问:
Proof of First Quartile in Uniform Distribution
For an uniform distribution between [a,b], find the first quartile Q1, as P(X>Q1)=0.75.

How do I proof that in this uniform distribution f(x)=1/(b-a), the Q1 is 0.25?

答:
First problem: Let $X \sim \mathsf{Unif}(a, b).$ In the first problem you seek $q_1$ such that
$P(X > q_1) = 0.75.$ That is
$$P(X > q_1) = \int_{q_1}^b \frac{1}{b-a}\,dx = \frac{1}{b-a}\int_{q_1}^b 1\,dx = \frac{b-q_1}{b-a}.$$
So $\frac{b-q_1}{b-a} - .75$ and $q_1 = .75a + .25b.$

Second problem: Then the second problem starts with
$$P(X \le q_1)= \int_a^{q_1} \frac{1}{b-a}\,dx = .25.$$

Solve for $q_1$ in terms of $a$ and $b.$ Do you get the same answer as above?

Numerical computations and graph. Consider the specific case with $a=2, b=6.$ In R a uniform CDF is denoted punif and a quantile function (inverse CDF)
is denoted qunif. Then we have:

```<code>a = 2;  b = 6
q.1 = qunif(.25, a, b)l  q.1
[1] 3
.75*a + .25*b
[1] 3

1 - punif(q.1, a, b)
[1] 0.75     # first problem

punif(q.1, a, b)
[1] 0.25     # second problem
</code>```

```<code>curve(dunif(x, a, b), a-.2, b+.2, lwd=2, n=10001,
      ylab="Density", main="UNIF(2, 6)")
 abline(h=0, col="green2")
 abline(v=3, col="red", lwd=2, lty="dotted")
</code>```

Body:
First problem: Let $X \sim \mathsf{Unif}(a, b).$ In the first problem you seek $q_1$ such that
$P(X > q_1) = 0.75.$ That is
$$P(X > q_1) = \int_{q_1}^b \frac{1}{b-a}\,dx = \frac{1}{b-a}\int_{q_1}^b 1\,dx = \frac{b-q_1}{b-a}.$$
So $\frac{b-q_1}{b-a} - .75$ and $q_1 = .75a + .25b.$

Second problem: Then the second problem starts with
$$P(X \le q_1)= \int_a^{q_1} \frac{1}{b-a}\,dx = .25.$$

Solve for $q_1$ in terms of $a$ and $b.$ Do you get the same answer as above?

Numerical computations and graph. Consider the specific case with $a=2, b=6.$ In R a uniform CDF is denoted punif and a quantile function (inverse CDF)
is denoted qunif. Then we have:

a = 2;  b = 6
q.1 = qunif(.25, a, b)l  q.1
[1] 3
.75*a + .25*b
[1] 3

1 - punif(q.1, a, b)
[1] 0.75     # first problem

punif(q.1, a, b)
[1] 0.25     # second problem

curve(dunif(x, a, b), a-.2, b+.2, lwd=2, n=10001,
      ylab="Density", main="UNIF(2, 6)")
 abline(h=0, col="green2")
 abline(v=3, col="red", lwd=2, lty="dotted")


问:
Proving convergence of a sequence for all integers
Having difficulty understanding how to approach the following sequence.

Let $f$ be differentiable on $\mathbb {R}$ with $a := \sup\{ |f'(x)| : x\in \mathbb {R}\} <1$. Select $s_0 \in \mathbb {R}$ and let
$s_1 = f(s_0)$, $s_2 = f(s_1)$ , ..., and, in general,
$s_n = f(s_{n-1})$ for $n = 1, 2, \ldots $.

Prove that $(s_n)_{n\in \mathbb {N}}$ is a convergent sequence.

答:
By the mean value theorem, $\forall x, y \in \mathbb{R}$, $|f(x)-f(y)|\leq|x-y||f'(\theta)|\leq a|x-y|$ (where $\theta$ is some value in the open interval bounded by $x$ and $y$.) We call such an $f$ a contraction. The proof of this theorem is identical to the proof of the more general "contraction mapping theorem" in a complete metric space, the statement and proof of which can be readily found in many places. But the following gives a good outline:

$1)$ We aim to show that $s_n$ is a Cauchy sequence and so converges since $\mathbb{R}$ is complete. Let $\Delta = |s_1-s_0|$. For any $N$, for all $n,m\geq N$ with $n < m$ we have:

$$|s_n-s_m|\leq|s_n-s_{n+1}|+\dots+|s_{m-1}-s_m|$$
$$|s_n-s_m|\leq|s_N-s_{N+1}|+\dots+|s_{m-1}-s_m|=\Delta(a^N+\dots+a^m)$$
$$|s_n-s_m|\leq\Delta a^N\bigg(\sum_{i=0}^{\infty}a^j\bigg)=\frac{\Delta a^N}{1-a}$$

This final upper bound tends to $0$ as $N \rightarrow \infty$ so the sequence is Cauchy, and thus it converges.

Some related results (and easier to prove, too) which you might like to think about:

$1)$ If we call the limit of the sequence $s$, show that $f(s)=s$ ("$s$ is a fixed point of $f$")

$2)$ Show $s$ is the unique fixed point of $f$.

ContentLicense: CC BY-SA 3.0


问:
Vieta's formulae - looking for a plain English explanation of the general-case formulae
I am working with a set of equations within which arise functions of the coefficients of a polynomial expressed in terms of the roots. The expressions get rather complicated as the order of the polynomial goes above 3, and so I want to write these coefficients using Vieta's formulae:
\begin{equation}
\sum_{1\leq i_1 < i_2 \cdots i_k \leq n}\left(\prod_{j=1}^k \alpha_{i_j}\right)
\end{equation}
However, I am struggling to disentangle this notation, in particular the expression beneath the summation and the notation subscripts for $\alpha$. I know what the result should be for a polynomial of order 1, 2, 3 4 and so on, but I would appreciate a plain-language description of how the notation describes this - and any suggestions as to how the expression could be made more straightforward to interpret?

答:
This particular formula is for a particular $k$.

The summation is for every strictly increasing sequence of $k$ elements, in which each element ranges from $1$ to $n$ inclusive. The inequality $1\le i_1 < i_2 < \ldots < i_k \le n$ denotes that the sequence $(i_1, i_2, \ldots, i_k)$ should have $k$ elements, the elements are strictly increasing, and the elements are between $1$ and $n$ inclusive.

There are $\binom n k$ such sequences, so there are $\binom nk$ summands. Each summand is a product, which may be written as

$$\prod_{j=1}^k \alpha_{i_j} = \prod_{i\in\{i_1, i_2,\ldots,i_k\}}\alpha_i.$$

The product means that for each unique sequence, the sequence would choose $k$ of the $n$ roots, and take their product.

For example, $k = 4$ and $n=6$, one of the sequences is $(1,2,3,5)$, and the sequence corresponds to the product $\prod_{i\in\{1,2,3,5\}} \alpha_i = \alpha_1\alpha_2\alpha_3\alpha_5$. There are $\binom64 = 15$ such products to sum.

So I may reword the summation and product sign as

$$\sum_{S : k-\text{sequence which is strictly increasing between }1\text{ and }n} \left(\prod_{i\in S}\alpha_i\right)$$

ContentLicense: CC BY-SA 4.0


问:
5 points in 3D space: how many planes and planes intersections?
Given 5 points in space such that no three of them are colinear and no four of them are coplanar.
If we consider all the planes containing any 3 of these 5 points, and the intersections of all these planes taken two by two, how many lines will we obtain, at most?

__

At first, my idea was just to say that there are $C(3,5) = 10$ (binomial coefficient) different planes.
And then any combination of 2 planes must give a different intersection line: therefore $C(2,10)=45$ should be the answer?

However, I sense that this is naive and overly simple. Could I be missing something?

答:
You are right that there are $\binom{5}{3}= \color{red}{10}$ planes BUT ...

The $3$ planes $(1,2,3),(1,2,4),(1,2,5)$ will intersect in the line $(1,2)$.

Can you fix things from here ?


问:
Let the matrix $A=[a_{ij}]_{n×n}$ be defined by $a_{ij}=\gcd(i,j )$. How prove that $A$ is invertible, and compute $\det(A)$?
Let $A=[a_{ij}]_{n×n}$ be the matrix defined by letting $a_{ij}$ be the rational number such that $$a_{ij}=\gcd(i,j ).$$ How prove that $A$ is invertible, and compute $\det(A)$? thanks in advance </p>

答:
Let $F\colon \{1, \ldots, n\}\to \mathbb{R}$ be a function.  The $n\times n$ matrix $A=(a_{ij}) = (F(\gcd(i.j)))$ is symmetric. It it possible to give its Cholesky decomposition. Indeed, let $g\colon \{ 1, \ldots, n\}\to \mathbb{R}$ such that
$$F(m) = \sum_{d \mid m} G(d)$$
The function $g$ is uniquely determined by the equation
$$G(m) = \sum_{d\mid m} \mu(d) F(\frac{m}{d})$$F
where $\mu$ is the Mobius function. We have
$$F(\gcd(i.j)) = \sum_{d=1}^n u_{d i} u_{d j}\, G(d)$$
where $u_{d i} = 1$ if $d\mid i$ and $0$ otherwise.

We see that we have the equality
$$A = U^t \cdot D \cdot U$$
where $U$ is the upper triangular matrix $(u_{ij})$, and $D$ is the diagonal matrix $(G(1), \ldots, G(n))$.

We conclude that $\det A= G(1) \cdots G(n)$.

Note: if $F(m) = m^{\alpha}$, with $\alpha > 0$, then $G(m)$ is always positive ( easy to see using the Mobius inversion formula). We conclude that all the Hadamard positive powers of the matrix $A$ are positive definite, with determinants that can be calculated explicitly.

This method works for other partial orders on the set $\{1, \ldots, n\}$, in particular the standard order. The bordered determinants are therefore easy to calculate.


问:
How to calculate rotation of arrow to point along a vector in 3D?
I have a 3D vector Starting at some point A and ending at some point B. I have an arrow sitting at point A. I need to rotate the arrow such that it is pointing along the vector from point A to point B. The arrow starts pointing along the y axis (take a look at the image to see this better). I'm really not much of a mathematician and I'm struggling with this. Any help would be much appreciated.

答:
Let's call your first vector--the one from A to B-- $\vec x=\langle x_1,x_2,x_3\rangle$. Let us call the other vector $\vec y=\langle y_1,y_2,y_3\rangle$. We follow these steps:

Find the angle between $\vec x$ and $\vec y$. To calculate the angle between vectors, use the dot product. The dot product has the property that $\vec x\cdot\vec y=\Vert \vec x\Vert \Vert \vec y\Vert\cos\theta$, where $\theta$ is the angle between vectors. Translated to vector components, this means $$x_1y_1+x_2y_2+x_3y_3=\sqrt{(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)}\cos\theta.$$ This allows you to calculate the angle $$\theta=\cos^{-1}\left(\frac{x_1y_1+x_2y_2+x_3y_3}{\sqrt{(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)}}\right).$$

Find the direction of the axis of rotation. The axis of rotation is perpendicular to both of the vectors. To find a vector in that direction, use the cross product. The cross product is $\vec x\times\vec y=\langle x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1\rangle$. Let $z_1=x_2y_3-x_3y_2$, $z_2=x_3y_1-x_1y_3$, and $z_3=x_1y_2-x_2y_1$, with $\vec z=\vec x\times\vec y=\langle z_1,z_2,z_3\rangle.$ The next step will be easier if the vector pointing in the direction of the axis of rotation has length one. To this end, let $$\vec w=\langle\vec w_1,w_2,w_3\rangle=\left\langle \frac{z_1}{\sqrt{z_1^2+z_2^2+z_3^2}},\frac{z_2}{\sqrt{z_1^2+z_2^2+z_3^2}},\frac{z_3}{\sqrt{z_1^2+z_2^2+z_3^2}}\right\rangle.$$ $\vec w$ is also perpendicular to both $\vec x$ and $\vec y$, but has length $1$.

Use Rodrigues' Formula. Rodrigues' Formula tells us the components of the rotated vector based on $\vec w$ and $\theta$, and it uses the dot product and cross product again. It is: \begin{align*}\text{rotated vector}&=\vec y \cos\theta+(\vec w\times \vec y)\sin\theta+\vec w(\vec w\cdot\vec y)(1-\cos\theta)\\&=\langle y_1 \cos (\theta )-w_2 y_3 \sin (\theta )+w_3 y_2 \sin (\theta ),\\&w_1 y_3 \sin (\theta )-w_3 y_1 \sin (\theta )+y_2 \cos (\theta ),\\&-w_1 y_2 \sin (\theta )+w_2 y_1 \sin (\theta )+y_3 \cos (\theta )\rangle.\end{align*}

It is pretty complicated, but that is the simplest way to do it.

Edit: since $\vec w$ is perpendicular to $\vec y$, their dot product is zero, and so the result of Rodrigues' Formula is simpler. I changed it above.


问:
Find a monomorphism function $U_7\rightarrow S_5$
Find a monomorphism $f$ function $U_7\rightarrow S_5$.

I know that:

$\phi(7)=6$

$|S_5|=120$

I tried to assign permutations but I'm not sure how:

$f(1)=id$

$f(2)=(1 2)$

$f(3)=(3 4)$

$f(4)=f(2)f(2)$

$f(5)=f(3)^-1$

$f(6)=f(2)f(3)$

Thank you

答:
Any homomorphism would be determined by what it does to a generator of $U(7)$ (which happens to be cyclic). Choose an element of order six, say $(12)(345)$, to map the generator to.

The generators of $U(7)$ are $3$ and $5$.


