \langle x, y \rangle\\
&= 0.
\end{align*}

Therefore $ g(x+y) = g(x) + g(y) $ for all $ x,y $. Similarly we can show $ g(ax) = ag(x) $ for all $ a \in \mathbb{R} $ and $ x \in \mathbb{R}^n $.

So $ g(x) = Ax $ for some $ n \times n $ matrix $ A $. Notice $ g(e_j) = A_j $, where $ (e_1, \cdots, e_n) $ is the standard basis of $ \mathbb{R}^n $ and $ A_1, \cdots, A_n $ are the columns of $ A $. Since $ \langle g(e_i), g(e_j) \rangle = \langle e_i, e_j \rangle $, we see $ \langle A_i, A_j \rangle $ is $ 0 $ if $ i \neq j $ and $ 1 $ if $ i = j $. Therefore $ A^T A = I $, i.e. $ A $ is an orthogonal matrix.

To summarise : Let $ f $ be an isometry of $ \mathbb{R}^n $. Then $ f(x) = f(0) + Ax $ for some orthogonal matrix $ A $.

( So especially the $ f $ here is bijective, answering the original question )

Edit : Treil's "Linear Algebra done Wrong" mentioned in the comments seems to take the same approach ( which is also outlined in above answers ), making this quite redundant. Nevertheless I'll leave it undeleted.

Title: Isometries of $\mathbb{R}^n$


问:
basic example of unramified extension by take a root of a unit (not necessarily 1)
Let $K$ be a local field with residue characteristic $p$, and let $u \in O_K$ be a unit. Let $e$ be relatively prime to $p$.

Let $L=K(u^{1/e})$ why is $L/K$ an unramified extension? Do we need $(e ,p)=1$?

Let $l, k$ be residue fields of $O_L, O_K$.

EDIT: Here is a simple argument I came up with, is it correct:
Let me assume (even though it might not be the case) $X^e-u$ is irreducible over $K$. Then in $k[X]$ since $X^e-\overline{u}$ is separable (since $(e, p)=1$), then it is irreducible (because if it factored, we could lift the factorization to $K$ by Hensel).
So $[L:K]$ and $[l:k]$ are equal since they both equal $e$, and that is a definition of unramified.

答: We do need $(e, p) = 1$. Also, your argument is correct.


问:
Proving this map to be quotient map( Closed /open sets in more than 1 dimensions)

Let p :$ \mathbb{R}^2 \to [0,\infty)$ by p((x,y))= $x^2 +y^2$ and suppose $[0,\infty)$ has the subspace topology induced by usual topology on $\mathbb{R}$. Show that p is a quotient map.

There is a theorem tahtif p is continuous ,surjective, open or closed then p is quotient map.

I was trying to use this result here but the problem is that I am not sure on how to prove that If map is closed/open or not and that is arising as it's a 2-variable map.

So, can you please help me with the how maps are proved to be Open / closed in more than one dimensions? It will be hard if I take a open/closed map and try to find its range and checking the set wheter it's open or not?

So, Can you please tell of a more elegant stratergy?

OR If there is a different but less brute method of proving the map to be quotient map that is also welcome.

Thanks for your time!

答:
A map $f$ is open if the image of any open set $\mathcal{O}$ is open. I.e., if $f(\mathcal{O})$ is open.

The standard open sets in $\mathbb{R}^2$ are of the form $B_{r}((a,b)) = \{(x-a)^2 + (y-b)^2<r^2 \}$. I.e., a ball of radius r about the point $(a,b)$. These form a basis for the topology on $\mathbb{R}^2$, so it suffices to check that $f(B_r(a,b))$ is open for any choice of $r$, $a$ and $b$.

Now, the map $f$ takes in a point $(x,y)$ and outputs $x^2+y^2$. That is, the length from the origin (modulus) squared. So, $f$ will take in all the points of $B_r(a,b)$ and output all the possible lengths squared. So, we need to find the range of lengths. This is determined by the point in $B_r(a,b)$ closest to the origin and the point in $B_r(a,b)$ farthest from the origin.

These points are easy to find. Take a line passing through the origin and the point (a,b). This line will intersect the circle in two spots: $P_1$ and $P_2$. Let's say that $P_1$ and $P_2$ are distances $r_1$ and $r_2$ from the origin respectively. Without loss of generality, assume that $r_1<r_2$.

