问:
How to compute the canonical bundle of a generalized flag bundle?
If $X:=Gr_k(V)$ denotes the Grassmannian of $k$-dimensional subspaces of an $n$-dimensional vector space $V$, it is well-known that the canonical bundle $\omega_X$ is given by $$\omega_X:= \det \Omega_X^1= \det E \otimes Q^* = \mathcal{O}(-n),$$ where $\Omega_X^1$ denotes the cotangent bundle of $X$, $E$ is the tautological subbundle of rank $k$ of the Grassmannian and $Q$ is the corresponding quotient subbundle.

This may be generalized to Grassmann bundles (e.g. https://arxiv.org/pdf/0807.3296.pdf, prop. 1.5): If $V$ is a vector bundle over some scheme $Y$ and $X=Gr_Y(k,V)$ parametrizes rank-$k$-subbundles of $V$, then similarly
$$\omega_X=\det E \otimes Q^* = (\det V)^{-k} \otimes \mathcal{O}(-n).$$

Now I would like to consider Grassmannians (or, more generally, flag bundles) of other Lie types, especially in type $D$. There are notions of orthogonal flag bundles (and orthogonal Grassmann bundles), but I cannot find any references for the canonical bundle. For an $2n$-dimensional vector space $V$ equipped with a quadratic form, if $X=OG(n,V)$ parametrizes maximal isotropic subspaces of $V$, I know that the canonical divisor is given by $\omega_X = \mathcal{O}(-2n+2)$ (still I did not find any good reference).

My questions are:

Is there a nice way to see $\omega_X=\mathcal{O}(-2n+2)$?

How do I generalize to bundles? There should be involved some more terms coming from the base scheme, but I do not have a good intuition for this.

Many thanks in advance.

答:
I figured out the following: The tangent bundle of the orthogonal Grasmann bundle over a variety $X$ is given by
$$T \mathrm{OG}_X(n,V) = \wedge^2 E^* \otimes \pi^* \mathcal{L}$$
where $E$ denotes the pullback of the tautological bundle over $\mathrm{Gr}(n,2n)$ and $\mathcal{L}$ is the line bundle in which the symmetric bilinear form  has its values. Taking the determinant of the dual of it we get
$$\omega_{\mathrm{OG}_X(n, V)} = (E^*)^{n-1} \otimes \pi^*\mathcal{L}^{\frac{n(n-1)}{2}} = \mathcal{O}(-2n+2) \otimes \pi^*\mathcal{L}^{\frac{n(n-1)}{2}}.$$
Did I make a mistake?


问:
Solving a summation problem - I know the answer I think, but not why it works
This problem I intuitively know the answer, (I think) but I am trying to get a handle on why it works.

$$\sum_{n=0}^x n=\frac{\sum_\limits{n=0}^x n^3}{\sum_\limits{n=0}^x n^2} $$

We want to know what x is.
I figured the answer was x = 1, because the only number I can come up with where $\frac{n^3 + (n+1)^3 + ... }{n^2 + (n+1)^2 + ...}=n$ is one, when I did a few sums and realized that anything other than n=1 doesn't work. And I know it's because if I add a number I am not getting a cube over a square of n anymore, $\frac{(1^3 + 2^3)}{(1^2+2^2)}=\frac{9}{4}$ for example. But I feel like there is some basic piece of a proof/procedure I am missing, even though this intuition is correct. Anyhow, I am curious what the "real" reason might be -- I suspect this is some relatively simple algebraic principle that makes this work, I feel like I am just not quite seeing it. Or maybe I got it right and just overthought this.

答:
A more elementary way:  Multiply both sides by the denominator to get

$$(1+2+\cdots+x)(1^2+2^2+\cdots+x^2) = (1^3+2^3+\cdots+x^3).$$

See that on the left, if you mupltiply everything out, you get every cube plus some other junk.  If you subtract the cubes from both sides, you get

$$\mbox{junk} = 0$$

which means you'd better not have any "cross terms" which means that $x$ had better be $1.$

ContentLicense: CC BY-SA 4.0


答: The simplest proof that $\sum_ii\sum_jj^2=\sum_kk^3$ requires (zero terms or) one term only in each sum is to note the cases with $i=j$ already give us every term on the right-hand side, so we need to avoid positive cross terms $ij^2$ with $i\ne j$.

Title: Solving a summation problem - I know the answer I think, but not why it works

Body:
This problem I intuitively know the answer, (I think) but I am trying to get a handle on why it works.

$$\sum_{n=0}^x n=\frac{\sum_\limits{n=0}^x n^3}{\sum_\limits{n=0}^x n^2} $$

We want to know what x is.
I figured the answer was x = 1, because the only number I can come up with where $\frac{n^3 + (n+1)^3 + ... }{n^2 + (n+1)^2 + ...}=n$ is one, when I did a few sums and realized that anything other than n=1 doesn't work. And I know it's because if I add a number I am not getting a cube over a square of n anymore, $\frac{(1^3 + 2^3)}{(1^2+2^2)}=\frac{9}{4}$ for example. But I feel like there is some basic piece of a proof/procedure I am missing, even though this intuition is correct. Anyhow, I am curious what the "real" reason might be -- I suspect this is some relatively simple algebraic principle that makes this work, I feel like I am just not quite seeing it. Or maybe I got it right and just overthought this.


问:
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$

$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
  $$x^2y+y^2z+z^2x < \frac12$$

This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.

答:
$\color{green}{\textbf{Computer free version of 17.12.20.}}$

$\color{brown}{\textbf{Preliminary notes.}}$

We should prove the inequality$\;f(x,y,z)= x^2y+y^2z+z^2x \le \frac12\;$ under the conditions
$x+y^2+z^3=1,\;$ $x,y,z\in [0,1].$

In accordance with the @RiverLi answer,

$y\le y^2+\frac14,\tag1$

$2-4(x^2y+y^2z+z^2x)\ge 2-x^2-(4x^2+4z)(1-x-z^3)-4z^2x,\tag2$

where $(1)$ follows from the evident $\;\left(y-\frac12\right)^2 \ge0\;$.

At the same time, for $\;t\in[0,1]\;$
$$t^4-t+\frac3{4\sqrt[\large3]4} = \left(t^2-\sqrt[\large3]2t-\frac1{2\sqrt[\large3]2}\right)\left(t^2 +\sqrt[\large3]2 t +\frac3{2\sqrt[\large3]2}\right) $$
$$= \left(t-\frac1{\sqrt[\large3]4}\right)^2\left(t^2+\sqrt[\large3]2 t
+\frac3{2\sqrt[\large3]2}\right) \ge 0,$$

$$t\le t^4+H,\quad \text{where}\quad H=\frac3{4\sqrt[\large3]4} = \dfrac12\sqrt[\large3]{\dfrac{27}{32}}<\dfrac12,\quad H\approx0.47247.\tag3$$

$\color{brown}{\textbf{The edges of area.}}$

Taking in account $(1)$ and $(3),$ one can get
\begin{align}
&f(x,y,0) = x^2y \le (1-y^2)^2y\,\bigg|_{\large y=\frac1{\sqrt5}} = \dfrac{16}{25\sqrt5}<0.28622,  \\[4pt]
&f(x,0,z) = z^2x = (1-z^3)z^2 \le (1-z^3)z \le H, \\[4pt]
&f(0,y,z) = y^2z = (1-z^3)z \le H.
\end{align}

Thus, the given inequality is correct on the edges of the area.

$\color{brown}{\textbf{The inner stationary points.}}$

Taking in account $(2)$ and $(1)$ in the form of
$\;4z\le 4z^2+1,\;$
it suffices to prove the inequality $\;g(x,z)\ge0,\;$ where
$$g(x,z) = 2-x^2-(4x^2+1+4z^2)(1-x-z^3)-4z^2x,$$
$$g(x,z) = 1+4x^2z^3 + (4z^5+z^3-4z^2) + (4x^3-5x^2+x),\tag4$$
under the conditions $\;\;x,z\in(0,1).$

Stationary points should satisfy the system $\;g'_x = g'_z =0,\;$
or
\begin{cases}
g'_x = 8z^3x + 12 x^2 - 10x + 1 = 0\\[4pt]
g'_z = 12x^2z^2+20 z^4+3z^2-8z = 0,\tag5
\end{cases}

and then
\begin{cases}
12x^2z = 8-3z-20z^3\hspace{163mu} \left[z^{-1}\times(5.2)\right]\\
8z^4x +(8-3z-20z^3) - 10xz + z = 0,\qquad \left[z\times(5.1)\right]\\
\end{cases}
$$x = h_1(z) = \dfrac{4-z-10z^3}{5z-4z^4},\quad x^2 = h_2(z) = \dfrac{8-3z-20z^3}{12z}\tag6.$$

At the same time, from $(4),(5)$ should
\begin{align}
&5g(x,z) = 5g - xg'_x - z g'_z\\[4pt]
& = 5+20x^2z^3 + (20z^5+5z^3-20z^2) + (20x^3-25x^2+5x)\\[4pt]
& - x(8z^3x + 12 x^2 - 10x + 1) - z(12x^2z^2+20 z^4+3z^2-8z)\\[4pt]
& = 5 + 4 x + 8x^3 -15x^2 - 12 z^2 + 2 z^3,
\end{align}
$$5g(x,z) = 5 - 15h_2(z) + 4 h_1(z) (1+2h_2(z)) - 12 z^2 + 2 z^3.\tag7$$
Substitution of $(6)$ to $(7)$ represents the inequality $\;g(x,z) \ge 0\;$ in the form of
$$256 - 568 z + 501 z^2 - 1280 z^3 + 1180 z^4 - 300 z^5 + 1600 z^6 - 624 z^7 - 96 z^8\ge 0.\tag8$$

Since
$$h(z)-1=\dfrac{4z^4-10z^3-6z+4}{5z-4z^4}=\dfrac{2(1-2z)(2+z+2z^2-z^3)}{5z-4z^4},$$
then $\;h(z)>1,\;$ if $\;z\in\left(0,\frac12\right).\;$

Besides, $\;h(z)\;$ has a positive denominator and  decreasing numerator in the interval $\;z\in(0,1],\;$ wherein $\;h\left(\frac7{10}\right) = -\frac{325}{6349}<0.\;$
Therefore, $\;h(z)< 0,\;$ if $\;z\in\left[\frac7{10},1\right].\;$

Thus, all inner stationary points belong to the interval $\;z\in\left[\dfrac12,\dfrac7{10}\right).\;$

And now, substitution $\;z=\dfrac{6+t}{10},\;t\in[-1,1),\;$ transforms $(8)$ to the evident inequality
$$-3 t^8 - 339 t^7 - 6214 t^6 - 13083 t^5 + 1041140 t^4\\ + 12923432 t^3 + 64494426 t^2 - 898904 t + 13278632 \ge 0.$$

Done!

ContentLicense: CC BY-SA 3.0


问:
Does the equation $k^k+1=a^2$ have any positive integer solutions?
As the title of the question says, I want to know if the equation
$k^k+1=a^2$
has any integer positive solutions, (which means k and a are both positive integers.)

I'd really appreciate it if someone could provide the following:

If the given solutions exists, please provide the minimal solution. Are the solutions infinite or finite?

If the given solutions doesn't exist, please provide your proof.

答: $k^k+1$ itself has factors, one of which is $k+1$.  These would have to be coprime, or a common divisor of 2, because the common factor between $(a^k+1)$ and $(k+1)$ would divide 2k.


问:
Show that $\text{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ (morphism construction)
I'm curious how we can construct a morphism $\text{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \rightarrow S_3$ and show then that it is an isomorphism.

I know that $(12),(23)$ generate $S_3$ and that $([a]_2,[b]_2)$ with $a,b \in \mathbb{Z}/2\mathbb{Z}$ generate $\mathbb{Z}/2\mathbb{Z}$. But, I don't really see how we can construct it.

If you could give me some ''hints'' I would appreciate it.

Thank you in advance

答: Hint: Any automorphism of the Klein four group must permute the $3$ nonidentity elements.


问:
what is the sum of this?$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $
What is the sum of $$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $$</p>

Body:
Not really an interesting question, I fear ...

