问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答: I have answered this before by saying "I was always bad at ". But then I would tell them how I got better at something, and ask them why they think they are not good at something. For people who genuinely want to become better or that do have an appreciation for the subject matter, I use it as an opportunity for them to teach me something I don't know in exchange for me to teach them something I know. I like the concept that learning and teaching should come hand-in-hand.</p>


问:
Factoring quadratics - how to?
It has been some time since I have studied math (and even then it was base-level math), and I know this will probably be easy for most, but I need help with Simplifying Algebra equations.

Simple equations like so I can do: 3x + 7 + 9x -4 would (I really hope) equal 12x + 3, right?

I now have an equation I wish to get my head around, and it is:

Now I have done some searching for the guides on how to perform the above and have either ended up with base level answers which did not help the above, or more difficult examples I could not quite get my head around. I have attempted to get the answer then "work backwards" at it, but I cannot wrap my head around the following (Wolfram Alpha):

I would ask you, the wonderful users of this forum, to either help explain the above question, or provide me the ability to learn this independently.

Update

After looking into the FOIL method, I referred to my genius co-worker for assistance- I think he is able to explain things to an idiot (me) and it works. Basically we went through the quadratic formula method (which I understand!) and my workout was as follows:

This is how I have processed this equation (with great skill!):

I then apply the same method to the bottom and get the following:

With the top line, I understand we perform both 3+7 and 3-7 (which results in 5 and -2), however in the second option I assume (and this is merit to my lack of math knowledge), we performed the opposite function of minus due to the original sum being plus 5. Is this correct?
Following on, performing both the addition and subtraction we get 6+4 which results in 10 and six minus 4 which results in two, which then when divided is 1. I am just attempting to understand why the 1 in this function is negative.

Can you please either highlight anything I am doing incorrectly, advise if there is a incorrect step or let me know if all is okay? Much appreciated!

答:
I think factorization method would be easier in this case
$$\frac{x^2-3x-10}{x^2-6x-5}$$
Then you simplify making it:
$$\frac{x^2-5x+2x-10}{x^2-5x-1x+5}$$
After which you factorise it:
$$\frac{x(x-5)+2(x-5)}{x(x-5)-1(x-5)}$$
Which equates to:
$$\frac{(x-2) (x-5)}{(x-1)( x-5)} $$
Then $x-5$ cancels each other thereby making it:
$$\frac{x+2}{x+1}$$

ContentLicense: CC BY-SA 3.0

Title: Factoring quadratics - how to?

ContentLicense: CC BY-SA 4.0


问:
Show that lim f (x) exists in a Banach space
Hi I'm struggling with a proof.

Let E be a normalized vector space and F a Banach space. We know that A ⊂ E, x0 ∈ Acc (A) and
f: A → F is a uniformly continuous function. How would you show that lim f (x) exists?

I have some ideas to how to prove with, but I'm probably missing a lot.

I would like to prove (())is Cauchy, because, from there, I will be able to prove that (() converges because we are in a Banach space.

I also think that if x0 ∈ Acc (A), it exists a sequence () as lim () = x0.

From there I don't know what to do. I'm kinda lost as you can see!
Thank you for your help!

答:
Consider $(A,d),(F,e)$ as metric spaces, with $d(a,a')=\|a-a'\|_E$ and $e(g,g')=\|g-g''\|_F.$

Let $x_0\in E$ and let $(a_n)_n$ be a sequence in $A$ with $\lim_{n\to\infty}\|x_0-a_n\|=0.$ Then $(a_n)_n$ is a $d$-Cauchy sequence in the space  $(A,d).$

For metric spaces $(A,d),(F,e)$  and a uniformly continuous $f:A\to F$, if $(a_n)_n $ is a $d$-Cauchy sequence then $(f(a_n))_n$ is an $e$-Cauchy sequence.

Proof. By contradiction, suppose $r\in \Bbb R^+$ and $\lim_{n\to\infty}\sup_{n\le n'<n''}\,e(f(a_{n'}),f(a_{n''})\,)>2r.$ Then there exists an infinite $S\subset \Bbb N$ such that

$(1*)$ for each $n'\in S$ there exists $n''>n'$ such that $e(f(a_{n'}),f(a_{n''}))>r.$

Now $f$ is uniformly continuous so take $\delta >0$ such that

$(2^*)\;\forall x,y\in A\,(\,e(x,y)<\delta \implies e(f(a_{n'}),f(a_{n''}))<r).$

And $(a_n)_n$ is a $d$-Cauchy sequence so take $n_0\in \Bbb N$ such that

$(3^*)\;n_0\le n'<n''\implies d(a_{n'},a_{n''})<\delta.$

But take $n'\in S$ with $n'\ge n_0.$ By $(2^*)$ and $(3^*)$ we have $\forall n''>n'\,(\,e(f(a_{n'}),f(a_{n''})<r)\,),$ contradicting $(1*)$.

Now if $F$ is a Banach space then $e$ is a complete metric.


问:
Confused about partial derivatives
I am having some issues understanding what should I keep constant and what not in certain cases when I take partial derivatives. Specifically in this kind of situation: say we have a function

$$f(x,y) = x^3+7y^2$$

and we also know that $y=2x+1$ and we need to find the partial derivative with respect to $x$ at a given point, say $\frac{\partial f(2,3)}{\partial x}$. From what I understood, when taking partial derivatives with respect to a variable, you need to keep the other constant, in which case, if I do that above (keeping $y$ constant) I would get, $\frac{\partial f(x,y)}{\partial x}=3x^2$, so I get $12$. However, if I plug $x$ in $y$ explicitly I would get $$f(x,y) = x^3+7(2x+1)^2$$ so,

$$\frac{\partial f(x,y)}{\partial x}=3x^2+28(2x+1)$$

So, I get $152$. What should I do? Thank you!

答:
The answer is that it depends (slightly) on the intent of the problem.  When I read the problem as you've written it, I see this:

Given that $y = 2x+1$ and  $f(x,y) = x^{3} + 7y^{2},$ calculate $\frac{\partial f(2,3)}{\partial x}$.

In that statement, I interpret this as $f$ is a function of two variables, $x$ and $y$, and we are asked to compute $\frac{\partial f}{\partial x}(x,y)$ and evaluate the result at $(x,y) = (2,3)$.  In this case, the fact that $y$ depends on $x$ isn't relevant, because since $f$ depends explicitly on $x$ and $y$, $\frac{\partial f}{\partial x}(x,y)$ mean hold $y$ constant and differentiate with respect to $x$.  That gives us $$\frac{\partial f}{\partial x}(x,y) = 3x^{2}\implies \frac{\partial f(2,3)}{\partial x} = 12.$$

Now, if instead the problem was phrased as

Given that $y = 2x+1$ and  $f(x,y) = x^{3} + 7y^{2},$ calculate the total rate of change of $f$ with respect to $x$ at $(2,3)$.

In this case, I would interpret the problem as asking for the total derivative, which accounts for dependencies among the variables.  In this case, your answer would be $$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} = 3x^2+28(2x+1) \implies \frac{df}{dx}(2,3) = 152.$$

If the original problem does indeed ask for $\frac{\partial f(2,3)}{\partial x}$ then I would say the first interpretation is correct.  If you are adding that notation yourself, it may be the case that the second interpretation is correct.

Title: Confused about partial derivatives


问:
If $E$ has positive measure, then prove that there exists $h \in \mathbb{R}$ such that $|(E+h) \cap E| > 0$.
If $E$ has positive measure, then prove that there exists $h \in \mathbb{R}$ such that $|(E+h) \cap E| > 0$. I tried doing this by contradiction but cannot seem to find my way about the argument.

Any hints or help would be appreciated. Thanks.

答:
If $E$ has positive measure then there exists an interval, say $I$ such that $m(E\cup I)>\frac{2}{3}m(I)$.
(Of course much more can be said about points of density of $E$ but we do not need it here).
Fix this interval $I$. Now, if $E-E$ does not contain a neighborhood of $0$ then there is a decreasing to zero sequence, say $x_{n}$ such that $E+x_n$ is disjoint with $E$. In particular, the set $I\cap E$ would be disjoint with $(I+x_n)\cap (E+x_n)$. However, $E+x_n$ is just a shift of $E$ so $m((I+x_n)\cap(E+x_n))> \frac{2}{3}m(I+x_n)$. Clearly, $m(I+x_n)=m(I)$. Now, this is not possible since for $x_n$ sufficiently small (i.e. $x_n< \frac{1}{6}m(I)$ ) we would have two disjoint sets contained in a small perturbation of $I$ ( add and subtract $x_n$ from I boundaries) each having measure larger than $\frac{2}{3}$ of $I$.

So, $E-E$ contains a neighborhood of $0$ which is the same as to say that $m(E\cap(E+h)) > 0$ for $h$ small enough.


问:
Question on finding the value of algebraic identity if $a+b+c=0$
If $a,b,c\in \Bbb{R}$ and $a+b+c=0$, then evaluate
$$\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$$
I tried using the identity, $$\frac{b-c}{b+c}+\frac{a-c}{a+c}+\frac{a-b}{a+b}=-\frac{b-c}{b+c}\cdot\frac{a-c}{a+c}\cdot\frac{a-b}{a+b}$$However, I couldn't get anywhere. I also tried using normal algebraic methods but I couldn't get anywhere.

 Please help me out. Thanks a lot :)

Body:
$$X=\left(\frac{c-a}{b}+\frac{b-c}{a}+\frac{a-b}{c}\right) \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=$$
$$=\frac{a^3-a^2 (b+c)-a \left(b^2-3 b c+c^2\right)+(b-c)^2 (b+c)}{a b c}=$$
$$=\frac{a^3-a^2 b-a^2 c-a b^2+3 a b c-a c^2+b^3-b^2 c-b c^2+c^3}{a b c}=$$
$$=\frac{(a+b+c)^3-4 (a b+a c+b c) (a+b+c)+9 a b c}{abc}$$
as $a+b+c=0$ we have $X=9$


问:
Meet of two compact elements need not to be compact.
I am doing exercises from Universal Algebra by Clifford Bergman.

Question (Bergman 2.34.5) Let $a,b$ be compact elements in algebraic lattice $L$. (2) Must $a\wedge b$ be compact?

My thoughts I think I found a counterexample on web (see below) but it is rather involved and it seems to me that the point of the exercise is, that there are simpler counter-examples.

I would like to find some more natural counter-example to $(2)$.

Definitions Let $L$ be a complete lattice

(a) Element $a$ in $L$ is called compact if for every $X\subset L$:

$a\leq \bigvee X \rightarrow (\exists \text{ finite } Z\subset X) ~a\leq \bigvee Z $

(b) $L$ is called algebraic if every element is the join of compact elements.

"My" counter-example

Wikipedia lists an example of two compact sets $U,V$ in Hausdorf space $X$ whose intersection is not compact https://en.wikipedia.org/wiki/Compact_space#cite_note-17.

I came to the conclusion, that unlike for general closure operator from topology, for this particular topology the closer operator is indeed Agebraic. So the lattice of closed subsets of $X$ is an algebraic lattice and $U,V$ are the desired counterexample.

答:
So essentially you have the following lattice:

where $a$ and $b$ are compact but $a \wedge b$ is not compact.

Notice that you can define a topological space from this lattice as the one whose open sets are, up to isomorphism, the elements of the lattice.

Make $X = \{ u_0, u_1, u_2, \ldots, a, b \}$ and the open sets are
$$\varnothing, U_0, U_1, U_2, \ldots, U, V, W, X,$$
where $U_i = \{ u_0, \ldots, u_i \}$, $U = \bigcup_i U_i, V = U \cup\{a\}$ and $W = U \cup \{b\}$


问:
Which is the most accurate notation for function recomposition?
Having the definition of recomposition of functions as the composition of a function with itself (ex: $f\circ f\circ x$), which is the most accurate way to denote this type of operation?

Examples:

The power notation: $\underbrace{f\circ f\cdots\circ f}_n = f^n, \underbrace{f\circ f\cdots\circ f}_n\circ x = f^n(x)$

The power composition notation: $\underbrace{f\circ f\cdots\circ f}_n = f^{\circ n}, \underbrace{f\circ f\cdots\circ f}_n\circ x = f^{\circ n}(x)$

The recomposition notation: $\underbrace{f\circ f\cdots\circ f}_n = f\overset n\circ, \underbrace{f\circ f\cdots\circ f}_n\circ x = f\overset n\circ(x)$

Note: for this question, I'm not in need of very strict answering in regard of things like domain, image / co-domain definition.

答:
As there's thousands of ways things can be done in math, its very unusual to have a most accurate way of doing something.

For example, when dealing with algebraic transformations, like matrix multiplication, $f^n(x)$ can be very handy, as it is easy to write, and don't change the operation at all (it's all some sort of multiplication).

But when dealing with, for example, trigonometric functions, the easiness becomes ambiguity: is $\sin^2 x = (\sin x)^2$ or $\sin(\sin x)$? Is $\sin^{- 1} x = {1\over \sin x} = \csc x$ or $\arcsin x$?

So, when dealing with this type of functions (or, in general, whenever not dealing only with the first type of functions), the recomposition notation comes into hand:

Letting $f$ be $f\circ x$, here are some notable identities:

$f\circ f\circ x = f\overset 2\circ x$

"$x\circ f$" $ = f\overset {- 1}\circ x$

$f\overset {n + 1}\circ x = f\circ f\overset n\circ x = f\overset n\circ f\circ x = x$

$f\overset {n + m}\circ x = f\overset m\circ f\overset n\circ x = f\overset n\circ f\overset m\circ x = x$

$f\overset {- 1}\circ (f\circ x) = f\circ (f\overset {- 1}\circ x) = x$

$f\overset {- n}\circ (f\overset n\circ x) = f\overset n\circ (f\overset {- n}\circ x) = x$

Involutions obey $f\overset 2\circ x = x$ (like ${1\over x}$, $- x$, and $\bar z$)

$\operatorname{arc [something]}(x)$ can be expressed / defined as $\operatorname{[something]}\overset{- 1}\circ x$ (ex: $\arcsin x := \sin\overset{- 1}\circ x)$

$g = f\overset n\circ\iff f \overset 1 = g\overset{1\over n}\circ\Rightarrow g\overset m\circ = f\overset n\circ\iff f \overset 1 = g\overset{m\div n}\circ$

One use for the recomposition of functions, besides of dealing with iterations of functions, is expressing the growth of a function based in other functions².

