问:
Linearization of $\sum_k (y_{k}^2+x_k\,y_k)$
How can I linearize $\sum_{k} (y_{k}^2+x_k\,y_k)$? when $y$ is a variable and $y\ge0$, and $x$ is a binary variable.

答:
You can linearize the $x_k y_k$ exactly:
https://or.stackexchange.com/questions/39/how-to-linearize-the-product-of-a-binary-and-a-non-negative-continuous-variable?r=SearchResults

But for the linearization of $y_k^2$ you’ll have to settle for an approximation:
https://or.stackexchange.com/questions/180/how-to-linearize-the-product-of-two-continuous-variables?r=SearchResults


问:
The sum of $2$ consecutive primes, Are there more $\mod 6=0$?
All prime numbers greater than $3$ take the form of either $6x+1$ or $6y-1$.

So the only options for the sums of $2$ consecutive primes are:

$(6x+1 + 6y-1) \mod 6=0$

$(6x-1 + 6y+1) \mod 6=0$

$(6x+1 + 6y+1) \mod 6≠0$

$(6x-1 + 6y-1) \mod 6≠0$

According to the limited available options above, the chances for $(6x+1 + 6y-1) \mod 6=0$ or $(6x-1 + 6y+1) \mod 6=0$ to occur are only $50\%$,

However:

Out of the first $200000$ prime numbers, when adding a prime number with the following prime number, $115141$ resulted in $\mod 6=0$ which is about $57.5705...\%$

When I mention "following prime number", I mean it as:  $5+7$  ($7$ is following $5$). another example: $31+37$ ($37$ is following $31$),  another example: $43+47$ ($47$ is following $43$).

Why is it that there are more $\mod 6=0$ occurrences then $\mod 6≠0$ occurrences, when the probability shows that the chances for $\mod 6=0$ is just $50\%$

Am I expecting to see these kind of statistics (margins) continue infinitely? Can it be proven?

Or maybe it is just that my sample test of 200000 is too small, and I should expect a correction in larger samples?

Update: Per @Peter 's comment:

A slight decrease if we continue until [$3⋅10^9+19,3⋅10^9+37$] :

[$79805741,64643795,0.55248180928736247377077071400215505019$]

答:
This is not a proof, but reasoning to explain why the excess above $50\%$ is around $1/\ln N$, and shrinks very slowly.

Consider numbers near $N$ but only the ones of form $6n\pm1$.  Any random number has a chance of $1/\ln N$ to be prime, but since numbers of the other four remainders are never prime, the chance any one of ours is prime is $$p=\frac3{\ln N}$$

For any prime, assuming a random model for primes, the chance the next prime is opposite remainder, and none of the ones in-between is prime, is
$$p+(1-p)^2p+(1-p)^4p+\ldots\\=\frac p{1-(1-p)^2}\\=\frac1{2-p}$$
When $N=10^6$, this estimate is $56.09\%$.

ContentLicense: CC BY-SA 4.0


问:
In △ABC, BD : DC = CE : EA = AF : FB = 1:4. Let K, L and M are points of intersection of AD, BE and CF, as shown on the picture...
In $\triangle ABC$, $BD:DC = CE : EA = AF : FB = 1:4$. Let $K$, $L$ and $M$ are points of intersection of $AD$, $BE$ and $CF$, as shown on the picture. Suppose the area of $\triangle ABC$ is 1, find the area of $\triangle KLM$.

答:
I found it. It is Routh's theorem.

https://en.wikipedia.org/wiki/Routh%27s_theorem


问:
Let $R$ be a finite ring. Show that $R$ has unity
Let $R$ be a finite ring that contains an element $a$, which is not a left divisor of zero, and an element $b$, which is not a right divisor of zero. Show that $R$ has unity.

Hint is that I should define two bijective mappings from $R$ to $R$.

So I defined $f:R \rightarrow R$ as $f(r)=ar$ and $g:R \rightarrow R$ as $g(r)=rb$.

Check if $f$ is injection: Suppose $f(r)=f(k)$. Then $ar=kr \Rightarrow ar-ak=0 \Rightarrow a(r-k)=0 \Rightarrow r-k=0$ (since $a$ is not a left divisor of zero) $\Rightarrow r=k$. So, $f$ is injective. Also, since $R$ is finite and $f$ is injective $\Rightarrow$ $f$ is surjective, hence bijection.

Similarly we show that $g$ is bijection.

But how to continue? Did I so far OK?

答:
Since $f$ is surjective, there is some $e \in R$ such that $ae = a$. We claim that $e$ is the desired unity.

For any $x \in R$ we have
$$f(ex) = aex = ax = f(x)$$
so $ex = x$.

In particular $eb = b$ so for any $x \in R$ we have
$$g(xe) = xeb = xb = g(x)$$
so $xe = x$.


问:
Hatcher exercise 1.1.18

Using Lemma 1.15, show that if a space $X$ is obtained from a path-connected subspace $A$ by attaching a cell $e^n$ with $n\geq 2$, then the inclusion $A\hookrightarrow X$ induces a surjection on $\pi_1$.

Proof. Note that $X = A\sqcup_{x\sim \phi(x)} e^n$ where $x\in S^{n-1}$ and $\phi:S^{n-1}\to A$ is an attaching map. My plan to use Lemma 1.15 is to using the fact that $X = A\cup e^n$ and $A,e^n$ are both open path connected subspace of $X$ and $\pi_1(e^n)=0$ for $n\geq 2$. But the problem is choosing a base point. $A\cap e^n$ is empty so there we can't choose base point contained both in $A$ and $e^n$. I need some manipulation to have two path connected open cover of $X$ whose intersection is also path connected. Following my plan, how can I choose such spaces?

答:
It seems that you use Hatcher's edition from 2001. Other editions are not compatible with that.

Let $\xi \in e^n$, $A_1 = X \setminus \{\xi\}$, $A_2 = e^n$ and $x_0 \in e^n \setminus \{\xi\}$. Then $A$ is a strong deformation retract of $A_1$. I think you can complete the proof now.

Title: Hatcher exercise 1.1.18


问:
How to calculate the maximum volume of a conical frustum?
This is just a question for my research on the ideal cup size to minimize the amount of plastic used.

Assuming that the conical frustum is in a shape of a cup. How do I go about in solving the maximum volume of this cup using calculus to minimize the amount of plastic? I stumbled upon some sort of confusion between integration and optimization, can someone explain to me what would be the best method in approaching this? (And please involve calculus, thanks!)

答:
Let the upside down cup be a frustum formed from a cone with base radius $b$ and height $h=cb$, where $c$ is a dimensionless constant (the cone's slope). The upper radius is $a<b$.

Then the volume is given by subtracting the volume of two cones, each a third of base area times height, $$V=\frac{\pi}{3}c(b^3-a^3).$$
The surface area is also obtained by subtracting the area of two cones, each equal to $\pi rl$, and adding the top disc (assuming the cup has a wider mouth than base): $$A=\pi (b^2-a^2)\sqrt{1+c^2}+\pi a^2$$

The question asks to minimize $A$ given $V$. Without loss of generality, we can ignore the factor of $\pi$ in the formula for $A$, and take $c(b^3-a^3)=v=3V/\pi$ as the constraint.

Using the method of Lagrange multipliers, let $$f(a,b,c):=\sqrt{1+c^2}(b^2-a^2)+a^2-\lambda c(b^3-a^3)$$ Then differentiating with respect to $a,b,c$, we get $\nabla f=0$, \begin{align}
2\sqrt{1+c^2}a-2a&=3\lambda ca^2\\
2\sqrt{1+c^2}b&=3\lambda cb^2\\
c(b^2-a^2)&=\lambda\sqrt{1+c^2}(b^3-a^3)\\
c(b^3-a^3)&=v
\end{align}

(i) Solving these (with computing help), gives, assuming $a\ne0$, $$a=\sqrt[6]{\frac{44}{63 \sqrt{7}}-\frac{16}{63}} x\approx0.464x,\quad b=\sqrt[6]{\frac{13}{126}+\frac{44}{63 \sqrt{7}}} x\approx0.846x,\quad c=\sqrt{\frac{14+8 \sqrt{7}}{9}}\approx1.977,\quad \lambda=
   \frac{2^{2/3} \sqrt[3]{2+\sqrt{7}}}{3 x}$$ where $x=v^{1/3}$.
The height of the cup is $c(b-a)\approx0.755x$.

The volume is then $\frac{\pi v}{3}=V$ while the surface area is $$A=\sqrt[3]{\frac{2+\sqrt{7}}{2}} \pi  x^2\approx4.16x^2\approx4.03V^{2/3}$$

(ii) The alternative is $a=0$, for which the cup becomes an open cone. Then $$V=\frac{\pi}{3}cb^3,\qquad A=\pi b^2\sqrt{1+c^2}$$ A straightforward minimization of $A$ keeping $V$ fixed, gives $$0=\frac{dA}{db}=\frac{d}{db}\pi b^2\sqrt{1+v^2/b^6}\iff b=\left(\frac{3}{\pi\sqrt2}\right)^{1/3}V^{1/3}$$ $$A=\pi b^2\sqrt{1+c^2}=3^{7/6}\left(\frac{\pi}{2}\right)^{1/3}V^{2/3}\approx4.188V^{2/3}$$

So it is option (i) that minimizes the area.

The shape looks like this:


问:
Solving this PDE?
My goal is to solve this partial differential equation in $u-v$ coordinates:

$$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$

Some answers below (very helpful) change the coordinates, but I'd like to solve it without changing coordinates.

How do you solve it?

I would show a better attempt but I am really stuck and don't know how to proceed. If someone can give me a reference where this PDE has been solved I could understand the solution.

I tried looking online for the solution but haven't found anything yet.

答:
For a problem like this I would almost automatically go to "separation of variables".
Let $\phi(u,v)= F(u)G(v)$.  Then $u\frac{\partial \phi}{\partial u}= uG(v)\frac{dF}{du}$ so $u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right)= u^2G(v)\frac{d^2F}{du^2}+ uG(v)\frac{dF}{du}$.

$v\frac{\partial \phi}{\partial v}= vF(u)\frac{dG}{dv}$ so $v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)= v^2F(u)\frac{d^2G}{dv^2}+ vF(u)\frac{dG}{dv}$

The equation becomes $u^2G(v)\frac{d^2F}{du^2}+ uG(v)\frac{dF}{du}+ v^2F(u)\frac{d^2G}{dv^2}+ vF(u)\frac{dG}{dv}= 0$.

Divide byFG: $\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}+ \frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}= 0$ or
$\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}= -\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}$.

Since the left side depends on u only and the right side depends on v only, in order to be equal for all u and v they must be equal to the same constant.  Calling that constant "$\alpha$",
$\frac{u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}}{F(u)}= \alpha$ and
$\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}= -\alpha$.

Then $u^2\frac{d^2F}{du^2}+ u\frac{dF}{du}- \alpha F= 0$ and
$\frac{v^2\frac{d^2G}{dv^2}+ v\frac{dG}{dv}}{G(v)}+ \alpha G= 0$

Those are ordinary, linear differential equations.  They are specifically "Euler type" equations which can be converted to "constant coefficients" by change of variables.  Or we can look for solutions of the form $F= u^n$ and $G= v^m$ for some n and m.

The first equation becomes $u^2(n(n-2)u^{n-2})+ u(nu^{n-1})-\alpha u^n= u^n(n^2+ n- \alpha)= 0$.  If $u^n$ is not identically 0 (the "trivial solution") we must have $n^2+ n- \alpha= 0$.

The second equation becomes $v^2(m(m-1)v^{m-2}+ u(mu^{m-1})+ \alpha v^m= v^m(m^2+ m+ \alpha)= 0$ so we must have $m^2+ m+ \alpha= 0$.

The solutions to those, of course, depend on $\alpha$ and we would get possible values for  $\alpha$ from the boundary conditions.

Title: Solving this PDE?


问:
For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$

Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.

Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.

I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense
and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.

This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.

Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.

I think this problem has not been solved.

Thanks!

答:
A pictoral representation of mm-aops's proof. I also avoid use of: "the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$."

Let $\ \varepsilon = \frac{1}{200}.\ $ Then, we split the circle up into $201$ arcs of equal length. Think of these arcs as "boxes" in the sense of the pigeonhole principle: then there must be $\ n,m \leq 201\ $ such that $\ \mid a_{n} - a_m\mid \leq \frac{1}{201} <\varepsilon.$

Suppose the first two such $\ n,m\ $ are $\ a_{45}\ $ and $\ a_{162}\ .$

Any point on the circle must lie within $\ \varepsilon=\frac{1}{200}\ $ distance from $\ a_{45}\ $ or $\ a_{162}\ $ or $\ a_{279}\ $ or $\ a_{396}\ $ or ...

ContentLicense: CC BY-SA 3.0


问:
If $\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$, then the number of values of $x$ are:

If $\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$, then the number of values of $x$ are:

I have solved the above like this:

$\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$

Squaring:

$9^x - 6^x = 6^x - 4^x$

$9^x + 4^x = 2\times(6^x)$

$3^{2x} + 2^{2x} = 2\times (3\times2)^{x}$

$(3^{x} - 2^{x})^{2} = 0$

Giving us, $3^x = 2^x$, so $x=0$ is the only value which satisfies.

But in the answers, they have stated that total solutions are $3$. Where have I made a mistake? And at what step did a specific action lead to loss of roots?