问:
Why is learning modern algebraic geometry so complicated?
Many students - myself included - have a lot of problems in learning scheme theory. I don't think that the obstacle is the extreme abstraction of the subject, on the contrary, this is really the strong point of modern algebraic geometry. I'm reading many books such those written by Hartshorne, Gortz & Wedhorn, Liu, Vakil (notes), Gathmann (notes), Shafarevich, Perrin and Milne (notes) and in my humble opinion the learning problems arise from the following considerations:

It is enlightening to read about "the aim" of modern algebraic geometry, so I'm referring to: motivations behind schemes, the correpondence between algebraic and geometric entities (so the duality between  the category of affine schemes and the category of rings), the importance of sheaves (so the concept of admissible functions) etc. But, despite this, when one goes into the actual construction of the new objects, all theorems, lemmas and propositions are missing details (that are left to the reader). For example the verification that  certain presheaves are sheaves, functorial properties of assignments between categories and details about limit/colimit constructions are often missing. Even if the student has a solid background in algebra and geometry, generally they don't have the time or the capacity to complete all the statements. Basically taking a course in algebraic geometry implies that one must take many statements as acts of faith. I realize that authors and professors may have the same difficulties (especially lack of time) in writing down all the boring details, and moreover that a book with all proofs may include thousands of pages, but in this way students are encouraged (read discouraged) to simply memorize the most important results without really understanding the constructions. Finally, a book or a course characterized by explanations and by motivating as complete proofs is much more instructive than a book or a course which covers many advanced arguments IMHO.

In mathematics when two object are isomorphic, it is a common practise to "identify" them. Basically if $A\cong B$ but $A$ has a simple description we write $A$ instead of $B$, but formally we are thinking of $B$. This procedure is used very often in algebraic geometry, but in some cases without explaining the isomorphisms and in other cases the two objects in question are considered "really the same" even if this can provoke formal problems (look for example  here). This "abuse of identifications" often makes one lose sight of the essence of what one is studying and once again the "stupid student", exhausted, tends to simply memorize things. I point again that the problem is not the abstraction, but the fact that the excessive tendency to simplify notation, often leads to inconsistencies.

Enough importance is not given to the following: the process of successive generalizations, put in place by the great mathematicians across history, which marked the birth of modern algebraic geometry. This process is fundamental in learning because it probably represents the most natural way whereby the human mind can deal with the subject.

In summary, because of the above issues (principally the first two), rigorous mathematical statements, incredibly seem to be informal dissertations at the eyes of the student that is eager for formalism.

In your opinion, what are the most common difficulties that a student encounters during their learning process of algebraic geometry? If my problems do arise precisely from the above considerations, can you give me some advice to solve them?

答:
Postscript added much later: This answer addresses the title question but not the more specific questions that follow. (See the comments below.) The point is that a pattern that affects much of mathematical pedagogy is at work here. But people good at mathematics often manage not to be harmed by it in ways that they notice, in the form of difficulty in learning, until they hit some subjects such as this one, where the motivations often appear only some time after the definitions, proofs, etc. End of later postscript.

I think a part of the problem may be that the deductive aspects of mathematical reasoning are codified and well understood, but the inductive ones are not (indeed, one occasionally sees denials that the latter even exist (in the writings of professors of philosophy, I think)).  Thus sometimes mathematics is presented as follows: Here are the basic definitions.  (That part is done dogmaticly.)  Here's how we check that these concepts are well defined.  Now we present proofs of the following $500000000$ theorems:${}\ \cdots\cdots\cdots$.

Where would those definitions come from?  In the 19th century people observed various instances of what we now call "groups" and then formulated the concept of group, as a set with a binary operation satisfying certain laws.  Today it is considered licit to begin an account of group theory by saying: Here's the definition of a group.  From that we deduce etc.etc.etc.etc.  This doesn't mean that examples are not given; indeed a large number may be described in detail.  But the reasoning from examples to definitions is not at all treated in the same way as the reasoning found in proofs of theorems.  Nobody considers it a gap in one's logic to omit these nor to less-than-fully present the process of concept-formation.

Appendix: The term "abuse of identifications", which appears in the question above, is another thing that makes me suspect that logic has not yet advanced into certain areas.  I suspect that when it does, one will see that things now casually called abuses are correct.  I don't have a really good example of this on tap, but here's something similar.  In some contexts one defines the "density" of a set $A$ of positive integers as $\lim\limits_{n\to\infty}(|A\cap\{1,2,3,\ldots,n\}|)/n$.  So in some book somewhere I saw this definition following a definition that says "densities" are numbers in $[0,1]$.  And by present-day standards of logic, who can say that that is incorrect?  If all men were husbands, then the same standards that condone definitions like the one above would say that "man" is synonymous with "husband".


问:
Prove $\lim_{n\rightarrow \infty} n\int_0^{+\infty}{\frac{g\left( x \right)}{n^2+x^2}\text{d}x=\frac{\pi}{2}c}$
Let $g\in C[0,+\infty)$, and $\displaystyle \lim_{x\to +\infty }g(x)=c\in \mathbb R$. Prove that
$$
\lim_{n \to \infty}\left[n\int_{0}^{\infty}
\frac{g\left(x\right)}{n^{2} + x^{2}}\,\mathrm{d}x\right] =
\frac{\pi}{2}\,c
$$

I think the integral is related to the following
$$
\int_{0}^{\infty}
{\frac{c}{1 + x^{2}}\,\mathrm{d}x} = \frac{\pi}{2}\,c
$$
But I failed to find the precise relationship. Can anyone help $?$.

答:
Substitute $u=x/n$:
$$\int_0^\infty\frac{g(x)}{n^2+x^2}\,dx=\frac{1}{n^2}\int_0^\infty\frac{g(x)}{1+(x/n)^2}\,dx=\frac{1}{n}\int_0^\infty\frac{g(nu)}{1+u^2}\,du.$$
Now since $g$ is continuous on $[0,\infty)$ and has a finite limit at $\infty$, $g$ must be bounded. Therefore, by the dominated convergence theorem:
$$\lim_{n\to\infty}n\int_0^\infty\frac{g(x)}{n^2+x^2}\,dx=\lim_{n\to\infty}\int_0^\infty\frac{g(nu)}{1+u^2}\,du=\int_0^\infty\frac{1}{1+u^2}\lim_{n\to\infty}g(nu)\,du,$$
and now you can use the integral relationship that you mentioned.


问:
Second order homogenous differential equation with change of variable
I have this differential equation  $$x^2y''+3xy'+y=0$$ With x > 0

I'm told to use the change of variables $x=e^t$

How should I proceed?

Thanks

答:
HINT

\begin{align*}
x^{2}y'' + 3xy' + y = 0 & \Longleftrightarrow (x^{2}y'' +2xy') + (xy' + y) = 0\\\\
& \Longleftrightarrow (x^{2}y')' + (xy)' = 0\\\\
& \Longleftrightarrow x^{2}y' + xy = k
\end{align*}

Can you take it from here?


问:
If n and m are sums of two squares so is $\frac{n}{m}$
So I found the following post regarding this problem, but I don't find the solution to be helpful at all, so I decided to ask the question again. Furthermore, my question is a bit different than the one asked in the post shown below.

if m and n are sum of squares then so is $\frac{m}{n}$

I have to prove that if n and m are a sum of two squares and if $m\mid n$ then $\frac{n}{m}$  is also a sum of two fractions.

My attempt at a solution:
$n=a^2 + b^2 $
and
$m=x^2 + y^2$
Furthermore, the fraction $\frac{n}{m}=q$ where $q$ is the ratio between both numbers and we are trying to prove that $k$ is also a sum of two squares. I am thinking of also using the fact that  any natural number can be written in the form $n=k^2p_1...p_r$, but I am kind of confused how to do it. Can somebody give me any ideas regarding this problem. Thanks!

答:
As a hint to get you started:

Consider the special case of $\frac 1m$ with $m=a^2+b^2$.  Then we have $$\frac 1m=\frac {a^2}{m^2}+\frac {b^2}{m^2}=\left(\frac am\right)^2+\left(\frac bm\right)^2$$

Now try to adapt this to the general case.


问:
Continuity of a function from $\mathbb{R}^2$ to $\mathbb{R}^2$
I was doing some self studying and came across this question I'm having difficulties with:

Let $D = \{(x,y)\in\mathbb{R}^2: 0≤x≤1,\; y≥x\}\cup\{(x,y)\in\mathbb{R}^2:−1≤x≤1, \;y=x\}$,

$E=\{(x,y)\in\mathbb{R}^2:−1≤x≤1,\; y=x\}$.

Define $f: D\rightarrow E$ by $f(x,y) = (x,x)$. Is $f$ continuous, and can we prove if it is a closed or an open map?

I've worked with proving continuity for functions that map from $\mathbb{R}$ to $\mathbb{R}$ before, and now working in two dimensions is a bit confusing.

答:
Before presenting the proofs, we should remember that when we want to speak about continuity, we need the appropriate 'context'. This context is given by either some notion of distance (a metric), or by some available topology. When we deal with one dimension, we almost take for granted that we are using a metric, namely the modulus that appears in the epsilon-delta definition of continuity! But we are actually not restricted to just one metric or topology, and hence the property of whether or not a function is continuous is very much dependent on the 'context'. In the case of $\mathbb{R}^2$, it is quite common that we defer to the standard topology, or the euclidean metric, which is what I do here. But I think it's important to remember that these assumptions matter, and that they are just assumptions! (At least this is very helpful for me!)

Proof of continuity

To prove the function is continuous, let's first treat $D$ and $E$  as subspaces of the metric space $\mathbb{R}^2$. We will assume $\mathbb{R}^2$ has the standard euclidean metric (and this is just an assumption, if you change the metric, you may change the situation!). So now we are to prove the function $f$ between the metric spaces $(D,d_2|_D) $ and $(E, d_2|_E)$ is continuous. These metrics are simply the restriction of the euclidean metric to $D$ and $E$. First we note the definition of continuity of a point between metric spaces:

We say the function $f$ is continuous at the point $a$ in $D$ if $$\forall \epsilon > 0 \exists\delta>0 \text{ s.t. whenever }x\in D \text{ and } d(x,a) < \delta, \text{ then, } d'(f(x),f(a)) < \epsilon$$

Now we prove your given function $f$ is continuous. Suppose $a = (a_1,a_2) \in D$ and suppose $\epsilon > 0$. Now let $\delta = \frac{\epsilon}{\sqrt{2}}$. Suppose now that $x = (x_1,x_2) \in D$ and $d_2|_D (x,a) < \delta$. Then we have $$\sqrt{(a_1 - x_1)^2 + (a_2 - x_2)^2} < \frac{\epsilon}{\sqrt{2}}$$ But this means $$(a_1 - x_1)^2 + (a_2 - x_2)^2 < \frac{\epsilon ^2}{2}$$ Since $(a_2 - x_2)^2$ is necessarily positive, we have $$(a_1 - x_1)^2 < \frac{\epsilon ^2}{2}$$

and thus $$2(a_1 - x_1)^2 = (a_1 - x_1)^2 + (a_1 - x_1)^2< \epsilon^2$$ But notice that this is nothing but $$d_2|_E (f(x),f(a)) < \epsilon$$ Hence $f$ is continuous. (Here we note that the proof shows continuity because we assumed $a\in D$ to be arbitrary, and hence we have continuity everywhere in D.)

Proof that $f$ is not an open map

Now we show that $f$ is not an open map. To do this, I will assume $D$ and $E$ take the subspace topology, with respect to the standard topology on $\mathbb{R}^2$. I think it will be very helpful to draw an image of what the sets $D$ and $E$ look like to aid with visualisation.

We define the set $S = D\setminus E$, and note that it is open in the subspace topology of $D$. To see why the set $S$ is open, we note that for every point in $x \in S$, there exists an open ball $B_d(x,\epsilon)$ with $\epsilon$ sufficiently small such that $\big(D \cap B_d(x,\epsilon)\big) \subset S$ (for example, take $\epsilon$ to be any value greater than $0$ and less than the perpendicular euclidean distance to the line $y=x$ with domain $0 \leq x \leq 1$). All open balls are open in the standard topology on $\mathbb{R}^2$, and thus $D \cap B_d(x,\epsilon)$ is an open set in $D$ (by definition of the subspace topology), contained in $S$. The union of all such sets for every point in $S$ is equal to $S$ itself. Thus $S$ is open in the subspace topology $D$.

Now we consider the image of $S$. $$\begin{align} f(S) & = \{f(x,y) \in E: (x,y) \in S\} \\ & = \{(x,x) \in E: x\in [0,1] \}\end{align}$$ In other words, the set $S$ is the line $y=x$ from $0$ to $1$, inclusive. But lines are closed in $\mathbb{R}^2$, and hence $f(S)$ is closed in the subspace topology $E$ (the conclusion here makes use of this theorem). But in $\mathbb{R}^2$, the only simultaneously closed and open sets are $\mathbb{R}^2$ itself and $\emptyset$. Thus $f(S)$ is not open. Therefore the image of the open set $S$ is not open, and hence the map $f$ is not an open mapping.


问:
Does inscribing a circle, then a triangle, then ..., inside of an initial triangle telescope to some "center" of that triangle?
Start with a triangle, and construct its inscribed circle. Take the three points where the inscribed circle is tangent to the triangle, and construct a new triangle with those points as the vertices. Then construct the inscribed circle of this new triangle. Continuing this construction indefinitely gives us a sequence of inscribed circles and triangles that telescope down to a limit point. This is just the construction from this question, but we start with a triangle instead of a circle (and so no choice of points is being made).