Now, I claim that $f(B_r(a,b)) = (r_1,r_2)$, which is indeed open. Why might that be?


问:
Doubt regarding distance between parabola and a line(coordinate geometry)
Here is a neat answer by CY aries.

Here he has dropped a normal from parabola to line but I am unable to understand why not from line to parabola.

Original OP is inactive. So please do not close this question

答: It is easier to calculate the distance of given point from given line and that point can be in parametric form, but for calculating it other way around, you need to take point (t, t+1) on the line, draw a perpendicular to the line which passes through point (t, t+1) which would be x+y=2t+1, as you can see this will involve multiple lines depending on value of t and this has to satisfy the point on x=$y^2$ Which would make equation as $y^2 +y = 2t+1$ Solve for values of y and then you would have to calculate distance for each of them, which is nothing but simply complicating a very simple problem.


问:
How to find the genus of curve $x^2-\cos(y)=c$?
$x^2-\cos(y)=c$ is not an algebraic curve, and its genus may depend on the constant $c$, how to find them? In general, do we have any algorithms for transcendental curves?

答:
This curve $X$ is the Riemann surface of the multivalued function $y \mapsto \sqrt{c + \cos(y)}$. The function $\cos(y) $ is surjective, hence there are infinitely many branch points, given by $(x,y)= (0,y_0)$ where $y_0 \in \mathbb C$ is a solution of $\cos(y_0) = -c$.

It follows that the genus of the curve is infinite, independently of $c$, because for each pair of branch points $(y_0,y_1)$ you get a new non-trivial cycle. If you take the $y_i$'s all disjoint they will be independent in homology.


答:
We can write
\begin{align*}
\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi}  \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi}  \ dt & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{\frac{-\pi}{2}+j\pi}^{\frac{\pi}{2}+j\pi}|\cos(t)| \ dt\\
&\overset{\color{blue}{u=t-j\pi} }{=} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du\\
\end{align*}
and if $j$ is an integer, we know that
$$|\cos(u + j\pi)| = |(-1)^j \cos(u)| = |\cos(u)|
$$
so we get\begin{align*}
\frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du
\end{align*}
and on the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ we know that cosine is positive, so on this interval (which is the one we're integrating on) we get $|\cos(u)| = \cos(u)$. And lastly we see
\begin{align*}
 \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(u) \ du\\
& = \frac{1}{\frac{\pi}{2}+j\pi} \sin(u)\Biggr|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\
& = \frac{1}{\frac{\pi}{2}+j\pi} \left(1 - (-1)\right)\\
&  = \frac{2}{\frac{\pi}{2}+j\pi}
\end{align*}


问:
Computing the homology group of $\mathbb{R^n \setminus S^{n-1}}$
I am trying to compute the singular homology group of $\mathbb{R^n \setminus S^{n-1}}$. I tried using the simplices but it is getting complicated. I think I need to use Mayer-Vietoris sequence but can anyone help me with the solution.

I was thinking of using the fact that $H_0(\mathbb{R^n \setminus S^{n-1}})=H_0(*) \oplus \overline{H}_0(\mathbb{R^n \setminus S^{n-1}})$ and $H_j(\mathbb{R^n \setminus S^{n-1}})=\overline{H}_j(\mathbb{R^n \setminus S^{n-1}})$ for all $j >0$ , where $\overline{H}$ is the reduced homology group. But I am stuck in this way as well. Please help me with this idea.

Thanks

答:
$\newcommand\sm\setminus\newcommand\R{\mathbb R}\renewcommand\S{\mathbb S}\newcommand\redhom{\widetilde H}\newcommand\into{\hookrightarrow}\newcommand\D{\mathbb D}\newcommand\Z{\mathbb Z}$
Assuming you really mean to compute the homology of $\R^n\sm h(\S^{n-1})$ where $h:\S^{n-1}\into\R^n$ is an arbitrary embedding, I think this is a somewhat involved endeavor, so I am giving less than an answer but more than a comment.

If you look at the beginning of section 2.B of Hatcher, he proves that, for $h:\S^{n-1}\into\S^n$ an embedding,
$$\redhom_i(\S^n\sm h(\S^{n-1}))=\begin{cases}\Z&i=0\\0&\text{otherwise}\end{cases}$$
For brevity's sake, we'll take this as given. Going from this to homology where $\R^n$ replaces $\S^n$ is another (now simple) application of Mayer-Vietoris.

