问:
Intuitive understanding of Hall's marriage theorem
Could someone help me understand Hall's marriage theorem? I don't get it either intuitively or formally, and I certainly don't understand the proof. I do not get the number of neighbours part, and I also don't get the proof in Rosen's textbook of j and j+1. And I also don't get the explanation for the example in this site when I tried to search for an explanation for dummies. I'm not sure how you can match only 2 teams to 3 days when each day has exactly 3 winners for $m=3$ in a round-robin tournament.

Notation is certainly necessary, but if you could, please match them with English explanations, thank you.

答:
Imagine you have two dudes, Bill and Ted (set A), who want to marry a lady, and the goal is for each guy to marry one lady (a matching on A). If a lady is already married to one dude, she can't marry the other.

Of the dudes, there are exactly 4 possible subsets: {}, {Bill}, {Ted}, {Bill, Ted}, i.e., the empty set, only Bill, only Ted, and Bill and Ted. Let's take the only Bill set first. In order for him to marry someone, he should want to marry at least 1 Lady, i.e. he won't marry anyone he doesn't want to. In that case, N(X) = 1 = |X| = 1 (if he only desires 1 lady). He could desire two, in which case N(X) = 2 > |X| = 1, and Hall's theorem holds. The same goes for the only Ted set.

What wouldn't work, is if, say, Bill didn't want to marry any available ladies. In that case, |X| = 1, but N(X) = 0, leading us to N(X) $\ngeq$ |X|, which means you cannot find a "matching" that would make Bill and Ted each marry a lady.

Of course, if we take the Bill and Ted set, and say Ted would be ok to marry either one of 2 ladies, we would get |X| = 2 = N(X) = 2, and that's all fine and well. But because the Bill set doesn't fulfill the given condition, we know that there cannot be a matching.

Though it becomes increasingly difficult to imagine and time consuming to verify manually if |A| > 3, this holds for any number of elements in A.


问:
Summing $\sum_{k=1}^n \binom{2n-2k}{n-k}\frac{H_{2k} - 2H_{k}}{2n-2k-1} \binom{2k}{k}$
The sum in the question has a nice closed-form:
$$
\sum_{k = 1}^{n}\binom{2n - 2k}{n - k}
\frac{H_{2k} - 2H_{k}}{2n-2k-1}\binom{2k}{k} =
\frac{1}{n}\left[4^{n} - 3\binom{2n-1}{n}\right],$$
where $H_k$ is the harmonic number $H_n = \sum_{k=1}^n 1/k.$

My proof is excruciating:

It involved the specialization of a two-variable generating function ( $x$ and $y$ take different functions of $t$ ) so that I could use an identity that seems to have no mere
mortal-constructed proof ( a variant of the WZ method was used, I believe ).

In the process of the proof, I noticed that the expression in big parentheses on the RHS is a sequence of integers, and I searched the OEIS to find $A213119$.

As logarithms are involved, it was a matter of talking the arguments into the logarithm to find that my generating function was indeed equal to implied by the right hand side of the equation.

I seek a simpler, more direct proof.

A contributor wanted to know where the identity originated.  It originated as an attempt to prove the very pretty

$$
\sum_{k = 1}^{n}\binom{n}{k}^{2}H_{k} =
\binom{2n}{n}\left(\,{2H_{n} - H_{2n}}\,\right)
$$

It may be impossible not to use this in any attempted proof, but hopefully, the specialization of a two-variable generating function can be avoided.

答:
Here is another variation.

We observe the left-hand side of OPs identity is the coefficient of a Cauchy-product of series and we start with separating the factors. We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left(\sum_{k=1}^n\binom{2n-2k}{n-k}\frac{H_{2k}-H_k}{2n-2k-1}\binom{2k}{k}\right)z^n}\\
&=\left(\sum_{m=1}^\infty\binom{2m}{m}\left(H_{2m}-2H_m\right)z^m\right)
\left(\sum_{l=0}^\infty\binom{2l}{l}\frac{1}{2l-1}z^l\right)\tag{1}
\end{align*}

We will make use from the generating functions of the central binomial coefficients and of the closely related Catalan numbers.
\begin{align*}
\sum_{n=0}^\infty\binom{2n}{n}z^n&=\frac{1}{\sqrt{1-4z}}\\
C(z)=\sum_{n=0}^\infty\binom{2n}{n}\frac{1}{n+1}z^n&=\frac{1-\sqrt{1-4z}}{2z}=\frac{2}{1+\sqrt{1-4z}}\tag{2}
\end{align*}

We start with the easy part of (1) and consider the right-hand series. We obtain
\begin{align*}
\color{blue}{\sum_{l=0}^\infty}\color{blue}{\binom{2l}{l}\frac{1}{2l-1}z^{l}}
&=-1+2z\sum_{l=1}^\infty\binom{2l-1}{l-1}\frac{1}{2l-1}z^{l-1}\tag{3}\\
&=-1+2z\sum_{l=0}^\infty\binom{2l}{l}\frac{1}{l+1}z^l\tag{4}\\
&=-1+2z\left(\frac{1-\sqrt{1-4z}}{2z}\right)\tag{5}\\
&\,\,\color{blue}{=-\sqrt{1-4z}}\tag{6}
\end{align*}
and observe we have in (5) the generating function $C(z)$ of the Catalan numbers in disguise.

Comment:

In (3) we separate the first term and use $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

In (4) we use $\binom{p}{q}=\binom{p}{p-q}$ and shift the index to start with $l=0$.

In (5) we use $C(z)$ and simplify afterwards.

This was the warm-up section. Now we consider the left-hand series of (1). But prior to this we recall a nice polynomial identity in the following intermezzo.

Intermezzo: We introduce generalised harmonic numbers $H_n(x)=\sum_{j=1}^n\frac{1}{j+x}$ and consider the following polynomial in $x$ and $n,k$ non-negative integers:
\begin{align*}
\frac{d}{dx}\binom{n+x}{k}&=\frac{1}{k!}\frac{d}{dx}\prod_{j=0}^{k-1}(n+x-j)\\
&=\frac{1}{k!}\sum_{l=0}^{k-1}\prod_{{j=0}\atop{j\ne l}}^{k-1}(n+x-j)\\
&=\binom{n+x}{k}\sum_{l=0}^{k-1}\frac{1}{n+x-l}\\
&=\binom{n+x}{k}\left(H_n(x)-H_{n-k}(x)\right)\tag{7}\\
\end{align*}
Evaluating the identity (7) at $x=0$ results in
\begin{align*}
\left.\left(\frac{d}{dx}\binom{n+x}{k}\right)\right|_{x=0}=\binom{n}{k}\left(H_n-H_{n-k}\right)\tag{8}
\end{align*}

Now we are ready to tackle the left-hand series of (1). We obtain
\begin{align*}
\color{blue}{\sum_{m=1}^\infty}&\color{blue}{\binom{2m}{m}\left(H_{2m}-2H_m\right)z^m}\\
&=\sum_{m=1}^\infty\binom{2m}{m}\left(H_{2m}-H_m\right)z^m-\sum_{m=1}^\infty\binom{2m}{m}H_mz^m\\
&=\left.\left(\frac{d}{dx}\left(\sum_{m=1}^\infty\binom{2m+x}{m}z^m\right)\right)\right|_{x=0}
-\sum_{m=1}^\infty\binom{2m}{m}H_mz^n\tag{9}\\
&=\left.\left(\frac{d}{dx}\left(\frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^x\right)\right)\right|_{x=0}
-\sum_{m=1}^\infty\binom{2m}{m}H_mz^n\tag{10}\\
&=\frac{1}{\sqrt{1-4z}}\left.\left(\frac{d}{dx}\left(C(z)\right)^x\right)\right|_{x=0}
-\sum_{m=1}^\infty\binom{2m}{m}H_mz^n\\
&=\frac{1}{\sqrt{1-4z}}\left.\left(\left(C(z)\right)^x\ln C(z)\right)\right|_{x=0}
-\sum_{m=1}^\infty\binom{2m}{m}H_mz^n\tag{11}\\
&=\frac{1}{\sqrt{1-4z}}\ln C(z)-\frac{2}{\sqrt{1-4z}}\ln\left(\frac{1+\sqrt{1-4z}}{2\sqrt{1-4z}}\right)\tag{12}\\
&=\frac{1}{\sqrt{1-4z}}\ln C(z)+\frac{2}{\sqrt{1-4z}}\ln\left(C(z)\sqrt{1-4z}\right)\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4z}}\left(3\ln C(z)+2\ln\sqrt{1-4z}\right)}\tag{13}
\end{align*}

Comment:

In (9) we apply the identity (8) from the intermezzo.

In (10) we use a generalised generating function of the central binomial coefficients. A derivation of it can be found for instance at this MSE post.

In (11) we make the derivation.

In (12) we use the identity (9) stated in this paper by H. Chen, which is certainly a formidable source to derive many different variations of proofs of the current problem. We do some simplifications in the following lines.

Putting (13) and (6) in (1) we obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left(\sum_{k=1}^n\binom{2n-2k}{n-k}\frac{H_{2k}-H_k}{2n-2k-1}\binom{2k}{k}\right)z^n}\\
&=\left(\sum_{m=1}^\infty\binom{2m}{m}\left(H_{2m}-2H_m\right)z^m\right)
\left(\sum_{l=0}^\infty\binom{2l}{l}\frac{1}{2l-1}z^l\right)\\
&=\left(\frac{1}{\sqrt{1-4z}}\left(3\ln C(z)+2\ln\sqrt{1-4z}\right)\right)\left(-\sqrt{1-4z}\right)\\
&\,\,\color{blue}{=\ln\frac{1}{1-4z}-3\ln C(z)}\tag{14}
\end{align*}

Finally we calculate the generating functions from the right hand side of OP's identity.

We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^\infty}&\color{blue}{\frac{1}{n}\left(4^n-3\binom{2n-1}{n}\right)z^n}\\
&=\sum_{n=1}^\infty\frac{1}{n}\left(4z\right)^n+\sum_{n=1}^\infty\binom{2n-1}{n-1}\frac{1}{n}z^n\\
&=\ln\frac{1}{1-4z}-\frac{3}{2}\sum_{n=1}^\infty\binom{2n}{n}\frac{1}{n}z^n\\
&\,\,\color{blue}{=\ln\frac{1}{1-4z}-3\ln C(z)}\tag{15}
\end{align*}
in accordance with (14) and the claim follows.

The term $\ln C(z)$ in (15) follows from (2) by division by $z$ and integrating both sides of the generating function of the central binomial coefficients. Note, this is also stated in Chen's paper shortly after corollary 4.

ContentLicense: CC BY-SA 4.0


问:
Do there exist totally ordered sets with the 'distinct order type' property that are not well-ordered?
Define that the order type of an element $x$ in a totally ordered set $X$ is the order type of $\{w \in X\mid w < x\}$. Under this definition, distinct elements of a well-ordered set have distinct order types. Thus, every element of a well-ordered set has a 'canonical name' that uniquely specifies that element, namely its order type.

Anyway, lets call this the 'distinct order type' property, or simply the DOT property. So we'll say that a totally ordered set has the DOT property iff every two distinct elements of that set have distinct order types.

Now clearly, not every totally ordered set has the DOT property. For instance, the elements of $\mathbb{R}$ all have the same order type, so $\mathbb{R}$ certainly doesn't have the DOT property. The same goes for $\mathbb{Z}.$ My question is, do there exist totally ordered sets with the DOT property that are not well-ordered?

答:
There are even densely ordered sets with the DOT property.

Let $\langle X,\le\rangle$ be a linear order, and suppose that there are distinct $x,y\in X$ such that $(\leftarrow,x]\cong(\leftarrow,y]$. Without loss of generality assume that $x<y$, and let $f:(\leftarrow,y]\to(\leftarrow,x]$ be an isomorphism. Clearly

$$h:X\to X:z\mapsto\begin{cases}
f(z),&\text{if }z\le y\\
z,&\text{if }y<z
\end{cases}$$

is an isomorphism of $X$ into itself. Call a linear order rigid if it admits no non-trivial isomorphism into itself; it follows that every rigid linear order has the DOT property. The following theorem is proved in Brian M. Scott, ‘A characterization of well-orders’, Fundamenta Mathematicæ, Vol. $111$ ($1981$), Nr. $1$, pp. $71$-$76$, freely available here.

Theorem. Let $\kappa\ge\omega$ be such that $2^{<\kappa}=\kappa$. Then there is a rigid dense linear order of power $2^\kappa$.

Such a linear order, being rigid, has the DOT property, but since it’s dense, it is not a well-order.

Of course $2^{<\omega}=\omega$, so there is a rigid dense linear order of power $2^\omega$. With a little work to handle limit cardinals, the generalized continuum hypothesis implies that there is a rigid dense linear order of every uncountable cardinality.

Added: A nice explicit example of a scattered linear order that has the DOT property but is not a well-order isn’t too hard to construct. Let

$$X=\{\langle n,\alpha\rangle:n\in\omega\text{ and }\alpha\in\omega_n\}\;,$$

and define a linear order $\preceq$ on $X$ as follows: for $\langle m,\alpha\rangle,\langle n,\beta\rangle\in X$, $\langle m,\alpha\rangle\preceq\langle n,\beta\rangle$ iff either $m>n$, or $m=n$ and $\alpha\le\beta$.