In the comments Steven Stadnicki brought up a good question: does this construction telescope down to a defined center of the original triangle? By "defined center", I mean any of the many many points that have been given a name  like prefixcenter of a triangle. It's naturally not the incenter, since the limit point will literally be the limit point of the incenters of the sequence of nested triangles, which isn't constant. It can't be the orthocenter or circumcenter since those could easily lie outside the triangle, and I checked that it's not the centroid.

I suppose an equivalent question would be: does there exist an alternative way to construct this limit point using only finitely many steps that maybe someone has considered before?

答:
Not a complete answer, but the first inner triangle is called the intouch triangle or contact triangle of the large triangle; some more information is at Wolfram MathWorld here, where it is mentioned that the triangles in this procedure approach equilateral ones.

The Encylopedia of Triangle Centers may be useful here, though I didn't see anything relevant in the first 1000 by searching for the phrase "intouch". For what it's worth, a friend of mine knowledgeable about triangle centers didn't know of a common term, and expected that the resulting center wouldn't be very "nice" - not having barycentric coordinates expressible as a polynomial in terms of the side lengths or angles (or trig functions thereof) of the original triangle, for instance.


问:
Confusion Regarding the Quotient Set
Supposing I have a set $A$ and binary relation $R$ such that

$A = \{1,2,3,4,5\}$
$R = \{(1,1), (1,3), (1,4), (2,2), (2,5), (3,1), (3,3), (3,4), (4,1), (4,3), (4,4), (5,2), (5,5)\}$

I want to find the quotient set $A/R$, but I'm not totally sure I understood my professor's explanation of how to do so.

From what I gathered, the quotient set is the set of all equivalence classes of $A$ under the relation $R$. Then if $R$ is an equivalence relation, the equivalence class of some element $a$ belonging to $A$ is the set of all elements related to $a$. In that case, would the quotient set of $A$ and $R$ be

$$A/R = \{\{1,3,4\}, \{2,5\}\}$$

Or would I have to include binary pairs in the quotient set, i.e. $A/R = \{\{(1,1), (1,3) ... \}\}$ since the elements of $R$ are binary pairs?

答:
This is easiest to digest with a diagram:

Here $x$ points to $y$ iff $(x,y) \in R$. From this diagram you'll also see there are two disjoint such "directed graphs," if you will. Separating them out, we have

What is the significance of these, though? Essentially, all elements that are related to each other form their own equivalence class. In other words, generating graphs in this way will give you a set of (possibly disjoint) graphs, the nodes of which each represent their own equivalence class!

You can see this because an equivalence class will be all of the elements that are related to each other. You can see that you can "walk" from each of $1,3,4$ to any of the other members of that group -- likewise, you can walk from $2$ to $5$ and vice versa. This is because they're all related and thus pointing to each other. However, you can't walk from any of $1,3,4$ to $2,5$, or from the latter set to the former set. That's why they're disjoint.

Remember, a quotient set is essentially a "set of sets:" more precisely, it is the set of these equivalence classes, each of which is their own set. So you would have, for the above relation,

$$A/R = \Big\{ \{1,3,4\} \; , \; \{2,5\} \Big\}$$

Altogether, hopefully this makes it clear what a quotient set actually looks like, and an easy way to visualize them.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.

Title: Confusion Regarding the Quotient Set


问:
Does $n n^{1/n} =O(n)$?
I was asked does $n n^{1/n} =O(n)$ ?

I can see that the left hand side is always bigger than $n$ but how would you prove the equality is false?

答:
Alternatively, recall that:

$$f(n)=O(g(n)) \iff \lim_{n\to\infty}{\frac{f(n)}{g(n)}} = C \text{ , for some constant } C\ge0$$

Thus, to see why $nn^{1/n} = O(n)$, it suffices to notice that:

$$
\lim_{n\to\infty}{\frac{nn^{1/n}}{n}} = \lim_{n\to\infty}{n^{1/n}} = \lim_{n\to\infty}{e^{\ln{(n^{1/n})}}} = e^{\lim_{n\to\infty}{(\ln{n})/n}} = e^{\lim_{n\to\infty}{(1/n)/1}} = e^{0} = 1 \ge 0
$$

by using L'Hopital's rule.

Title: Does $n n^{1/n} =O(n)$?


问:
Formalization of an elementary statement in first-order logic
First, some motivation for this question. In a finite set $T$ of axioms in a first-order language $L$, we can always join the axioms by conjunction to get a 1-element set of axioms. By this method, we can get a non-redundant set of axioms of the theory generated by $T$. But this method is artificial and unsatisfactory. What we obviously want is for the axiomatization to consist of "elementary" statements. It is this notion of elementary statement that I want to make formal. Obviously, no conjunction can be elementary. But also, something like $((P \land Q) \vee (Q \land P))$ is a "cheating" way to get around this requirement of not being a conjunction. So, what exactly is the proper definition of an elementary statement, a statement that can't be broken down in a non-trivial way?

答:
I think this is a sufficiently vague notion that, while arguably natural, there isn't going to be a single "right way" to set it up. Let me describe, however, a general framework which I think may be fruitful:

The idea is to use a complexity measure on finite sets of formulas. By this I mean a map $\delta: \mathcal{P}_{fin}(Form)\rightarrow W$ for some well-founded partial order $W$, where we interpret $\delta(X)\le \delta(Y)$ as meaning "$X$ is at least as simple as $Y$." I'm being deliberately vague about exactly what I mean here, and certainly there are various basic axioms such a $\delta$ should satisfy (e.g. presumably we should have $X\subseteq Y\implies \delta(X)\le\delta(Y)$), but I don't want to get into that at the moment. I just want to give a high-level impression of what such a thing might do for us here.

For example, we could let $\delta$ be the "sum-of-lengths" function, taking values in $\mathbb{N}$ with the usual ordering. More naturally in my opinion, we could use the measure (also taking values in $\mathbb{N}$) which is implicit in arguments by "induction on formula complexity:" send atomic formulas to $0$, have Booleans and quantifiers increment complexity by $1$, and give a finite set of formulas the same complexity as its conjunction. And there are also approaches which are more context-specific. For example, if we're playing in arithmetic or set theory, the arithmetical or Levy hierarchy (together with some fixed "brute-force" prenexing algorithm) provides such a $\delta$ on individual formulas (which we'd then have to extend to finite sets of formulas) which takes values in a sort of "ladder" partial order: $\Sigma_m$ and $\Pi_m$ are incomparable (excepting that $\Sigma_0=\Pi_0$) but $\Sigma_m>\Pi_n$ and $\Pi_m>\Sigma_n$ whenever $m>n$.

In line with my above-mentioned deliberate vagueness, I don't want to get into the question of which measures are "right" in which contexts. My point in giving the examples above is simply that there are several natural notions of complexity measures floating around already, so it's not like we're introducing a new idea here.

In what follows, fix some $\delta$ which we'll agree measures complexity correctly.

There are two main ways we might measure the complexity of a given finitely axiomatizable theory $T$:

How simple can we make the formulas in an axiomatization of $T$?

How many sentences do we need to provide an axiomatization of $T$ which can't be "simplified?"

The first of these is easy to formalize - it's just asking for $\min\{\delta(A): A\equiv T\}$, or rather (since $\delta$ takes values in a partial, rather than total, order) for the set of elements of $W$ which equal $\delta(A)$ for some axiomatization $A$ of $T$ and are $W$-minimal with that property. Obviously we'd hope that in "natural" situations we'd get a single element as our answer, and this doesn't seem too unlikely. So let's consider this question more-or-less solved.

The second question is a bit more nuanced, but the following seems satisfactory to me. Say that a finite set $\Gamma$ of first-order formulas is irreducible iff for every $X\subseteq\Gamma$ and every finite set of formulas $Y$ such that $\Gamma\models Y$ and $(\Gamma\setminus X)\cup Y\models X$, we have $\delta(X)\le\delta(Y)$. That is, $\Gamma$ is irreducible if there's no way to take part of $\Gamma$ and reduce its complexity as measured by $\delta$ while keeping the same overall logical strength. Since $\delta$ takes values in a well-founded partial order, "reasonable" hypotheses on $\delta$ will imply that every finite set of formulas is equivalent to an irreducible set. This lets us talk about the essential size of a theory, namely the smallest size of any irreducible finite axiomatization of that theory.

A good test problem for both a particular $\delta$ and this idea as a whole is:

Does essential size $1$ have any particular significance?

Offhand I have no particular thoughts on the matter, but maybe someone else can shed light here.


问:
Classical Mechanics, Euler-Lagrange equation, Conservative Force.
I’m studying classical mechanics with Mathematical Methods of classical mechanics, by Arnold, doing some exercises lists and got stuck on this part.

Being . Consider a material point of
mass 1 moving under the constant conservative force  and linked to$$V=\{(x,y,z)\vert z=W(x,y)\}$$ This means that, the point is linked to the graph of the W function. Write the Lagrangian.

答:
As $F$ s conservative, then $F = -\nabla U\Rightarrow U = z + C_0$

taking $C_0=0$ and assuming unity mass, the movement kinetic energy is

$$
T = \frac 12(\dot x^2+\dot y^2+\dot z^2) = \frac 12\left(\dot x^2+\dot y^2+(W_x(x,y)\dot x+W_y(x,y)\dot y)^2\right)
$$

and the lagrangian

$$
L = T-U =\frac 12\left((1+W_x^2(x,y))\dot x^2+2W_x(x,y)W_y(x,y)\dot x\dot y+(1+W_y^2(x,y))\dot y^2\right)-W(x,y)
$$


问:
Apply product and chain rule to differentiate with respect to one variable
I have the following:

$$
u = \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}} \phi\left(\frac{x}{t}\right)
$$

I need to find $u_t$
This equation was found by solving the characteristic equation:
$$
\frac{dx}{xt} = \frac{dt}{t^2} = \frac{du}{-\left(\frac{1}{4}x^2 + \frac{1}{2}t\right)u}
$$

solving the equation from the first two yields:
$$
r = \frac{x}{t}
$$

solving from the second two yields:
$$
v = u\sqrt{t}e^{\frac{x^2}{4t}}
$$

we let $v = \phi(r)$
this gives us:
$$
u\sqrt{t}e^{\frac{x^2}{4t}} = \phi\left(\frac{x}{t}\right)
$$

that is where the equation for $u$ comes from.

Here is what I have:

\begin{align}
u_t & = \frac{-1}{2t^{\frac{3}{2}}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) + \frac{x^2}{4t^2}\frac{1}{\sqrt{t}}e^\frac{-x^2}{4t}\phi\left(\frac{x}{t}\right) + \left( \frac{-x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\
&  = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \left( \frac{x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\
\end{align}

However the solution given has:

$$
u_t = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \phi'\left( \frac{x}{t} \right) \left(\frac{xu}{t^2}\right)
$$

I can't figure out what I am doing wrong before I look at $u_{xx}$ (I already tried that and again, am making the same mistake).

答:
Your mistake is in the second term. It should be
$$\frac{x^2}{4t^{2}}\frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) = \frac{x^2}{4t^{2}}u$$
You just forgot the chain rule.


问:
What is the probability $P(Z<X+Y)$ given a function of $3$ variables?
What is the probability $P(Z<X+Y)$ given $f(x,y,z)=24xy^2z^3$ for $x,y,z\in(0,1)$?

I can't find the bounds of the triple integral. I keep getting the answer of $1/18$ whereas the book says it's $1/14$. My bounds for the innermost variable ($z$) are $0$ to $x+y$, for $y$: $0$ to $1-x$, and for $x$: $0$ to $1$.

I can't see how to find the region that describes $Z<X+Y$. Please help.

答: When you say $0\leq y\leq 1-x$, you're making the condition that $X+Y\leq1$, but that's not required.  For example, we could have $X=Y=\frac34$ and and value $0\leq Z\leq1$ would work.  In short, in addition to the integral you computed, you need $$\int_0^1\int_{1-x}^1\int_0^1f(x,y,z)dzdydx$$


问:
Construction of a $C^1$ function, which is (locally) larger than a given continous function

Let $u: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continous function which is differentiable at the point $x_o \in \mathbb{R^n}$. Find a function $v \in C^1(\mathbb{R^n})$ such that $v(x_o)=u(x_o)$, $\nabla v(x_o)=\nabla u(x_o)$ and $g=u-v$ has a strict (local) maximum at $x_o$.

I am not aware of any extension theorems I could apply to this problem

Would appreciate any help/hints.

答:
Suppose for simplicity $n=1,x_0=0$ and $u(0)=0=u'(0).$

Let $f$ be a $C^1$ function such that $f=0,x\le 0,$ $f=1,1\le x<\infty,$ and $f$ is strictly increasing on $[0,1].$ Let $D=\sup_{\mathbb R}f'.$

Let $a_k= 2^{-k},k=1,2,\dots.$ Set $m_k = \max_{[-a_k,a_k]}|u|. $ Then by continuity of $u,$ $m_1\ge m_2\ge \dots \to 0.$ We can assume no $m_k=0.$

I'll first build $v$ to the right of $0;$ we can then define $v$ to be even. Let $I_k = [a_{k+1},a_k).$ Define $f_1 =m_1\chi_{I_1}.$ For $k>1,$ define

$$f_k(x) =m_k +(m_{k-1}-m_k)f\left(\frac{x-a_{k+1}}{a_k-a_{k+1}}\right)\chi_{I_k}(x).$$

The idea is that $\sum_{k=1}^{\infty}f_k$ is a nice $C^1$ function on $[0,\infty)$ that bounds $u$ from above. At this point I hope it's clear that the $f_k$ fit smoothly together so that the sum of them is $C^1$ on $(0,\infty).$

Note that for $x\in I_k,$

$$\tag 1 f_k'(x) = (m_{k-1}-m_k)f'\left(\frac{x-a_{k+1}}{a_k-a_{k+1}}\right)\cdot \frac{1}{a_k-a_{k+1}}  \le m_{k-1}D2^{k+1}.$$

Now $m_{k-1}=u(c_{k-1})$ for some $c_{k-1}\in [-a_{k-1},a_{k-1}].$ Thus the right side of $(1)$ equals

$$|u(c_{k-1})|D2^{k+1} = \left|\frac{u(c_{k-1})}{c_{k-1}}\right|D|c_{k-1}|2^{k+1}$$ $$\le \left|\frac{u(c_{k-1})}{c_{k-1}}\right|D|c_{k-1}|2^{k+1} \le \left|\frac{u(c_{k-1})}{c_{k-1}}\right|D\cdot 4.$$

As $x\to 0^+$ the last expression above $\to 0.$ That's because $u'(0)=0.$ This imples $g=\sum_{k=1}^{\infty}f_k$ belongs to $C^1[0,\infty)$ as desired.

