问:
Smallest value for P(A)=P(B)=P(C) s.t. P(A and B and C) always exceeds 0.95
What is the smallest value for P(A)=P(B)=P(C) s.t. P(A and B and C) always exceeds 0.95?

I have made some attempts:

P(A and B and C)=P(A)P(B|A)P(C|A and B)>0.95. But how should the conditinal probabilities be handled?

or

P(A and B and C)=P(A)+P(B)+P(C)-P(A and B)-P(A and C)-P(B and C)+P(A or B or C)>0.95. It is unknown whether the events are independent or not e.g. P(A and C) is not necassary zero, so how to proceed in this case?

Thanks!

答:
$\min P(A \cap B) = P(A) + P(B) - 1$ if not negative or $0$ otherwise

Similarly $\min P(A \cap B \cap C) = P(A) + P(B) + P(C) - 2$ for same condition as above.

Here let $P(A) = P(B) = P(C) = x$, and minimum intersection is given: 0.95

So, solving equation $( 0.95 = 3x - 2 )$ gives $x = 0.983$


问:
can we use the limit to find the maximum of a function?
I am just wondering if we can find the maximum of function in [0,+inf] if this function is monotonically increasing by just finding the limit ( lim(x-->+inf) )</p>

答:
If you truly mean the function to be defined on $[0,\infty]$, then the hypothesis of being weakly increasing implies that its maximum on $[0,\infty]$ is just $f(\infty)$.

If you mean the function to be defined on $[0,\infty)$, then $\lim_{x\to\infty}f(x)$ will be the least upper bound of $f$ over $[0,\infty)$, which is its maximum on $[0,\infty)$ if and only if $f$ is eventually constant as $x\to\infty$.

ContentLicense: CC BY-SA 4.0


问:
Trying to track down an anecdote about the nature of proof
I recall (maybe incorrectly) a story about a mathematician trying to be a tutor to some nobleman (French, i think).  Nobleman got impatient with the intricacies of mathematical proof and asked "Why do you bother with this stuff?"  Tutor said "so you will know for sure that what I say is true."  Nobleman said "If that is the point, this is all unnecessary--i would not dream of doubting your word."

Who were the nobleman and tutor?  and what's the source?  or am i hallucinating this story?

thanks

herblin@stanford.edu

答:
It's almost surely "Le bourgeois gentilhomme", one of the foremost play of Molière (1670).

It is not a gentleman (gentilhomme) but a snobbish wealthy man (bourgeois) who hires private teachers in order to acquire "good manners" in different domains: dance, litterature, philosophy (here, its aristotelian logic component)...

The integral text can be found here. The part you mention is in pages 13-14.

I don't know if you are fluent in French ; I have given a translation below.

MAÎTRE DE PHILOSOPHIE, en raccommodant son collet.— Venons à notre leçon.

MONSIEUR JOURDAIN.— Ah! Monsieur, je suis fâché des coups qu'ils vous ont donnés.

MAÎTRE DE PHILOSOPHIE.— Cela n'est rien. Un philosophe sait recevoir comme il faut les choses, et je vais composer contre eux une satire du style de Juvénal, qui les déchirera de la belle façon. Laissons cela. Que voulez-vous apprendre?

MONSIEUR JOURDAIN.— Tout ce que je pourrai, car j'ai toutes les envies du monde d'être savant, et j'enrage que mon père et ma mère ne m'aient pas fait bien étudier dans toutes les sciences, quand j'étais jeune.

MAÎTRE DE PHILOSOPHIE.— Ce sentiment est raisonnable, Nam sine doctrina vita est quasi mortis imago. Vous entendez cela, et vous savez le latin sans doute.

MONSIEUR JOURDAIN.— Oui, mais faites comme si je ne le savais pas. Expliquez-moi ce que cela veut dire.

MAÎTRE DE PHILOSOPHIE.— Cela veut dire que sans la science, la vie est presque une image de la mort.

MONSIEUR JOURDAIN.— Ce latin-là a raison.

MAÎTRE DE PHILOSOPHIE.— N'avez-vous point quelques principes, quelques commencements des sciences?

MONSIEUR JOURDAIN.— Oh oui, je sais lire et écrire.

MAÎTRE DE PHILOSOPHIE.— Par où vous plaît-il que nous commencions? Voulez-vous que je vous apprenne la logique?

MONSIEUR JOURDAIN.— Qu'est-ce que c'est que cette logique?

MAÎTRE DE PHILOSOPHIE.— C'est elle qui enseigne les trois opérations de l'esprit.

MONSIEUR JOURDAIN.— Qui sont-elles, ces trois opérations de l'esprit?

MAÎTRE DE PHILOSOPHIE.— La première, la seconde, et la troisième. La première est, de bien concevoir par le moyen des universaux. La seconde, de bien juger par le moyen des catégories: et la troisième, de bien tirer une conséquence par le moyen des figures. Barbara, celarent, darii, ferio, baralipton, (see (*)) etc.

MONSIEUR JOURDAIN.— Voilà des mots qui sont trop rébarbatifs. Cette logique-là ne me
revient point. Apprenons autre chose qui soit plus joli.

Translation (Adapted from an initial one, rather accurate, from Google Translate):

MASTER OF PHILOSOPHY. (mending his collar) Let us come to our lesson.

MONSIEUR JOURDAIN. Oh! Sir, I am sorry for the blows they gave you.

MASTER OF PHILOSOPHY. That is nothing. A philosopher knows how to receive things properly, and I am going to compose a satire in the style of Juvenal against them, which will tear them apart in a beautiful way. Let’s leave that. What do you want to learn?

MONSIEUR JOURDAIN. All that I can, for what I desire the most in the world is to be a scholar, and I am enraged that my father and my mother did not make me study well all sciences, when I was young.

MASTER OF PHILOSOPHY. This feeling is understandable, Nam sine doctrina vita est quasi mortis imago. You hear that, and you probably know Latin.

MONSIEUR JOURDAIN. Yes, but pretend I don't know. Explain to me what that means.

MASTER OF PHILOSOPHY. This means that without science, life is almost an image of death.

MONSIEUR JOURDAIN. That Latin is right.

MASTER OF PHILOSOPHY. Have you been introduced to some principles, some bits of science?

MONSIEUR JOURDAIN. Oh yes, I can read and write.

MASTER OF PHILOSOPHY. Where would you like us to start? Do you want me to teach you logic?

MONSIEUR JOURDAIN. What is this logic?

MASTER OF PHILOSOPHY. It teaches the three operations of the mind.

MONSIEUR JOURDAIN. Who are these three operations of the mind?

MASTER OF PHILOSOPHY. The first, the second, and the third. The first is to conceive properly by means of universals. The second, to judge well by means of categories: and the third, to draw proper consequences by means of figures: Barbara, celarent, darii, ferio, baralipton, (see (*)) etc.

MONSIEUR JOURDAIN. These are words that are too off-putting. This logic does not appeal to me. Let's learn something else that is prettier.

(*) Personal note: these pseudo-latin words were mnemonics helping students to remember the main correct ways to construct syllogisms ; for example the voyels in "barbara", i.e., A A A, translated into our modern quantifiers meant $\forall$ and $\forall$, then $\forall$, for example:

All cats are felines, all felines are mammals, then all cats are mammals.


问:
Calculate limit of Floor $\Rightarrow (n^2+n+1)*[\frac {2019}{n^2+n+1}]$
I can't figure out how to solve it.

The sequence is :
$(n^2+n+1)*[\frac {2019}{n^2+n+1}]$

I tried using the formula $x-1\lt [x] \le x$ and find the limit using the squeeze theorem.

so I get $\frac {2019}{n^2+n+1} -1 \lt [\frac {2019}{n^2+n+1}] \le \frac {2019}{n^2+n+1}$ , multyipling each side by ($n^2+n+1$).

$2019-(n^2+n+1) \lt (n^2+n+1)[\frac {2019}{n^2+n+1}] \le 2019$.

well , $\lim_{n\to \infty} (2019-(n^2+n+1)) = -\infty $.

$\lim_{n\to \infty} 2019 = 2019$

No common limit between them , so I can't use the squeeze theorem.

Thank you

答:
You say ceiling, but you use the definition of floor. Ok, so:

if you meant floor

The floor of $[\frac{2019}{n^2+n+1}]=0$ is identically zero for $n$ big enough (say $n>2019$). So your sequence looks like $a_1, \dots, a_{2019}, 0, 0, 0, 0, 0, \dots$

In particular, then, its limit for $n \to \infty$ is $0$.

if you meant ceiling

Here things are more interesting. Note that $n \leq [n]$. Now you have an indeterminate form because $[\frac{2019}{n^2+n+1}] =1$ but $[\lim_{n \to \infty} \frac{2019}{n^2+n+1}] = [0] = 0$.

What you can do is consider the following:

$$ (n^2+n+1) \frac{2019}{n^2+n+1} \leq (n^2+n+1) \left[\frac{2019}{n^2+n+1}\right]$$

which tells you that the limit cannot be zero, as the sequence is uniformly bigger than $2019$.

Now note that for big enough $n$ (just as before, take $n>2019$),
$$(n^2+n+1) \left[\frac{2019}{n^2+n+1}\right] = n^2+n+1$$

since the quantity inside the brackets is nonzero, but certainly smaller than $1$, so its ceiling is $1$. In particular then your sequence gets arbitrarily big: for any number $M>0$, there is $N$ such that for all $n>N$ you have that $a_n > M$, so the limit is $+\infty$.


问:
Probability recursion on normal distribution
I'm currently reading this paper. I have one question on some probability part.
In the paper they have the following equations.
$$ P(A)\sim \mathcal {N}(\Pi,\tau \Sigma), r_A \sim \mathcal{N}(P(A),\Sigma) \tag{11}\label{11}$$
$$ r_A \sim \mathcal{N}(\Pi,(1+\tau)\Sigma)  \tag{12}\label{12}$$
My question is that I have absolutely no idea how they pass from $\eqref{11}$ to $\eqref{12}$. It is of course some lack in probability theory for something like recursive functions maybe?. I don't know. Can somebody explain this relationship? Why $\mathcal{N}(\mathcal {N}(\Pi,\tau \Sigma),\Sigma)$ equals $\mathcal{N}(\Pi,(1+\tau)\Sigma)$? (if I'm not mistaken)

答:
I believe the context of $r_A \sim \mathcal{N}(P(A),\Sigma)$, should be interpreted as the conditional distribution of $r_A$ given $P(A)$ and $r_A \sim \mathcal{N}(\Pi , (1+\tau)\Sigma))$ is then the marginal distribution $r_A$.

We can derive the relation in the following way. If $r_A | P(A) \sim \mathcal{N}(P(A),\Sigma)$, then $$(r_A - P(A))|P(A) \sim N(0,\Sigma),$$
but since this conditional distribution does not depend on $P(A)$ we may in fact conclude that $r_A - P(A)$ and $P(A)$ are independent random variables/vectors. Writing $r_A$ as
$$r_A = (r_A - P(A)) + P(A)$$
we may conclude, that $r_A$ is the sum of two independent variables/vectors with distributions $\mathcal{N}(0,\Sigma)$ and $\mathcal{N}(\Pi,\tau\Sigma)$ respectively, and thus conclude that
$$r_A \sim \mathcal{N}(\Pi , (1+\tau)\Sigma).$$


答:
Answer is Zero.
start with inside of brackets.as $n$ goes to infinity $\frac{2019}{n^2+n+1}$ goes to zero and is positive so its floor is absolute zero and we know that multiply of each number by absolute zero is zero.

Title: Calculate limit of Floor $\Rightarrow (n^2+n+1)*[\frac {2019}{n^2+n+1}]$

Body:
I can't figure out how to solve it.

The sequence is :
$(n^2+n+1)*[\frac {2019}{n^2+n+1}]$

I tried using the formula $x-1\lt [x] \le x$ and find the limit using the squeeze theorem.

so I get $\frac {2019}{n^2+n+1} -1 \lt [\frac {2019}{n^2+n+1}] \le \frac {2019}{n^2+n+1}$ , multyipling each side by ($n^2+n+1$).

$2019-(n^2+n+1) \lt (n^2+n+1)[\frac {2019}{n^2+n+1}] \le 2019$.

well , $\lim_{n\to \infty} (2019-(n^2+n+1)) = -\infty $.

$\lim_{n\to \infty} 2019 = 2019$

No common limit between them , so I can't use the squeeze theorem.

Thank you


问:
Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points
Well, this is a homewrok question (which I know I should not be asking, but I cannot find an answer to this anywhere):

The exercise is as follows:
i) Find the equation of the plane of $\mathbb{R}^3$ that contains the following three points:
 (0,0,1),(-1,3,5) and (1,-1,0)
ii) Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points

So I have managed to solve (i) using a lot of Google and a third party tool (my notes are not all that good), but I cannot find (ii) anywhere. Can someone prove this providing an example and explaining it properly so that I can understand the reason this is impossible?

Thanks in advance!

P.S: This is an exercise from a course on Matrixes and how to solve Systems using matrices, in case anyone wants to know.

答: Let $ax+by+cz=d$ be the equation of a plane (where $\{a,b,c\}\not=\{0\})$. Show that $(0,0,0),(0,0,1),(0,1,0),(0,0,1)$ cant satisfy the equation $ax+by+cz=d$ simultaneously. </p>

ContentLicense: CC BY-SA 3.0


问:
Finding CDF of a transformation of a random variable
Let $Y=F_X^{-1}(U)$ a random variable:

$U$ it is a random variable with a probability density function (pdf) $$f_U (u) = 1$$ for  $0 \leq u < 1$;

$X$ it is a random variable and $F^{-1}_X$ is the inverse of the cumulative distribution function  (cdf) of $X$.

Prove that the CDF of $Y$ is the same of X

I am really lost with this problem and I don't know how to start it so I can't show my work yet , so please can you give some hint of what I have to do?