One thing that could be called an abuse of notation is: $f\overset{2}{\overline{\circ\circ}} x=f\overset{f\circ x}\circ x$.

Notes:

There certainly are cases where there are multiple solutions to these equations.

No article for reference.


问:
Find all integers such that there is a number and all elements more than it can be written using the elements of a set in exactly that many ways.
Find all integers $r$ such that $\exists$ $p \in \mathbb{Z^{+}}, A \subset \mathbb{Z^{+}}$ and every $q > p$ can be written using distinct elements from $A$ in no more or less than $r$ ways.

I think it is not possible. But I could not generate an example for any number either. Can I get a few examples so I can verify it further?

答:
Here is a solution for $r=2$ (which might make the general pattern clear).  Let $$A=\{1,2,3,4,8,16, \cdots , 2^n, \cdots\}$$

That is, $A=\{2^n\}\cup \{3\}$.

In that case, of course, $1,2$ can only be written in one way.  But any greater number has one decomposition that uses $3$ and one which does not.


问:
Eigenvalues and Eigenvectors - Proof
Hey! I’m quite stuck on the question below. Thanks!

Let $v$ be an eigenvector of the $n \times n$ matrix $M$ corresponding to the eigenvalue $\lambda$. Set $Q = [v,a_2,a_3,\ldots,a_n]$ where ${v,a_2,a_3,\ldots,a_n}$ is a set of $n$ linearly independent vectors. Show that the first column of $Q^{-1}MQ$ is $(\lambda,0,0,\ldots,0)$

答:
Hint: consider $M$ as the matrix representation of a linear function (say $f$) in a particular basis. Then, $Q^{-1}MQ$ corresponds to the matrix representation of this function in another basis, namely $(\vec v, \vec a_2, \ldots, \vec a_n)$. I think from there you can conclude quite easily.
Good luck.</p>

Title: Eigenvalues and Eigenvectors - Proof


问:
How can I solve the equation $\frac{dy}{dt} = \frac{1}{e^y - t}$?
How can I solve the following equation?

$$\frac{dy}{dt} = \frac{1}{e^y - t}$$

I feel as if I am missing something obvious with this question but have been trying to figure it out for too long now. The only hint given is to try and treat $y$ as the dependent variable, however I don’t understand what that means for me to do.

答:
Hint:
$$y'(e^y-t)=1$$
Rewrite it as:
$$t'=e^y-t$$
$$t'+t=e^y$$
Where $t'=\dfrac {dt}{dy}$. And solve.


答:
If $y$ is dependent on $t$, then we're treating $t$ as a function of $y$ (i.e. $t = t(y)$), and it makes more sense for you to find $\frac{dt}{dy}$.

$\frac{dt}{dy} = \frac{1}{dy/dt} = \frac{1}{1/(e^y - t)} = e^y - t$

And it should be much easier to solve from that point.

Title: How can I solve the equation $\frac{dy}{dt} = \frac{1}{e^y - t}$?

Body:
How can I solve the following equation?

$$\frac{dy}{dt} = \frac{1}{e^y - t}$$

I feel as if I am missing something obvious with this question but have been trying to figure it out for too long now. The only hint given is to try and treat $y$ as the dependent variable, however I don’t understand what that means for me to do.


问:
Show that the sequence defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}$ converges
Show that the sequence $(x_n)$ defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}, x_0=3,x_1=2$ converges. Here i wil try to prove that $(x_n)$ converges considering all details. So. i would like to have a critical feedback on my proof and would like to know if i missed something or did much more than it should be.. Here $\mathbf{R}^*$ will denote $\mathbf{R}\ \ \not \ \{0\}$.

Consider a function $f(x)=\sqrt[3]{x}$. First, we show that it's defined on $\mathbf{R}$ and then we will show that $f(x)$ is an increasing function on $\mathbf{R}$.

We remark that $f(x)$ is differntiable on $\mathbf{R}^*$. Let $a \in \mathbf{R}^*$. By definition of differentiable function the $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ msut exist:

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\frac{1}{3\sqrt[3]{a^2}}$ and so it exists $\forall a \in \mathbf{R}^*$ as we chose an arbitrary $a$. So $f(x)$ is differentiable on $\mathbf{R}^*$ and so is continious on $\mathbf{R}^*$.

We show now that $f(x)$ is continious on $0$. To show that we pass by the definition :

$\forall \varepsilon>0 \ \exists \delta>0 \ \forall x \in \mathbf{R}$: $|x|<\delta \implies |f(x)|=|\sqrt[3]{x}|<\varepsilon$

To satisfy the definition, one can take $\delta=\varepsilon^3$, so it holds and we can conclude that $f(x)$ is continious on $\mathbf{R}$.

Let $a,b \in \mathbf{R}$ such that $a<b$. As $f(x)$ is continious on $\mathbf{R}$ we can write:

$a<b \iff \sqrt[3]{a}<\sqrt[3]{b} \iff f(a)<f(b)$

As $a,b$ are arbitrary numbers (satisfying $a<b$), we can conclude that $f(x)$ is strictly incrasing function.

Now, as all preliminary results are proven, we can show that $(x_n)$ is a convergent sequence. For that, we show that $(x_n)$ is monotonic and bounded.

$\bullet$ $(x_n)$ is decreasing: Suppose $x_{n+1}<x_n$. The previous inequality holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We show nowthat it holds for $n+1$:

$x_{n+2}<x_{n+1} \iff \sqrt[3]{x_{n+1}+x_n}<\sqrt[3]{x_n+x_{n-1}}$. But, as $f(x)=\sqrt[3]{x}$ is strictly increasing function and by induction hypothesis $x_{n+1}<x_n$, we can conclude that $x_{n+2}<x_{n+1}$ and so $x_{n+1}<x_n$ holds $\forall n\in\mathbf{N}^*$. So $(x_n)$ is decreasing.

$\bullet$ $(x_n)$ is bounded below: Suppose $x_n>1$. It holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We want to show now that t holds for $n+1$:

$x_{n+1}>1 \iff \underbrace{x_{n}}_{>1}+\underbrace{x_{n-1}}_{>1}>1$ which is true, because $(x_n)$ is decreasing. So we conclude that $x_n>1 \ \forall n\in \mathbf{N}^*$.

So, we have shown that $(x_n)$ converges.

答: After showing that the limit exists and call it $x$. We show next that: $x \ge \sqrt{2}$. Using the decreasing property just established we have: $x_n = \sqrt[3]{x_{n-1} + x_{n-2}} > \sqrt[3]{2x_n} \implies x_n^3 > 2x_n \implies x_n(x_n^ 2 - 2) > 0 \implies x_n > \sqrt{2}, \forall n \ge 1 \implies x \ge \sqrt{2} \implies x = \sqrt{2}$, because it’s shown that $x = 0$ or $x = \sqrt{2}$.


问:
If you have four complex numbers and divide each of them by the others then will one of the results have to have non negative real and imaginary parts
Say you have four distinct nonzero complex numbers and you divide each one by the other three (ie you take $\frac{z_k}{z_j}$ for each $k \neq j$). Is there a way to prove that at least one of these quotients must produce a complex number with nonnegative real and imaginary parts? I've been trying to come up with a counter example all day and I can't find one so I'm wondering if this is something that can actually be proven.

答: Assume not. If any $z_i/z_j$ is in the fourth quadrant, then $z_j/z_i$ is in the first. So all quotients must be in the 2nd and 3rd quadrant. Then of the three $z_1/z_i$, two must be in the same quadrant, i.e., their quotient $z_i/z_j$ must be in the 4th or 1st, contradiction.


问:
Does the parameter change during data generation in Bayesian Inference?
Let's assume that we have the following graphical model:

This graph encodes the joint distribution $P(p,x_1,x_2,x_3,x_4) = P(p)\prod_{i=1}^{4}P(x_i|p)$. In the Bayesian inference, if we know $x_1,x_2,x_3$ then a full Bayesian predictive posterior for $x_4$ is given as:

$$ P(x_4|x_3,x_2,x_1) = \int_{p}P(x_4|p)P(p|x_3,x_2,x_1)dp$$

I used to interpret a model like in the above as follows: Some probabilistic system picks a $p$ from $P(p)$. Then this picked $p$ generates data as $x_1\sim P(x_1|p),x_2\sim P(x_2|p),x_3\sim P(x_3|p)$. We do not know the true value of $p$ and therefore we integrate over all possible values of $p$ using its posterior distribution given the data in order to obtain the posterior predictive distribution of $x_4$.

In this interpretation, the value of $p$ stays the same once it is picked from the prior, during the data generation. But our lack of knowledge about it leads us to integrate over its all possible values in order to infer $x_4$. This can be considered as a generative model.

My question is, is my interpretation correct here? I am asking this, because I have seen some sources on web which imply that $p$ changes during generating $x_i$s. If it were that way, shouldn't each of these different values be named as different random variables like $p_1$ for $x_1$, $p_2$ for $x_2$, etc. ?

This has greatly confused me. I appreciate any comments.