To finish, define $g$ on all of $\mathbb R$ by making it even. Finally, let

$$v(x) = x^2 + g(x).$$

Then $v>u$ in a deleted nighborhood of $0$ and $v(0)=0=v'(0).$ We are done in the case we started with. The higher dimensional case should follow easily from this.


问:
Factored form of a quadratic
Is there a way to formulate a factored form formula (not the roots) for $ax^2+bx+c$ in terms of $a, b,$ and $c \,?$

Starting from the AC method of factoring, we derive these two equations:

$$\frac{ac}m=n \quad,\quad  m+n=b.$$

I solved the systems by substituting $\frac{ac}{m}$ into $n.$

$$m^2 - bm + ac = 0.$$

After completing the square, I found $m, n$ , which equals:

$$ \frac{b±\sqrt{b^2 -4ac}}2$$

So now $ax^2+bx+c$ is equal to
$$ax^2 + \frac{b+\sqrt{b^2 -4ac}}{2}x + \frac{b-\sqrt{b^2 -4ac}}2x + c$$

This is where I got stuck. My thinking led me to believe that I needed to derive a formula for the greatest common integer to factor out any divisor that is divisible by $a,b,c.$ Any thoughts?

答:
Here is an alternate approach based on attempting to find a perfect square "hidden" in the expression
$$ ax^2+bx+c $$

In order to have a perfect square in the expression, we would like the first term to be a perfect square. So we begin by multiplying and dividing the expression by $4a$. [There is a reason that $4$ works best and I will give a hint: try multiplying by $ak^2$ and try to see why $k$ should be equal to $2$.]

$$ \frac{1}{4a}\cdot(4a^2x^2+4abx+4ac) $$

Now notice that the square of $2ax+b$ equals $4a^2x^2+4abx+b^2$. So we can re-write the expression as

\begin{eqnarray}
\frac{1}{4a}\cdot(4a^2x^2+4abx+b^2 -b^2+4ac)&=&\frac{1}{4a}\cdot[(2ax+b)^2-(b^2-4ac)]
\end{eqnarray}

Now the second expression can be written as the difference of two squares:

$$ \frac{1}{4a}\cdot{\large[}(2ax+b)^2-(\sqrt{b^2-4ac})^2{\large]}$$

This can be factored into the product of a sum and difference to give

$$ \frac{1}{4a}\cdot{\large[}(2ax+b)-\sqrt{b^2-4ac}{\large]}\cdot{\large[}(2ax+b)+\sqrt{b^2-4ac}{\large]} $$

This is one way to factor $ax^2+bx+c$. If you set it equal to $0$ and solve for $x$ you will quickly arrive at the quadratic formula.

Title: Factored form of a quadratic


问:
can $(x^3 , y)$ be a chart if $(x , y)$ is a chart on $M$?
Let $M$ be a $2$-dimensional manifold and $(U , \varphi)$ be a chart on $M$ with $\varphi = (x , y)$. Can $(U , \psi)$, with $\psi = (x^3 , y)$, be a chart on $M$? I thought no because if it were a chart, the map $\psi = g \circ \varphi : U \to {\mathbb{R}}^2$ should be a diffeomorphism, with $g : {\mathbb{R}}^2 \to {\mathbb{R}}^2$ given by $g(a , b) = (a^3 , b)$. But it implies that $g$ is also a diffeomorphism, but the first partial derivative of $g^{- 1} : (a , b) \mapsto (a^{\frac13} , b)$ does not exist.

My argument should be wrong, as on can be read on $\textbf{Lee's Smooth Manifold}$ page 65, that $(\tilde{x},\tilde{y})$ in $\mathbb{R}^2$ related by
$$
\tilde{x} = x, \qquad \tilde{y} = y+x^3
$$
is a chart if $(x , y)$ denotes the standard coordinates on ${\mathbb{R}}^2$. Where is my mistake?

答:
These things do not contradict each other. In your example, the transition maps for $\phi$ and $\psi$ are $g$ and $g^{-1}$, and $g^{-1}$ is not smooth, as you observed.

But there is no such problem for the other example. The forward transition map is $g: (x,y)\to (x, y+x^3)$ and the inverse is $g^{-1}:(\tilde{x}, \tilde{y})\to (\tilde{x}, \tilde{y}-\tilde{x}^3)$. Both are perfectly nice smooth maps.


问:
Suppose $f(z)$ is an entire function that is not a polynomial. What kind of singularity can $f(z)$ have at $\infty$?
I know that if $f(z)$ is a nonconstant entire function, then $e^{f(z)}$ has an essential singularity at $z = \infty$,

Now suppose $f(z)$ is an entire function that is not a polynomial. What kind of singularity can $f(z)$ have at $\infty$?

My attempt is since $f(z)$ is entire,
$f(z) = \sum_{i=0}^{\infty} a_nz^n$ around $0$,
since $f$ is not polynomial, the series has infinite terms.
Then $f(\frac{1}{w})= \sum_{i=0}^{\infty} a_n\frac{1}{w^n}$. But I don't know how to continue from there

答:
As you observed, we can classify the singularity of $f$ at $\infty$ by classifying the singularity of $g(z)=f(1/z)$ at $0$. We know that if $g(z)$ has a removable singularity at $0$, then $f$ has a constant limit at $\infty$, meaning that $f$ is bounded, and therefore it is constant, so that cannot happen. Now suppose $g$ has a pole at the origin, then there is some $n$ such that $z^n g(z)$ is holomorphic in a neighborhood of the origin, so $$g(z)=\frac{a_{n}}{z^n}+\cdots+\frac{a_{1}}{z}+a_0.$$ But then $f(z)=a_0+a_1z+\cdots+z_nz^n$, a polynomial, so this cannot happen either. Therefore, the singularity must be essential.

This leads to a useful characterization of an essential singularity: a singularity at $z_0$ is essential iff the Laurent series about $z_0$ has infinitely many negative powers. The singularity of an entire function at $\infty$ is essential iff the Laurent series has infinitely many positive powers.


问:
$\prod_{n=1}^{\infty} (1- z/ a_n) $ is entire iff $\sum_{n=1}^{\infty} 1/(z-a_n) $ is meromorphic
This question was asked in a masters exam previous year paper and I was unable to prove it.

Show that the $$\prod_{n=1}^{\infty} (1- \frac{z}{a_{n}}) $$ is entire iff $$\sum_{n=1}^{\infty} \frac{1}{z-a_n} $$ is meromorphic.

$\prod_{n=1}^{\infty}( 1- \frac{z}{a_{n}})$ is entire if Convergence of infinite product is uniform and that happens if: $$\sum_{n=1}^{\infty} \frac{z}{a_{n}}$$ converges for all z. But I am not able to correlate it with:  $$\sum_{n=1}^{\infty} \frac{1}{z-a_n} $$.

Can you please help by telling which result should be helpful.

Thank you!

答:
You probably meant if $a_n\ne 0,f_N(z)=\sum_{n\le N} \frac1{z-a_n}$ and $\lim_{N\to \infty}f_N(z)$ converges locally uniformly away from the $a_n$ to a meromorphic function, then $\prod_n (1-z/a_n)$ is entire.

In which case it is obvious:

$\lim_{N\to \infty}\exp(\int_0^z f_N(s)ds)$ converges locally uniformly to an analytic function away from the $a_n$.

Around the $a_n$ we use the maximum modulus principle, uniform convergence on $|z-a_n|=r$ implies uniform convergence on $|z-a_n|\le r$.


问:
Non-Noetherian ring with a single prime ideal

What are the most simple examples of a commutative ring $R$ satisfying both of the following two properties:

  1. $R$ is not Noetherian.

  2. $R$ has exactly one prime ideal.

答: If you're willing to ask for one nonzero prime ideal, then for any prime $p$ the ring of integers of ${\mathbf C}_p$ (the $p$-adic complex numbers) is an example. The only nonzero prime ideal in that ring is its maximal ideal, which is not finitely generated.</p>


问:
Is $\{a + bi: a, b \in \mathbb{F}_3\}$ a field?
Is $P = \{a + bi: a, b \in \mathbb{F}_3\}$ a field?  This is the question I am having.

First, I list the elements in $\mathbb{F}_3$, which consists of $\overline{0}$, $\overline{1}$ and $\overline{2}$ modulus 3.  Then, I form the quick idea whether that set is a field or not.

Suppose that $P$ is a field.

We need to show that $(P, +)$ is an abelian group.  Clearly, the addition of any elements, belonging to $(P, +)$, is always associative and also commutative.  The identity element is $\overline{0} \in (P, +)$.  The additive inverse of the element, say $a + bi \in (P,+)$ is $-a - bi$.  Realize that $-a$ and $-b$ modulo 3 are congruent to some elements in $\mathbb{F}_3$, which implies that $-a - bi \in (P, +)$.  So elements in $(P,+)$ have their additive inverses, in which $c_0 + (-c_0) \equiv 0$ for any $c_0$ in the form $\{a + bi : a,b \in \mathbb{F}_3\}$.

We then need to prove that $(P \backslash \{\overline{0}\}, \times)$ is also an abelian group.  The product of the elements is also commutative and associative.  $\overline{1} = 1 + 0i$ is the identity element belonging to $(P \backslash \{\overline{0}\},\times)$.  Next, we need to show that each element in $(P \backslash \{\overline{0}\}, \times)$ has the multiplicative inverse.  That is: If $p \in (P \backslash \{\overline{0}\}, \times)$, then $p^{-1} \in (P \backslash \{\overline{0}\}, \times)$ also.  If this happens, then this means that the product of nonzero elements $pp^{-1} = 1$, where $p$ and $p^{-1}$ are in the form $a + bi$ for $a,b \in \mathbb{F}_3$.  Observe that if $a + bi$ has its multiplicative inverse in $(P \backslash \{0\}, \times)$, then the multiplicative inverse is

$$\dfrac{1}{a + bi} \cdot \dfrac{a - bi}{a - bi} = \dfrac{a - bi}{a^2 + b^2} = \dfrac{a}{a^2 + b^2} - \dfrac{b}{a^2 + b^2}i$$

where $a, b, a^2 + b^2 \text{ ($a^2 + b^2$ being nonzero)} \in \mathbb{F}_3$, and $a$ and $b$ are both not $\overline{0}$}

If we check the inverse of each element $a + bi \in (P \backslash \{\overline{0}\}, \times)$, then each element has its multiplicative inverse.  This means that $a^2 + b^2$ is never $\overline{0}$.    Thus, $(P \backslash \{\overline{0}\}, \times)$ is an abelian group.

Clearly, $a(b + c) = ab + ac$ holds for any $a,b,c \in P$ since multiplication and addition are inherited from $(P, +)$ and $(P, \times)$.

Thus, $P$ is a field.

Any advices or comments?

答: I think there's a far quicker way of doing this. Consider $F_{3}[x]$, the ring of polynomials over $F_{3}$. Notice that $x^{2}+1$ has no roots in $F_{3}$ and therefore, since its degree is 2, it is irreducible in $F_{3}[x]$. It is quite easy to show that this implies that $A:=F_{3}[x] /(x^{2}+1)$ is a field (you can do this by hands, finding an explicit inverse for each non zero element of that quotient, or arguing about maximality of ($x^{2}+1$)), where $(x^{2}+1)$ is the ideal generated by $x^{2}+1$ in $F_{3}[x]$. Now it should not be difficult to show that the linear map $A\longrightarrow P$ which sends $[a]$ to $a$ for $a\in F_{3}$ and $[x]$ to $i$ is an isomorphism (of rings). Since $A$ is a field, you get by this that also $P$ is such.</p>


问:
Convergence in Probability of $ Z = \frac{\sqrt{n}(\hat{p}-0.2)}{\sqrt{0.16}} $
I have the following quantity
$$
Z = \frac{\sqrt{n}(\hat{p}-0.2)}{\sqrt{0.16}}.
$$
I have already proven by the WLLN that $\hat{p} \stackrel{P}{\longrightarrow} p = 0.2$. Given that, can I claim that $Z \stackrel{P}{\longrightarrow} 0$? What is troubling me is that, in any limiting process, this $\sqrt{n}$ would blow up. So even though ($\hat{p} - 0.2$) converges to $0$, I'm afraid the $\sqrt{n}$ would "ruin" the converge.

For some context, we have that $\hat{p} = (1/n)\sum_{i=1}^{n}X_i = \bar{X}_n,$ where $X_i \sim Bernoulli(p)$. We can assume that the real value of $p$ is $p=0.2$

答:
This is central limit theorem. In general, for i.i.d. random variables with finite second moment, we have
$$
\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\longrightarrow Z\sim N(0,1).
$$
In this case, each $X$ is a Bernoulli r.v. so $\mu=p$ and $\sigma=\sqrt{p(1-p)}$.