So say now we have $h:\S^{n-1}\into\R^n$ and write $\S^n=\R^n\cup\{\infty\}$. Since $h(\S^{n-1})\subset\R^n$ is compact (in particular, bounded), we can find an open disk $U\subset\S^n$ around $\infty$ so that $U\cap h(\S^{n-1})=\emptyset$. Now, let $A=\R^n\sm h(\S^{n-1})$ and $B=U\sm h(\S^{n-1})=U\simeq*$, so $A\cup B=\S^n\sm h(\S^{n-1})$ and $A\cap B=U\cap\R^n=U\sm\{\infty\}\simeq\S^{n-1}$. Applying Mayer-Vietoris to $A\cup B=\S^n\sm h(\S^{n-1})$ very quickly gives
$$\redhom_i(\R^n\sm h(\S^{n-1}))=\begin{cases}\Z\oplus\Z&i=0\text{ and }n=1\\\Z&i=0,n-1\text{ and }n>1\\0&\text{otherwise}\end{cases}$$
In the end, you can find the answer to your question in Proposition 2B.1 of Hatcher, and then you have to do a little more work to replace $\S^n$ with $\R^n$.


问:
Boltzmann Distribution With Constraints
I have a problem with showing the existence of Boltzmann distribution given some constraints.

Consider $p_1,...,p_n$ a Boltzmann distibution, where $p_i=\frac{\epsilon^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{-\beta \cdot E_j}}$.

Define the expected value $E = \sum_{i}^{} p_i \cdot E_i$.

Whether Boltzmann distribution exists in the three following cases?

$E_1<E<\frac{\sum_{i}{} E_i}{n}$

$E>\frac{\sum_{i}{} E_i}{n}$

$E=\frac{\sum_{i}{} E_i}{n}$

Let's consider the first case, when $E_1<E<\frac{\sum_{i}{} E_i}{n}$.

$E= \sum_{i}^{} p_i \cdot E_i  = \sum_{i}^{} \frac{e^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{- \beta \cdot E_j}}$.

Whether $E_1<\sum_{i}^{} \frac{e^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{- \beta \cdot E_j}}<\frac{\sum_{i}^{} E_i}{n}$?

The problem is so far I don't have an idea how to approach the solution.

Would appreciate any help.

Addendum: I would like to thank joriki for his great answer.

Let's try to make rigorous proof.

When $E_1<E_2$, by definition of Boltzmann probability $p_1>p_2$. Show that $E<\frac{E_1+E_2}{2}$. When $p_1>p_2$ and $p_1+p_2=1$ then $p_1>0.5$ and $p_2<0.5$.

$E_1p_1 + E_2p_2 $? $0.5E_1 + 0.5E_2$

$E_1p_1 + E_2(1-p_1) $? $0.5E_1 + 0.5E_2$

$E_1(p_1-0.5) + E_2(0.5-p_1) $? $0$

Where $p_1-0.5>0$ and $0.5-p_1<0$, because $p_1>0.5$.

I am really stuck here.

At least let's show that $E_1<E$

$E_1p_1 + E_2p_2$?$E_1$

$E_1(p_1-1) + E_2p_2$ ? 0

$-E_1p_2 + E_2p_2$ ?0

$p_2(E_2-E_1)>0$

Let's try to it show for the general case $\sum_{i}^{} E_ip_i > E_1$

[sorry for the last attempt]

$\sum_{i}^{} E_ip_i $?$ E_1$

$E_1p_1 + \sum_{i=2}^{} E_ip_i$?$ E_1$

$E_1(p_1-1) + \sum_{i=2}^{} E_ip_i$?$0$

$p_1+...+p_n =1$

$E_1(-\sum_{i=2}^{} p_i) + \sum_{i=2}^{} E_ip_i$?$0$

$\sum_{i=2}^{} p_i (\sum_{i=2}^{} E_i - E_1) > 0$

In addition, finally I get the idea in 2) the trick with derivative, unfortunately I cannot get it equal to negative variance, but I am keep trying, the computations get very long. But one minor point, by showing that derivative is negative we conclude that $E$ is a concave function with maximum at $\beta = 0$, which is exactly like in the 3) case where the distribution exists, so why Boltzmann distribution is impossible in 2)?

答:
Take $E_1\lt E_2$; then $p_1\gt p_2$ and thus $E_1\lt E\lt(E_1+E_2)/2$.