答:
I am posting my opinion about this, because I had the same question. I looked into page 83 of "Statistical Rethinking: A Bayesian Course with Examples in R and Stan," by Richard McElreath (1st Edition), and found the following code:

```<code>sample_mu <- rnorm(1e4, 178, 20)
sample_sigma <- runif(1e4, 0, 50)
prior_h <- rnorm(1e4, sample_mu, sample_sigma)
dens(prior_h)
</code>```

Basically this code samples 10,000 heights from a normal distribution with mean mu and standard deviation sigma, where mu follows a normal prior, and sigma also follows a uniform prior.

The way the heights prior_h are sampled, indicates that for each observation of height, a brand new mu and sigma is drawn from the prior.

So to answer your question, I think that yes, each draw from the prior should be labeled with a subscript i=1,2,3,... But I guess this is just not done by convention.

Body:
I am posting my opinion about this, because I had the same question. I looked into page 83 of "Statistical Rethinking: A Bayesian Course with Examples in R and Stan," by Richard McElreath (1st Edition), and found the following code:

sample_mu <- rnorm(1e4, 178, 20)
sample_sigma <- runif(1e4, 0, 50)
prior_h <- rnorm(1e4, sample_mu, sample_sigma)
dens(prior_h)

Basically this code samples 10,000 heights from a normal distribution with mean mu and standard deviation sigma, where mu follows a normal prior, and sigma also follows a uniform prior.

The way the heights prior_h are sampled, indicates that for each observation of height, a brand new mu and sigma is drawn from the prior.

So to answer your question, I think that yes, each draw from the prior should be labeled with a subscript i=1,2,3,... But I guess this is just not done by convention.


答:
The idea is to use a pigeonhole principle type argument: Two of the four numbers must have arguments which differ by at most $\pi/2$.

Let $\alpha_j$ be the arguments of the four numbers in the range $[0, 2 \pi)$ and sort the numbers by their arguments in increasing order. The non-negative differences
$$
 \alpha_2 - \alpha_1, \,\alpha_3-\alpha_2, \,\alpha_4-\alpha_3, \,(2\pi+\alpha_1) -\alpha_4
$$
add up to $2 \pi$, so that at least one of them must $\le \pi/2$. Then the corresponding quotient, that is, one of
$$
 z_2/z_1,\, z_3/z_2, \,z_4/z_3, \,z_1/z_4 \,
$$
has an argument between $0$ and $\pi/2$, which means that it has non-negative real and imaginary part.

Title: If you have four complex numbers and divide each of them by the others then will one of the results have to have non negative real and imaginary parts

Body: Say you have four distinct nonzero complex numbers and you divide each one by the other three (ie you take $\frac{z_k}{z_j}$ for each $k \neq j$). Is there a way to prove that at least one of these quotients must produce a complex number with nonnegative real and imaginary parts? I've been trying to come up with a counter example all day and I can't find one so I'm wondering if this is something that can actually be proven.


问:
Alternative interpretation of $\mathbb{CP}^n/\mathbb{CP}^{n-1}$
I saw from some algebraic topology paper like J.F. Adams' vector field on sphere claiming that $\mathbb{RP}^n/\mathbb{RP^{n-1}}$ is homeomorphic to $S^n$. So, I have a similar question regarding $\mathbb{CP}^n/\mathbb{CP}^{n-1}$. Is it just $S^{2n}$?

I hope someone can have a more pictorial explanation on that!

Thanks in advance for answering.

答:
$\mathbb{CP}^n$ is homeomorphic to a CW complex with a single cell in each even dimension between $0$ and $2n$ (inclusive), i.e. $\mathbb{CP}^n = e^0 \cup e^2\cup\dots\cup e^{2n-2}\cup e^{2n}$. The union of the cells up to and including dimension $2k$ can be identified with $\mathbb{CP}^k \subset \mathbb{CP}^n$. When we quotient out by such a subspace, all these cells collapse to a single point, so $\mathbb{CP}^n/\mathbb{CP}^k = e^0\cup e^{2k+2}\cup e^{2k+4}\cup\dots e^{2n-2}\cup e^{2n}$. In particular, $\mathbb{CP}^n/\mathbb{CP}^{n-1} = e^0\cup e^{2n}$ which is a CW complex homeomorphic to $S^{2n}$.

Said another way, we can obtain $\mathbb{CP}^n$ from $\mathbb{CP}^{n-1}$ by attaching a $2n$-cell. When we quotient by $\mathbb{CP}^{n-1}$, the resulting space is obtained by attaching a $2n$-cell to a point, which results in a sphere of dimension $2n$.


问:
If 3 six faced fair dice are thrown together, then the probability that the sum of the numbers $ 9\leqslant k \leqslant 14$
If 3 six faced fair dice are thrown together, then the probability that the sum of the numbers appearing on the dice is k where $ 9\leqslant k \leqslant 14$ is  :

My approach : Possible sum values are from 3 to 18 and our cases of interst are $9,10,11,12,13,14 \therefore$ probability should be $\frac{6}{16}$ but the answer given is $\frac{21k-k^{2}-83}{216}$ How can arrive at the given answer ?

答:
Every roll of three dice corresponds to an ordered triple $(D_1, D_2, D_3)$, where $1 \leq D_i \leq 6$ for each $1 \leq i \leq 3$. There are $6^3 = 216$ such ordered triples, which make up our sample space: $$\Omega := \{(D_1, D_2, D_3):\text{for each } 1 \leq i \leq 3, D_i \in \Bbb{N} \text{ and } 1 \leq D_i \leq 6 \} .$$ Therefore the probability we want to find, $$\Bbb{P}( D_1 + D_2 + D_3 = k ) \text{ for } k \in \Bbb{N} \text{ and } 3 \leq k \leq 18,$$ must have a denominator of $216$, if we write it as $$\Bbb{P}( D_1 + D_2 + D_3 = k ) = \frac{|E_k|}{|\Omega|},$$ where $E_k \subset \Omega$ is the set $$E_k := \{(D_1, D_2, D_3):\text{for each } 1 \leq i \leq 3, D_i \in \Bbb{N}, 1 \leq D_i \leq 6, \text{ and } D_1 + D_2 + D_3 = k \}.$$

Because $$(D_1, D_2, D_3) \in E_k$$ is equivalent to $$(7 - D_1, 7 - D_2, 7 - D_3) \in E_{21 - k},$$ we have $|E_k| = |E_{21-k}|$, so it suffices to evaluate $$E_3, E_4, ..., E_9, E_{10},$$ as we will then know the values of $$E_{18} = E_3, E_{17} = E_4, ..., E_{12} = E_9, E_{11} = E_{10}.$$ An application of stars-and-bars then gives us $$E_k = \binom{k-1}{2} - 3\binom{k-7}{2} = \frac{-2k^2+42k-166}{2} = -k^2+21k-83 = k(21-k) - 83$$ for $3 \leq k \leq 10$.

(The second term comes from excluding partitions of $k$ that have at least one part $>6$. We can partition $k-6$ into a sum of three positive integers in $\binom{k-7}{2}$ ways, and then we have $3$ choices of which positive integer to add $6$ back to to get a partition of $k$ where one part is strictly greater than $6$. This gives a total of $3\binom{k-7}{2}$ such partitions of $k$.)

This matches the original numerator when $k = 9, 10$. Since the quadratic $f(x) = x(21-x) - 83$ satisfies $f(x) = f(21-x)$, we know that this formula works for $k = 11, 12, 13, 14$ as well. Done.


问:
Geometry of the space $C[a, b]$ respect to the norm $\lVert x \rVert_{\infty} = \max_{t\in [a,b]}\lvert x(t)\rvert$.
I have studied that the space $C[a, b]$ of all scalar-valued (real or complex)
continuous functions defined on $[a, b]$ is a Banach space with respect to the norm $\lVert x \rVert_{\infty} = \max_{t\in [a, b]}\lvert x(t)\rvert$.

I want to understand geometry of the space $C[a, b]$ for the norm defined above. Is this possible? Because I am not able to visualize this space with respect to the norm defined above. Could anybody explain me?

Thanks

答:
This questions is rather broad. As soon as you go from the finite dimensional space to the infinite dimensional visualizing becomes much more complicated in general. One needs to develop certain intuition which comes from practice. There is nothing special about $C[a,b],$ you have to get used to infinite dimensional spaces first. When you are dealing with concrete problem you can try to apply your "euclidean" intuition first. Roughly speaking, there are two major differences to keep in mind:

$1.$ There is no Pythagorean theorem $\|x\|^2+\|y\|^2=\|x+y\|^2$ whenever one $x,y$ are "orthogonal." The reason is that we cannot define orthogonality in the usual way because $C[a,b]$ is not a Hilbert space. One can however, choose "almost" orthogonal vectors if necessary.

$2.$ The unit ball is not compact, but this is a common thing for all infinite dimensional spaces. One has to be very careful when choosing convergent subsequences etc.


问:
What does path-connected property correspond to in Gelfand duality?
A commutative C* algebra $A$ correspond to algebra of continuous function from locally compact Hausdorff space, where that space is the spectrum of $A$. Many properties of the space can be obtained from properties of $A$. I am wondering if spectrum of $A$ is path connected, what property of $A$ should have?

答:
This is not an answer, but the following paper, Section 5, deals with a slightly weaker property, approximate path connectedness:

Hadwin, Don; Shulman,
Tatiana, Tracial stability for
$C^*$-algebras,
Integral Equations Oper. Theory 90, No. 1, Paper No. 1, 35 p. (2018).
ZBL1396.46045.


问:
sum of the roots of unity
$$\sum_{k=0}^{N-1} e^{-j\frac{2\pi}{N}(i + n)k} = \begin{cases} N & \text{for } (i + n) = 0,N,2N,3N, \ldots \\ 0 & \text{otherwise} \end{cases} = N\delta[(i + n)\text{ mod }N]$$

Can someone provide an explanation why the sum is equal to zero if (i+n) mod N is not equal to zero?

答: For any polynomial where the coefficient of the leading term $=1$, then the coefficient of the second term is the negative sum of the roots.  Therefore for $z^n-1=0$, the sum$=0$.

Title: sum of the roots of unity


问:
How was this simplification done?

\begin{align*} \begin{pmatrix}
     -2\\
     n\end{pmatrix}&=\frac{(-2)(-3)(-4)\dots(-2-n+1)}{n!}\\[5pt] &= (-1)^n\frac{2\times3\times4\times\dots n(n+1)}{n!}\\[5pt]
 &=(-1)^n(n+1)\\[5pt] \end{align*}

While reviewing the binomial theorem, I came across this example problem and I feel like this simplification was not clearly explained. It seems like -1 was factored out but what happens to the expression it is multiplied by. How is it written in such a way? Also, why is $-1$ raised to the power of $n$...$(-1)^n$. I do not find this intuitive and felt like a step was skipped although that might not be the case.

Overall, I just want to understand what the process is here.

答:
Suppose $m>n$. We can write simplify
$$\dbinom mn=\dfrac{m!}{n!\,(m-n)!}=\frac{m(m-1)(m-2)\dots(m-n+1)}{n!}.$$

By analogy, if $m<n$, we can set
$$\dbinom mn=\frac{m(m-1)(m-2)\dots(m-n+1)}{n!}.$$
In particular, if $m<0$, the numerator has all its factors negative, so we can write it  as \begin{align}&-|m|\cdot -\bigl(|m|+1\bigr)\cdot -\bigl(|m|+2\bigr)\dots -\bigl(|m|+n-1\bigr)=\\[1ex]
&(-1)^n|m|\bigl(|m|+1\bigr)\bigl(|m|+2\bigr)\dots\bigl(|m|+n-1\bigr)
\end{align}
since there are $n$ factors.

Title: How was this simplification done?


问:
$n$ particles hunting each other
If $n$ particles $A_1,A_2,...,A_n$ are situated at the vertices (at time $t=0$ $s$ of an $n$ sided regular polygon $\overline{A_1A_2...A_n}$ such that the velocity of the $n^{th}$ particle is always along the side $\overline{A_{n-1}A_n}$ of the polygon. Each particle has a constant velocity $v$.

I want to model a general equation to find the time at which these particles would collide with each other. One example is that of an equilateral triangle's case. Any suggestions would be helpful. (One idea is to take the angle-component of one particle toward another side, and use relative velocity)

答:
Because the particles are situated on an n-sided regular polygon at $t=0$ they admit a radial symmetry at an angle of $\alpha = \frac{2\pi}{n}$. The velocity vectors preserve this symmetry, so the particles will stay on an n-sided regular polygon.

Therefore it suffices to consider one point $(x, y)$, or in the complex plane $z=x+yi$, if we take $0$ as the center of the polygon, the next point is at $e^{-\alpha i} z$. So the differential equation to figure out the path of $z$ is $$\frac{dz}{dt} = z - e^{-\alpha i} z = (1 - e^{-\alpha i}) z.$$

We won't care about the velocity of the point for now, because this differential equation is easy to solve. The solutions are $$z(t) = C e^{(1-e^{-\alpha i})t}$$ for some $C$ in $\mathbb{C}$ which is the equation for a spiral. We can rewrite this by noticing $e^{-\alpha i} = \cos(\alpha) - i\sin(\alpha)$ so that
$$z(t) = C e^{(1-\cos(\alpha))t}(e^{i\sin(\alpha)t}).$$
In polar coordinates $$r(\theta) = R e^{\frac{1-\cos(\alpha)}{\sin(\alpha)}\theta}.$$ Let's introduce the variable $\beta = \frac{1-\cos(\alpha)}{\sin(\alpha)} = \tan(\frac{\alpha}{2})$, then we see that the trajectory of the point is a classic logarithmic spiral $r(\theta) = R e^{\beta \theta}$.

The arc length parametrization of this function is known (check Wikipedia for a full explanation). This also explains the way to calculate the arc length to the origin, it requires the following calculation.