It's really in the form
$$
Z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}
$$


答:
Let $f(x) = x^4 + 2 x^3 - 6 x^2 - 6 x + 25$. We have to find a perfect square $(x^2+ax+b)^2$ that is less or equal to $f(x)$, but greater than $0$ for all $x \in \mathbb R$.

Comparing coefficients, $2a x^3 = 2 \Rightarrow a = 1$, and for safety, $(2b + a^2) x^2 = -7 \Rightarrow b = -4$.

Expanding gives $g(x) = (x^2+x-4)^2 = x^4 + 2x^3 - 7x^2 - 8x + 16$. We have to be careful about the $-8x$ term as it increases the value of $g(x)$ when $x$ is negative. However, we can find a suitable $c$, such that $f(x) - g(x) = x^2+2x+(9 - c) > 0$. For example, $c = 7$ works as $x^2+2x+2$ $ = (x+1)^2 + 1 > 0$, since $u^2 ≥ 0, u \in \mathbb R$.

Thus $f(x) ≥ (x^2+2x-4)^2 + 7 > 0$.

Now just verify that when $x = 0, x^2+x-5=0$, the function is greater than $0$. Hence there are no real roots, and $4$ complex roots by the fundamental theorem of algebra.

Title: how how can I work ? $x^4 + 2 x^3 - 6 x^2 - 6 x + 25=0$

Body:
Solve $\frac{x^2+x-5}{x}+\frac{3x+4}{x^2+x-5}=0 $

we know $x(x^2+x-5)\neq 0$ so we have $(x^2+x-5)^2+3x^2+4x=0 \to x^4 + 2 x^3 - 6 x^2 - 6 x + 25=0$
now how how can I work ?


问:
p-adic cyclotomic character on Frobenius at p
Let $p$ be a prime. From what I understand, the p-adic cyclotomic character $\varepsilon:Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to \mathbb{Z}_p^\times$ is given by choosing a compatible sequence of $p$-power roots of unity $(\zeta_{p^n})_n$ with $\zeta_{p^n}^p=\zeta_{p^{n-1}}$ in $\overline{\mathbb{Q}}$ then for $\sigma$, $$\sigma(\zeta_{p^n}) = \zeta_{p^n}^{\varepsilon(\sigma) \bmod p^n}$$

If we take $F=Frob_p$ - an arithmetic Frobenius element of at $p$, i.e an element of the decomposition group $Gal(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ at p such that  $F(\zeta_{p^n}) = \zeta_{p^n}^p \bmod p$, so what is  $\varepsilon(F) \bmod p$?

More generally, can anyone explain in detail the underlined sentence in the following paragraph (taken from Breuil's ICM 2010 paper, at the end of the intro where he sets the basic notations/conventions)

答:
There are many possible Frobenius elements (and decomposition groups). There is a canonical one on $\Bbb{Q}(\zeta_\infty)$ which is $\phi(\zeta_{p^n-1}) =\zeta_{p^n-1}^p,\phi( \zeta_{p^n})= \zeta_{p^n}$ which gives $\varepsilon(\phi)=1$ (then extend $\phi$ to $\overline{\Bbb{Q}}$ as you wish). The others are of the form $\sigma \phi$ for some $\sigma\in Gal(\overline{\Bbb{Q}}/\Bbb{Q}(\zeta_{p^n-1}))$  and $\varepsilon(\sigma\phi)=\varepsilon(\sigma)$ doesn't have to be $1$.

About the sentence underlined in red:

The previous $\phi$ (the one on $\Bbb{Q}(\zeta_\infty)$) extends naturally to $\Bbb{Q}_p(\zeta_\infty)$.

$Gal(\Bbb{Q}_p^{ab}/\Bbb{Q}_p)$ is non-trivial because of the non-cyclotomic wildly ramified abelian extensions (linked to abelian formal group laws linked to elliptic curves), in contrary to $Gal(\Bbb{Q}_p(\zeta_\infty)/\Bbb{Q}_p)$ which is $=Gal(\Bbb{Q}_p(\zeta_\infty)/\Bbb{Q}_p(\zeta_{p^\infty-1}))\times Gal(\Bbb{Q}_p(\zeta_\infty)/\Bbb{Q}_p(\zeta_{p^\infty})) \cong \Bbb{Z}_p^\times\times \phi^{\hat{\Bbb{Z}}} $ then $\Bbb{Z}_p^\times\times \phi^{\Bbb{Z}}$ is dense and this latter group is isomorphic to $\Bbb{Q}_p^\times$, they are saying a canonical choice is to send $(a, \phi^n)$ to  to $a p^{-n}$. Your character is sending  $Gal(\Bbb{Q}_p^{ab}/\Bbb{Q}_p)\to Gal(\Bbb{Q}_p(\zeta_\infty)/\Bbb{Q}_p)\to \Bbb{Z}_p^\times\times \phi^{\hat{\Bbb{Z}}} \to \Bbb{Z}_p^\times $.


问:
Maclaurin series is the geometric series, question
I have a question on the geometric series being represented by the Maclaurin series.

Wiki defines the Taylor series as:
$\sum_{n=0}^\infty \frac {f^{(n)}(a)} {n!}(x-a)^n$

where $a = 0$ is the Maclaurin series.

Wiki then states "The Maclaurin series for $\frac 1 {1-x}$ is the geometric series $1 + x + x^2 + ...$

My first question is, what is $f(a)$ for the Taylor Series that results in this MacLaurin series?

Wiki then goes into the Taylor series for $\frac 1 x$ at $a = 1$ is $1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + ...$

Does anyone know how this is derived?

答:
To start with, let us consider the geometric sequence whose first term equals one with ratio $|x| < 1$.

The we can express the sum of its first $n+1$ terms by
\begin{align*}
s_{n}(x) = 1 + x + x^{2} + \ldots + x^{n} & \Rightarrow xs_{n}(x) = x + x^{2} + x^{3} + \ldots + x^{n+1}\\\\
& \Rightarrow (1-x)s_{n}(x) = 1 - x^{n+1}\\\\
& \Rightarrow s_{n}(x) = \frac{1 - x^{n+1}}{1-x}
\end{align*}

Since $x^{n}\to 0$ for $|x| < 1$ as $n$ approaches infinity, one concludes that
\begin{align*}
\lim_{n\to\infty}s_{n}(x) = \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x} = \frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \ldots
\end{align*}

Consequently, the desired function is given by $f(x) = \dfrac{1}{1-x}$.

Based on such results, it is possible to give an answer to your second question as well.

More precisely, if $|x - 1| < 1$, we can conclude that

\begin{align*}
\frac{1}{x} = \frac{1}{1 - (1-x)} = 1 + (1 - x) + (1-x)^{2} + (1-x)^{3} + \ldots
\end{align*}

Hopefully this helps!


Title: Maclaurin series is the geometric series, question

Body:
I have a question on the geometric series being represented by the Maclaurin series.

Wiki defines the Taylor series as:
$\sum_{n=0}^\infty \frac {f^{(n)}(a)} {n!}(x-a)^n$

where $a = 0$ is the Maclaurin series.

Wiki then states "The Maclaurin series for $\frac 1 {1-x}$ is the geometric series $1 + x + x^2 + ...$

My first question is, what is $f(a)$ for the Taylor Series that results in this MacLaurin series?

Wiki then goes into the Taylor series for $\frac 1 x$ at $a = 1$ is $1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + ...$

Does anyone know how this is derived?

Body:
If we take the function $f(x) = 1/(1-x)$ for $|x|<1$, then
$f^{(n)}(x) = n! /(1-x)^n$ for all $n$ [by induction], so
$f^{(n)}(0) = n!$ for all $n$, so the Maclaurin series is
$$
\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n
\sum_{n=0}^\infty \frac{n!}{n!}(x-0)^n= \sum_{n=0}^\infty x^n.
$$

If we take $f(x) = \frac{1}{x}$ for $x>0$, then $f^{(n)}(x)
= \frac{(-1)^n n!}{x^{n+1}}$ and $f^{(n)}(1) = (-1)^n$.
The Taylor series at $a=1$ is
$$
\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n =
\sum_{n=0}^\infty \frac{(-1)^n n!}{n!}(x-1)^n = \sum_{n=0}^\infty (-1)^n(x-1)^n .
$$


问:
Question about the proof of Theorem D.5, Introduction to Smooth Manifolds by Lee
I am trying to understand the proof of the following theorem from Lee's "Introduction to Smooth Manifolds":

The following statement is in the proof:

Question: What I don't see is how $\frac{2\epsilon e^{CT}}{E}\cdot(e^{ET}-1)$ can be made as small as desired by choosing $h$ and $\tilde{h}$ sufficiently small. How does the expression depend on $h$ and $\tilde{h}$? To me it looks like an upper bound.

The only idea I have is the following:

There is the following statement

Let $k\in\mathbb{N}$ such that $k>\frac{2e^{CT}}{E}\cdot(e^{ET}-1)$ and $\tilde{\epsilon}:=\frac{\epsilon}{k}$,  and $\epsilon$ as above.
By the above statement I can find a $\tilde{\delta}$ such that $|y_{1}-y_{2}|<\tilde{\delta}$ implies
\begin{align}
|\frac{\partial V^{i}}{\partial y^{k}}(y_{1})-\frac{\partial V^{i}}{\partial y^{k}}(y_{2})|<\tilde{\epsilon}=\frac{\epsilon}{k}
\end{align}

Now the book says

I would instead say "Suppose that $h$ and $\tilde{h}$ are both less than $\frac{\tilde{\delta} e^{-CT}}{n}$".

From here on I would proceed exactly as in the book to end up with the inequality
\begin{align}
|\Delta_{h}(t,x)-\Delta_{\tilde{h}}(t,x)|\leq...\leq\frac{2\tilde{\epsilon} e^{CT}}{E}\cdot(e^{ET}-1)=\frac{2\epsilon e^{CT}}{k\cdot E}\cdot(e^{ET}-1)<\epsilon
\end{align}

Does the statement "choosing $h$ and $\tilde{h}$ sufficiently small  " refer to the line "$h$ and $\tilde{h}$ are both less than $\frac{\delta e^{-CT}}{n}$" ?

I hope this makes any sense. Thank you very much in advance!

答:
I'm not surprised that you're perplexed by this -- it was very poorly stated. The sentence

Since the expression on the right can be made as small as desired by choosing $h$ and
$\tilde  h$ sufficiently small, this shows ...

should be replaced by

Thus the expression on the left can be made as small as desired by choosing $h$ and $\tilde h$
sufficiently small. This shows ...

And you're right, the "sufficiently small" just refers to the stipulation that $|h|$ and $|\tilde h|$ are both less than $\delta e^{-CT}/n$.

I've added a correction to my online list.


问:
Proving convergence of a sequence for all integers
Having difficulty understanding how to approach the following sequence.

Let $f$ be differentiable on $\mathbb {R}$ with $a := \sup\{ |f'(x)| : x\in \mathbb {R}\} <1$. Select $s_0 \in \mathbb {R}$ and let
$s_1 = f(s_0)$, $s_2 = f(s_1)$ , ..., and, in general,
$s_n = f(s_{n-1})$ for $n = 1, 2, \ldots $.

Prove that $(s_n)_{n\in \mathbb {N}}$ is a convergent sequence.

答:
From the mean value theorem, $\forall x,y\in\mathbb R$ we have that there exists $\xi$ between $x$ and $y$ such that
$$|f(x)-f(y)|=|f'(\xi)|\cdot |x-y|\leq a|x-y|\Rightarrow$$
$$|f(x)-f(y)|\leq a|x-y|,\forall x,y\in\mathbb R$$

Now take $$|s_{n+1}-s_n|=|f(s_n)-f(s_{n-1})|\leq a|s_n-s_{n-1}|=$$
$$a|f(s_{n-1})-f(s_{n-2})|\leq a^2|s_{n-1}-s_{n-2}|=\dots$$
$$\dots \leq a^n|s_1-s_0|$$
Therefore, $$|s_{n+p}-s_n|=|s_{n+p}-s_{n+p-1}+s_{n+p-1}-\dots-s_{n+1}+s_{n+1}-s_n|\leq$$
$$|s_{n+p}-s_{n+p-1}|+\dots +|s_{n+1}-s_n|\leq$$
$$|s_1-s_0|\cdot \sum \limits_{k=n}^{n+p-1} a^k\leq $$
$$|s_1-s_0|\sum\limits_{k=n}^\infty a^k<|s_1-s_0|\cdot\varepsilon$$
Where the last inequality holds, since the sum converges($a<1$), and therefore its "tail" tends to zero.

The above relation proves that $\{s_n\}$ is a Cauchy sequence, and therefore converges.


问:
Multivariable calculus - partial derivative
I got this question:

Let $z=\frac{2x}{y}+\ln{(xy)}$ where $x$ and $y$ are functions of $s$ and $t$ defined as $s=x-y^2$ and $t=x^2+y$.
Find $\frac{\partial z}{\partial s}$ when $x=1$ and $y=1$.

Does this question make sense? To me it seems quite ill-posed, I can't figure out how to apply the usual chain rule (is it even possible to do that?)

答:
I'm going to take this opportunity to introduce a non$-$conventional notation for partial derivatives I learned while tutoring a student attending MIT. This notation may clear up some of the confusion you're having when the variables are all jumbled up like they are here.