This is impossible for non-negative $\beta$. To see this, form $\partial E/\partial\beta$, show that it's the negative variance of $E$ and thus non-positive, and infer that $E$ is never greater than its value at $\beta=0$, which is the mean.

This occurs either if all $E_i$ are equal, or for $\beta=0$, corresponding to $T\to\infty$.

[Edit in response to addendum:]

For 1., you started out on the right path; then you can write

\begin{align}
p_1E_1+p_2E_2
&=
p_1\left(\frac{E_1+E_2}2+\frac{E_1-E_2}2\right)+p_2\left(\frac{E_1+E_2}2-\frac{E_1-E_2}2\right)
\\
&=
(p_1+p_2)\frac{E_1+E_2}2+(p_1-p_2)\frac{E_1-E_2}2
\\
&=
\frac{E_1+E_2}2+(p_1-p_2)\frac{E_1-E_2}2
\\
&\lt
\frac{E_1+E_2}2\;.
\end{align}

I'm not sure what you're trying to do to show that the mean is $\gt E_1$. The mean of elements that are not all equal is greater than the least of the elements; I doubt that you're expected to prove that ab initio.

Title: Boltzmann Distribution With Constraints


问:
Show convergence of this series of exponentials
For a result in percolation theory, namely showing that in the subcritical regime we can expect to have finite clusters ($\mathbb{E}_p[|C_0|]<\infty$), I need to show the following:

Show: For any integer $d\geq1$ we have that
$$
\sum_{n=0}^\infty ne^{-n^{1/d}}
$$
converges.

Proofidea: For $d=1$ we have that by swapping limit and differentiation
$$
\sum_{n=0}^\infty ne^{-n^{1/d}}=\sum_{n=0}^\infty ne^{-n}=e^{-1}\sum_{n=0}^\infty\frac{d}{dx}\mid_{x=e^{-1}}x^n=e^{-1}\frac{d}{dx}\mid_{x=e^{-1}}\frac{1}{1-x}=\frac{e^{-1}}{(1-e^{-1})^2}<\infty.
$$
Sadly for $d>1$ this procedure doesn't work anymore. I got a hint that I should consider
$$
\sum_{n=0}^\infty ne^{-n^{1/d}}=\sum_{n=0}^\infty (ne^{-n^{1/d}/2})e^{-n^{1/d}/2}\leq C_k+\sum_{n=k}^\infty e^{-n^{1/d}/2}
$$
as for $k$ large enough $ne^{-n^{1/d}/2}\leq 1$ for all $n\geq k$ and $C_k<\infty$ some constant. But why is $\sum_{n=k}^\infty e^{-n^{1/d}/2}$ converging, or equivalently $\sum_{n=0}^\infty e^{-n^{1/d}/2}$?

答:
Hint

The series $u_n = e^{-\beta n^\alpha}$ converges for any $\alpha, \beta \gt 0$. To prove it, remember that

$$e^{-y} \le \frac{1}{y^\gamma}$$ for any $\gamma \gt 0$. Then substitute $y$ with a well suited function of $x$.


问:
Find a recurrence relation for the number of regions created by n mutually intersecting circles
Find a recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point).

I need help understanding the answer to this

My work:

So after looking a while on firefox to help me figure this out. I've eventually found that the number of regions for the problem is

$n^2-n+2$ which I think is because of Euler's formula (r = e - v + 2). Am I incorrect? I'm not sure why it's $n^2$ can someone please explain?

I know the above is right for the number of regions because when I test with $n=2, n=3, n=4$ I get $ r = 4, r=8, r=14$ which when I double check by drawing the circles manually and seeing how many regions there are, I get the answers above.

so then I've found online that the recurrence relation to be $a_n=a_{n-1}+2(n-1)$ which is equivalent to $a_n=n^2-n+2$

But I don't know why this is.

Does $a_{n-1}$ represent the previous regions of $n-1$ circles? and does $2(n-1)$  represent the new regions added with each new circle?

Is the proper answer just $a_n=n^2-n+2$?

I'd really like to learn why $a_n=a_{n-1}+2(n-1)$ works and is (probably) the answer? In order to improve my understanding.

答:
Consider n mutually intersecting circles and no three have common intersecting points. This implies that any two will have exclusive 1/2 point of intersection.