The exact answer is $\frac{A}{B}$, where $A$ is the 873-digit number

$$\begin{align}
&1457823020375738882085285521838680376351317016179578495942597370\\
&0378320415510098781465535340716404121275266248327296358362232678\\
&6278734102622894043296263865840984434831046940807020935913518007\\
&6610170430184975025171489501039855136340044946766172114345187347\\
&2525258101947297211093973306792764671036527339494431750637074683\\
&4230708772438305435131071598225161542024364751281758482910493532\\
&5272679486529134147964088721658417270988342669017294670746222437\\
&5396289282709392060556292768642859506675716675390362111468228302\\
&0782368060066857220428567636667351228472222194509213532517098318\\
&4215312313922564589890501941681764939250636021442764267189354097\\
&6967312781103861891868999835897977944251765509632397992463764707\\
&2757539640418390812471213879592790834646976169124993489059261606\\
&2602876187541263449739774491663332258026609790766133246441424484\\
&757471204138450579265132408964711222835023\\
\end{align}$$

and $B$ is the 872-digit number

$$\begin{align}
&2029024882178588859070537995606394892063138617868032497494729352\\
&4603117588071109954985856100075037278586690682862226370908784076\\
&1373758182460385685570815580907584983469652245087586518224049939\\
&8538189200052467764823885503628456029563567766917663813400872806\\
&6581413837126629349013539786812278235204246553266267204666411453\\
&3233291465507116087957721742791662582248630820924043188884944781\\
&6407535614283438262641159101052916955096435138537168813836363531\\
&9456214006726533268868764071437503549723903396960519957683391782\\
&6358829094665518541761643350701837468138883360957730534259985433\\
&0461170299850901340867381633962516193737332035551187788058506648\\
&9862570396408121482924662574281092579439686439440266363727224229\\
&3000091507088457965475315993866302318289137980710843830365091268\\
&4178987345171765323904440193241153257507552225072663118382054707\\
&65887615406383472808692605376427780480000\\
\end{align}$$

Also, the first digits are 7.184845455 ...


问:
Do we have this symbol in mathematics?

It is the symbol I used to show left hand side of inequality isn't greater than right hand side.

More precisely I used this to show the below matrix is not diagonally dominant:

$$\begin{bmatrix}5&6&7\\
2&-4&2\\
3&2&-5\\
\end{bmatrix}$$

So at the first row I want to emphasize that $|5|$ is not greater than $|6|+|7|$  therefor instead of writing $|5|<|6|+|7|$ I used the symbol in the image above.

Is it ok to use this symbol in mathematics?

答:
To me, it is OK. Putting a bar/cross on a pre-existing maths symbol just means the negation of it.

I'm not sure it is standardized, though.

Edit

As shown by @Dietrich Burde in his answer, these symbols are $\LaTeX$ symbols, so that's clearly OK you can use them.

Title: Do we have this symbol in mathematics?


问:
For 52 cards in a line, what is the probability that the card immediately to the right of the first queen is a queen or a seven?
For a standard deck of 52 cards, lay down all 52 cards in a line. What is the probability that the card immediately to the right of the first
queen is a queen or a seven?

I tried 51!*28/52! but I dont think that I'm accounting for finding the first queen.

51! ways to order the items so that queen is next to another queen or seven (treat as 1 item).

28 ways to rearrange first queen with another queen or seven

52! possible outcomes.

答:
Consider a related question: What is the probability that the card following the first queen is the queen of hearts?

The entire deck can be arranged in $52!$ equally likely ways. We would like to count the number of these in which the card following the first queen is the queen of hearts. To do so, suppose we remove the queen of hearts from the deck and deal out the remaining $51$ cards.  This can be done in $51!$ ways.  There is only one place where the queen of hearts can be inserted if it immediately follows the first queen, so there are $51!$ such arrangements.  So the probability that the card following the first queen is the queen of hearts is
$$\frac{51!}{52!} = \frac{1}{52}$$

The same argument applies to any of the other queens or to any of the sevens: $8$ cards in all, including the queen of hearts. So the probability that the card following the first queen is a queen or a seven is
$$8 \cdot \frac{1}{52} = \frac{2}{13}$$


问:
Dirichlet series $\sum a_n n^{-x}$ where $(a_1+\dots+a_n)/n \rightarrow A$
I am working on some Dirichlet series and I am familiar with the proofs of convergence of the functions $\zeta$ and $\eta$.

And I was wondering if $g(x)=\sum a_n n^{-x}$ was normally convergent on $[a,b]$ where $1<a<b$ under the assumption that $(a_1+\dots+a_n)/n$ converges to a given value A.

Thus I would be able to prove that $g$ is infinitely differentiable on $]1,+\infty[$ (It is my final goal and I can clearly see how to do all the other steps).

答:
Partial summation
$$\sum_{n\ge 1} a_n n^{-s}=\lim_{N\to \infty}( A(N)N^{-s}+\sum_{n=1}^{N-1}A(n)(n^{-s}-(n+1)^{-s}))$$
For $s > 1$ using that $n^{-s}-(n+1)^{-s}=\int_n^{n+1} s t^{-s-1}dt\le sn^{-s-1}$ it is clear that it converges if $A(N)/N$ is bounded.


问:
Fourier coefficients and series of functions
We define $f_n\in C(\mathbb{T})$ by

$$f_1=f, \quad f_{n+1}=f_{n}*f$$

where $f$ is a function s.t

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx \leq 1$$

Now we define

$$I_n=\frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx$$

and we want to prove that either $I_n\rightarrow 0$ or $I_n\rightarrow 1$. Using Parsevals equality, I get that

\begin{align}
I_n &= \frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^{2n} \\
&\leq \left(\sum_{k\in\mathbb{Z}}|\tilde f(k)|^2 \right)^n
\end{align}

where I used $\tilde f(k)$ to denote the $k$th fourier coefficient, and that $\tilde{(f_1*f_2)}(k) = \tilde f_1(k) \tilde f_2(k)$.

Using this, if

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx < 1$$

then showing $I_n\rightarrow 0$ is quite easy. However, I'm having troubles handling the case where

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx = 1$$

I tried proving by induction that for each $n\in\mathbb{N}$, $I_n=1$. For the case $n=1$ it's obvious from the assumption, and by assuming for $n$ I want to conclude that it happens for $n+1$. But, how do I go from

$$\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \rightarrow \sum_{k\in\mathbb{Z}}|\tilde f_{n+1}(k)|^2$$

? Is it just using Cauchy's convolution between $\sum_{k\in\mathbb{Z}}|\tilde f_{n}(k)|^2$ and $\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^2$?

答:
As per OP, we consider the case $I_1=1$ or $\sum_{k \in \mathbb Z} |a_k|^2=1$, where $a_k=\hat f_1(k)$

There are two cases; if there is $m, |a_m|=1$ then $a_k=0, k \ne m$ and clearly all $I_n=1$ so done!

Assume now $|a_k|<1, k \in \mathbb Z$; then pick a fixed $m$ for which $0<|a_m|<1$; then since $\sum_{k \ne m} |a_k|^2=A<1$ we get $\sum_{k \ne m} |a_k|^{2n} \le A^{2n} \to 0$, while of course $|a_m|^{2n} \to 0$ also, so we are done too!


答:
I would also say that we have both relations
$$
x \not\le y, \quad x\not \ge y.
$$
There are variants, of course, like
$$
x \lneq y,\quad x \gneq y
$$
and
$$
x\nless y, x\ngtr  y
$$

Title: Do we have this symbol in mathematics?

Body:
It is the symbol I used to show left hand side of inequality isn't greater than right hand side.

More precisely I used this to show the below matrix is not diagonally dominant:

$$\begin{bmatrix}5&6&7\\
2&-4&2\\
3&2&-5\\
\end{bmatrix}$$

So at the first row I want to emphasize that $|5|$ is not greater than $|6|+|7|$  therefor instead of writing $|5|<|6|+|7|$ I used the symbol in the image above.

Is it ok to use this symbol in mathematics?


问:
Do all p.d.f. with same c.d.f have same expected value?
Let $F$ be a c.d.f. over reals and $\{R_i\}$ be the set of all real random variables with c.d.f. $F$. Do all $R_i$ have the same expected value? i.e. is the following statement true?

\begin{equation}
\mathbb{E}[R_i] = \mathbb{E}[R_j] \quad \forall i, j
\end{equation}

If yes, why? If not, could you show me a counter-example?

答:
Yes, and you do not need the distribution to be continuous or have a density. You can say

$$\mathbb E[R_i] = \int\limits_0^{\infty}(1- F(x))\, dx-\int\limits_{-\infty}^0 F(x)\, dx$$ if the two integrals are finite; if both are infinite then the expectation is not defined.

So the same cumulative distribution function gives the same expectation.


答:
Note that $\hat f_n(k)=\hat f_1(k)^n$, so
$\newcommand{\pint}{\frac1{2\pi}\int_{-\pi}^\pi}$ $$I_n=\sum_k|\hat f_1(k)|^{2n}.$$ In particular $$\sum|\hat f_1(k)|^2=I_1\le1,$$so $|\hat f_1(k)|\le1$ for all $k$.

Case 1: $f=0$.

I'll leave this case to you...

Case II: There exist two distinct values of $k$ with $\hat f_1(k)\ne0$.

In this case it follows that $|\hat f_1(k)|<1$ for every $k$, and since $\hat f_1(k)\to0$ there exists $c<1$ with $|\hat f_1(k)|\le c$ for all $k$. Hence $$I_n\le c^{2(n-1)}I_1\to0.$$

Case III: There exists $k_0$ such that $\hat f(k)=0$ for all $k\ne k_0$.

In this case $$I_n=|\hat f(k_0)|^{2n},$$which tends to $0$ or$1$.

Title: Fourier coefficients and series of functions

Body:
We define $f_n\in C(\mathbb{T})$ by

$$f_1=f, \quad f_{n+1}=f_{n}*f$$

where $f$ is a function s.t

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx \leq 1$$

Now we define

$$I_n=\frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx$$

and we want to prove that either $I_n\rightarrow 0$ or $I_n\rightarrow 1$. Using Parsevals equality, I get that

\begin{align}
I_n &= \frac{1}{2\pi}\int^{\pi}_{-\pi}|f_n(x)|^2 dx \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \\
&=\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^{2n} \\
&\leq \left(\sum_{k\in\mathbb{Z}}|\tilde f(k)|^2 \right)^n
\end{align}

where I used $\tilde f(k)$ to denote the $k$th fourier coefficient, and that $\tilde{(f_1*f_2)}(k) = \tilde f_1(k) \tilde f_2(k)$.

Using this, if

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx < 1$$

then showing $I_n\rightarrow 0$ is quite easy. However, I'm having troubles handling the case where

$$\frac{1}{2\pi}\int^{\pi}_{-\pi}|f(x)|^2 dx = 1$$

I tried proving by induction that for each $n\in\mathbb{N}$, $I_n=1$. For the case $n=1$ it's obvious from the assumption, and by assuming for $n$ I want to conclude that it happens for $n+1$. But, how do I go from

$$\sum_{k\in\mathbb{Z}}|\tilde f_n(k)|^2 \rightarrow \sum_{k\in\mathbb{Z}}|\tilde f_{n+1}(k)|^2$$

? Is it just using Cauchy's convolution between $\sum_{k\in\mathbb{Z}}|\tilde f_{n}(k)|^2$ and $\sum_{k\in\mathbb{Z}}|\tilde f_1(k)|^2$?


问:
The closes point to a curve in space.
I am working on the following problem.

Find the point closest to the origin, of the curve of intersection of the plane $2y+4z =5$ and the cone $z^2 = 4(x^2+y^2)$

I was able to see that the intersection will be a tilted ellipse with coordinates $(x,y,4(x^2+y^2))$.

So I was thinking that I could try putting $F(x,y) = \sqrt{x^2+y^2+16(x^2+y^2)^2}$ and optimize it. But this seems to be too long a calculation and I want to somehow use vectors, such as dot products and gradients because that seems to be easier.

Can anyone help me get started ?

答:
I think it is not necessary to use Lagrange multipliers as it is can be treat a simple geometry problem.

you already know that there is a ellipse and there are two special points $P$ and $Q$.

you can see the $P$ and $Q$ in on the plane $x=0$, if you consider two rings on the cone, one is through $P$, another is through $Q$, all the points on the two rings to original point will have same distances, one is $PO$, another is $QO$,it is trivial that $PO$ is min,$QO$ is max for the ellipse. so you only need to find $P,Q$, recall $x=0$ plane, the  problem become a simple one. $P(.5,1)$, so $PO=\sqrt{(0.5)^2+1}=\dfrac{\sqrt{5}}{2}$

Title: The closes point to a curve in space.


问:
Prove integral inequality $\int_{0}^{\pi} e^{-A\sin\theta}d\theta < \frac{\pi}{A}$
Given the Jordan's inequality

$$\frac{\sin \theta}{ \theta} \geq \frac{2}{ \pi}
$$ $\text{ for }0\leq\theta\leq \frac{\pi}{2}$, I have to prove that

$$\int_{0}^{\pi} e^{-A\sin\theta}d\theta < \frac{\pi}{A}$$

But I'm not sure how they are related, or how I can use the first to prove the second. Any help is appreciated.

Thank you.