We know that $\beta = \tan(\frac{\alpha}{2}) = \tan(\frac{\pi}{n})$ so that the formula for the arc length between a point on the curve $(r, \theta)$ and the origin is equal to $\frac{r}{\sin(\frac{\pi}{n})}$. Meaning that the time it takes to reach the origin from a point at distance $r$ is equal to $\frac{r}{v\sin(\frac{\pi}{n})}$

Some values of $\frac{1}{\sin(\frac{\pi}{n})}$ for specific values of $n$.

For $n=3$ the value is $\frac{2}{\sqrt 3} = 1.1547...$

For $n=4$ the value is $\sqrt 2 = 1.4142...$

For $n=5$ the value is around 1.7013,

For $n=6$ the value is exactly 2, etc...

Title: $n$ particles hunting each other


问:
Prove that for all bijective functions $\pi:\mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$, $\sum_{n=1}^{\infty} a_{\pi (n)}$ is convergent.
I am tasked with this question:

Let $$\sum_{n=1}^{\infty} a_{(i, n)} = a_{(i, 1)}+a_{(i, 2)}+a_{(i, 3)}+\dots$$
Such that $\sum_{n=1}^{\infty} a_{(i, n)}$ is absolutely convergent for each $i$, and $\sum_{n=1}^{\infty} |a_{(i, n)}| = b_i.$ Furthermore, $\sum_{n=1}^{\infty} b_i$ is also convergent. Prove that for all bijective functions $\pi:\mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$, $\sum_{n=1}^{\infty} a_{\pi (n)}$ is convergent.

My attempt:

Since $\pi$ is bijective, every $a_{(i,n)}$ is represented in $\sum a_{\pi (n)}$. Thus $\sum a_{\pi (n)} = \sum a_{(i,n)}$.

We take $\sum |a_{\pi (n)}| = \sum b_i$, which is convergent. Since every absolutely convergent series is also just convergent, it follows that $\sum a_{(i,n)} = \sum a_{\pi (n)}$ is convergent.

Is that just it? The weightage of the question is rather high, and I can't help but feel that I've made a mistake in reasoning. Anyway I can further 'flesh out' the proof in case something is missing?

答:
Let $b=\sum_{n\ge 1}b_n$; the real work is in proving that

$$\sum_{n\ge 1}|a_{\pi(n)}|=b\,.\tag{1}$$

Let $\epsilon>0$; there is an $n_\epsilon\in\Bbb Z^+$ such that $\sum_{k>n_\epsilon}b_k<\frac{\epsilon}2$. For each $i\in\{1,\ldots,n_\epsilon\}$ there is an $m_i\ge n_\epsilon$ such that $\sum_{k>m_i}|a_{(i,k)}|<\frac{\epsilon}{2^{i+1}}$; let

$$m=\max\{m_i:i=1,\ldots,n_\epsilon\}\,.$$

Finally, there is an $\ell_\epsilon\in\Bbb Z^+$ such that

$$\{1,\ldots,m\}\times\{1,\ldots,m\}\subseteq\pi[\{1,\ldots,\ell_\epsilon\}]\,.$$

Then

$$\begin{align*}
\sum_{k=1}^{\ell_\epsilon}|a_{\pi(k)}|&\ge\sum_{i=1}^{\ell_\epsilon}\sum_{n=1}^{\ell_\epsilon}|a_{(i,n)}|\\
&\ge\sum_{i=1}^{\ell_\epsilon}\left(b_i-\sum_{n>\ell_\epsilon}\left|a_{(i,n)}\right|\right)\\
&>\sum_{i=1}^{\ell_\epsilon}\left(b_i-\frac{\epsilon}{2^{i+1}}\right)\\
&\ge\sum_{i=1}^{n_\epsilon}b_i-\sum_{i\ge 1}\frac{\epsilon}{2^{i+1}}\\
&>b-\frac{\epsilon}2-\sum_{i\ge 1}\frac{\epsilon}{2^{i+1}}\\
&=b-\epsilon\,.
\end{align*}$$

Clearly $\sum_{k=1}^n|a_{\pi(k)}|<b$ for all $n\in\Bbb Z^+$, so $(1)$ holds.


问:
How to get the angle between the inner common tangents to two general ellipse
I have been trying to find the angle made by inner tangents to two general ellipse by following the method described by Futurologist to find four homogeneous equations of tangents, this method is presented here and uses the concept of dual conics and degenerate conics. Using that method, I am able to get four tangent equations in homogeneous coordinates. However, I am looking to find the angle between the inner common tangents. For that purpose, currently, I am solving for the intersection of the four tangent lines with the ellipse and then tracing those points back to get the angle made by inner common tangents. This process is a bit computationally expensive and I am looking for a less expensive algorithm.

My question here is that if it is possible to obtain that angle directly instead of going through the hassle of finding the intersection points? I am not interested in getting the equation of tangents. I just need the angle between them.

Graphically in the diagram below, that angle can be represented as $\theta$:

Note: Using the slopes of the lines obtained from above method do not return correct answer because we do not know if that angle is greater one or smaller one  (two angle made by exterior or interior to those line), it depends on the relative position of the two conics. I want a generalize solution which will work for every configuration of two given conics anywhere in the 2D space.

Body:
Too long for a comment. In an example in Macaulay2

R=QQ[x,y,MonomialOrder=>Lex]

Two ellipses

I=ideal(-3*x^2+2*x*y-2*y^2-17*x+14*y-30,-2*x^2-y^2+14*x+y-18)

And their duals

dI=ideal((71*y^2)/4-59*x*y+25*y+11*x^2-20*x+5,(-13*y^2)+7*x*y+2*y+(71*x^2)/4+14*x+2)

matrix {{131274351*y^4-96177240*y^3-254150568*y^2-116827680*y-15474064, 4357738320*x+65593417383*y^3-77753736324*y^2-93563362092*y-17649191264}}

R=QQ[y,x,MonomialOrder=>Lex]

matrix {{131274351*x^4-57576960*x^3-86987952*x^2+15882240*x+1105856, 11193976176*y-37500706269*x^3-4585249998*x^2+14300319336*x+3683892688}}

So we have the solutions $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ and the relations  $131274351y^4-96177240y^3-254150568y^2-116827680y-15474064=0$ that the $y_i$ satisfy and also $131274351x^4-57576960x^3-86987952x^2+15882240x+1105856=0$ that the $x_i$ satisfy

then

$(x x_1+y y_1+1)(x x_2+y y_2+1)(x x_3+y y_3+1)(x x_4+y y_4+1)=0$

or using

ideal((x*x1+y*y1+1)*(x*x2+y*y2+1)*(x*x3+y*y3+1)*(x*x4+y*y4+1),131274351*y4^4-96177240*y4^3-254150568*y4^2-116827680*y4-15474064, 4357738320*x1+65593417383*y1^3-77753736324*y1^2-93563362092*y1-17649191264,131274351*y1^4-96177240*y1^3-254150568*y1^2-116827680*y1-15474064, 4357738320*x2+65593417383*y2^3-77753736324*y2^2-93563362092*y2-17649191264,131274351*y2^4-96177240*y2^3-254150568*y2^2-116827680*y2-15474064, 4357738320*x3+65593417383*y3^3-77753736324*y3^2-93563362092*y3-17649191264,131274351*y3^4-96177240*y3^3-254150568*y3^2-116827680*y3-15474064, 4357738320*x4+65593417383*y4^3-77753736324*y4^2-93563362092*y4-17649191264,131274351*x1+131274351*x2+131274351*x3+131274351*x4-57576960,-131274351*x1*x2-131274351*x1*x3-131274351*x2*x3-131274351*x1*x4-131274351*x2*x4-131274351*x3*x4-86987952,131274351*x1*x2*x3+131274351*x1*x2*x4+131274351*x1*x3*x4+131274351*x2*x3*x4+15882240,-131274351*x1*x2*x3*x4+1105856,(131274351*y1+131274351*y2+131274351*y3+131274351*y4-96177240),(-131274351*y1*y2-131274351*y1*y3-131274351*y2*y3-131274351*y1*y4-131274351*y2*y4-131274351*y3*y4-254150568),(131274351*y1*y2*y3+131274351*y1*y2*y4+131274351*y1*y3*y4+131274351*y2*y3*y4-116827680),-131274351*y1*y2*y3*y4-15474064)

the four tangents are

$1105856x^4+5396992x^3y-5326656x^2y^2-31177472xy^3-15474064y^4-15882240x^3+31083840x^2y+174435840xy^2+116827680y^3-86987952x^2-234816192xy-254150568y^2+57576960x+96177240y+131274351=0.$

If you can find the correct factorization into quadratics, you can hope to use the formula $\tan \theta = {2 \sqrt{h^2-ab} \over a+b}, $ where $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents your line pair.


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
Serious math involves serious abstraction.  Most people aren't very good at that.  It takes practice.  But when people say that they're no good at math, I'm not sure that's the problem, because people who say that probably have never gotten that far.

When I look at a page of symbols in an area of math about which I know nothing, it looks like nonsense, it looks impressive, and it looks impenetrable.  But I know from experience that that's how math always looks when you don't know anything about it.  So I'm not afraid of the mysterious pages of symbols.  I know that I can learn it--at least if I find the right textbook, and go slowly, and it doesn't turn out that I will need 7 years of study to get to where I can understand this area of math.  (I also think that the mysterious symbols look cool, even when I don't understand them, but that's just my perversion.)

I think that many people just see the impenetrable symbols and shudder.  And they do this even when they know what each symbol means.  Their eyes glaze over.  I think the key is to stop, take a deep breath, and worth through the symbols one by one.  One probably needs to learn that that's a possibility.

About pride:  I once worked in a menial office job with someone who was obviously good at careful, complex reasoning.  Any time I would point that out, he'd insist that I was wrong, and that he was no good at logic.  No, he was an artistic type.  Math is probably like that.  Math seems rational, rigid, inhuman to some people.  So if you're good at math, there's something a little bit wrong with you, from that point of view.


问:
$2$-norm of a rank-$1$ matrix
I want to prove that $\|A\|_2 = \|x\|_2\|y\|_2$ given that $A = xy^T$ is a rank one matrix. This is my incomplete attempt so far, I get stuck when I need to take into account the spectral radius of the symmetric matrix:
\begin{align}
\|A\|_2 &= \sqrt{\rho(A^TA)} \\
                  & = \sqrt{\rho(xy^Tyx^T)} \\
                  & = \sqrt{\rho((y^Ty)xx^T)} \\
                  & = \sqrt{\rho(\|y\|_2^2xx^T)} \\
                  & \qquad\quad \vdots\\
                  & = \sqrt{x^Tx } \cdot\sqrt{y^Ty}\\
                  & = \|x\|_2\|y\|_2
\end{align}

I know $xx^T$ is a symmetric matrix and its got a scalar $\|y\|_2^2$ out in front of it. Wondering if there's some property of a rank one matrix here that I could use, or perhaps my knowledge on spectral radii is limited and that is why I cannot get to the next step? I tried to look up the spectral radius/largest eigenvalue for a symmetric matrix but what I found was beyond the scope of what I am currently studying. Not sure where to go from here.

答: Hint: The spectral radius of an operator is by definition the magnitude of its largest eigemvalue. What can you say about the eigenvalues of a rank-one matrix?</p>

Title: $2$-norm of a rank-$1$ matrix


问:
Interior and closure operators iterated
Let $X$ be a topological space and for arbitrary $A \subseteq X$ define a sequence of subsets of $X$ as follows: $A_0 := A$, be $A_{n+1}:= Int(A_n)$ (the interior of $A_n$ for $n$ even) and $A_{n+1}:= Cl(A_n)$ (the closure of $A_n$ for $n$ odd). May this sequence in general have infinitely many different elements? Is there a characterization of  those spaces in which the sequence stabilizes after finitely many iterations for every $A$?

答: It always stabilises as $\operatorname{int} \overline{A}$ is regular open for any $A$, and $\overline{\operatorname{int}(A)}$ is regular closed for any $A$. So after int cl int cl we’re back at int cl again, and also cl int cl int = cl int as well. So $A_4=A_2$ always (if I've counted the iterations correctly).

Title: Interior and closure operators iterated


答:
You need to find an integrating factor, and remember the product rule.

Before we do that, let's get this into a bit of a nicer form.

$$
\frac{dy}{dt} = \frac{1}{e^y - t} \\
e^y\,dy - t\,dy = dt \\
$$

Now, the tough term is $-t\,dy$ because it mixes variables.  However, the product rule can help with mixed variables.  However, it needs two parts to work, and one of them needs to be with $dt$.  So, we will move this term over to be on the same side as $dt$.

$$ e^y\,dy = dt + t\,dy $$

Now, the right-hand side is almost in a form that will work for the product rule.  Remember, the product rule is $d(jk) = k\,dj + j\,dk$.  However, there is not a function of $y$ next to the $dt$.  However, if you multiply by $e^y$, then, since $e^y$ is both the value and the derivative, it can work in both settings.  Therefore, multiplying the whole equation by $e^y$ gives you:

$$ e^y\cdot e^y\,dy = e^y\,dt + e^y\,t\,dy \\
e^{2y}\,dy = e^y\,dt + e^y\,t\,dy
$$
The product rule tells us that the right-hand side is merely the differential of $t\,dy$.  Therefore, we can rewrite the equation as
$$ e^{2y}\,dy = d(t\,e^y) $$.
This is then very easy to integrate:
$$ \int e^{2y}\,dy = \int d(t\,e^y) \\
\frac{1}{2}e^{2y} = t\,e^y + C
$$


问:
Given a 3D box, a ray direction and a point on a face, find the other point on the box intersected by the ray
I have a 3D program which only renders the rear faces of a box. Given a ray intersection with one of the rear faces (point A), I want to obtain the point B on the box which would be the point of intersection with the front face. I understand that I can move in the negative direction of the ray from point A some distance D to arrive at point B. Is that the best approach, and if so how do I calculate that distance?

答:
Imagine that you have your box like this, placed on the x y z axis. (Note this is a left-handed system.) Suppose that the point on the rear corresponding to the ray is (x0,y0,z0) and the ray has direction (a,b,c). Suppose the box has dimensions length (l), width (w), and height (h). We want to find the point (not labeled) on the front of the box such that the ray originated.