If we take those two circles and pass a circle through both of them such that none of those two points lies on the third circle, we get 2 new regions, one with first and one with the second circle. For n mutually intersecting circles, there will be n-1 pairs of such two circles (as we can choose only those which are intersecting to each other). So total regions that can be formed are 2*(n-1)

Hence U(n+1) = U(n) + 2*(n-1)


问:
If $x=(x_1, x_2, x_3, x_4)$ is majorized and weakly log-majorized by $y=(y_1, y_2, y_3, y_4)$, then $x=y$?
More specifically, let $x_1\geq x_2\geq x_3\geq x_4>0$, and $y_1\geq y_2\geq y_3\geq y_4>0$. Suppose $x_1\leq y_1$, $x_1+x_2\leq y_1+y_2$, $x_1+x_2+x_3\leq y_1+y_2+y_3$, $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4$; also, $x_1x_2\leq y_1y_2$, $x_1x_2x_3\leq y_1y_2y_3$, $x_1x_2x_3x_4\leq y_1y_2y_3y_4$. Is it true that $x_i=y_i$ for all $i=1,2,3,4$? In general? Thanks

答:
(Fill in the details as needed. If you're stuck, show your work and explain what you've tried.)

Claim: If $ y $ majorizes $x$, then $ \prod x_i \geq \prod y_i$, with equality iff $x_i = y_i$.

Proof:

Since $y$ majorizes $x$, so $x$ is in the convex hull of $y$ and it's permutations. So we have $ x = \sum_i \alpha_i y^{(i)}$, where $ y^{(i)}$ are permutations of $y$ and $ 0 \leq \alpha \leq 1$.

Specifically, for each coordinate $j$, $x_j = \sum_i \alpha_i y_j ^{(i)}$.

Since $ \log x$ is concave, so $ \log x_j \geq \sum_i \alpha_i \log y_j ^{(i)}$.

Hence, $ \sum_j \log x_j \geq \sum_j \sum_i \alpha_i \log y_j^{(i)} = \sum_j \log y_j$. Thus, $ \prod x_i \geq \prod y_j$.

Check the equality cases to show that equality holds iff $x_i = y_i$.  Sight care has to be taken, esp when some coordinates of $y$ are equal.

Corollary: If $ \prod x_i \leq \prod y_i$, then we're in the equality case, and the desired result follows.


问:
$[G:Z(G)] = n$ prove that each conjugacy class has at most n elements
$[G:Z(G)] = n$ prove that each conjugacy class has at most n elements.

what i tried -

I know from the orbit stabiliser theorem that

$|G| = \sum_{x_i} |G:C(x_i)| + |Z(G)|$

Because Z(G) < G i also know from Lagrange theorem that:

$|G| = |Z(G)| * [G:Z(G)] = |Z(G)| * n$

i know that the size of each conjugacy class is $|G:C(x_i)|$

here I'm stuck.. any help will be appreciated

答: $\textbf{Proof:} \, \text{We know} $ $G/Z(G) \cong \text{Inn}(G)$. Since $G$ is finite, $n=|G:Z(G)|= \frac{|G|}{|Z(G)|}=|\text{Inn}(G)|$ by Lagrange's Theorem. So, there are $n$ distinct inner automorphisms $\theta_{g_i}$ where $g_i \in G$ for $i=1,...,n$. Let $x \in G$, then $\text{cl}(x)=\{g^{-1}xg | g \in G\}=\{\theta_{g_i}(x) | g_i \in G \,\, \text{for}\,\, i=1,...,n\}$. Then $|\text{cl}(x)| \leq n$.


问:
Alignment of multiple bodies on circular paths
Below is a problem that I came across in which I don't know the answer. It's as follows:

$\\$

Problem:

Suppose there are $n-2$ planets currently at rest (stationary) - all positioned on the positive side of the x-axis (line) of the euclidean plane. They will soon orbit on a circular motion around the sun. Each planet $k$ will travel at the same speed and have an orbit of length $k$ where $k=3,4,...,n$ and $n>4$. All orbits will lie on the euclidean plane and will be centred at the origin (where the sun is positioned). Suppose now that all planets leave their initial position of rest at the same time. From now on, two or more planets are considered to be aligned if they lie on the positive side of the x-axis (line) at the same time in their respective orbits. Now consider all the planets that won't be aligned with planet $n$ during its first orbit. How many of them will be aligned with at least one planet that will be aligned with planet $n$ (during the first orbit of planet $n$)?