Suppose that $\langle m,\alpha\rangle\prec\langle n,\beta\rangle$. If $m=n$, it’s easy to see that $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$. If $m>n$, the interval $\big[\langle m,\alpha\rangle,\langle n,\beta\rangle\big)$ has order type $\omega_m+\beta$, where $\beta<\omega_n<\omega_m$. Suppose that $\langle k,\gamma\rangle\prec\langle m,\alpha\rangle$, and let $I=\big[\langle k,\gamma\rangle,\langle m,\alpha\rangle\big)$. If $k=m$, the order type of $I$ is less than $\omega_m\le\omega_m+\beta$. If $k>m$, then the order type of $I$ is at least $\omega_{m+1}+\alpha>\omega_m+\beta$. Thus, the order type of $I$ cannot be $\omega_m+\beta$, and $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$.

Thus, $\langle X,\preceq\rangle$ has the DOT property. A rough sketch of $X$:

$$\cdots\underset{\omega_3}\longrightarrow\underset{\omega_2}\longrightarrow\underset{\omega_1}\longrightarrow\underset{\omega}\longrightarrow$$

ContentLicense: CC BY-SA 3.0


问:
What is the name for a magma which is neither a quasigroup nor a semigroup yet has both an identity and inverses?
Is there a name which is more specific than `unital magma' for a magma whose only requirements are that it should have both an identity and (L/R symmetric) inverses for all elements?

The following Cayley Table is the simplest non-trivial object of the type I am interested in, though of course you could make much more complicated examples:

```<code>  | 0 1 2
--+------
0 | 0 1 2
1 | 1 0 1
2 | 2 1 0
</code>```

This object it is not a quasigroup (or a loop) as the table is not a Latin Square.

This object is not a semigroup as it is not associative.  For example: (1*1)*2 = 0*2 = 2 while 1*(1*2) = 1*1 = 0.

It does, however, have an identity, 0.  Furthermore, every element has a unique inverse (which is itself).

For what it's worth the multiplication rule is a*b = |a-b| on the set {0,1,2}.

A part of an older question ( Magma with inverse and identity yet not a quasigroup ) could be interpreted as requesting an answer to the same question I am asking today. However, the one answer which that question received did not answer the bit which I would like to see answered. I have therefore judged that it's helpful to ask the missing part here more succinctly to maximise the chance that it gets an answer.

答:
This paper, for example, uses the terms

inverse property magma

magma with inverses

for properties defined in two different ways. The second one means a magma where each element has a unique inverse element.

Also, beware because the finite groupoid as in the Cayley table in your question is definitey not cancellative:
$$1*0=1*2=2*1=0*1,$$
but $2\ne 0$.


问:
Sine Wave with Oscillating Half-wavelength
What's the equation of a sine wave where the length of $\frac{λ}{2}$ oscillates, say between 2 and 3?

Thanks.

答:
Multiplying by a constant inside the brackets stretches in the $x$ direction

e.g. $y=sin (2x)$ has wavelength 180 degrees instead of 360 degrees.

If we work in radians you'll need $sin({\pi} x)$ to make a wavelength of 2 and half the wavelength will be 1.  By chance, since the wave repeats, it's also between 2 and 3.

Here is a graph of it on Desmos.

https://www.desmos.com/calculator/aqfzl4g1p4


问:
Prove $z + 1/z = 2 \cos(α) ⇒ z^n + 1/z^n = 2 \cos(nα)$
Please prove $z + 1/z = 2 \cos(α) ⇒ z^n + 1/z^n = 2 \cos(nα)$

答: Let $S_n = z^n + z^{-n}\implies S_n = \left(z+ z^{-1}\right)S_{n-1}-S_{n-2}=2\cos \alpha\cdot S_{n-1}-S_{n-2}$, and from this a strong induction proof can be used to establish the said identity.


问:
Pell's equation has high solutions when d is a member of OEIS A008784
Let

$x(d),\, y(d),\, d$ and $n$ be natural numbers, such that

$(x(d))^2-d(y(d))^2=1$.

I denote the $x$ solution $x(\textrm{dmax})$, if $x(\textrm{dmax}) > x(d)$ in the interval $n^2<d<(n+1)^2$ (I have some difficulties in the description and I  don't know the natural name for this type of  series, but in a temporal series it would be called a series of all time high – between two squares). This means that $x(\textrm{dmax})$ constitutes an increasing series between squares.

We get the results for $\textrm{dmax} = 2, 5, 10, 13, \ldots$. Note that $d=n^2+1$ is included in the $\textrm{dmax}$ series by definition, being the first case of for $d>n^2$. For $d=2$ to $d=113$, I get the 30 numbers of $\textrm{dmax}$. I compare this sequence $2, 5, 10, 13, \ldots$ with OEIS A008784 – “Numbers n such that sqrt(-1) mod n exists; or, numbers n that are primitively represented by $x^2+y^2$”.
A008784 has 27 numbers $\le 113$ of which only 25 are relevant here since the sequence contains $1$ and $25$, both squares. These two series have 21 numbers in common. The less obvious cases (other than the 9 cases where $d=n^2+1$ which is by definition both a member of A008784 and $\textrm{dmax}$ are still impressive.
I therefore speculate that the requirement
$d=a^2+b^2$ or $x^2 = (ay)^2+(by)^2+1$ may (often) rule out small solutions.

I note after a  puzzle with $x^2, (ay)^2$ and $(by)^2\ (\mathrm{mod} 8)$ that $x^2=1 (\mathrm{mod} 8)$ but I find that $x$ is only $1$ or $3$ and not $5$ or $7\ (\mathrm{mod} 8)$ in my cases ($d=2-106$). Do you think this is enough to explain the large number of common terms of the two sequences?

答:
EDIT: found Mordell's proof I mention below, number theory - approximate square root of a prime number

ORIGINAL: I will do more fiddling with this later, but I think there is a simple answer: for the numbers $d$ for which there exists a square root of $-1,$ there is often a solution to negative Pell $$  u^2 - d v^2 = -1.  $$ Then the first solution to positive Pell is $$  (u^2 + d v^2)^2 - d (2uv)^2 = 1.  $$ For your $d=n^2 +1,$ you get $u=n,v=1$ and so $(2n^2 + 1)^2 - (n^2 + 1)(2n)^2 = 1.$ So, there is a mechanism that will tend to enlarge the entries in the positive Pell solution.

Other example

==========

```<code>jagy@phobeusjunior:~$ cd "old drive"
jagy@phobeusjunior:~/old drive$ cd home
jagy@phobeusjunior:~/old drive/home$ cd jagy
jagy@phobeusjunior:~/old drive/home/jagy$ cd Cplusplus
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
85

0  form   1 18 -4   delta  -4
1  form   -4 14 9   delta  1
2  form   9 4 -9   delta  -1
3  form   -9 14 4   delta  4
4  form   4 18 -1   delta  -18
5  form   -1 18 4   delta  4
6  form   4 14 -9   delta  -1
7  form   -9 4 9   delta  1
8  form   9 14 -4   delta  -4
9  form   -4 18 1   delta  18
10  form   1 18 -4

 disc   340
Automorph, written on right of Gram matrix:
-6805  -123984
-30996  -564733

 Pell automorph
-285769  -2634660
-30996  -285769

Pell unit
-285769^2 - 85 * -30996^2 = 1

=========================================

Pell NEGATIVE
378^2 - 85 * 41^2 = -1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
</code>```

==============

Now, this happens often but not always. the first $d$ you have in mind for which the negative Pell equation is impossible is $d=34.$ The smallest such odd (nonsquare) numbers are $d=205$ and $d=221.$ Note that for a prime $p \equiv 1 \pmod 4,$ $p$ is always (primitively) the sum of two squares, and the negative Pell equation $u^2 - p v^2 = -1$ is always possible. There is a short proof in Mordell's book Diophantine Equations. If you want, I typeset Mordell's proof for another MSE answer, if i cannot find that I can paste the proof here or in another answer from my Latex original file.

So, my advice on this one is to also check negative Pell for all your numbers.