Please help me to go about the same. Thanks!

答: Your answer is correct. Here is a slightly different way of getting the same answer:\begin{align}9^x-6^x=6^x-4^x&\iff\left(\frac32\right)^{2x}-2\left(\frac32\right)^x+1=0\\&\iff\left(\left(\frac32\right)^x-1\right)^2=0\\&\iff\left(\frac32\right)^x=1\\&\iff x=0.\end{align}


问:
points matching minimizing sum of square distances
I have $2$ sets of $n$ points in $\mathbb{R}^d$ $A$ and $B$.

I am trying to find a bijective function $f:[n]\rightarrow[n]$ such that
$$\sum_{i=1}^n{||A_i-B_{f(i)}}||_2^2$$

is minimized, i.e., to find a pairing between the sets that minimize the sum of square distances.

I thought of using the Hungarian Method which takes $O\left(n^3\right)$ time but is there any way to do it faster? I was looking for papers that try to do it faster but all I could find is papers that talks about minimizing sum of distances instead of square distances.

Is there any way to do it faster than $O\left(n^3\right)$ and either get an optimal solution or a solution with a provable cost of at most $(1+\epsilon)$OPT, where OPT is the optimal cost?

答:
For the linear assignment problem, the cost function is arbitrary and specified by an $n \times n$ matrix with entries $c_{i,j}$ that can represent distances or squared distances or anything else. The binary decision variable $x_{i,j}$ indicates whether point $i$ is matched to point $j$, and the problem is to minimize $\sum_{i,j} c_{i,j} x_{i,j}$ subject to
\begin{align}
\sum_j x_{i,j} &= 1 &&\text{for all $i$}\\
\sum_i x_{i,j} &= 1 &&\text{for all $j$}\\
\end{align}
This problem is linear even if the input costs arise from highly nonlinear functions.


问:
Proving number of matches played in a knock-out tournament with $n$ players is $n-1$ using induction
I am supposed to prove that if $n$ players play in $n-1$ ways. Clearly the base case is trivially true, whence only $1$ match will be played. Also the case when the number of players is of the form $2^k$ where $k\in \mathbb{Z}^{+}$, can be easily formulated as follows. $$2^{k-1}+2^{k-2}+2^{k-3}+\ldots+2+1=2^{k}-1$$

I am not sure how to use induction on it in the general case, including the case when $n$ is odd. Any hints are appreciated. Thanks.

答:
It's clearly true, because everyone but the winner loses one game, and each game has one loser.

For an inductive proof, say there are $n>1$ players and the theorem is true for a knockout tournament with $n-1$ players.  Consider the very first game played.  The winner is then is a knockout tournament with the remaining $n-2$ players.


问:
Does this integral converge when $\frac{1}{p}+\frac{1}{q}\geq1$
Fix a small $\delta>0$ and let $p,q>1$. Consider the integral

$$I(p,q):=\int\limits_{1-\delta}^{1+\delta}
\int\limits_{y/2}^{2y}\frac{1}{|y-x|^{\frac{1}{p}}|1-x|^{\frac{1}{q}}}
\,\mathrm{d}x\,\mathrm{d}y.
$$

I am trying to show that $I(p,q)$ diverges if $\frac{1}{p}+\frac{1}{q}\geq 1$. I am not sure this is even the case ? Any hints on how to handle this?

Remark: This seems to be related to the failure of the Hardy-Littlewood-Sobolev  inequality (HLS) at the endpoint $p=1$. HLS reads:

If $1<p,q<\infty$, $f\in L^p$ and

$$Tf(x):=\int_{\mathbb{R}^n} \frac{f(y)}{|x-y|^{\gamma}}dy$$

Then   $$\|Tf\|_q\leq \|f\|_p$$
if and only if
$$\frac{1}{p}-\frac{1}{q}=1-\frac{\gamma}{n}.$$

Many thanks.

答:
Split the $x$ integral into two

$$ \int_{y/2}^{2y} = \int_{y/2}^{(1+y)/2} + \int_{(1+y)/2}^{2y} $$

If $y < 1$:

$$ \int_{y/2}^{(1+y)/2} \lesssim |y/2|^{1-1/p} |1-y|^{-1/q} $$

and

$$ \int_{(1+y)/2}^{2y} \lesssim |1-y|^{-1/p} |2y-1|^{1-1/q} $$

If $y > 1 $:

$$ \int_{y/2}^{(1+y)/2} \lesssim |1 - y|^{-1/p} |1-y/2|^{1-1/q} $$

and

$$ \int_{(1+y)/2}^{2y} \lesssim |y|^{1-1/p} |1-y|^{-1/q} $$

Note: Each of these integrals put the singular term in $L^1$ and the other term in $L^\infty$. E.g. for the first one, since $y < 1$ we have $(1+y)/2 < 1$ so $|1-x| > |1-y|/2$. So the integral becomes
$$ \leq 2^{1/q}|1-y|^{-1/q} \int_{y/2}^{(1+y)/2} |x-y|^{-1/p}~ dx $$
and the interior integral can be evaluated explicitly. The other terms are treated similarly.

So the integral is bounded by

$$ \int_{1-\delta}^{1+\delta} |y|^{1-1/p} |1-y|^{-1/q} + |2y-1|^{1-1/q} |1-y|^{-1/p} ~dy $$

and so your $I(p,q)$ should converge.


问:
Proof of Real Number Property
In his introduction to Calculus, Apostol gives a foundation for the properties
 of real numbers. After laying down the field and order axioms but before stating the least-upper-bound axiom, the author uses the set $T = \{x \ : \ 0 \leq x < 1\}$ as an example of a set which has no maximum element, yet has a least-upper-bound, namely, 1.

Is the author relying on intuition to make these assertions, or can we really prove these statements even before introducing the least-upper-bound axiom?

答: The least upper bound axiom is not needed to prove that this specific, concretely-given set of real numbers has a least upper bound: all that’s needed is to prove that $x<1$ for each $x\in T$, which is a matter of definition, and that if $u<1$, then there is an $x\in T$ such that $u<x$. The latter can be done by taking $x=\frac12(u+1)$: that $u<x<1$ can be proved from the field and order axioms.</p>

Title: Proof of Real Number Property


问:
How to find limits of an integral in spherical and cylindrical coordinates if you transform it from cartesian coordinates
If you have a volume integral in Cartesian coordinates with given limits of x,y and z and you want to transfer it to another coordinate system like spherical and cylindrical coordinates.
I can easily determine how the integral will look like but how to find the limits of the integral(for example, limits of $r,\theta$ and $\phi$ in spherical system )?

For example :

To calculate a triple integral in spherical coordinates for the volume inside $z^{2}=x^{2}+y^{2}$ between the planes z=1 and z=2.
Now I have the integral in Cartesian with limits :
$$\int_{z=1}^{2}\int_{x=-z}^{z} \int_{y=-\sqrt{z^2-x^2}}^{\sqrt{z^2-x^2}} d x d y d z$$
Now if I turned it into a spherical coordinate [$x=r\sin\theta\cos\phi$,$y=r\sin\theta\sin\phi$,$z=r\cos\theta$].

The integral will be:
$$\int\int\int r^{2}\sin\theta dr d\theta d\phi$$
But how do I calculate the limits of $r$,$\theta$ and $\phi$?

I'm not asking how to solve this particular question, I want to know a general method to find limits when changing coordinates like this without using any computer programs like graph plotters.

答: That region is what you get rotating the red region from the figure below arounf the $z$ axis (the vertical axis). So, $\varphi$ can take any value from $0$ to $2\pi$. Furthermore, $\varphi$ can take any value from $0$ to $\frac\pi4$. Finally, you have $z=r\cos\theta$, and so $1\leqslant z\leqslant2\iff\frac1{\cos\theta}\leqslant r\leqslant\frac2{\cos\theta}$. So, compute$$\int_0^{2\pi}\int_0^{\pi/4}\int_{1/\cos(\theta)}^{2/\cos(\theta)}r^2\sin(\theta)\,\mathrm dr\,\mathrm d\theta\,\mathrm d\varphi.$$


Body:
If $\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$, then the number of values of $x$ are:

I have solved the above like this:

$\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$

Squaring:

$9^x - 6^x = 6^x - 4^x$

$9^x + 4^x = 2\times(6^x)$

$3^{2x} + 2^{2x} = 2\times (3\times2)^{x}$

$(3^{x} - 2^{x})^{2} = 0$

Giving us, $3^x = 2^x$, so $x=0$ is the only value which satisfies.

But in the answers, they have stated that total solutions are $3$. Where have I made a mistake? And at what step did a specific action lead to loss of roots?

Please help me to go about the same. Thanks!

Body:
Factorize the equation

$$\sqrt{9^x - 6^x} - \sqrt{6^x - 4^x}
=\sqrt{3^x(3^x - 2^x)}- \sqrt{2^x(3^x - 2^x)}\\
= \sqrt{3^x - 2^x}\left(\sqrt{3^x}-\sqrt{2^x} \right)=0
$$


问:
For a measurable cardinal $κ$, show that $cf(γ)≠κ$ implies $j_U(γ)=\sup\{j_U(δ):δ<γ\}$ ($U$ $κ$-complete ultrafilter, $j_U$ associated embedding)
For :

$κ$ a measurable cardinal,

$U$ a $κ$-complete ultrafilter over $κ$

$j_U$ the elementary embedding of $V$ into the ultrapower of $V$ to $U$

How to show that :  If $\operatorname{cf}(γ)≠κ$ Then $j_U(γ)= \sup \{j_U(δ)\mid δ<γ\}$ ?

答:
Let us divide the cases. Let $\langle \gamma_\zeta\mid \zeta<\mu\rangle$ be a cofinal sequence converging to $\gamma$. (Here $\mu=\operatorname{cf}\gamma$.)

$\operatorname{cf}\gamma<\kappa$: Let $[f]_U<j_U(\gamma)$. Without loss of generality we may assume that $f(\xi)<\gamma$ for all $\xi<\kappa$. (Why?) For each $\xi$, choose $\eta_\xi<\mu$ such that $f(\xi)<\gamma_{\eta_\xi}$. Then there is $\delta<\mu$ such that $\{\xi<\kappa\mid \eta_\xi=\delta\}\in U$. Can you see how to proceed from there?

$\operatorname{cf}\gamma>\kappa$: Let $[f]_U<j_U(\gamma)$ again. Still, we may assume that $f(\xi)<\gamma$ for all $\xi<\kappa$. Since $\kappa<\mu$, $\delta:=\sup_{\xi<\kappa}f(\xi)<\gamma$. Then completing the proof is easy, so I will leave it to you.


答:
Certainly $T$ has an upper bound, namely $1$. Now to show that $1$ is the least upper bound, let $a$ be any upper bound of $T$. Can $a$ be less than $1$? No because $T$ contains elements arbitrarily close to $1$, i.e. for any $x < 1$ there is an element $y \in T$ such that $x < y$ which is to say that $x$ is not an upper bound for $T$. Therefore the upper bound $a$ is greater than or equal to the upper bound $1$, and since $a$ was arbitrary we conclude that $1$ is the least upper bound of $T$.

Note that the exact same argument works for $T' = \{x \in \mathbb Q \mathrel| 0 \leq x < 1\}$ even though $\mathbb Q$ doesn't have the least upper bound property.

The least upper bound property of $\mathbb R$ is there to guarantee least upper bounds of "weird" sets like $T_\xi = \{r \in \mathbb Q \mathrel| r < \xi\}$. Those sets don't have least upper bounds in the rationals when $\xi$ is irrational, but they do in the reals.

Title: Proof of Real Number Property

Body:
In his introduction to Calculus, Apostol gives a foundation for the properties
 of real numbers. After laying down the field and order axioms but before stating the least-upper-bound axiom, the author uses the set $T = \{x \ : \ 0 \leq x < 1\}$ as an example of a set which has no maximum element, yet has a least-upper-bound, namely, 1.

Is the author relying on intuition to make these assertions, or can we really prove these statements even before introducing the least-upper-bound axiom?


答:
I don't think that my proof is by induction, but I'll give it anyway.

The tournament starts with $n$ players.

Each match eliminates exactly one player from the tournament.

There is exactly $1$ winner.

Therefore, there must be $n-1$ matches.

$[n - 1 - 1 - 1 ... - 1 = 1.$ There must be $(n-1)\ $ "$-1$'s", and each $-1$ represents a match (someone losing a match, in fact).]

Title: Proving number of matches played in a knock-out tournament with $n$ players is $n-1$ using induction

Body:
I am supposed to prove that if $n$ players play in $n-1$ ways. Clearly the base case is trivially true, whence only $1$ match will be played. Also the case when the number of players is of the form $2^k$ where $k\in \mathbb{Z}^{+}$, can be easily formulated as follows. $$2^{k-1}+2^{k-2}+2^{k-3}+\ldots+2+1=2^{k}-1$$

I am not sure how to use induction on it in the general case, including the case when $n$ is odd. Any hints are appreciated. Thanks.


问:
Does a linear map change the basis of the target vector-space? If no: Is a linear map clearly defined just by how it transforms the basis vectors?
I am learning linear algebra on my own (I only have high school experience). I am not able to find an answer to this seemingly easy question - perhaps because I lack the mathematical lingo to correctly express myself.

If a linear map is defined as V -> W between two vector spaces, are the transformed basis vectors of V also basis vectors of W?