You're looking to compute $\Big(\frac{\partial z}{\partial s}\Big)_t$ which in words represents "the partial derivative of $z$ with respect to $s$ while keeping $t$ fixed." From chain rule we have $$\Big(\frac{\partial z}{\partial s}\Big)_t=z_x\Big(\frac{\partial x}{\partial s}\Big)_t+z_y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$\Big(\frac{\partial s}{\partial s}\Big)_t=\Big(\frac{\partial x}{\partial s}\Big)_t-2y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$\Big(\frac{\partial t}{\partial s}\Big)_t=2x\Big(\frac{\partial x}{\partial s}\Big)_t+\Big(\frac{\partial y}{\partial s}\Big)_t$$ Since $\Big(\frac{\partial s}{\partial s}\Big)_t=1$ and $\Big(\frac{\partial t}{\partial s}\Big)_t=0$ we have that $$\Big(\frac{\partial z}{\partial s}\Big)_t=z_x\Big(\frac{\partial x}{\partial s}\Big)_t+z_y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$1=\Big(\frac{\partial x}{\partial s}\Big)_t-2y\Big(\frac{\partial y}{\partial s}\Big)_t$$ $$0=2x\Big(\frac{\partial x}{\partial s}\Big)_t+\Big(\frac{\partial y}{\partial s}\Big)_t$$ As I said in my comments, the last two equations provide us with a system which we can solve for $\Big(\frac{\partial x}{\partial s}\Big)_t$ and $\Big(\frac{\partial y}{\partial s}\Big)_t$ explicitly. Doing so gives us an expression for $\Big(\frac{\partial z}{\partial s}\Big)_t$: $$\Big(\frac{\partial z}{\partial s}\Big)_t=\frac{z_x-2xz_y}{4xy+1}$$ You might ask: How would we compute something like $\Big(\frac{\partial z}{\partial s}\Big)_x$? In words this represents "the partial derivative of $z$ with respect to $s$ while keeping $x$ fixed." Chain rule gives us a similar set of equations to work with: $$\Big(\frac{\partial z}{\partial s}\Big)_x=z_x\Big(\frac{\partial x}{\partial s}\Big)_x+z_y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial s}{\partial s}\Big)_x=\Big(\frac{\partial x}{\partial s}\Big)_x-2y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial t}{\partial s}\Big)_x=2x\Big(\frac{\partial x}{\partial s}\Big)_x+\Big(\frac{\partial y}{\partial s}\Big)_x$$ Because $\Big(\frac{\partial s}{\partial s}\Big)_x=1$ and $\Big(\frac{\partial x}{\partial s}\Big)_x=0$ the three equations above boil down to $$\Big(\frac{\partial z}{\partial s}\Big)_x=z_y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$1=-2y\Big(\frac{\partial y}{\partial s}\Big)_x$$ $$\Big(\frac{\partial t}{\partial s}\Big)_x=\Big(\frac{\partial y}{\partial s}\Big)_x$$ Combining the first two give $\Big(\frac{\partial z}{\partial s}\Big)_x=-\frac{z_y}{2y}$. For practice, see if you can compute $\Big(\frac{\partial z}{\partial y}\Big)_t$ and $\Big(\frac{\partial z}{\partial t}\Big)_x$. This notation does take some getting used to, but if you can get past its ugliness, I guarantee you will find it helpful.


问:
Prove $\lim_{n\rightarrow \infty} n\int_0^{+\infty}{\frac{g\left( x \right)}{n^2+x^2}\text{d}x=\frac{\pi}{2}c}$
Let $g\in C[0,+\infty)$, and $\displaystyle \lim_{x\to +\infty }g(x)=c\in \mathbb R$. Prove that
$$
\lim_{n \to \infty}\left[n\int_{0}^{\infty}
\frac{g\left(x\right)}{n^{2} + x^{2}}\,\mathrm{d}x\right] =
\frac{\pi}{2}\,c
$$

I think the integral is related to the following
$$
\int_{0}^{\infty}
{\frac{c}{1 + x^{2}}\,\mathrm{d}x} = \frac{\pi}{2}\,c
$$
But I failed to find the precise relationship. Can anyone help $?$.

答:
Here's a proof that doesn't use the DCT:

Consider the quantity
$$Q=\Bigg|n\int_0^{\infty}\frac{g(x)}{n^2+x^2}dx-\frac{\pi c}{2}\Bigg|$$

This can be cleverly rewritten as follows, after substituting $x=nt$ and $\frac{\pi}{2}=\int_{0}^{\infty}\frac{dt}{1+t^2}$

$$Q=\Bigg|\int_{0}^{\infty}\frac{g(nt)-c}{1+t^2}\Bigg|$$

Now we know that there exists $x_0$ s.t $|g(nt)-c|<\frac{2\epsilon}{\pi}, n>x_0/t$, since the limit of $g$ exists. To apply this inequality successfully, we bound the quantity of interest as follows
$$Q\leq\int_{0}^{M}\frac{|g(nt)-c|}{1+t^2}+
\int_{M}^{\infty}\frac{|g(nt)-c|}{1+t^2}$$
where we arbitrarily split the integral in two pieces. The inequality above can only be applied on the second piece, with $n>\frac{x_0}{M}$. We put a bound on the first piece by using the boundedness of $g$ (inferred by it's continuity and the finite limit at infinity) which implies that $|g(nt)-c|<\Delta ~\forall~ n,t$.

All in all we have that
$$Q<\Delta\int_{0}^M\frac{dt}{1+t^2}+\frac{2\epsilon}{\pi}\int_{M}^\infty\frac{dt}{1+t^2}=\left(\Delta-\frac{2\epsilon}{\pi}\right)\arctan M +\epsilon$$
$Q$ doesn't depend on M so we can conclude that
$$Q<\inf_{M}\left(\Delta-\frac{2\epsilon}{\pi}\right)\arctan M +\epsilon=\epsilon$$
where we made the choice that $\epsilon<\frac{\pi\Delta}{2}$ has to be small enough. Hence we proved by the definition of the limit itself that
$$\lim_{n\to\infty}n\int_{0}^{\infty}\frac{g(x)}{n^2+x^2}dx=\frac{\pi c}{2}$$


问:
sum of relatively prime numbers
Suppose $\gcd(a,b)=\gcd(c,d)=\gcd(a,c)=1$. Then, $\{an+b:n\in\mathbb{N}\}\cap\{cn+d:n\in\mathbb{N}\}\neq\emptyset$. Why is that? I think I missed something silly...

答:
Assume $a> 0$ and $c> 0$. Since $\gcd(a,c) = 1$, a multiple of Bézout's identity is

$$ax - cy = d-b$$

for some $x,y\in \mathbb Z$.

To obtain positive $x',y'$ based on $x,y$, note that

$$\begin{align*}
ac - ca &= 0\\
a(x+kc) - c(y+ka) &= d-b\\
a(x+kc) + b &= c(y+ka)+d
\end{align*}$$

So regardless of whether $x,y\in\mathbb N$, for sufficiently large $k$, both $x'=x+kc \in \mathbb N$ and $y'=y+ka \in \mathbb N$.

Title: sum of relatively prime numbers


问:
Showing the energy is conserved in the shallow water equations
The shallow water equations are:

\begin{align}
\frac{Dv_x}{Dt}-fv_y&=-g\frac{\partial h}{\partial x}\\
\frac{Dv_y}{Dt}-fv_x&=-g\frac{\partial h}{\partial y}\\
\frac{D(h-h_b)}{Dt}&=-(h-h_b)(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y})
\end{align}

Where $\mathbf{v}=(v_x(x,y,t),v_y(x,y,t))$, $f$ and $h_b$ are functions of $x$ and $y$ and $h$ is a function of $x,y$ and $t$.

I want to show that the energy is a conserved quantity, I know energy takes the form
$$E=\frac{1}{2}(h-h_b)(v_x^2+v_y^2)+\frac{1}{2}gh^2$$

I've seen that

$$\frac{\partial E}{\partial t}+\nabla\cdot F=0$$

With $F$ being the energy flux.

My question is how do i show this?
I've tried taking the material derivative of $E$ and working through and that gets me that

$$\mathbf{v}\nabla\cdot E=\nabla\cdot F$$

I can't find a way to have the divergence of something on the RHS.
The only other thing i can think to do is guess the form of $F$ which since it is a energy flux it should be similar to $\mathbf{v}E$.

答:
Note that with $q = v_x$ or $q = v_y$ we have

$$\tag{1}q \frac{Dq}{Dt} = q \frac{\partial q}{\partial t} + qv_x \frac{\partial q}{\partial x} + qv_y \frac{\partial q}{\partial y}= \frac{\partial }{\partial t}\left(\frac{1}{2}q^2   \right) + v_x \frac{\partial }{\partial x}\left(\frac{1}{2}q^2\right)+  v_y \frac{\partial }{\partial y}\left(\frac{1}{2}q^2\right)$$

Multiplying each of the first two shallow water equations by $v_x$ and $v_y$, respectively, and using (1) we get

$$\tag{2}\frac{\partial }{\partial t}\left(\frac{1}{2}v_x^2   \right) = fv_xv_y+ v_x \frac{\partial }{\partial x}\left(\frac{1}{2}v_x^2\right)+  v_y \frac{\partial }{\partial y}\left(\frac{1}{2}v_x^2\right)- g v_x \frac{\partial h}{\partial x},$$
$$\tag{3}\frac{\partial }{\partial t}\left(\frac{1}{2}v_y^2   \right) = fv_xv_y+ v_x \frac{\partial }{\partial x}\left(\frac{1}{2}v_y^2\right)+  v_y \frac{\partial }{\partial y}\left(\frac{1}{2}v_y^2\right)- gv_y \frac{\partial h}{\partial y}$$

Define $K = \frac{1}{2}(v_x^2 + v_y^2)$ and add (2) and (3) to obtain

$$\tag{4}\frac{\partial K}{\partial t} = 2fv_xv_y + v_x\frac{\partial K}{\partial  x}+  v_y \frac{\partial K}{\partial y}- g v_x \frac{\partial h}{\partial x}- g v_y \frac{\partial h}{\partial y}$$

The third shallow water equation can be rewritten as

$$\tag{5} \frac{\partial h}{\partial t} = \frac{\partial}{\partial t}(h-h_b) = -v_x \frac{\partial }{\partial x}(h-h_b) - v_y \frac{\partial }{\partial y}(h-h_b) -(h-h_b) \frac{\partial v_x}{\partial x}- (h-h_b) \frac{\partial v_y}{\partial y}\\=-\frac{\partial }{\partial x}[(h-h_b) v_x] - \frac{\partial }{\partial y}[(h-h_b) v_y]$$

SInce $E = (h-h_b)K + \frac{1}{2}gh^2$, we have

$$\tag{6} \frac{\partial E}{\partial t} = (h-h_b)\frac{\partial K}{\partial t} + K \frac{\partial }{\partial t}(h- h_b)+gh \frac{\partial h}{\partial t}$$

Multiplying (4) by $(h-b)$ and (5) by $K$, yields

$$\tag{7}(h-h_b)\frac{\partial K}{\partial t} = 2fv_xv_y(h-h_b)+  (h-h_b)v_x\frac{\partial K}{\partial  x}+  (h-h_b)v_y \frac{\partial K}{\partial y} - g(h-h_b) v_x \frac{\partial h}{\partial x}- g v_y(h-h_b) \frac{\partial h}{\partial y},$$

$$\tag{8} K\frac{\partial }{\partial t}(h- h_b)= -K\frac{\partial }{\partial x}[(h-h_b) v_x] - K\frac{\partial }{\partial y}[(h-h_b) v_y]$$

Adding (7) and (8), we get

$$\tag{9}(h-h_b)\frac{\partial K}{\partial t} + K \frac{\partial }{\partial t}(h- h_b) = 2fv_xv_y(h- h_b) - \frac{\partial}{\partial x} [(h-h_b)Kv_x] -  \frac{\partial}{\partial y} [(h-h_b)Kv_x] - g(h-h_b) v_x \frac{\partial h}{\partial x}- g v_y(h-h_b) \frac{\partial h}{\partial y},$$

Using (5) we get

$$\tag{10}gh\frac{\partial h}{\partial t} = -gh\frac{\partial }{\partial x}[(h-h_b) v_x] - gh\frac{\partial }{\partial y}[(h-h_b) v_y] \\= -\frac{\partial }{\partial x}[gh(h-h_b) v_x] - \frac{\partial }{\partial y}[gh(h-h_b) v_y] + g(h-h_b)v_x\frac{\partial h }{\partial x}+ g(h-h_b)v_y\frac{\partial h }{\partial y}  $$

Substituting with (9) and (10) back into (6) yields

$$\frac{\partial E}{\partial t} = 2fv_xv_y(h- h_b) - \frac{\partial}{\partial x} [(h-h_b)Kv_x] -  \frac{\partial}{\partial y} [(h-h_b)Kv_x] - g(h-h_b) v_x \frac{\partial h}{\partial x}- g v_y(h-h_b) \frac{\partial h}{\partial y}-\frac{\partial }{\partial x}[gh(h-h_b) v_x] - \frac{\partial }{\partial y}[gh(h-h_b) v_y] + g(h-h_b)v_x\frac{\partial h }{\partial x}+ g(h-h_b)v_y\frac{\partial h }{\partial y}$$

The fourth, fifth , eighth and ninth terms on the RHS cancel, and we obtain

$$\frac{\partial E}{\partial t} + \nabla \cdot \mathbf{F} = fv_xv_y(h-h_b),$$

where

$$\nabla \cdot \mathbf{F} = \nabla \cdot [(h-h_b)(K +gh) \mathbf{v}]$$


问:
Monte Hall Problem - Question on Decision Tree Construction (Conditional Probability)

(The Monte Hall Problem, also known as the 3 door problem):  On the game show Let's Make a Deal, there are 3 doors hiding 3 prizes.  Two of the prizes are goats, 1 prize is an expensive new automobile.  Naturally you would like the car, but only Monte knows which door it is behind.  You select one of the doors.  Rather than opening the door you chose, Monte opens one of the two other doors to reveal a goat.  You are now given the option to stick with your original selection or to switch to the other opened door.