Body:
Hint:
$$P(F^{-1}_X(U) \leq y) = P(U \leq F_X(y))$$


问:
How close is the sample mean $\mu_N$ for $N$ random variables to the sample mean $\mu_{N-1}$ for $N-1$ random variables?
Define $\mu_{N}$ to be the sample mean for $N$ i.i.d. random variables that each have mean $\mu$ and variance $\sigma^2$:
$$
\mu_{N} := \frac{1}{N} \sum_{i=1}^N X_i.
$$

Now consider the sample mean if we removed one of these random variables. That is consider, $\mu_{N-1}$ where w.l.o.g. we remove the last random variable from the sum:
$$
\mu_{N-1} := \frac{1}{N-1} \sum_{i=1}^{N-1} X_i.
$$

I am interested in describing how close $\mu_{N-1}$ is to $\mu_{N}$ in an appropriate sense..it is this 'sense' that I am unsure of as I haven't been working with probability that long.

Subtracting $\mu_{N-1}$ from $\mu_{N}$ we have
\begin{align}
\mu_{N} - \mu_{N-1} &= \frac{1}{N} \sum_{i=1}^N X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\
& = \frac{1}{N}X_N + \frac{N-1}{N}\frac{1}{N-1} \sum_{i=1}^{N-1} X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\
& = \frac{1}{N}X_N + \bigg(\frac{N-1}{N}-1\bigg)\frac{1}{N-1} \sum_{i=1}^{N-1} X_i.
\end{align}

So as $N\to \infty$, heurstically it 'looks like' both terms go to zero. But as the $X_i$ are random variables I imagine we need to set some conditions on them for this to hold. For example, we could be unlucky and draw $X_N$'s that are sufficiently large such that $\frac{1}{N} X_N$ does not go to zero as $N \to \infty$.

I feel like the variance $\sigma^2$ could have an effect on the convergence but I'm not sure.

So my questions are

In the finite sample case can we say something more informative than what I have already derived above regarding how close $\mu_N$ is to $\mu_{N-1}$?

What conditions are needed on the $X_i$ to ensure $\mu_{N} - \mu_{N-1}$ converges to zero as $N \to \infty$?

What is an appropriate sense of convergence for $\mu_{N} - \mu_{N-1}$? I am thinking maybe we can say it converges with 'high probability'. Most importantly, is it possible to find an exact rate of convergence?

答:
You need to read about the different concepts of convergence of random variables.

In this case, calling $Z_n =  \frac{1}{N} \sum_{i=1}^N X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i$, you want to know if and (in what sense ) $Z_n$ converges to zero. Notice that $Z_n$ is a random variable.

Given that $X_i$ are iid with finite mean and variance, you can deduce that $E[Z_n]=0$ and $\sigma_{Z_n}^2 =   \frac{1}{n(n+1)}\sigma_{X}^2$.

This says that $\sigma_{Z_n}^2 \to 0$ (and quite quick).

Then you can apply  Chebyshev's_inequality : for any $\epsilon >0$

$$P(|Z_n| > \epsilon) \le \frac{1}{\epsilon}\sigma_{Z_n}^2=\frac{1}{\epsilon \, n(n+1)}\sigma_{X}^2 < \frac{\sigma_{X}^2}{\epsilon \, n^2} $$

Then $P(|Z_n| > \epsilon) \to 0$ as $n\to \infty$

The above amounts to prove convergence in probability from convergence in $L^2$-norm. These are quite strong measures of convergence for random variables, though not so strong as almost sure convergence.


问:
Sequence of the ratio of two successive terms of a sequence
If $(a_n)_{n\in N}$ is a strictly decreasing sequence of real number converging to $0$ and s.t. $\forall n\in N$, $0<a_n<1$, does the following limit:
$$
\lim_{n}\frac{a_{n+1}}{a_n}
$$
exists?

When does the limit is strictly greater than $0$ (it can be zero if for example the sequence is defined as $a_n=\frac{1}{n!}$)?

Body:
Examine this sequence
$$1/2, 1/4, 1/16, 1/64, 1/128, 1/256, 1/1024, 1/4096,\cdots$$


问:
How do you go about learning mathematics?
I really like mathematics, but I am not good at learning it.  I find it takes me a long time to absorb new material by reading on my own and I haven't found a formula that works for me.  I am hoping a few people out there will tell me how they go about learning math so I can try out their systems.

I need to know basic things.

Should I use one book at a time or should I be reading many books on the same topic at once?

Do you stop reading when you hit on a fact that you don't understand or do you keep reading?

Do you read all in one go or do you do a little bit and for how long (1 hr, 2 hr or more?)

Do you read all the chapters or do you do all the exercises before moving on from a chapter?

Do you adjust your technique in Calculus (calculation heavy) vs. Analysis (proof heavy)? If so, how?

When you make notes, what do you make notes about? Do you make notes while you read or after?

If you think these decisions all depend, can you say what they depend on?

I am really lost here.  I would appreciate any input.

答: Maybe it would help if I told you a very short story:  a long time ago I would look out my window and wonder why about a lot of things.  Then I started studied differential equations.  I no longer wonder why about a lot of things.  My point is, most people live their lives in a cloud of ignorance, never understand why things are the way they are.  This causes an incalculable amount of suffering in the world.  But to understand why, now that  grants a great deal of peace of mind.  So my best recommendation for you is to keep working at it no matter how difficult it may seem, realizing with time, this understanding will emerge and with it, peace of mind and happiness that only mathematics can provide.</p>


问:
Expected value in term of characteristic function
Let $X$ be a random variable. It is known that for a function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+,$ of class $C^1,$ increasing, $f(0)=0,$
$$E[f(|X|)]=\int_0^{+\infty}f'(x)P(|X|>x)dx=\int_0^{+\infty}f'(x)(1-F_{|X|}(x))dx \ \ \ \ \ (1)$$ In particular, if $X \in L^1,$ then $E[X]=\int_0^{+\infty}(1-F_X(x)-F_X(-x))dx.$

Also $E[|X|]=\frac{2}{\pi}\int_{]0;+\infty[}\frac{1-\Re(\varphi_X(x))}{x^2}dx.$

In $(1), E[f(X)]$ is expressed in term of the distribution function $F_{|X|},$ is there an expression for $E[f(X)]$ in term of $\varphi_X$ ? In particular, if $X \in L^1,$ how can we write $E[X]$ in term of $\varphi_X$? To do so, is it possible to find $\varphi_{X^+}$ and $\varphi_{X^-}$ in term $\varphi_X$?

答:
Well, let $\varphi = \varphi_X$ for convenience, and then consider

$$ \begin{align}
\varphi' (0)  & = \lim_{h \to 0} \frac{ \varphi(h) - \varphi(0) }{h} \\
& = \lim_{h \to 0} \int_\mathbb{R} \frac{e^{ihx} -1}{h} dP(x) \\
& = \int_\mathbb{R} \lim_{h \to 0} \frac{e^{ihx} -1}{h} dP(x) \\
& = \int_\mathbb{R} ix dP(x) \\
& = i E(X)
\end{align} $$

so $E(X) = -i\varphi'(0)$. You need $|X|$ to be integrable to move the limit inside the integral using dominated convergence. I'm not sure what is true in general.

Whatever $f$ is, there's some new distribution $P_f$ and you have $-i \varphi_{f(X)}'(0) = E(f(X))$, with the same $L^1$ caveat.

Hope that helps.


问:
Permutation and Combination of 4 letters from a $11$ letter word

Find the number of combinations and permutations of $4$ letters taken from the word EXAMINATION.

-from Arihant Algebra

My work

There are $11$ letters in the word. E,X,A,M,I,N,A,T,I,O,N

Case $1$. When all the letters are different,

But there are $2$ A's, $2$ I'S, and $2$ N'S. (Consider them as 3 different letters).

Now other $5$ letters are repeating once only. Therefore there are $8$ distinct letters.

Then, the number of combinations=$C(8,4)$=$\frac{8!}{4!4!}$=$70$

And number of permutations=$P(8,4)$=$\frac{8!}{4!}$=$1680$

Case $2$. Two letters are alike of one kind and other two letters of another kind'

Like AAII (just for example)

Number of combinations=$C(3,2)$=$3$

Number of permutation=$\frac{4!}{2!2!}$=$6$

Case $3$: Two letters are alike and the other two different

Like AXAO (just for example)

Number of combinations=$C(3,1) \times C(7,2)$=$63$

Number of permutations=$\frac{4!}{2!}$=$12$

Adding the results of Cases $1,2,3$

Number of combinations=$70+3+63$=$136$

Number of permutations=$1680+6+12$=$1698$

Now, according to the answer given in the book, the number of combinations is correct but the number of permutations is wrong, it is $2454$.

Can you give me a hint?

答:
Your mistake is that in case $1$, you are finding permutation correctly but in other cases you are not. What you are finding in other cases is just the ways to arrange letters once you have chosen them. To find permutation, you should multiply combinations with the ways to arrange chosen letters.

Take case $1$.

Number of ways to choose $4$ letters = $8 \choose 4$

Number of ways to arrange = $4!$ (as all letters are different).

So permutation for case $1 = 4! \times {8 \choose 4} = 1680$ (this one is correct).

Where are you doing that for case $2$ and case $3$? Does that explain?


问:
How to Construct a matrix with requisite properties or how to show why no such matrix exists?
Construct a matrix  with requisite properties or why no such matrix exists.

The null space and column space both have the basis \begin{bmatrix}
1
\\ 0
\\
0
\end{bmatrix}

By rank nullity theorem rank of this matrix will be 1 and nullity is 1 and so the number of column is this matrix will be 2 and since rank of this matrix is 1 then the other column is the scalar multiple of \begin{bmatrix}
1
\\ 0
\\
0
\end{bmatrix}

But no such matrix exists since the number of column of this matrix is not equal to 3. But I am not sure about my reasoning. Please help.

答: Your reasoning is good. Think of it this way: If the column space and the null space are to have the same basis, they have to be subspaces of the same vector space, which forces the matrix to be square (see why)? Now, since the particular basis in question consists of a vector in $\mathbb{R}^3$, we are talking about a $3 \times 3$ matrix. However, as you point out, by the rank-nullity theorem, the number of columns would have to be 2, which is a contradiction.


问:
Solution of Lagrange PDE: $(z^2-2yz-y^2)p+(xy+zx)q= xy-zx$
This question has already been asked. But in that question, user was having difficulty in finding the second integral. My question is different. It says that if the solution to the above PDE represents a sphere, what will be the coordinates of that sphere? I am taking the solution as
$ x^2+y^2+z^2= f(z^2+2yz-y^2)$.

Could I say that the centre will be $(0,0,0)$? Or there is more to it than that meets the eyes? Please suggest.

答:
$$(z^2-2yz-y^2)z_x+(xy+zx)z_y= xy-zx$$
Charpit-Lagrange :
$$\frac{dx}{z^2-2yz-y^2}=\frac{dy}{xy+zx}=\frac{dz}{xy-zx}$$
First characteristic equation from solving $\frac{dy}{xy+zx}=\frac{dz}{xy-zx}\quad;\quad \frac{dz}{dy}=\frac{y-z}{y+z}$ :
$$z^2+2yz-y^2=c_1$$
Second characteristic equation from

$\frac{dx}{z^2-2yz-y^2}=\frac{dy}{xy+zx}=\frac{dz}{xy-zx}=\frac{xdx+ydy+zdz}{x(z^2-2yz-y^2)+y(xy+zx)+z(xy-zx)}=\frac{xdx+ydy+zdz}{0}\implies xdx+ydy+zdz=0$
$$x^2+y^2+z^2=c_2$$
General solution of the PDE on the form of implicit equation $c_2=f(c_1)$ with arbitrary function $f$ :
$$\boxed{x^2+y^2+z^2=f(z^2+2yz-y^2)}$$
In the particular case $f=$constant$=R^2$ the solutions are spheres of any radius and center $(0,0,0)$.

The question is : Is there other(s) sphere(s) which are solution of the PDE ? This would imply that a non-constant function $f$ exists such as $f(z^2+2yz-y^2)=(x^2+y^2+z^2)$ be equivalent to $(x-a)^2+(y-b)^2+(z-c)^2=R^2$.  This is not possible since $f(z^2+2yz-y^2)$ is function of $(yz)$ while $(x-a)^2+(y-b)^2+(z-c)^2$ is not function of $(yz)$.

A more direct proof consists in putting $(x-a)^2+(y-b)^2+(z-c)^2=R^2$ into the original PDE with $z_x=-\frac{x-a}{z-c}$ and $z_y=-\frac{y-b}{z-c}$. After a booring task of simplification the unique result is $a=b=c=0$.


问:
Show that $n | \sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$
I must show that $n$ divides $\sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$.

When $n$ is prime, Fermat's theorem gives us the answer easily because then $\sum_{k=1}^n m^{\textrm{gcd}(k, n)} = m^n + (n-1)m \equiv 0  [n]$.

In the general case, how to show this? I would like to use mostly integers/arithmetics arguments, not too high level theories.

答:
Not an answer. Too long for a comment. The below might be helpful.

$$\sum_{k=1}^n m^{\gcd(k,n)} = \sum_{d \mid n} \sum_{\substack{1 \le k \le n \\ \gcd(k,n) = d}} m^d = \sum_{d \mid n} m^d \sum_{\substack{1 \le k \le n \\ d \mid k \\ \gcd(\frac{k}{d},\frac{n}{d}) = 1}} 1 = \sum_{d \mid n} m^d \sum_{\substack{1 \le r \le n/d \\ \gcd(r,\frac{n}{d}) = 1}} 1$$ $$ = \sum_{d \mid n} m^d \sum_{1 \le r \le n/d} \sum_{s \mid \gcd(r,\frac{n}{d})} \mu(s) = \sum_{d \mid n} m^d \sum_{1 \le r \le n/d} \sum_{\substack{s \mid r \\ s \mid n/d}} \mu(s) = \sum_{d \mid n} m^d \sum_{s \mid n/d} \mu(s) \sum_{\substack{1 \le r \le n/d \\ s \mid r}} 1$$ $$ = \sum_{d \mid n} m^d \sum_{s \mid n/d} \mu(s)\frac{n}{ds} = \left(\sum_{ds \mid n} \frac{m^d}{d}\frac{\mu(s)}{s}\right)n.$$


答:
(Not a complete answer, but too long for a comment.)