========================

d = 34, 205, 221  NO negative Pell

```<code>jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
34

0  form   1 10 -9   delta  -1
1  form   -9 8 2   delta  4
2  form   2 8 -9   delta  -1
3  form   -9 10 1   delta  10
4  form   1 10 -9

 disc   136
Automorph, written on right of Gram matrix:
5  54
6  65

 Pell automorph
35  204
6  35

Pell unit
35^2 - 34 * 6^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
205

0  form   1 28 -9   delta  -3
1  form   -9 26 4   delta  6
2  form   4 22 -21   delta  -1
3  form   -21 20 5   delta  4
4  form   5 20 -21   delta  -1
5  form   -21 22 4   delta  6
6  form   4 26 -9   delta  -3
7  form   -9 28 1   delta  28
8  form   1 28 -9

 disc   820
Automorph, written on right of Gram matrix:
881  24948
2772  78497

 Pell automorph
39689  568260
2772  39689

Pell unit
39689^2 - 205 * 2772^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
221

0  form   1 28 -25   delta  -1
1  form   -25 22 4   delta  6
2  form   4 26 -13   delta  -2
3  form   -13 26 4   delta  6
4  form   4 22 -25   delta  -1
5  form   -25 28 1   delta  28
6  form   1 28 -25

 disc   884
Automorph, written on right of Gram matrix:
-97  -2800
-112  -3233

 Pell automorph
-1665  -24752
-112  -1665

Pell unit
-1665^2 - 221 * -112^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ date
Wed May 29 12:43:50 PDT 2013
</code>```

=================

Body:
EDIT: found Mordell's proof I mention below, number theory - approximate square root of a prime number

ORIGINAL: I will do more fiddling with this later, but I think there is a simple answer: for the numbers $d$ for which there exists a square root of $-1,$ there is often a solution to negative Pell $$  u^2 - d v^2 = -1.  $$ Then the first solution to positive Pell is $$  (u^2 + d v^2)^2 - d (2uv)^2 = 1.  $$ For your $d=n^2 +1,$ you get $u=n,v=1$ and so $(2n^2 + 1)^2 - (n^2 + 1)(2n)^2 = 1.$ So, there is a mechanism that will tend to enlarge the entries in the positive Pell solution.

Other example

==========

jagy@phobeusjunior:~$ cd "old drive"
jagy@phobeusjunior:~/old drive$ cd home
jagy@phobeusjunior:~/old drive/home$ cd jagy
jagy@phobeusjunior:~/old drive/home/jagy$ cd Cplusplus
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
85

0  form   1 18 -4   delta  -4
1  form   -4 14 9   delta  1
2  form   9 4 -9   delta  -1
3  form   -9 14 4   delta  4
4  form   4 18 -1   delta  -18
5  form   -1 18 4   delta  4
6  form   4 14 -9   delta  -1
7  form   -9 4 9   delta  1
8  form   9 14 -4   delta  -4
9  form   -4 18 1   delta  18
10  form   1 18 -4

 disc   340
Automorph, written on right of Gram matrix:
-6805  -123984
-30996  -564733

 Pell automorph
-285769  -2634660
-30996  -285769

Pell unit
-285769^2 - 85 * -30996^2 = 1

=========================================

Pell NEGATIVE
378^2 - 85 * 41^2 = -1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

==============

Now, this happens often but not always. the first $d$ you have in mind for which the negative Pell equation is impossible is $d=34.$ The smallest such odd (nonsquare) numbers are $d=205$ and $d=221.$ Note that for a prime $p \equiv 1 \pmod 4,$ $p$ is always (primitively) the sum of two squares, and the negative Pell equation $u^2 - p v^2 = -1$ is always possible. There is a short proof in Mordell's book Diophantine Equations. If you want, I typeset Mordell's proof for another MSE answer, if i cannot find that I can paste the proof here or in another answer from my Latex original file.

So, my advice on this one is to also check negative Pell for all your numbers.

========================

d = 34, 205, 221  NO negative Pell

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
34

0  form   1 10 -9   delta  -1
1  form   -9 8 2   delta  4
2  form   2 8 -9   delta  -1
3  form   -9 10 1   delta  10
4  form   1 10 -9

 disc   136
Automorph, written on right of Gram matrix:
5  54
6  65

 Pell automorph
35  204
6  35

Pell unit
35^2 - 34 * 6^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
205

0  form   1 28 -9   delta  -3
1  form   -9 26 4   delta  6
2  form   4 22 -21   delta  -1
3  form   -21 20 5   delta  4
4  form   5 20 -21   delta  -1
5  form   -21 22 4   delta  6
6  form   4 26 -9   delta  -3
7  form   -9 28 1   delta  28
8  form   1 28 -9

 disc   820
Automorph, written on right of Gram matrix:
881  24948
2772  78497

 Pell automorph
39689  568260
2772  39689

Pell unit
39689^2 - 205 * 2772^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell
Input n for Pell
221

0  form   1 28 -25   delta  -1
1  form   -25 22 4   delta  6
2  form   4 26 -13   delta  -2
3  form   -13 26 4   delta  6
4  form   4 22 -25   delta  -1
5  form   -25 28 1   delta  28
6  form   1 28 -25

 disc   884
Automorph, written on right of Gram matrix:
-97  -2800
-112  -3233

 Pell automorph
-1665  -24752
-112  -1665

Pell unit
-1665^2 - 221 * -112^2 = 1

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ date
Wed May 29 12:43:50 PDT 2013

=================


问:
Does $\int_0^{2\pi} \sqrt{(1-a\cos x)^2 + b} \mathrm{d}x$ evaluate in terms of elliptic integrals?
Consider the integral

$$\int_0^{2\pi} \mathrm{d}x \sqrt{(1-a\cos x)^2 + b}$$

with $a,b>0$. I believe I can somehow evaluate this integral by incomplete elliptic integrals. This would be obvious, for example, if $a=1$. But, in the general case, this integral appears rather challenging. Mathematica, and SymPy have not been useful. And the Weierstrass substitution returns polynomials of 4th degree.

答:
If you make the $z=tan(x/2)$ substitution, Mathematica quickly evaluates the indefinite integral, with this horrible expression:

$$\frac{1}{2} z \sqrt{\frac{\left(a
   \left(z^2-1\right)+z^2+1\right)^2}{\left(z^2+1\right)^2}+b}-\frac{i \left(z^2+1\right)
   \sqrt{1-\frac{z^2 \left(a-i \sqrt{b}+1\right)}{a+i \sqrt{b}-1}} \sqrt{1-\frac{z^2
   \left(a+i \sqrt{b}+1\right)}{a-i \sqrt{b}-1}} \sqrt{\frac{\left(a
   \left(z^2-1\right)+z^2+1\right)^2}{\left(z^2+1\right)^2}+b} \left(2 \left(-i a
   \sqrt{b}+a+b+1\right) F\left(i \sinh ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i
   \sqrt{b}+1}} z\right)|\frac{-a^2+2 i \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b}
   a+b+1}\right)+\left(a^2+2 i a \sqrt{b}-b-1\right) E\left(i \sinh
   ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i \sqrt{b}+1}} z\right)|\frac{-a^2+2 i
   \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b} a+b+1}\right)-4 a \Pi \left(\frac{-a+i
   \sqrt{b}+1}{a+i \sqrt{b}+1};i \sinh ^{-1}\left(\sqrt{\frac{a+i \sqrt{b}+1}{-a+i
   \sqrt{b}+1}} z\right)|\frac{-a^2+2 i \sqrt{b} a+b+1}{-a^2-2 i \sqrt{b}
   a+b+1}\right)\right)}{2 \sqrt{\frac{a+i \sqrt{b}+1}{-a+i \sqrt{b}+1}} \left(a^2
   \left(z^2-1\right)^2+2 a \left(z^4-1\right)+(b+1) \left(z^2+1\right)^2\right)}$$


问:
Let be $E = \{ (x, y, z) ∈ \mathbb{R}^{3} \ | \ z = 0 \} ⊂ \mathbb{R}^3$. Prove that...
Let be $E = \{ (x, y, z) ∈ \mathbb{R}^{3} \ | \ z = 0 \} ⊂ \mathbb{R}^3$.
Prove that the endomorphism on $\mathbb{R}^3$ given by $f(x, y, z) = (−x − 3y, −3x − y + 2z, 2z)$ let $E$ invariant. Caluculate a basis of $E$ and the marix of restriction $f_{|E}$ of $f$ in the same basis. Calculate all the endomorphism on $\mathbb{R}^3$ that can be diagonalizable, leave $E$ invariant and match with $f$ over $E$.

My attemp,

$ \forall u \in E$, $f(u) = (a,b,0) \in E$, with $a,b\in \mathbb{R}$ $\Rightarrow$ $f(E)⊂E $

Let $\mathscr{B}$ a basis of E, then $\mathscr{B}= \{ (1,0,0), (0,1,0) \} $.

Now, for the matrix of restriction we need the transformed vector of $\mathscr{B}$ on the same basis.

\begin{equation}
A=
\begin{pmatrix}
-1 & -3 & 0\\
-3 & -1 & 2\\
 0 & 0  & 2
\end{pmatrix}
\end{equation}

$ [ f(1,0,0)]_{\mathscr{B}} = (-1,-3) $

$ [ f(0,1,0)]_{\mathscr{B}} = (-3,-1) $

\begin{equation}
A_{f_{|E}} =
\begin{pmatrix}
-1 & -3\\
-3 & -1
\end{pmatrix}
\end{equation}

We can see that $f$ is not diagonalizable.

So now, how can I calculate all the endomorphism on $\mathbb{R}^{3}$ with the given conditions?

答: Indeed, it is not diagonalizable. And endomorphisms of $\Bbb R^3$ that can be diagonalizable, leave $E$ invariant and match with $f$ over $E$ are those whose matrix with respect to the standard basis is of the form$$\begin{bmatrix}-1&-3&0\\-3&-1&0\\0&0&t\end{bmatrix},$$for some $t\in\Bbb R$.


问:
Find a power series representation for $\frac{1}{2+x}$

Find a power series representation for $\frac{1}{2+x}$

The question did not specify a real number $c$ and an expansion on the form

$$f(x)=\sum_{n\geq 0} a_n(x-c)^n.$$

I did this problem as follows, but I seem to be off by a factor of $1/3$.

$$\frac{1}{2+x} = \frac{1}{1+(x+1)}=\frac{1}{1-[-(x+1)]}.$$

Putting this into the geometric form of:

$$\sum_{n=0}^{\infty} \frac{a}{1-r},$$ I get $a=1$ and $r=-(x-1).$ So, this could be written as $$\sum_{n=0}^{\infty} [(-1)(x-1)]^n = \sum_{n=0}^{\infty} (-1)^n(x-1)^n.$$

Wolfram's widget gets the following:

$$\frac{1}{2+x}=\sum^\infty_{n=0}(-1+x)^n(-1)^n3^{-1-n}\quad \text{for}|-1+x|\lt3$$

which looks like my answer except for the $3^\left(-1-n\right)$ term.

Can you assist me in finding my error?  I don's see where the $1/3$ term gets introduced.

Body:
$$f(x)=\frac{1}{2+x}=\frac{1}{3+(x-1)}=\frac{1}{3}~~\frac{1}{[1+(x-1)/3]}$$
Use IGP
$$\implies f(x)=\frac{1}{3}\sum_{k=0}^{\infty} (-1)^k [(x-1)/3] ^k, ~if~ |x-1|<3. $$


答:
Rewrite $z + 1/z = 2 \cos α$ as a quadratic equation $z^2 - 2 z\cos α+1=0$ and solve to get $z=e^{\pm i \alpha}$. Then

$$ z^n + 1/z^n = e^{i n\alpha} + e^{-i n\alpha} = 2 \cos(nα)$$

Title: Prove $z + 1/z = 2 \cos(α) ⇒ z^n + 1/z^n = 2 \cos(nα)$

Body: Please prove $z + 1/z = 2 \cos(α) ⇒ z^n + 1/z^n = 2 \cos(nα)$


问:
Euler characteristic of the zero set of $z_1^d+z_2^d+z_3^d=0$ in $CP^2$
I'm doing this qualifying exam problem from my university and got stuck with this one:

Let $Z$ be the zero set of the polynomial $z_1^d+z_2^d+z_3^d=0$ in $CP^2$, here $z_1,z_2,z_3$ are complex numbers; what is the Euler characteristic of $Z$?

I really have no clue what to do with this. I know that the Euler characteristic is the same as the self intersection number for a compact oriented manifold. However, I'm not sure how to deform the zero set to be transversal. Can anyone get me started please? Thank you! (This question might be related to algebraic geometry, however algebraic geometry is not in our syllabus.)

Edit: the available results from my syllabus are those in Lee ,Hatcher, and Guillemin and Pollack only.

答: The magic words are "Brieskorn variety". You can look at the original papers to see how people worked with them (Milnor's singular points of complex hypersurfaces is highly recommended, too). See this recent-ish paper: https://arxiv.org/abs/1310.0343</p>


问:
If $L$ is regular, prove that $\sqrt{L}=\left\{ w : ww\in L\right\}$ is regular

Let $L$ be a regular language. Prove that $\sqrt{L}:=\left\{ w : ww\in L\right\}$ is also a regular language.

I suppose I need to modify state machine for $L$ to accept $\sqrt{L}$, but I've been thinking how to do that for a few hours and still don't know. I would be very grateful for help.

答:
Let $D=(Q,\Sigma,\delta,q_0,F)$ be the DFA which recognizes $L$.

We need to construct an automaton which can simultaneously start from the initial ($q_0$) as well as "middle" state ($q_m$) of $D$, and parallelly parse two occurrences of $w$, one from $q_0$ to $q_m$ and another from $q_m$ to some accepting state.

When the first half is parsed, we also need to remember what this "middle" state $q$ was. Thus the machine will work on $Q\times Q\times Q$ state space, two to parse both halves of string, and third to remember the middle state.

The NFA which accepts $\sqrt{L}$ is: $N=(Q\times Q\times Q, \Sigma, \Delta, Q_0, F')$, where

$Q_0=\{(q_0,q_m,q_m)|q_m\in Q\}$

$F'=\{(q_m,q_f,q_m)|q_f\in F, q_m\in Q\}$

$\Delta=\{((q_1,q_2,q_m),a,(q_1',q_2',q_m))|\delta(q_1,a)=q_1', \delta(q_2,a)=q_2'\}$

Using this, it can easily be shown: $x\in \sqrt{L}\iff\exists$ an accepting run of $N$


问:
Power tower inequality
I want to prove the following power tower inequality:

$$
3 \uparrow \uparrow 100 > 4 \uparrow \uparrow 99
$$

but I don't know how to do this. I think that induction will not work, because I think there will be an $N$ for which

$$
3 \uparrow \uparrow N < 4 \uparrow \uparrow (N-1)
$$

Could anyone help me in the right direction? Please don't answer a full solution, but I do need a hint.

答:
Try to prove