If the answer is yes: could we uniquely define a specific linear map V -> W just by how it transforms the basis vectors of V to the basis vectors of W? Because any vector in the W vector-space would just be a linear combination of these (transformed) basis vectors?

答:
I'm not sure exactly what you are asking in the first part of your question, so I'll address the second part first.

It is true that that a linear transformation is completely determined by its action on a basis.  If the transformation is non-singular, then the images of the basis vectors will be linearly independent.  They may or may not form a basis for $W$.  If the dimension of $W$ is equal to the dimension of $V$ they will, but if it is greater, they will not.  (It's not possible to have a non-singular transformation to a space of smaller dimension.)

If the transformation isn't non-singular, then the images of the basis vectors won't be linearly independent, so they certainly won't form a basis.  If this is what you are asking in the first part, then the answer is "No".  In particular, the $0$ transformation maps everything to the $0$ vector, which isn't an element of any basis.


答:
Let $V$ and $W$ be vector spaces, let $B$ be a subset of $V$, and let $f\colon B \to W$ be a function (of sets, ignoring any vector space structure). Then:

If $B$ spans $V$, then there is at most one linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$. (However, if $B$ isn't linearly independent, there might be no such linear map, because the elements of $B$ might satisfy a linear relation that the elements $f(b) \in W$ don't satisfy.)

If $B$ is linearly independent, then there is at least one linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$. (However, if $B$ doesn't span $V$, there is more than one such linear map, because you can extend $B$ to a basis and make many choices for where to send the other basis vectors.)

Putting the above two facts together: If $B$ is a basis, then there is a unique linear map $T\colon V \to W$ such that $T(b) = f(b)$ for all $b \in B$.

In other words, given a basis of a vector space, there's a one-to-one correspondence between maps from the basis to another vector space and linear maps to another vector space; to define a linear map, it suffices to define it on a basis. (This is essentially what the matrix representation of a linear map is: you're just taking the image of each vector in a chosen basis of the domain, representing it as a list of numbers via a chosen basis of the codomain, and assembling these lists of numbers into a rectangular array.)

However, given a basis $B$ of $V$ and a linear map $T\colon V \to W$, the image $T(B) = \{T(b) : b \in B\}$ does not have to be linearly independent, nor does it have to span $W$. For example, if you take the map $(x, y, z) \mapsto (x + y, x + y) : \mathbb{R}^3 \to \mathbb{R}^2$, the image of the standard basis $\{(1, 0, 0), (0, 1, 0), (0,0,1)\}$ under this map is $\{(2, 2), (2, 2), (0, 0)\}$, which is neither linearly independent nor spans $\mathbb{R}^2$.

The key here is that these are related to the notions of surjective (onto), injective (one-to-one), and bijective (one-to-one and onto) functions:

A linear map $T\colon V \to W$ is surjective if and only if, for some basis $B$ of $V$, the image set $T(B)$ spans $W$.

$T$ is injective if and only if, for some basis $B$ of $V$, the image set $T(B)$ is linearly independent.

Putting the above two together: $T$ is bijective if and only if, for some basis $B$ of $V$, the image set $T(B)$ is a basis of $W$.

Also, you can replace "for some basis" with "for every basis" and the above will still be true (the choice of basis doesn't matter here).

I recommend working out examples of all of these (and drawing pictures, say in 2 or 3 dimensions, if possible), this will help a lot with the intuition.

Title: Does a linear map change the basis of the target vector-space? If no: Is a linear map clearly defined just by how it transforms the basis vectors?

Body:
I am learning linear algebra on my own (I only have high school experience). I am not able to find an answer to this seemingly easy question - perhaps because I lack the mathematical lingo to correctly express myself.

If a linear map is defined as V -> W between two vector spaces, are the transformed basis vectors of V also basis vectors of W?

If the answer is yes: could we uniquely define a specific linear map V -> W just by how it transforms the basis vectors of V to the basis vectors of W? Because any vector in the W vector-space would just be a linear combination of these (transformed) basis vectors?


答:
There is no general formula. You must have a good understanding of how spherical coordinate works if you are converting to it. You also should have a rough visualization of the surface in your mind. Even a rough $2D$ sketch helps. Then use the limits in cartesian coordinates to find the limits in spherical coordinates.

Now if we work through the example you have written down,

As you mentioned, $x = r \cos \phi \sin \theta, y = r \sin \phi \sin \theta, z = r \cos \theta$.

Now for the cone $z^2 \geq x^2 + y^2$, $r^2 \cos^2 \theta \geq r^2 \sin^2 \theta \implies \theta \leq \frac{\pi}{4}$.

Now bounds of $z$ is $1 \leq z \leq 2$. Given $z = r \cos \theta$, we obtain,

$\sec \theta \leq r \leq 2 \sec \theta$.

Limits of $\phi$ is simply $0$ to $2 \pi$.

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_{\sec \theta}^{2 \sec \theta} r^2 \sin \theta \ dr \ d\theta \ d\phi$.

Title: How to find limits of an integral in spherical and cylindrical coordinates if you transform it from cartesian coordinates

Body:
If you have a volume integral in Cartesian coordinates with given limits of x,y and z and you want to transfer it to another coordinate system like spherical and cylindrical coordinates.
I can easily determine how the integral will look like but how to find the limits of the integral(for example, limits of $r,\theta$ and $\phi$ in spherical system )?

For example :

To calculate a triple integral in spherical coordinates for the volume inside $z^{2}=x^{2}+y^{2}$ between the planes z=1 and z=2.
Now I have the integral in Cartesian with limits :
$$\int_{z=1}^{2}\int_{x=-z}^{z} \int_{y=-\sqrt{z^2-x^2}}^{\sqrt{z^2-x^2}} d x d y d z$$
Now if I turned it into a spherical coordinate [$x=r\sin\theta\cos\phi$,$y=r\sin\theta\sin\phi$,$z=r\cos\theta$].

The integral will be:
$$\int\int\int r^{2}\sin\theta dr d\theta d\phi$$
But how do I calculate the limits of $r$,$\theta$ and $\phi$?

I'm not asking how to solve this particular question, I want to know a general method to find limits when changing coordinates like this without using any computer programs like graph plotters.


问:
Showing that a level set is not a submanifold
Is there a criterion to show that a level set of some map is not an (embedded) submanifold? In particular, an exercise in Lee's smooth manifolds book asks to show that the sets defined by $x^3 - y^2 = 0$ and $x^2 - y^2 = 0$ are not embedded submanifolds.

In general, is it possible that a level set of a map which does not has constant rank on the set still defines a embedded submanifold?

答:
It is certainly possible for a level set of a map which does not have constant rank on the set to still be an embedded submanifold.  For example, the set defined by $x^3 - y^3 = 0$ is an embedded curve (it is the same as the line $y=x$), despite the fact that $F(x,y) = x^3 - y^3$ has a critical point at $(0,0)$.

The set defined by $x^2 - y^2 = 0$ is not an embedded submanifold, because it is the union of the lines $y=x$ and $y=-x$, and is therefore not locally Euclidean at the origin.  To prove that no neighborhood of the origin is homeomorphic to an open interval, observe that any open interval splits into exactly two connected components when a point is removed, but any neighborhood of the origin in the set $x^2 - y^2$ has at least four components after the point $(0,0)$ is removed.

The set $x^3-y^2 = 0$ is an embedded topological submanifold, but it is not a smooth submanifold, since the embedding is not an immersion.  There are many ways to prove that this set is not a smooth embedded submanifold, but one possibility is to observe that any smooth embedded curve in $\mathbb{R}^2$ must locally be of the form $y = f(x)$ or $x = f(y)$, where $f$ is some differentiable function.  (This follows from the local characterization of smooth embedded submanifolds as level sets of submersions, together with the Implicit Function Theorem.)  The given curve does not have this form, so it cannot be a smooth embedded submanifold.


问:
Prove that if $X$ has $n$ points then $C(X,Y)$ is homeomorphic to $Y^n$
Let $X$ be a finite set with $|X|=n$ equipped with the dicrete topology.Show that $C(X,Y)$ is homeomorphic to $Y^n$ where $Y^n=\overbrace{Y\times Y \times Y \cdots \times Y}^{n-times}$.

In the book there are a theorem that can help me

Theorem:Let Let $X, Y, Z$ be spaces, and assume all function
spaces have the compact-open topology. There exists a natural one-to-one map
$\gamma:C((X \times Y),Z) \longrightarrow C(X,C(Y \times Z))$

Futhermore:
If $Y$ is locally compact and Hausdorff $\gamma$ is bijective.

If $X$ is locally compact and Hausdorff $\gamma$ is one to one.

If $X,Y$ are both locally compact and Hausdorff $\gamma$ is homeomorphism.

The problem here is that I need have the $CO$(Compact-open) topology and I don´t have it, thus I don´t know how to apply this result to the demostration of our result.

I don´t have a idea about how to I should attack this problem.If someone may provide me a idea about how to attack the problem in this case, was useful.Thanks in advice

答:
First note that if $X$ has the discrete topology then $C(X,Y)$ is just the set of all functions from $X$ to $Y$ (every function is continuous on a discrete space).

Say, $X=\{x_1,\ldots,x_n\}$ and we can identify the function $f: X \to Y$ with $\Phi(f)=(f(x_1), \ldots, f(x_n)) \in Y^n$, in that $\Phi$ is a bijection between these sets.

So we only need to check that $\Phi$ is open and continuous when we give the domain $C(X,Y)$ the compact-open topology (with subbasic open sets $[K,U] =\{f \in C(X,Y)\mid f[K] \subseteq U\}$, where $K \subseteq X$ is compact (but in our case all subsets of $X$ are finite so compact) and $U \subseteq Y$ is open.

If $\prod_{i=1}^n U_i$ is basic open in $Y^n$ (so all $U_i$ are open in $Y$)
then $$\Phi^{-1}[\prod_{i=1}^n U_i] = \{f \in C(X,Y)\mid (f(x_1), \ldots f(x_n)) \in \prod_{i=1}^n U_i\} = \bigcap_{i=1}^n [\{x_i\}, U_i]$$

and the latter is a finite intersection of subbasic open sets in C(X,Y)$, hence open.

And also if $[K, U]$ is subbasic open then $$\Phi[[K,U]]= \prod_{i = 1}^n V_i$$

where $V_i = U$ for when $x_i \in K$ and $V_i = Y$ otherwise, so $\Phi$ sends subbasic open sets to open sets and as $\Phi$ is a bijection (!) this implies $\Phi$ is an open map.

This concludes the proof that $\Phi$ is the required homeomorphism.


问:
Why does the function $\sin(\sin(x) \cdot x)$ not represent FM?
I was trying to figure out a way to represent frequency modulation (FM) with sinusoids, and thus graphed the function $f(x) = \sin(\sin(x) \cdot x)$. Needless to say, the graph was not what I wanted. I couldn't understand why the graph is shaped as such. Since $\sin(x)$ oscillates between $-1$ and $1$, I would've thought that the wavelength of $f(x)$ would not go below $2\pi$. Is there something I'm missing? I could always confirm it mathematically using derivatives and such, but I'm looking for intuition.

P.S. I would also appreciate it if anyone could also share and explain the equation representing FM with sinusoids, but I'll probably ask another question just for that.

答:
Answering here so I will understand later on when I revisit this topic:

Imagine $f(x)=x\sin(x)$ running through the values from $-x$ to $x$ in one period (as $\sin(x)$ traverses through $-1$ and $1$, but here $x$ also acts as a variable amplitude). It will traverse many points where $x\sin(x)=k\pi$. As $x$ increases down the $x$-axis, the "faster" $f(x)$ traverses the $x$-values as it takes the same amount of time to run through all the values of $x$ (sharper gradient).

When plugging $f(x)$ into another sine function, i.e. $\sin(x\sin x)$, it could be thought as traversing through multiple wavelengths faster and faster as $x$ increases. This can be proven knowing that every time $f(x) = k\pi$, $\sin(f(x)) = 0$. As $x$ increases, the number of points where $f(x) = k\pi$ increases as well, so the frequency of $\sin(x\sin x)$ must increase with $x$.

To model FM, we need to instead set a variable that will represent the change in frequency and vary it in accordance to the signal wave. This takes away the modulated frequency's direct dependence on the $x$-values and solves the problem.


问:
Integration methods for stiff systems of ordinary differential equations
I am reading this paper by Liniger and Willoughby on Integrating Stiff ODE. I am hoping that I am providing enough information for an answer. We are looking at an system,

$\dot{x} = -\lambda x$

The paper breaks down the integration as follows,
$x(t+h) - x(t) - h [(1 - \mu) \dot{x}(t+h) + \mu \dot{x} (t)] = e_1(t)$,

where

$e_1 = - h^2 \int_0^1(\theta - \mu) \ddot{x}(t + \theta h) d\theta$

I have no idea how this came about, if someone could clue me in I would much appreciate it. I tried both the integral and Legrange remainder theorems, not sure what I am doing wrong but I could not arrive at the result.

答:
Consider for $t$ and $h$ fixed the function
$$
g(θ)=x(t+θh)-h(θ-μ)\dot x(t+θh)
$$
Then
\begin{align}
e_1(t)=g(1)-g(0)&=\int_0^1g'(θ)\,dθ\\
&=\int_0^1\left[h\dot x(t+θh) -h\dot x(t+θh)-h^2(θ-μ)\ddot x(t+θh)\right]\,dθ\\
&=-h^2\int_0^1(θ-μ)\ddot x(t+θh) \,dθ
\end{align}
as claimed.