Sometimes you can write $$\mathrm E[f(X)] = \frac{1}{2\pi} \int_{-\infty}^\infty \hat f(t) \varphi_X(t) dt,$$ where $f$ is the Fourier transform of $f$, thanks to Parseval's identity. Even more often, you can write it in the distributional sense: $$\mathrm E[f(X)] = \frac{1}{2\pi} \langle \hat f , \varphi_X\rangle.$$ For example, the Fourier transform of $x^2$ is $-2\pi\delta''_0(t)$, so $$\mathrm E[X^2] = -\langle \delta_0'',\varphi_X\rangle = -\varphi''_X (0),$$ which we know does hold provided that $\mathrm E[X^2]<\infty$.

Title: Expected value in term of characteristic function

Body:
Let $X$ be a random variable. It is known that for a function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+,$ of class $C^1,$ increasing, $f(0)=0,$
$$E[f(|X|)]=\int_0^{+\infty}f'(x)P(|X|>x)dx=\int_0^{+\infty}f'(x)(1-F_{|X|}(x))dx \ \ \ \ \ (1)$$ In particular, if $X \in L^1,$ then $E[X]=\int_0^{+\infty}(1-F_X(x)-F_X(-x))dx.$

Also $E[|X|]=\frac{2}{\pi}\int_{]0;+\infty[}\frac{1-\Re(\varphi_X(x))}{x^2}dx.$

In $(1), E[f(X)]$ is expressed in term of the distribution function $F_{|X|},$ is there an expression for $E[f(X)]$ in term of $\varphi_X$ ? In particular, if $X \in L^1,$ how can we write $E[X]$ in term of $\varphi_X$? To do so, is it possible to find $\varphi_{X^+}$ and $\varphi_{X^-}$ in term $\varphi_X$?


问:
Best way to integrate modulus functions?
Consider
$ \int_{-2}^{3} |1-x^2| dx$

if we find part of domains where function  is positive, then we write

$ \int_{-2}^{-1} (x^2 -1) dx + \int_{-1}^{1} (1-x^2)dx + \int_{1}^{3} (x^2 -1) dx$

But this is very painful to evaluate, is there any faster way to do this?

答:
After more than five months of this question sitting in the back of my mind, I have finally come to a 'fast way' to solve such questions.

Observe that the function we are integrating is even and hence it is deal to split over even bound,

$$  \int_{-2}^{3} |1-x^2| dx = \int_{-2}^2  |1-x^2|dx + \int_2^3 |1-x^2| dx$$

Now, $$ \int_{-2}^2 |1-x^2|dx = 2 \int_0^2 |1-x^2|dx$$

Hence,

$$  \int_{-2}^3 |1-x^2|  dx =2 \int_0^2 |1-x^2|dx+ \int_2^3 |1-x^2| dx$$

Now, notice the definition of our function,

$$ q = \begin{cases} x^2-1, x>1\\ 1-x^2, 0<x<1 \end{cases}$$

Hence our integral now becomes,

$$ \int_{-2}^3 |1-x^2| dx = 2 [\int_0^1 1-x^2 dx + \int_1^2 x^2 -1 dx ] + \int_2^3 x^2 -1 dx $$

Call $f=1-x^2$ and the anti derivative of it as $F$, then by fundamental theorem of calculus:
$$ \int_{-2}^3 |1-x^2| dx = 2 \left[- [F(1) - F(0)] + F(2) - F(1) \right] + F(3) - F(2)= -4F(1) +2F(0)  + F(3) - F(2) = F(3) +F(2)- 4F(1) + 2F(0) = F(3) +F(2)- 4F(1) $$

Which is a much simpler task. This method is a much more efficient one only if you do the parts it consists of in your head.

Tl;dr: Use even -odd property, split the integral, use fundamental theorem , drop off terms which have zeros.


问:
Linear transformation range question
I am trying to show that If $V, W$ are finite-dimensional vector spaces over R and $T1 : V → W$ and $T2 : V → W$ then are linear $rank(T_1+T_2)\leq rank(T_1) + rank(T_2)$, but I cannot connect my steps to the statement

Here is my attempt:

suppose $y \in ran(T_1+T_2)$  then $y=(T_1+T_2)(x)$

By linearity, $y=T_1(x)+T_2(x) \in ran(T_1)+ran(T_2)$

and here is the part I stuck on. Is there a theorem that says that $ran(T_1+T_2) \subseteq ran(T_1)+ran(T_2)$ ? Searching for my linear transformation book and I didn't see any theorems that have this property. Thanks in advance.

答:
If $y \in ran(T_1+T_2)$, then $y=(T_1+T_2)(x)$ for some $x \in V.$

Since $(T_1+T_2)(x)=T_1(x)+T_2(x)$ and $T_1(x) \in ran(T_1)$ and $T_2(x) \in ran(T_2)$ we get $y \in ran(T_1)+ran(T_2).$

This shows that $ran(T_1+T_2) \subseteq ran(T_1)+ran(T_2).$

Title: Linear transformation range question


问:
Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$
The exercise

Let $n\geq 3$ and $a_1, \dots, a_n \in \mathbb{U}$. Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$.

My try

Writing $z = e^{i \theta}$ and $a_k = e^{i \theta_k}$, I get that the module of the product is $2^n\prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$, using the fact that $e^{i \theta} - e^{i \theta_k} = 2 i \sin(\frac{\theta - \theta_k}{2}) e^{i \frac{\theta+\theta_k}{2}}$. I then need to find a value of $\theta$ that would make this number equal to $1$. Clearly, the function $f : \theta \mapsto 2^n \prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$ is continuous over $\mathbb{R}$. I would like to use the intermediate value theorem, but I have to show first $f$ takes at least one value $<1$ and another $> 1$.

Your help is welcome!

答:
Without holomorphic functions, one can use the well-known fact (and easy to prove by the doubling formula, noting that the function is integrable by calculus considerations) that $\int_0^{\pi}\log(2|\sin \theta|)d \theta=0$.

Changing variables we get $\int_0^{2\pi}\log(2|\sin \theta/2|)d \theta=2\int_0^{\pi}\log(2|\sin t|)dt=0$ and $\log(2|\sin \theta/2|)$ is periodic with period $2\pi$, so this means $\int_0^{2\pi}\log(2|\sin (\theta-\theta_n)/2|)d \theta=0$ for any fixed $\theta_n$

But now if $f(\theta)=2^n\prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) | \ge 0$ integrable and periodic on $[0,2\pi]$, the above means $\int_0^{2\pi} \log f(\theta)d\theta=0$ and since $f(\theta)$ is not identically $1$ (as it is zero near $n$ points and continuous) we get that there is $\psi, \log f(\psi) >0$ hence $f(\psi)>1$, so by the intermediate value theorem ($f$ is zero at $\theta_k$) there is $\theta, f(\theta)=1$ and we are done!


问:
Value of surface Integral
Let S be the part of the surface $z = x^2+y^2$ which lies under the plane $z=4$.

What is the value of the surface integral $\iint_S z \, ds \, \,$?

What I've done so far:

I have used the surface integral formula however I do not know if I need to convert to polar coordinates for the boundaries when integrating? i can't seem to figure out the boundaries.

I have set up the interval as follows:

$\int \int (x^2+y^2) \sqrt{1+4(x^2+y^2)} \, dx \, dy$

答:
$\displaystyle \iint_S z \, dS = \iint (x^2+y^2) \sqrt{1+4(x^2+y^2)} \, \,dx \,dy \, \,$ is correct.

$x^2 + y^2 = r^2 = z \leq 4 \,$ so $ \,0 \leq r \leq 2$. Your integral becomes

$\displaystyle  =\int_0^{2\pi} \int_0^2 r^2 \sqrt{1+4r^2} \, \, r \, dr \, d \theta = \int_0^{2\pi} \int_0^2 r^3 \sqrt{1+4r^2} \, dr \, d \theta$

Can you take it from here?

Title: Value of surface Integral


问:
What is the least possible radius for graph built on 100 vertices with maximal degree equals to 4
So, what I did is I wrote a definition of radius. $R = \min\;\epsilon(x)$, $x \in V(x)$. By $\epsilon(x)$ I mean eccentricity of vertex $x$. So, to minimize radius we, obviously, want to minimize minimal eccentricity. After that in order to translate it to even more easier terms let us write its definition.

$$\epsilon(x) = \max d(x, y), \;\;y \in V(x)$$

So, in order to minimize the least eccentricity we should minimize maximal distance from vertex $x$ to any other vertex.

Can you, please, suggest your ideas for continuation of this proof?

答:
Suppose $G$ is a graph with maximum degree $\Delta(G)\le4$. Fix a vertex $x$ and define $S_k=\{y:d(x,y)=k\}$. Observe that $|S_0|=1$, $|S_1|\le4$, and for $k\ge1$ we have $|S_{k+1}|\le3|S_k|$; the last inequality because each vertex in $S_k$ (where $k\ge1$) has at least one neighbor in $S_{k-1}$ and at most three neighbors in $S_{k+1}$. It follows that
$$|\{y:d(x,y)\le n\}|=|S_0|+|S_1|+|S_2|+\cdots+|S_n|\le1+4+4\cdot3+\cdot+4\cdot3^{n-1}=2\cdot3^n-1.$$
Since $|\{y:d(x,y)\le3\}|\le1+4+4\cdot3+4\cdot3^2=53$, a graph with maximum degree $4$ and radius $3$ can have no more than $53$ vertices.
Since $1+4+4\cdot3+4\cdot3^2+4\cdot3^3=161$, a graph with maximum degree $4$ and radius $4$ may have as many as $161$ vertices. So the answer to your problem is $4$. We can obviously construct a tree with maximum degree $4$ and radius $4$ which has $161$ vertices, namely $1+4+12+36=53$ vertices of degree $4$ and $108$ leaves. To get a tree with maximum degree $4$, radius $4$, and exactly $100$ vertices, delete $61$ of the leaves.


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答: Numpy, scipy and matplotlib are amazing packages in Python. They basically handle all kinds of numerical analysis and basically replace matlab. </p>

Title: Free online mathematical software


问:
Intuitive vs Formal (model-theoretic) definition of consistency of a theory
A theory $T$ of $L$-sentences is said to be consistent if there exists a model of $T$. More precisely, there exists an $L$-structure $\mathcal{M}$ such that $M$ satisfies $\sigma$ for each $\sigma \in T$.

How does this definition of consistent compare with the more 'intuitive' definition of consistent, that is a theory $T$ is said to be consistent if and only if for all $L$-sentences $\sigma$, $T$ does not contain both $\sigma$ and $\lnot \sigma$. I think they should be equivalent but I am unable to come up with a proof.

I thought that if $T$ contains both, then any model of $T$ will satisfy both, so there cannot exist a model of $T$. But I am not sure how to show that there cannot exist a model of $T$.

Also, if $T$ is a consistent theory, then it has a model $\mathcal{M}$ and if there exists $\sigma, \lnot \sigma \in T$, then $\mathcal{M}$ satisfies both $\sigma$ and $\lnot \sigma$, which means $\mathcal{M}$ is not a valid $L$-structure. Again, I am not sure how to show that it is not a valid $L$-structure from the definitions..

答:
Assume that $\sigma,\lnot \sigma \in T$.

A model of $T$ is a structure $\mathcal M$ such that:

$\mathcal M \vDash T$.

Thus: $\mathcal M \vDash \sigma$ and $\mathcal M \vDash \lnot \sigma$.

But one of the clause of the definition of the satisfaction relation is:

$\mathcal M \vDash \lnot \varphi \text {  if and only if it is not the case that  } \mathcal M \vDash \varphi$.

The usual terminology is: a theory $T$ is satisfiable iff it has a model.

Thus, the above argument is: if $T$ is inconsistent, then it is unsatisfiable, i.e. if $T$ is satisfiable, then is consistent. [Thanks to David.]

The other part of the equivalence: $T$ is satisfiable iff it is consistent, is the so-called Model existence lemma, a more general version of Gödel's completeness theorem.


问:
Derivative of a function with respect to a specific change of another function
Let's consider two functions $f$ and $g$ such that $f:\mathbb{R}^+ \rightarrow\mathbb{R}^+$ and $g:\mathbb{R}^+ \rightarrow\mathbb{R}^+$. Function $f$ is defined as:

$f(x)=\frac{e^{-g(x)}}{\int_{0}^{+\infty} e^{-g(y)} \,\text{d}y}$

I would like to determine the sign of the derivative of $f$ regarding to a change of function $g$ as follow:

$\text{d}\,g(x)=0$ for $x\leq X$, and $\text{d}\,g(x)>0$ for $x>X$.

With such change, I would expect:

$\text{d}\,f(x)>0$ for $x\leq X$ (via a change in the denumerator), and $\text{d}\,f(x)<0$ for $x<X$ (via a change in the numerator).

I can solve $\frac{\text{d}\,f(x)}{\text{d}\,g(x)}$. But I don't know how to deal with such specific changes of $\text{d}\,g$. Could you please give me a hand?

PS: I am not a mathematician by training. Sorry if the notations are a bit clumsy.