答:
The given inequality $\sin \theta\geq \frac{2}{ \pi}\theta$ leads to $e^{-A\sin\theta } \le e^{- \frac{2A}\pi\theta}$. Then
$$\int_{0}^{\pi} e^{-A\sin\theta}d\theta
=2\int_{0}^{\pi/2} e^{-A\sin\theta}d\theta <2 \int_{0}^{\pi/2} e^{- \frac{2A}{\pi}\theta} d\theta = \frac{\pi}A(1-e^{-A})<\frac{\pi}A $$


问:
Evaluate $\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right)$
Evaluate the limit
$$
\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right)
$$

I got stuked when trying to evaluate it and I even don't know where to begin. Mathematica tells me that
$$
\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}=\frac{2^{1/n} \left(-\left(2^{1/n}\right)^n \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)+\left(2^{1/n}\right)^{n+1} \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)-2^{1/n} \Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+\Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+n \left(2^{1/n}\right)^n-n\right)}{\left(2^{1/n}-1\right) n^2}
$$
However it doesn't give the result of the limit either. And I tried $n=10000$ numerically, the answer is $1.44274..$.

So how to calculate the limit?

答:
We have $$
\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n} = \frac1n\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{nk}+1}$$ and now
$$\frac1n\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{nk}+1}=\frac1n\sum _{k=1}^n 2^{k/n}-\frac1n\sum _{k=1}^n \frac{2^{k/n}}{nk+1}$$
where the first term converges to $\int_0^1 2 ^xdx$ and the second term vanishes.


问:
Let $A_1,A_2\in L(V,V)$, $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Is $\dim\ker (A_2A_1)=n_1+n_2$ even if $\dim V=\infty$?
Let $V$ be a vector space that is not necessarly finite dimensional. Consider two surjective linear operators $A_1,A_2\in L(V,V)$ with finite-dimensional kernel: $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Can it be proven that $\dim\ker (A_2A_1)=n_1+n_2$?

In words: Is the dimension of the kernel of the composition of surjective linear operators with finite dimensional kernels the sum of the dimensions of the kernels?

Looking at this formula, one might think that the formula $\dim\ker A_2A_1=n_1+_2$ is valid, but I have looked at the answers and they are based on the rank nullity theorem, which I can't use, since $V$ is not finite dimensional.

Motivation: I want to prove that the solution space of homogeneous linear differential equations of order $n$ is $n$-dimensional by writing the $n$-th order differential operator as composition of first-order differential order with one-dimensional kernel. Actually, the answer to this question completes my proof.

答:
The result is true, and it only requires that $A_1$ is surjective, $A_2$ doesn't have to be.

Suppose that $$\ker A_1=\operatorname{span}\{a_1,\ldots,a_n\},\qquad \qquad\!\!\!\! \ker A_2=\operatorname{span}\{b_1,\ldots,b_m\},$$with $\dim \ker A_1=n$, $\dim\ker A_2=m$. Since $A_1$ is surjective, there exist $x_1,\ldots,x_m$ with $A_1x_k=b_k$. The set $\{a_1,\ldots,a_n,x_1,\ldots,x_m\}$ is linearly independent: if
$$\tag1
\sum_j\alpha_j\,a_j+\sum_k\beta_k\,x_k=0,
$$
applying $A_1$ we get $\sum_k\beta_kb_k=0$, which gives $\beta_1=\cdots=\beta_m=0$. And then going back to $(1)$ gives us that $\alpha_1=\cdots=\alpha_n=0$.

By construction, $\operatorname{span}\{a_1,\ldots,a_n,x_1,\ldots,x_m\}\subset\ker A_2A_1$. Conversely, if $x\in \ker A_2A_1$, then $A_1x\in\ker A_2$, so there exist $\beta_1,\ldots,\beta_m$ with
$$
A_1x=\sum_k\beta_k\,b_k=\sum_k\beta_kA_1x_k=A_1\Big(\sum_k\beta_kx_k\Big).
$$
So $x-\sum_k\beta_kx_k\in\ker A_1$, implying that there exist $\alpha_1,\ldots,\alpha_n$ such that
$$
x-\sum_k\beta_kx_k=\sum_j\alpha_ja_j,
$$
and this shows the reverse inclusion. Thus $\dim\ker A_2A_1=n+m$.


问:
How many people must be present at a party to guarantee that at least three of their birthdays are in the same month?
I keep getting 36 as an answer since if 36 people were present at a party, 36/12 (for the months in a given year) = 3. But this is incorrect. Any help would be appreciated.</p>

答:
This can be done using the Pigeonhole principle which says that if you have $m$ cages and you want to put $n>m$ pigeons in  those cages, then at least one of the cages must contain more than one pigeon. Let me give you an example. If you have $3$ cages, then at least how many pigeons should you have such that if you put them in those $3$ cages then at least $2$ pigeons must be in the same cage$?$ it can be easily answered, and the answer is $4$. Because if you try to put exactly one pigeon in those cages then after putting three pigeons in three cages one pigeon is left, and you have to put that in one of the cages which already contain one pigeon.

Similar is the case here. You have $12$ months, so if there are $24$ people in the party then it is possible that every two them has birthday in the same month. But if you take $25$ people then by Pigeonhole principle birthday of $3$ of them must be in the same month.


问:
Any commutative associative operation can be extended to a function on nonempty finite sets
This is a fact we use very frequently in general mathematics when we write such notations as $1+2+3+4$: since we know that $+$ is commutative and associative, we can just "drop the parentheses" and not worry about order of operations. Of course I believe this, but how does one prove this in full generality? Even stating it is giving me trouble. Here's my attempt:

Assume an operation $\oplus:S\times S\to S$ is provided satisfying $x\oplus y=y\oplus x$ and $x\oplus(y\oplus z)=(x\oplus y)\oplus z$ for all $x,y,z\in S$.

Claim: Given any finite set $\emptyset\subset A\subseteq S$, there exists a unique $z\in S$ such that for any function $f:{\cal P}(A)\to{\cal P}(A)$ which satisfies $\emptyset \subset f(B)\subset B$ for all $B\subseteq A$ with $|B|\ge 2$ and any function $g:{\cal P}(A)\to S$ which satisfies $g(\{x\})=x$ for all $x\in S$ and $g(B)=g(f(B))\oplus g(B-f(B))$ for all $|B|\ge 2$, $g(A)=z$.

The operation $\oplus$ does not necessarily have an identity element, so we do not attempt to define an empty sum. Intuitively, this element $z$ represents the finite sum of the elements in $A$, so if $A=\{1,2,3\}$ and $f(\{1,2,3\})=\{1\}$ and $f(\{2,3\})=\{3\}$, then
$$z=g(\{1,2,3\})=g(\{1\})\oplus g(\{2,3\})=g(\{1\})\oplus (g(\{3\})\oplus g(\{2\}))=1\oplus(3\oplus 2).$$
There has got to be a better way to say that, but this is the only way I can think of to capture all the possibilities of parenthesization, and still be amenable to a formal proof. And now that I've stated it, how should I prove it? I suppose I should induct on something, but I've no idea what.

Edit: The goal here is to be able to define an operation $F$ such that $F(\{x_1,\dots,x_n\})=x_1\oplus\cdots\oplus x_n$ and be assured that the operation is well defined and satisfies $F(A\cup B)=F(A)\oplus F(B)$, when $A$ and $B$ are disjoint finite nonempty subsets of $S$.

答:
You can state it as follows: Given is a set $S$ with a binary $\oplus$ operation on it which is associative and commutative.

For every finite nonempty set $A\subseteq S$, there exists an element $S_A\in S$ such that: 1) If $A=\{a\}$, then $S_A=a$. 2) If $|A|=n$ then for any bijective function $f:\{1,2,3,4,\cdots ,n\}\to A$, it holds that $S_A = f(1)\otimes S_{\{f(2),f(3),\cdots, f(n)\}}$.

Moreover, this assignment $A\mapsto S_A$ is the only one with these properties.

The proof of both existence and uniqueness is done by a standrad proof by induction.


问:
How do I find the equilibrium points of a system with a free variable
For the following system, $$\dot{x_1} = 3(x_1-x_2) \\ \dot{x_2} = x_1(k-2-x_3) \\ \dot{x_3}=x_1x_3$$ I want to determine its equilibrium points together with their stability.

By the third equation I get $x_1=0$ or $x_3 = 0$. If $x_1 =0$ then $x_2=0$ so that I found an equilibrium point $(0,0,0)$ as $x_1=x_2$ by the first equation. However, if $x_3=0$ then $x$ can be anything unless $(b-1-x_3)=0$ which means that $b=1$. So that if $b \neq 1$, I find equilibrium points $(x_1,x_2,0)=(x_1,x_1,0)$ where $x_1 \neq 0$ which looks unusual.

How can I find the equilibrium points of this system?

答:
From $\dot{x}_3 = x_1x_3$ we have that either $x_1=0$ or $x_3=0$. Let us start with $x_1=0$. Then from $\dot{x}_1=3(x_1-x_2)$ we conclude $x_2=0$. For $x_1=0$ we have $\dot{x}_2=0$, and thus we obtain the equilibrium set $\left\{x: x_1=x_2=0, \ x_3\in \mathbb{R}\right\}$. Any point on this line is an equilibrium.

Next, consider $x_3=0$. From the equation for $\dot{x}_1$ we have $x_1=x_2$, and if $x_1=0$ we have the origin that belongs to the equilibrium set we have already found. Let us consider $x_1\ne 0$ to check if there is something else besides that line.

From the equation for $\dot{x}_2$ we conclude that the only solution satisfying $x_3=0$ and $x_1\ne 0$ is when $k=2$ that yields $x_1\in\mathbb{R}$.

To summarize, you have the equilibrium set (line) $\left\{x: x_1=x_2=0, \ x_3\in \mathbb{R}\right\}$. And for $k=2$ you have additionally the equilibrium set $\{x:x_1\in \mathbb{R}, x_2=x_1, x_3=0\}$ that is another line.


问:
Prove that a metric space $M$ is compact iff for every continuous function $f:M \rightarrow \mathbb{R}$...
Prove that a metric space $M$ is compact if, and only if, for every continuous function $f:M \rightarrow \mathbb{R}$, such that $f(x) > 0$ for all $x \in M$, we have $\inf\limits_{x\in M} f(x) >0$

The proof I made is in accordance with the corollary that says: Let $M$ be a compact metric space, and let $f:M\rightarrow \mathbb{R}$ be continuous. Then $f[M]$ is delimited, and there is $x \in M$ such that $f(x) =\inf\limits_{x\in M}f(x)>0$.
$\textbf{If anyone can give me a feedback on my demo. I would be very grateful.}$

Proof: $f[M]$ is a compact subset of $\mathbb{R}$, hence closed and bounded. Now, any bounded set $A\subseteq \mathbb{R}$ has a least upper bound $\sup$ $A$ and a greatest lower bound $\inf A$,
and these two points belong to the closure $\overline{A}$ (why?). But applying this to $A = f[M]$, which
is closed, we conclude that $\sup$ $f[M]$ and $\inf$ $f[M]$ belong to $f[M]$ itself, which is exactly what is being claimed (why?). \

$\square$

答:
If $A$ is a bounded subset of $\Bbb R$, and if $s=\sup A$ then for each $\varepsilon>0$, $s-\varepsilon$ is not an upper bound of $A$ (since $s$ is the least upper bound of $A$) and thereore, there is some $a\in A$ such that $a>s-\varepsilon$. But, since $s$ is an upper bound of $A$, $a\leqslant s$. So, $|s-a|<\varepsilon$. Since this occurs for every $\varepsilon>0$, $s\in\overline A$. For a similar reason, $\inf A\in\overline A$.

And, since $\inf f(M)\in f(M)\subset(0,\infty)$, there is some $m\in M$ such that $f(m)=\inf f(M)$ and, since $f(m)>0$, this proves that $\inf f(M)>0$.

Not this only proves that if $M$ is compact, then $\inf f(M)>0$; it doesn't prove the reverse implication. In order to prove it, suppose that $M$ is not compact. Then there is an unbounded continuous function from $M$ into $\Bbb R$ (you will find a proof here). Let $f$ be such a function. If $f$ has no lower bound, the $\exp\circ f$ is a continuous function from $M$ into $(0,\infty)$ whise infimum is $0$. And if has has a lower bound then,since $f$ is unbounded, you can apply the same argument to $-f$.


问:
Why is $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\left<(1,2,4)\right>$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$?
I understand that this is isomorphic to either $\mathbb{Z}_4 \times \mathbb{Z}_8$ or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_8$, but I can't think of any way to show that it is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$.</p>

答:
You can also play directly with the definitions:

\begin{alignat}{1}
&\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle= \\
&= \{(a,b,c)+\left<(1,2,4)\right>\mid a,b,c\in \Bbb Z \} \\
&= \{(a,b,c)+\{(l,2l,4l)\mid l\in\Bbb Z\}\mid a,b,c\in \Bbb Z \} \\
&= \{(a+l,b+2l,c+4l)\mid a,b,c,l\in\Bbb Z \} \\
&\stackrel{m:=a+l}{=} \{(m,b+2m-2a,c+4m-4a)\mid a,b,c,m\in\Bbb Z \} \\
&= \{\{(m,2m,4m)\mid m\in \Bbb Z\}+(0,b-2a,c-4a)\mid a,b,c\in \Bbb Z \} \\
&= \{(0,b-2a,c-4a)+\langle(1,2,4)\rangle\mid a,b,c\in \Bbb Z \} \\
\end{alignat}

Therefore, there is a complete set of representatives, isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$. So $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle$ is, too.