问:
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.

Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.

Here's what I have done so far:

$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$

How do I continue the problem from here? Is it even possible?

答:
It seems the expression $S$ needs to be revised somewhat. If we take a look at the first two summands
\begin{align*}
\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\cdots
\end{align*}
we see the left-hand factor in the denominator is odd and the right-hand factor is even. Taking $n$ summands of this kind we have
\begin{align*}
S_{1}=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)2n}\tag{1}
\end{align*}
where the last summand has an odd and even factor of the given form in the denominator.

Here we calculate $S_1$. We obtain from (1)
\begin{align*}
\color{blue}{S_1=\sum_{k=1}^n\frac{1}{(2k-1)2k}}
&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\
&=\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{2}\\
&=\left(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\right)-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{3}\\
&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\tag{4}\\
&\,\,\color{blue}{=H_{2n}-H_{n}}\tag{5}
\end{align*}

Comment:

In (2) we split the sum into two sums and factor out the constant $\frac{1}{2}$ from the right-hand sum.

In (3) we add and subtract summands with even denominator to the left-hand sum for convenient calculations in the further steps. This does not change the value of the sum since we are adding zero only.

In (4) we simplify and collect the sums.

In (5) we write the sums using the symbol for Harmonic numbers.


问:
Lift of a shortest loop is length minimizing
Let $(M,g)$ be a $2$-dimensional oriented complete Riemannian manifold without boundary. Let $f:\big(\Bbb S^1,1\big)\to (M,x_0)$ be a smooth loop such that $f$ has minimum length among all admissible loops
in the given free homotopy class of loops. That is, if $f':\big(\Bbb S^1,1\big)\to (M,x_0')$ is another admissible loop such that there is a continuous map $H:\Bbb S^1\times [0,1]\to M$ with $H(-,0)=f, H(-,1)=f'$, then length of $f$ is no more than length of $f'$.

Consider the following commutative diagram obtained from a smooth lifting.
$\require{AMScd}$
\begin{CD}
 (\Bbb R,0) @>\displaystyle\ell>> (\widetilde M,\widetilde {x_0})\\
@V\text{Universal Cover}  V V @VV  \text{Universal Cover}V\\
(\Bbb S^1,1) @>>\displaystyle f> (M,x_0)\\
\end{CD}

Let $p:=\ell(t_0)$ and $q:=\ell(t_1)$ be two distinct points, and
$\lambda:=\ell\big|[t_0,t_1]$. Is it true that
$\text{Length}_{\widetilde g}(\lambda)\leq \text{Length}_{\widetilde g}(\lambda')$ for all
admissible curves $\lambda'$ with the same endpoints $p$ and $q$?

$\bullet$ Always consider the pull-back metric on every covering.

$\bullet$ In a Riemannian manifold, every minimizing curve is a geodesic
when it is given a unit-speed parameterization, i.e. $\text{im}(\lambda)$ is the image of some geodesic.

$\bullet$ If needed, one may assume $f$ is not null-homotopic.

$\textbf{Definition:}$ On a smooth manifold $M$ a curve $\gamma:[a,b]\to M$ is said to be admissible if there is a partition $a=a_0<a_1<...<a_{n-1}<a_n=b$ of $[a,b]$ such that $\gamma\big|[a_{k-1},a_k]$ is smooth with non-vanishing velocity for each $k=1,...,n$.

答:
I just saw the "orientability" hypothesis on your question, which kills this answer. But I'm going to leave the answer anyway, because there are orientable generalizations in higher dimensions.

For a (nonorientable) counterexample, take $\widetilde M \to M$ to be the universal covering of the projective plane
$$f : S^2 \mapsto \mathbb R P^2
$$
where $S^2$ has its usual metric and the (unique) nontrivial deck transformation is the antipodal map of $S^2$. The group $\mathbb R P^2$ is cyclic of order $2$. There is a constant speed, smooth map $f : S^1 \to \mathbb R P^2$ which represents that element and has minimal length, that minimum being equal to $\pi$ (so $f$ has speed $1/2$). The corresponding lift $\ell : \mathbb R \to S^2$ is a constant speed universal covering map of a great circle in $S^2$; in fact, the covering map restricted to that great circle gives a 2-to-1 covering of the circle $\text{image}(f)$. But $p = \ell(0)=\ell(2\pi) = q$ and so the length of the restriction $\lambda = \ell \mid [0,2\pi]$ is strictly greater than the constant path $\lambda'$ with the same endpoints as $\ell$, namely the point $p = q$.

But there is a restricted situation in which the statement you want is true, namely that the Gaussian curvature of $M$ is non-positive at each point. There is a theorem in differential geometry which says that on any nonpositively curved complete simply connected Riemannian manifold, every locally geodesic path is globally geodesic (this is not the same as, but is closely connected to, the Cartan-Hadamard Theorem). Since each of your maps $f$ is locally geodesic, it follows that each of their lifts $\lambda$ is locally geodesic, and by applying that theorem it follows that $\lambda$ is globally geodesic.


问:
Question about notation in proof of limit comparison test
In a book I came across the following:

Theorem: Limit Comparison Test

Suppose that $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ have positive terms and that $\frac{\color{blue}{ a_n }}{\color{blue}{ b_n }} \rightarrow L$ as $n \rightarrow \infty$, where $L \not = 0$.

If $\sum_{n=1}^{\infty} b_n$ is convergent, then $\sum_{n=1}^{\infty} a_n$ is convergent.

I'm getting a bit lost in the notation details. Question: What are $\color{blue}{ a_n }, \color{blue}{ b_n }$ in the part where it says " ... $\frac{\color{blue}{ a_n }}{\color{blue}{ b_n }} \rightarrow L$ as $n \rightarrow \infty$ ..." ?

To my understanding, $a_n, b_n$ are the n-th partial sums of the two series, which is not the same as the n-th term of the respective series.

If then the ratio between the two partial sums is tending towards  a constant, it makes sense that they have both the same convergence status. And this allows to deduce the convergence status of the series.

Is that what is meant here? Or do we actually check the ratio between the n-th terms?

答: No, that is not what is meant there. In the expression $\lim_{n\to\infty}\frac{a_n}{b_n}=L$, $a_n$ is the $n$th term of the sequence $(a_n)_{n\in\Bbb N}$, and $b_n$ is the $n$th term of the sequence $(b_n)_{n\in\Bbb N}$. There are no partial sums here.


问:
Equation for a cone that is "upside down"?
The equation of a cone who was its tip on the xy plane is $x^{2}+y^{2}=z^{2}$. How can I describe a cone whose base is on the xy plane, and tip on a point on the z-axis, i.e. the cone $x^{2}+y^{2}=z^{2}$ but upside down?

答:
For a right circular cone having height $H$, radius $R$ on the $z = 0$ plane, and apex at $z = H$:
$$\left( x^2 + y^2 \right) \left( \frac{H}{R} \right)^2 = \left( z - H \right)^2, \quad 0 \le z \le H$$
i.e.
$$\left( x^2 + y^2 \right) \left( \frac{H}{R} \right)^2 - \left( z - H \right)^2 = 0, \quad 0 \le z \le H$$
or as a function $z = f(x, y)$ (solving above for $z$),
$$z = H \left( 1 - \sqrt{\frac{ x^2 + y^2 }{R^2}} \right), \quad x^2 + y^2 \le R^2$$


问:
What is a good asymptotic for $f_n = f_{n-1}+\ln(f_{n-1})$?
Let $f_0=2$ and $f_n=f_{n-1}+\ln(f_{n-1})$. What is a good asymptotic to the sequence $f_n$? With good I mean much better than $f_n \sim \dfrac{3n \ln(2)\ln(n)}{2}$.</p>

答:
$$\lim_{n\to\infty}\left(\frac{f_n}n-\log n-\log\log n\right)=-1$$

To prove this, consider $g_n=n\log n+n\log\log n-n$ and $h_n=f_n-g_n$. One wants to prove that $h_n=o(n)$. The identity $f_{n+1}=f_n+\log f_n$ is equivalent to
$$
h_{n+1}=g_n+h_n+\log(g_n+h_n)-g_{n+1}.
$$
Using simple properties of the logarithm, one can show that this implies
$$
h_{n+1}=h_n+\log\left(1+\frac{\log\log n}{\log n}-\frac1{\log n}+\frac{h_n}{n\log n}\right)+O\left(\frac1{\log n}\right).
$$
In particular, if $h_n=o(n\log n)$, the logarithm in the RHS goes to zero hence $h_{n+1}=h_n+o(1)$, which implies $h_n=o(n)$. Thus, our task is to prove the easier statement that
$$
f_n=n\log n+o(n\log n).
$$
To do so, first note that $f_n\geqslant2$ for every $n\geqslant0$ yields $f_{n+1}-f_n\geqslant\log2$ hence $f_n\geqslant n\log2$ for every $n\geqslant0$. Plugging this once again in the recursion $f_{n+1}=f_n+\log f_n$ yields $f_{n+1}-f_n\geqslant\log n+\log\log2$ hence, summing up, $f_n\geqslant n\log n+o(n\log n)$.

In the other direction, $f_{k+1}-f_k=\log f_k\leqslant\log f_n$ for every $k\leqslant n$ hence $f_n\leqslant f_0+n\log f_n$, which can be seen to imply $f_n\leqslant n\log n+2n\log\log n$ for every $n$ large enough. This completes the proof.


问:
why isn't my proof valid? system of linear equations
The problem was: given the matrix $A \in M_{m\times n}$ and let $b \in F^{m}_{col}$.
$y \in F^{m}_{col}$ is a solution of the system of equations $AX=b$.
prove: every solution of the system of equations $AX=b$ can be represented as $y+x$, where $x \in F^{m}_{col}$ is the solution of the homogeneous system $AX=0$.

so my proof was like that:
because $y \in F^{m}_{col}$ is a solution of the system of equations $AX=b$ we can conclude:
$$(1) A \cdot y=b$$
and because $x \in F^{m}_{col}$ is a solution of the system of equations $AX=0$ we can conclude:
$$(2) A \cdot x=0$$
from (1) and (2) we get:
$$ A \cdot y+A \cdot x=0+b$$
according to the rules of matrix multiplication we get:
$$(3) A \cdot ( y+ x)=0+b=b$$
and therefore according to (3) the solution which is represented by $z=x+y$ is also a solution of the system of equations $AX=b$

my instructor gave me only 3 points for that telling that "it isn't a valid proof". Why is that? what's the problem with it? and what should I say in order to appeal his decision.
thank in advance.

答:
This is the part that you missed: let $\;z\;$ be a solution, any solution, to $\;Ax=b\;$ , then

$$A(z-y)=Az-Ay=b-b=0\implies \;z-y\;\text{is a solution to the hom. system}\;\;Ax=0$$

and thus there exists $\;t\in\;$ space of all solutions to the hom. sytem, such that

$$z-y=t\implies z=t+y$$

with $\;t\;$ a solution to the hom. system, and $\;y\;$ a particular solution to the non-hom. system


问:
Process of graphing a function $\arccos(\ln{(\frac{x+1}{x-3})})$
$$\arccos\left(\log{\left(\frac{x+1}{x-3}\right)}\right)$$

We  need to graph this function. I can graph $g(x)=\log{\left(\frac{x+1}{x-3}\right)}$ but I get trouble when I want to graph  $\arccos(g(x))$.

What is the easiest thought process for graphing such functions.

答:
$$f(x)=\arccos\left(\ln{\frac{x+1}{x-3}}\right)$$
domain of $\arccos$ is $[-1,1]$ thus we must have
$$-1\le \ln{\frac{x+1}{x-3}}\le 1;\;\frac{x+1}{x-3}> 0$$

$$ x\leq -\frac{3+e}{e-1}\lor x\geq \frac{3 e+1}{e-1}$$

$$\lim_{x\to\infty}\arccos\left(\ln \left(\frac{x+1}{x-3}\right)\right)=\frac{\pi}{2}$$
Horizontal asymptote $y=\frac{\pi}{2}$
$$\lim_{x\to \frac{-3-e}{e-1}}\arccos\left(\log \left(\frac{x+1}{x-3}\right)\right)=\pi$$
$$\lim_{x\to \frac{3 e+1}{e-1}}\arccos\left(\log \left(\frac{x+1}{x-3}\right)\right)=0$$
Derivative

$$y'=\frac{4}{(x-3) (x+1) \sqrt{1-\log ^2\left(\frac{x+1}{x-3}\right)}}$$
$y'>0$ for $x<-1\lor x>3$


问:
Normal distribution equality proving
I proved that for $\xi \sim \mathcal{N}(0,1): \mathbb{E}(\xi)=0$ and $\text{Var}(\xi)=1$. How can I prove the following:

Let $\xi \sim \mathcal{N} (0, 1), a, b \in \mathbb{R}$. Prove that $b\xi + a \sim \mathcal{N}(a, b^2)$.

答:
Let $ a,b \in \mathbb{R}$ and $X \sim N(0,1)$

$E[bX+a] = bE[X] + E[a] = b\cdot 0 + a$

$Var[bX+a] = E[(bX+a)^2] - E^2[bX+a] = E[b^2X^2 +2Xa + a^2] - (b\cdot 0 +a)^2] = b^2E[X^2] +2aE[X] + a^2 -a^2 = b^2Var[X] = b^2\cdot 1 $

Set $Y = bX+a$ then

$P(bX+a \le x) = P(bX \le x-a) = P(X \le \frac{x-a}{b}) $ so $F_Y(y) = F_X(\frac{x-a}{b}) = \Phi(\frac{x-a}{b})$

Find the density from here and you have it.