$\\$

Additional Information:

In the previous version of this problem, I realised I had made the mistake of letting planet $n$ orbit the sun more than once which made it quite obvious that every other planet would be aligned with planet $n$ after a certain number of orbits - hence you have the second comment below and the answer (from 1st January) that were driven by this mistake. So I've now corrected the mistake by letting planet $n$ orbit the sun once only - which was my original intent.

答:
As I understand the question, at the end of the single orbit for planet $n$ that we are considering, it will be alignment with all planets $\{k_i\}$ where each $k_i\mid n$ - that is all the factors of $n$ greater than $2$, since the orbital period is here directly proportional to the radius (all speeds are equal).

So these planets may also align with other planets not in this set, in earlier passes through the $x$-axis - for example, $n=28$ means that planet $n$ then aligns with planets $4$, $7$ and $14$, but on earlier orbits planet $4$ has also aligned with planets $8, 12, 16, 20$ and $24$ and planet $7$ with planet $21$.

For odd $n$ the missing planet $2$ will introduce no complications. Essentially there we can count all multiples of prime factors of $n$ and subtract off the actual factors of $n$.

For even $n$ we have to work around the missing planet $2$, but the concept is similar. Note, we can't treat $4$ as an "honorary prime" though since $n$ may be divisible by 8.


问:
How can I solve this ODE using a predictor-corrector method?
I'm looking to solve an ODE system for $y$ over time using the predictor-corrector method below. I'm not sure how predictor-corrector for ODEs work. I know other numerical methods for ODEs, like RK and forward Euler.

$\ y = 0$

Timestep (h) is 0.001s.

$\ \frac{dy}{dt}  = PI-QB \cos\beta\, $

Predictor:

$\ y(n+1) = y(n)+(\frac{h}{2}) [3dy/dt(n)-dy/dt(n-1)]$ is used to calculate $\ dy/dt(n+1)$

Corrector:

$y(n+1) = \frac{1}{3} [4y(n)-y(n-1)+2h dy/dt(n+1)]$

For the corrector, do I just substitute in the predictor instead of the $\ dy/dt(n+1)]$ on the right hand side of the equation?

Then $\ y(n+1)$ is $\ y(n)$  added to the corrector.
Is that right?

Link to my spreadsheet,  with my Predictor corrector formulas on the right hand side, that I think are wrong. - Google Sheet

答:
Yes, for an ODE system $\frac{dy}{dt}=f(t,y)$ the predictor extrapolates the existing data to the new point, here with
$$
\tilde y_{n+1}=y_n+\frac{h}{2}(3f_n-f_{n-1}),~~~f_k=f(t_k,y_k).
$$
Then the next solution point is refined with the corrector equation, here with the backward differentiation formula
$$
y_{n+1}=\frac13[4y_n-y_{n-1}+2h\tilde f_{n+1}],~~~\tilde f_{n+1}=f(t_{n+1},\tilde y_{n+1}).
$$
This correction step can also be repeated to get closer to the solution of the implicit step equation
$$
y_{n+1}=\frac13[4y_n-y_{n-1}+2hf(t_{n+1},y_{n+1})].
$$
This implicit equation can also be solved faster with a Newton-like method employing the Jacobian of $f$ or a locally valid approximation of it. This is done by most modern implementations of such implicit methods.


问:
Weighted initial ideal versus lex or graded reverse lex initial ideal
By imposing certain weights $\mathbf{w}$ on the variables, say, of a polynomial ring $k[x_1,\ldots, x_n]$, I read that we may obtain the initial ideal $in_{\mathbf{w}}(I)$ of an ideal $I$ with respect to this weight.

Is it true that for any fixed weight $\mathbf{w}$ imposed on the variables, we have $in_{\mathbf{w}}(I)=in_{>}(I)$ for some monomial ordering $>$, where $>$ is thought to be lex or graded reverse lex, with the variables shuffled around?

I thought I read this somewhere but I cannot find the reference. Thanks for your time.

Edit: The initial ideal $in(I)$ of an ideal $I$ is thought to be an ideal generated by the leading terms of $f$, where $f\in I$.

答:
I know this is an old post, but I stumbled upon it trying to answer my own question, and I found it interesting, so I thought I should chime in.

The short answer would be yes.  The problem is with ties on your weighting.  For example, in $k[x,y,z]$ and $\omega = (3,4,5)$, the monomials $x^3$ and $yz$ have the same weight, so you have to pass to a second term order to break the tie.  So what you need is not a single weighting, but actually 3 linearly independent weight vectors to guarantee all ties can be broken.