Body:
Let $c:=\int_{0}^{+\infty} e^{-g(y)} dy.$ Then $f(x)= \frac{1}{c}e^{-g(x)} .$

Hence

$$f'(x)=\frac{-g'(x)}{c}e^{-g(x)}.$$

Thus

$$ f'(x) >0 \iff \frac{g'(x)}{c} <0$$

and

$$ f'(x) <0 \iff \frac{g'(x)}{c} >0.$$


问:
My trial for showing that $K[[x]]$ over a field is a local ring.
Here is the question I want to answer letter $(b)$ in it:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$ In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$

$(b)$ Show that, for a field $K,$ $R = K[[x]]$ is a local ring.

Hint: According to part $(a),$ $\mathfrak{m} = R\setminus R^{*}$ and you know what $R^*$ is.

My questions are:

I already know the proof for letter $(a).$ Also, I have proved thoroughly before that:

If $R$ be an integral domain and let $R[[x]]$ be the corresponding ring of formal power series,then $R[[x]]$ is an integral domain. and $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$

And I have the following hint given to solve my question:

Hint: According to part $(a),$ $\mathfrak{m} = R\setminus R^{*}$ and you know what $R^*$ is.

1-But I do not understand how to use it. Could anyone show me how can I use this hint please?

Also, I understood that I should prove that $K[[x]]$ has a unique maximal ideal.

And according to the hint given here by Arthur:

The set of formal power series over a fieldis a local ring? which is:

" Hint: take an element with non-zero constant term, and construct an explicit inverse, degree by degree (or at least show that it can be done, by finding the first three or so terms of the inverse and point out that you can keep going indefinitely). This shows that $(x)$ is the only maximal ideal."

I should construct an explicit inverse of an element $x$ with a non-zero constant term and it will be the only maximal ideal $<x>$.

2-I do not know what is explicityly the form of this ideal and I do not know how to prove that this is the only maximal ideal, could anyone show me the proof of this please?

Here is my detailed proof for $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$

Let $R$ an integral domain (commutative division ring with no zero divisors), and let $R[[x]]$ be the corresponding ring of formal power series. i.e. $$\displaystyle R[[x]]=\left\{\sum_{n=0}^{\infty}a_n x^n\;\middle\vert\; a_n\in R\right\}$$ With addition and multiplication as defined for polynomials.

\textbf{First: showing that if $a_0\in R$ is a unit, then  $\displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$}

Let $a=\sum_{n=0}^\infty a_nx^n\in R[[x]]$, where $a_0$ is a unit. We want to construct some $b=\sum_{n=0}^\infty b_nx^n\in R[[x]]$ such that $ab=1$, or after expanding,
$$ab=a_0b_0+(a_1b_0+a_0b_1)x+\cdots=1+0x+0x^2+\cdots \quad \quad (1)$$
We therefore need $b_0=a_0^{-1}$ (recall that $a_0$ is a unit by the given). We want to have $a_1b_0+a_0b_1=0$, so our only choice for $b_1$ is $$b_1=\frac{-a_1b_0}{a_0}=-a_1a_0^{-2}.$$Also, We want $a_2b_0+a_1b_1+a_0b_2=0$, so we must have $$b_2=\frac{-a_2b_0-a_1b_1}{a_0}=-a_2a_0^{-2}+a_1^2a_0^{-3}.$$
So, to find a recursive definition of $b_{n}$ we will use the definition of multiplication in the ring of formal power series, we have that $$\sum_{n = 0}^{\infty}a_n x^n  . \sum_{n = 0}^{\infty}b_n x^n = \sum_{n\geq 0} (\sum_{i=0}^{n} a_{i} b_{n-i})x^n =  a_{0}b_{0} + (a_{0}b_{1} + a_{1}b_{0})x + \cdots +  (\sum_{i=0}^{k}a_i b_{k-i}) x^k + \cdots .$$
Now, we need in our problem here $ab = 1,$ i.e. $$ a_{0}b_{0} + (a_{0}b_{1} + a_{1}b_{0})x + \cdots +  (\sum_{i=0}^{k}a_i b_{k-i}) x^k + \cdots  = 1, $$
So we need all terms except the constant term to vanish.

Assume that for some natural number $n,$ we know that the coefficients of $b$ are nonzero up to $(n-1),$  then the $n^{th}$ coefficient of $ab$ is zero. So, we can write $$0 = a_{0}b_{n} + a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0},$$Or equivalently, $$ a_{0}b_{n} = -( a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0}),$$Hence, $$ b_{n} =  \frac{-1}{a_{0}}( a_{1}b_{n-1} + \cdots + a_{n-1}b_{1} + a_{n}b_{0}) = \frac{-1}{a_{0}} (\sum_{i=1}^n a_{i} b_{n-i}). $$And this is the recursion relation describing the coefficients $b_{n}$ of $b$ that will make $b$ an inverse of $a.$

\textbf{Second: showing that if $a = \displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$ then $a_0\in R$ is a unit}

Assume that $a = \displaystyle\sum_{n=0}^{\infty}a_n x^n$ is a unit in $R[[x]]$ and we want to show that $a_0\in R$ is a unit.

Since $a$ is a unit, then $\exists b =  \displaystyle\sum_{n=0}^{\infty}b_n x^n\in R[[x]]$  such that $ab =1.$ But this means that $(a_0 + a_1 x+\cdots)(b_0 + b_1 x+ \cdots)=1+0x+\cdots$ so $a_0b_0+ (a_1b_0+a_0 b_1)x+\cdots=1+0x+\cdots$ therefore $a_0b_0=1$ and hence $a_{0}$ is a unit as required.

答: Using your hint, $M=K[[x]]\setminus K[[x]]^*$ is the candidate for the unique maximal ideal. Now, $M$ is the set of formal power series with constant term equal to zero. That means that your elements in $M$ have the form $$a=a_1x+a_2x^2+\cdots=x(a_1+a_2x+\cdots)=xb,$$ so they are in the ideal generated by $x$, which is denoted by $(x)$. Actually, $M=(x)$. This ideal is maximal since the quotient $K[[x]]/(x)=K$ is a field (you can think of the quotient, as the ring $K[[x]]$ with the condition that $x=0$, so you get $K$. Another way to see this, is taking the evaluation map $e_0:K[[x]]\rightarrow K$ that takes a formal power series and evaluate it at $0$. That is a homomorphism with kernel $(x)$). Finally, any ideal (maximal or not) which is not the entire ring, does not have invertible elements, that is, it is contained in $M$. Hence, $M$ is the only maximal ideal.


问:
Ergodic theory question about the support of a measure.
I am trying to work through some problems from  " Ergodic theory with view towards number theory" but I am finding it a tad more difficult than expected. In particular, I have the following question:

For $X$ a compact metric space, $T:X\to X$ continuous and $\mu$ a $T$-invariant ergodic probability measure defined on the Borel sets of $X$. Prove that for $\mu$ almost every $x\in X$ and $y\in \operatorname{Supp}(\mu)$ there exists a sequence $n_k\nearrow\infty$ such that $T^{n_k}(x)\to y$. Recall
  $$\operatorname{Supp}(\mu)=X-\bigcup U$$
  where $U\subset X$ is open and has null measure.

This problem screams the Poincaré Recurrence Theorem to me, but I am not sure where to use it. I could really use some guidance for this problem.

NOTE: This is not homework for a class, I just really want to learn ergodic theory.

答:
It is enough to prove that for each $y\in\operatorname{Supp}\mu$, we have
$$\mu\{x\in X\mid\exists n_k\uparrow +\infty, T^{n_k}x\to y\}=1.$$
Indeed, assume we have done that. The support of $\mu$ is a closed set in a separable metric space, hence it is separable. Let $\{y_j,j\in\Bbb N\}$ be a countable dense subset. As a countable union of negligible sets is negligible, we have
$$\mu\left\{x\in X\mid\forall j\in\Bbb N,\exists n_k\uparrow +\infty, T^{n_k}x\to y_j\right\}=1.$$
Let $y\in \operatorname{Supp}(\mu)$. Then for each $i$, consider $y_{j_i}$ for which $d(y,y_{j_i})<i^{-1}$, and construct $(n_i)$ by induction, taking $d(y_{j_i},T^{n_i}x)<i^{-1}$.

Fix now $y\in\operatorname{Supp}\mu$, and call $C:=\left\{x\in X\mid\exists n_k\uparrow +\infty, T^{n_k}x\to y\right\}$. We can check that $C$ is almost invariant, in sense that $\mu(C\Delta T^{-1}C)=0$. As $T$ is ergodic, the measure of $C$ is either $0$ or $1$.

As $y$ is in the support of $\mu$, for each $l$, $\mu\left(B\left(y,l^{-1}\right)\right)$ is positive. By Poincaré recurrence theorem,  $\mu\left( \left\{x\in B\left(y,l^{-1}\right) \mid T^nx\in  B\left(y,l^{-1}\right)\mbox{ for infinitely many }n\mbox{'s}   \right\}  \right)$ is positive hence, by ergodicity, its value is $1$. Therefore, the measure of $C$ is $1$.


问:
How close is the sample mean $\mu_N$ for $N$ random variables to the sample mean $\mu_{N-1}$ for $N-1$ random variables?
Define $\mu_{N}$ to be the sample mean for $N$ i.i.d. random variables that each have mean $\mu$ and variance $\sigma^2$:
$$
\mu_{N} := \frac{1}{N} \sum_{i=1}^N X_i.
$$

Now consider the sample mean if we removed one of these random variables. That is consider, $\mu_{N-1}$ where w.l.o.g. we remove the last random variable from the sum:
$$
\mu_{N-1} := \frac{1}{N-1} \sum_{i=1}^{N-1} X_i.
$$

I am interested in describing how close $\mu_{N-1}$ is to $\mu_{N}$ in an appropriate sense..it is this 'sense' that I am unsure of as I haven't been working with probability that long.

Subtracting $\mu_{N-1}$ from $\mu_{N}$ we have
\begin{align}
\mu_{N} - \mu_{N-1} &= \frac{1}{N} \sum_{i=1}^N X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\
& = \frac{1}{N}X_N + \frac{N-1}{N}\frac{1}{N-1} \sum_{i=1}^{N-1} X_i - \frac{1}{N-1} \sum_{i=1}^{N-1} X_i \\
& = \frac{1}{N}X_N + \bigg(\frac{N-1}{N}-1\bigg)\frac{1}{N-1} \sum_{i=1}^{N-1} X_i.
\end{align}

So as $N\to \infty$, heurstically it 'looks like' both terms go to zero. But as the $X_i$ are random variables I imagine we need to set some conditions on them for this to hold. For example, we could be unlucky and draw $X_N$'s that are sufficiently large such that $\frac{1}{N} X_N$ does not go to zero as $N \to \infty$.

I feel like the variance $\sigma^2$ could have an effect on the convergence but I'm not sure.

So my questions are

In the finite sample case can we say something more informative than what I have already derived above regarding how close $\mu_N$ is to $\mu_{N-1}$?

What conditions are needed on the $X_i$ to ensure $\mu_{N} - \mu_{N-1}$ converges to zero as $N \to \infty$?

What is an appropriate sense of convergence for $\mu_{N} - \mu_{N-1}$? I am thinking maybe we can say it converges with 'high probability'. Most importantly, is it possible to find an exact rate of convergence?

答:
$Z_n\to 0, n\to\infty,$ almost surely iff $\mathrm{E}[|X_1|]<\infty$.

Proof. Sufficiency: by the Kolmogorov SLLN, $\mu_n \to 0$ and $\mu_{n-1} \to 0$, $n\to\infty$, almost surely.

The necessity is proved similarly to the proof of SLLN: since $\mu_{n-1}$ and $X_n$ are independent, it is clear that the both terms in the expression you wrote for $\mu_n - \mu_{n-1}$ must converge to $0$, in particular, $X_n/n \to 0, n\to\infty$, a.s. Then, with probability $1$, $|X_n|/n > 1$ only finitely many times. By the reverse Borel–Cantelli lemma, we then must have
$$
\sum_{n=1}^\infty \mathrm P(|X_n|>n) = \sum_{n=1}^\infty \mathrm P(|X_1|>n)<\infty,
$$
which is equivalent to integrability of $X_1$.

Using the strong law of large numbers for non-integrable random variables (I can elaborate if you wish), it is also possible to prove that $Z_n \to 0$ in probability provided that $\mathrm{E} [|X_1|^{1/2}]<\infty$, though I have a doubt that this is also a necessary condition (it is necessary for the second term in your expression to vanish almost surely; it is also possible to formulate a necessary and sufficient conditions in terms of behavior of characteristic function at zero).


问:
Metric assuming the value infinity
If we instead define a metric as $d:X\times X \rightarrow [0,\infty]$, do we lose any nice properties of metric spaces?

The reason I ask is that I saw this theorem: Given a finite measure space $(X,\mathcal{M},\mu )$, define equivalence relation $A \sim B$ if $\mu(A \mathbin{\Delta} B)=0$ for $A,B\in \mathcal{M}$. Then $d: \mathcal{M} /{\sim} \rightarrow [0,\infty)  $ defined by $d(A,B)=\mu(A \mathbin{\Delta}  B)$ is a metric.

But the same argument would go through if $X$ wasn't a finite measure space, the only place where it seems to fail is because metrics are defined not to take the value $\infty$.

答:
We can use infinite metrics to complete metric spaces in an easy manner:

Given any set $X$, let $\mathbb{R}^X$ be the space of functions $X\to \mathbb{R}$, equipped with the (infinite-valued) uniform metric $d_\infty(f,g)=\sup_{y\in X}|f(y)-g(y)|$. The usual proof shows that $\mathbb{R}^X$ is complete.