$3 \uparrow \uparrow n+1 > \log_3(4) \cdot 4 \uparrow \uparrow n$ for all $n \in \mathbb{N}$

and use the following facts about $\log$

$\log_a (x) = \frac{\log_b(x)}{\log_b(a)}$

$\ln : \mathbb{R}_+ \rightarrow \mathbb R$ is strictly increasing

Title: Power tower inequality


答:
The confusion comes from the fact that your answer is wrong. See the answer from Z Ahmed for a correct solution. In your solution you somehow transformed $x+1$ into $-x+1$. This happened when you wrote $r$.

Note that the different answers write a series around different points:

yours is around $-1$, with small $x+1$

the given answer, and the one from ZA is around $1$, with small $x-1$

hm2020 expands around $0$, with small $x$

Body:
Find a power series representation for $\frac{1}{2+x}$

The question did not specify a real number $c$ and an expansion on the form

$$f(x)=\sum_{n\geq 0} a_n(x-c)^n.$$

I did this problem as follows, but I seem to be off by a factor of $1/3$.

$$\frac{1}{2+x} = \frac{1}{1+(x+1)}=\frac{1}{1-[-(x+1)]}.$$

Putting this into the geometric form of:

$$\sum_{n=0}^{\infty} \frac{a}{1-r},$$ I get $a=1$ and $r=-(x-1).$ So, this could be written as $$\sum_{n=0}^{\infty} [(-1)(x-1)]^n = \sum_{n=0}^{\infty} (-1)^n(x-1)^n.$$

Wolfram's widget gets the following:

$$\frac{1}{2+x}=\sum^\infty_{n=0}(-1+x)^n(-1)^n3^{-1-n}\quad \text{for}|-1+x|\lt3$$

which looks like my answer except for the $3^\left(-1-n\right)$ term.

Can you assist me in finding my error?  I don's see where the $1/3$ term gets introduced.


问:
interpretation of angle bracket $\langle \circ ; \circ \rangle$ notation appeared in linear algebra
I'm confused around the angle bracket $\langle \circ ; \circ \rangle$ notation appeared in my book, where no definition of that is given. So, it is really nice if someone tell me how to interpret this. The context that the notation is used is as follows:

Definition 2.28 (Dual of a linear map)

If $A: U \to V$ is a linear map between finite-dimensional vector
space $U$ and $V$, then the linear map $A^*: V^* \to U^*$ defined by

$\langle A^*(\alpha) ; u \rangle = \langle \alpha; A(u)\rangle, \alpha \in V^*, u \in U$

is a dual of $A$.

The book name is Bullo, Francesco, and Andrew D. Lewis. "Geometric Control of Mechanical Systems." (2005).

答:
The notation $\langle A^*(\alpha) ; u \rangle$ is another notation for $A^*(\alpha)(u)$.

It is a notation to highlight duality.


答:
Here is another answer: For this rational function it is possible to give an elementary construction of a powerseries and an "equality"

$$ \frac{1}{2+x}=\sum_{n \geq 0} a_n(x-a)^n$$

for any $a\neq -2$:

$$\frac{1}{2+x}=\frac{1}{2(1-(-x/2))} =\frac{1}{2}\sum_{n\geq 0}(-\frac{x}{2})^n=$$

$$ \sum_{n\geq 0}(-1)^n\frac{x^n}{2^{n+1}}.$$

I used

$$\frac{1}{1-t}=\sum_{n\geq 0} t^n$$

with $t:=-\frac{x}{2}$.

We can generalize as follows (let $a\neq -2$):

$$ \frac{1}{2+x}=\frac{1}{2+a-(-(x-a))} =$$

$$ \frac{1}{2+a}\frac{1}{1-(-(\frac{x-a}{2+a}))}=$$

(here we let $t:=\frac{x-a}{2+a}$)

$$ \frac{1}{2+a}\sum_{n\geq 0}(-(\frac{x-a}{2+a}))^n.$$

We get a power series

$$\frac{1}{2+x}= \frac{1}{2+a}\sum_{n\geq 0}(-1)^n(\frac{x-a}{2+a})^n.$$

With $a:=1$ we get

$$\frac{1}{2+x}=\frac{1}{3}\sum_{n \geq 0}(-1)^n(\frac{x-1}{3})^n.$$

When I write $=$ in the above equation this means for all $x$ where the powerseries converge. You should be able to calculate this radius of convergence yourself.

In general: If $\frac{f(x)}{g(x)}$ is a rational function over the reals, there is a product decomposition

$$P1.\text{  }g(x)=\prod_{i} (x-c_i)^{l_i}\prod_j p_j(x)^{k_j}$$

where $c_i$ are real numbers and $p_j(x)=b_j^2+(x-a_j)^2$ where $a_j,b_j\neq 0$ are real numbers. For a rational function on the form $\frac{1}{p_j(x)}$
you get a similar calculation:

$$ \frac{1}{p_j(x)}=\frac{1}{b_j^2+(x-a_j)^2}=$$

$$\frac{1}{p_j(x)}"="\frac{1}{b_j^2}\sum_{n\geq 0}(-1)^n(\frac{x-a_j}{b_j})^{2n} = \sum_{n \geq 0} \frac{(-1)^n}{b_j^{2(n+1)}}(x-a_j)^{2n}.$$

Hence you can "in principle" construct an explicit formal power series expansion for any rational function $\frac{f(x)}{g(x)}$ using the geometric series and partial fraction decomposition of the product in P1.

Title: Find a power series representation for $\frac{1}{2+x}$


问:
Understanding mathematics imprecisely
For a long time, it has been a complete mystery to me how any of my peers understood any math at all with anything short of filling in every detail, being careful about every set theoretic detail down to the axioms.  That's a slight exaggeration, but I certainly did much worse in courses where I attempted to replicate this myself by not reading every proof to save time.

It's only recently, and only within the field of probability theory, that I've developed the ability to do this myself.  I am currently following Grimmett's book on Percolation theory.  There are far too few details for someone of my level to fill it in completely, but I am getting more than nothing out of it.

Question 1: I would like to learn how to get even more out of such "incomplete" studying.

What tends to happen even now, and more before, is that as soon as I don't understand something, I lose focus and everything just flies over my head.  I imagine this is partly psychological, since from a logical perspective if I have to accept proposition $P$ to derive $Q$, I could just think of myself as having proven merely that $P$ implies $Q$, and then no "acceptances" are being made.

Most of the time, professors simply stare blankly at me, wondering how I could persist like this, and all they say is to stop.  But it's not that simple, because it appears my intuition is also primarily symbolic.  Sure, I think of some geometric pictures when they're called for, but most of my problem-solving creativity comes from pattern matching methods and tricks with situations.

Question 2: How does one distill out the important ideas of a mathematical reading, such as a proof or paper?

Grimmett's book is very helpful in this regard.  He will always tell me what's important, and as long as I'm willing to believe him, then I don't have to do anything.  But what if I need things that are different than he emphasizes?  I always worry that by not understanding everything, I will eventually reach some point in my life when I need to use some fact/method I glossed over and forgot, and that it could be framed in such a way that I would not even be aware of what is missing.  That way I wouldn't even be able to do a huge review to rescue the fact from the depths of my ignorance.  My current way of thinking about this subproblem of question 2 is that mathematicians always take this risk by not studying everything.  So it's a risk-minimization game with time as the constraint.  If so, how do I make smart choices with regard to this game?

Question 3: With my recent ability to learn imprecisely in probability, I've started to see many connections, even with outside fields.  Many of them are probably fictitious.  Many of the questions that I think are highly-motivated might actually be not really worth answering.  How does one decide what questions are interesting?  As a graduate student who has barely popped out above what the traditional classroom has to offer, I am very lost in this regard.

Question 4: The revelations that enabled me to understand math imprecisely came all at once.  A similar comment about the abruptness of my coming upon the ability to proof-check without significant error could be made 2 years ago.  Most of my peers seem to learn rather continuously, but the evolutions of my way of thinking seem to come all at once.  Is there anything bad or good about this?  If so, how do I minimize the bad and maximize the good?

As always, answers to subsets are appreciated.

答:
Full understanding is illusory. If you pursue it, you will find yourself trying to say what a number is, or a set, and digressing into the problem of making language, which for math is a meta-language, precise. And, of course, that can't be done.

So regarding your first question, it might help to observe how futile that innate wish of yours is, and how much you understand without full understanding (or compunction) in all other aspects of your life.

Imagine trying to learn biology and studying the chemical processes in the body, then asking "what is a chemical". You are given an answer that has to do with molecules, a term which you then inspect for precision's sake. Atoms come up, then electrons. Eventually you are learning quantum physics when all you wanted to do was understand how allergens work, or some such thing.

You must operate at the appropriate level for a specific problem. It's no use to reinvent the wheel and do everything from first principles. That would be like writing every program in machine code.

One day, our brains may be augmented with enhancements that allow us to have enough knowledge to understand everything down to our "axioms". Until then, it is a matter of becoming comfortable with our limitations and trying to work with what we have to be awesome.

In terms of knowing which questions are interesting, I think that is one of the harder parts of research. One almost has to be prescient.

And as for getting the important ideas of a proof, my first answer is that sometimes you can't really. Some proofs are just a confluence of numerical estimations and limit results and don't give any real insight into what is going on. Since you seem to be a probabilist, I would point to the proof that a random walk in dimension $n$ is recurrent for $n=1,2$ and transient otherwise. One feels there should be an intuitively understandable reason, but all one gets is Stirling's formula.

For other proofs it is a matter of becoming comfortable enough with the terminology and techniques used in the proof (by re-reading) to see the forest for the trees. In Kung Fu one talks about "learning to forget". You learn the movements carefully so you can perform them without thinking about them when the time comes. You do the same when you learn to integrate or differentiate - you don't want to be doing this from the limit definition when crunch time comes (exam say).

Title: Understanding mathematics imprecisely


问:
How to make an hyperellipsoid touch a hyperplane
Let's say we have an hyperellipsoid $\mathcal{E}$ defined by the symmetric, positive definite matrix $P$, i.e.,

$$\mathcal{E} = \{ x \in \mathbb{R}^n \mid x^\top P x = c \}$$

where $c>0$. I would like to find the biggest $c$ possible to make $\mathcal{E}$ touch the hyperplane $\{x \in \mathbb{R}^n \mid a^\top x = b\}$. According to a German book I read, this problem can be solved using Lagrange multipliers. Stated as an optimization problem, this should read
$$
\begin{align}
&\max_{x \in \mathbb{R}^n} x^\top P x = c \\
&\mathrm{s.t.} \: a^\top x = b
\end{align}
$$
The corresponding Lagrangian is $\mathcal{L} = x^\top P x - \lambda (a^\top x-b)$ with $\lambda$ being the Lagrange multiplier. The necessary condition for optimality leads to the linear system
$$
\underbrace{\begin{bmatrix}
-a & 2P \\
0 & a^\top
\end{bmatrix}}_{ =A}
\begin{pmatrix}
\lambda \\
x
\end{pmatrix}
=
\begin{pmatrix}
0 \\
b
\end{pmatrix}
$$
and I have no idea how to invert $A$. Am I even on the right track? Any hint/help would be greatly appreciated.

Remark: The aforementioned book states that the solution is $c = {b^2 \over a^\top P^{-1} a }$ if that helps.

答:
Ok, it's much easier than I initially thought. Should I delete the question (and the following solution)?

By switching $\lambda$ and $x$ in the above linear system, we can use the ansatz
$$
\begin{bmatrix}
2P & -a \\
a^\top & 0
\end{bmatrix}
\begin{bmatrix}
D & e \\
f^T & g
\end{bmatrix}
=
\begin{bmatrix}
I & 0 \\
0 & 1
\end{bmatrix}
$$
where $D \in \mathbb{R}^{n\times n}$, $e,f \in \mathbb{R}^n$, $g \in \mathbb{R}$ and $I$ is the $n\times n$ identity matrix.
Solving this provides us with $e=\frac{P^{-1}a}{a^\top P^{-1} a}$ and $x=b e$. Inserting the so found $x$ into $c=x^\top P x$ yields the desired result.


问:
Getting intuition for the coordinates of a gradient vector.
Let's say I have some function $f(x, y)$ whose gradient at $(x_0, y_0)$ is $$\nabla f(x_0, y_0) = \langle 4, 1\rangle$$

In this 3Blue1Brown video, Grant says something akin to

the change in $f$ is $4\times$ more sensitive to changes in $x_0$ then it is to changes in $y_0$.