问:
Interchanging limits of integration for an even function
I wanted to know is this true for an even function $\mathbf f(x)$:

$\int_a^b$$\mathbf f(x)$ $\mathbf =$$\int_b^a$$\mathbf f(x)$

That is, there is no requirement of a negative sign while interchanging the limits of integration.

答: No, it is not true, since, by definition, if $b<a$, then $\displaystyle\int_b^af(x)\,\mathrm dx=-\int_a^bf(x)\,\mathrm dx$.


问:
How many different $ 8$ bit strings begin with $000$, end with $001$, or have $10$ as the $4th$ and $5th$ bits, respectively?
I keep obtaining $127$ as my final answer yet I know this is incorrect. I start with the # of combinations for $8$ bit strings that begin with $000$, which is $2^5$. Then calculate the # of $8$ bit strings that end with $001$, which is also $2^5$, then attempt to calculate the # of $8$ bit strings that have $1$0 as the $4$th and $5$th bits (this is where I think I am messing up), which I get as $2^6$. $2^5 + 2^5 + 2^6 = 128$, but I know we must subtract the case where a string has all three of these conditions, of which there is only 1 (a string that looks like $00010001$). I therefore obtain $128-1 = 127$, yet this is incorrect. Any ideas?

答:
Let's do this step by step. As you correctly noted, there are $2^5$ strings that begin with $000$, and $2^5$ strings that end in $001$. The key is to correctly count how many we counted twice. In other words we're looking for all strings of the form $000xx001$ -- but the $x$'s could be anything, so there are $2^2$ of those. This gives us a partial total of $2^5 + 2^5 - 2^2$. Next we're looking at all strings of the form $xxx10xxx$. As before, there are $2^6$ of those but we need to know how many we counted twice, i.e. subtract all strings of the form $00010xxx$ and $xxx10001$. There are $2^3 + 2^3$ but we must take care here: there is in fact a single string, $00010001$, that we're subtracting twice because it satisfies both formats. So the expression becomes: $$(2^5+2^5-2^2)+(2^6-2\cdot2^3+1).$$ I leave it to you to compute the total.

P.S. As InterstellarProbe notes in the comments, and if you're familiar with set theory, the procedures above may be formalised (as far as event spaces go) with the following principle:
$$\#\left(A \cup B \cup C\right) = \#A + \#B + \#C - \#(A\cap B) - \#(A\cap C) - \#(B\cap C) + \#(A\cap B\cap C)$$


问:
How to form a cubic equation with the substitution method?
I had this question:

"Find the cubic equation whose roots are twice the roots of the equation $3x^3 - 2x^2 + 1 = 0$"

In my first attempt, I solved it through the use of simultaneous equations, where I let the cubic equation be: $x^3 + bx^2 + cx + d$. However, I've been told that there is a more efficient method for higher degree polynomials - and it has been labelled as "The Substitution method".

It involved something like "Let $y = x^2$" then form a new equation using that.
Could someone solve this question and explain how to do it for me?

Any help would be greatly appreciated, thanks!

The solution with the substitution method worked like this:

Let $y=2x$,
$x=y/2$
then it was substituted back into the given equation.

Can someone explain this?

答:
Let the roots of the given equation be $a_{1},a_{2},a_{3},a_{4}$.The new roots will be $2a_{1},2a_{2},2a_{3},2a_{4}$
Now use Vieta' formula. Vieta's Formulas

Sum of roots = $2(\sum a_{i})=\dfrac{4}{3}$ and product of roots = $8a_{1}a_{2}a_{3}=-8$

The equation will be : $$x^{3}-\frac{4}{3}x^{2}+\frac{8}{3}=0$$ or, $$3x^{3}-4x^{2}+8=0$$


答: No, as José Carlos Santos points out. But it is true that $$\int_a^bf(x)\,dx=\int_{-b}^{-a}f(x)\,dx$$ for even functions

Title: Interchanging limits of integration for an even function

Body:
I wanted to know is this true for an even function $\mathbf f(x)$:

$\int_a^b$$\mathbf f(x)$ $\mathbf =$$\int_b^a$$\mathbf f(x)$

That is, there is no requirement of a negative sign while interchanging the limits of integration.


问:
Confusion about linearly independent vectors
Let $K$ be a field an $V$ a vector space. Let $v_1,\ldots,v_n$ be linearly independent vectors in $V$.

Are the following two statements equivalent? If not, why not?

for all $a_1,\ldots,a_n$ in $K\setminus\{0\}$ is $\sum\limits_{i=1}^na_iv_i\neq 0$

for all $(a_1,\ldots,a_n)^t$ in $K^n\setminus\{0\}$ is $\sum\limits_{i=1}^na_iv_i\neq 0$

Does it make a difference whether we talk about the vector $(a_1,\ldots,a_n)^t$ (2.) or if we look at $a_i$ as elements in $K$ only (1.)?

Thanks in advance

答:
It does not matter whether you view the scalars $a_1, \ldots, a_n$ as elements of $K$ or as a vector in $K^n$.

In your second statement the $0$ is the zero vector.


答:
If $A$ is set of all strings starting with $000$,

$B$ is set of all strings ending with $001$ and,

$C$ is set of all strings with $4$th and $5$th places being $10$

Then,

$|A| = |B| = 2^5, |C| = 2^6$

$|A \cap B| = 2^2, |B \cap C| = |A \cap C| = 2^3$

$|A \cap B \cap C| = 1$

So your answer should be

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

$= 109$

Title: How many different $ 8$ bit strings begin with $000$, end with $001$, or have $10$ as the $4th$ and $5th$ bits, respectively?

Body: I keep obtaining $127$ as my final answer yet I know this is incorrect. I start with the # of combinations for $8$ bit strings that begin with $000$, which is $2^5$. Then calculate the # of $8$ bit strings that end with $001$, which is also $2^5$, then attempt to calculate the # of $8$ bit strings that have $1$0 as the $4$th and $5$th bits (this is where I think I am messing up), which I get as $2^6$. $2^5 + 2^5 + 2^6 = 128$, but I know we must subtract the case where a string has all three of these conditions, of which there is only 1 (a string that looks like $00010001$). I therefore obtain $128-1 = 127$, yet this is incorrect. Any ideas?


问:
Checking that a particular set is a $\sigma$-algebra
I'm working on an interesting project regarding general measure theory. I'm wondering if you would be interested in hearing about it. Suppose $S$ is a set which is equinumerous to $\mathbb{R}$. We know that this means there exists a function $\phi$: $\mathbb{R}$ $\rightarrow$ $S$ that is one-to-one and onto. Now what I do is let $\mathcal{M}_\phi$ = {$E$ $\subset$ $S$ $\mid$ $\phi^{-1}$$(E)$ $\in$ $\mathcal{M}$}. I want to check that $\mathcal{M}_\phi$ is a $\sigma$-algebra. I'm clear with $\varnothing$ $\in$ $\mathcal{M}_\phi$. How do I show the corresponding properties for the complements of subsets of $\mathcal{M}_\phi$ and countable unions? </p>

答:
Let $E \in \mathcal{M}$.  Then $\overline{E} \in \mathcal{M}_\phi$ iff $\phi^{-1}(\overline{E}) \in \mathcal{M}$,and, since $\phi$ is a bijection, $\phi^{-1}(\overline{E}) = \overline{\phi^{-1}(E)}$ (as an element of $S$ is not in $E$ iff the corresponding $\phi^{-1}(s)$ is not in $\phi^{-1}(E)$).  But, as $\mathcal{M}$ is a $\sigma$-algebra, $\overline{\phi^{-1}(E)} \in \mathcal{M}$.  This proves the first part.

The second part follows with basically the same ease as the first- since $\phi$ is a bijection, it commutes with the relevant set-theoretic operations (complements and unions, in this case), so that the properties of $\mathcal{M}$ carry over.

Also, the result is expected, because the existence of such a bijection $\phi$ means that, from the point of view of elementary set theory (which is all that the definition of $\sigma$-algebra uses), $S$ and $\mathbb{R}$ are "the same" (i.e., isomorphic as sets), so it is to be expected that the set of correspondents of elements of $\mathcal{M}$ would be a $\sigma$-algebra too.


答:
We want the roots $y_i$ to be $2x_i$. So substitute $\frac{y}{2}$ everywhere for $x$, and simplify if you wish. </p>

<p>Afterwards, if you want the variable name to be $x$, replace $y$ by $x$. </p>

Title: How to form a cubic equation with the substitution method?

Body:
I had this question:

"Find the cubic equation whose roots are twice the roots of the equation $3x^3 - 2x^2 + 1 = 0$"

In my first attempt, I solved it through the use of simultaneous equations, where I let the cubic equation be: $x^3 + bx^2 + cx + d$. However, I've been told that there is a more efficient method for higher degree polynomials - and it has been labelled as "The Substitution method".

It involved something like "Let $y = x^2$" then form a new equation using that.
Could someone solve this question and explain how to do it for me?

Any help would be greatly appreciated, thanks!

The solution with the substitution method worked like this:

Let $y=2x$,
$x=y/2$
then it was substituted back into the given equation.

Can someone explain this?

Body:
We want the roots $y_i$ to be $2x_i$. So substitute $\frac{y}{2}$ everywhere for $x$, and simplify if you wish.

Afterwards, if you want the variable name to be $x$, replace $y$ by $x$.


问:
Challenging integral with root in denominator
I'm wondering if anyone has ideas to get started with this integral:

$$
I = \frac{b}{\pi}\int_{-\infty}^{\infty}dx\,\frac{1}{\sqrt{\left[(x-1)^2+b^2\right]\left[(x+1)^2+b^2\right]}}
$$

for $b>0$. Mathematica 12 can't get it done (except for $b=1$) and the most obvious methods haven't worked for me. Ideas? Can it be done exactly?

Ben

Edit: Here is a variation that I also would like to solve:

$$
I_2 = \frac{\sqrt{b_+ b_-}}{\pi}\int_{-\infty}^{\infty}dx\,\frac{1}{\sqrt{\left[(x-1)^2+b_+^2\right]\left[(x+1)^2+b_-^2\right]}}
$$

I'm not sure that ComplexYetTrivial's solution will work for this one. There is a linear term appearing in the root that will mess up the square completion. Ideas?

Edit 2: It seems like it was solved here using Maple, but the solution looks quite messy:

https://stats.stackexchange.com/questions/243943/is-it-correct-to-compute-bhattacharyya-distance-for-cauchy-like-bell-shaped-fun

答:
I found an answer to my suggested generalization for $b_+ \neq b_-$. The result is:

$$I = \frac{4}{\pi}\sqrt{\frac{b_+ b_-}{A_+}} K\left(\frac{A_-}{A_+}\right).$$

Here, I have introduced:

$$A_\pm = 4 + (b_+ \pm b_-)^2,$$

and $K$ is the complete elliptic integral of the first kind as defined by Mathematica's EllipticK[].


问:
When to say that a set is above bounded or below bounded?
for example the intervals $M = (2,6]$, $H=[1,5]$, $J=[8,\infty)$. How to specify which set is bounded and if it is bounded how to say whether it's bounded from above?

答: A set $M$ is bounded from above if there exists some $C$ such that for any $x \in M$, we have $x \leq C$.  For intervals on the real line, I'd start by drawing the interval and looking for a potential bound $C$.  That will lead you to understand that any interval $(a,b)$ with a finite upper bound $b$ is bounded above (by $b$).