Title: Normal distribution equality proving


问:
Deriving the general formula of alkane using graph theory
With context as per this post, it turns out that if $n$ represents the number of skeleton's in any one of the graphs which OP has drawn, then the total hydrogen required to fulfil valency of carbon (I.e: give each carbon node 4 attachments) is $2n+2$. How do we prove this?

It is often a formula given in chemistry classes as a fact to memorize, I hope that using mathematics there can be an understanding on why the relation between number of hydrogen and carbon must be such

Ideally looking for a non induction based proof.

答:
There's not going to be any way to avoid using induction some way or another.  Induction is pretty much the only way you can ever prove statements about finite structures of arbitrary size.  (Ultimately, this is because "finiteness" is defined in terms of induction, with the natural numbers defined as the smallest inductive set.)

Here is probably the simplest way to prove it.  Every time you attach a new carbon to your alkane, you replace a hydrogen (that used to be in the spot where you attached your new carbon) with a $\mathrm{CH_3}$, giving a net increase of $2$ hydrogens.  So every new carbon you add increases the number of hydrogens by $2$.  Building this up inductively from the base case of $\mathrm{CH_4}$, this means there will always be $2n+2$ hydrogens.  (Here I am glossing over one important detail: how do you know you can build up every alkane by repeatedly replacing hydrogens with $\mathrm{CH_3}$?  How to prove this depends on exactly what your definitions are, but it crucially uses the fact that the carbons must form a tree, without any cycles.)

Or, here's a more high-tech way of thinking about it using a little machinery of graph theory.  Consider the carbons as a tree.  By a basic theorem in graph theory, a tree with $n$ vertices has $n-1$ edges (this is proved by induction, so you still aren't avoiding induction here).  So, there are $n-1$ bonds between the carbons.  Each carbon must have $4$ bonds, giving a total of $4n$ bonds where you count each bond twice (once for each end).  Of these $4n$ bonds, $2(n-1)$ of them are accounted for by carbon-carbon bonds.  Thus there must be $4n-2(n-1)=2n+2$ carbon-hydrogen bonds, and thus $2n+2$ hydrogens.


问:
$n^5-n$ is divisible by $10$?
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>

答: Since $n$ and $n^5$ have the same parity, $f(n):=n^5-n$ is divisible by $2$. It is also divisible by $5$, since $f(0)=0$ and $f(n+1)-f(n)=\dotsc=5 n (n^3 + 2 n^2 + 2 n + 1)$. More generally, for every prime $p$, $f(n):=n^p-n$ is divisible by $p$, this is Fermat's little theorem. In fact, $f(n+1)-f(n)=\sum_{0<k<p} \binom{p}{k} n^k$ and $p \mid \binom{p}{k}$ for $0 < k < p$.</p>

Title: $n^5-n$ is divisible by $10$?


问:
On notation of the generalized probability union formula
Let $(A_i)$ be a finite family of events of a probability space. Then

$$P(\bigcup A_i) = \sum_{i=1}^m P(A_i) - \sum_{i=1}^{m-1}\sum_{j=i+1}^m P(A_i \cap A_j) + ... +(-1)^{m-1} P(A_1 \cap ... \cap A_m)$$

This theorem is pretty intuitive. However the way it's stated is not very elegant. Is there a better way to notate this? Intuitively very sum will be over the probabilities of the intersections of a finite number of events modulo commutativity, but trying to express that seems even more complicated than the current formula.

The writter also has difficulty proving this theorem. Even though the general process is intuitive, trying to write it out rapidily gets stuck in a wall of notation. Are there any tips on how to write proofs for these notation-heavy (but not deep) theorems without getting lost in messy notation?

答:
Be aware that using indicator functions for a finite index set $I$ we have:$$\mathbf{1}_{\bigcup_{i\in I}A_{i}}=\sum_{\varnothing\neq J\subseteq I}(-1)^{|J|-1}\mathbf{1}_{\bigcap_{i\in J}A_{i}}$$

See the proof of the theorem stated in this answer.

Taking expectations with respect to probability measure $P$ on both side we arrive at:$$P\left(\bigcup_{i\in I}A_{i}\right)=\sum_{\varnothing\neq J\subseteq I}(-1)^{|J|-1}P\left(\bigcap_{i\in J}A_{i}\right)$$That is the same expression in your answer for $I=\{1,\dots,m\}$.


问:
If $f$ is total recursive and $A \subseteq \Bbb N$ is recursively enumerable then $f(A), f^{-1}(A)$ are recursively enumerable.
Question: If $A \subseteq \Bbb N$ and $f:\Bbb N \to \Bbb N$ is a function, then let $f(A)=\{ n \in \Bbb N \mid \exists m \in A \text{ such that } f(m)=n \}$ and $f^{-1}(A)=\{ n \in \Bbb N \mid f(n) \in A \}$.
Prove that if $f$ is a total recursive function and $A \subseteq \Bbb N$ is recursively enumerable then $f(A)$ and $f^{-1}(A)$ are recursively enumerable.

I tried:
Definition. Let $B$ be a set of $\Bbb N$. $B$ is recursively enumerable if there exists a total recursive function $g$ such that $B= \{g(0),g(1),g(2),g(3),\ldots \}$.
So a set is recursively enumerable if it is the range of some total recursive function.

$\operatorname{rng}(f(A))=\Bbb N$ so $f(A)$ is recursively enumerable.
$f^{-1}(A)=A$ so $f^{-1}(A)$ is recursively enumerable.

Hints?

答:
So your definition says of $A$ that there is a total recursive function $g$ that enumerates $A$, i.e., such that $A =\{ g(0), g(1), g(2), \ldots \}$. Let us fix some such function $g$, then we have:
$$
f(A) = (f(g(0)), f(g(1)), f(g(2)), \ldots \}
$$
so the total recursive function $f \circ g$ that composes $f$ with $g$ enumerates $f(A)$.

To show that $f^{-1}(A)$ is recursively enumerable, it is easiest to use an alternative characterisation of recursively enumerable set: a set is recursively enumerable if it is the domain of some partial recursive function. (See here. If you don't know about this, please add a comment to say so and I can say more).
To see that $f^{-1}(A)$ is recursively enumerable according to this alternative characterisation, define a partial recursive function $h$, that given an argument $x$ first calculates $y = f(x)$ and then (with $g$ as in the previous paragraph) calculates $g(0)$, g(1), $g(2), \ldots$, halting when it finds an $n$ with $y = g(n)$. The domain of $h$ is then $f^{-1}(A)$.

As a small aside: it it usual to consider the empty set to be recursively enumerable. So the definition in terms of an enumerating function should say $B$ is recursively enumerable if $B$ is empty or if there exists a total recursive function that enumerates $B$.


问:
Why is $\sup_{x\in X}\sup_{y:x = g(y)}f(y) = \sup_{x\in X}f(x)$
Let $f$ be a function from $X$ to $\mathbb R$ and $g$ be a function from $X$ to $X$. Furthermore, define $h(x) = \sup_{y:x=g(y)}f(y)$.

Then it apparently holds that $$\sup_{x\in X}h(x) = \sup_{x\in X}\sup_{y:x=g(y)}f(y) = \sup_{x\in X}f(x)$$

This is used in the prove of the invariance property of MLEs (c.f. https://stats.stackexchange.com/questions/333737/maximum-likelihood-is-not-re-parametrization-invariant-so-how-can-one-justify-u). I don't understand why the second equality follows. My professor said it is "obvious", however, I don't see why.

答:
Kavi Rama Murthy's objection can be gotten around either by changing the statement to $$\sup_{x \in g(X)} h(x) = \sup_{x \in X} f(x)$$ or defining
$h : X \to \overline {\Bbb R}$, with $\sup_{x \in \emptyset} f(x) := -\infty$.

In either case, the statement is true.

Now let $M = \sup_{x\in X} f(x)$. For any $m < M$, there is a $y_m\in X$ with $f(y_m) > m$. Let $x = g(y_m)$. Then $h(x) \ge f(y_m) > m$. Therefore $\sup_{x \in X} h(x) > m$ for all $m < M$. Since $h(x) \le M$ for all $x$, we must have
$$\sup_{x \in X} h(x) = M$$


答:
There are multiple ways to prove this. Here is one using Moment generating functions (MGF). MGF of $X \sim\mathcal{N}(a, b^2)$ (normal with mean$=a$, and variance$=b^2$) is $M_{a,b}(t):=\mathbb{E}(\exp(tX))=\exp\left(at+\frac{b^2t^2}{2}\right)$, for all $t\in \mathbb{R}$. Now MGF of $\xi$ is therefore
$$
M_{0,1} = \exp\left(\frac{t^2}{2}\right).
$$
Thus MGF of $b\xi +a$ is
$$
\mathbb{E}(\exp(t(b\xi +a))) = \exp(ta)\mathbb{E}(\exp(tb\xi))= \exp(ta)\exp\left(\frac{b^2t^2}{2}\right),
$$
which is the MGF of $\mathcal{N}(a, b^2)$. Since MGF uniquely determines a distribution (when exists), therefore $b\xi +a \sim \mathcal{N}(a, b^2)$.

Title: Normal distribution equality proving

Body:
I proved that for $\xi \sim \mathcal{N}(0,1): \mathbb{E}(\xi)=0$ and $\text{Var}(\xi)=1$. How can I prove the following:

Let $\xi \sim \mathcal{N} (0, 1), a, b \in \mathbb{R}$. Prove that $b\xi + a \sim \mathcal{N}(a, b^2)$.


问:
Taylor's Theorem applied to Newton's method
Consider we're applying Newton's method to a complex function (I don't think the function matters ...) $f(x) = x^n-pw x^{n-1}- p w^{-1} x -1 + 2p$ with $0<p<1$ and $w=e^{2\pi i /n}$ (we want to find $\lambda$ such that $\lambda^n-pw \lambda^{n-1}- p w^{-1} \lambda -1 + 2p=0$):
$$x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$$
with $x_0=1$.

By Taylor's theorem,
\begin{equation*}
    |f(x_{i+1})| \leq \frac{1}{2} \max_{0 \leq u \leq 1} |f''(u x_i + (1-u) x_{i+1})| \ \left|\frac{f(x_i)}{f'(x_i)}\right|^2.
\end{equation*}

Where does this bound come from?

What I've done:

Suppose the method converges to a root $x^*$. This root can be expressed as a convex combination of $x_i$ and $x_{i+1}$, that is, $x^* = u x_i + (1-u) x_{i+1}$. Let $\delta_i = |x^* - x_{i+1}| \leq |x_{i+1}-x_i| = \left| \frac{f(x_i)}{f'(x_i)}\right|.
\\
$By Taylor's theorem, since $f(x^*)=0$,
$$ f(x_{i+1})=f'(x^*)\delta_i + \frac12 f''(x^*)\delta_i^2 \leq f'(x^*)\delta_i + \frac12 f''(x^*)\left| \frac{f(x_i)}{f'(x_i)}\right|^2.$$
It is very close from what I want, but I don't know how to get rid of $f'(x^*)$.

I don't know if it helps, but this comes from Lemma $3$ of this paper.

答:
You are right that the specific function doesn't matter here, you only need it to be regular enough (and your function is smooth because it's a polynomial).

Fix $i\geq 0$. Taylor's theorem with Lagrange's remainder states that
$$f(x_{i+1}) = f(x_i) + f'(x_i)(x_{i+1}-x_i) + \frac 1 2 f''(\xi_u)(x_{i+1}-x_i)^2$$
for some $\xi_u = ux_i+(1-u)x_{i+1}$ and some $u \in (0,1)$. But since $f'(x_i) \neq 0$ we get
$$f(x_i) + f'(x_i) (x_{i+1} -x_i) = f(x_i) + f'(x_i)\left(- \frac{f(x_i)}{f'(x_{i})} \right) =0, $$
so
$$|f(x_{i+1})| = \frac 1 2 |f''(\xi_u)| |x_{i+1} - x_i|^2 \leq \frac1 2M\left|\frac{f(x_i)}{f'(x_{i})}\right|^2, $$
where $M := \max_{0\leq u \leq 1} |f''(\xi_u)|$. Is this what you were looking for?


问:
prove or disprove in $\mathbb{Z}[i]$ decomposition of real integer to 2 elements with the same norm is unique up to multiplying by unit
let $a,b,c\in\mathbb{Z}[i]$such that a=bc and say $|b|=B$ and $|c|=C$ . prove or disprove, there is no other decomposition such: $a=de ,|d|=B,|e|=C$
up to multiplying b and c by unity.

Edit: this claim is not true for complex number, but is it for reals? prove it!

答:
No it doesn't hold for real number either:
counterexample: $25=(4+3i)(4-3i)=5*5$ yet $(4+3i)$ has the same norm as $5$, as well $(4-3i)$ and $5$.


问:
Closed form for limit of a sum $\lim\limits_{m\to\infty}(\ln m-4\sum_{n=1}^{m}\frac{(-1)^{n}}{2n+1}\sum_{k=1}^{2n}(-1)^{k}k\ln k)\approx-0.092$
I'm looking for a closed form of

$$\lim\limits_{m\to\infty}\left(\ln m-4\sum_{n=1}^{m}\frac{\left(-1\right)^{n}}{2n+1}\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k\right)\approx-0.092$$

for even $m$.