You want to scale the direction of the ray so that it can basically be superimposed on the box so we can use the pythagorean theorem. Focus on the x direction, which spans the length of the box. Divide l by a to find r, the scaling factor, that you will need to multiply the components of the ray to find the scaled ray "superimposed" on the box. From here, you can directly use the pythagorean theorem to get

$$\begin{split}d&=\sqrt{a^2r^2+b^2r^2+c^2r^2}\\
&=r\sqrt{a^2+b^2+c^2}\\
&=\frac l a\sqrt{a^2 + b^2+c^2}\end{split}$$

Note: the letter "l" looks like a 1, but it's not. Or, going step by step, look at the above picture and see that the base is $base=\sqrt{a^2r^2+b^2r^2}$. Now there is one more step, we take the base and factor in the height of the ray.

And we find that the distance is $distance=\sqrt{c^2r^2+\left(\sqrt{a^2r^2+b^2r^2}\right)^2}$ as before. To find the point, simply move a distance d in the opposite direction of the ray. This is done by finding the unit vector in the direction opposite the ray and adding that times the distance to the point you were at, effectively moving you the desired distance in 3-D space.

$$\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}-\frac{1}{\sqrt{a^2+b^2+c^2}}\begin{pmatrix}a\\b\\c\end{pmatrix}*\frac l a\sqrt{a^2 + b^2+c^2}\\
=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}-\begin{pmatrix}a\\b\\c\end{pmatrix}*\frac l a\\
=\begin{pmatrix}x_0-l\\y_0-b*\frac l a\\z_0-c*\frac l a\end{pmatrix}$$

So you see we didn't actually have to do all that work, the simplified form shows that it is attained by simply finding the scaling factor along the length of the box, and then moving in the direction of the ray by the scaling factor from the original point. Hope this helps!


问:
How can I solve the equation $\frac{dy}{dt} = \frac{1}{e^y - t}$?
How can I solve the following equation?

$$\frac{dy}{dt} = \frac{1}{e^y - t}$$

I feel as if I am missing something obvious with this question but have been trying to figure it out for too long now. The only hint given is to try and treat $y$ as the dependent variable, however I don’t understand what that means for me to do.

答:
$\dfrac{dy}{dt}=\dfrac1{e^y-t}$

$\big(e^y-t\big)\dfrac{dy}{dt}=1$

$e^y\dfrac{dy}{dt}-t\dfrac{dy}{dt}-1=0$

$2e^{2y}\dfrac{dy}{dt}-2te^y\dfrac{dy}{dt}-2e^y=0$

$\dfrac d{dt}\left(e^{2y}-2te^y\right)=0$

$e^{2y}-2te^y=c\;,\quad$ where $\;c\in\mathbb{R}\;,$

$e^y=t\pm\sqrt{t^2+c}$

$y(t)=\ln\left(t\pm\sqrt{t^2+c}\right)\;.$


答:
Here is another approach for your own entertainment and understanding.

If $v\in \mathbb{R}^n$ is arbitrary, then we have from Cauchy Schwartz that $$\|Av\|_2=\big|y^Tv\big|\|x\|_2\leq\Big(\|x\|_2\|y\|_2\Big)\|v\|_2$$ So for any $v \neq  0$ $$\frac{\|Av\|_2}{\|v\|_2}\leq \|x\|_2\|y\|_2$$ This shows $\|A\|_2\leq \|x\|_2\|y\|_2$. We get equality by noting that this upper bound is attained whenever we take $v\in \text{span}(y)-\{0\}$.

Title: $2$-norm of a rank-$1$ matrix

Body:
I want to prove that $\|A\|_2 = \|x\|_2\|y\|_2$ given that $A = xy^T$ is a rank one matrix. This is my incomplete attempt so far, I get stuck when I need to take into account the spectral radius of the symmetric matrix:
\begin{align}
\|A\|_2 &= \sqrt{\rho(A^TA)} \\
                  & = \sqrt{\rho(xy^Tyx^T)} \\
                  & = \sqrt{\rho((y^Ty)xx^T)} \\
                  & = \sqrt{\rho(\|y\|_2^2xx^T)} \\
                  & \qquad\quad \vdots\\
                  & = \sqrt{x^Tx } \cdot\sqrt{y^Ty}\\
                  & = \|x\|_2\|y\|_2
\end{align}

I know $xx^T$ is a symmetric matrix and its got a scalar $\|y\|_2^2$ out in front of it. Wondering if there's some property of a rank one matrix here that I could use, or perhaps my knowledge on spectral radii is limited and that is why I cannot get to the next step? I tried to look up the spectral radius/largest eigenvalue for a symmetric matrix but what I found was beyond the scope of what I am currently studying. Not sure where to go from here.


问:
Show a connection $\nabla$ is compatible with a metric $\langle \cdot, \cdot \rangle$ of $\mathbb{R}^3$

We introduce in $\mathbb{R}^3$ with the usual Euclidean metric $\langle \cdot, \cdot \rangle$, the connection defined in Cartesian ccoordinates $(x^1, x^2, x^3)$ by
$$\Gamma_{jk}^i = \omega \varepsilon_{ijk}$$
where $\omega: \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth function and
$$\varepsilon_{ijk} = \begin{cases} +1 \quad \text{if }(i,j,k) \text{ is an even permutation of }(1,2,3)\\
-1 \quad \text{se }(i,j,k) \text{ is an odd permutation of }(1,2,3)\\
0 \quad \text{ otherwise}
\end{cases}$$
Show that $\nabla$ is compatible with $\langle \cdot, \cdot \rangle$;

I've taken a local coordinate system $(x_1, x_2, \dots, x_n)$ and used the

We know that a connection $\nabla$ is compatible with a metric $\langle \cdot, \cdot \rangle$ if and only if for every differentiable curve $c: I \rightarrow M$ and every pair $V, \,W$ of vector fields differentiable along $c$ we have
$$\frac{d}{dt} \langle V, W \rangle = \left\langle \frac{DV}{dt}, W \right\rangle +\left\langle V, \frac{DW}{dt} \right\rangle, \quad t \in I$$

Let $(x_1(t), x_2(t), x_3(t))$ be the parametrization of $c(t)$ in the local coordinate system, and let $X_i = \frac{\partial}{\partial x_i}$, and write $V$ as $V =\sum_{j=1}^3 v^j X_j$ (and similarly for $W$), where $v_j = v_j(t)$. We can express the covariant derivative as

$$
\begin{align}\frac{DV}{dt} &= \sum_{k=1}^3 \left[ \frac{dv^k}{dt} + \sum_{i,j} v^j \frac{dx_i}{dt} \Gamma_{ij}^k\right] X_k\\
&= \sum_{k=1}^3 \left[ \frac{dv^k}{dt} + \sum_{i,j} v^j \frac{dx_i}{dt} \omega \varepsilon_{ijk}\right] X_k
\end{align}$$

I thought about calculating $\frac{d}{dt} \langle V, W \rangle$ and $ \left\langle \frac{DV}{dt}, W \right\rangle +\left\langle V, \frac{DW}{dt} \right\rangle$ in these terms and showing they are equal, but it would be too messy. I think there must be a smart solution that doesn't require too many calculations.

答: Since $\nabla\langle \cdot,\cdot\rangle$ is a tensor, it suffices to check that $(\nabla_{\partial_k}\langle\cdot,\cdot\rangle)(\partial_i,\partial_j) = 0$ for all $i,j,k \in \{1,2,3\}$. In other words, we must show that $$\partial_k \langle \partial_i,\partial_j\rangle = \langle \nabla_{\partial_k}\partial_i,\partial_j\rangle + \langle \partial_i, \nabla_{\partial_k}\partial_j\rangle$$for all $i,j,k \in \{1,2,3\}$. The left side is obviously zero, so the situation doesn't look so bad. By the definition of $\nabla$, we have that $$\begin{align} \langle \nabla_{\partial_k}\partial_i,\partial_j\rangle + \langle \partial_i, \nabla_{\partial_k}\partial_j\rangle &= \langle \omega \varepsilon^r_{~ki}\partial_r,\partial_j\rangle + \langle \partial_i, \omega \varepsilon^r_{~kj}\partial_r\rangle \\ &= \omega \varepsilon^r_{~ki} \delta_{rj} + \omega \varepsilon^r_{~kj}\delta_{ir} \\ &= \omega(\varepsilon^j_{~ki} + \varepsilon^i_{~kj}) \\ &= 0,\end{align}$$because since $(j,k,i) \mapsto (i,k,j)$ is an odd permutation, we have that $\varepsilon^j_{~ki} =- \varepsilon^i_{~kj}$.


问:
Prove $\sum_{n \leq x} \tau (n) = 2(\sum_{n \leq \sqrt{x}} [\frac{x}{n}]) - [\sqrt{x}]^2$
Here $\tau(n)$ is the number of positive integers dividing $n$ and $[x]$ is the floor of $x$.

So I know if $n$ is not a perfect square, then half the positive integers dividing $n$ are less than $\sqrt{n}$, and the other half are greater than $\sqrt{n}$. Also, the amount of perfect squares that are less than or equal to $x$, is $[\sqrt{x}]^2$.

So let $d_n(k) = 1$ if $k|n$ and $d_n(k) = 0$ otherwise. Then
$\sum_{n \leq x} \tau (n) = \sum_{n \leq x} \sum_{k \leq n} d_n(k) = \sum_{n \leq x} 2(\sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2 = 2(\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2$.

So all I have to so is some how prove $\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) =\sum_{n \leq \sqrt{x}} [\frac{x}{n}]$

答:
$$\begin{align}\sum_{n=1}^x\tau(n)&=\sum_{n=1}^x\#\{\,(a,b):ab=n\,\}\\
&=\#\{\,(a,b,n):ab=n\le x\,\}\\&
=\#\{\,(a,b):ab\le x\,\}\\&
=\#\{\,(a,b):ab\le x, a\le b\,\}+\#\{\,(a,b):ab\le x, a\ge b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\#\{\,(a,b):ab\le x, a\le b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\sum_{a\le\sqrt x}\#\{\,b:ab\le x, a\le b\,\}-\#\{\,a\mid a^2\le x\}\\&
=2\sum_{a\le\sqrt x}\left(\#\{\,b:ab\le x\,\}-\#\{\,b:ab\le x, a> b\,\}\right)-\#\{\,a\mid a^2\le x\}\\&
=2\sum_{a\le\sqrt x}\left(\left\lfloor\frac xa\right\rfloor-(a-1)\right)-\lfloor\sqrt x\rfloor\\&
=2\left(\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\frac{\lfloor \sqrt x\rfloor(\lfloor \sqrt x\rfloor-1)}2\right)-\left\lfloor\sqrt
x\right\rfloor\\
&=2\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\left\lfloor\sqrt
x\right\rfloor^2
\end{align}$$


答:
There is a way easier formula for the sum of $\tau(n)$, which is very intuitive too:
$$\sum_{n\leq x} \tau(n) = \sum_{n\leq x} \left\lfloor\frac{x}{n}\right\rfloor$$

Title: Prove $\sum_{n \leq x} \tau (n) = 2(\sum_{n \leq \sqrt{x}} [\frac{x}{n}]) - [\sqrt{x}]^2$

Body:
Here $\tau(n)$ is the number of positive integers dividing $n$ and $[x]$ is the floor of $x$.

So I know if $n$ is not a perfect square, then half the positive integers dividing $n$ are less than $\sqrt{n}$, and the other half are greater than $\sqrt{n}$. Also, the amount of perfect squares that are less than or equal to $x$, is $[\sqrt{x}]^2$.

So let $d_n(k) = 1$ if $k|n$ and $d_n(k) = 0$ otherwise. Then
$\sum_{n \leq x} \tau (n) = \sum_{n \leq x} \sum_{k \leq n} d_n(k) = \sum_{n \leq x} 2(\sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2 = 2(\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) ) - [\sqrt{x}]^2$.

So all I have to so is some how prove $\sum_{n \leq x} \sum_{k \leq \sqrt{n}} d_n(k) =\sum_{n \leq \sqrt{x}} [\frac{x}{n}]$


问:
Let $f(z)=\frac{z^3}{(z-\pi)^3(z+5)^2}$ and let $C$ be $|z|=3$.Then $\int_C{f(z)dz}=0$ because :
Which of the following options are true?

Let $f(z)=\dfrac{z^3}{(z-\pi)^3(z+5)^2}$ and let $C$ be $|z|=3$.Then $\int_C{f(z)dz}=0$ because :

(A) the residue is $0$ at its only pole within $C$.

(B) the sum of the residues at its two poles within $C$ is $0$.

(C) $f(z)$ is analytic on and within $C$.

(D) $|f(z)|< \frac {M}{|z|^2}$ for some constant $M$.

答: $C$ is the correct answer here because the given fuction has poles on $\pi$ and $5$ but none of them are within and on the circle</p>


问:
Proof of Clairaut’s theorem in Terence Tao Analysis 2
I'm self-studying the book Real Analysis 2 by Terence Tao.

I'm having some trouble working with vector-valued functions. Particularly, I'm stuck at his proof of Clairaut's theorem for a vector-valued function while having no math professor to ask :(

screenshot 1
Theorem 6.5.4 (Clairaut's theorem). Let $E$ be an open subset of $\mathbf{R}^{n}$, and let $f: E \rightarrow \mathbf{R}^{m}$ be a twice continuously differentiable function on
E. Then we have $\frac{\partial}{\partial x_{j}} \frac{\partial f}{\partial x_{i}}\left(x_{0}\right)=\frac{\partial}{\partial x_{i}} \frac{\partial f}{\partial x_{j}}\left(x_{0}\right)$ for all $1 \leq i, j \leq n$.

In his proof, he proves for the case $\fbox{m=1}$ and $x_0 = 0$. The beginning of the proof is easily to understood:

screenshot 2
Let $a$ be the number $a:=\frac{\partial}{\partial x_{j}} \frac{\partial f}{\partial x_{i}}(0),$ and $a^{\prime}$ denote the quantity $a^{\prime}:=\frac{\partial}{\partial x_{i}} \frac{\partial f}{\partial x_{j}}(0) .$ Our task is to show that $a^{\prime}=a$.
Let $\varepsilon>0$. Because the double derivatives of $f$ are continuous, we can find a $\delta>0$ such that
$$
\left|\frac{\partial}{\partial x_{j}} \frac{\partial f}{\partial x_{i}}(x)-a\right| \leq \varepsilon
$$
and
$$
\left|\frac{\partial}{\partial x_{i}} \frac{\partial f}{\partial x_{j}}(x)-a^{\prime}\right| \leq \varepsilon
$$
whenever $|x| \leq 2 \delta$.

Then comes the difficult part that I can't understand his notation:

screenshot 3
Now we consider the quantity
$$
X:=f\left(\delta e_{i}+\delta e_{j}\right)-f\left(\delta e_{i}\right)-f\left(\delta e_{j}\right)+f(0)
$$
From the fundamental theorem of calculus in the $e_{i}$ variable, we have
$$
f\left(\delta e_{i}+\delta e_{j}\right)-f\left(\delta e_{j}\right)=\int_{0}^{\delta} \frac{\partial f}{\partial x_{i}}\left(x_{i} e_{i}+\delta e_{j}\right) d x_{i}
\qquad\color{red}{(*)}$$
and
$$
f\left(\delta e_{i}\right)-f(0)=\int_{0}^{\delta} \frac{\partial f}{\partial x_{i}}\left(x_{i} e_{i}\right) d x_{i}
$$
and hence
$$