答:
Your result $R(n)$ in the last line can be approximated in Mathematica using the function

```<code>Series[R[n],Infinity,1]
</code>```

I got $$\frac{1}{2 n}+\frac{1}{\log 2}\to \frac{1}{\log 2}$$

Title: Evaluate $\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right)$

Body:
Evaluate the limit
$$
\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right)
$$

I got stuked when trying to evaluate it and I even don't know where to begin. Mathematica tells me that
$$
\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}=\frac{2^{1/n} \left(-\left(2^{1/n}\right)^n \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)+\left(2^{1/n}\right)^{n+1} \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)-2^{1/n} \Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+\Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+n \left(2^{1/n}\right)^n-n\right)}{\left(2^{1/n}-1\right) n^2}
$$
However it doesn't give the result of the limit either. And I tried $n=10000$ numerically, the answer is $1.44274..$.

So how to calculate the limit?

Body:
Your result $R(n)$ in the last line can be approximated in Mathematica using the function

Series[R[n],Infinity,1]

I got $$\frac{1}{2 n}+\frac{1}{\log 2}\to \frac{1}{\log 2}$$


问:
Imaginary interest rates
If you have watched lockdown math series by "3 blue 1 brown",

or if you are willing to watch this video, then I have a doubt.

At  38:02 of the video why is the number of years equal to the distance travelled on the circle?

How do we know that the full circle represents $\ 2\pi\ $ years?

I have asked the question in the video comments but don't expect any reply there so can anyone help me out here?

答:
It really is only equal to the distance if you start with precisely 1 dollar on your bank account, because then the circle becomes the unit circle (radius 1). This means that the distance traveled along the edge of the circle has the same magnitude as the angle of the vector that rotates around origo (in this case z(t) describing the money as a function of time). For example, one whole rotation $z(2\pi)$ yields a circle with circumference $2\pi$. If you start with any other amount, the length traveled does not correspond to the time (years). It is the angle that corresponds to time. Watch the whole video and this should become clear (around he 55 min mark is when he presents the function z(t), where t is time). It might also be helpful to refresh radians in general. The wiki page on radians is sufficient.

If you wanted to visualize time in a graph, you would need to add another dimension to the graph. The result would be a spiral coming out of the screen as time increases.

Title: Imaginary interest rates


问:
Prove that $\int_{B^2(0,1)}||D(\Phi)||=\int_{\Phi(B^2(0,1))}||D(\Phi^{-1})||$
Let $\Phi:\Bbb R^2 \to \Phi(\Bbb R^2)\subseteq \Bbb R^2$ be a diffeomorphism. Prove that

$\int_{B^2(0,1)}||D(\Phi)||=\int_{\Phi(B^2(0,1))}||D(\Phi^{-1})||$,

where $||A||=(\sum_{i,j=1}^2a_{ij}^2)^{1/2}$ is the Hilbert-Schmidt norm of the matrix and $D$ is the differential.

I have tried in this way:
Let us consider $I=\int_{B^2(0,1)}||D(\Phi)||$. Now substitute $\underline y=\Phi(\underline x)$ then $\underline{x}=\Phi^{-1}(\underline{y})$.

Then, $I=\int_{\Phi(B^2(0,1))}||D(y)||Jacobian(D(\Phi^{-1}))$.

If I am not wrong here $D(y)$ will be the identity matrix so $||D(y)||=n$. Here I am confused again that $D(y)=(\frac{\partial y}{\partial x})$, the matrix or not!

Here is my problem, is it true that $Jacobian(D(\Phi^{-1}))=\frac{1}{||D(y)||}||D(\Phi^{-1})||$? If it is true then we are done i.e.
$$I=\int_{\Phi(B^2(0,1))}||D(1)||Jacobian(D(\Phi^{-1})=\int_{\Phi(B^2(0,1))}||D(\Phi^{-1})||$$ but how to prove that portion?

Please help

答:
By the Inverse Function Theorem, as $\Phi$ is a diffeomorphism we have that $D\Phi^{-1}(\Phi(x,y)) = (D\Phi(x,y))^{-1}$. Using this and the fact given by Ted Shifrin that for a $2\times 2$ matrix $A$, under the Hilbert Schmidt norm,
$$||A^{-1}|| = \frac{1}{|\text{det}A|} ||A||$$
we can see that,
$$||D\Phi^{-1}(\Phi(x,y))|| = ||(D\Phi(x,y))^{-1}|| = \frac{1}{|J_\Phi(x,y)|}||D\Phi(x,y)|| $$ which of course implies that $|J_\Phi(x,y)| \cdot||D\Phi^{-1}(\Phi(x,y))|| = ||D\Phi(x,y)||$
Thus by the Change of Variables formula,
$$\int_{\Phi(B^2(0,1))} ||D\Phi^{-1}(x,y)|| = \int_{B^2(0,1)}|J_\Phi(x,y)|\cdot||D\Phi^{-1}(\Phi(x,y))|| = \int_{B^2(0,1)} ||D\Phi(x,y)|| $$


问:
Is a square matrix similiar to its row echelon form?
Let $A, B$ be two square matrices with real entries. We say that $A$ and $B$ are similar if an invertible matrix $P$ exists such that $A = P^{-1}BP$.

My question is:  if we consider the row echelon form of $A$, it is similar to  $A$ itself?

I think that there are some cases in which it could be true (I think about matrices with $LDU$ factorization), but in general it is not true.

Could anyone please help me to understand if it is true or not, and present some example/counterexamples?

答:
The result of the QR algorithm is a matrix that is similar to the original one and in row-echelon form, at least if all eigenvalues are real. Conversely, if $P^{-1}AP$ is upper triangular, then the diagonal elements are the eigenvalues, so the computation of this form can not be simpler than any eigenvalue algorithm, so LU, QR or any other simple decompositions will not give such a result.

The best you get with a simple, non-iterative algorithm is the Hessenberg form of the matrix, which has non-zero elements only above the first sub-diagonal. This form is also the recommended starting point for the QR algorithm and its variants.


问:
Prove that volume of a cylinder section cut off by a plane is $\frac16$ of circumscribing parallelepipe
I was given the following question:

Given a cylinder circumscribed within a parallelepiped with a square base that has a plane going through the center of the base circle and through one side of the square on the top of the parallelepiped. Use calculus to prove that the volume of the segment of the is $\frac16$ the volume of the rectangular parallelepiped circumscribing the cylinder.

Here is what I've done:

I created a diagram of the described cylinder on GeoGebra (see below). Given that the base is a square, the volume of the parallelepiped can be given as $a^2b$ (if necessary this can be shown using calculus by doing $\int^a_0 ab\ dx$). The radius of the cylinder is $\frac a2$ and the height is $b$. The volume would therefore be: $\frac{\pi a^2}{4}b$. The only problem is that I need to factor in the plane. I would take the integral of the equation of the plane minus the area of each cross section of the cylinder ($\frac{\pi a^2}{4}$) but I cannot figure out the equation of the plane! I know it goes through the point $(\frac a2,\frac a2,0)$ and the line $z=b,x=a$ (I'm not sure if I wrote that line right). But I don't know how to get the equation of that plane.

Am I going about this problem in the right way? How can I find that equation?

答: This is the so-called 'hoof of Archimedes," which was solved over 2000 years ago. For a lengthy exposition of the problem in modern terms please see Gray, S.B., Ding, D.Y., Gordillo, G., Landsberger, S., and Waldman, C. (2015). The Method of Archimedes: Propositions 13 and 14, Notices of the AMS, 62, 9, pp, 2-6.</p>


问:
Proving angle congruence using circles and triangles

The above picture is of an ellipse with foci G and F. AC and BD are tangent to the ellipse at the two ends of the major axis and CD is tangent to the ellipse at E. There are two circles, circle I and circle J. $\angle CFD=\angle DGC=90^{\circ}$.

I was asked to prove that $\angle ACF=\angle DCG$ and $\angle CDF=\angle BDG$.

Here is what I know: $\angle EDH=\angle EGH$ because they intersect the same arc. $\angle ECH=\angle EFH$ because they intersect the same arc. $\angle DBG$ and $\angle CAF$ are right angles because they are tangents of a point at the end of an axis.

If I can prove that $\angle CFA=\angle GDC$, I can show that $\angle ACF=\angle DCG$ because they would be congruent parts of similar triangles. If I can show that $\angle EFH=\angle FCA$ I can show that $ACF=DCG$. But I am stuck at this point. Any ideas of how to proceed?

答:
$DGFC$ is cyclic. So $$\angle CDF = \angle CGF = \angle BDG$$

Also $$\angle DCG = \angle DFG = \angle ACF$$

Thus it follows that $HG$, $HF$, $HE$ are angle bisectors and H is incenter of $\triangle EGF$.

Explanation :

$D, G, F, C$ lie on a circle since $\angle DGC=90=\angle DFC$ subtended by chord $DC$. Thus $DGFC$ is cyclic. $DC$ is diameter.

$\angle CDF=\angle CGF$ - angles subtended by chord $CF$ of semicircle $DGFC$.

$\angle BDG + \angle BGD=90$ Also $\angle BGD + \angle CGF=90$ $\Rightarrow \angle BDG=\angle CGF$

$\angle CDF=\angle EDH=\angle EGH=\angle CGF \Rightarrow HG$ is angle bisector of $\angle EGF$.

Can you complete this now?


答:
Suppose you have a root $x$ of $$3x^3-2x^2+1=0$$

If $y=2x$, then $x=\cfrac y 2$ and $$3(\frac y 2)^3-2(\frac y 2)^2+1=0$$

Clear the fractions and you obtain an equation with integer coefficients satisfied by $y$.

The key to the substitution method is to find an expression for $x$ (a root of the original equation) and substitute it into the original equation. It depends on doing the same thing to each of the roots.


答:
To avoid the inclusion-exclusion principle, we can count the complement.

If an $8$-bit string does not begin with $000$, does not end in $001$, and does not have $01$ as its $4^{\text{th}}$ and $5^{\text{th}}$ bits, then there are $2^3-1 = 7$ possibilities for the first three bits, $2^2-1 = 3$ possibilities for the next two bits, and $2^3-1=7$ possibilities for the last two.

Therefore there are $7 \cdot 3 \cdot 7 = 147$ strings that don't satisfy the condition, and $2^8 - 147 = 109$ strings that do.


答:
Another approach (which is kind of fun I think) is to count the number of string that don't have any of the three properties.

For $8$ it strings, there are $2^8=256$ possibilities.

Now - to count the strings that don't have any of these properties:

For the first three bits, there are 7 possibilities (2^3=8 except for 000)

For the next two, there are 3 possibilities (2^2-4 except for 01)

For the lat three, there are again 7 possibilities

So, overall there are $7 \times 3 \times 4 = 147$ strings that don't have any of the three criteria which leaves $256 - 147 = 109$ that have at least one.


问:
The "tree property" and dividing types
Recall that a family of formulas $\{\varphi_i(v)\}_{i\in I}$ is said to be "$k$-inconsistent" if every conjunction of $k$ distinct members of the family is inconsistent. We say that an $\mathcal{L}$-formula $\varphi(v,w)$ has the "tree property" for $k$ if there is a collection of parameters $\{b_s:\emptyset\neq s\in\omega^{<\omega}\}$, where $\omega^{<\omega}=\bigcup_{n\in\omega}\omega^n$, such that (a) $\{\varphi(v,b_{si})\}_{i\in\omega}$ is $k$-inconsistent for every $s\in\omega^{<\omega}$, and (b) $\{\varphi(v,b_s):\emptyset\neq s\subseteq\sigma\}$ is consistent for every $\sigma\in\omega^\omega$. In other words, the children of every node of the tree are $k$-inconsistent, and every path through the tree cuts out a consistent family of formulas. A theory $T$ is simple if there is no formula with the tree property, modulo $T$.

I'm having a bit of trouble with proposition 7.2.5. in Tent and Ziegler, which proves that $T$ is simple if and only if there is a cardinal $\kappa$ such that, for all models $M$ and all $p\in S_n(M)$, there is some $X\subseteq M$, of size $\leqslant\kappa$, such that $p$ does not divide over $X$. I'm having trouble in particular with their proof of the backwards direction; I've copied it verbatim below. We work in a monster model $\mathfrak{C}$.

If $\varphi(v,w)$ has the tree property, there are $\varphi$-$k$ dividing sequences $(\varphi(v,b_\alpha))_{\alpha<\kappa^+}$. It is easy to construct an ascending sequence of models $(M_\alpha)_{\alpha<\kappa^+}$ such that each $\varphi(v,b_\alpha)$ divides over $M_\alpha$ and such that $b_\beta\in M_\alpha$ for all $\beta<\alpha$. Extend the collection $\{\varphi(v,b_\alpha)\}_{\alpha<\kappa^+}$ to a type $p(v)\in S(M)$, where $M=\bigcup_{\alpha<\kappa^+}M_\alpha$. Then $p$ divides over each $M_\alpha$.

(A "$\varphi$-$k$" dividing sequence over $X$, of length $\lambda$, is a consistent family $(\varphi(v,b_\alpha))_{\alpha<\lambda}$ such that each $\varphi(v,b_\alpha)$ divides over $X\cup\{b_\beta\}_{\beta<\alpha}$ with respect to $k$. Tent and Ziegler have proved that every formula with the tree property for $k$ has a $\varphi$-$k$ dividing sequence over any parameter set, of any length.)