So once you have all your tiebreaking vectors, you simply have to finagle the equations until you have $x_{i_1}<x_{i_2}<\cdots<x_{i_n}$, and since you have a term order, you know that $x_{i_j}M<x_{i_k}M$ for all $j<k$ and all monomials $M$.  The only 'danger' in thinking this way might be that one could be interested in the weighting itself; if you just want to know the order of the monomials, then you are all set.


答:
The identification of $V''$ with $V$ is confusing. If you're interested, I have a long answer about just this. With that in mind, I will first answer without the $[\cdot,\cdot]$ notation. For any element $x \in V$, I will use $\alpha_x$ to denote the associated element of $V''$. Note that $\alpha_x:V' \to \Bbb F$ is defined by
$$
\alpha_x(f) = f(x).
$$
Proof: By definition, $\eta^{00}$ is the set of all elements $\alpha \in V''$ for which $\alpha(f) = 0$ holds for all $f \in \eta^0$. Note that every $\alpha \in V''$ is identified with the unique $x \in V$ for which $\alpha = \alpha_x$. Thus, the set $\eta^{00}$ is identified with the set of all $x \in V$ such that for all $f \in \eta^0$,
$$
\alpha_x(f) = 0 \implies f(x) = 0.
$$
Now, by the definition of $\eta^0$, we can see that if $x \in \eta$, then the element of $V''$ associated with $x$ is an element of $\eta^0$. That is, $\eta \subseteq \eta^{00}$. On the other hand, with finite-dimensionality we have
$$
\dim(\eta^{00}) = \dim(V) - \dim(\eta^0) = \dim(V) - [\dim(V) - \dim(\eta)] = \dim(\eta).
$$
Thus, we indeed have $\eta = \eta^{00}$.

So what does this look like in terms of the brackets? The point is that $[x,y]$ is really $\alpha_x(y)$, and $[y,x] = y(x)$. The identification of $V$ with $V''$ is defined such that $\alpha_x(y) = y(x)$, which is to say that $[x,y] = [y,x]$.

Title: Question regarding the simple proof involving dual space and annihilator

Body:
If $\eta$ is a suspace in a finite-dimensional vector spce $v$, then
$\eta^{00}=\eta$

Proof: Observe that we use here the convention, established at the end
of 16, that identifies $v$ and $v''$. By definition, $\eta^{00}$ is
the set of all vectors $x$ such that $[x,y]=0$ for all $y$ in
$\eta^{0}\dots$

16 explains that in convention, we can identify $v$ with its double dual space $v''$, because they are isomorphic. But I'm not sure how the second sentence in the above proof follows.
From my understanding,
by definition, $\eta^{00}$ is the set of all vectors $x$ in $v''$ such that $[y, x]=0$ for every $y\in\eta^{0}$. I'm not sure how we can switch the position of $x$ and $y$ from the isomorphism of $v$ and $v''$.


问:
Why aren't the values true and false part of logic, which is concerned only with the preservation of truth in arguments?
I quoted from the author's "Introduction for Instructors" on p. 15 where he's trying to distinguish his book from standard logic books that "did not address
significant problems and confusions—those that were common among the
students in those classes and those that appeared among students in other
philosophy courses, even when such students had taken more than one class
in logic and critical thinking."

I'm trying to decide if this book will introduce me comfortably to logic. But I don't understand the emboldened sentences. Can someone explain them so that a 16 year old high schooler can understand?

How can logic be "concerned only with the preservation of truth in arguments", if the values true and false aren't part of logic? Then again, I don't understand what the authors means by "minimalist account of the values true and false" or "preservation of truth".

      Critical thinking, as understood in this book, includes logic—formal and
informal—but in addition all rigorous methods for reliably arriving at true
propositions. Hence the book contains a minimalist account of the values true
and false. Such an account is not part of logic, which is concerned only with the
preservation of truth in arguments. The account of truth and falsity given here
is minimal in the sense that it includes all schemas of the form "It is true that P
if and only if P," where that P is a proposition which may be expressed in some
language and P picks out the state of affairs which the proposition purports
to describe. This account is stated less formally in the text by distinguishing
the state of affairs picked out by the proposition "P" from the proposition
"P." While this account is close to deflationary accounts such as those given
by Horwich$^1$, it is not meant to be a philosophically adequate account of truth
and falsity, but rather, in keeping with the pedagogical purposes of the book,
is meant to help students avoid common confusions by distinguishing the
language with which we talk about the world from the world about which
we talk. Moreover, this account of truth and falsity is not intended to rise to
the level of a correspondence theory of truth, since it proposes no detailed
articulation of states of affairs in accord with the articulation of propositions.