(a)  What is the conditional probability that you win the car if you stick with the original choice of doors?

(b)  What is the (conditional) probability that you win the car if you change yoour mind and select the other opened door?

I read about the Monte Hall problem and understand the principle behind it:  The door you choose is random, but the door Monte chooses is NOT.  This is why switching doors gives you a higher probability.  What I really had a question on is the construction of the decision tree for part (b).

(a):  I understand this solution is 1/3.

(b)  This question is asking you $Pr(win_{auto}|\Delta doors)=\cfrac{Pr(win_{auto}) \cap Pr(\Delta doors)}{Pr(\Delta doors)}$.  I was just confused on the decision tree provided as a solution.  Originally, I thought a decision tree gives you ALL possible outcomes in the sample space.  But in this problem, a decision tree is drawn ONLY IF you change doors.  The solutions tree is below:

Above you can correctly calculate  the Probability by summing the branches:  (1/3)*1*1 *2 = 2/3.

Now, lets say I said draw the tree assuming that you always keep your initial door choice.  What is the probability that you win an auto given that you always stay with your initial door (I know its 1-2/3 just go with me for a second)?  I thought I would draw something like this:

Above you see the probability of winning is = 2*(1/3* 0) + 1/3 * 1 = 1/3 (which is the correct answer).

Is it correct to state that decision trees do not necessarily show the entire sample space of a problem?  Instead, sometimes they only show a specific scenario or subset of the entire sample space when calculating probabilities?

Thank you!

答: This decision tree seems to be showing the decision tree given that you change doors.  By including the "given that you change doors" shrinks your sample space.</p>


问:
Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$
Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable

Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$

答:
HINT

\begin{align*}
y'' + y = \frac{1}{y^{3}} & \Longleftrightarrow y''y' + yy' = \frac{y'}{y^{3}}\\\\
& \Longleftrightarrow (y')^{2} + y^{2} = -\frac{1}{y^{2}} + c\\\\
& \Longleftrightarrow (y')^{2} = \frac{cy^{2} - y^{4} - 1}{y^{2}}\\\\
& \Longleftrightarrow y' = \pm\sqrt{\frac{cy^{2} - y^{4} - 1}{y^{2}}}
\end{align*}


问:
Line bisecting the angle between the coordinate curves?
Prove that the lines that at each point bisect the angles between the coordinate lines are given by the differential equations $$\sqrt{E}\,du \pm \sqrt{G}\,dv = 0$$

I calculated the angle between the coordinate curves: $cos\varphi=\frac{F}{\sqrt{EG}}$.

Next, I tried to substitute a half angle into the formula for the angle between the curves and prove the original statement, but the result does not match.

Any help would be appreciated.

答: You can consider the unit vector along u direction $\frac{1}{\sqrt{E}}\frac{\partial}{\partial u}$ and the unit vector along v direction $\frac{1}{\sqrt{G}}\frac{\partial}{\partial v}$. Now the tangent vector of the solution curve is $v=\sqrt{G}\frac{\partial}{\partial u}\pm \sqrt{E}\frac{\partial}{\partial v}$. You will get the same projection of $v$ onto the two directions $\sqrt{EG}\pm F$.


问:
How many ways to put n objects into n bins, with object i not in bin i for 1<=i<=n ??
How many ways are there to arrange n objects into n bins, subject to the n constraints that the ith object can't go in the ith bin for any 1 <= i <= n ?  Each bin has to contain exactly one object.</p>

答:
You do not state that each bin can only contain $1$ object, if this were so then the answer would be derangements (as JW comments)

Each of the $n$ objects can go into $n-1$ bins ... there are $\color{red}{(n-1)^n}$ ways to do this.


问:
Range of error in truncated division $\lfloor\frac{a\pm 1}{b\pm 1}\rfloor$
With $a,b\in\mathbb{N}^+$, $b > 1$ and $2\lfloor\log_2(b+\Delta_b)\rfloor\ge\lfloor\log_2(a+\Delta_a)\rfloor$ where $\Delta_n\in\{-1,0,1\}$ is my guess that
$$x + \Delta_x = \left\lfloor\frac{a+\Delta_a}{b+\Delta_b}\right\rfloor$$
correct? If the difference would be restricted to the numerator it would be
$$\left\lfloor \frac{a - 1}{b} \right\rfloor = \left\lfloor \frac{a}{b} + \frac{b - 1}{b} \right\rfloor -1$$
and
$$\left\lfloor \frac{a + 1}{b} \right\rfloor = \left\lfloor \frac{a}{b} + \frac{1}{b} \right\rfloor$$
respectively if I remember it right but the denominator is involved, too. So, how do I (dis)prove my assumption?

Edited for clarity:

$\Delta_n$ is meant to be a variable, so with all possible variations of the division the outcome is one of $x + 0$, $x - 1$, or $x + 1$ where $x + 0$ is the outcome of $\lfloor (a + 0 / (b + 0) \rfloor$. My assumption is that for all of the variations
$\lfloor(a+1)/(b+1)\rfloor$,
$\lfloor(a+1)/(b+0)\rfloor$,
$\lfloor(a+1)/(b-1)\rfloor$,
$\lfloor(a+0)/(b+1)\rfloor$,
$\lfloor(a+0)/(b-1)\rfloor$,
$\lfloor(a-1)/(b+1)\rfloor$,
$\lfloor(a-1)/(b+0)\rfloor$,
$\lfloor(a-1)/(b-1)\rfloor$, the result $x'$ is at most one unit away from  $x = \lfloor (a + 0 / (b + 0) \rfloor$

答:
The largest and smallest numbers that can be expressed as $\frac{a+\Delta_1}{b+\Delta_2}$ with $\Delta_1,\Delta_2\in\{-1,0,1\}$ are $\frac{a+1}{b-1}$ and $\frac{a-1}{b+1}$, respectively. We can calculate that
$$\frac{a+1}{b-1}-\frac{a}{b}=\frac{a+b}{b(b-1)}$$
and
$$\frac{a}{b}-\frac{a-1}{b+1}=\frac{a+b}{b(b+1)}.$$
If these differences are both at most $1$, that is, if $a\leq b^2-2b$, then the floors of any of these and $\frac{a}{b}$ can't differ by more than $1$; if $\lfloor y\rfloor\geq\lfloor x\rfloor+2$, then
$$y\geq \lfloor y\rfloor\geq \lfloor x\rfloor+2>x+1.$$

In general, it's a bad idea to use the same letter to refer to different variables (even if they vary among the same set); I've used $\Delta_1$ and $\Delta_2$ to clarify what's happening here.


答:
Another Way to look at it

$$\frac{d^2y}{d^2x} = \frac{1}{y^3} - y  =>  \frac{d^2y}{d^2x} = \frac{1-y^4}{y^3}   =>$$
$$\frac{y^3}{1-y^4} d^2y = d^2x $$

Now this can be double integrated using the integral of a division formula

Title: Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$

Body:
Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable

Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$


问:
Why doesn't this approach to solving this integral (with composite Dirac delta functions) work?
I recently encountered this integral:

$\mathbf J_1 = \int_{0}^{2\pi} I \delta (z) \delta (y-a(1-\cos t))\delta (x-a\sin t) (\mathbf {\hat x} \sin t + \mathbf {\hat y} \cos t) adt$

The way I attempted to solve it was by noting that $\delta (x-a(1-\cos t))  \delta (y-a\sin t)$ is nonzero iff $t = \cos^{-1}(1 - \frac{x}{a}) = \sin^{-1}(\frac{y}{a})$, resulting in the integral evaluating to the following expression

$\mathbf J_1 = Ia\delta (z) (\mathbf {\hat x} \frac{y}{a}+ \mathbf {\hat y} (1 - \frac{x}{a}))$

However, in the context of the problem, that does not make sense. $\mathbf J_1$ is supposed to be the current density of a circular current loop in the xy-plane. So I know that I solved the integral incorrectly. But how? What is $\delta (x-a(1-\cos t))  \delta (x-a\sin t)$  supposed to evaluate to?

答: Assuming that the third Delta is $\delta(y-a\sin t)$ instead of $\delta(x-a\sin t)$. $J_1$ is already in the simplest form, it is the current density going through the xy-plane $z=0$ in that circle. You can use it to get a current flux through any surface, the t parameter can then help you check all the points on the circle to see if there's an intersection with the surface. If there is one that you take the dot product of the tangent vector to the circle with the surface normal there.


问:
Solving $\sin\left(2x\right)+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+1=0$
I attempted to do this trig problem I have written below with steps I took to reach an answer. Did I do this correctly?
$$\sin\left(2x\right)+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+1=0$$

using the well known identity: $\sin^2\left(x\right)+\cos^2\left(x\right)=1$, I get:

$$\color{red}{\sin\left(2x\right)}+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)+\color{red}{\sin^2\left(x\right)+\cos^2\left(x\right)}=0$$

Then I see that the "red" is just $\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)^2$, so
$$\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)^2+5\bigl(\sin\left(x\right)+\cos\left(x\right)\bigr)=0$$

Then I let $u=\sin\left(x\right)+\cos\left(x\right)$, I get
$$u^2+5u=0$$
$$u=0,\:\text{and}\:u=-5$$

removing $u=-5$ as no solution, leaving just $u=0$ as the only solution, plugging back into the defined "$u$",

$$0=\sin\left(x\right)+\cos\left(x\right)$$
$$\sin\left(x\right)=-\cos\left(x\right)$$
$$\tan\left(x\right)=-1$$

the solution I get is
$$x=\frac{3\pi }{4}+\pi n$$

No need to see if there are any extraneous solutions since the function is continuous.

Thanks for any feedback!

答:
Overall, looks great, though I've a few things to comment on.

My first criticism is that you should perhaps justify the following:

$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + \sin 2x$. This is easily done by squaring the binomial by noting $\sin 2x = 2 \sin x \cos x$.

$u = -5$ generates no solution. This follows easily since $\sin x, \cos x$ are, at minimum, $-1$ each.

Granted, as noted, these are simple justifications (clearly you already know them), but perhaps would be appreciated by your instructor for completeness (if this is for a class).

Next, towards the end, you divide by $\cos x$ in order to get $\tan x$. This results in a problem if $\cos x = 0$, though. Thus, you need to handle that case separately, and need to determine whether or not $x$ can be such that both sine and cosine are equal to zero.

(Obviously, they're not, and you have the complete solution set -- this is just, again, in the interest of completeness.)


问:
Linear Transformation Orthogonality

True or False:

If $T$ is a linear transformation from $R^n$ to $R^n$

such that

$$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots,  T\left(\vec{e_n}\right)  $$

are all unit vectors

then $T$ must be an orthogonal transformation

The answer is ?
I know a linear transformation is orthogonal if it preserves the length of vectors.
And I understand a linear transformation is orthogonal if $$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots,  T\left(\vec{e_n}\right)  $$  form an orthonomal basis.

But how do I prove that withough knowing what the $T$, transformaton, does to the elementary column vectors...?
Could anybody help me prove this?

答: False.  Let $T$ be the linear transformation represented by the matrix whose first row is all 1's, and the other rows are all 0's.  $T(e_i)=e_1$ for all $i$.</p>

Title: Linear Transformation Orthogonality


问:
Smith normal form of a polynomial matrix
I have the following matrix

$$P(s) := \begin{bmatrix}
s^2 & s-1 \\
s   & s^2
\end{bmatrix}$$

How does one compute the Smith normal form of this matrix? I can't quite grasp the algorithm.

答:
One easy way is by looking at fitting ideal.

If \begin{bmatrix} d_1 & 0\\ 0 & d_2 \end{bmatrix}
is the required smith normal form then,

check that ,
$Fitt_1(P) = (d_1)=(1)$ and $Fitt_2(P) = (d_1d_2)=(x^4-x^2+x)$.


问:
Why does $ab = ba$ imply $(ab)^{n} = a^{n}b^{n}$ for $a, b\in G$?
I found the following post: Let $a,b$ be in a group $G$. Show $(ab)^n=a^nb^n$ $\forall n\in\mathbb{Z}$ if and only if $ab=ba$..

It means for $a, b\in G$, if $ab = ba$, then $(ab)^{n} = a^{n}b^{n}$. But if we let $G = (\mathbb{Z}, +)$, wouldn't this not be true? For example, take $a = b = 5$ so that $(a + b)^{2} = 100 \neq 625= a^{2}b^{2}.$

I'm new to abstract algebra. Maybe my understanding of what the theorem is saying is off.

答:
This is one of those annoying notation issues. Bear in mind that $a^n$ in a group $G$ with operation $\ast$ means

$$a^n = \underbrace{a \ast a \ast a \ast \cdots \ast a}_{n \; a's}$$

If $G$ is $\Bbb Z$ and your operation is addition, then the notation $a^n$ means

$$a^n = \underbrace{a + a + a + \cdots + a}_{n \; a's} = na$$

It's a bit confusing. In general, though, for abelian groups (those where $a \ast b = b \ast a$ for every $a,b \in G$), this is why we instead adopt the additive notation $na$ instead of the multiplicative notation $a^n$.