Unfortunately the construction of the models is unclear to me. We seem to have two choices; the naive approach is to choose a single dividing sequence $(\varphi(v,b_\alpha))_{\alpha<\kappa^+}$ at the onset, say over $\emptyset$, and then try to construct the desired models with this dividing sequence fixed. I run into problems with this approach, however. Constructing $M_0$ is fine; we can use, eg, exercise 7.1.1. of Tent and Ziegler to find a model $M_0$ such that $\varphi(v,b_0)$ divides over $M_0$. (This is just an application of the "standard lemma" on indiscernible sequences.) However, I struggle with the higher stages:

For simplicity, I focus on constructing $M_1$, which is representative of the problem I'm having in the general case too. We know that $\varphi(v,b_1)$ divides over $b_0$, and that $\varphi(v,b_0)$ divides over $M_0$. If we could show that $\varphi(v,b_1)$ divides over $\{b_0\}\cup M_0$, we would then be done; we could find $\mathcal{I}$ an $M_0\cup\{b_0\}$-indiscernible sequence such that $b_1\in\mathcal{I}$ and $(\varphi(v,b))_{b\in\mathcal{I}}$ is $k$-inconsistent, and then use (the argument of) exercise 7.1.1. to find a model $M_1\succeq M_0$ such that $b_1\in M_1$ and such that $\mathcal{I}$ remains indiscernible over $M_1$. Then $\varphi(v,b_1)$ would divide over $M_1$, witnessed by $\mathcal{I}$, as desired. However, I'm not convinced that $\varphi(v,b_1)$ will divide over $\{b_0\}\cup M_0$ in general... to show such a thing, we'd want to use a sort of transitivity property of the following shape:

If $\varphi(v,b)$ divides over $\{a\}$, and $\varphi(v,a)$ divides over $X$, then $\varphi(v,b)$ divides over $X\cup\{a\}$. (Subject to some appropriate conditions on $X$, $a$, and $b$, eg that $\varphi(v,a)\wedge\varphi(v,b)$ is consistent.)

However, this is of different form than all the transitivity properties of dividing that we've proved so far; for instance, in exercise 7.1.7. we showed that

If $\operatorname{tp}(a/X\cup\{b\})$ does not divide over $X$, and $\varphi(v,b)$ divides over $X$, then $\varphi(v,b)$ divides over $X\cup\{a\}$.

Most transitivity properties have been of this form, obtaining a conclusion of dividing with one hypothesis of dividing and one hypothesis of non-dividing. As far as I can tell, this is of fundamentally different shape than obtaining a conclusion of dividing from two hypotheses of dividing, so I'm not even convinced the desired result is true. (My attempts to prove it, even in the specific case above, have not led anywhere, although I haven't been able to come up with a counterexample either.)

So, this approach doesn't appear fruitful to me. The second approach is to construct not only the models but also the parameters $(b_\alpha)_{\alpha<\kappa^+}$ inductively. However, we seem to run into a different problem here; in this approach, for the rest of Tent and Ziegler's argument to work, we need to ensure that $\varphi(v,b_\alpha)$ is consistent with $\{\varphi(v,b_\beta)\}_{\beta<\alpha}$ at each stage, and I don't see how to do this. For simplicity, again suppose that we have constructed $M_0$ and $b_0$ as above and are trying to construct $M_1$ and $b_1$. We can find a $\varphi$-$k$ dividing sequence over $M_0\cup\{b_0\}$; call it $(a_\alpha)_{\alpha<\kappa^+}$. Unfortunately, this alone is of no help, since there is no guarantee that $\{\varphi(v,b_0),\varphi(v,a_\gamma)\}$ is consistent, for any $\gamma$.

So I'm at a bit of a loss. Tent and Ziegler do say this construction should be easy, so I'm wondering if I'm overthinking something or missing something obvious; any insight (or hints!) would be appreciated. Once we've actually constructed the models, then I don't have any trouble with finishing the argument. Indeed, if $X\subseteq M$ is such that $|X|\leqslant \kappa$, then for each $x\in X$ let $\alpha(x)$ be minimal such that $x\in M_{\alpha(x)}$. Then $\gamma:=\bigcup_{x\in X}\alpha(x)$ is a union of $\leqslant\kappa$ sets, each of cardinality $\leqslant\kappa$, so $\gamma<\kappa^+$. Then we have $X\subseteq M_\gamma$, and so, since $p\ni\varphi(v,b_\gamma)$ divides over $M_\gamma$, it divides over $X$ as well.

答:
To be clear, the statement you want to prove is:

If $T$ has the tree property, then for every cardinal $\kappa$, there is model $M$ and a type $p\in S(M)$ such that for all $X\subseteq M$ with $|X|\leq \kappa$, $p$ divides over $X$.

Note that if we drop the condition that $M$ is a model, the proof is easy,  given  that we know that the  tree property gives rise to dividing sequences. If $\varphi(x,y)$ has the tree property, then there is a $\varphi$-$k$-dividing sequence $(\varphi(x,b_\alpha))_{\alpha<\kappa^+}$. Let $B = \{b_\alpha\mid \alpha<\kappa^+\}$. By the definition of dividing sequence, $\{\varphi(x,b_\alpha)\mid \alpha<\kappa^+\}$ is consistent, so we can take $p\in S(B)$ to be any completion of this partial type. Now for any $X\subseteq B$ with $|X|\leq \kappa$, since $\kappa^+$ is regular there is some $\beta<\kappa^+$ such that $X\subseteq B_\beta = \{b_\alpha\mid \alpha<\beta\}$. By the definition of dividing sequence, $\varphi(x,b_\beta)\in p$ divides over $B_\beta$, and hence over $X$, so $p$ divides over $X$.

Now we want to "upgrade" $B$ to a model. There's a standard  trick for turning behavior over arbitrary sets into behavior over models: Skolemization.  It's entirely possible that there's a straightforward way to carry out the proof in T&Z  without Skolemizing, but I agree with you that it's not obvious.

Let $\mathfrak{C}^+$ be an expansion of the monster model $\mathfrak{C}$ by Skolem functions, let $L^+$ be the expanded language, and let $T^+$ be $\mathrm{Th}_{L^+}(\mathfrak{C}^+)$. Then $T^+$ has the tree property, witnessed by the same $L$-formula $\varphi(x;y)$ and the same tree.

[Aside: This is one of the benefits of the tree property characterization of (non)-simplicity - it is "purely local" in the sense that it only depends on the behavior of a single formula $\varphi(x,y)$, so it is preserved under expansions and reducts which contain $\varphi$. Notice that, in contrast, it is not obvious that the existence of a $\varphi$-$k$-dividing chain is preserved under expansion, because an instance $\varphi(x,b)$ which divides over $A$ may no longer divide over $A$ after expanding the language. The same goes for characterizing (non)-stability by the order property, and many other properties in model theory.]

Now we carry out the proof above in $T^+$. We have an $L$-formula $\varphi(x,y)$ with the tree property, so there is a $\varphi$-$k$-dividing sequence $(\varphi(x,b_\alpha))_{\alpha<\kappa^+}$. Let $M = \langle \{b_\alpha\mid \alpha<\kappa^+\}\rangle$, the substructure of the monster generated by all the parameters in the sequence. Since $T^+$ has Skolem functions, $M\models T^+$. Similarly, for all $\beta<\kappa^+$, if we define $M_\beta = \langle \{b_\alpha\mid \alpha<\beta\}\rangle$, then $M_\beta\models T^+$. Note that $M = \bigcup_{\beta<\kappa^+} M_\beta$.

By the definition of dividing sequence, $\{\varphi(x,b_\alpha)\mid \alpha<\kappa^+\}$ is consistent, so we can take $p\in S(M)$ to be any completion of this partial type. Now for any $X\subseteq M$ with $|X|\leq \kappa$, since $\kappa^+$ is regular there is some $\beta<\kappa^+$ such that $X\subseteq M_\beta$. By the definition of dividing sequence, $\varphi(x,b_\beta)\in p$ divides over $\{b_\alpha\mid \alpha<\beta\}$, so it divides over $M_\beta$ (if a formula divides over $A$, then it also divides over $\langle A \rangle$) and hence over $X$, so $p$ divides over $X$.

It remains to show that the conclusion is preserved when we take the reduct back to $L$. The key thing to check is that if an $L$-formula divides over $X$ in $\mathfrak{C}^+$, then it divides over $X$ in $\mathfrak{C}$.

Title: The "tree property" and dividing types


问:
$2019$ Canadian Mathematical Olympiad Problem $3$

You have a $2m$ by $2n$ grid of squares coloured in the same way as a standard checkerboard. Find the total number of ways to place $mn$ counters on white squares so that each square contains at most one counter and no two counters are in diagonally adjacent white squares.

Solution-  Split the entire grid into $2 \times 2$ squares. The first key observation to be made is that none of these squares contain more than $1$ counter. In fact, since there are $mn$ counters, there must be exactly $1$ counter per $2 \times 2$ square. First, observe that if a counter is placed in the bottom right cell of a $2 \times 2$ square, then all counters to the right and under the cell are fixed; they must be in the bottom right position. Otherwise, we are free to decide.

Now, view the entire grid as an $m \times n$ board, where we color a square blue if it's counter is in the bottom right position, and red otherwise. Observe that the amount of ways to color this grid bijects to a path taken from the bottom left corner to the top right corner, so the answer is ${m + n \choose n}$ as requested

I don't see the bijection between the two ,pls someone guide me with this.

thankyou

答:
Here is an example that may make the things clear. (It is not a solution by example, but a way to visualize the solution using a special example.)

Note: In an initial version of the question, the solution was linked to

AOPS Problem + Solutions .

This page may have a shorter or longer life, but next days it will be there, so the reader may want to look at the many solutions having one common idea.

Let us build a configuration, by placing some stars on the white fields of the following empty board, already divided in $2\times2$-squares.

The fill-in-rule wants that no two (at a corner) touching white fields
(from the same and/or touching $2\times2$-squares)
get stars in the same time. And each $2\times2$-square has exactly one star (in a white field).

$$
\begin{array}{|cc|cc|cc|cc|cc|}
\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
\end{array}
$$
In this choice, the left upper corner is white.
If the checker board is "the other one", just use a vertical reflection to reduce to this case (since the right upper corner is than white, so the reflection fits the above pattern).

When we use the left/upper field, than we also use the red color:
$$
\begin{array}{|cc|}
\hline
\color{red}*& \blacksquare \\
\blacksquare &
\\\hline
\end{array}
\text{ and abbreviate this as }
\color{red}{\blacksquare}
$$

When we use the right/lower field, than we also use the blue color:
$$
\begin{array}{|cc|}
\hline
& \blacksquare \\
\blacksquare & \color{blue}*
\\\hline
\end{array}
\text{ and abbreviate this as }
\color{blue}{\blacksquare}
$$

Now we will fill in first the first row only, one $2\times2$-square after the next one, starting from left to right.

A first observation is that if somewhere on this row we use the blue pattern, than all (right) following $2\times2$-squares are blue.

So the first row may be

$$
\begin{array}{|cc|cc|cc|cc|cc|}
\hline
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & \color{blue}* &
\blacksquare & \color{blue}* &
\blacksquare & \color{blue}*
\\\hline
\end{array}
$$
and we encode it for short simply as:
$$
\color{red}{\blacksquare \blacksquare }\color{blue}{\blacksquare \blacksquare \blacksquare }
$$
This "simple scheme" to represent the filling is the key to the solution.

Let us fill now the first two rows, and observe the constraints.
First of all we must have in the simple scheme blue under blue, i.e.
$$
\color{red}{\blacksquare \blacksquare }\color{blue}{\blacksquare \blacksquare \blacksquare }
\\
\color{gray}{\blacksquare \blacksquare }\color{blue}{\blacksquare \blacksquare \blacksquare }
$$
and the gray squares have some choices. But each choice should give connected intervals of $\color{red}{\blacksquare}$ and of $\color{blue}{\blacksquare}$ pieces in the "simple scheme".

With this convention, the possible solution
$$
\begin{array}{|cc|cc|cc|cc|cc|}
\hline
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
\color{red}*& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare &
\\\hline
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*
\\\hline
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & &
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*
\\\hline
\color{red}*& \blacksquare &
\color{red}*& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & &
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*
\\\hline
\color{red}*& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & &
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*
\\\hline
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare &
& \blacksquare
\\
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*&
\blacksquare & \color{blue}*
\\\hline
\end{array}
$$