(For context this is a continuation of my ongoing work from this answer.)

Here's what I've tried so far. Wolfram alpha gives gives me

$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\zeta^{(1,0)}\left(-1,n+\frac{1}{2}\right)+2\zeta^{(1,0)}\left(-1,n+1\right)+n\ln2+3\ln A-\frac{1}{12}\ln2-\frac{1}{4}$$

From Wolfram's function site I get

$$\zeta^{(1,0)}(-1,x)=-\frac{1}{12}\left(-6x^{2}+18x-12\left(x-1\right)\ln\left(x-1\right)+12\ln A+6\left(x-1\right)\ln2\pi-13\right)+\psi^{(-2)}\left(x-1\right)$$

Combining these gets me

$$\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k=-2\psi^{(-2)}\left(n-\frac{1}{2}\right)+2\psi^{(-2)}\left(n\right)-\left(2n-1\right)\ln\left(n-\frac{1}{2}\right)+2n\ln n+n+n\ln2-\frac{1}{2}\ln2\pi+3\ln A-\frac{1}{12}\ln2-1$$

I could keep trying to find more expressions, but it feels like this is making things more complicated and possibly going to bring me back where I started.

Body:
We have
$$
\zeta ( - 1,n + 1) =  - \frac{1}{2}n^2  - \frac{1}{2}n - \frac{1}{{12}}
$$
and
\begin{align*}
\zeta ^{(1,0)} ( - 1,n + 1) = \zeta ^{(1,0)} ( - 1,n) & + n\log n =  - \frac{1}{4}n^2  + \left( {\frac{1}{2}n^2  + \frac{1}{2}n + \frac{1}{{12}}} \right)\log n + \frac{1}{{12}} \\ & - \int_0^{ + \infty } {\left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} e^{ - nt} dt}
\end{align*}
(cf. https://doi.org/10.1098/rspa.2017.0363). Noting that
$$
\zeta ^{(1,0)} \left( { - 1,n + \tfrac{1}{2}} \right) = \frac{{\log 2}}{2}\zeta ( - 1,2n + 1) + \frac{1}{2}\zeta ^{(1,0)} ( - 1,2n + 1) - \zeta ^{(1,0)} ( - 1,n + 1),
$$
we derive
\begin{align*}
 & \sum\limits_{k = 1}^{2n} {( - 1)^k k\log k}  = n\log (2n) + \frac{1}{4}\log n + 3\log A - \frac{{\log 2}}{{12}} \\ & + \int_0^{ + \infty } {\left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \left( {e^{ - 2nt}  - 4e^{ - nt} } \right)dt} .
\end{align*}
Thus, after some tedious calculations, we arrive at
\begin{align*}
& \mathop {\lim }\limits_{m \to  + \infty } \left( {\log (2m) - 4\sum\limits_{n = 1}^{2m} {\frac{{( - 1)^n }}{{2n + 1}}\sum\limits_{k = 1}^{2n} {( - 1)^k k\log k} } } \right)
\\ &
 = \mathop {\lim }\limits_{m \to  + \infty } \left[ {\log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n} } \right] \\ & + \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n \log n}}{{2n + 1}}}  + \left( {3\log A - \frac{{7\log 2}}{{12}}} \right)(4 - \pi )
\\ &
 + \int_0^{ + \infty } \!\! \left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \! \left( {16e^{\frac{t}{2}} \arctan (e^{ - \frac{t}{2}} ) - 4e^t \arctan (e^{ - t} ) - 12} \right)dt .
\end{align*}
Therefore, it remains to obtain a closed expression for
$$
\mathop {\lim }\limits_{m \to  + \infty } \left[ {\log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n} } \right].
$$
Some algebraic manipulation and Stirling's formula give
\begin{align*}
& \log (2m) - 2\sum\limits_{n = 1}^{2m} {( - 1)^n \log n}  =  - \log (2m) + 2\sum\limits_{n = 1}^{m - 1} {\log \left( {\frac{{2n + 1}}{{2n}}} \right)} \\ & =  - \log (2m) + 2\log \left( {\frac{{(2m - 1)!!}}{{(2m - 2)!!}}} \right) \to  \log \left( {\frac{2 }{\pi}} \right).
\end{align*}
Accordingly, a closed form expression for your limit is given by
\begin{align*}
&  \log \left( {\frac{2 }{\pi}} \right) + \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n \log n}}{{2n + 1}}}  + \left( {3\log A - \frac{{7\log 2}}{{12}}} \right)(4 - \pi )
\\ &
 + \int_0^{ + \infty } \!\! \left( {\frac{1}{{e^t  - 1}} - \frac{1}{t} + \frac{1}{2} - \frac{t}{{12}}} \right)t^{ - 2} \! \left( {16e^{\frac{t}{2}} \arctan (e^{ - \frac{t}{2}} ) - 4e^t \arctan (e^{ - t} ) - 12} \right)dt .
\end{align*}
The numerical value I got form this formula is $-0.09222744\ldots$. Using the formula
$$
\log n = \int_0^{ + \infty } {\frac{{e^{ - t}  - e^{ - nt} }}{t}dt} ,
$$
one can derive the quickly convergent integral representation
$$
\sum\limits_{n = 1}^\infty  { \frac{{( - 1)^n\log n}}{{2n + 1}}}  = \int_0^{ + \infty } {\frac{1}{t}\left( {\left( {\frac{\pi }{4} - 1} \right)e^{ - t}  - e^{\frac{t}{2}} \arctan  ( e^{ - \frac{t}{2}}  ) + 1} \right)dt} .
$$


问:
Replacing the value of a function with the value of the limit - is this a standard construction?
Consider a partial function $f : X \rightarrow Y$ where $X$ and $Y$ are topological spaces and $Y$ is Hausdorff. Note that, although the source of $f$ is $X$, the actual domain of $f$ is a (not necessarily proper) subset thereof.

Given this data, we can define a new partial function $f^\lambda : X \rightarrow Y$ by asserting

$$f^\lambda(a)=\lim_{x\rightarrow a}f(x)$$

and taking the domain of $f^\lambda$ to be the set of all $a \in X$ such that the above limit makes sense. This may include elements $a \in X$ that aren't in the domain of $f$.

Notice that $f^\lambda$ may have a larger or a smaller domain than $f$. Indeed, their domains may be incomparable. Furthermore, it is surprising to note that $f^\lambda$ needn't always be continuous - see Jonas Meyer's answer.

Anyway, my question: is $f^\lambda$ a standard construction? If so, what's the usual terminology/notation?

答:
You have a uniqueness problem. Consider the signum function

$$\operatorname{sgn}:\Bbb R\to\Bbb R:x\mapsto\begin{cases}
-1,&\text{if }x<0\\
0,&\text{if }x=0\\
1,&\text{if }x>0\;.
\end{cases}$$

You can construct $f^\lambda(x)\equiv-1$ with domain $(\leftarrow,0]$, but you can also construct $f^\lambda(x)\equiv1$ with domain $[0,\to)$. Each of these satisfies your definition, but they can’t be combined: there is no unique maximum function of the desired type.


问:
Normal Distribution - percentiles
i have this closed question,
I don't know how to solve it, I have only the answer but I would like to understand the solution and know how to solve them,
Thank you!

1.From a population that is normally distributed in which the 44th percentile is 44, a sample of 44 participants is taken. The probability that the sample's average will be higher than 44 is:

28%

51%

56%

84% (The right answer)

Thanks

答:
Let's say $X$ is a member of the population which is normally distributed with mean $\mu$ and standard deviation $\sigma$, neither of which are known.  But we know
$$P(X < 44) = 0.44$$
and $$X < 44 \iff \frac{X-\mu}{\sigma} < \frac{44-\mu}{\sigma}$$

and $(X-\mu)/\sigma$ has a Normal(0,1) distribution, so we know from tables of the distribution that
$$\frac{44-\mu}{\sigma} = -0.151$$
The average of the sample of size $44$, $\bar{X}$, has a mean of $\mu$ and a standard deviation of $\sigma / \sqrt{44}$.  Now
$$\bar{X} > 44 \iff \frac{\bar{X} - \mu}{\sigma / \sqrt{44}} > \left( \frac{44-\mu}{\sigma} \right) \sqrt{44}=-0.151 \sqrt{44} = -1.00$$
so $P(\bar{X} > 44) = P(Z> -1.00)$, where $Z$ has a Normal(0,1) distribution.  Refering to a table of the Normal distribution, we have
$$P(\bar{X} > 44) = \boxed{0.84}$$

Title: Normal Distribution - percentiles


问:
How do we justify that $|A+h\setminus A|\to 0$ as $|h|\to 0$?
Let $A\subset \Bbb R^d$ be a measurable set and note its translation $A+h= \{a+h: a\in A\}$ for $h\in \Bbb R^d$.

Let $|A|= \lambda(A)$ be the Lebesgue measure of $A$.
It is well known that $|A+h|= |A|$.

Assume $|A|<\infty$. How we show that $|A+h\setminus A|\to 0$ as $|h|\to 0$ and $|A\setminus A+h|\to 0$ as $|h|\to 0$?

How to apply the translation invariance of the Lebesgue measure here?

答:
Note that $\bigcup_n (A_n+h)=(\bigcup_n A_n)+h$

You must prove that $|A \cap A+h| \to |A|$ as $h \to 0$

Since $A$ has finite measure then $|A|+|A+h|-|A \cap A+h|=|A \cup (A+h)|$

So if we prove that $|A \cup (A+h)| \to |A|$ ,we are done.

Assume firstly that $A= Q$ a cube in $\Bbb{R}^d$.This is easy to prove.

If $A$ is open then $A=\bigcup_n I_n$,a disjoint union of dyadic cubes $I_n$

Now if $\epsilon>0$ then exists $N \geq 1$ such that $\sum_{n>N}^{\infty}|I_n \cup I_n+h|<\epsilon$ because $\sum_n|I_n \cup (I_n+h)| \leq 2|A|<+\infty$

Also $A \cup A+h \subseteq \bigcup_n(I_n \cup I_n+h)$

So $|A \cup A+h| \leq \sum_{n=1}^N |(I_n+h) \cup I_n| +\epsilon$

Thus $$\limsup_{h \to 0}|A \cup A+h| \leq \sum_{n=1}^N|I_n| +\epsilon \leq |A|+\epsilon$$

From this you can derive the conclusion since $|A \cup A+h| \geq |A|,\forall h \in \Bbb{R}^d$


问:
Proof by induction that a sequence is bounded
Let $x_1=1$ and $x_{n+1}=(2+x_n)^{1/2}$ for $n \in \mathbb{N}$. Prove by induction that the sequence $(x_n)$ is monotone and bounded, and determine its limit.

For the base case, we have that n=1, so $x_{1+1}=x_2=(2+1)^{1/2}=\sqrt(3)\ne x_1$, so how does this show the base case is satisfied? For the inductive step, I think what we have to prove is $x_{n+1}=(2+x_n)^{1/2}$ implies $x_{n+2}=(2+x_{n+1})^{1/2}$. I am not quite sure how to do the inductive step since I am bad at recursion. Also, how does this induction prove that the sequence is monotone and bounded? I don't see how induction proves this.

答:
Show that the sequence's bounded above by $\;2\;$. Here is the inductive step, assuming $\;x_n\le 2\;$ :

$$x_{n+1}=(2+x_n)^{1/2}\le(2+2)^{1/2}=2$$

and we're done.


问:
Properties of limits when dealing with functions and parentheses .
My calculus instructor recently mentioned some odd properties of limits that I don't recall ever seeing, and seem alien to me. He says that the following statements are allowed:

$$\lim_{n\to\infty}\sin\left(\frac{1}{n}\right)=\sin\left(\lim_{n\to\infty}\frac{1}{n}\right)=\sin(0)=0$$

What rules allow for this? The only ones I've been made aware of previously were the following:

$$\lim_{n\to\infty}cf(n)=c\lim_{n\to\infty}f(n)$$
$$\lim_{n\to\infty}f(n)\pm g(n)=\lim_{n\to\infty}f(n)\pm\lim_{n\to\infty}g(n)$$
$$\lim_{n\to\infty}f(n)\cdot g(n)=\lim_{n\to\infty}f(n)\cdot\lim_{n\to\infty}g(n)$$
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac{\lim_{n\to\infty}f(n)}{\lim_{n\to\infty}g(n)},\lim_{n\to\infty}g(n)\neq 0$$

答:
The relation $$\lim_{n\to\infty}\sin\left(\frac{1}{n}\right)=\sin\left(\lim_{n\to\infty}\frac{1}{n}\right)$$
is valid due to the continuity of $\sin$ function.</p>


问:
Frobenius method to solve $xy''+(1-x)y'-y=0$

Frobenius method to solve this equation: $$xy''+(1-x)y'-y=0$$

I am trying to find the solution for the DE but I'm stuck at the step where to find the equation to find the terms of the sequence.Attached is my working.Therefore,I would like to ask that how can I get the equation to calculate the terms?

working picture

答:
1.Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$.
Plugging in the appropriate series expansions, we get that
$$\sum_{n=2}^{\infty}n(n-1)a_n x^{n-1}+(1-x)\sum_{n=1}^{\infty}na_nx^{n-1}-\sum_{n=0}^{\infty}a_n x^n=0.$$
Hence, we conclude that
$$a_{n+1}=\frac{a_n}{n+1},\quad\forall\ n\in\mathbb{N} \implies a_n=\frac{a_0}{n!}.$$

Thus we get a solution
$$y_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x,$$
and it is easy to check this solution.