when discussing the gradient of a cost function in his explanation of neural networks. Why does this hold?

It seems intuitive to me that changing $x_0$ would result in a larger change in $f$ then changing $y_0$ since the gradient more closely lines up with the $x$-axis but why is it exactly $4\times$?

答:
The assertion$$\nabla f(x_0,y_0)=(4,1)$$means that, near $(x_0,y_0)$, $f(x,y)$ behaves as $f(x_0,y_0)+4(x-x_0)+y-y_0$. So, near $(x_0,y_0)$, a small change in the value of $x$ is multiplied approximately by $4$, whereas a small change in the value of $y$ is creates a change of approximately the same size. So, yes, near $(x_0,y_0)$ $f$ is $4$ times more sensitive to changes in $x$ than to changes in $y$.

This would still hold if $\nabla f(x_0,y_0)=(4t,t)$, for some $t\ne0$.


问:
How to solve $x \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+x y=0$
How to solve $$x \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+x y=0$$
First of all i could not find the two linearly independent solutions $y_1(x),y_2(x)$.

I tried polynomials, but in vain.

答:
Let set $\begin{cases}a(x)=x\\b(x)=4\\c(x)=x\end{cases}\quad$ then $\quad a(x)y''+b(x)y'+c(x)y=0$

Can be reduced according to Sturm-Liouville theory via the integration factor

$\displaystyle\mu(x)=\dfrac 1{a(x)}\exp\left(\int \frac{b(x)}{a(x)}\right)=\frac 1x\exp\left(\int\frac 4x\right)=\frac 1x\exp(4\ln(x))=\frac {x^4}x=x^3$

Therefore let substitute $y(x)=\dfrac {u(x)}{x^3}$

After reduction to the same denominator and eliminating the superfluous $x^2$ factor

$$xu''-2u'+xu=0$$

It may seem we have not made much progress, but in fact this ODE is simpler to solve.
Unfortunately not as simple as $(xu)''=xu''+2u'$ (with a $+$ sign instead) would have been much more appealing.

Cheating at the result given by CAS we can set $u=xv'-v$, notice that $u'=xv''$.

We get $\, x(xv'''+v'')-2xv''+x^2v'-xv=0$ which can be rewritten

$$(v''+v)=x(v''+v)'$$

Solving for $\quad w=v''+v\quad $ we get $\quad w=xw'\iff w=C_1x$

And then $\quad v''+v=C_1x\iff v=C_1x+C_2\sin(x)+C_3\cos(x)$

Substituting back in $u$ gives (the term $C_1x$ get cancelled):

$$u=C_2(x\cos(x)-\sin(x))-C_3(x\sin(x)+\cos(x))$$

Just divides by $x^3$ to get the expression for $y$.

Rem: If someone knows the reason why the substitution $u=xv'-v$ works well in this context, I'm interested about it.


问:
Moving circles, passing through 3 points
I would like to know if there is an unique solution or at least a good approximation for the following problem:

I have three points in a space forming a triangle. There is nothing special for this triangle, it is not equilateral, but if we find a solution for an equilateral triangle, we can approximate it to that.

The points are (0,18.27), (24.58,34.37) and (24.82,0) and I have the exactly time when a moving circle touches each point for the first time.

The assumptions are:

The circle moves at a constant speed at a constant direction;

I only know the first time it touch the points, not the time it touch with the other half;

The radius is, at least 20 (the circle that circumscribe the points) and at most 200;

It can come from any direction;

The main objective:

To have some characteristic length scale of those moving circles, e.g. its radius.

I am looking for a general solution for that. The thing is: I have a spatial data from three points and I could identify a particular thing happening at in the time series from the three points. From the theory behind this natural phenomena, I could approximate it to a circle. So, I am assuming that this is a moving circle that pass through three points.

I added a figure to illustrate the problem. T_pi is the time the circle touches each point.

I also made another figure to ilustrate what I think are the variables of the problem: radius, angle, offset and speed. I am assuming that they are constant for each circle.

Feel free to ask for more information about the problem and if we can make any other assumption about that.

答:
This isn't an answer yet, just an attempt to formulate the problem.  It really should be a comment, but it won't fit in a comment box.

The center of the circle moves along a straight line with constant velocity.  We can say that at time $t$ it is at the point$$(x_0+at, y_0+bt)$$ where $x_0,y_0,a,\text{ and }b$ are unknown constants.  At times $t_1<t_2<t_3$ the point is a some fixed, but unknown distance $r$ from the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, respectively, where $t_i,x_i,y_i$ are given for $i=1,2,3$.

That gives us $3$ equations in $5$ unknowns:
$$(x_0+at_i-x_i)^2+(y_0+bt_i-y_i)^2=r^2,\ i=1,2,3$$

I've been trying to eliminate one or more unknowns.  Of course, we can eliminate $r$, by just saying the the distances are equal, but this eliminates an equation, so we have $2$ equations in $4$ unknowns.  I don't see how to eliminate any other variables.  $(x_0, y_0)$ is the position of the moving point at time $t=0$.  We can reset the origin of the time scale, say by setting $t_i'=t_i-t_1$, but that doesn't eliminate the unknowns $x_0, y0$.  Similarly, it's only the ratio of $b$ to $a$ that's really fixed, but $\sqrt{a^2+b^2}$ is the distance the point moves in one unit of time, and I don't see any way around this.

Since we have more unknowns that equations, I tend to believe that there is not a unique solution to the system.  If the OP would gives us the values of $t_1,t_2,t_3$, I think I could demonstrate this.

EDIT

I recognize that I haven't taken cognizance of the requirements that the times are the first time the circle passes through the points, or the restrictions on the radius, but I don't think that this will change matters.  This is just a gut feeling.  I think there are infinitely many solutions to the system, and we'll be able to pick out more than one that meet the requirements.  (Of course, I'm wrong a lot.)


问:
Why are Neumann Boundary conditions imposed weakly?
let's consider a mixed boundary problem $-\Delta u = f$, $u=0$ on $ \Gamma_D$ and $\nabla u \cdot n = 0 $ on $\Gamma_N$.
Then the weak formulation is to find $u \in H^1_D = \{u \in H^1 | u = 0 \text{ on } \Gamma_D\}$ such that $\int_\Omega \nabla u \cdot \nabla v = \int_\Omega f \ v $ in $\Omega$. That means the Neumann bc were imposed weakly. Why is it not possible to look for a solution in the space $H^1_{D,N} = \{u \in H^1 | u = 0 \text{ on } \Gamma_D\ ,\nabla u \cdot n = 0 \text{ on } \Gamma_N\}$?

答:
This is a rather techincal issue. A priori, we can not talk about $\nabla u \cdot n$ in an $H^1$, i.e. trace-sense. If our domian $\Omega$ has $C^2$-boundary, we can say that our weak solution $u \in H^1_0$ (that we are granted by Lax-Milgram-Lemma) even lives in $H^2(\Omega)$. This is Gilbarg and Trudinger Theorem 8.12. It is sometimes referred to as Friedrichs-shift-theorem.

Then, traces can be defined for $\nabla u$, and we can make sense of the Neumann BC.

So we only choose $H_0^1$ as test and solution space, because no further regularity is needed.

Since Neumann-BCs are integrated into the weak formulation (in our case, since the Neumann-values are $0$, they do not appear), they are often referred to as natural BCs or do-nothing BCs.


问:
how find minimum value of this function ? $f=\prod_{i=1}^{20}b_i(x)$
let $0\lt a_1\lt a_2\lt....\lt a_{20}$ ($a_i\in\mathbb R$)and for all i $\in\{1,2,...,20\}$ define $$b_i:(0,\infty)\to \mathbb R$$$$b_i(x)=\left\{
\begin{array}{c}
\frac{a_i}{x} ;x\le a_i \\
 \\
\frac{x}{a_i};x\gt a_i
\end{array}
\right.$$
how find minimum value of  $f$ ?  such that $f(x)=\prod_{i=1}^{20}b_i(x)$

Thanks in advance

答:
I use the fact that the function f is continuous.

For $x<a_{10}$, $f(x)$ has the form of $K/x^n$, so the function is decreasing. The minimum cannot be there

For $x>a_{11}$, $f(x)$ has the form of $K*x^n$, so the function is increasing. The minimum cannot be there

For $a_{11} >x>a_{10} $, $f(x) = \frac{a_{20}*...*a_{11}}{a_{10}*...*a_1}$ is constant. Here is the minimum


问:
Eliminating unwanted branches of algebraic curves related to triangle centres
Lately I have become fascinated with triangle centres. To that end, I have written a small Python module that can compute explicit positions of centres for arbitrary triangles in the plane to arbitrary precision, based on the trilinear and barycentric coordinates in Kimberling's Encyclopedia of Triangle Centers (ETC).

For certain combinations of triangles and centres, the sum of barycentric coordinates evaluates to $0$ and I have a point at infinity. To study those cases where a triangle centre leaves the friendly confines of the Euclidean plane for the line at infinity, I invented the notion of infinity curves.

The infinity curve $\infty_n$ associated with the Kimberling triangle centre $X_n$ is the locus of points $(x,y)$ with $y\ne0$ and $(x,y)\ne(0,\pm\sqrt3)$ where the centre $X_n$ of the triangle with vertices $(-1,0),(1,0),(x,y)$ is a point at infinity. $\infty_n$ may be empty; I will ignore these cases here, as well as points that always lie on the line at infinity such as $X_{30}$.

For algebraic centres, which make up the vast majority of centres in the ETC, the associated infinity curves are also algebraic, and the problem then is to find an exact bivariate polynomial describing them. They can be derived by setting the sum of barycentric coordinates equal to zero, then expanding, simplifying and factoring. For example, the Exeter point $X_{22}$ has triangle centre function $a(b^4+c^4-a^4)$, from which I derived
$$\infty_{22}(x,y):3x^4 + 2x^2y^2 - 6x^2 - y^4 - 6y^2 + 3=0$$
Infinity curves can be visualised (or determined not to exist) for a given centre by making a complex plot of the centre's location for the triangles $(-1,0),(1,0),(x,y)$ with varying $z=x+yi$. Here is such a plot for $X_{22}$; the infinity curve is white and marks discontinuities in the hue.

The file infcurves has polynomials describing infinity curves up to $X_{70}$ (also the upper limit of the centres that can be computed by the Python module at the time of posting this question). However, I have been unable to fully reduce some of the polynomials to match the real appearance of their corresponding curves, and this is why I ask this question.

Take for example $X_{29}$ with centre function $\frac{\sec A}{\cos B+\cos C}$. Its complex plot looks like this, the infinity curve being a simple W-shape on the upper half-plane:

The polynomial I have computed as $\infty_{29}$, on the other hand, produces that and more:
$$27 x^{16} + 144 x^{14} y^{2} - 216 x^{14} + 306 x^{12} y^{4} - 864 x^{12} y^{2} + 756 x^{12} + 324 x^{10} y^{6} - 1404 x^{10} y^{4} + 2160 x^{10} y^{2} - 1512 x^{10} + 171 x^{8} y^{8} - 1180 x^{8} y^{6} + 2430 x^{8} y^{4} - 2880 x^{8} y^{2} + 1890 x^{8} + 36 x^{6} y^{10} - 540 x^{6} y^{8} + 1480 x^{6} y^{6} - 1800 x^{6} y^{4} + 2160 x^{6} y^{2} - 1512 x^{6} - 132 x^{4} y^{10} + 530 x^{4} y^{8} - 600 x^{4} y^{6} + 270 x^{4} y^{4} - 864 x^{4} y^{2} + 756 x^{4} - 16 x^{2} y^{12} + 92 x^{2} y^{10} - 124 x^{2} y^{8} - 140 x^{2} y^{6} + 324 x^{2} y^{4} + 144 x^{2} y^{2} - 216 x^{2} + 4 y^{10} - 37 y^{8} + 116 y^{6} - 126 y^{4} + 27=0$$

This curve appears to be decomposable into four parts, one of which is the one I want. The problem is that the polynomial is absolutely irreducible, something I have checked with my implementation of Gao's factorisation algorithm (one of whose steps I inquired about in this question). I want to eliminate the unwanted branches of this curve, but factoring does not seem to be the way.

Given an irreducible algebraic curve with several distinct branches like $\infty_{29}$ above, how can I select a specific branch and remove the others?

答:
In general if you are given an explicit formula for a polynomial $f(x,y):=\sum_{(i,j)} a_{i,j}x^iy^j \in k[x,y]$ where $k$ is a subfield of the field $\mathbb{C}$ of complex numbers, it is difficult to determine if $f$ is an irreducible polynomial in $k[x,y]$.
If the ideal $I:=(f)\subseteq k[x,y]$ is non-prime, it is difficult to calculate the irreducible components $f=f_1\cdots f_k$ of $f$. I dont think there are any "general methods". How do you check that the polynomial is "absolutely irreducible" on a computer?

Note: A computer has "finite memory", hence when you calculate on a computer you calculate in a finite field with large characteristic. So if you have an algorithm on your computer that claims to "prove a polynomial is irreducible", this algorithm proves it is irreducible in a polynomial ring over a finite field.