X=\int_{0}^{\delta}\left(\frac{\partial f}{\partial x_{i}}\left(x_{i} e_{i}+\delta e_{j}\right)-\frac{\partial f}{\partial x_{i}}\left(x_{i} e_{i}\right)\right) d x_{i}.
$$

Where $e_i$ is a $n-dimensional$ vector with all zeros except the $1$ in the $i^{th}$ position.

My question concerns the $\color{red}{(*)}$ equations:

How to use the Fundamental theorem of calculus to lead to the (*) equation?

Why the limit of the integral is from $0$ $\to$ $\delta$ and NOT from ($\delta \, e_j$ $\to$ $\delta \, e_i + \delta \, e_j$) ? (like in the case of single variable integration: $F(b) - F(a) = \int \limits_{(a,b)} f $ ) ?

($i.e$ In my opinion, we consider $b = \delta \, . e_i + \delta \, .e_j$ and $a= \delta \, .e_j$, so the limit of the integration should be from $a$ to $b$ which is different from what is written in the text: $0$ $\to$ $\delta$)

Why the "main" function to integrate is $\frac{\partial f}{\partial x_j}$? Why is it evaluated at the point ($x_i*e_i + \delta * e_j$)?

Why the differential part in the integral is $dx_i$?

Could you please suggest to me a $\fbox{document}$ or some $\fbox{keywords}$ so that I can read more to further understand the things related to the $\color{red}{(*)}$ equation?

In the case where $m > 1$, does the equation $\color{red}{(*)}$ still hold ? Is there any definition of integration of a vector-valued function?

($i.e.$ $\int f dx$ where $f : \mathbb{R^n} \to \mathbb{R}^m$ with $m > 1$ and $x$ is a single variable. In other words, does an integration always take scalar- value or it can be a vector) ?

Thank you very much for your help!

答:
Let me share following thoughts:

Fundamental theorem of calculus leads to the equation, because (*), and equality after it, is exactly fundamental theorem applied to specific functions and bounds.

Antiderivative of $\frac{\partial f}{\partial x_i}(x_i e_i + \delta e_j)$ with respect to $x_i$ is $f(x_i e_i + \delta e_j) +C$, for some $C$ constant and it is used for bounds $0$ and $\delta$ and gives by fundamental theorem $f(x_i e_i + \delta e_j)\big|_{x_i=0}^{x_i=\delta} = f( \delta e_i + \delta e_j) - f( \delta e_j)$.

and 4. have same answer: it is proved, in equality (*) and equality after it, that quantity $X$ can be represented in integral form.


问:
Show that the sequence defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}$ converges
Show that the sequence $(x_n)$ defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}, x_0=3,x_1=2$ converges. Here i wil try to prove that $(x_n)$ converges considering all details. So. i would like to have a critical feedback on my proof and would like to know if i missed something or did much more than it should be.. Here $\mathbf{R}^*$ will denote $\mathbf{R}\ \ \not \ \{0\}$.

Consider a function $f(x)=\sqrt[3]{x}$. First, we show that it's defined on $\mathbf{R}$ and then we will show that $f(x)$ is an increasing function on $\mathbf{R}$.

We remark that $f(x)$ is differntiable on $\mathbf{R}^*$. Let $a \in \mathbf{R}^*$. By definition of differentiable function the $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ msut exist:

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\frac{1}{3\sqrt[3]{a^2}}$ and so it exists $\forall a \in \mathbf{R}^*$ as we chose an arbitrary $a$. So $f(x)$ is differentiable on $\mathbf{R}^*$ and so is continious on $\mathbf{R}^*$.

We show now that $f(x)$ is continious on $0$. To show that we pass by the definition :

$\forall \varepsilon>0 \ \exists \delta>0 \ \forall x \in \mathbf{R}$: $|x|<\delta \implies |f(x)|=|\sqrt[3]{x}|<\varepsilon$

To satisfy the definition, one can take $\delta=\varepsilon^3$, so it holds and we can conclude that $f(x)$ is continious on $\mathbf{R}$.

Let $a,b \in \mathbf{R}$ such that $a<b$. As $f(x)$ is continious on $\mathbf{R}$ we can write:

$a<b \iff \sqrt[3]{a}<\sqrt[3]{b} \iff f(a)<f(b)$

As $a,b$ are arbitrary numbers (satisfying $a<b$), we can conclude that $f(x)$ is strictly incrasing function.

Now, as all preliminary results are proven, we can show that $(x_n)$ is a convergent sequence. For that, we show that $(x_n)$ is monotonic and bounded.

$\bullet$ $(x_n)$ is decreasing: Suppose $x_{n+1}<x_n$. The previous inequality holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We show nowthat it holds for $n+1$:

$x_{n+2}<x_{n+1} \iff \sqrt[3]{x_{n+1}+x_n}<\sqrt[3]{x_n+x_{n-1}}$. But, as $f(x)=\sqrt[3]{x}$ is strictly increasing function and by induction hypothesis $x_{n+1}<x_n$, we can conclude that $x_{n+2}<x_{n+1}$ and so $x_{n+1}<x_n$ holds $\forall n\in\mathbf{N}^*$. So $(x_n)$ is decreasing.

$\bullet$ $(x_n)$ is bounded below: Suppose $x_n>1$. It holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We want to show now that t holds for $n+1$:

$x_{n+1}>1 \iff \underbrace{x_{n}}_{>1}+\underbrace{x_{n-1}}_{>1}>1$ which is true, because $(x_n)$ is decreasing. So we conclude that $x_n>1 \ \forall n\in \mathbf{N}^*$.

So, we have shown that $(x_n)$ converges.

答:
You asked for feedback on your proof, here are some thoughts:

To prove $f$ is continuous, you say that it's differentiable. But you do not prove that $f$ is differentiable directly - you only state the derivative. You could show that the derivative exists with this value, but that proof would be very similar to proving $f$ is continuous directly (e.g. using the binomial theorem). So it would be simpler to just do that.

To prove $f$ is increasing, you just say "$f$ is continuous, therefore [$f$ is increasing]". There is no proof in your question that $f(x) = \sqrt[3]{x}$ is an increasing function.

Your proofs that $(x_{n})$ is decreasing and bounded below by 1 appear to be correct. Note that they only rely on the fact that $f$ is increasing and that $f(1) = 1$. You haven't used the fact that $f$ is continuous anywhere.

However, the property that $f$ is continuous does allow you to find the limit. In general, if $x_n \to x$ and $f$ is continuous, then $f(x_n) \to f(x)$. Applying this to the equation $x_{n+1} = f(x_n+x_{n-1})$ shows that the limit satisfies $x = f(2x)$. In other words, $x^3 = 2x$, so either $x=0$, $x=-\sqrt{2}$ or $x=\sqrt{2}$. But since you argued already that $x > 1$, the limit must be $\sqrt{2}$.


问:
Geometry of the space $C[a, b]$ respect to the norm $\lVert x \rVert_{\infty} = \max_{t\in [a,b]}\lvert x(t)\rvert$.
I have studied that the space $C[a, b]$ of all scalar-valued (real or complex)
continuous functions defined on $[a, b]$ is a Banach space with respect to the norm $\lVert x \rVert_{\infty} = \max_{t\in [a, b]}\lvert x(t)\rvert$.

I want to understand geometry of the space $C[a, b]$ for the norm defined above. Is this possible? Because I am not able to visualize this space with respect to the norm defined above. Could anybody explain me?

Thanks

答:
All you can "visualize" in this space is the convergence as the notion of distance is fairly easy to visualize here. So, $f$ and $g$ are close if their graphs are within a certain distance of each other.
This is will help if you study the space of $n-$times continuously differentiable functions also.</p>


问:
Number of labelled spanning trees in the following graph.

Hi, I have this question in one of my assignments. Normally I would use an adjacency matrix (a very big one) to solve these kinds of questions but the hint makes me believe the professor wants me to solve this in a different way. I see no mention in the lectures of why cycles of length 4 have any significance in counting the number of spanning trees so I'm wondering if anyone can help me out with maybe a theorem or lemma that I'm missing? I'm not looking for the answer, just the tools to each the answer because right now I am at a loss. The hint is throwing me off a lot.

Thanks for any help.

Edit: The picture is a bit blurry. The hint says "Consider 5 cases depending on how many edges of the cycle of length 4 you keep."

答:
I’ll expand on the hint a bit.

Suppose that you remove all $4$ edges of the $4$-cycle. What’s left is a $16$-cycle; how many labelled spanning trees does it have?

Now suppose that you remove only $3$ of those $4$ edges: you have a $5$-cycle and a $13$-cycle that share one edge. How many more edges must you remove to get a spanning tree? How many labelled spanning trees can you get in this case?

There are three more cases, and they’re a bit harder, but the principle is the same for all of them: you have to remove just enough edges to break every cycle.


问:
Finding $\frac{DK}{DI}$
In triangle $ABC,$ where $AB = 8, AC = 7,$ and $BC = 10,$ $I$ is the incenter. If $AI$ intersects $BC$ at $K$ and the circumcircle of $\triangle ABC$ at $D,$ find $\frac{DK}{DI}.$

I first drew a diagram, but I was unsure where to go from here.

答:
I have doubts that you correctly compute the result, since the value $IK$ given in comments is incorrect. The correct result is:
$$\frac{DK}{DI}=\frac23.
$$
The details are given below.

Let $x,y,z$ being the distances from the vertices $A,B,C$ to the tangent points of the incircle. From the equations $x+y=c, y+z=a, z+x=b$ one obtains $x=\frac{b+c-a}2=s-a$, where $s$ is the semiperimeter. Then:
$$AI=\sqrt{x^2+r^2}=\sqrt{(s-a)^2+\frac{(s-a)(s-b)(s-c)}s}=\sqrt {\frac {bc (s-a)}{s}}=2\sqrt{\frac{14}5}.$$
The angle bisector length is:
$$AK=\sqrt{bc\left[1-\left(\frac  a{b+c}\right)^2\right]}=\frac {\sqrt{4bc (s-a)s}}{b+c}=\frac{10}3\sqrt{\frac{14}5}.$$
By the power of point $K$ and angle bisector theorem:$$\color {red}{DK}=\frac{BK\cdot KC}{AK}=\frac{a^2\frac{bc}{(b+c)^2}}{AK}=\frac{a^2}{b+c}  \sqrt {\frac{bc}{4 (s-a)s}}=\frac83\sqrt{\frac{14}5},$$
so that:
$$\color {red}{DI}=DK+AK-AI=4\sqrt{\frac{14}5}.$$

Title: Finding $\frac{DK}{DI}$


问:
Infinite set notation

Is it acceptable to write "$X$ is an infinite set" as "$|X| = \infty$"?

By "acceptable" I mean that I can use it in a research paper or in a textbook, and a reasonable person won't be against it. Possible arguments against it:

$\infty$ is not actually a value, so it doesn't fit into a standard set-size notation.

$|\mathbb R| = \infty$ and $|\mathbb N| = \infty$, so one may think that $|\mathbb R| = |\mathbb N|$.

(I looked for duplicates with some variations of the title, and found none)

答: $\infty$ is a bit ambiguous/misplaced in the context of set cardinality, something like an Aleph Number (https://en.wikipedia.org/wiki/Aleph_number) might be more appropriate, i.e. $|X| \ge \aleph_0$, which states "X has the cardinality of at least the smallest possible infinite set".

Title: Infinite set notation


问:
Principal Ultrafilter contains a finite set
Prove that an ultrafilter is principal if and only if it contains a finite set.

I can't quite figure out why and it may be because I don't have a clear understanding of a principal ultrafilter. I have read a few different definitions that don't seem to line up. Could someone clear up what a principal ultrafilter is?

答:
Let $X$ be a set, and let $x\in X$. The principal ultrafilter at $x$ is

$$\mathscr{U}_x=\{U\subseteq X:x\in U\}\,.$$

The principal ultrafilters on $X$ are precisely these ultrafilters $\mathscr{U}_x$ for $x\in X$.

Note that it’s always true that $\{x\}\in\mathscr{U}_x$, so clearly every principal ultrafilter on $X$ contains a finite set, even a singleton. Your real task is to show that if $F$ is a finite subset of $X$, $\mathscr{U}$ is an ultrafilter on $X$, and $F\in\mathscr{U}$, then $\mathscr{U}=\mathscr{U}_x$ for some $x\in X$. In other words, you must show that there is some $x\in X$ such that $\{x\}\in\mathscr{U}$.

HINT: In fact there is an $x\in F$ such that $\{x\}\in\mathscr{U}$.


问:
An application of Brouwer fixed point Theorem

Let $\zeta:\mathbb{R}\to [0,+\infty)$ be a continuous non-negative function such that $\zeta(0)=0$ and $\tau\mapsto \zeta(\tau)\tau$ is a non-decreasing differentiable function whose derivative is bounded on every compact subset of $\mathbb{R}$.

Let $\{\phi_{k}, \lambda_{k}\}_{k \in \Bbb N}$ be the the Dirichlet eigenpairs of the n Laplace operator on an open bounded set $\Omega\subset \Bbb R^N$, i.e.,  $\phi_k\in H_0^1(\Omega)$  and $-\Delta \phi_k= \lambda_k\phi_k$. Recall $\{\phi_k\}_{k}$ is an orthonormal basis in $L^2(\Omega)$.

Question: Let $\mathcal{V}_{k}= \operatorname{span}\{\phi_1,\dotsc, \phi_{k}\}$. According to page 5 Eq (3.3) of Starovoitov - Boundary value problem for a global in time parabolic equation, the Brouwer fixed-point theorem implies the existence of a vector $v_k\in \mathcal{V}_k$ such that
\begin{equation}\label{Star-3.3}
\int_\Omega \nabla v_{k}\cdot \nabla \phi dx + \int_\Omega\zeta(v_k)v_k\phi dx=\int_\Omega f\phi dx\quad\text{for all}\quad\phi\in \mathcal{V}_{k}.
\end{equation}

How can one justify this claim?