2.Let $y(x)=y_1(x)v(x)$, put it in the equation, then we have
$$xv''+(1+x)v'=0.$$
Denote $u(x)=v'(x)$ and assume $u\neq0$. Solve the 1st ODE
$$\frac{u'}{u}=-(1+\frac{1}{x})\implies u(x)=C_1\frac{e^{-x}}{x},\quad v(x)=C_1\int \frac{e^{-x}}{x}dx+C_2.$$
Thus all solutions are
$$y(x)=e^x(C_1\int \frac{e^{-x}}{x}dx+C_2).$$


问:
Understanding a lemma about prime numbers and divisors
In the MCS book from MIT (p. 170) they give the following lemma:

For any prime $p$ and any positive integers $n,x_1, x_2, \ldots,  x_n$, if $p \mid x_1\cdot x_2\cdot \ldots \cdot x_n$,
then $p\mid x_i$ for some $1\le i\le n$.

I have two questions about this lemma though...

(1) What even is the point of it? Consider the prime $p = 7$. $p$ only divides $7$ and $1$, so the if statement doesn't even really apply to primes $p$ unless you write $7 = 7 \cdot 1 \cdot 1$ or something similar, which brings me to (2)

(2) Why restrict it to primes $p$? It seems infinitely more useful to talk about composite numbers.

答: Composite numbers don't fulfill that condition. For example, $\;6\,\mid 3\cdot4\;$,  yet $\;6\nmid 3\;$ and $\;6\nmid 4\;$ ...


答:
Very often, that is the definition of a prime number. This is not always true for composites. For example, $4 | 2 \cdot 2$ but $4 \nmid 2$.

In your question, I think you confused "for all" and "for some". The statement says that $p | x_i$ for some i. In your example, $p | 7$ so that is enough.

Title: Understanding a lemma about prime numbers and divisors

Body:
In the MCS book from MIT (p. 170) they give the following lemma:

For any prime $p$ and any positive integers $n,x_1, x_2, \ldots,  x_n$, if $p \mid x_1\cdot x_2\cdot \ldots \cdot x_n$,
then $p\mid x_i$ for some $1\le i\le n$.

I have two questions about this lemma though...

(1) What even is the point of it? Consider the prime $p = 7$. $p$ only divides $7$ and $1$, so the if statement doesn't even really apply to primes $p$ unless you write $7 = 7 \cdot 1 \cdot 1$ or something similar, which brings me to (2)

(2) Why restrict it to primes $p$? It seems infinitely more useful to talk about composite numbers.


问:
Count the specific amount of steps in a sorting algorithm with given pseudo code
I need to count the specific amount of steps which are used by sorting algorithm pseudo code.

Here are some guide lines given:

a:=1 is considered 1 step

a:= b + c is 1 step

a:= b + c - d is 2 steps

A[i+j] = a + b is 2 steps

for i := 1 to k do is k(2+1) steps

My given code is as follows

A[1 ; n]

```<code>1//m := n - 1
2//for i := 1 to m do
3//   min := A[i]
4//   k := i
5//   for j := i + 1 to n do
6//       if A[j] < min then
7//           min := A[j]
8//           k := j
9//   A[k] := A[i]
10//  A[i] := min
</code>```

I marked the lines of the code because I have quite a few of them figured out, however I need some help with some of them.

1// 1 step

2// 2(m+1) = 2n

3// m = n-1

4// m = n-1

5//  $\sum_{j=2}^n j$   + 2(n-1)  I am not sure about this one.

6//  $\sum_{j=2}^n j$  I am not sure about this one as well.

I really can't figure out how many steps the 7th or 8th lines should be.

9// m = n-1

10// m = n-1

I really need help with lines 5-8 because I can figure out how many steps these of code these lines would take.

Thank you in advance for the help!

答:
The outer $\texttt{for}$ loop

```<code>for i := 1 to m do
</code>```

amounts to
\begin{align*}
2\sum_{i=1}^m 1+2=2m+2
\end{align*}
operations.

The inner $\texttt{for}$ loop

```<code>for j := i+1 to n do
</code>```

amounts to
\begin{align*}
2\sum_{j=i+1}^n 1+2&=2\sum_{j=1}^{n-i} 1+2\\
&=2(n-i)+2=2(n-i+1)\tag{1}
\end{align*}
operations.

Both $\texttt{for}$ loops

```<code>for i := 1 to m do
    for j := i+1 to n do
</code>```

amount according to (1) to
\begin{align*}
\color{blue}{2\sum_{i=1}^m}&\color{blue}{\left(2\sum_{j=i+1}^n 1+2\right)+2}\\
&=2\sum_{i=1}^m2\left(n-i+1\right)+2\\
&=4n\sum_{i=1}^m 1 - 4\sum_{i=1}^m i+4\sum_{i=1}^m 1+2\\
&=4nm-4\frac{m(m+1)}{2}+4m+2\\
&\,\,\color{blue}{=2m(2n-m+1)+2}
\end{align*}
operations.

Body:
The outer $\texttt{for}$ loop

for i := 1 to m do

amounts to
\begin{align*}
2\sum_{i=1}^m 1+2=2m+2
\end{align*}
operations.

The inner $\texttt{for}$ loop

for j := i+1 to n do

amounts to
\begin{align*}
2\sum_{j=i+1}^n 1+2&=2\sum_{j=1}^{n-i} 1+2\\
&=2(n-i)+2=2(n-i+1)\tag{1}
\end{align*}
operations.

Both $\texttt{for}$ loops

for i := 1 to m do
    for j := i+1 to n do

amount according to (1) to
\begin{align*}
\color{blue}{2\sum_{i=1}^m}&\color{blue}{\left(2\sum_{j=i+1}^n 1+2\right)+2}\\
&=2\sum_{i=1}^m2\left(n-i+1\right)+2\\
&=4n\sum_{i=1}^m 1 - 4\sum_{i=1}^m i+4\sum_{i=1}^m 1+2\\
&=4nm-4\frac{m(m+1)}{2}+4m+2\\
&\,\,\color{blue}{=2m(2n-m+1)+2}
\end{align*}
operations.


答:
The general form is $\lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty}a_n)$.

You can use this if the limit $\lim_{n \to \infty} a_n = a$ exists and $f$ is continuous at $a$. (Which is the case in your example).

In fact, this is one of the equivalent definitions of continuity at a point for real valued functions.

Title: Properties of limits when dealing with functions and parentheses .

Body:
My calculus instructor recently mentioned some odd properties of limits that I don't recall ever seeing, and seem alien to me. He says that the following statements are allowed:

$$\lim_{n\to\infty}\sin\left(\frac{1}{n}\right)=\sin\left(\lim_{n\to\infty}\frac{1}{n}\right)=\sin(0)=0$$

What rules allow for this? The only ones I've been made aware of previously were the following:

$$\lim_{n\to\infty}cf(n)=c\lim_{n\to\infty}f(n)$$
$$\lim_{n\to\infty}f(n)\pm g(n)=\lim_{n\to\infty}f(n)\pm\lim_{n\to\infty}g(n)$$
$$\lim_{n\to\infty}f(n)\cdot g(n)=\lim_{n\to\infty}f(n)\cdot\lim_{n\to\infty}g(n)$$
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac{\lim_{n\to\infty}f(n)}{\lim_{n\to\infty}g(n)},\lim_{n\to\infty}g(n)\neq 0$$


问:
Satisfying two universal quantification propositions

Find non empty $X$ and $Y$ that satisfy the following: $\forall x \in X \;\exists y\in Y p(x,y) \;\; and \;\; \forall y \in Y \; \exists x\in X \; not \; p(x,y)$, where $p(x,y)$ is defined as the propositional function $x-y < 10$.

I cannot think of an $X$ and $Y$ that could satisfy such a condition. If for all $y$ in $Y$ there exists some $x$ in $X$ such that $not \; p(x,y)$, then it must be the case that there exists some $x \in X$ such that for all $y \in Y$ not $p(x,y)$, no? But, this is the negation of the first conjunct.

答: Let both $X$ and $Y$ be the set of all real numbers. Check that this works.  Then go back to your reasoning for why such $X$ and $Y$ lead to a contradiction, and pinpoint where that reasoning goes wrong in this example. (This pinpointing is not strictly required by the question, but I think it's important to prevent similar errors in future questions.)


答:
What your teacher said is correct. A standard theorem, although more difficult than the four other properties of limits you listed is:

Theorem: If $f$ and $g$ are functions such that $f$ is continuous at $a$ and
  $g(x)\rightarrow a$ as $x\rightarrow b$ then:

$$\lim_{x\to b} f(g(x))=f\left(\lim_{x\to b} g(x)\right)=f(a).$$


问:
Let $f(x)=2|x|+1$ if $x\in\mathbf{Q}$ and $f(x)=0$ if not. Show that $f$ is not Riemann integrable
Let $f(x)=2|x|+1$ if $x\in\mathbf{Q}$ and $f(x)=0$ if $x\in \mathbf{R}/\mathbf{Q}$. Show that $f$ is not Riemann integrable on $[-2,3]$. I would like to have a feedback on my proof and to know if it holds, please.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$. We know as well that $\int_{-2}^{0}f(x)+\int_{0}^{3}f(x)=\int_{-2}^{3}f(x)$. So, if we show that $f$ is not integrable on $[-2,0]$ or on $[0,3]$, we could conclude that $f$ is not integrable on $[-2,3]$. In the following proof I will work with the interval $[0,3]$.

By density of irrational numbers, the lower Darboux sum is always equal to $0$ whatever the partition. So, we have to shiw that $\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\epsilon$. Let $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$. We have then the following:

$\overline{S}_{\sigma}(f)=\sum_{i=0}^{n}M_i(x_{i+1}-x_i)=\sum_{i=0}^{n}(2x_{i+1}+1)(x_{i+1}-x_i)\ge \int_{0}^{3}2x+1=12$.

Thus, there is no partition  such that $\overline{S}_{\sigma}(f)<\epsilon$ for $0<\epsilon<12$. We conclude that $f$ is not intergable on $[0,3]$ and so not integrable on $[-2,3]$.

答: The computations are correct (although I think that it would be more interesting to do it without computing the Riemann integral of another function). However, there is a problem when you state that$$\int_{-2}^3f(x)\,\mathrm dx=\int_{-2}^0f(x)\,\mathrm dx+\int_0^3f(x)\,\mathrm dx.$$Your goal is to prove that $f$ is not integrable, but here you are acting as if it was. You could simply clam that if $f$ is not Riemann integrable on a subinterval of $[-2,3]$, then it is also not Riemann integrable on $[-2,3]$.


问:
intersection of analytic set : codimension?
Hello. Let $A$ and $B$ be analytic sets of pure dimension in an analytic set X of $\mathbb{C }^{n}$. Suppose $A$ is a local complete intersection (i.e. for every point $a$ of $A$, there exists $\Omega$ an open neighbourhood of $a$ such that  $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorphs distincts and $p = codim_{X}(A)$).
Then, if $A \cap B \neq \emptyset$ why do we have $codim_{X}(A \cap B) \leq codim_{X}(A) + codim_{X}(B)$?

I tried to solve this by induction on $dim(A)$; here is the proof I first made (I realised afterwards the step $*$ is false). It's trivial if $dim(A) = 1$. Then, if $dim(A) > 1$, I choose $a \in A \cap B$ and I wright $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorph distincts and $p = codim_{X}(A)$ and $B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p'}(x) = 0\}$ where $g_{i}$ are holomorph and distincts. Since the problem is located in $a$, we will wright $dim_{a} = dim$ and the same for $codim$. Notice that I have $p' \leq codim_{X}(B) $. We can wright $A \cap B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\}$ where $p'' \leq p'$ is such that there is no repetition between $g_{j}$ and $f_{i}$. Now let's have a look at : $A' := \{x \in \Omega | g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\} \subset A$. We wright $B' := \{x \in \Omega | g_{1}(x) = ... = g_{p''-1}(x) = 0\}$. By looking$*$ the local irreductible components of $A$( which are all of dimensions $dim(A)$ since $A$ is of pure dimension) in $a$ we have $A'$ is of dimension $dim(A) - 1$ so of codimension $codim_{X}(A) + 1$. Thus, $A'$ is a local complete intersection and so by induction $codim_{X}(A \cap B) = codim_{X}(A' \cap B') \leq codim_{X}(A') + codim_{X}(B') = codim_{X}(A) + 1 + codim_{X}(B')$. But $1 + codim_{X}(B')$ doesn't necessary have to be $\leq codim_{X}(B)$ (imagine $B$ is an irreductible component of $B'$ in $a$).

$*$Here I use the (6.2) Theorem page 106 of this curse https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf . But it dosen't work since $g_{p''}$ can be constant on some irreductible components of $A$ in $a$.

Thank you for your help.

I wish you a good day.