is translated in simple scheme as
$$
\begin{array}{ccccc}
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{red}\blacksquare
\\
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare
\\
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare
\\
\color{red}\blacksquare &
\color{red}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare
\\
\color{red}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare
\\
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare &
\color{blue}\blacksquare
\end{array}
$$
and the social distancing of red and blue translates further as a record or the right most red square, its column number, considered row after row.

So the above simple scheme translates as the tuple:
$$
(5,3,3,2,1,0)\ .
$$
There are $6$ entries, so many as the rows, and the entries are non-increasing.

Since we hate repetitions in the tuple, we will do the following new translation. Add on its last entry $\color{magenta}1$, on the forelast entry $\color{magenta}2$, and so on, $\color{magenta}3,\color{magenta}4,\color{magenta}5,\color{magenta}6$, so that the new tuple becomes
$$
(5+\color{magenta}6, 3+\color{magenta}5,3+\color{magenta}4,2+\color{magenta}3,1+\color{magenta}2,0+\color{magenta}1)=
(11, 8,7,5,3,1)\ .
$$
This last tuple has six different entries between $1$ and $5+\color{magenta}6$.

By following this encoding process in reverse order, we observe that this makes sense (injectivity of the map), and decoding each tuple obtained from decreasingly rearranging an element of the appropriate $\binom {5+6}6$ combinations also makes sense (surjectivity of the encoding, when the codomain is made out of these combinations).

The general case works along the same lines. Just use $m,n$ instead of $6,5$.


答:
(2) says that $\sum\limits_{i=1}^na_iv_i\neq 0$ except in the case where all the $a_i=0$.

(1) on the other hand only says that $\sum\limits_{i=1}^na_iv_i\neq 0$ when all the $a_i=0$

So (2) always implies (1).

Also (2) is just the definition of the linear independence of the set $\{v_1,\dots, v_n\}$.

If the set $\{v_1,\dots, v_n\}$ is linearly independent, then both (2) and (the "weaker") (1) are both true.

If the set $\{v_1,\dots, v_n\}$ is linearly dependent, then (2) is false. But (1) may be true. For example take the vectors $(1,0), (1,0), (0,1)$ in $\mathbb{R}^2$. Then $a(1,0)+b(1,0)+c(0,1)\not=(0,0)$ for all $a,b,c\in\mathbb{R}\setminus \{0\}$.

Title: Confusion about linearly independent vectors

Body:
Let $K$ be a field an $V$ a vector space. Let $v_1,\ldots,v_n$ be linearly independent vectors in $V$.

Are the following two statements equivalent? If not, why not?

for all $a_1,\ldots,a_n$ in $K\setminus\{0\}$ is $\sum\limits_{i=1}^na_iv_i\neq 0$

for all $(a_1,\ldots,a_n)^t$ in $K^n\setminus\{0\}$ is $\sum\limits_{i=1}^na_iv_i\neq 0$

Does it make a difference whether we talk about the vector $(a_1,\ldots,a_n)^t$ (2.) or if we look at $a_i$ as elements in $K$ only (1.)?

Thanks in advance


问:
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

This is how i started solving this limit:

$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$

$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$

$\left(1- \frac {x+1}{4x+3}\right)^x$

$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$

$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$

$e^{\infty} = \infty$

answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!

答:
We have:

$$\lim_{x\to\infty}\bigg(\frac{3x + 2}{4x + 3}\bigg)^{x} = \lim_{x\to\infty}\bigg(\frac{3 + \frac{2}{x}}{4 + \frac{3}{x}}\bigg)^{x} = \bigg(\frac{3}{4}\bigg)^{\infty} = \boxed{0}$$

Your mistake was in assuming that $\frac{4x+3}{x + 1}\to\infty$ as $x\to\infty$, which meant you could apply the definition of $e$. However, $\frac{4x + 3}{x+1}\to 4$. Remember that a rational function only diverges if the degree of the numerator is greater than the degree of the denominator.


答: Now @Infinity_hunter has explained your error, note that $\frac34-\frac{3x+2}{4x+3}=\frac{1}{4(4x+3)}>0$, so $0<\left(\frac{3x+2}{4x+3}\right)^x<\left(\frac34\right)^x$ proves the limit is $0$ by squeezing.

Title: Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

Body:
Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

This is how i started solving this limit:

$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$

$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$

$\left(1- \frac {x+1}{4x+3}\right)^x$

$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$

$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$

$e^{\infty} = \infty$

answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!


问:
Calculate bounds of the inner rectangle of the polygon based on it's constant bounds along with border
Yesterday I asked this question seems not clear. So, I am writing new question with mathematical terms only.

Here is some points about shapes I used:

Square: Rectangle that all angles are 90° & all sides are equal length. So, consider Square also a Rectangle.

Triangle: I used an equilateral triangle is a triangle in which all three sides have the same length.

Polygon: A finite number of straight line segments connected to form a closed.

So, Here after I use the term Polygon to describe triangle & rectangle shapes when it's required to mention consider this shape as polygone. Becuase, I used triangle (regular polygon) to simplfy the work to find out the formula.

Variables used in this problem:

b = Border width

Bounds of Outer Rectangle (r) that enclosing the polygon(p):

x = Origin x

y = Origin y

w = Width

h = Height

So, If r = (x,y,w,h) then, Bounds of Inner Rectangle that encloses the inner polygon(p1) (r1) = (x1,y1,w1,h1)

Here, inner polygon = Polygon by subtracting the border width.

What is my final goal:

I should able to calculate inner rectangle boundary of any polygon like below image, I have angles, outer rectangle bounds & border width. The problem is to find out the inner rectangle bounds across to the outer rectangle based on the polygon at that enclosed by inner rectangle.

Problem 1:

Consider a Polygon(p) as a shape of rectangle(r) which has width(w) is 100 & height(h) is 100. The border width of the polygon is 10. The what is the inner boundary of the rectangle that encloses the inner polygon?

Note: I solved it by mind calculation(I didn't took angles in the calculation). But we can't apply this same
formula for triangle. So, this is not solution for my problem.

Problem 2:

Consider the polygon (don't consider regulard) with three edges and three vertices & all angles are equal in measure.

If the rectangle(r) boundary is zero & size(w,h) is (100,87) that encloses the polygon(p) along with polygon border width as 10, then

What is the bounds of the inner rectangle(r2) that encloses the inner polygon(p2) ?

Hope I clearly explained the problem.  If anything I am missed then please let me know. Thanks in advance.

答:
Let us denote width and height of the inner rectangle as $w_\mathrm{in}$ and $h_\mathrm{in}$ respectively. Then, we have,
$$w_\mathrm{in} = w-2\sqrt{3}b. \qquad\mathrm{and}\qquad h_\mathrm{in} = h\space–\space3b.$$

Using these results, we can easily derive the coordinates of the four corners of the sought rectangle $P$, $Q$, $R$, and $S$ as given in the diagram.

Frankly, I am not sure whether these are the results you were looking for. If you have doubts you can always comment on this answer no holds barred.


问:
Why are these two double integrals different ? A question on Dirac delta distribution
Consider the following two integrals:

\begin{align}
I_1&:=\iint_{0\leq x,\, y\leq 1,\,x=y}\,dx\, dy,\\
I_2&:=\int_0^1\int_0^1 \delta(x-y)\,dx \,dy.
\end{align}

I believe $I_1=0$ because it is the measure "surface area" of the line segment $y=x$ inside a two dimensional rectangle.

From the properties of the delta function, we have

$$I_2=\int_0^1\int_0^1 \delta(x-y)\, dx\, dy=\int_0^1 1 \,dy=1$$

as
$$\int_0^1 \delta(x-y)\, dx =1$$ for any $y\in ]0,1[.$

Question: I understand very well that the integrands are different in $I_1$ and $I_2$. In $I_1$, the integrand is
the characteristic function
$\chi_{0\leq x,y\leq 1,x=y}(x,y)$. But these integrands are different on a negligible set... I am confused!

答: What happens is that the delta "function" is not a function. Integration against $\delta\, dx$ simply means integration against the delta measure $$\delta(E)=\begin{cases}1,&\ 0\in E\\ 0,&\ 0\not\in E\end{cases} $$


答: Note that $$\lim_{u\to \infty}\left( 1-\frac{1}{u}\right)^{u}=\frac{1}{e}$$


问:
Lipschitz continuous
Let $\delta$ be an interval in $\mathbb{R}$. Recall that a function $f$ is called Lipschitz continuous on $\delta$ with Lipschitz constant $L$ if there holds
 $|f(x) - f(y)| \leq L|x-y|$ for all $x,y$ in $\delta$.

(a) Show that the composition of Lipschitz continuous functions is again Lipschitz continuous.

(b) Is the pointwise maximum of two Lipschitz continuous functions necessarily Lipschitz
continuous?

答:
Part (a) is obvious. For (b) let $\max\{f,g\}=:h$. Then
$$f(y)\leq f(x)+|f(y)-f(x)|\leq h(x)+L_f|y-x|\ ,$$
and similarly  $g(y)\leq h(x)+L_g|y-x|$. It follows that
$$h(y)\leq h(x)+\max\{L_f,L_g\}|y-x|\ ,$$ and by symmetry we conclude that
$h$ is Lipschitz continuous with Lipschitz constant $L_h\leq \max\{L_f,L_g\}$.</p>

Title: Lipschitz continuous


答: First of all let investigate terms of inside the parentheses. When $x$ approaches to infinity $\displaystyle \frac{3x+2}{4x+3}=\frac{3}{4}$ then the limit transformed into $\displaystyle \lim_{x \to \infty} \left(\frac{3}{4}\right)^{x}$. Rewritre $\frac{3^x}{4^x}$. Obviously denomiter is greater than numerator. So $\lim$ approaches to $0$.


问:
Predicate logic difficulty
If $∀x∃y P(x, y)$ is true, can the following be true?

$∀x∀yP(x, y)$
or

$∃x∀yP(x, y)$
or

$∃x∃yP(x, y) $

I understand the order of $X$ and $Y$ matters when the quantifiers are different but can the quantifier rather than the variable change, and can a 'there exists' be upgraded to a 'for all'  or vice versa as in the first and third question?

答:
If $∀x∃y P(x, y)$ is true, can the following be true?

Of course, all three statements can be true given the truth of the given statement. If we consider a domain with a single object that stands in relation $P$ to itself, then all four statement are true.

... this seems too obvious ... are you sure you phrased that question correctly? Shouldn't you be asking which other statements must be true given the truth of the given statement?  Indeed, that is a more typical question for logic ...

So, let's change the can into must ... or, what is the same thing,let's ask which, if any, of the three statements logically follows from the given statement.

To that, the answer is only statement 3 ... and that is assuming (as most standard logics do), that the domain cannot be empty. Indeed, if the domain is not empty, then any universal statement implies its existential counterpart, i.e. $\forall x \ \varphi(x)$ implies $\exists x \ \varphi(x)$, and therefore  $\forall x \ \exists y \ P(x,y)$ implies $\exists x \ \exists y \ P(x,y)$ as a specific instance of that.  However, an existential statement does not imply its universal counterpart (i.e. an existential cannot be 'upgraded' to a universal)

Title: Predicate logic difficulty


问:
Hartshorne Chapter V Proposition 1.5
This proposition states as follows,

What confuses me is the last sentence. Notice that the Lemma 1.3 states as follows.

We also have the fact that,

The pairing  Div $X \times \operatorname{Div} X \rightarrow \mathbf{Z},$ only depends on the linearly equivalence class.

If $C$ and $D$ are nonsingular curves meeting transversally, then $C . D=$ $\#(C \cap D),$ the number of points of $C \cap D.$

So my qusetion is

(1) In the case of Adjunction formula, how to guarantee  there exists   $H\in|C+K|$ such that $H$ meets $C$ transversally?

(2) As mentioned in the statement of self-intersection, for a nonsingular curve $C$ on the smooth projective variety, we also have $C^2=\text{deg}\mathcal O_C(C)$.

So, it seems that we always have $C . D=\text{deg}\mathcal O_C(D)$ for any nonsingular irreducible curve $C$? Is it right? Why?

Any help would be appreciated. Thanks a lot!

答:
Let me answer (2) in my way (and afterwards how Hartshorne possibly wanted to do it):

My way:

In Proposition 1.4 Harthorne shows that we have $C \cdot D = h^0(\mathscr{O}_{C \cap D})$ for all curves $C$ and $D$ not having common irreducible components. Therefore it suffices to show $$\text{deg}(\mathscr{O}_C(D)) = h^0(\mathscr{O}_{C \cap D}).$$ For this we tensor the short exact sequence $$0 \rightarrow \mathscr{O}_C(-D) \rightarrow \mathscr{O}_C \rightarrow \mathscr{O}_{C \cap D} \rightarrow 0$$