Example: Let $A:=\mathbb{Z}[x_1,..,x_n]$ be the polynomial ring in $n$ variables over the integers and assume $p>0$ is the largest prime expressible on your computer. Let $k:=\mathbb{Z}/p\mathbb{Z}$.

Assume $F(x_1,..,x_n):=f_1(x_i)\cdots f_k(x_i)+pf(x_i)\in A$ is an irreducible polynomial. Your computer is working in the polynomial ring $B:=k[x_1,..,x_n]$ and in this ring (when reducing $F$ modulo the prime $p$) you get the polynomial

$$ \overline{F}=\overline{f_1}(x_i)\cdots \overline{f_k}(x_j),$$

and the polynomial $\overline{F}$ is not irreducible in $B$. Hence the original polynomial $F$ is irreducible in $A$ but your computer does not detect this since it is working with the polynomial ring $B$.

Comment: "This curve appears to be decomposable into four parts, one of which is the one I want. The problem is that the polynomial is absolutely irreducible, something I have checked with my implementation of Gao's factorisation algorithm (one of whose steps I inquired about in this question). I want to eliminate the unwanted branches of this curve, but factoring does not seem to be the way."

In general: A computer cannot detect if a polynomial is irreducble in $\mathbb{Z}[x_i]$ or $\mathbb{Q}[x_i]$ because it has "finite memory" and is working with polynomials in a polynomial ring over a finite field.

Example: Let $p>0$ be a prime number and consider the polynomial $$f(x):=x^2+p\in A:=\mathbb{Q}[x].$$

Let $p$ be the largest prime expressible on your computer.
You can yourself verify that $f(x)$ is irreducible in $A$ but your computer believes there is an equality $f(x)=x^2$ which is a reducible polynomial, since your computer calculates in the field $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$ and the polynomial ring $\mathbb{F}_p[x]$.

Here is a post on the problem of studying the number of solutions to polynomial equations and computer algebra:

When does a system of polynomial equations have infinitely many solutions?


答:
The gradient vector of a function $f:\mathbb{R}^n\to\mathbb{R}$  at a point
$(x_1,\dotsc, x_n)$ has the partial derivatives $\frac{\partial f}{\partial x_j}$ as its entries. The j-th partial derivative precisely measures the change of $f$ in the j-th direction (with respect to the standard basis), it is defined as the derivative of the funciton $\mathbb{R}\to\mathbb{R}, t\mapsto f(x + te_j)$ where
$x = (x_1,\cdots, x_n)$ and $e_j$ is the j-th standard basis vector. That is in your case  partial derivative in the direction of $x$ is 4 and the one in the direction of $y$ is 1, implying that the rate of change in the direction of $x$ is 4 times the rate of change in the direction of $y$.

Title: Getting intuition for the coordinates of a gradient vector.

Body:
Let's say I have some function $f(x, y)$ whose gradient at $(x_0, y_0)$ is $$\nabla f(x_0, y_0) = \langle 4, 1\rangle$$

In this 3Blue1Brown video, Grant says something akin to

the change in $f$ is $4\times$ more sensitive to changes in $x_0$ then it is to changes in $y_0$.

when discussing the gradient of a cost function in his explanation of neural networks. Why does this hold?

It seems intuitive to me that changing $x_0$ would result in a larger change in $f$ then changing $y_0$ since the gradient more closely lines up with the $x$-axis but why is it exactly $4\times$?


问:
Big $O$ notation on both the sides of equality
If $$ f(x)+O(g(x))=O(g(x)),$$

is it possible to write $$ f(x)=O(g(x))$$

and if yes, how does that work?

答:
$$f(x)+O(g(x))=O(g(x))$$
I'm not familiar with the notation of writing $O()$ on the LHS of an equation. But presumably the $O(g(x))$ on the LHS represents a functions $h(x)=O(g(x))$ such that $|h(x)|\leqslant A g(x)$ for sufficiently large $x$.
The $O(g(x))$ on the RHS represents a functions $j(x)=O(g(x))$ such that $|j(x)|\leqslant B g(x)$ for sufficiently large $x$.

So we have
$$f(x)=j(x)-h(x)$$
$$|f(x)|=|j(x)-h(x)|$$
and so by the triangle inequality
$$|f(x)|\leqslant |j(x)|+|h(x)| \leqslant (A+B)g(x)$$
and so
$$f(x)=O(g(x))$$


问:
Converse to Properness of Group Action
I'm working on Lee's Manifolds problem 21-1, which states the following:

Suppose a Lie group $G$ acts continuously on a manifold $M$ . Show that if the map $G \times M \rightarrow M$ defining the action is proper, then the action is a proper action. Give a counterexample to show that the converse need not be true.

I think I've got the first part sorted out by playing around with definitions and projection mappings, but I still don't have much intuition for what this really means. That lack of intuition has given me some issues trying to come up with a counterexample for the converse. A comment in this thread [https://math.stackexchange.com/questions/2434982/why-is-the-definition-of-a-proper-group-action-the-way-it-is] states that:

Also note that map $f:G \times M \rightarrow M$ is proper iff $f^{-1}[K]$ is compact for every compact , while the mentioned map $G \times M \rightarrow M \times M$ is proper iff $f^{-1}[K] \cap (G \times K')$ is compact for every compact $K$, $K'$

which feels like it should be enough to find a counterexample, but my lack of intuition for what these actions really look like makes it impossible for me to try to apply this comment to come up with an example. Any help regarding this topic (and the converse example specifically) would be extremely welcome!

答:
If $f: X \to Y$ is a map between metric spaces there is an easier way to define the notion of properness which ought to give some intuition.

A map $f: X \to Y$ is proper if, for any sequence $x_n \in X$ diverging to $\infty$ in $X$, the sequence $f(y_n)$ also diverges to $\infty$ in $Y$.

Here we say a sequence diverges to $\infty$ if it has no convergent subsequence. While I tend to prove things about properness using the general notion this gives me an intuitive handle on the concept. Exercise: prove that this is equivalent to $f$ being proper.

Hint for your question: take $G = M$. To be even more explicit, take $G = M = \Bbb R$.

Title: Converse to Properness of Group Action


问:
Two circles inscribed in a semicircle
Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)

答:
It's not easy to believe, but
by pure luck, the solution is found as:

\begin{align}
R&=\frac67\sqrt{187\sqrt2}\approx 13.939
,\\
r_1&=\frac13\sqrt{187\sqrt2}\approx 5.42
,\\
r_2&=\frac38\sqrt{187\sqrt2}\approx 6.0983
.
\end{align}

Edit

$\require{begingroup} \begingroup$
$\def\i{\cdot\mathbf{i}}$

I just started to build the construction from
two radii of inscribed circles,
and it was a pure stroke of luck: a pair
$r_1=8,\ r_2=9$ looks nice to start with.
When the inradii are known,
the radius of semicircle $R$ can be easily found as

\begin{align}
R&=
\frac{2r_1 r_2(r_1+r_2+2\sqrt{2r_1 r_2})}{6r_1 r_2-r_1^2-r_2^2}
.
\end{align}

Using complex numbers to express coordinates of the points, we have:

\begin{align}
O&=0
\\
,\quad
O_1&=
O-\sqrt{(R-r_1)^2-r_1^2}+r_1\i
=-\tfrac{48}7\sqrt2+8\i
,\\
O_2&=
O+\sqrt{(R-r_2)^2-r_2^2}+r_2\i
=\frac{36}7\sqrt2+9\i
,\\
D&=O_{1x}=-\frac{48}7\sqrt2
,\\
E&=O_{2x}=\frac{36}7\sqrt2
,\\
F&=O_1+r_1\cdot\frac{O_1-O}{|O_1-O|}
=-\frac{864}{77}\sqrt2+\frac{144}{11}\i
,\\
G&=O_2+r_2\cdot\frac{O_2-O}{|O_2-O|}
=\frac{64}7\sqrt2+16\i
,\\
T&=O_1+r_1\cdot\frac{O_2-O_1}{|O_2-O_1}
=-\frac{144}{119}\sqrt2+\frac{144}{17}\i
.
\end{align}

Given these known points, the corresponding areas are

\begin{align}
s_1=S_{TFD}&=\frac{10368}{187}\sqrt2
=\frac{288}{187}\sqrt2\cdot 36
,\\
s_2=S_{TDE}&=\frac{864}{17}\sqrt2
=\frac{288}{187}\sqrt2\cdot 33
,\\
s_3=S_{TEG}&=\frac{1152}{17}\sqrt2
=\frac{288}{187}\sqrt2\cdot 44
,\\
s_4=S_{TGF}&=\frac{11520}{187}\sqrt2
=\frac{288}{187}\sqrt2\cdot 40
,
\end{align}

so the solution was obvious.

$\endgroup$

Title: Two circles inscribed in a semicircle


问:
The closure of a totally bounded set in a metric space is also totally bounded
Let $(X, d)$ be any metric space, and let $E \subset X$ such that
$$
E \subset \cup_{i = 1}^k B \left( x_i, \epsilon \right),
$$
for some points $x_1, \ldots, x_k \in X$ and for some $\epsilon > 0$. Then how to show that
$$
\overline{E} \subset \cup_{i = 1}^k B \left( x_i, 2\epsilon \right)?
$$

My Attempt:

Let $p$ be any point of $\overline{E}$. Suppose that  $d \left( p, x_i \right) \geq 2 \epsilon$ for each $i = 1, \ldots, k$. Then since,  for any point $x \in E$, there is a point $x_{i_0}$ such that $d \left( x, x_{i_0} \right) < \epsilon$, therefore we obtain
$$
d (p, x) \geq d \left( p, x_{i_0} \right) - d\left( x_{i_0}, x \right) \geq 2 \epsilon - \epsilon = \epsilon.
$$
So for any real number $\delta$ such that $0 < \delta < \epsilon$, we have
$$
B (p, \delta) \cap E = \emptyset,
$$
which contradicts the assumption that $p \in \overline{E}$. Hence our supposition is wrong and thus there exists an $i = i_1$ such that $d \left( p, x_{i_1} \right) < 2 \epsilon$.

Is this proof correct and clear enough? Or, are there any errors?

答:
Yes, this is correct, but a direct proof is more elegant: Let $x \in \overline{E}$ and $y \in E$ such that $d(x, y) < \varepsilon$. Furthermore let $j \in \lbrace 1, ..., k \rbrace$ such that $y  \in B(x_j, \varepsilon)$. Then:
$$
d(x, x_j) \leq d(x, y) + d(y, x_j) < \varepsilon + \varepsilon = 2\varepsilon
$$
Therefore $x \in B(x_j, 2 \varepsilon) \subseteq \displaystyle \bigcup_{i = 1}^k B(x_i, 2 \varepsilon)$.


问:
A question on euclid's lemma
Euclid's lemma:
Let $A$ be a principal ring with $x$ and $y$ belonging to $A$, and $p$ an irreducible element of $A$. If $p$ divides
the product $xy$, then $p$ divides $x$ or $p$ divides $y$.

Definition of irreducible element: A non zero element p is said to be irreducible in A (where A is an integral domain) if p is non invertible in A and all divisors of p in A are trivial. In other words a nonzero element p of A is irreducible iff p is non invertible and the only divisors of p in A are the invertible elements of A and the elements associated to p.

Question: Should $A$ be an integral domain too? And I need a justification why please. I'm not quite grasping why an integral domain can help with such lemmas.

答:
You don't need the assumption that $A$ is integral.

In fact, the following proof works for any commutative principal ring $A$ (understood as every ideal being generated by one element).

Proof: Suppose $p$ divides $xy$ with $p$ irreducible.

Since $A$ is principal, there exists $u\in A$ such that $(u) = (p, x)$.

It follows that there exists $t, r \in A$ such that $p = ut$ and $x = ur$. But $p$ is irreducible, hence either $u$ or $t$ is a unit.

If $t$ is a unit, then we have $x = ur = t^{-1}pr$ which means that $p$ divides $x$.

If $u$ is a unit, then we have $(p, x) = (u) = A$ and hence there exists $a, b \in A$ such that $pa + xb = 1$. It follows that $p$ divides $pay + xyb = y$.

Title: A question on euclid's lemma


问:
Finding closed form for $\sum_{j = \lceil n/2 \rceil}^n \binom{n}{j} \binom{j}{n-j}$
I'm having some trouble finding a closed formula for this quantity:

\begin{align*}
\sum_{j} \binom{n}{j}\binom{j}{n-j}
\end{align*}
I know that this size is equal to
\begin{align*}
\sum_{j} \binom{n}{j}\binom{n-j}{j}
\end{align*}
And also
\begin{align*}
\sum_j \binom{n}{2j}\binom{2j}{j}
\end{align*}
All these sums are over all allowed $j$, that is, so the lower index in the binomial coefficient is not greater than the upper, and is not negative. Can someone help me? Thanks :)

Edit: It seems like people agree that there might not be a simple closed form for this sum.

Edit: Apparently, this sum is the n-th term in the sequence of central trinomial coefficients, which do not have a closed form that are on my level. I did not know that at the time. Thanks for your answers!

答:
Mathematica says (for Jean Marie's version):

$$\binom{n}{\frac{n}{2}} \,
   _2F_1\left(-\frac{n}{2},-\frac{n}{2};\frac{1}{2};\frac{1}{4}\right)$$

If this is really the best that can be done, then there is no "closed form".


答:
Actually, there are no integer solutions to
$$ s^2 - 156 t^2 = -3$$

Which tells us that there are no integral solutions to
$3m^2 - 3 m + 1 = 13 n^2$

The solutions   to
$$ u^2 - 12 v^2 = -3$$
have $v$ in the sequence $1, 13, 181, 2521, 35113, \ldots$ with
$$ v_{k+2} =14 v_{k+1} - v_k$$
so that there are infinitely many solutions to
$3m^2 - 3 m + 1 =  n^2$

The bit about 156 amounts to this list