Now let $(M,d)$ be a metric space. Set $D\colon M\to \mathbb{R}^M$ as $D(x)=d_x=d(x,\cdot)\colon y\mapsto d(x,y)$ for all $x\in M$. Then $D$ is an isometry:
$$|d_x(y)-d_{x'}(y)|=|d(x,y)-d(x',y)|\leq d(x,x')$$
and
$$|d_x(x)-d_{x'}(x)|=d(x,x'),$$
which shows that $d_\infty(D(x)-D(x'))=d(x,x')$. The closure $\overline{D(M)}$ is a completion of $M$ (which is unique and satisfies the usual universal property as per general theory).

Note this avoid instances of AC as in the usual construction of a completion (as a quotient of the space of Cauchy sequences on $M$, etc.)

Title: Metric assuming the value infinity


问:
What is the flux of $\mathbf{f}$ through S along its normal vector?
Let $\mathbf{f}:\mathbb{R}^3\to\mathbb{R}^3$ be a continuous vector field which is parallel to the tangent plane at each point of a piecewise-smooth simple surface S. What is the flux of $\mathbf{f}$ through S along its normal vector?

What I have done so far:

The flux of the integral is $\iint\limits_{S}\mathbf{f\cdot \mathbf{n}}dS$ and it's normal vector will be perpendicular to it at every point. Thus making $\mathbf{f}\cdot \mathbf{n}=0$ Hence I think the answer is $0$.

答: You are right, the dot product with $\mathbb{n}$ takes the normal component of your field, which is $0$. You can think that, at each point of your surface, there is a base of $\mathbb{R}^3$ with two vectors in the tangent plane at that point and the third vector, the normal $\mathbb{n}$. Hence, you can decompose your field $f$ at each point of the surface and the flux takes only the normal component.


问:
$X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find $\sigma$
$X \sim N(12, \sigma^2)$ and $P(3<X<15) = 0.3$. Find the standard deviation $\sigma$.

If the question was: find $P(9<X<15) = 0.3$, then this would be easy using the inverse normal function. But I don't see how to use the inverse normal function for $P(3<X<15)$. I'm fairly sure the value of $\sigma$ is unique. If I were to guess, I would say $\sigma \approx 15$ or $20$.

It's not a book question: it's a question I came up with. And I don't see how to do it without maths I'm not familiar with, but I may just be lacking imagination for how to solve this.

答:
This is one case where we can do much without numerical methods.

For the normal distribution, you are looging for the zero of function
$$f(\sigma)=\Phi\left(\frac3\sigma\right)-\Phi\left(-\frac{9}\sigma\right)-\frac 3{10}$$ Using the error function, this means
$$f(\sigma)=\frac{1}{2} \text{erf}\left(\frac{3}{\sqrt{2} \sigma}\right)+\frac{1}{2}
   \text{erf}\left(\frac{9}{\sqrt{2} \sigma}\right)-\frac{3}{10}$$ Let $\sigma=\frac{3}{x\sqrt{2} }$ and, after simplication, we need to find the root of
$$g(x)=5 \,\text{erf}(x)+5\, \text{erf}(3 x)-3$$ Using Taylor expansions around $x=0$, ths gives
$$g(x)=-3+\frac{40 }{\sqrt{\pi }}x-\frac{280 }{3 \sqrt{\pi }}x^3+\frac{244 }{\sqrt{\pi   }}x^5+O\left(x^7\right)$$

Using now series reversion
$$x=t+\frac{7}{3} t^3+\frac{307 }{30}t^5+O\left(t^7\right) \quad \text{where} \quad t=\frac{\sqrt{\pi }}{40}\,\,  (g(x)+3)$$
Making $g(x)=0$ as desired, this gives as an estimate
$$x=0.13884 \implies \sigma=15.2789$$ which is not too bad.


问:
Range of values for $y$ given $-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$
I want to find the range of values of $y$ for which

$-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$

Now initially I thought to subtract the inequalities, but that gives $8 \leq y \leq 11$

However, $y$ cannot be any of these values because then $\max x + \max y = 3 + 11 = 14$ and this is bigger than the maximum value of $x + y$.

Any help is appreciated. Probably being a bit stupid!

Body:
We have $$9-x \le y \le 11-x$$

Since $$-3 \le -x \le 2,$$

$$9-3 \le 9-x \le y \le 11-x \le 11+2$$

$$6 \le y \le 13$$


Title: Range of values for $y$ given $-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$

Body:
I want to find the range of values of $y$ for which

$-2 \leq x \leq 3$ and $9 \leq x + y \leq 11$

Now initially I thought to subtract the inequalities, but that gives $8 \leq y \leq 11$

However, $y$ cannot be any of these values because then $\max x + \max y = 3 + 11 = 14$ and this is bigger than the maximum value of $x + y$.

Any help is appreciated. Probably being a bit stupid!

Body:
WLOG $x=2.5\cos t+0.5$

$$9\le2.5\cos t+0.5+y\le11$$

$$\iff8.5-2.5\cos t\le y\le 10.5-2.5\cos t$$

Now $-\cos t\ge-1\implies y\ge8.5+2.5(-\cos t)\ge8.5-2.5$

Similarly, $-\cos t\le1, y\le10.5+2.5(-\cos t)\le10.5+2.5$


问:
what is the jacobian of the rotation function applied to angles?
I'm trying to come up with an elegant way to compute the Jacobian of a function but having some hard time doing so.
I'll first give a intuitive description of the function and then I'll try to make it more formal.

The Jacobian I come up with has some fundamental problems (biggest example: when there is no rotation i.e $R=I$ I get that the partial derivative does not exist). It seem to me this problem should have an easier solution other than brute force using sympy...

verbal description (i.e hand waving):

The function gets 3 Euler angles (yaw, pitch and roll) and uses some rotation matrix $R$ to change the CS in which these angles are given. (Compute rotation matrix $\hat{R}$ from the angles and extract the Euler angles from $R\cdot \hat{R} \cdot R^T$)

Attempt to be more formal:

Let $$\psi=\{R \ |\ R\in\mathbb{R}^{3x3}, R\cdot R^T=I \\\wedge \\\text{det}(R)=1\}$$

$$f: \mathbb{R}^3\times\psi \rightarrow \mathbb{R}^3$$
$$g: \mathbb{R}^3\ \rightarrow \psi$$
$$h:  \psi\ \rightarrow \mathbb{R}^3$$
$$f(v, R)=h(R\cdot g(v) \cdot R^T) $$

to be clear - I'm looking for $J_R(f_R)$ where $f_R(v)=f(v, R)$

I omitted the formal definition of $g$ and $h$. they are basically the functions that compose (g) and decompose (h) rotation matrices from Euler angles.

h is something along this:

$$h(R)^{(1)}=atan2(R_{2, 3}, \sqrt{R_{1, 3}^2 + R_{3, 3}^2})$$
$$h(R)^{(2)}=atan2(R_{1, 3}, R_{3, 3})$$
$$h(R)^{(3)}=atan2(-R_{2, 1}, R_{2, 2})$$

g is something like this:

$$
\left[\begin{matrix}\cos(y) &0& \sin(y)\\ 0& 1& 0\\ -\sin(y)&0&  \cos(y)\\\end{matrix} \right]
\times
\left[\begin{matrix}1 &0& 0\\ 0& \cos(p)& \sin(p)\\ 0& -\sin(p)& \cos(p)\\\end{matrix} \right]
\times
\left[\begin{matrix}\cos(r) &\sin(r)& 0\\ -\sin(r)& \cos(r)& 0\\ 0& 0& 1\\\end{matrix}\,\right] $$

Hope it's clear and that I made no typing mistakes..

Thanks.

答:
I was trying to solve the same problem for yaw-pitch-roll rotations but gave up since it's too complicated.
Instead, I assumed that the rotation is small, and used the transformation matrix which converts the rotation rates in the global frame to the inertial frame

$$
\begin{bmatrix}
\delta \phi \\
\delta \theta \\
\delta \psi
\end{bmatrix} =
\begin{bmatrix}
1&0&-sin(\theta_1)\\
0&cos(\phi_1)&sin(\phi_1)cos(\theta_1)\\
0&-sin(\phi_1)&cos(\phi_1)cos(\theta_1)
\end{bmatrix}\begin{bmatrix}
\phi_2-\phi_1\\
\theta_2-\theta_1\\
\psi_2-\psi_1\
\end{bmatrix}
$$
and then apply jacobian to that matrix.
It seems like your rotation sequence is pitch-roll-yaw.
You may find the appropriate transformation matrix by following:
https://ethz.ch/content/dam/ethz/special-interest/mavt/robotics-n-intelligent-systems/rsl-dam/documents/RobotDynamics2016/RD2016script.pdf

Hope it helps.


问:
Does $\sup\limits_{0 \leq t \leq 1}X_{t}$ have the same law as $\sup\limits_{0 \leq t \leq 1}Y_{t}$
Let two stochastic processes $(X_{t})_{t \geq 0}$ and $(Y_{t})_{t \geq 0}$ that are equal in distribution. For simplicity let's say both are Gaussian processes.

Is it necessarily true that

$\sup\limits_{0 \leq t \leq 1}X_{t}$ has the same law as $\sup\limits_{0 \leq t \leq 1}Y_{t}$?

We know that for any $A \in \mathcal{B}(\mathbb R)$ we obtain $$ \mathbb P(X_{t} \in A)=\mathbb P(Y_{t} \in A)$$

I am unsure what operations we are allowed to consider on $\sup\limits_{ 0 \leq t \leq 1}X_{t}$ and $\sup\limits_{ 0 \leq t \leq 1}Y_{t}$ since $[0,1]$ is an uncountable set.

Any ideas?

答:
Under your assumption of sample continuity, these two objects are equal in law.

There are two ways to see it:

Either you see that  $$\max_{ t \in [0,1]} X_t = \sup_{t \in [0,1] \cap \mathbb{Q}^+} X_t$$

Or, you can argue that $$\begin{align} H :\mathcal{C}([0,1]) &\quad \longrightarrow &   \mathbb{R}\\ \quad f& \quad \longmapsto & \max_{t \in [0,1] } f(t)\end{align}$$
is a continous function.


问:
Polynomial ring is not a field.
I read the following post on stack exchange: Show that $(R/I)[x]\cong R[x]/I[x]$..

However, I’m now confused. I know that polynomial rings are not fields. But suppose $I[X]$ is maximal in $R[X]$ where $I$ is an ideal in $R$ then by the above post $R/I[X]$ is a field. But this is a contradiction. So where is the flaw in this argument.

答: ... therefore $I[X]$ is never maximal.

Title: Polynomial ring is not a field.

Body: ... therefore $I[X]$ is never maximal.


问:
Can $ x^2+\frac{x^2}{(x+1)^2}=3 $ be solved by 'completing square method'?
On using the quartic formula, there are two zeroes of this equation

$$
x^2+\frac{x^2}{(x+1)^2}=3
$$

So, I was wondering if this equation could be solved like a quadratic equation.

Is it possible. If yes, how?

Body:
Write $
x^2+\frac{x^2}{(x+1)^2}=3
$ as
$$x^2(x+1)^2+x^2=3(x+1)^2$$
or
$$x^2(x+1)^2+2x(x+1)+1= 4(x+1)^2$$
which allows the completion of square
$$[x(x+1)+1]^2=[2(x+1)]^2$$
Then, factorize with $a^2-b^2=(a+b)(a-b)$
$$(x^2-x-1)(x^2+3x+3)=0$$


问:
Historical meaning of the word 'class' as used by Frege and Russell
When reading mathematical writing from the time of Frege and Russell, I frequently see the use of the word 'class', and as far as I can tell they mean it in the sense modern mathematics use the word 'set'.

When I try to look up the meaning of the word 'class', I can only find it in terms of modern theories like ZFC.

Is it correct to interpret the word 'class' as used by Frege and Russell as meaning the same thing as 'set'?

答:
Pick up a copy of Russell's "Principles of Mathematics" to know what Russell said.  You will find that he distinguishes between classes, class-concepts, and collections. Collections are associated with mathematical contexts. They are, in effect, enumerations of individuals. By contrast,
classes are  aggregates of individuals. In so far as the modern notion of set is sometimes said to be a class that is an element of another class, Russell's classes ought not be compared with modern sets. Class-concepts require an understanding of the difference between intensional logic and extensional logic as it had been inherited from Leibniz.

When Frege declared that logicians ought concern themselves with truth, he effectively introduced semantics as opposed to epistemology. By invoking the "function concept" of mathematics, he introduced a distinction between an intensional name for a  concept and the extension of a concept. Individuals are said to "fall under" concepts. This is where naive comprehension arises in the logicism of Frege.

With regard to. modern paraphrasing, Frege simply recognized the importance of changing a statement that uses a proper name into a formula where the proper name is replaced by a variable. Eventually this leads to the specification of symbols into syntactic categories when describing some particular formal system. One must at least distinguish between proper names and variables.

The influence of Leibniz lies with his portrayal of logic using an inversion of the order in the Aristotelian class hierarchy.  Aristotle had considered grammatical subjects. The names of individuals correspond with "primary" substances. Other subjects, now associated with the Russellian class-concept or some modern accounts of types, correspond with "secondary" substances. Our understanding of the subset ordering corresponds with this "directionality."

Leibniz recognized that logic could be inverted because orders have converses.  His influence for this appears to have been St. Thomas Aquinas. St. Thomas argued that God could know individual souls because individuals are the smallest (extensional) species. This is the source of Leibniz' identity of indiscernibles. Today we speak of filters and ideals.

Frege saw classes and extensions of classes.


问:
Ideal $(2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$

Consider the ideal $J = (2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $J \neq (1)$.

I am a bit stuck. If $J = (1)$, then $J = \{1 \cdot r \mid r \in \mathbb{Z}[\sqrt{-5}\}$, so $J = \mathbb{Z}[\sqrt{-5}]$, so I would only need to find one element of $\mathbb{Z}[-5]$. I feel this isn't the correct strategy, and instead I should in some way use the norm function and the fact that the norm function is multiplicative.

Any help would be appreciated.