In fact, that $\phi_k\in L^\infty (\Omega)$ is the only important property needed from $\phi_k$. So that by assumption the function $\phi(v_k)v_k$ is  bounded.
Since we are in finite-dimensional space and $\int_\Omega \nabla \phi_{i}\cdot \nabla \phi_j dx=\lambda_i\delta_{ij}$, the above equation reduces into finding $v_k=(v_{k,1}\phi_1+ \dotsb+v_{k,k}\phi_k)$ satisfying
\begin{equation}\label{Star-3.v}
  \sum_{i=1}^k v_{k,i}\lambda_i + \zeta(v_k)v_k = f_k\quad\text{in} \quad \mathcal{V}_{k},
\end{equation}
where  $f_k=(f_{k,1}\phi_1+ \dotsb+f_{k,k}\phi_k)$ is the projection of $f$ on $\mathcal V_k$ and by abuse of notation, we write again $\zeta(v_k)v_k$ to represent its own projection on $\mathcal V_k$.

Recall Brouwer fixed-point theorem: Every continuous function from a closed ball of a Euclidean space into itself has a fixed point.

答:
First using Brouwer

Let $w\in \mathcal{V}_k$, necessarily $\zeta(w)$ is a bounded function since $\phi_k $'s are also bounded. The Lax-Milgram lemma implies there is a unique function $\widehat{w}\in  \mathcal{V}_k$
such that
\begin{align}
\int_\Omega\nabla\widehat{w}\nabla\phi +\zeta(w)\widehat{w}\phi -f\psi dx=0\quad\text{for all}\quad\phi\in \mathcal{V}_{k}.
\end{align}

The Poincar'{e}--Friedrichs inequality yields
\begin{equation}
\int_\Omega|\nabla\widehat{w}|^2dx  + \int_\Omega \zeta(w )\widehat{w}^2\, d x\leq \|f\|_{L^{2}(\Omega)}\,\|\widehat{w}\|_{L^{2}(\Omega)}\leq  C\|f\|_{L^{2}(\Omega)}\, \|\widehat{w}\|_{H_0^1(\Omega )}
\end{equation}

Thus, letting $R=C\,\|f\|_{L^{2}(\Omega)}$, since $\varphi\geq0 $ we obtain the following estimates
\begin{align}\label{eq:boundedmapT}
&\|\widehat{w}\|_{H_0^1(\Omega )}\leq R
\quad\text{ and }\quad
 \int_\Omega \zeta(w )\widehat{w}^2\,dx\leq R^2.
\end{align}
We let $ \mathcal{B}_R=\big\{ w\in \mathcal{V}_k: \|w\|_{H_0^1(\Omega)} \leq R\big\}$, be the closed ball in $\mathcal{V}_k$ of radius $R$ centered at the origin.  Clearly, the mapping $T:\mathcal{V}_k\to \mathcal{B}_R$ with $Tw=\widehat{w}$  is well defined.

It remains to prove that $T$ is a continuous mapping. Indeed, let
$\{w_n\}$ be a sequence in $\mathcal{V}_k$ with $w_n= \lambda_{1,n}\phi_1+\cdots+  \lambda_{k,n}\phi_k$ converging in $\mathcal{V}_k$ to a function $w= \lambda_1\phi_1+\cdots+  \lambda_k\phi_k$; i.e., $\lambda_{\ell,n}\xrightarrow{n\to\infty } \lambda_\ell$, $\ell=1,2,\cdots,k$.
By continuity we have $ \varphi(w_n)\xrightarrow{n\to \infty}\varphi(w)$ almost everywhere. In addition, the convergence in $L^2(\Omega)$ also holds, since the continuity gives $\sup_{n\geq 0} \|\varphi(w_n) \|_{L^\infty(\Omega)} <\infty$ because $\sup_{n\geq 0}\|w_n\|_{L^\infty(\Omega) } <\infty$.

On the other side, in virtue of  the first estimate above the sequence
$\{Tw_n\}$ is bounded in finite dimensional space $\mathcal{V}_k$ and thus converges $\mathcal{V}_k$ up to a subsequence  to some $w_*\in \mathcal{V}_k$. Altogether, it follows that, for all  $\phi\in \mathcal{V}_k\subset L^\infty(\Omega)$
%
\begin{align*}
(f,\phi)= \lim_{n\to \infty} \int_\Omega\nabla\widehat{w}_n\nabla\phi
+\zeta(w_n)\widehat{w}_n\phi dx=
\int_\Omega\nabla\widehat{w}\nabla\phi  +\zeta(w)w_* \phi.
\end{align*}

The uniqueness of $\widehat{w}$ entails  $w_*=\widehat{w}=Tw$ and hence  the whole sequence $\{Tw_n\}$ converges in $Tw$ in $\mathcal V_k$, which gives the continuity of $T$.

Therefore, by the Brouwer fixed-point theorem,  $T$ has a fixed point $v_k\in \mathcal V_k$, i.e., $v_k=Tv_k$ which clearly satisfies  the announced relation.

An alternative.

The given problem is equivalent to the minimization problem
\begin{align}
 \mathcal{J}(v_k)=  \min_{v\in \mathcal{V}_k}  \mathcal{J}(v)\quad \text{with}\quad  \mathcal{J}(v):=  \frac12 \int_\Omega|\nabla v|^2 dx + \int_\Omega G(v)d x + \int_\Omega fvd x
 \end{align}
and we define the function $G(v)=  \int_0^v\zeta(\tau)\tau d \tau $.  Note that $G$ is non-negative since $\zeta(\tau)\geq 0$ and that $\mathcal J$ is continuous on $\mathcal V_k$. Using the Poincaré-Friedrichs inequality we find that $\mathcal{J}(v)\to \infty$, as $\|v\|_{L^2(\Omega)} \to \infty$ and $v\in \mathcal{V}_k$. Which implies the existence of a minimizer $v_k\in \mathcal{V}_k$ of \mathcal{J}, since we are in finite dimension space.


问:
Cylindrical Shell method conceptual question
I am self-teaching myself calculus for the summer to get ready for the actual class. Let us say that we have a region bounded above by the curve $y = 2 - x^2$ and below by the curve $y=x^2$ from $x=0$ to $x=1$. Suppose that the region is revolved around the $y-$axis.

Why do we use this formula instead of the washer method?:

$ 2 \pi \int_a^b x(f(x) - g(x)) dx$

The surface area of a cylinder is $2\pi rh$ (without the top and bottom) and the volume is $\pi r^2 h$, so how can the volume of a cylinder be $2 \pi rh dr$? I don't understand that way of thinking.

I thought that we're adding infinitesimally thin cylinders as we go up and down, so why not use the washer method?

答:
If we want, we could use the washer method (in fact, this is the disk method since the $y$-axis is a boundary of the region) and integrate with respect to $y$. However, this would involve two separate integrals, since the radius of each disk is defined by two different curves. Thus, the volume using the disk method is:
$$
V= \left[ \pi \int_0^1 (\sqrt{y})^2 dy \right] + \left[ \pi \int_1^2 (\sqrt{2-y})^2 dy \right] = \dfrac{\pi}{2} + \dfrac{\pi}{2} = \boxed{\pi}
$$
The computations in this example actually turned out to not be that bad, but sometimes it just isn't feasible this way if the inverse functions of the boundaries can't be easily found.

Instead, it would be easier to add infinitesimally thin cylindrical shells ("hollow pipes"). If "unrolled" into a rectangular prism, the dimensions of an arbitrary shell would be $2\pi r \times h \times dx$, where the radius is $r=x$ and the height is $h=(2-x^2)-x^2$. Note that $2\pi r$ corresponds to the circumference of the original cylindrical shell's circular base and $dx$ corresponds to the shell's infinitesimally thin thickness. This yields the integral:
$$
V= \int_0^1 2\pi x ((2-x^2)-x^2)dx = \boxed{\pi}
$$


问:
Golden functions?
If one seeks the functional solutions of $f^2(x)=f(x)+1=f(f(x))$

Are they called 'golden functions' ?

Are they always of the form of rational functions ?

答:
At least on the (unbounded and infinite!) set $\operatorname{im}(f)$, we see that $f(t)=t+1$. Hence the only rational function solution is just $f(x)=x+1$.

In fact, the following will describe all solutions $f\colon \Bbb R\to\Bbb R$ of the functional equation
$$ \forall x\in \Bbb R\colon f(f(x))=f(x)+1$$

Pick a non-empty set $A\subseteq \Bbb R$ that is closed under "$+1$". In other words, we have $A+1\subseteq A$, or  $A$ can be written as $A=\bigcup_{a\in A_0}(\Bbb N+a)$.
Let $g\colon \Bbb R\setminus A\to A$ be arbitrary and define
$$ f(x)=\begin{cases}x+1&x\in A\\g(x)&x\notin A\end{cases}$$
Indeed, with such $f$, we have $f(x)\in A$ for all $x$ and hence $f(f(x))=f(x)+1$.

Here's a non-trivial example of a smooth such function:
$$ f(x)=\begin{cases}x+1&x\ge 0\\x^2+x+1+e^{1/x}&x<0\end{cases}$$

Title: Golden functions?


问:
Find the cube roots of -8i. Express them in algebraic form.
Would you do this by translating to trig form and then simplifying?</p>

答:
Let $z\in \mathbb{C}$ then we can write $z=|z|(\cos \theta +i\sin \theta)$ this representation recives the name of polar form of the complex number.

Exists a theorem that speak about the roots of some complex number and say that the $n$-th roots of the number $z$ above are given by

$$w_k=|z|^{1/n}\left( \cos \left(\frac{\theta+2\pi k}{n}\right)+i \sin \left(\frac{\theta+2\pi k}{n}\right) \right)$$

and where $k=0,1,2,3\cdots n-1$

One example for this  get the roots of $z=1+0i$ notice that $\theta=0$ and $|z|=1$ and we want $\sqrt[3]{z}$.

The theorem say

$$w_k=|z|^{1/n}\left( \cos \left(\frac{\theta+2\pi k}{n}\right)+i \sin \left(\frac{\theta+2\pi k}{n}\right) \right)$$

Applying the theorem

$$w_0=1\left( \cos \left(\frac{0}{3}\right)+i \sin \left(\frac{0}{3}\right) \right)=1+0i$$
$$w_{1}=\left( \cos \left(\frac{2\pi}{3}\right)+i \sin \left(\frac{2\pi}{3}\right) \right)=\frac{-1}{2}+\frac{\sqrt{3}}{2}$$
$$w_{3}=\left( \cos \left(\frac{4\pi}{3}\right)+i \sin \left(\frac{4\pi}{3}\right) \right)=\frac{-1}{2}-\frac{\sqrt{3}}{2}$$

Notice that the values of the binomial form are get from the value of $\cos(\text{angle})$ for the real part and $\sin\text{angle}$ for the imaginary part.

I wait that you can apply the theorem from here

Hint:
In your case $z=0-8i$ and $\theta=\frac{3 \pi}{2}$, $|z|=8, n=3, k=0,1,2$


问:
Show that $a=c\mathrm{e}^{\mathrm{i}\beta}+b\mathrm{e}^{-\mathrm{i}\gamma}$ in a triangle $ABC$ with side lengths $a$, $b$ and $c$
The question goes like this:

In triangle $ABC$, the lengths of the sides $BC$, $CA$, $AB$ are $a$, $b$, $c$ respectively, and the radian measures of the angles at $A$, $B$, $C$ are $\alpha$, $\beta$, $\gamma$ respectively. Prove that
$$a=c\mathrm{e}^{\mathrm{i}\beta}+b\mathrm{e}^{-\mathrm{i}\gamma}.$$

My attempt at a solution:

For convenience sake, let $z^{}_1$, $z^{}_2$ and $z^{}_3$ denote the complex numbers represented by the points $A$, $B$ and $C$ respectively in an Argand diagram.

Referring to my drawing above, since $\overrightarrow{CB}$ is $\overrightarrow{CA}$ stretched by a factor of $\dfrac{a}{b}$ and rotated through $\gamma$ radians anticlockwise (about $C$), I conclude that
$$z_2-z_3=\frac{a}{b}e^{i\gamma}(z_1-z_3).$$

Using similar arguments for the directed line segment pairs $\overrightarrow{BC}, \overrightarrow{BA}$ and $\overrightarrow{AB}, \overrightarrow{AC}$, I obtain
$$z_1-z_2=\frac{c}{a}e^{i\beta}(z_3-z_2), \quad z_3-z_1=\frac{b}{c}e^{i\alpha}(z_2-z_1).$$

How do I proceed from here onwards to establish the given result?

答: Generally with problems like this, you want to pick a nice coordinate system which makes the calculations become as trivial as possible.  In this case, let's pick coordinates so that $B=0$, $C$ is on the positive real axis, and $A$ is in the upper half plane.  This means that as a complex number, $C$ is just $a$.  Travelling from $B$ to $A$, we find that $A=ce^{i\beta}$.  On the other hand, travelling instead from $C$ to $A$, we get $A=a+be^{i(\pi-\gamma)}=a-be^{-i\gamma}$.  Setting these two expressions for $A$ equal gives the desired equation.


答:
I'm personally strongly against this - we have more exact notation for describing infinite sets, so why not use it?

That said, I have seen this notation used when describing objects which are subsets of some fixed countable set (e.g. sets of natural numbers). Here there's really no possible confusion. However, I still consider this bad practice.

Title: Infinite set notation

Body:
Is it acceptable to write "$X$ is an infinite set" as "$|X| = \infty$"?

By "acceptable" I mean that I can use it in a research paper or in a textbook, and a reasonable person won't be against it. Possible arguments against it:

$\infty$ is not actually a value, so it doesn't fit into a standard set-size notation.

$|\mathbb R| = \infty$ and $|\mathbb N| = \infty$, so one may think that $|\mathbb R| = |\mathbb N|$.

(I looked for duplicates with some variations of the title, and found none)


问:
Prove that ever $\Sigma_0$ sentence is provable from PA (Peano Arithmetic, with exponentiation)
I am trying to study the intuition and steps leading up to proving Godel's Incompleteness theorems. In a text I am studying it asserts that every true $\Sigma_0$ sentence is provable from PAE (where this system is the peano axioms with exponentiation). I am confused about how I get this fact; I don't know if it follows from constructing proof predicates or using Godel numbering in a similar way. Any help and guidance would be really appreciated. Thank you.

答:
We want to prove that for all sentences $\psi \in \Sigma_0$,   $\mathbb{N} \models \psi \implies PAE \vdash \psi$.

By completeness, it suffices to prove the following:

For all models $M$ of $PAE$ and for all sentences $\psi \in \Sigma_0$, $\mathbb{N} \models \psi \iff M \models \psi$.

Fix a model $M$ of $PAE$ and proceed by induction on the structure of $\psi$:

$\psi$ atomic: $\mathbb{N}$ is a term (minimal) model of PAE, so there is an embedding $\mathbb{N} \to M$ given by $n \mapsto \bar n^M$. Embeddings preserve atomic formulas, and so $\mathbb{N} \models \psi \iff M \models \psi$.

$\psi \equiv \chi \land \phi$: Since $\psi \in \Sigma_0$, $\chi ,\phi \in \Sigma_0$ and so this follows easily by the induction hypothesis.

$\psi \equiv \lnot \chi$: Same idea