```<code>624    factored   2^4 * 3 *  13

    1.             1          24         -12   cycle length             2
    2.            -1          24          12   cycle length             2
    3.             3          24          -4   cycle length             2
    4.            -3          24           4   cycle length             2
    5.             5          22          -7   cycle length             4
    6.            -5          22           7   cycle length             4
    7.             7          22          -5   cycle length             4
    8.            -7          22           5   cycle length             4

  form class number is   8
</code>```

or this cycle

```<code>0  form   1 24 -12   delta  -2
1  form   -12 24 1   delta  24
2  form   1 24 -12
</code>```

or this presentation of the continued fraction for $\sqrt{156}$

$$  \sqrt { 156} = 12 +     \frac{  \sqrt {156} - 12 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {156} - 12 }  =  \frac{    \sqrt {156} + 12 }{12 } = 2 +  \frac{    \sqrt {156} - 12 }{12 } $$
$$    \frac{ 12 }{   \sqrt {156} - 12 }  =  \frac{    \sqrt {156} + 12 }{1 } = 24 +  \frac{    \sqrt {156} - 12 }{1 } $$

Simple continued fraction tableau:

$$
 \begin{array}{cccccccccc}
 & & 12 & & 2 & & 24 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 12 }{ 1 }   & &   \frac{ 25 }{ 2 }    \\
  \\
 & 1 & & -12 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 156 \cdot 0^2 = 1 &     \mbox{digit}  &  12  \\
  \frac{ 12 }{ 1 }   & 12^2 - 156 \cdot 1^2 = -12 &     \mbox{digit}  &  2  \\
  \frac{ 25 }{ 2 }   & 25^2 - 156 \cdot 2^2 = 1 &     \mbox{digit}  &  24  \\
 \end{array}
 $$

Title: Is there a way to prove that $A^3 - (A-1)^3 = 13^x$ is false when $x$ is an odd positive integer greater than $1$?

Body:
In the case of $A^3 - (A-1)^3 = B^x$ we can find some rare examples, such as:

$8^3-7^3 =13^2$

$28712305723921^3−28712305723920^3=49731172316281^2$

But according to the Beal's Conjecture $A^3 - (A-1)^3 = B^x$ should be false if $x$ is an odd positive integer greater than $1$.

So I was trying to prove the specific case of $A^3 - (A-1)^3 = 13^x$.

For $13^x$ when $x$ is an odd positive integer greater than $1$, we have:

$13^x = C$

$13^3=2197$

$13^5=371293$

$13^7=62748517$

$13^9=10604499373$

$13^{11}=1792160394037$

$13^{13}=302875106592253$

...

As you can see $C$ always ends in $7$ or $3$.

I have then listed the following:

$A^3 - (A-1)^3 = D$:

$1^3-0^3=1$

$2^3-1^3=7$

$3^3-2^3=19$

$4^3-3^3=37$

$5^3-4^3=61$

$6^3-5^3=91$

$7^3-6^3=127$

...

As you can see:

$D$ grows by $+6+12+18+24+30+36...$

$D$ never ends in $3$, therefor all the results of $C$ ending in $3$ are automatically excludes.

$C$ always ends in $7$ and $3$, therefor all the results in $D$ ending in $1$ or $9$ are automatically excluded.

The list resulting in $D$ is now left with only numbers ending in: $7$:

$7, 37 ,127 ,217,397, 547,817, 1027 ,1387, 1657, 2107...$

We have two growing sets here:

$7, 127 , 397 ,817,1387,2107..$ growing at $+120,+270,+420,+570,+720...$ (growing at $+150$)

$37,217,547,1027,1657...$  growing at $+180,+330,+480,+630...$ (growing at $+150$)

So we can now take any resulting $C$ and subtract either $-120-270-420-570-720-...$ to check whether it will land on $7$ OR either $-180-330-480-630-...$ to check whether it will land on $37$

So these are the results:

$13^3=2197$

$2197-120-270-420-570-720-...=97$

$2197-180-330-480-630-...=577$

$13^7=62748517$

$62748517-120-270-420-570-720-...=52687$

$62748517-180-330-480-630-...=134977$

$13 ^ {11} = 1792160394037$

$1792160394037-120-270-420-570-720-...=7020817$

$1792160394037-180-330-480-630-...=20933137$

...

As you can see the results always grow further and further away from possibly being equal to $7$ or $37$.

Assuming it will prove the specific case of $A^3 - (A-1)^3 = 13^x$, is there a way to prove that these results can only grow further away?

Body:
Actually, there are no integer solutions to
$$ s^2 - 156 t^2 = -3$$

Which tells us that there are no integral solutions to
$3m^2 - 3 m + 1 = 13 n^2$

The solutions   to
$$ u^2 - 12 v^2 = -3$$
have $v$ in the sequence $1, 13, 181, 2521, 35113, \ldots$ with
$$ v_{k+2} =14 v_{k+1} - v_k$$
so that there are infinitely many solutions to
$3m^2 - 3 m + 1 =  n^2$

The bit about 156 amounts to this list

624    factored   2^4 * 3 *  13

    1.             1          24         -12   cycle length             2
    2.            -1          24          12   cycle length             2
    3.             3          24          -4   cycle length             2
    4.            -3          24           4   cycle length             2
    5.             5          22          -7   cycle length             4
    6.            -5          22           7   cycle length             4
    7.             7          22          -5   cycle length             4
    8.            -7          22           5   cycle length             4

  form class number is   8

or this cycle

0  form   1 24 -12   delta  -2
1  form   -12 24 1   delta  24
2  form   1 24 -12

or this presentation of the continued fraction for $\sqrt{156}$

$$  \sqrt { 156} = 12 +     \frac{  \sqrt {156} - 12 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {156} - 12 }  =  \frac{    \sqrt {156} + 12 }{12 } = 2 +  \frac{    \sqrt {156} - 12 }{12 } $$
$$    \frac{ 12 }{   \sqrt {156} - 12 }  =  \frac{    \sqrt {156} + 12 }{1 } = 24 +  \frac{    \sqrt {156} - 12 }{1 } $$

Simple continued fraction tableau:

$$
 \begin{array}{cccccccccc}
 & & 12 & & 2 & & 24 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 12 }{ 1 }   & &   \frac{ 25 }{ 2 }    \\
  \\
 & 1 & & -12 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 156 \cdot 0^2 = 1 &     \mbox{digit}  &  12  \\
  \frac{ 12 }{ 1 }   & 12^2 - 156 \cdot 1^2 = -12 &     \mbox{digit}  &  2  \\
  \frac{ 25 }{ 2 }   & 25^2 - 156 \cdot 2^2 = 1 &     \mbox{digit}  &  24  \\
 \end{array}
 $$


问:
Changing order of limit and complex derivative
Assume $f(z,t)$ is complex differentiable at $z=z_0$, $t$ is a real parameter. Under which condition is $\lim_{t \rightarrow t_0} f(z,t)$ also complex differentiable at $z = z_0$?

I know some theorems and their proofs from real analysis, but I was curious whether I could prove the corresponding theorems in the complex domain but I am unsure what the condition of limit switching theorem might be!

EDIT

How does the result extends to interchangeability of derivatives? I.e. assume $f(z,t)$ is both $z \in \mathbb{C}$ and $t \in \mathbb{R}$ differentiable in some neighborhood of $z_0$ and $t_0$. Under which conditions we have that $\frac{\partial f}{\partial t}$ is also $z$ differentiable?

ADDENDUM
(Comment to @Bobby Ocean answer)
Consider a following: Somehow you have proven that $z^t$ is $z$ complex differentiable for each real $t$ (except branches). Then you define a function $g(z) = \lim_{t \rightarrow 0} (z^t - 1)/t$, how do you prove this is also complex differentiable? I would like to define the logaritmic function this way, and proving $g'(z)=1/z$ would require knowing whether I can interchange the derivatives.

答:
My answer is in response to your second question: "Under which conditions we have that $\frac{\partial f}{\partial t}$ is also $z$ differentiable?"

Suppose $U$ is open in $\mathbb C$ and $g:U\to \mathbb C.$ Assume $g\in C^1(U).$ Define the operator

$$D=\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}.$$

Then $g$ is analytic in $U$ iff $Dg\equiv 0$ in $U.$ This follows easily from the Cauchy-Riemann equations.

Now suppose $f:U\times I\to \mathbb C$ with $f\in C^2(U\times I).$ (Here $U$ is open in $\mathbb C$ and $I$ is open in $\mathbb R.$) This is your $f(z,t)$ here. Think of $z=x+iy.$

Assume that for fixed $t,$ $f$ is complex differentiable in $z.$ The $C^2$ hypothesis implies switching the order of derivatives in $x,y,t$ leaves the result unchanged. Thus

$$D\frac{\partial}{\partial t} f(x+iy,t) =  \frac{\partial}{\partial t} Df(x+iy,t).$$

By hypothesis, the inner partial derivative on the right is $0$ for all $(z,t)\in U\times I.$ Hence the left side vanishes identically. This proves $\dfrac{\partial}{\partial t} f(z,t)$ is analytic in $U$ for each fixed $t.$

Thus $C^2$ smoothness (in the real sense) is a sufficient condition for $\partial f/\partial t $ to be complex differentiable with respect to $z.$


问:
Expectation of a game where player doubles the bet every time he wins
Player P plays a game whereby he bets on an event E occuring.

Event E occurs with probability q. So ¬E occurs with probability (1 - q)

The payoff per round is 1:1 -> For every 1 unit P bets on E , if E occurs, E collects 1 unit as the payoff, and keeps the 1 unit he bet; and if ¬E occurs, P loses the 1 unit he bet.

He adopts a strategy of playing agressive when he is winning, and playing safe when he is losing:

P plays exactly 10 rounds of the game

P always bets on E in every round

P will start the first round with a bet of 1 unit

After each and every loss, P will bet 1 unit on the next round, if there is a next round

If P wins the i^th round, he will increase his bet for the next round ( (i + 1)^th round ) by 2 times of the amount he bet in the i^th round.

ie, P bets 1 unit in i^th round and won. He will bet 2 units in the (i+1)^th round. If he wins again, he will bet 4 units in the (i+2)^th round.

After a triple consecutive win, P returns to betting 1 unit on the following round.

Rules 1 to 6 applies to every round of the game.

An example of a game,

Outcomes
E
E
E
E
¬E
¬E
E
E
¬E
¬E

Bets
1
2
4
1
2
1
1
2
4
1

Question: What is the expectation of the payoff if P adopts this strategy?

Note: No use of markov chain

##################################################

I need help with this question. I was thinking of using the usual way of calculating expectation, until I realised that the payoffs for games with the same number of wins and losses differs with the sequence of occurances of the wins and losses.

I tried to calculate the expectation of playing 1 round, 2 rounds, 3 rounds ... individually to try to find a relationship between them, but to no avail. I could not find any perceivable relation. In fact, the expectation became too tedious to calculate beyond 3 rounds.

Is there any intuitive way of solving this problem?

答:
Suppose we have $n$ rounds to go and P's stake is going to be $x$ on the first of those rounds.  $x$ can be 1, 2 or 4.  If it's 1 or 2 and they win they double their stake, but if it's 4 and they win, that signals three wins and their stake returns to 1.

Let $E(n,x)$ be the expected payout on the $n$ rounds assuming the stake of $x$.

The following three equations then follow (by considering the probability of a win then the probability of a loss):

$E(n+1,1) = q(1+E(n,2))+(1-q)(-1+E(n,1)) = 2q-1 + (1-q)E(n,1)+qE(n,2)$

$E(n+1,2) = q(1+E(n,4)) + (1-q)(-2+E(n,1))=2(2q-1)+(1-q)E(n,1)+qE(n,4)$

$E(n+1,4) = q(1+E(n,1)) + (1-q)(-4+E(n,1))=4(2q-1)+E(n,1)$

Initial conditions for these recurrence relations are:

$E(1,1) = q + (1-q)(-1) = 2q - 1$

Similarly, $E(1,2) = 2(2q-1)$

and $E(1,4) = 4(2q-1)$

If we express this using matrices, then:

Define $E_n = \begin{pmatrix}
    E(n,1) \\
    E(n,2) \\
    E(n,4) \\
    \end{pmatrix}$

Define $u = (2q-1)\begin{pmatrix}
    1 \\
    2 \\
    4 \\
   \end{pmatrix}$

Define $Q = \begin{pmatrix}
 1-q & q & 0 \\
 1-q & 0 & q \\
 1 & 0 & 0 \\
 \end{pmatrix}$

Then $E_1 = u$

and $E_{n+1} = u + Q E_n$

The solution to this takes the form

$E_n=Q^n a + b$

where $b = (I-Q)^{-1} u$ and $a = Q^{-1} (u-b)$

Now all that remains is to calculate $Q^{-1}$ and $Q^{10}$, use those to calculate $a$ and $b$ and so calculate $E_{10}$ from which $E(10,1)$ can be deduced.


答:
Some simple manipulation shows that the answer is this sequence, which doesn't seem to have a nice closed formula.

The "simplest" formula indeed involves hypergeometric functions. See reference in the linked page.

Title: Finding closed form for $\sum_{j = \lceil n/2 \rceil}^n \binom{n}{j} \binom{j}{n-j}$

Body:
I'm having some trouble finding a closed formula for this quantity:

\begin{align*}
\sum_{j} \binom{n}{j}\binom{j}{n-j}
\end{align*}
I know that this size is equal to
\begin{align*}
\sum_{j} \binom{n}{j}\binom{n-j}{j}
\end{align*}
And also
\begin{align*}
\sum_j \binom{n}{2j}\binom{2j}{j}
\end{align*}
All these sums are over all allowed $j$, that is, so the lower index in the binomial coefficient is not greater than the upper, and is not negative. Can someone help me? Thanks :)

Edit: It seems like people agree that there might not be a simple closed form for this sum.

Edit: Apparently, this sum is the n-th term in the sequence of central trinomial coefficients, which do not have a closed form that are on my level. I did not know that at the time. Thanks for your answers!


问:
Complex number to fractional power
I am trying to write $(-i)^\frac{1}{2}$ in the form $c=a+ib$ where $a,b\in\mathbb{R}$.

I am using De Moivre's formula and have $\cos(\frac{3\pi}{4}+\frac{k\pi}{2})+i\sin(\frac{3\pi}{4}+\frac{k\pi}{2})$.

I am not sure if this gives me a principal argument of $\frac{3\pi}{2}$ or $\frac{3\pi}{8}$. Any help appreciated.

答:
The $n$-th complex root is not unique, and hence it doesn't make sense to talk about the principal argument here.

For instance, you have
$$(-1)^{\frac 12}=\pm \mathrm i,$$
both roots may be written as
$$\cos\frac\pi 2+\mathrm i\,\mathrm\sin \frac\pi2$$
and
$$\cos\frac{-\pi} 2+\mathrm i\,\mathrm\sin \frac{-\pi}2.$$

Both arguments are principal.

Title: Complex number to fractional power


答:
Remark on interpretation

The equals sign in big O notation is an unfortunate but prevalent abuse of notation which makes it harder to interpret relationships such as $f(x) + O(g(x)) = O(g(x))$.

In particular, note the asymmetry: it is true that $O(n) = O(n^2)$, but it is not true that $O(n^2) = O(n)$. A clearer, but rarely used notation for $f(x) = O(g(x))$ is $f(x) \in O(g(x))$ which highlights the asymmetry. It also shows that the appropriate interpretation of $f(x) + O(g(x)) = O(g(x))$ is:

For every $h(x) = O(g(x))$, we have $f(x) + h(x) = O(g(x))$.

Proof

The relationship can be proven using the "limit" definition of big-O which says

$$
a(x) = O(b(x))\iff \limsup_{x\to\infty} \frac{|a(x)|}{b(x)} \lt +\infty.
$$

Now, $h(x) = O(g(x))$ means that

$$
\limsup_{x\to\infty} \frac{|h(x)|}{g(x)} \lt +\infty
$$

and $f(x)+h(x)=O(g(x))$ says that

$$
\limsup_{x\to\infty} \frac{|f(x) + h(x)|}{g(x)} \lt +\infty.
$$

Finally, observe that

$$
\limsup_{x\to\infty} \frac{|f(x)|}{g(x)} \le \limsup_{x\to\infty} \frac{|f(x)+h(x)|}{g(x)}+\limsup_{x\to\infty} \frac{|h(x)|}{g(x)} \lt +\infty
$$

which means that $f(x)=O(g(x))$.

Title: Big $O$ notation on both the sides of equality

Body:
If $$ f(x)+O(g(x))=O(g(x)),$$

is it possible to write $$ f(x)=O(g(x))$$