答:
You can show that if $a+b\sqrt{-5}\in(2,1+\sqrt{-5})$ then $a+b$ is even. Thus, $1=1+0\sqrt{-5}$ is not in the ideal.

More generally:

Theorem: Given a square-free integer $D\neq 1$ and $a+b\sqrt{D}\in \mathbb Z[\sqrt D]$ and positive integer $d\mid a^2-Db^2$ then the ideal $I=(d,a+b\sqrt D)$ has the property that if $u+v\sqrt D\in I$ then $d\mid u^2-Dv^2.$

In particular, then if $d\neq\pm 1,$ you have $I\neq (1).$

I’ll leave the reader to prove this.

When $D$ is odd, and $a=b=1,$ then we can choose $d=2\mid 1-D,$ and $u^2-Dv^2$ is even iff $u+v$ is even.

Your case above is $D=-5,\,a=b=1,\,d=2.$


问:
Linear approximation using the generalised formula
I have been reviewing my analysis course and there is something that seems odd to me.
My professor has wrote that the linear approximation for a function of two variables is :

$ x  = x_{0} + h$

$ y  = y_{0} + k$

$ h  = x - x_{0}$

$ k  = y - y_{0}$

$$ f(x_{0}+h,y_{0}+k) = f(x_{0},y_{0}) + \nabla{f}(x_{0},y_{0})*\begin{pmatrix}
h  \\
k  \\
\end{pmatrix} +\text{negligible}.$$

Then, he has explained that in dimension $n$, the formula is generalised as follows :

$$ f(x^{0}+H) = f(x^{0}) + \nabla{f}(x^{0})*H + \frac{1}{2}*H^{T}*\nabla{^2f}(x^0)*H + \text{negligible}.$$

With $x^0$ being :
$\begin{pmatrix}
x_{1}^{0}  \\
...  \\
x_{n}^{0}  \\
\end{pmatrix}$
and $ H = \begin{pmatrix}
h_{1}  \\
...  \\
h_{n}  \\
\end{pmatrix} $

I am a bit confused because he gave us an example for a function $(x,y)\mapsto x^2+y^2x^2y+4$ to compute :
$$f(1+0.0001,2+0.0006)$$
and he has used the generalised expression instead of the first one.

Will the result be different ? Because the expressions seem different...

答:
After reviewing my course again and looking at some stuff over the internet I come to the following conclusions :
An approximation can be computed using the n-order approximation and in my coures, we have covered the first and second order approximation.

If you are dealing with a one-variable function :

1st order approximation would be $ f(x_{0}+h) = f(x_{0})+f'(x_{0})*h + 0(h^2) $

with h the small floating number you want to add to a random integer

2nd order approximation would be $ f(x_{0}+h) = f(x_{0})+f'(x_{0})*h + \frac{1}{2}*f''(x_{0})*h^2 $

If you are dealing with a two-variables function :

1st order approximation would be :
$ f(x_{0}+h,y_{0}+k) = f(x_{0},y_{0}) + \nabla{f}(x_{0},y_{0})*
\begin{pmatrix}
h  \\
k  \\
\end{pmatrix} +\text{negligible}. $

if you are dealing with a n-variables function :


问:
Branch cuts for $\ln (z^2-1)$
Consider the function $$f(z) = \ln (z^2-1),$$ which as we know is a multi-valued function. Now doing $$f(z)=\ln[(z-1)(z+1)]=\ln(r_1r_2 e^{i(\theta_1+\theta_2)}),$$ where I let $z-1=r_1e^{\theta_1}$, $z+1=r_2e^{\theta_2}$.

To make single-valued, I let $$-\frac{3\pi}{2}<\theta_1\leq \frac{\pi}{2}, \, -\frac{3\pi}{2}<\theta_2\leq \frac{\pi}{2}.$$ To find its branch cut, I need to find where $f(z)$ is discontinuous. So $$-3\pi<\theta_1+\theta_2\leq \pi \, \Rightarrow \, -\pi<\theta_1+\theta_2\leq \pi.$$ So the discontinuous location should be the negative real axis.

Am I making any sense?   Thanks!

答:
In case this link Defining a holomorphic branch of $\log(z^2-1)$ does not address your specific question:

If you simply cut along the negative real axis, the resulting domain will still include the point $z=+1$ at which no branch   of $\log(z^2-1)$ can be defined.  Any cut must eliminate bot $z=1$ and $z=-1$.

But there is another way to approach obtaining a single valued holomorphic version of $\log f(z)$, where for simplicity we will take $f(z)$ to be holomorphic except at a finite number of zeros and / or poles.  We can require a domain $\Omega$ in which every path $\gamma \subset \Omega$ from $z_0$ to $z$ gives the same value,
$$ \log z = \log(z_0) + \int_\gamma \frac{f'(z)}{f(z)} ~ dz. $$  This can only arise if all paths joining $z_0$ to $z$ are homotopic in $\Omega$, that is each can de deformed continuously into  the other without encountering poles of $f'(z)/f(z)$.

That will be the case if, as above, the domain excludes all the zeros of $f(z)$ and the paths $\gamma$ cannot circulate the poles of $f'(z)/f(z)$.  There may be exceptional cases where the sum of their residues is zero, and then there may be more flexibility, but we will momentarily exclude this case. Then, to avoid circulation, a sufficient condition is that the domain be simply connected.

In our case $f(z) = z^2-1$ and the residue sum for $f'(z)/f(z)$ is non-zero, so our cut must prevent a circuit to include $z =\pm 1$.  That still leaves many choices.  One is the domain that is cut along the real axis from $z=-\infty$ to $z=1$, so that $\Omega = \mathbb C \setminus (-\infty, 1]$.  Another, used in the link, is to take $\Omega = \mathbb C \setminus \Big( (-\infty, -] \cup [1,\infty) \Big).$  Both give different holomorphic versions of the $\log$ function.

Any cut can be made to work if it meets these criteria:  choose the domain $\Omega$, take a point $z_0 \in \Omega$ and its corresponding value $\log z_0$ and define,
$$ \log(z) = \log(z_0) + \int_{z_0}^z \frac{2w}{w^2-1}~dw = \log(z_0) + \int_{z_0}^z \frac{1}{w+1} + \frac{1}{w-1}~dw. $$  This will give a single valued holomorphic branch.  The value is independent of the path by Cauchy's theorem.

Coming back to an example where the residues do cancel, consider $\displaystyle \log\frac{z-1}{z+1}$.

Here $f'(z)/f(z) = 2/(z^2-1)$, the residues are $\pm1$ and the corresponding cut for $\log z$ need only exclude jointly the points $z = \pm 1$. A commonly used domain in this case is $\Omega = \mathbb C \setminus [-1,1]$, although you could again use $\Omega = \mathbb C \setminus \Big( (-\infty,-1] \cup [1,\infty) \Big)$.

Addition:  I should have mentioned that for $f'(z)/f(z)$ every zero of $f$ has residue 1 and every pole residue -1, so the sum of the residues is simply the number of zeros less the number of poles.

Title: Branch cuts for $\ln (z^2-1)$


问:
finding $f(x,y)$ that satisfies tangent-plane equation
Q: Give an example for a function $f(x,y)$ continious partial derivatives that satisfies that following:

$z=f(x,y)$ is not a plane

$z=f(x,y)$ passes through the point $(6,4,3)$

The tangent plane to the surface $z=f(x,y)$ at the point $(6,4,3)$ is
$$
 -4(x-6) +1(y-4) +z -3 = 0
\\~\\
$$

My take:
By looking at the tangent plane equation we can tell that:
$$
G(x) = f(x,y) -z \\ ~\\
\nabla G(6,4,3) = (-4,1,1)
$$
And I don't know how to procceed any further than that.
I can think about countless random options, but what is the way to tackle this problem?

答:
From algebraic geometry I'm used to the tangent cone strategy for finding the tangent plane (or line) of a variety given implicitly by one (or, say, two) polynomial equation(s). In its simplest form this method amounts to writing a polynomial around a point (or translating to the origin) and investigating the polynomial there. If your point is on the variety (read surface) the constant term vanishes. The zero set of the linear terms give the tangent plane, etc. The flip side of this is that if you have the tangent plane you can add higher order terms around the point to get surfaces that have this plane as tangent plane. And if you don't add terms of higher order in $(z-3)$ it will still be expressible in polynomials as $z=f(x,y).$

One of the simpler ways of answering your question is then

$$(x-6)^2+(y-4)^2-4(x-6) +1(y-4) +1(z-3) = 0.$$


问:
How to find the first term given the common ratio and the sum

The ratio of a GP is $r$ = $2$ and the sum to eight terms is $1785$. Find the first term.

I have no real idea on how to approach this problem, so far I tried:

Since the $r$ is given I tried dividing $1785$ by $2$ and then divided that answer so on and so forth by $2$ 8 times, but is there an easier way to do this?

答:
May help to write down the equations. The terms of the GP are:

$$ \{ a_o , a_o r , a_o r^2, a_o r^3, a_o r^4, a_o r^5 , a_o r^6 , a_o r^7 \}$$

Where $a_o$ is first term

It is said in the question that,

$$ a_o \left[ 1 + r + r^2 + r^3 + r^4 + r^5 + r^6 + r^7 \right]= 1785$$

Can you finish?


问:
Using $p\supset q$ instead of $p\implies q$
I saw that a use for the notation $p\supset q$ instead of $p\implies q$
that got me a bit confused.

One occurrences is in this Wikipedia link.

It seems to me opposite than what it should be, let me explain what
I mean:

If $A,B$ are sets s.t $A\subset B$, $p$ is for $x\in A$, and $q$
is for $x\in B$ then we can identify (in some way) $$A\text{ with }p$$
$$B\text{ with }q$$

We have it that $p\implies q$, since $A\subset B$, but in the above
notation we have $q\subset p$ which to me looks like $B\subset A$

which is the opposite of what we wanted to express.

Can someone please explain to me the logic behind this notation ?

答:
Actually, historically, the reason we use $\supset$ is that Peano originally wrote $p C q$ for "$p$ is a consequence of $q$", and wrote a backward "$C$" for "$p$ has as a consequence $q$". Eventually, just as the "$\epsilon$" became "$\in$", so too did the backward "$C$" become "$\supset$".  So it doesn't actually have anything to do with set theory, per se, though maybe he used the same letter for both superset and consequence.

But as Lord_Farin notes, one way to look at things might be in terms of "informational content" or even (very informally) "provability power" (if "$p \supset q$" is true, then accepting "$p$" requires accepting everything that would go with accepting "$q$"; though what I just said might be very philosophically contentious!). And yes, it seems more natural to think of it semantically, and so think that the "$\supset$" should be a "$\subset$". That's math for you.


问:
Calculate The Following Probability
Given the following Question:

The possibility to win contest $A$ is $P1$, and the possibility to win contest $B$ it's $P2$.

We buy tickets for contest $A$ until we win for the first time, then We buy tickets for contest $B$ until we win for the first time, then again we but tickets for contest $A$ until we win the the first time and so on.

In other words, We try $A$ multiple times until we win once, then We go $B$ until we win etc.

Let $P(n)$ be the possibility that the $n$th ticket purchased is for contest $A$, find a recursive function for $P(n)$.

What I have done:

First, since we are talking about first win we can use the Geometric distribution.

I figured out that in order for ticket $n$ to be for contest $A$ then before that we are supposed to win even number in contest $A$ (Not sure if this helps at all)

答:
Say probability of winning contest $A$ is $p$ and contest $B$ is $q$ (instead of using $P1, P2)$.

$P(1) = 1$

$P(n) = P(n-1)(1-p) + (1-P(n-1))q \,$ ($n \gt 1$).

I validated this for a few low values of $n$.

$P(2) = (1-p)$

$P(3) = (1-p)^2 + p q$

$P(4) = (1-p)^3 + p(1-q)q + (1-p)pq + pq(1-p) = (1-p)^3 + p(1-q)q + 2 (1-p)pq$

...

Title: Calculate The Following Probability


答:
An alternative strategy is to work out the quotient ring ${\mathbb Z}[\sqrt{-5}]/(2,1+\sqrt{-5})$ and see that it is not the trivial ring.

It is probably easiest to do see this ring as ${\mathbb Z}[x]/(x^2+5)/(2,1+x)$. After a short and straightforward calculation, this turns out to be ${\mathbb F}_2$.

Title: Ideal $(2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$

Body:
Consider the ideal $J = (2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $J \neq (1)$.

I am a bit stuck. If $J = (1)$, then $J = \{1 \cdot r \mid r \in \mathbb{Z}[\sqrt{-5}\}$, so $J = \mathbb{Z}[\sqrt{-5}]$, so I would only need to find one element of $\mathbb{Z}[-5]$. I feel this isn't the correct strategy, and instead I should in some way use the norm function and the fact that the norm function is multiplicative.

Any help would be appreciated.


问:
Sequence of the ratio of two successive terms of a sequence
If $(a_n)_{n\in N}$ is a strictly decreasing sequence of real number converging to $0$ and s.t. $\forall n\in N$, $0<a_n<1$, does the following limit:
$$
\lim_{n}\frac{a_{n+1}}{a_n}
$$
exists?

When does the limit is strictly greater than $0$ (it can be zero if for example the sequence is defined as $a_n=\frac{1}{n!}$)?

答: For the first question (existence of a limit), if you set $u_n\stackrel{\rm{}def}{=}\ln \frac{a_{n+1}}{a_n}$, what can you say about  the sequence $(u_n)_{n\in\mathbb{N}}$?</p>


问:
Can every even number $n \geq 8$ be written as a sum of a prime $p \geq \frac{n}{2}$ and a positive odd that is a non-multiple of $3$? (Part 2)
This is part 2 of another question posted before.

Problem

Let $n$ be any even number equal to or greater than $8$. Then there exists a prime $p \geq \frac{n}{2}$

and odd $x>1$ that doesn't have $3$ as a prime factor, such that $n=p+x$.