The Elements of Arguments: An Introduction to Critical Thinking and Logic by Philip Turetzky. I screen shot the whole p. 16 if you like more context.

答:
For what it is worth, here is how I look at the notions of true and false in formal logic:

To formally state that proposition $A$ is true, we simply write: $A$

To formally state that proposition $A$ is false, we simply write:
$\neg A$

Why then do we have $T$ for true, and $F$ for false in truth tables? Consider, for example, the truth table for $\neg A\lor B$:

Source: https://www.erpelstolz.at/gateway/TruthTable.html

This is just a convenient representation of the following with each line corresponding to a line in the above truth table:

$A \land B \implies (\neg A \lor B)$

$A \land \neg B \implies \neg (\neg A \lor B)$

$\neg A \land B \implies (\neg A \lor B)$

$\neg A \land \neg B \implies (\neg A \lor B)$

In each line, the antecedent corresponds to columns 1 and 2 of the truth table. And the consequent corresponds to column 3.


问:
Finding $a_{1996}$ if $\prod_{n=1}^{1996}(1+nx^{3^n})= 1+a_1x^{k_1} + a_2x^{k_2} + \cdots + a_mx^{k_m}$
I need to find the coefficient $a_{1996}$

$$\prod_{n=1}^{1996}(1+nx^{3^n})= 1+a_1x^{k_1} + a_2x^{k_2} + \cdots + a_mx^{k_m}$$

$a_1, a_2, ... , a_m$ are non zero.

$k_1 < k_2 <... < k_m$

So if $x=1$ you can find the sum of all coefficients, but I am not sure how to use this fact

Do I need to simplify the product somehow plug different values of x and check if something can be approximated that way, or would that be a waste of time?

Thanks,

答:
What powers of $x$ appear as a term in the expression? Of course $x^0$. And $x^3$. More generally, any positive integer that can be built from summing powers of $3$ with no repeats.
$$x^0,\overset{1\text{st}}{x^3},\overset{2\text{nd}}{x^9},\overset{3\text{rd}}{x^{12}},\overset{\cdots}{x^{27}},x^{30},x^{36},x^{39},\ldots$$
In base 3, these are numbers that use only 0 and 1 for their digits, with a 0 in the "ones" place.
$$x^0,x^{10},x^{100},x^{110},x^{1000},x^{1010},x^{1100},x^{1110},\ldots$$
So what would be the 1996th term's power of $x$? In binary, 1996 is $$11111001100$$

So we are looking for $x^{111110011000}$ but the exponent is in base 3. It is built from multiplying $$x^{3^{11}}x^{3^{10}}x^{3^{9}}x^{3^{8}}x^{3^{7}}x^{3^{4}}x^{3^{3}}$$

So its coefficient is $$11(10)(9)(8)(7)(4)(3)=665280$$


问:
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

I tried many method, such as AM-GM, but all of them failed.

Thank you.

答:
The equation in fact is the surface of a ball at center $C_0=(-1/2,-1,-3/2)$ with radius $r=3\sqrt{3}/2$.

Then consider when there is only one point on the intersection between sphere and plane $x+y+z=t$


问:
What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?
At a certain customer service line the waiting time for the next call follows exponential distribution with expected time $3$ minutes. What is the probability that the next call come in $3$ minutes provided that in the past $2$ minutes there have been no calls?

From the given datils we know $\lambda=\dfrac{1}{3}$ so the distribution

$P(X=x)=\frac{1}{3}e^-\frac{x}{3}$

We have to find $P(0<x<3/x>2)=\dfrac{P(2<x<3)}{p(x>2)}=\frac{\int_{2}^3 \frac{1}{3}e^-\frac{x}{3}\,dx}{\int_{2}^\infty \frac{1}{3}e^-\frac{x}{3}\,dx}$

but I am not sure this answer I think we have to find $P(2<x<3)$
Can someone help me to understand?

答:
The numerator of the fraction is $P(0<x<3 \cap x>2)$. Thus you have to find the intersection of both. A number line helps to find it.

So the required probability is indeed $P(2<x<3)$. Thus the bounds are $2$ and $3$.