答: I've got it : in the case where $A \cap B \subset B_{sing}$, I choose $a \in (A \cap B)_{reg}$ (which it's always possible since this set is dense in $A \cap B$) and I suppose $codim_{X, a}(A \cap B) > codim_{X, a}(A) + codim_{X, a}(B)$. Then, by corolary $(4.33)$ we know that $\mathcal{J}_{A \cap B, a}$ is generated by at least $codim_{X, a}(A) + codim_{X, a}(B) + 2$ elements as a $\mathcal{O}_{B, a}$ module which is absurde because $\mathcal{J}_{A \cap B, a}$ (which is prime since $A\cap B$ is regular in $a$) is generated as a $\mathcal{O}_{a}(\mathbb{C}^{n})$ (and so as a $\mathcal{O}_{B, a}$ module) by $codim_{X, a}(A) + codim_{X, a}(B)$ elements by using and applying the (6.3) Theorem page 106 to  $A \cap B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\}$.


问:
Infinite intersection with real numbers
I cannot wrap my head around something I did read in a math book. Maybe someone can tell me at which point I went wrong.

Background

The author defines the infinite intersection as follows:

$$\bigcap_{n \in M} A_{n} := \{ x\,|\,x \in A_{n}\:\text{ for all } n \in M \}$$

Later on the author brings up an example:

$$
M = \mathbb{N}, A_{n} := \Big\{ x \in \mathbb{R}\, \Big| \,0 < x < \frac{1}{n} \Big\}$$

And for this example the author states, that

$$
\bigcap_{n \in M} A_{n} = \emptyset
$$

Problem

I do not understand why the infinite intersection results in the empty set.

My thoughts

$n$ ranges from $1$ to infinity (never reaching infinity) $\implies$ No $A_{n}$ should be empty, as it should contain at least $\frac{1}{n+1}$

$A_n$ is the set of real numbers between $0$ and $\frac{1}{n}$ (both exclusive)

$A_{n} \subsetneq A_{n-1}$ should be true

So for any $j$, $k\in \mathbb{N}$ with $k \leq j$: $A_k$ contains at least $\frac{1}{j+1}$. Therefore the intersection of $A_n$ should contain at least this lowest element.

Example:

$A_1$ from $0$ to $1$

$A_2$ from $0$ to $\frac{1}{2}$

$A_3$ from $0$ to $\frac{1}{3}$

All of them contain the element $\frac{1}{4}$.

So basically my problem is, that I do not understand where I did make the mistake. Since my thoughts conflict with the statement, that the infinite intersection equals the empty set. Because in my imagination it contains at least one element which goes towards 0 but never reaches 0.

Thanks in advance for your time and effort.

答: Yes, but $A_5$ doesn't contain $\frac14$.  For any $x>0$ there is an integer $N$ such that $x>\frac1N$.  Then for $n>N$, $x\notin A_n$.  Thus, there is no number in all of the $A_n$.

Title: Infinite intersection with real numbers


答:
Let's suppose there was at least one number $x$ in the infinite intersection
$$
x \in \bigcap_{n \in M}A_n
$$
by the Archimedean property of real numbers, there is some natural number $N$ such that
$$
\frac{1}{N} < x
$$
and so ${x \notin A_N}$. Meaning we are forced to conclude
$$
x \notin \bigcap_{n \in M}A_n
$$
$x$ was arbitrary. This argument shows that the infinite intersection could not possibly have any elements inside, and so
$$
\bigcap_{n \in M}A_n = \emptyset
$$

TL;DR: for every number you believe is in the intersection, you can find some $A_N$ such that this number is not in $A_N$. Hence the intersection must be empty.

EDIT: maybe another way to think about it is this. In the above answer, I gave you a proof that the infinite intersection must be empty, and in the comments I tried to sort of "play to your intuition". Instead, let's look at how we might go about defining some of these things. Expressions such as
$$
\bigcap_{n=1}^{\infty}\left(0,\frac{1}{n}\right)
$$
(which is the same as in your question) don't have any "self-evident" answer. We are required to expand upon what we mean by this sort of expression in order to try and conclude a "reasonable" answer, just like with infinite sums, where we are required to define limits and things to try to make sense of it. If we take everything we know about real numbers as "given" (including, as my proof above uses, the Archimedean property) - and expand the infinite intersection as meaning "any element in the above intersection must belong to each set in the intersection", then the proof I did is correct, and the intersection is empty. Since as we said, for any number $x$, we can always find a set ${\left(0,\frac{1}{N}\right)}$ in the intersection such that $x$ is not in this set, and so couldn't be in the intersection.

If the answer was non-empty, it would be a contradiction to my assumptions. Meaning ${(1)}$ what we know about the behaviour of real numbers is broken, or ${(2)}$ I expanded what we meant by the infinite intersection incorrectly. And I don't think there is anything wrong with either assumption. Of course, you could disagree with me, and use alternative definitions - and then you could arrive at different answers. But if you follow the assumptions I made, the only logically sound result is ${\emptyset}$. You could override this if you like, and say "the result is non-empty, there is some number in there" - but then things become inconsistent, since we would start contradicting ourselves.

I think should also help explain your confusion in the comments, as you say

I can find ${A_N}$ so that ${x_1 \notin A_N}$, but I can also find ${x_2 \in A_N}$. Why is one stronger than the other?

and the answer is because the latter is not how we define an infinite intersection. Not only that, but I don't think this would really be helpful at all in trying to define what we are trying to represent when we say ${\bigcap_{n=1}^{\infty}\left(0,\frac{1}{n}\right)}$. Sure, ${x_2}$ is in ${A_N}$ - but intuitively we can then find a set for a bigger ${N'>N}$ so that ${x_2}$ is once again not there. So that would not be a helpful definition.

Also, I want to further clarify my example in the comments of
$$
1 + 2 + 3 + \dots
$$
I am definitely not saying infinity is a number when I say "we would call the result of this ${\infty}$" <- infinity is not a number, and in the comment I do further state "this is not a positive integer". The ONLY point I was trying to make with this example is that if we follow our definition of the infinite sum, and try ${1 + 2 + 3 + \dots}$ - we end up with something that couldn't possibly be a number. In the same way - we followed our definition of an infinite intersection of a bunch of sets that were subsets of eachother, and ended up with something that was not one of the sets (none of the ${A_N}$ are ${\emptyset}$). Perhaps this example was not helpful and could add more confusion.

Title: Infinite intersection with real numbers

Body:
I cannot wrap my head around something I did read in a math book. Maybe someone can tell me at which point I went wrong.

Background

The author defines the infinite intersection as follows:

$$\bigcap_{n \in M} A_{n} := \{ x\,|\,x \in A_{n}\:\text{ for all } n \in M \}$$

Later on the author brings up an example:

$$
M = \mathbb{N}, A_{n} := \Big\{ x \in \mathbb{R}\, \Big| \,0 < x < \frac{1}{n} \Big\}$$

And for this example the author states, that

$$
\bigcap_{n \in M} A_{n} = \emptyset
$$

Problem

I do not understand why the infinite intersection results in the empty set.

My thoughts

$n$ ranges from $1$ to infinity (never reaching infinity) $\implies$ No $A_{n}$ should be empty, as it should contain at least $\frac{1}{n+1}$

$A_n$ is the set of real numbers between $0$ and $\frac{1}{n}$ (both exclusive)

$A_{n} \subsetneq A_{n-1}$ should be true

So for any $j$, $k\in \mathbb{N}$ with $k \leq j$: $A_k$ contains at least $\frac{1}{j+1}$. Therefore the intersection of $A_n$ should contain at least this lowest element.

Example:

$A_1$ from $0$ to $1$

$A_2$ from $0$ to $\frac{1}{2}$

$A_3$ from $0$ to $\frac{1}{3}$

All of them contain the element $\frac{1}{4}$.

So basically my problem is, that I do not understand where I did make the mistake. Since my thoughts conflict with the statement, that the infinite intersection equals the empty set. Because in my imagination it contains at least one element which goes towards 0 but never reaches 0.

Thanks in advance for your time and effort.


问:
Find number of terms and coefficient of $x^5$ in $(1+x+x^2)^7$
I tried using the multinomial theorem to find general term ie.

$$T_n =\frac{7!}{a!b!c!} 1^a \cdot x^b \cdot x^{2c} $$

Now $b+2c=5$ and $a+b+c=7$

The second equation can be interpreted as distributing $7$ things in $3$ groups, ie $\binom {9}{2}$ but I don’t know what to do with the second condition, and how there solutions will overlap

How should I proceed?

答:
The simplest way to solve the problem is finding the partitions of the number $5$ into positive parts, each not exceeding the power of the polynomial - in our case it is $2$. There are only three such partitions:
$$
(1,1,1,1,1),(1,1,1,2),(1,2,2)
$$
of the length $5,4,3$, respectively.  (In fact looking for the partitions we should have restricted the number of parts to at most $7$, but since $5\le7$ this restriction played no role.) To make the length equal to 7 we have to fill the rest positions with zeros, so that the final result is:
$$
c_5=\frac{7!}{2!5!0!}+\frac{7!}{3!3!1!}+\frac{7!}{4!1!2!}.
$$

(The factorials in the denominator are those of the counts of $0,1,2$, respectively.)


问:
Norm of tensors on a Hermitian manifold
On a Riemannian manifold $(M, g)$, given a $(l, k)$ tensor $T  = T_{m_1 \ldots m_{l}} dx^{1} \otimes \ldots \otimes dx^{l}$, we know its norm is given by $g^{m_1 n_1} \ldots g^{m_l n_l} T_{m_1 \ldots m_{l}} T_{n_1 \ldots n_{l}}$. If instead $(M, g)$ is a Hermitian manifold with an $(l, l)$ tensor $T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} dz^{m_1} \otimes d \bar z^{p_1} \otimes \ldots \otimes dz^{m_l} \otimes d \bar z^{p_l}$, the norm then will be $g^{m_1 \bar{q_1}} \ldots g^{m_n \bar{q_n}} T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} \overline{T_{n_1 \bar {q_1} \ldots n_{l} \bar q_{l}}}$.

This confuses me because if we are to imitate the definition in the real case we would want $g^{m_{1} n_{1}}g^{\bar p \bar q_1} \ldots T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} T_{n_1 \bar {q_1} \ldots n_{l} \bar q_{l}}$. However apparently this does not make sense beacuse $g^{m_1 n_1} = 0$.

Maybe in order to compare it with the real case one needs to transform the coordinates into real coordinates?

答:
Part of your confusion may be down to some of the definitions involved, so let me clear up a few things.

"On a Riemannian manifold $(M,g)$, given a $(l,k)$ tensor $T = T_{m_1\dots m_l} dx^1\otimes\cdots\otimes dx^l$"

That is not a $(l, k)$-tensor, that is a $(0, l)$-tensor. Likewise, on a complex manifold, $T_{m_1 \bar {p_1} \ldots m_{l} \bar p_{l}} dz^{m_1} \otimes d \bar z^{p_1} \otimes \cdots \otimes dz^{m_l} \otimes d \bar z^{p_l}$ is not an $(l, l)$-tensor, but rather a $(0, 2l)$-tensor.

If $T = T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\cdots\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\cdots\otimes d\bar{z}^{p_l}$ is a $(0, k + l)$ tensor on a hermitian manifold, then the square of the norm of $T$ is given by

$$g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g^{q_1\bar{p_1}}\cdots g^{q_l\bar{p_l}}T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{T_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}}.$$

This follows from the definition of the inner product $\langle\cdot, \cdot\rangle$ on $(0, k + l)$-tensors induced by the hermitian metric. You're viewing a hermitian metric as a complex bilinear pairing $g$ on $TM\otimes_{\mathbb{R}}\mathbb{C}$, so one obtains a metric on $T^{1,0}M$ by considering the pairing $(v, w) \mapsto g(v, \overline{w})$; note, that $g(\overline{w}, v) = g(v, \overline{w})$ as $g$ is symmetric. The pairing $g$ induces pairings on $T^*M\otimes_{\mathbb{R}}\mathbb{C}$ and $\bigotimes^{k+l}_{\mathbb{C}}T^*M\otimes_{\mathbb{R}}\mathbb{C}$ which I will also denote by $g$. With this in mind, we have

\begin{align*}
& \small{\langle T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}dz^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{n_k}\otimes d\bar{z}^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{q_l}\rangle}\\
=&\ \small{g(T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, \overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}dz^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{n_k}\otimes d\bar{z}^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{q_l}})}\\
=&\ \small{g(T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}dz^{m_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{m_k}\otimes d\bar{z}^{p_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{p_l}, \overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}}d\bar{z}^{n_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes d\bar{z}^{n_k}\otimes dz^{q_1}\otimes\!\cdot\!\cdot\!\cdot\!\otimes dz^{q_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g(dz^{m_1}, d\bar{z}^{n_1})\cdots g(dz^{m_k}, d\bar{z}^{n_k})g(d\bar{z}^{p_1}, dz^{q_1})\cdots g(d\bar{z}^{p_l}, dz^{q_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g(dz^{q_1}, d\bar{z}^{p_1})\cdots g(dz^{q_l}, d\bar{z}^{p_l})}\\
=&\ \small{T_{m_1\dots m_k\bar{p_1}\dots\bar{p_l}}\overline{S_{n_1\dots n_k\bar{q_1}\dots\bar{q_l}}} g^{m_1\bar{n_1}}\cdots g^{m_k\bar{n_k}}g^{q_1\bar{p_1}}\cdots g^{q_l\bar{p_l}}.}
\end{align*}

Title: Norm of tensors on a Hermitian manifold