(here we assume that no irreducible component of $C$ is contained in $D$ to get exactness on the left side) with the line bundle $\mathscr{O}_X(D)$ (which is an exact functor) and get the short exact sequence $$0 \rightarrow \mathscr{O}_C \rightarrow \mathscr{O}_C(D) \rightarrow \mathscr{O}_{C \cap D} \rightarrow 0.$$ Now we have $h^0(\mathscr{O}_{C \cap D}) = \chi(\mathscr{O}_{C \cap D}) = \chi(\mathscr{O}_C(D)) - \chi(\mathscr{O}_C) = \text{deg}(\mathscr{O}_C(D)),$ where the second equality comes from the additivity of the Euler characterstic and the third equality comes from Riemann-Roch for curves.

Harthorne's way (I guess):

I guess Hartshorne wants to combine 1.3 and 1.4 in the following way. Due to 1.3 we have that the degree $\text{deg}(\mathscr{O}_C(D))$ counts the intersection points of $C$ and $D$ and by 1.4 we can compute $C \cdot D$ in terms of the local intersection multiplicities. These local intersection multiplicities at some point $p$ are $\geq 1$ if and only if the point $p$ lies in the intersection. Therefore one has the inequality $$\#(C \cap D) \leq \sum_p (C \cdot D)_p,$$ where $p$ runs over all points in the intersection $C \cap D$. If we now can assure that we never have local intersection multiplicity $\geq 2$ by some assumptions on $C$ and $D$, we have the wanted equality. This is the case if we have transversal intersections, which is the assumption Hartshorne states.

Title: Hartshorne Chapter V Proposition 1.5


答:
There's no need to make this about double integrals.  Your question would apply just as well to the two integrals  $$\int_{[0,1]}\chi_{\{0\}}(x)\,dx=0$$ and $$\int_{[0,1]}\delta(x)\,dx=1,$$ which are different even though $\delta$ and $\chi_{\{0\}}$ are the same everywhere except at $0$.  The explanation for this is simply that $\delta$ is not actually a function and its "integral" is not actually an integral in the usual sense.  It's a theorem that if two integrable functions agree except on a set of measure zero, then their integrals are the same.  But $\delta$ is not a function at all, so that theorem doesn't apply.  Instead, $\int_{[0,1]}\delta(x)\,dx$, though written as an "integral", is actually simply defined as the evaluation of the distribution $\delta$ on the constant function $1$.  Since $\delta$ is by definition the distribution that takes a function and outputs that function evaluated at $0$, this gives the answer of $1$.

Intuitively, $\delta$ is a "function" that gets infinitely large at the point $0$, such that even though it's just one single point, the "area" under it somehow manages to be positive.  But it's not actually literally the function $\mathbb{R}\to[-\infty,\infty]$ that is $\infty$ at $0$ and $0$ elsewhere; that is an actual genuine integrable function whose integral is $0$.

Title: Why are these two double integrals different ? A question on Dirac delta distribution

Body:
Consider the following two integrals:

\begin{align}
I_1&:=\iint_{0\leq x,\, y\leq 1,\,x=y}\,dx\, dy,\\
I_2&:=\int_0^1\int_0^1 \delta(x-y)\,dx \,dy.
\end{align}

I believe $I_1=0$ because it is the measure "surface area" of the line segment $y=x$ inside a two dimensional rectangle.

From the properties of the delta function, we have

$$I_2=\int_0^1\int_0^1 \delta(x-y)\, dx\, dy=\int_0^1 1 \,dy=1$$

as
$$\int_0^1 \delta(x-y)\, dx =1$$ for any $y\in ]0,1[.$

Question: I understand very well that the integrands are different in $I_1$ and $I_2$. In $I_1$, the integrand is
the characteristic function
$\chi_{0\leq x,y\leq 1,x=y}(x,y)$. But these integrands are different on a negligible set... I am confused!


问:
The weights of pears are normally distributed with mean 118g and standard deviation 4.6g. 5 pears are chosen at random.

Question:

The weights of pears are normally distributed with mean 118g and standard deviation 4.6g. 5 pears are chosen at random.
Find the probability that their combined weight is less than 570 grams.

I have solved these questions before, but I'm doing them after 1 year hence not being able to create a certain link. [Have Covid, suffering from apparent brain fog]

What I'm doing:

Consider one pear to be denoted by P, which is distributed as: P~N(118,4.6^2)

Hence 5P will have:

mean:-

118*5 = 590

variance:
5^2 * 4.6^2 = 529

Hence 5P distributed as: 5P~N(590, 529)

Finding probability using calculator function:

p = 0.192

Answer in the book:

0.0259

Additional information:

I'm pretty sure that my variance calculation is wrong. I am confused as to when is the coefficient of a combination of random variables supposed to be squared as well. In the book itself it has been explained that variance is a^2*Var(x), hence I have squared a (5) and squared the standard deviation to give the variance.

答:
A simple way to illustrate the issue is to consider the following situation.  Suppose $X$ is a Bernoulli random variable with probability mass function $$\Pr[X = 0] = \Pr[X = 1] = 1/2.$$  Then the expectation is $\operatorname{E}[X] = 0 \cdot \Pr[X = 0] + 1 \cdot \Pr[X = 1] = 0(1/2) + 1(1/2) = 1/2$, and $$\operatorname{Var}[X] = (0 - 1/2)^2 \Pr[X = 0] + (1 - 1/2)^2 \Pr[X = 1] = (1/4)(1/2) + (1/4)(1/2) = 1/4.$$  What is $\operatorname{Var}[5X]$?  The random variable $Y = 5X$ has probability mass function $$\Pr[Y = 0] = \Pr[Y = 5] = 1/2$$ and its support is $Y \in \{0, 5\}$.  So $$\operatorname{Var}[Y] = (0 - (5/2))^2 \Pr[Y = 0] + (5 - (5/2))^2 \Pr[Y = 5] = (5/2)^2 (1/2) + (5/2)^2 (1/2) = 25/4.$$  Therefore $$\operatorname{Var}[Y] = 5^2 \operatorname{Var}[X].$$

Now suppose we have $W = X_1 + X_2 + X_3 + X_4 + X_5$, where for each $i \in \{1, 2, \ldots, 5\}$, $X_i$ is independent and identically Bernoulli distributed as $X$; i.e., $\Pr[X_i = 0] = \Pr[X_i = 1] = 1/2$.  Then $$W \sim \operatorname{Binomial}(n = 5, p = 1/2)$$ with $$\Pr[W = w] = \binom{5}{w} \frac{1}{2^5}, \quad w \in \{0, 1, \ldots, 5\}$$ and $$\operatorname{Var}[W] = np(1-p) = 5(1/2)(1/2) = \frac{5}{4} \ne \operatorname{Var}[Y].$$

Why are these two variables different?  Well, you can immediately see they are different because they do not have the same support:  $Y \in \{0, 5\}$, whereas $W \in \{0, 1, 2, 3, 4, 5\}$.

Suppose you flip a fair coin.  If the coin lands heads, I pay you a dollar.  If it lands tails, you win nothing.  Then $X$ is a random variable that describes your winnings after a single coin toss.  $W$ would describe your winnings after five coin tosses.  $Y$ would describe your winnings if you only tossed the coin once, but instead of the payout for heads being one dollar, it is five dollars.

Since in this last scenario (with random variable $Y$) does not admit a possibility of winning $1, 2, 3$, or $4$ dollars, the variance of the winnings is larger than the case where you get to flip the coin five times and allowing outcomes in which there are some wins and some losses.

Now to translate this situation to your original question, we can immediately see that when we take a random sample of $5$ pears and weigh them, their total weight is the sum of five independent normally distributed random variables.  Some of the pears might be lighter than the others, and some may be heavier.  This is in contrast to a situation in which you might pick a single pear and multiply its weight by $5$.  Consequently, the variance of the total weight of five randomly selected pears is not the same as the variance of five times the weight of a single pear.


答:
It all depends on what $P(x,y)$ is. I think that perhaps you may be misunderstanding the point of the question, it seems like you are thinking about this too syntactically (i.e. you are focusing on whether 1. 2. and 3. are allowed under the language of predicate logic, which they all are) rather than semantically (i.e. thinking about what each sentence means and deducing whether they can follow from the premise of $\forall x\exists y P(x, y)$).

So to answer your question, what really needs to be done is to analyze the meaning of each of the sentences. Looking at the sentence $\forall x\exists y P(x, y)$, all it is saying is that for every $x$ you pick there will be some $y$ such that $P(x,y)$ is true. I assume that that's all the information you have about $P(x,y)$, and it's the only information you have to deduce whether the other ones can be true. I emphasize can because it is not about determining whether each sentence is actually true, just whether they're consistent with the premise. Make sure you think about $x,y$ and $P(x,y)$ abstractly, we don't actually care what each of them are, only what we can deduce based on the above sentence. Now let's analyze each of the options:

$\forall x\forall yP(x, y)$ This just says every $x$ can be combined with any $y$ to make $P(x,y)$ true. This is different from the original statement, which doesn't assert something as strong, but they are still compatible with each other. Although $\forall x\exists y P(x, y)$ only says that every $x$ has some $y$ that makes $P(x,y)$ true, it could moreover be true that any choice of $y$ would make $P(x,y)$ true. So $\forall x\forall yP(x, y)$ can be true.

This is just a weaker statement than 1., so if 1. can be true so can 2.!

This can be true, but as was pointed out in the comments below is not always true. If the universe from which these variables are quantified over is empty, then while $\forall x\exists y P(x, y)$ is vacuously true, $\exists x\exists y P(x, y)$ must be false. If on the other hand the universe is non-trivial then this must be true. There does exist an $x$ for which there exists a $y$ that makes $P(x,y)$ true. In fact, by our assumption, any choice of $x$ satisfies this. In general you can look at $\exists$ as a downgrade of $\forall$. If something is true for all choices of $x$ then it must be true for some single choice of $x$, as long as choices for $x$ actually exist (i.e. we're not talking about the empty set). The converse is not always true on the other hand, if something is true for some single choice of $x$, you can't necessarily conclude it's true for all $x$.

Title: Predicate logic difficulty

Body:
If $∀x∃y P(x, y)$ is true, can the following be true?

$∀x∀yP(x, y)$
or

$∃x∀yP(x, y)$
or

$∃x∃yP(x, y) $

I understand the order of $X$ and $Y$ matters when the quantifiers are different but can the quantifier rather than the variable change, and can a 'there exists' be upgraded to a 'for all'  or vice versa as in the first and third question?


问:
Finding the index number of normalizer of a non-normal subgroup when the index of subgroup is prime
Suppose $H$ is a non-normal subgroup of a group like $G$, where $[G:H]=p$ and $p$ is a prime(in other words $o(G)/o(H)=p$).
prove that $N(H)=H$. in other words $[G:N(H)]=p$.

appreciate it if you help me with this one.

答: Hint: $H \subseteq N_G(H) \subseteq G$ and $|G:H|=|G:N_G(H)| \cdot|N_G(H):H|$. $H$ is normal iff $G=N_G(H)$.


问:
A question about convergence in measure
I have been working on this question for a while now, and I would really like to get some help.

Let $(X,\mathscr{M}, \mu)$ be a measure space and let $A_1 \subset A_2 \subset A_3 \cdots$ be an increasing sequence of measurable sets such that $\mu(A_i)< \infty$ for every $i = 1,2,\cdots $ Denote $A=\cup_{i=1}^{\infty}A_i$. Prove that $\mathbb{1}_{A_i} \xrightarrow{measure} \mathbb{1}_{A}$ if and only if $\mu(A) < \infty$.

My current approach is using the fact that $\mu(A) \leq \lim_{t\xrightarrow{}\infty}\sum_{i=1}^{T}\mu(A_i)< T\mu(A_T) <\infty$ for the first direction. But it is not so clear how should we approach the reverse direction.

Thanks!

答:
I'm not sure I understand your argument, since it looks like your captial "$T$" should be tending to $\infty$.

Suppose $\mu(A)<\infty$.  Then for any $\varepsilon>0$, by Markov's inequality, $$\mu(|\mathbb 1_{A}-\mathbb 1_{A_n}|>\varepsilon) \leq \frac{1}{\varepsilon} \int |\mathbb 1_{A}-\mathbb 1_{A_n}|d\mu $$$$=\frac{1}{\varepsilon} \int \mathbb 1_{A}-\mathbb 1_{A_n}d\mu = \frac{1}{\varepsilon}\mu(A\setminus A_n)\to 0,$$ as $n\to\infty$.

Now, suppose $\mathbb 1_{A_n}\stackrel{measure}\to \mathbb 1_A$.  Since
$$\mathbb 1_{A}(x)-\mathbb 1_{A_n}(x)
=\begin{cases}
1& x\in A\setminus A_n\\
0& x\in A_n \cup A^c
\end{cases}
$$

Then for any $\varepsilon \in(0,1)$, $x\in\{|\mathbb 1_{A}-\mathbb 1_{A_n}|>\varepsilon\}$ if and only if $x\in A\setminus A_n$, i.e. $\{|\mathbb 1_{A}-\mathbb 1_{A_n}|>\varepsilon\}=A\setminus A_n$.  So,

$$
\mu(A) = \mu(A_n) + \mu(A\setminus A_n) = \mu(A_n) + \mu(|\mathbb 1_{A}-\mathbb 1_{A_n}|>\varepsilon)<\infty,
$$
since $\mu(A_n)<\infty$ and $\mu(|\mathbb 1_{A}-\mathbb 1_{A_n}|>\varepsilon)\to 0$.

Title: A question about convergence in measure