Could anyone show a proof for this problem? Thanks

答:
If $n$ is large enough, yes.

Dirichlet's theorem gives a prime in any reasonable congruence class in large enough intervals, so in particular a prime $\equiv n-1$ or $n+1 \pmod 6$ in $[\frac n2, n-2]$. (Depending on which of $n-1, n+1$ is not divisible by $3$.)


问:
Fermat's Little Theorem for Eisenstein primes

Prove that if $\alpha \in \mathbb{E}$ is an Eisenstein integer and
$\pi$ is an Eisenstein prime, than $\pi \mid \alpha^{N(\pi)}-\alpha$.

$\mathbb{E} = \mathbb{Z}[\varepsilon] = \{ a+\varepsilon b \mid a, b \in \mathbb{Z} \}$ is the ring of the Eisenstein integers (where $\varepsilon = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$). An Eisenstein prime is a prime in this ring. $N(z) = z\overline{z} = a^2 -ab + b^2 \ \ \ \ \ \forall z = a+\varepsilon b \in \mathbb{E}$

Hi! I saw this problem in an algebra textbook, but I think it has more things to do with number theory. The statement strongly reminds me of Fermat's Little Theorem, but I haven't been able to "translate" its proof to ring theory yet.

I know the characterization of the Eisenstein primes:

ordinary primes congruent to $2$ modulo $3$

$\lambda = i\sqrt{3} = 1 + 2\varepsilon \in \mathbb{E}$

any ordinary prime congruent to $1$ modulo $3$ factors as $\pi\pi^*$ where each of $\pi$ and $\pi^*$ are primes in $\mathbb{E}$

I thought it could be a good idea to prove the proposition in these 3 cases, but I don't know how to proceed.

Any help will be appreciated!

答: Fermat's little theorem says that $a^{p-1}=1$ in the cyclic group $\left(\mathbb{Z}/p\mathbb{Z}\right)^*\cong \mathbb{Z}/(p-1)\mathbb{Z}$ (the order of an element divides the order of the group). For Eisenstein integers, the analogous statement would be $\left(\mathbb{Z}[\varepsilon]/\pi\mathbb{Z}[\varepsilon]\right)^*\cong \mathbb{Z}/(N(\pi)-1)\mathbb{Z},$ since, in that case, you would have $\alpha^{N(\pi)-1}=1$. For example, if $p\in\mathbb{Z}$ is prime in $\mathbb{Z}[\varepsilon]$, then $\mathbb{Z}[\varepsilon]/(p)\cong \frac{\mathbb{Z}/p\mathbb{Z}[x]}{(x^2+x+1)}$ is a field with $p^2=N(p)$ elements. Hence, the group of units is cyclic of order $p^2-1.$


问:
Positive integer solution for $2n^3 + 5 \mid n^4+n+1$
I am starting to study number theory and came across this question on the book I'm reading:

Give all positive integer values for n such that $2n^3 + 5 \mid n^4+n+1$.

I came up with an attempt but it doesn't seem right, can anyone help me figure it out?

We know $2n^3 +5 \mid n^4+n+1$ and trivially, $2n^3 + 5\mid 2n^3 + 5$.

From $a \mid b, a \mid c \implies a \mid xb + yc, x,y \in \mathbb{Z}$ we can show $2n^3+5 \mid n(2n^3+5) - 2(n^4 + n + 1) = 3n-2$

We also know that $a\mid b \implies a = 0$ or $|a| \le |b|$. From the first part, $3n-2=0 \implies n \notin
 \mathbb{Z}$. From the second part, $|2n^3+5| \le |3n-2|$. I didn't prove this but this inequality doesn't seem to have any positive integer solutions

Any help is appreciated.

答:
You're almost there, though your last bullet point is not quite correct; it should be
$$a\mid b\quad\implies\quad b=0\ \text{ or }\ |a|\leq|b|.$$
Of course $3n-2=0$ is impossible, so indeed you get $|2n^3+5|\leq|3n-2|$.

The left hand side of this inequality clearly grows faster than the right hand side, so there can only be finitely many solutions. Now it suffices to find an upper bound for solutions, and then to check all values of $n$ up to that bound. Can you find an upper bound for $n$?

Here's example of such an upper bound (try to find one by yourself first):

 For example, if $n\geq2$ then $n^3\geq4n$ and so $$2n^3+5\geq8n+5>3n-2,$$ and then it suffices to check the inequality for $n=1$.


问:
Prove that there are infinitely many pairs of positive integer $(p,q)$ such that given functions have distinct integer roots
$f(x)=x^3 -px^2 + qx$ and $g(x)=3x^2-2px+q$

I want to prove that there are infinitely many pairs of positive integer $p,q$ such that $f(x)$ has three distinct integer roots and $g(x)$ has two distinct integer roots. I can show that if $f,g$ have the required roots, then $p$ is multiple of $3$, $q$ is multiple of $9$, $p^2-3q$ is positive perfect square and $p^2-4q$ is positive perfect square. But i do not know how to show there are infinitely many required pairs. We also require that $gcd(p,q)=3$

答:
There are an infinite number of solution pairs. That's a general result for quadratic Diophantine equations: they either have no solution (except perhaps for trivial solutions involving zero), or they have an infinite number of solutions, although those solutions will fall into a finite number of families. IIRC, Lagrange gave the first proof of that result (along with various other important proofs on Pell's equation and quadratic forms).

As mentioned by Daniyar, for both equations to have integer solutions, both $p^2-4q$ and $p^2-3q$ must be perfect squares.

Let
$$p^2-4q=r^2$$
$$p^2-3q=s^2$$
Subtracting,
$$q=s^2-r^2$$
and substituting
$$p^2=4s^2-3r^2$$

Using Brahmagupta's identity, products of integers of the form $4s^2-3r^2$ are also of that form if the second terms are even:

$$(4s^2-3r^2)(4x^2-3y^2)$$
$$=(16s^2x^2+9r^2y^2) - 3(4s^2y^2+4r^2x^2)$$
$$=(16s^2x^2+9r^2y^2 \pm 24sxry) - 3(4s^2y^2+4r^2x^2 \pm 8syrx)$$
$$=(4sx \pm 3ry)^2 - 3(2sy \pm 2rx)^2$$

If either of $r$ or $y$ are even, $4sx \pm 3ry$ will also be even.

Now we could search for numbers of the form $4s^2-3r^2$, looking for perfect squares. But we can use a minor modification of the above identity to construct such squares.

Let
$$s=(u^2+3v^2)/4$$
where $u$ & $v$ are coprime and both odd, hence
$$u^2\equiv v^2\equiv 1\pmod 4$$
and
$$u^2+3v^2\equiv 0\pmod 4$$
(so $s$ is an integer), and let
$$r=uv$$

So
$$4s^2-3r^2$$
$$=(u^2+3v^2)^2/4-3u^2v^2$$
$$=(u^4+9v^4+6u^2v^2 - 12u^2v^2)/4$$
$$=(u^4+9v^4-6u^2v^2)/4$$
$$=(u^2-3v^2)^2/4$$

Note that $(u^2-3v^2)\equiv 2 \pmod 4$, i.e., it's twice an odd number, so $(u^2-3v^2)^2$ is an even square, and
$$p=(u^2-3v^2)/2$$
is an odd integer. We do need to choose appropriate $(u, v)$ pairs to ensure that $p$ and $q$ are positive.

Now sometimes $\gcd(p,q)=3$, but if that's not the case they are coprime, so we can easily multiply them both by 3 to achieve the desired GCD constraint.
I'll leave the proof of that as an exercise for the reader. ;)

Here are some solutions constructed using the above procedure.

$$\begin{array}{|r|r|}
\hline
p & q \\
\hline
33 & 72 \\
39 & 360 \\
69 & 360 \\
111 & 3024 \\
141 & 840 \\
177 & 2520 \\
183 & 5616 \\
213 & 2640 \\
219 & 11880 \\
249 & 5040 \\
291 & 12240 \\
321 & 3168 \\
327 & 22176 \\
363 & 32760 \\
393 & 10920 \\
429 & 8568 \\
429 & 15120 \\
501 & 18480 \\
537 & 23760 \\
573 & 7920 \\
681 & 28728 \\
717 & 19872 \\
753 & 43680 \\
789 & 51480 \\
897 & 15960 \\
897 & 62832 \\
933 & 72072 \\
1041 & 59400 \\
1149 & 94248 \\
1221 & 118560 \\
1257 & 131040 \\
\hline
\end{array}$$

You can see more solution pairs using this small interactive Python script. The script itself is encoded into the URL, and it runs on a SageMath, Inc, server. I use a Farey sequence generator to produce coprime $(u, v)$ pairs. The script parameter $m$ selects the Farey sequence.

Body:
There are an infinite number of solution pairs. That's a general result for quadratic Diophantine equations: they either have no solution (except perhaps for trivial solutions involving zero), or they have an infinite number of solutions, although those solutions will fall into a finite number of families. IIRC, Lagrange gave the first proof of that result (along with various other important proofs on Pell's equation and quadratic forms).

As mentioned by Daniyar, for both equations to have integer solutions, both $p^2-4q$ and $p^2-3q$ must be perfect squares.

Let
$$p^2-4q=r^2$$
$$p^2-3q=s^2$$
Subtracting,
$$q=s^2-r^2$$
and substituting
$$p^2=4s^2-3r^2$$

Using Brahmagupta's identity, products of integers of the form $4s^2-3r^2$ are also of that form if the second terms are even:

$$(4s^2-3r^2)(4x^2-3y^2)$$
$$=(16s^2x^2+9r^2y^2) - 3(4s^2y^2+4r^2x^2)$$
$$=(16s^2x^2+9r^2y^2 \pm 24sxry) - 3(4s^2y^2+4r^2x^2 \pm 8syrx)$$
$$=(4sx \pm 3ry)^2 - 3(2sy \pm 2rx)^2$$

If either of $r$ or $y$ are even, $4sx \pm 3ry$ will also be even.

Now we could search for numbers of the form $4s^2-3r^2$, looking for perfect squares. But we can use a minor modification of the above identity to construct such squares.

Let
$$s=(u^2+3v^2)/4$$
where $u$ & $v$ are coprime and both odd, hence
$$u^2\equiv v^2\equiv 1\pmod 4$$
and
$$u^2+3v^2\equiv 0\pmod 4$$
(so $s$ is an integer), and let
$$r=uv$$

So
$$4s^2-3r^2$$
$$=(u^2+3v^2)^2/4-3u^2v^2$$
$$=(u^4+9v^4+6u^2v^2 - 12u^2v^2)/4$$
$$=(u^4+9v^4-6u^2v^2)/4$$
$$=(u^2-3v^2)^2/4$$

Note that $(u^2-3v^2)\equiv 2 \pmod 4$, i.e., it's twice an odd number, so $(u^2-3v^2)^2$ is an even square, and
$$p=(u^2-3v^2)/2$$
is an odd integer. We do need to choose appropriate $(u, v)$ pairs to ensure that $p$ and $q$ are positive.

Now sometimes $\gcd(p,q)=3$, but if that's not the case they are coprime, so we can easily multiply them both by 3 to achieve the desired GCD constraint.
I'll leave the proof of that as an exercise for the reader. ;)

Here are some solutions constructed using the above procedure.

$$\begin{array}{|r|r|}
\hline
p & q \\
\hline
33 & 72 \\
39 & 360 \\
69 & 360 \\
111 & 3024 \\
141 & 840 \\
177 & 2520 \\
183 & 5616 \\
213 & 2640 \\
219 & 11880 \\
249 & 5040 \\
291 & 12240 \\
321 & 3168 \\
327 & 22176 \\
363 & 32760 \\
393 & 10920 \\
429 & 8568 \\
429 & 15120 \\
501 & 18480 \\
537 & 23760 \\
573 & 7920 \\
681 & 28728 \\
717 & 19872 \\
753 & 43680 \\
789 & 51480 \\
897 & 15960 \\
897 & 62832 \\
933 & 72072 \\
1041 & 59400 \\
1149 & 94248 \\
1221 & 118560 \\
1257 & 131040 \\
\hline
\end{array}$$

You can see more solution pairs using this <a href="https://sagecell.sagemath.org/?z=eJxdUstugzAQvPMVo0hVjOs8GnKqitRTf6LtAbChFmCMDVT5-9omhSQWgmV2Zmd3gYsSZWbEhaj4NYI7m80GHx6BFf0oVCHQdlyWUnAMHTrVXHCRouHoOIfOpLFeE7QZQ85QMHCkODK8hEuF3O-PbAQKvEHNRv7UjkcUnpHHOBzAl8RdJf9gqGmBnU_UlLsgX7iyBMk8EuMJJ6TOenXwZ27Xl4yid6kGYbJiiLgbvc2kIm2aHK_D266x6ed3iMvOYGIYIdV1RW281rW-85GOrveETnQK_Z-XtHHpkU7Lu_6n727op3Vca4UZoKn27Z-ppTYwDTULp3clZvwWdcP3bqcv9xPnRmT1glROWRWcaIY-vlVW3u1B6kksSBLq4oT2_lY9thqkyboOt7h9prVQnMw-cbRsdG87M5DZWRv3AQj1MHN_mE63X2ob_wG3ypa4&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="nofollow noreferrer">small interactive Python script. The script itself is encoded into the URL, and it runs on a SageMath, Inc, server. I use a Farey sequence generator to produce coprime $(u, v)$ pairs. The script parameter $m$ selects the Farey sequence.


问:
How do I calculate the work done in a vector field across a line segment?

Let C be the line segment from $(2, 3, 0)$ to $(5, 7, 9)$ followed by the line segment from $(5, 7, 9)$ to $(1, 1, 2).$ Consider a force field
$F(x, y, z) = <3x^2+yz+y,xz+x+z^2,xy+2yz+2z>$

Calculate the work done by $F$ on a particle taking the path $C$.