and if yes, how does that work?


问:
Relative growth, adding a constant rate
Hey I was just confused on how I should be approaching this question I have for homework.
I have
$$\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right).$$

Where $k$ is relative growth, $P$ population, $M$ carrying capacity. If I were to add a constant rate for example, $Z$, that would increase the population growth of this function, would I add $Z$ within the brackets or outside of the brackets?

*edit: thanks for the help everyone, sorry I am a newb !

答:
It depends on what you are going for.

Case 1 - Add inside the brackets

\begin{align*}
\frac{dP}{dt} &= kP\left(1-\frac{P}{M}+Z\right)\\
&=  kP-\frac{kP^2}{M}+kZP\\
&= (1+Z)kP-\frac{kP^2}{M}\\
&=(1+Z)kP\left(1-\frac{P}{(1+Z)M}\right).
\end{align*}
Note that this basically corresponds to scaling both the relative growth rate and the carrying capacity by the same factor, which I don't think is what you are trying to do.

Case 2 - Add outside the brackets

$$\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right) + Z.$$

Here we are increasing the overall growth rate of the population by a constant of $Z$ individuals per unit time, which I think is probably what you are trying to accomplish.

Title: Relative growth, adding a constant rate


问:
Let $W = \{A ∈ M(3, \mathbb{R}) | A = A^T \text{ and } \operatorname{tr}(A) = 0\}.$ Write down a basis for $W.$
I asked a question about basis and dimension earlier, and that arose from this question. Apparently, the dimension is 5 but, I can't wrap my head around why the dimension exceeded 3.</p>

答:
Every matrix $A\in W$ can be written in the following form

$$
A(x, y, z, w, t) = \begin{pmatrix}
x & y & z \\
y & w & t \\
z & t & - x - w
\end{pmatrix}
$$

giving us a total of five independent parameters. It is easy to check that the set $$\{A(1, 0, 0, 0, 0),\\ A(0, 1, 0, 0, 0),\\ A(0, 0, 1, 0, 0),\\ A(0, 0, 0, 1, 0),\\ A(0, 0, 0, 0, 1)\}$$

is a basis of $W$.


问:
Equal annihilator implies associates
Let $\mathbb Z_n$ be the ring of integers modulo $n$. Let $x,y \in \mathbb Z_n$ be such that $ann(x) = ann(y)$ where $ann(x) = \{a \in \mathbb Z_n : ax =0 \text{ in } \mathbb Z_n\}$. Then $(x,n) = (y,n)$ and hence $x = u y$ for some unit $u \in \mathbb Z_n$.

is this result is true in an arbitrary finite commutative ring with unity?

Thank you.

答:
Counterexample

In $R=F_2[x,y]/(x,y)^2$, $ann_R(x)=ann_R(y)=(x,y)$ but $xR\cap yR=\{0\}$.

An extension

If $R$ in any ring with the annihilator hypothesis that you wrote for $x,y\in R$, we have, by homomorphism theorems, that the left multiplication homomorphism factors through the right multiplication homomorphism to produce a homomorphism $g:xR\to yR$ such that $g(xr)=yr$ for all $r\in R$.

Because $R$ is finite and the annihilators are equal, $|xR|=|yR|$, and since $g$ is necessarily onto, this means that $g$ is an isomorphism between $xR$ and $yR$.

If $R$ is additionally right self-injective, $g$ extends to $\tilde{g}$ from $R\to R$. At this point, $\tilde{g}(1)xr=yr$ for all $r\in R$.

We claim that $\tilde{g}(1)$ is not a zero divisor.  If it were, and $r\neq 0$ was such that $\tilde{g}(1)r=0$, we'd have that $0=\tilde{g}(1)xr=g(xr)=yr$, but this contradicts the injectivity of $g$ on $xR$.  We must conclude $\tilde{g}(1)$ is not a zero divisor.

Now in a finite ring, each element is either a unit or a zero divisor, so we must conclude that $\tilde{g}(1)$ is a unit. In the special case of $r=1$ that reads that $ux=y$ where $u=\tilde{g}(1)$.

So the theorem holds for finite commutative self-injective rings. All of the rings $\mathbb Z/(n)$ are such rings.

Title: Equal annihilator implies associates


问:
Second homology group of Hirzebruch surface
Let $L_0$ denote the trivial complex line bundle over $S^2$ and let $L_n$ denote the complex line bundle over $S^2$ with $c_1[L_n]=n \in \Bbb Z$. Let $\Bbb H_n$ denote the $n$-th Hirzebruch surface, which is the projectivization $\Bbb P(L_n\oplus L_0)$. Thus $\Bbb H_n$ is a $S^2$-bundle over $S^2$. Let $s,f\in H_2(\Bbb H_n;\Bbb Z)$ denote the homology class of the zero section  and a fiber, respectively. I want to show that $H_2(\Bbb H_n;\Bbb Z)$ is freely generated by $s$ and $f$ (without using algebraic geometry if possible).

What I've tried is as follows: Since $\Bbb H_n$ is simply-connected, $H_1(\Bbb H_n)=0$ and hence $H_3(\Bbb H_n)=0$. Decomposing the base $S^2$ to upper and lower hemispheres, we can decompose $\Bbb H_n$ into union of two copies of $S^2\times D^2$ with intersection $S^2\times S^1$. Using Mayer-Vietoris, we get $0\to H_2(S^2\times S^1)=\Bbb Z \to H_2(S^2\times D^2)\oplus H_2(S^2\times D^2)=\Bbb Z^2 \to H_2(\Bbb H_n)\to H_1(S^2\times S^1)=\Bbb Z\to 0$. Thus $\text{rank} H_2(\Bbb H_n)=2$. But what I can't show are:

How do we know that $H_2(\Bbb H_n)$ is torsionfree?

How do we know that $s$ and $f$ generate $H_2(\Bbb H_n)$?

答:
There is a vast generalization of this due to the Leray-Hirsch theorem. Namely, if $P$ is the projectivization of a rank $n$ bundle $E \rightarrow X$ then the cohomology of $P$ as a module over $H^* (X)$ (action given by the pullback) is $H^*(X) \otimes H^*(\mathbb{C}P^n)$.

Here is a refrence written by Stephan Stolz, see pages 53-55.


Title: $R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$

Body:
So the question is:

$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.

I don't really know what to do from here. If I substitute

$$R_n = 3(2^n)-4(5^n)$$

into

$$Rn = 7R_{n-1}-10R_{n-2}$$

I end up getting

$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$

Dont know what to do...

EDIT: Thanks to Zev, what I did was:

$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$

Body:
Let $\,S\,$ be the shift operator: $\,S f_n = f_{n+1}.\,$  Note $\,(S\!-\!a)(ca^n)\! = ca^{n+1}\!-ca^{n+1} = 0.\,$ The recurrence is $\, 0 = (S^2\! - 7S + 10)f_n = (S\!-\!5)(S\!-\!2) f_n,\,$ which is easily verified as follows

$$\begin{eqnarray}(\color{#0a0}{S\!-\!2)2^n\! = 0}\\ \color{#c00}{(S\!-\!5)5^n\! = 0}\end{eqnarray}\!\Rightarrow (S\!-\!5)(\color{#0a0}{S\!-\!2})(\color{#0a0}{a2^n}\!+\!b5^n) = (S\!-\!5)(S\!-\!2)(b5^n) \stackrel{(\star)}= (S\!-\!2)(\color{#c00}{S\!-\!5})(\color{#c00}{b5^n}) = 0$$

$(\star)$ Key here is that that $\,S\!-\!5\,$ commutes with $\,S\!-\!2\,$ because they have constant coefficients, where $\,c\,$ constant means $\,Sc = cS,\,$ i.e. $\, S cf_n = cSf_n = cf_{n+1}$.


问:
Is there a difference between one or two lines depicting the norm?
I know that the  |  stands for the norm, however I see it sometimes with two lines, sometimes with one. Is there a difference? Do I need to take care when writing a formula?

So basically, is there a difference between

$|a|$ and $\|a\|$ ?

Is it the same when it comes to vectors instead of just numbers?

答:
If I posted this as a question it would be a duplicate, so I decided to post it as an answer in the hope that it either gets confirmed or rejected - if the former, it will stay as a valid answer; if the later, it may stay as a counter-example, or be deleted.

Even though it seems as harping on $\vert \cdot \vert$ versus $\Vert \cdot \Vert$ is as pedantic as useless, there is some sense in which it reinforces the concept, and that is in the difference in the space of continuous functions between the absolute value of the function $\vert \cdot \vert,$ as in, for example for $f(x)=\sin x:$

$f(x) = \vert \sin x \vert$

and different norms, e.g.

$$\Vert f(x) \Vert_1 = \int_0^{2\pi}\vert \sin x \vert dx =4$$

vs

$$\Vert f(x) \Vert_2 = \left( \int_0^{2\pi}\vert \sin x \vert^2 dx \right )^{1/2} =\sqrt \pi$$

or

$$\Vert f(x) \Vert_\infty = \underset{0\leq x \leq 2\pi}{\max}\left(\vert \sin x \vert \right ) =1$$


问:
Show (inner) product of two infinite sums is convergent or divergent.

Show that if ${ \sum_{n=1}^{\infty}{b_n}}$ absolutely convergent and ${ \sum_{n=1}^{\infty}{a_n} }$ convergent then the inner product ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ is also convergent.

Introductory Real Analysis by F. Dangello and M. Seyfried.

I feel like this could be proved using the Dirichlet's test but I did not know how to approach it from that direction. I also read something about Hadamard products on power series but I don't know if I could use it to prove my question. Furthermore, I read this question Convergence of product of convergent series but since I only have $b_n \ge 0$ the question is different.

My approach was to try to use the Cauchy condition stating that:

Since ${ \sum_{n=1}^{\infty}{a_n}}$ convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^na_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.

Furthermore, ${ \sum_{n=1}^{\infty}{b_n}}$ (absolutely)convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^nb_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.

Then, $\vert\sum_{n=1}^{\infty}{a_n}\vert \vert\sum_{n=1}^{\infty}{b_n} \vert = \vert \sum_{n=1}^{\infty}{a_n}{b_n} \vert \lt \sqrt\varepsilon\sqrt\varepsilon= \varepsilon$. Hence ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ convergent.

I feel like I made a mistake in the last line where I multiplied both series, and that my approach is too simplistic. So is my approach correct? Or where did I make a mistake? And how would I go about to prove this?

答: That proof is not correct. It is not true in general that $$\left\vert\sum_{n=1}^{\infty}{a_n}\right\vert\left\vert\sum_{n=1}^{\infty}{b_n}\right\vert=\left\vert \sum_{n=1}^{\infty}{a_n}{b_n}\right\vert.$$It is false even for finite sums:$$1=|1+0||0+1|\ne|1\times0+0\times1|=0.$$


答:
$\{a_n\}$ is eventually bounded by one as $\sum a_n$ is convergent. Hence for $N$ large enough and $n \ge N$

$$\vert a_n b_n \vert \le \vert b_n\vert$$

Which implies that $\sum a_n b_n$ is absolutely convergent as $\sum b_n$ is absolutely convergent.

Title: Show (inner) product of two infinite sums is convergent or divergent.

Body:
Show that if ${ \sum_{n=1}^{\infty}{b_n}}$ absolutely convergent and ${ \sum_{n=1}^{\infty}{a_n} }$ convergent then the inner product ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ is also convergent.

Introductory Real Analysis by F. Dangello and M. Seyfried.

I feel like this could be proved using the Dirichlet's test but I did not know how to approach it from that direction. I also read something about Hadamard products on power series but I don't know if I could use it to prove my question. Furthermore, I read this question Convergence of product of convergent series but since I only have $b_n \ge 0$ the question is different.

My approach was to try to use the Cauchy condition stating that:

Since ${ \sum_{n=1}^{\infty}{a_n}}$ convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^na_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.

Furthermore, ${ \sum_{n=1}^{\infty}{b_n}}$ (absolutely)convergent thus $\forall \varepsilon \gt0, \exists n_0 \in \mathbb{N} : \vert s_n -s_m\vert = \vert \sum_{k=m+1}^nb_k \vert \lt \sqrt\varepsilon$ if $ n \gt m \gt n_0$.

Then, $\vert\sum_{n=1}^{\infty}{a_n}\vert \vert\sum_{n=1}^{\infty}{b_n} \vert = \vert \sum_{n=1}^{\infty}{a_n}{b_n} \vert \lt \sqrt\varepsilon\sqrt\varepsilon= \varepsilon$. Hence ${ \sum_{n=1}^{\infty}{a_n}{b_n} }$ convergent.

I feel like I made a mistake in the last line where I multiplied both series, and that my approach is too simplistic. So is my approach correct? Or where did I make a mistake? And how would I go about to prove this?


答: A principal argument $\phi \in (-\pi, \pi]$. This implies that of the two possibilities you mentioned, only $\phi = \dfrac{3\pi}{8}$ works. In other words, $c = 3\cos \frac{3\pi}{8}+3i\sin \frac{3\pi}{8}$. Note here that there are a total of $4$ solutions to this equation, but if you are looking for a solution with a principal argument, then the $\phi = \dfrac{3\pi}{8}$ does the job !

Title: Complex number to fractional power

Body:
I am trying to write $(-i)^\frac{1}{2}$ in the form $c=a+ib$ where $a,b\in\mathbb{R}$.

I am using De Moivre's formula and have $\cos(\frac{3\pi}{4}+\frac{k\pi}{2})+i\sin(\frac{3\pi}{4}+\frac{k\pi}{2})$.

I am not sure if this gives me a principal argument of $\frac{3\pi}{2}$ or $\frac{3\pi}{8}$. Any help appreciated.


问:
Converting parametric equations to xy equation
I have these parametric equations:
$$
x=1+t \\
y=1-t
$$

How do I make this to the xy equation?

David

答:
Use the first equation to find an expression for $t$ in terms of $x$.

Substitute into the second equation to get an equation only involving $x$ and $y$.

Rearrange if desired.

Update: This method will work for the example you gave, but might not work for a general example.  For example, the following is a particularly common parametrization:

\begin{align}
x &= \cos t \\
y &= \sin t
\end{align}

Here, you can't get a unique expression for $t$ in terms of $x$: for example, if $x$ is $1$, we could have $t=0$ or $t=2\pi$.  In this case, the solution is to square both equations to give:

\begin{align}
x^2&=\cos^2 t\\
y^2&=\sin^2 t
\end{align}

Then add them together to get

$$
x^2+y^2=\cos^2t+\sin^2t=1
$$

This equation ($x^2+y^2=1$) is the equation for a circle of radius $1$ about $(0,0)$.

Edit: (answering the question in comment below) Sometimes you're presented with a parametrization like

\begin{align}
x&=1+t\\
y&=1-t
\end{align}

and you want to convert that into a vector equation.  A mathematician would immediately recognize that system of parametric equations as a vector equation: $x$ and $y$ are themselves coordinates of the vector $\begin{pmatrix}
x\\
y
\end{pmatrix}$.  The equations then become:

$$
\vec{r}=\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}1+t\\1-t\end{pmatrix}
$$

and you can rearrange that into whatever (vector) form you like using the usual rules for addition of vectors.


问:
Intuition for multiplying vectors by complex scalars
There is a very pleasing interpretation of multiplying a vector by a real scalar. I can ''stretch or shrink'' and also ''flip over the origin (in the case of negative scalars)''.

Is there some analogous way to think of scalar multiplication when the vector field is over $\mathbb{C}$ instead? Intuitively I feel like its the same idea, just with more freedom over where scalars can point the vector (in the same way that complex number multiplication rotates and stretches other complex numbers) but I can't seem to write down anything concrete.

答:
The problem you're having is that a function with a complex input and a complex output requires four dimensions to graph.  The best we can do is consider the complex multiplier in the form $re^{i\theta}$ so that $r$ is the scalar multiplier you are used to, and the $e^{i\theta}$ part we have to imagine as a rotation through some angle $\theta$ in some invisible dimension.
This is often considered as a "phase" at every point, a function.