Thus, in this light, we have

$$(ab)^n = a^n b^n \text{ means } n(a+b) = na + nb$$


问:
Finding the number of non-neg integer solutions?
How would I find the number of non negative integer solutions to this problem?
$$x_1 + x_2 + x_3 + x_4 = 12$$ if $0 \leq x_1 \leq 2$.</p>

Body:
$$x_1 + x_2 + x_3 + x_4 = 12, 0 \leq x_1 \leq 2$$
According to http://oeis.org/wiki/User:Adi_Dani_/Restricted_compositions_of_natural_numbers
there is derived formula
$${\binom{m}{k}}_{s}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+k-si-1}{m-1}\,$$
for number of solutions of $$x_0+x_1 + x_2 +...+ x_{m-1} =k,0\leq x_i\leq s-1$$

in our case $k=12,s=3,m=4$
$${\binom{4}{12}}_{3}=\sum_{i=0}^{4}(-1)^{i}\binom{4}{i}\binom{15-3i}{3}=$$
$$=\binom{4}{0}\binom{15}{3}-\binom{4}{1}\binom{12}{3}+\binom{4}{2}\binom{9}{3}-\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{3}{3}=$$

$$=\binom{15}{3}-4\binom{12}{3}+6\binom{9}{3}-4\binom{6}{3}+\binom{3}{3}$$


问:
Further questions about probability: the hunter and the rabbit
The question is stated as follows:

Suppose there are $m$ hunters and $n$ rabbits. Each hunter shoots one out of the $n$ rabbits uniformly at random independently. Let X be the number of rabbits that are shot.

*we can assume that there are m shots in total,i.e each hunter take exactly one shot at one rabbit

Suppose $r$ a non-negative integer. Compute $\text{Prob}[X = r]$ in terms of $m$ and $n$.

Compute $\mathbb E[X]$ in terms of $m$ and $n$.

There are similar questions asked. But the ones I stated has a subtle difference since we are asked to give a probability of X = $r$ and calculate Expected Value.

My approach to this problem is as follows:
Since each hunter will shoot a rabbit at one time, and that two hunters may shoot the same rabbit, the total ways of shooting r rabbit is $r^m - C(n,1) \times r^{m-1} + C(n,2) \times r^{m-2} - ... $ (Inclusion-Exclusion Principle) and the total shootings are $r^m$. And hence we derive the probability.

My question is:

Is my approach correct?

The expected values are a bit tricky. One way is to use the probability I calculated above and multiply each by r. Another way is to set another set of random variables.(Supposingly makes it easier, but I don't think it can be used here since, though the hunter shoots independently, the rabbit doesn't die independently)

How to calculate the expected value here?

答:
Various questions relating to this problem have been asked before on math.stack.exchange. Here is an answer to part $2$ of your question.  Note, however, that in this answer, $\ m\ $ is the number of rabbits, not the number of hunters, and $\ n\ $ is the number of hunters.

There doesn't appear to be an answer giving the distribution of the number of rabbits killed for general values of $\ m\ $ and $\ n\ $.   While the principle of inclusion-exclusion is certainly the way to go to calculate this distribution, I don't understand how you arrived at your formula for the "number of ways of shooting $\ r\ $ rabbits". Also, the total number of possible ways the rabbits can be shot is $\ n^m\ $, not $\ r^m\ $.  This just the number of functions from the set of $\ m\ $ hunters to the set of $\ n\ $ rabbits.

The total number of ways that a specific set of $\ r\ $ rabbits can be shot is just the number of surjective functions from the set of hunters onto that set of rabbits.  By the principle of inclusion-exclusion, this is
\begin{align}
\cases{\displaystyle\sum_{i=0}^{r-1}(-1)^iC^r_i(r-i)^m&if $ r\le m\ $, or\\
0&if $\ r>m\ $.}
\end{align}
However, there are $\ C^n_r\ $ distinct sets of $\ r\ $ rabbits, so the total number of ways the hunters can shoot exactly $\ r\ $ rabbits is
$$
C^n_r\sum_{i=0}^{r-1}(-1)^iC^r_i(r-i)^m\ ,
$$
for $\ r\le m\ $, and the probability of their doing so is therefore
$$
\text{Prob}[X=r]=\frac{\displaystyle C^n_r\sum_{i=0}^{r-1}(-1)^iC^r_i(r-i)^m}{n^m}\ .
$$


问:
Variables vs. Constants, how are they defined in ZFC, and how they differ?
The other day I was thinking that whenever I'm presented in Uni with constants my teachers use letters like $c$, $k$, $n$, $a$, $b$, etc, and when encountered with variables they use letters like $x$, $y$, $z$, $w$, etc. However, these are only labels, there's nothing stopping me from using $c$ as a variable, or $x$ a constant; this led me to ask myself how are constants vs. variables formally defined.

So far, I have only come up with the definition of a constant in a set by reading "Introduction to Set" Theory by Jech and Hrbáček:

According what they say, given a set $A$ and a $0$-ary operation $R$ on $A$, $(∅,a)$, such that $(∅,a)\in R$ and $a\in A$, is a constant and we can call it just $a$.

Nonetheless, I haven't found anything on the book about the formal definition of a variable.

Is my understanding of the definition of a constant correct? Do you know what's the definition of a variable? (if so, could you share it with me?)

Thanks in advance.

答:
"The" language of first-order logic, which ZFC is based on, is a family of languages.

All languages share the same logical symbols:

individual variables: $x_n$ for each $n \in \mathbb{N}$

connectives: $\neg, \land, \lor, \to$

quantifiers: $\exists, \forall$

and auxiliary symbols:

parentheses: $), ($

comma: $,$

Different languages differ in their inventory of non-logical symbols:

individual constants

function symbols

predicate symbols

One defines a formal language by listing all members of each category.

For the language of ZFC, it is sufficient to limit oneself to:

individual constants: none

function symbols: none

predicate symbols: $\in$

One can add some useful additional symbols, e.g. *

individual constants: $\emptyset$

function symbols: $\cap, \cup, \wp$

predicate symbols: $\subset, \subseteq$

Variables and constants are thus defined extensionally by specifying which elements they encompass in a given language; but there is no formal definition of the underlying concept of variables and constants. Remember that variables and constants themselves are just symbols, not mathematical objects; and there are no global regulations on what these symbols need to look like. Though it makes sense in practice to stick to conventions, such as using $x, y, z$ (possibly with indices) for variables and $a, b, c, \ldots$ for constants.

Your understanding of a constant is not entirely correct, though the way it is presented in the text also makes it kind of difficult: We have constant/function/predicate symbols which are interpreted as elements/functions/relations on the domain. The object $a \in A$ would be the interpretation of a constant symbol such as $c$.

Which constants there exist is fixed by specifying the formal language; how they are interpreted is fixed by specifying a structure consisting of a domain of objects and an interpretation function which maps the constant/function/predicate symbols to elements/functions/relations on that domain. This interpretation again has to be defined by explicitly listing which formal symbols map to which object: $\mathcal{I}(c) = a$. A constant is a symbol which by itself is meaningless; its meaning is given by fixing an interpretation which maps it to an element of the domain.

The meaning of variables is given by an assignment function which maps each variable to an element of the domain; again, this assignment needs to be explicitly specified.

The difference between variables and constants is that a variable can undergo different assignments and thus take different values in the same structure under which the formula is evaluated, whereas the meaning of a constant is fixed within a structure.

* though typically these would not be treated as elementary symbols of the formal language whose interpretation needs to be specified in a structure, but rather as abbreviations for some complex expression, e.g. $y = \wp(x)$ is taken as shorthand for $\forall z (z \in y \leftrightarrow \forall u \in z (z \in x)))$ (meaning that the elements of $\wp(x)$ are exactly those sets all of whose members are elements of $x$).


问:
Prove that These Families of Level Curves are Orthogonal
From p. 79 in Brown's and Churchill's "Complex Variable and Application":

Let the function $f(z) = u(x, y)+iv(x, y)$ be analytic in a domain $D$, and consider the family of level curves $u(x, y) = c_1$ and $v(x, y) = c_2$.  Prove that these families are orthogonal.

Specifically, show that if $z_0 = (x_0, y_0)$ is a point in $D$ which is common to two particular curves $u(x, y) = c_1$ and $v(x, y) = c_2$, and if $f'(z_0) \ne 0$, then the lines tangent to the curves at $(x_0, y_0)$ are perpendicular to each other.  Then, the question gave a clue that:

$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx} = 0$ and  $\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}\frac{dy}{dx} = 0$ (*)

At first, I thought that the authors meant $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}\frac{dx}{dx}+\frac{\partial u}{\partial y}\frac{dy}{dx} = 0$, $\frac{\partial v}{\partial x} = \frac{\partial v}{\partial x}\frac{dx}{dx}+\frac{\partial v}{\partial y}\frac{dy}{dx} = 0$.  However, that cannot be the case, since we were told that $f'(z_0) = u_x(x_0, y_0)+iv_x(x_0, y_0) \ne 0$.  Furthermore, $\frac{\partial u}{\partial x} = 0$ implies that $u(x, y)$ is constant in a direction parallel to the $x$-axis, but that certainly is not the case in general.  So, how do you get the two equalities in (*)?

Furthermore, what is the significance in $f'(z_0) \ne 0$?  In the next question, we are asked to show that with $f(z) = z^2$, the level curves $u(x, y) = x^2-y^2=0$ and $v(x,y)= 2xy = 0$ are not orthogonal.  But a straight-forward computation $\nabla u · \nabla v = u_xv_x + u_yv_y = 4xy - 4xy = 0$, a constant zero.  What did I gloss over?

答:
One sec. I think I realize how to use the hint (trying to answer my own question in comments).

If $u_x + u_y y' = 0$, then $v_y - v_x y'=0$ by Cauchy-Riemann. Here, we must have $v_x\neq 0$ for otherwise $v_y=0$ too, contradicting the nonzero assumption. Hence, we get $y'=v_y / v_x$ for the first level curve. Similarly, the second level curve gives $y'=-v_x/v_y$. Multiplying these gradients gives $-1$, proving orthogonality as desired.

Is this OK?


答: Let $T e_k = \frac{1}{\sqrt{k}} \sum_{i=1}^k e_i$. Then $T$ is invertible, and $\|T e_k \| = 1$ for all $k$.</p>

Title: Linear Transformation Orthogonality

Body:
True or False:

If $T$ is a linear transformation from $R^n$ to $R^n$

such that

$$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots,  T\left(\vec{e_n}\right)  $$

are all unit vectors

then $T$ must be an orthogonal transformation

The answer is ?
I know a linear transformation is orthogonal if it preserves the length of vectors.
And I understand a linear transformation is orthogonal if $$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots,  T\left(\vec{e_n}\right)  $$  form an orthonomal basis.

But how do I prove that withough knowing what the $T$, transformaton, does to the elementary column vectors...?
Could anybody help me prove this?


问:
Eigenvalues of the unit matrix

Consider the $n \times n$ unit matrix $A$ where $A_{ij} = 1$ for all $i$ and $j$ over a field $F$. Find the eigenvalues of $A$ and their geometric and algebraic multiplicities.

After having done some work, I found that the eigenvalues of $A$ are $\lambda = 0$ and $\lambda = n$. In the first case, the eigenvectors have the property that their components sum to $0$. In the second case, the components of the eigenvectors are all equal (and non-zero).

I cannot understand the geometric and algebraic multiplicities. Given eigenvalue $\lambda$, the geometric multiplicity is the dimension of $\text{null}(A - \lambda I)$ and the algebraic multiplicity is the dimension of $\text{null}(A - \lambda I)^n$. I am, in particular, struggling with algebraic multiplicity, without knowing the characteristic polynomial.

Any help would be appreciatied.

答:
You don't need the characteristic polynomial, as the spectral theorem guarantees that $A$, being a symmetric matrix ($A = A^t$), is diagonalizable. In fact, it can be diagonalized with an orthogonal matrix (i.e. we can find a orthonormal basis of eigenvectors for $A$). That means that all the geometric and algebraic multiplicities are equal.

In this case, it seems pretty clear that a basis of eigenvectors could be given by an orthogonal basis for the space $\{ (x_1, ..., x_n) \in \Bbb{R}^n : x_1 + ... + x_n = 0 \} \cong \Bbb{R}^{n-1}$, together with a unit vector in the direction of $(1, ..., 1)$. That means the multiplicity (geometric or algebraic) of $\lambda = 0$ is $(n-1)$ and the multiplicity (geometric or algebraic) of $\lambda = n$ is $0$.

Title: Eigenvalues of the unit matrix


问:
Calculus Transforming Functions (Horizontal Compression) Question
so I'm taking an online Calculus I course and I am currently working through some of the practice problems on the transforming functions section. It's set up to tell me right away if my answer is right or wrong and I have been getting the same type of question wrong over-and-over and was hoping for an explanation since they do not provide one.

The question goes Suppose that y= x^2+4x+1. What is the equation of this function compressed horizontally by a factor of 4? And then it asks me for the new equation.

The way I looked at it was the course taught us that a horizontally compressed function when c>1 is y=f(cx) so applying this to the question I thought it should be y= cx^2+4x(c)+1 am I wrong here?

So evaluated... the answer would be y= (4)x^2 + (4)4x + 1

```<code>                             y= 4x^2 + 16x + 1
</code>```

The reason I am confused is because the course keeps telling me the answer is...

```<code>                             y=16x^2 + 16x + 1
</code>```

But I don't see how we get 16x^2 from an original x^2 when all we are doing is cx^2 when c=4. Is there something I am missing here or is the course answer wrong...I'd love an explanation as simple as this might be.

答:
You misunderstood the substitution. The substitution is $x\mapsto \color{blue}{cx}$, rather than multiplying the function by $c$. The compressed function should be
$$
y= (\color{blue}{4x})^2 + 4(\color{blue}{4x}) + 1 = 16x^2 + 16x + 1
$$
rather than
$$
y= (4)x^2 + (4)4x + 1
$$