How would I solve this? Here's what I have so far:

Calculating the curl for the force field, I get = $(0,0,0)$ which means it's conservative. So I know then the integral must be:

$$\int _CF\cdot \:dr\:=\:f\left(1,\:1,\:2\right)-\:f\left(5,\:7,\:9\right)$$

Taking their respective partial derivatives I get:

$\frac{\partial }{\partial x}\left(3x^2+yz+y\right)$

$\frac{\partial }{\partial y}\left(xz+x+z^2\right)$

$\frac{\partial \:}{\partial \:z}\left(xy+2yz+2z\right)$

$$f(x,y,z)=6x+2y+2$$

So my final answer is:

$$\left(6\left(1\right)+2\left(1\right)+2\right)-\left(6\left(5\right)+2\left(7\right)+2\right)=-36$$

Is this the right process and answer?

答:
Hint:

It is a conservative vector field but the potential function you have set up is not correct.

The Potential function is $f(x,y,z) = x^3 + xyz + xy + (y+1)z^2$.

$\frac{\partial f}{\partial x} = 3x^2 + yz + y$ which is part of your vector field. You can check other partial derivatives similarly.

One way to find the potential function is take the first component ($i$) of the vector field and integrate wrt $x$, similarly take the second component ($j$) wrt $y$. From there on, you should get the function.

Pls take only the end points for your calculation $(2, 3, 0)$ to $(1, 1, 2)$ because the integral will be path independent.


答:
Let $c_n$ be an arbitrary sequence of numbers in $(0,q)$ with $q<1$, esp. it may converge to $0$, to $q$, to some value inbetween, or not converge at all.
Then $a_n:=\prod_{k=1}^{n-1} c_k$ is strictly decreasing, converging to $0$ and $\frac{a_{n+1}}{a_n}=c_n$.

Title: Sequence of the ratio of two successive terms of a sequence

Body:
If $(a_n)_{n\in N}$ is a strictly decreasing sequence of real number converging to $0$ and s.t. $\forall n\in N$, $0<a_n<1$, does the following limit:
$$
\lim_{n}\frac{a_{n+1}}{a_n}
$$
exists?

When does the limit is strictly greater than $0$ (it can be zero if for example the sequence is defined as $a_n=\frac{1}{n!}$)?


问:
Hilbert space and quantum mechanics
If we consider a vector space $V$ over the complex field $\mathbb{C}$, and a linear application $f$
$f:V\rightarrow C$
The set of all functionals $f$ form a vector space $V^{\ast}$, the dual of $V$. And $f(v)$ is noted as $\langle f│v\rangle$, where
$\langle f|\in V^{\ast}$. So  $f(v)=\langle f│v\rangle$, and it is important to note that $\langle f│v\rangle$ is not a scalar product; it is $f$ operating on $v$.
In quantum mechanics books, where $V=\mathbb{R}^3$, and $|\phi\rangle \in L^2$   where $L^2$ is Hilbert space of square summable functions, we see with no explanation that $\langle \phi|r\rangle=\phi(r)^{\ast}$ where $|r\rangle\in\mathbb{R}^3$.
Now $\langle \phi|r\rangle$ is not a scalar product, since $\mathbb{R}^3$ and $L^2$ are two different vector spaces. A scalar product is between two elements of the same vector space.
But then, if $\phi$ is operating on $r$ to yield $\phi(r)^{\ast}$, $L^2$ should be the dual of $\mathbb{R}^3$; which is not.
I would be really grateful for any clarification.

答:
The notation you allude to is a very useful, but not rigorous notation used by physicists.

With $|r\rangle$ it is meant the delta function at $r$ so

$$
|r\rangle = \delta(r-x)
$$

Now the above is not a function but a distribution. A distribution applied to a test function $\psi$ is written

$$
\langle r , \psi\rangle
$$

Now the above can be formally written as

$$
\int r(x) \psi(x) dx.
$$

In this way the above looks like the $L^2$ scalar product.

In any case, using the property of the delta function, one has

$$
\langle r , \psi\rangle = \int \delta(r-x) \psi(x) dx = \psi(r).
$$

Title: Hilbert space and quantum mechanics


问:
If $X$ is Stein and $H^2(X,\mathbb{Z})\neq 0$ then there are no principal divisors?
There is a result that claims: If $X$ is a Stein manifold and $H^2(X,\mathbb{Z})=0$ then for any divisor $D$ there is a meromorphic function $f$ such that $(f)=D$.

What happens if $H^2(X,\mathbb{Z})\neq 0$ ? Is there any divisor that can not be of the form $(f)$ for some meromorphic function?

Thanks

答:
It is well known as the Cousin problem II.

If $M$ is a Stein manifold of dimension $n$, the exact sequence of sheaves
$$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$
yields a morphism of groups in cohomology (part of a long exact sequence) $$H^1(M,\mathcal O_M)\to H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)\to H^1(M,\mathcal O_M).$$

We have $H^1(M,\mathcal O_M)$ and $H^2(M,\mathcal O_M)$ both vanish thanks to Cartan's theorem B. Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the isomorphism $$c_1:Pic(M)\to H^2(M,\mathbb Z).$$

We then absolutely have $Pic(M)=0,$ thus any line bundle $\mathcal O(D)$ corresponding to the divisor $D$ is trivial.

Due to a classical result, a divisor $D$ is principal iff $\mathcal O (D)$ is trivial (A), we are done.

If $H^2(M, \mathbb Z)\neq 0$, we absolutely can find a non-principal divisor $D$ when $\mathcal O(D)$ is not trivial thanks to (A).


问:
provability and theorem
I am studying first order logic and I have a hard time understanding the link between provable formulas and theorem.
In the book by Shoenfield, the predicate $ Pr_{T}(a,b)$ of is defined as the set of (a,b) such that there exists formulas $A_{1},A_2,...,A_k$ which are a proof in T and $<Num(A_1),...,Num(A_k)>=b$ and a=$Num(A_k)$. This predicate is recursive and thus representable in the theory N by a formula $B^T$ with x,y distinct variables. Then we define $A^T$ as $\exists y B^T$.

So if I understood well, $T|-A^T_{x}[ \underline{Num(E)}] $ means that there exists a proof of E in T.
We also have 3 derivability conditions, one of which interests me here:

if $T|- D $ then $N|-A^T_{x}[ \underline{Num(D)}] $

My issue is that I don't see why this relation is not an equivalence. Because if $T|-A^T_{x}[ \underline{Num(D)}] $ then it means that $D$ is provable in T and thus that it is a theorem.

Can anyone help me with this matter ?

Thanks!

答:
Long comment

The so-called arithmetization of syntax is relative to a theory $T$ with language $L$ [page 123].

The author defines a collection of numerical predicates and relations that mimic the properties of theory $T$: $\text {Term}, \text {Ax}, \text {For}$ and so on.

Specifically, predicate $\text {Pr}_t$ is a binary predicate holding of numbers $a$ and $b$ iff $b$ is the number corresponding to a proof [a derivation in $T$] ending with formula $A$ and $a$ is the number corresponding to $A$, i.e. $a= \ulcorner A \urcorner$.

Then the author defines anew unary predicate: $\text {Thm}_T(a) \leftrightarrow \exists x\text {Pr}_T(a,x)$, that means: there is a proof [in $T$] of formula $A$, i.e. $A$ is a theorem of theory $T$.

Now, to say that a numerical (unary) predicate $P$ is representable in theory $\mathsf N$ means that there is a formula $A_P$ of the theory such that:

if $P(a)$ holds, then $\mathsf N \vdash A_P[k_a/x]$, where $k_a$ is the numeral corresponding to $a$, and

if $P(a)$ does not hold, then $\mathsf N \vdash \lnot A_P[k_a/x]$.

Now we have the condition that $\text {Thm}(a)$ is representable in $\mathsf N$.

This means that:

if $\text {Thm}(\ulcorner \varphi \urcorner)$ holds, i.e. $\mathsf N \vdash \varphi$, then $\mathsf N \vdash A_T[k_a/x]$,

where $a= \ulcorner \varphi \urcorner$.

The condition: if $\mathsf N \vdash \varphi$, then $\mathsf N \vdash A_T[k_a/x]$, is the first derivability condition.

But we have to note that the second clause of representability is: if $P(a)$ does not hold, then $\mathsf N \vdash \lnot A_P[k_a/x]$.

When we apply it to $\text {Thm}$ we have: if $\mathsf N \nvdash \varphi$, then $\mathsf N \vdash \lnot A_T[k_a/x]$ that is stronger than:

if $\mathsf N \nvdash \varphi$, then $\mathsf N \nvdash A_P[k_a/x]$.

Title: provability and theorem


问:
System of equations involving 4 variables

If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$

I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.

答:
Use succssive eliminations.

From $(1)$, $b=6-a$. PLug in $(2)$ to get $y=\frac{a x-10}{a-6}$. Plug in $(3)$ to get $a=\frac{22}{3 x^2-10 x+12}$. Plug in $(4)$ and obtain
$$22\frac{x^2-3x+1}{3 x-5}=0 \implies x= ???$$

Now, go backward.


问:
Lifting a map to the total space of a circle bundle
Let $\pi:P \to M $ be a smooth circle bundle, so $S^1$ is the fibre, $f:N \to M$ a smooth map. I would like to know what are the necessary and sufficient conditions for $f$ to lift to a map $g:N \to P$ covering $f$.

There are related questions reading which I think the answer is that $f$ lifts if and only if $f^*:H^2(M,\mathbb{Z})\to H^2(N,\mathbb{Z})$ is the zero map. However my understanding of algebraic topology is very basic and I cannot quite follow the arguments presented, so I would like for someone to confirm or  correct what I wrote. Incidentally, advice for the most accessible introduction to obstruction theory you know of would be appreciated.

In fact I am interested in a related but slightly different question: with $\pi:P\to M$ as above when does a diffeomorphism $f:M\to M$ lifts to a map $h:P\to P$? If the above criterium is correct $f$ lifts if and only if $\pi^*: H^2(M) \to H^2(P)$ is the zero map. Can anything be said about when that happens given that $P$ is a circle bundle?

答:
There are a few issues I'd like to address.  First, is a distinction that must be made between a circle bundle and a principal circle bundle.  What you described is a circle bundle: a fiber bundle where the fiber happens to be diffeomorphic to $S^1$.  In contrast, a principal circle bundle is a circle bundle for which there is a consistent choice of free,transitive $S^1$ action itself.

There are circle bundles which are not principal; the simplest example is probably the Klein bottle as an $S^1$ bundle over $S^1$.

A principal bundle over a base $B$ gives rise to a distinguished element $e$ (called the Euler class) which lives in $H^2(B;\mathbb{Z})$.  The condition on lifting is not that $f^\ast$ be the zero-map, but rather, that $f^\ast(e) = 0$.  Tsemo gave a fine proof that this condition is sufficient; the necessity follows from the fact that the kernel of the map $H^2(B;\mathbb{Z})\rightarrow H^2(P;\mathbb{Z})$ is generated by $e$.

But I want to stress that the argument involving the Euler class does not work for non-principal bundles (because the don't have an Euler class).  And, indeed, the result is false in this case.

To see this, consider the unit tangent bundle of $\mathbb{R}P^2$, $T^1\mathbb{R}P^2$.  This is diffeomorphic to the lens space $L(4,1)$, as argued in Proposition 4.2 of this paper.

Now, consider the natural projection $\pi:S^2\rightarrow \mathbb{R}P^2$.  Because $H^2(\mathbb{R}P^2;\mathbb{Z})$ is torsion, while $H^2(S^2;\mathbb{Z})$ is not, the induced map on $H^2$ must be the trivial map.

On the other hand, $\pi$ induces an isomorphism $\pi_2(S^2)\rightarrow \pi_2(\mathbb{R}P^2)$.  Because lens spaces are covered by $S^3$, they have trivial $\pi_2$, so the map $\pi:S^2\rightarrow \mathbb{R}P^2$ cannot lift to a map into $L(4,1)$.


问:
How do I calculate an integral of $F$ where $F = \langle y,x^2\rangle$?

I have to calculate the above integral and I'm not sure how to do it. First I find the curl of $\langle y,x^2\rangle$ and I get:

$$\langle0,0,2x-1\rangle$$

Therefore it's not conservative because it's not all $0$s and I can't solve the integral.

Is that right? I feel like that isn't right.

答: You can parametrize the path using $r(x)=(x,2+\frac{x^2}{2}),\,x\in[0,2]$. Hence, $r'(x)=(1,x),\,F(r(x))=F(x,2+\frac{x^2}{2})=(2+\frac{x^2}{2},x^2)$, and then proceed by definition of the line integral.


问:
Canonical disjunctive normal form in Boolean algebra
In my notes on canonical disjunctive normal form in Boolean algebra I have written:

A monom is a finite conjunction of literals $L_1,...L_m$. The universe $U$ is a degenerate monom. A formula is in disjunctive normal form if it is a finite disjunction of monism $M_1,..., M_n$. The empty set $\emptyset$ is in DNF degenerately.

Why is the universe $U$ a degenerate monom? Why is the empty set in DNF degenerately? What does this mean?

答:
A Disjunctive Normal Form (DNF) is also called Sum of Products. The products (or monoms) are AND expressions of literals (inverted or non-inverted inputs). If the sum or disjunction of products does not contain any products, this is an empty set. The expression is not fulfilled for any combination of input values.

The opposite case of an empty set of products is a set which contains all $2^n$ products. They constitute a full truth-table. In this case, the expression is fulfilled for every combination of input values. The complete set of all possible products is sometimes called universe. If literals of a product are dropped one-by-one by merging adjacent products with just one different literal, the remaining product value in the degenerate case without any literals left is $1$.