$\psi \equiv \exists v \leq \bar n \chi(v)$: We have that $\mathbb{N} \models \psi \iff \mathbb{N} \models \lor_{m=0}^n \chi(\bar m) \iff M \models \lor_{m=0}^n \chi(\bar m)$ by the induction hypothesis as $\chi(\bar m)$ is a $\Sigma_0$ sentence. Finally, $M \models \lor_{m=0}^n \chi(\bar m) \implies M \models \psi$ trivially. For the other direction we have that since $M \models PAE$, $\{ m \in M : m\leq_M \bar n^M\}=\{\bar 0, \bar 1, ..., \bar n\}$ (Can you see why?). Therefore $M \models \psi \implies M\models \lor_{m=0}^n \chi(\bar m)$.

Hence if $\mathbb{N} \models \psi$, $M \models \psi$ for all models of $PAE$, so $PAE \models \psi$ and by completeness $PAE \vdash \psi$.


答: Consider the open disc of radius $3.1$. Since $C$ lies in this disc and $f(z)$ is analytic there (the poles are at $\pi$ and $-5$, and so lie outside the disc), we can conclude using Cauchy's theorem that the integral is zero. </p>

Title: Let $f(z)=\frac{z^3}{(z-\pi)^3(z+5)^2}$ and let $C$ be $|z|=3$.Then $\int_C{f(z)dz}=0$ because :

Body:
Which of the following options are true?

Let $f(z)=\dfrac{z^3}{(z-\pi)^3(z+5)^2}$ and let $C$ be $|z|=3$.Then $\int_C{f(z)dz}=0$ because :

(A) the residue is $0$ at its only pole within $C$.

(B) the sum of the residues at its two poles within $C$ is $0$.

(C) $f(z)$ is analytic on and within $C$.

(D) $|f(z)|< \frac {M}{|z|^2}$ for some constant $M$.


问:
Manifolds with Euler characteristic equal to $\pm 1$
A compact connected smooth surface has Euler characteristic equal to $\pm 1$ if and only it is homeomorphic to the real projective plane or the connected sum of $3$ real projective planes. What are other examples of connected manifolds (of even dimension) that have Euler characteristic equal to $\pm 1$? For instance, one could take products of the above surfaces.

答: The union of a projective plane with any number of tori also has euler characteristic $\pm 1$.


问:
Find posterior distribution and find marginal distribution
We have
z = H s + $\epsilon$
with $\epsilon \sim N(0, R)$ and $s \sim N(\mu, Q)$ where R and Q are symmetric and positive definite. I'm trying to derive the posterior distribution. I have p(s) and p(z|s). However I'm unsure of how to get p(z), and the marginal distribution of y so I can get the posterior distribution p(s|z) using bayes formula. Also R nxn. Q is nxn. H is nxn. s is nx1 and $\epsilon$ is nx1.

p(s) = $\frac{e^{-1/2(s-\mu)^TQ^{-1}(s-\mu)}}{(2*\pi)^{n/2}|Q|^{1/2}} $

p(z|s) = $\frac{e^{-1/2(z-Hs)^TR^{-1}(z-Hs)}}{(2*\pi)^{n/2}|R|^{1/2}} $

Any help would be appreciated.

答:
When you find a posterior, it is true that $p(s|z)=\frac{p(z|s)p(s)}{p(z)}$ by Bayes rule. To do this, you would need to calculate the integral $p(z)=\int_{\Omega}p(z|s)p(s)ds$, however this turns out to be a constant (does not depend on s because it has been marginalized out) that will be part of the normalizing factor; consequently, it is more common to use the fact that $p(s|z)\propto p(z|s)p(s)$ to find the posterior density. Here, we can write

$$\begin{split}p(s|z)&\propto p(z|s)p(s)\\
&\propto\exp-\frac 1 2 (s^T-\mu^T)Q^{-1}(s-\mu)\exp-\frac 1 2 (z^T-s^TH^T)R^{-1}(z-Hs)\end{split}$$

Focus on the part inside the exponentiation

$$\begin{split}s^TQ^{-1}s-2\mu^TQ^{-1}s+s^TH^TR^{-1}Hs-2z^TR^THs=\\
s^T\underbrace{(Q^{-1}+H^TR^{-1}H)}_{=\Sigma^{-1}}s-2\underbrace{(\mu^TQ^{-1}+z^TR^TH)}_{=A^T}s\end{split}$$

We want to write this in the form $(s-m)^T\Sigma^{-1}(s-m)=s^T\Sigma^{-1}s-2m^T\Sigma^{-1}s+m^T\Sigma^{-1}m$
. Let $A^T=m^T\Sigma^{-1}$, then $\Sigma A=m$. So we add and subtract (subtraction is not shown because it is part of the normalizing constant) $A^T\Sigma A$ from the expression.

$$s^T\Sigma^{-1}s-2A^Ts+A^T\Sigma A=(s-\Sigma A)^T\Sigma^{-1}(s-\Sigma A)$$

Thus the posterior of $s$ follows a multivariate normal with mean $\Sigma A=(Q^{-1}+H^TR^{-1}H)^{-1}(Q^{-1}\mu+H^TRz)$ and variance covariance matrix $(Q^{-1}+H^TR^{-1}H)^{-1}$.


问:
If A is semisimple algebra then every A-module is completely reducible
A semisimple Algebra means the regular module is completely reducible. Let $A= \sum_{i}A_{i}$ where $A_{i}$ is irreducible sub modules of A. Suppose V is A-module. I tried to define a map between the $A_{i}$ and V.
$$\theta_{i}: A_{i} \longrightarrow V$$
$$ x_{i}\mapsto vx_{i} $$
the map is homomorphism so
$A_{i}\setminus ker(\theta_{i})$  is isomorphic to $\theta (A_{i}) \subseteq V$
the kernel is $\{0\}$ since $A_{i}$ is irreducible, so $A_{i}$  is isomorphic to $\theta_{i} (A_{i}) \subseteq V$.

$\theta_{i} (A_{i})$ is irreducible because of the isomorphic. Now, $\oplus \sum_{i} \theta_{i} (A_{i})$ is the direct sum of irreducible V submodules.

$\oplus \sum_{i}A_{i}$  is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$, i.e  $A$ is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$

If the steps are right, how can I show that $\oplus \sum_{i} \theta_{i} (A_{i})=  V$

答:
I tried to define a map between the $A$ and V.

For some v $\in$ V define the following map:

$$\theta: A \longrightarrow V$$
$$ x\mapsto vx $$
the map is homomorphism so
$A\setminus ker(\theta)$  is isomorphic to $\theta (A) \subseteq V$

$\theta (A) = vA$ $\subset V$

Then I proved that $A\setminus ker(\theta)$  is completely reducible so is $vA$

So $vA$ can be written as sum of irreducible submodules $V_{\alpha}$

$V= \sum_{v \in\ V\setminus {0}} vA = \sum_{s=1}^{n-1} \sum V_{\alpha}$

So $V$ is written as sum of irreducible submodules so $V$ is completely reducible


问:
Is $x-\sqrt{x}< p_{\pi(x)}\leq x$?
I am wondering if anyone is aware of any results related to the following? I need a few good updates/ideas/hints ...

For large enough $x$'s $$x-\sqrt{x}< p_{\pi(x)}\leq x \tag{1}$$

A few explanatory notes:

$\pi(x)$ is the prime counting function

$p_n$ - is the $n$'s prime with $p_1=2$

one example, $p_{\pi(\sqrt{12})}=p_{\pi(3)}=p_2=3$

if $x=p_n$ then $\pi(p_n)=n$ and $p_{\pi(p_n)}=p_n$.

Examples and progress ...

Observation 1. $(1)$ is true for the prime numbers

$$p_n - \sqrt{p_n}< p_{\pi(p_n)}=p_n \leq p_n$$
It also works for $x=p_n+1$ because $p_{\pi(x)}=p_{\pi(p_n)}=p_n$. Given $x-p_n=1<\sqrt{x}$ we conclude
$$x-\sqrt{x}<p_n=p_{\pi(x)}<p_n+1=x$$

Observation 2. RHS of $(1)$ is true, i.e. $$p_{\pi(x)}\leq x$$

From the definition of $\pi(x)$, if $\pi(x)=n$ then $p_1<p_2<...<p_n\leq x$, thus $p_{\pi(x)}\leq x$.

Observation 3. Heuristics for LHS of $(1)$ $$\lim\limits_{x\to\infty} \frac{p_{\pi(x)}}{x-\sqrt{x}}=1$$

From $$\frac{p_{\pi(x)}}{x-\sqrt{x}}=\frac{p_{\pi(x)}}{x}\cdot \frac{x}{x-\sqrt{x}}$$
$\frac{x}{x-\sqrt{x}}\to 1$ and $\frac{p_{\pi(x)}}{x}\to 1$ (an application of PNT). As a result, for some very small $\varepsilon>0$ and suficiently large $x$'s
$$p_{\pi(x)}>(1-\varepsilon)\cdot (x-\sqrt{x})>x\cdot(1-\varepsilon)-\sqrt{x} \tag{2}$$
Also a few examples
$$2-\sqrt{2}<p_{\pi(2)}=p_1=2\leq 2$$
$$3-\sqrt{3}<p_{\pi(3)}=p_2=3\leq 3$$
$$4-\sqrt{4}<p_{\pi(4)}=p_2=3< 4$$
$$5-\sqrt{5}<p_{\pi(5)}=p_3=5\leq 5$$
$$6-\sqrt{6}<p_{\pi(6)}=p_3=5< 6$$
$$...$$
$$14-\sqrt{14}<p_{\pi(14)}=p_6=13< 14$$

To some extent, this is a re-formulation of the Oppermann's conjecture, with $(2)$ as a faint light at the end of the tunnel. $\varepsilon$ is spoiling everything.

Another application is, from $(1)$:
$$x-p_{\pi(x)}<\sqrt{x}$$
for $x=p_n - 1$ we have $\pi(p_n - 1)=n-1$, thus
$$p_n - 1 - p_{n-1}<\sqrt{p_n - 1} \Rightarrow \\
p_n - p_{n-1}<\frac{\sqrt{p_n - 1}}{\sqrt{p_{n-1}}}\cdot \sqrt{p_{n-1}}+1<\sqrt{2}\cdot\sqrt{p_{n-1}}+1$$
and this is (a little bit better than) Andrica's conjecture.

答:
If this is true, then Legendre’s Conjecture would be true for large enough $n.$ Legendre’s Conjecture is currently unsolved, even for large enough $n.$

This is because when $x=(n+1)^2,$ your conjecture says there is a prime between $(n+1)^2-(n+1)=n^2+n>n^2$ and $(n+1)^2.$

Title: Is $x-\sqrt{x}< p_{\pi(x)}\leq x$?


问:
Is the triangle equilateral? (resolve without using trigonometry)
I have been asked the following question. Take a circle, and three arcs  of magnitude $\dfrac{\pi}{3}$ (or you could say the corresponding centered angles are of $60$ degrees) each, that do not intersect with each other, say arc  $AB$, arc  $CD$ and arc $EF$.

For each of the three remaining arcs take the corresponding chords $BC$, $DE$ and $FA$. Let $M, N, P$ respectively be the middles of these three line segments.   Is the triangle $MNP$ equilateral?

The person that asked me this question (a mathematician) did not know if the statement is true. I asked him where did he find it and he told me that his barber asked him. I tried to solve it using Euclidean geometry and I failed. I found a positive solution using trigonometry (I may present it here later).
I tried repeatedly to solve it again only using Euclidean geometry and I failed.   Any idea? (I mean no analytic geometry or complex numbers, just to prove $MN=NP=PM$ using Euclidean geometry)

Edit :   I will present the solution that I have, that is based on trigonometry.
Let $O$ be the center of the circle,  $R$ its radius and a,b,c respectively the magnitudes of the angles BOC,  DOE and FOA respectively.
Then $OM=R\cos\frac{a}{2}$,  $ON=R\cos\frac{b}{2}$, $OP=R\cos\frac{c}{2}$.

The angle  $MON$ is equal to $MON=\frac{a}{2}+\frac{\pi}{3}+\frac{b}{2}$ and taking into account that $a+b+c=\pi$ we get that $MON=\frac{5\pi}{6}-\frac{c}{2}$.

Thus $\cos(MON)=\cos(\frac{5\pi}{6}-\frac{c}{2})= \cos\frac{5\pi}{6}\cos\frac{c}{2}+\sin\frac{5\pi}{6}\sin\frac{c}{2}$
from which we deduce that
$$ \cos(MON)=
-\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}$$

Since $\frac{c}{2}=\frac{\pi}{2}-(\frac{a}{2}+\frac{b}{2})$ we have that
$\sin\frac{c}{2}=cos(\frac{a}{2}+\frac{b}{2})
=\cos\frac{a}{2}\cos\frac{b}{2}-\sin\frac{a}{2}\sin\frac{b}{2}$.
We get   $$ \sin\frac{c}{2}-\cos\frac{a}{2}\cos\frac{b}{2}=-\sin\frac{a}{2}\sin\frac{b}{2}. $$

Apply now the law of cosines in the triangle $MON$, in order to compute $MN^2$. Our main concern is to find a formula that is symmetric to any permutation of $a,b,c$, which will imlpy that the same formula is valid for $NP^2$ and $PM^2$.

$$ MN^2 =  OM^2+ON^2-2 OM\cdot ON \cos(MON)  \\
       =  R^2 \cos^2\frac{a}{2}+ R^2 \cos^2\frac{b}{2} -2 R\cos \frac{a}{2}R\cos\frac{b}{2} \cos(MON)\\
=R^2\Big(\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2} \cos(MON)\Big)\\
=R^2\Big[\cos^2\frac{a}{2}+ \cos^2\frac{b}{2} -2 \cos \frac{a}{2} \cos\frac{b}{2}\big (-\frac{\sqrt{3}}{2}\cos\frac{c}{2}+\frac{1}{2}\sin\frac{c}{2}\big) \Big]\\
=  R^2 \Big( \sqrt{3}  \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2}
-\cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\
=R^2 \Big( \sqrt{3}  \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}-1 +\sin^2\frac{c}{2}- \cos \frac{a}{2}\cos\frac{b}{2}\sin\frac{c}{2}\Big)\\
=R^2[-1+\sqrt{3}  \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}\\
\qquad\qquad\qquad\qquad\qquad\qquad
+\sin\frac{c}{2}(\sin\frac{c}{2}-\cos \frac{a}{2}\cos\frac{b}{2})\Big]\\
=R^2\Big[-1+\sqrt{3}  \cos\frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}
+\cos^2\frac{a}{2} + \cos^2\frac{b}{2} +\cos^2\frac{c}{2}
-\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}\Big]
$$

It seems to me impossible to translate this trigonometric proof to a pure euclidean geometry proof. This is the reason that I made the post, for someone to suggest a pure proof based on euclidean geometry.

Edit3:  As pointed out by @Blue the formula is simlpified to
$$MN^2=  R^2(1+\sqrt{3}\cos\frac{a}{2}  \cos\frac{b}{2}\cos \frac{c}{2}+
\sin \frac{a}{2}  \sin \frac{b}{2} \sin \frac{c}{2}).$$

答:
It is not difficile but something tedious solve this problem without trigonometry. Let a circle of equation $x^2+y^2=r^2$ and put $A=(x_A,y_A),\cdots,F=(x_F,y_F)$. The  triangles $\triangle{OAB},\triangle{OCD}$ and $\triangle{OEF}$ are clearly equilateral so we have $$\overline{AB}^2=r^2=(x_A-x_B)^2+(y_A-y_B)^2\Rightarrow 2(x_Ax_B+y_Ay_B)=r^2\space\space (1)$$ and similarly$$\overline{CD}^2=r^2\Rightarrow2(x_Cx_D+y_Cy_D)=r^2\space\space (2)\\\overline{EF}^2=r^2\Rightarrow2(x_Ex_F+y_Ey_F)=r^2\space\space (3)$$ Besides$$M=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2}\right)\\N=\left(\frac{x_D+x_E}{2},\frac{y_D+y_E}{2}\right)\\P=\left(\frac{x_A+x_F}{2},\frac{y_A+y_F}{2}\right)$$
It follows $$4\overline{PM}^2=(x_A+x_F-x_B-x_C)^2+(y_A+y_F-y_B-y_C)^2\\4\overline{PN}^2=(x_A+x_F-x_D-x_E)^2+(y_A+y_F-y_D-y_E)^2\\4\overline{NM}^2=(x_B+x_C-x_D-x_E)^2+(y_B+y_C-y_D-y_E)^2$$
We know by trigonometry that the request is true and then what we have to do is simply check the equalities $\overline{PM}=\overline{PN}=\overline{NM}$ using $(1),(2),(3)$ and the fact that  $A,B,C,D,E,F$ are in the circle.(Easily we must have for example $4\overline{PM}^2-4\overline{PN}^2=0$ and $4\overline{PM}^2-4\overline{NM}^2=0$)