问:
Questions about parameter estimation for the Cauchy Distribution.
$X_{1} ... X_{n}$ are i.i.d.  according to the Cauchy distribution centered at θ. The the density function is given by:

$\frac{1}{\pi} \cdot\frac{1}{[1 + (x-\theta)^2]}$

and the cdf is given by:

$\frac{1}{2} + \frac{1}{\pi}arctan(x - \theta)$

Find the estimator of $\theta$. Using a concept similar to the method of moments, $\hat p_{n}$ denotes the proportion of negative observations which is close to its expected value of $F_{\theta}(0)$. Demonstrate that the estimator is consistent, and find the limiting distribution.

Couple questions here though:

For determining the estimator we set $\hat p_{n}$ equal to its expected value of $F_{\theta}(0)$ and solve for theta, correct? Here I obtained tan($\frac{\pi}{2} - \pi\hat p)$ = $\hat \theta$

How do we demonstrate the consistency of the estimator though? Can we argue that $X_1,\ldots,X_n\stackrel{iid}{\sim}\mathrm{Ber}(p)$ because p is a proportion of negative observations and there can only be two outcomes. Thus $\hat p_{n}$ = $\frac{\sum\limits_{i=1}^n X_i}{n}$ which means it converges to p using the weak law of large numbers? And because $\hat p_n$ converges to p, $\hat \theta$ must converge to $\theta$ as well due to the continuous mapping theorem?

I believe we can find the asymptotic distribution using the delta method. Is this correct as well?

Thanks, wondering here if I am on the right track. I don't know if I am off base here, to be honest.

答:
See the estimation section of Wikipedia on Cauchy.

For the center, the sample median works, but the 38% trimmed mean is said to have slightly smaller variability.

Simulation for the special case of Student's t distribution with degrees of freedom $\nu=1$ and samples of size $n=100.$

[Note that the conditions of the CLT for
medians are met, but conditions of the CLT for means are not.]

```<code>set.seed(111)
h = replicate(10^5, median(rt(100,1)))
summary(h);  sd(h)

      Min.    1st Qu.     Median       Mean    3rd Qu.       Max.
-0.7395925 -0.1059586  0.0005111 -0.0000415  0.1050140  0.7057973
[1] 0.1585002

a.38 = replicate(10^5, mean(rt(100,1),trim=.38))
summary(a.38);  sd(a.38)

      Min.    1st Qu.     Median       Mean     3rd Qu.       Max.
-0.7270404 -0.1025867 -0.0009389 -0.0006116  0.1006515  0.7054522
 [1] 0.1530064  # slightly smaller SD
</code>```

Body:
See the estimation section of Wikipedia on Cauchy.

For the center, the sample median works, but the 38% trimmed mean is said to have slightly smaller variability.

Simulation for the special case of Student's t distribution with degrees of freedom $\nu=1$ and samples of size $n=100.$

[Note that the conditions of the CLT for
medians are met, but conditions of the CLT for means are not.]

set.seed(111)
h = replicate(10^5, median(rt(100,1)))
summary(h);  sd(h)

      Min.    1st Qu.     Median       Mean    3rd Qu.       Max.
-0.7395925 -0.1059586  0.0005111 -0.0000415  0.1050140  0.7057973
[1] 0.1585002

a.38 = replicate(10^5, mean(rt(100,1),trim=.38))
summary(a.38);  sd(a.38)

      Min.    1st Qu.     Median       Mean     3rd Qu.       Max.
-0.7270404 -0.1025867 -0.0009389 -0.0006116  0.1006515  0.7054522
 [1] 0.1530064  # slightly smaller SD


问:
Asymptotic Symbol
I am reading a paper and trying to understand what the following statement means

$R(\hat{G}_{\text{chime}})-R_{\text{opt}}(\hat{G}_{\theta})\asymp\frac{\text{log}p}{n}$

I understand the left hand what I am trying to understand is $\asymp\frac{\text{log}p}{n}$

The paper is trying to find asymptotic bounds for misclassification errors of a new proposed clustering method.

答:
My guess it would either mean that the ratio tends to one or that the ratio has a lower and upper bound
similar to $\Theta$.

ContentLicense: CC BY-SA 4.0

Title: Asymptotic Symbol


问:
Confusion regarding Fermat's Theorem and Closed Interval Method
In James Stewart's Calculus (8th edition) Fermat's Theorem is stated as

If $f$ has a local maximum or minimum at $c$, and if $f'(c)$ exists, then $f'(c) = 0$.

The author cautions that the converse of the statement is generally false. In other words, finding $x$-values such that $f'(x) = 0$  does not necessarily imply that a local max or min occurs there. The functions $f(x) = x^3$ and $f(x) = |x|$  are provided as examples.

The author then introduces the idea of critical numbers and defines them as

A critical number of a function $f$ is a number $c$ in the domain of $f$ such that either $f'(c) = 0$ or $f'(c)$ does not exist [$f'(c)$ is undefined]

and then modifies Fermat's Theorem as

If $f$ has a local maximum or minimum at $c$, then $c$ is a critical number of $f$.

My confusion begins when the author describes how to find absolute max and min.

To find an absolute maximum or minimum of a continuous function on a closed interval, we note that it is either local (in which case it occurs at a critical number by [Fermat's modified Theorem]) or it occurs at an endpoint of the interval[...]

The wording to me seems like the author is suggesting to use critical points to find or identify local maxes and mins of a function. This notion seems to be further reinforced in the first step of the Closed Interval Method

Find the values of $f$ at the critical numbers of $f$ in $(a, b)$

I don't think this is what the author meant since he goes to great length to caution against the converse of Fermat's Theorem, but I'm not entirely sure if rephrasing Fermat's Theorem in terms of critical numbers now renders the converse true.

Just to be safe I will ask the following questions:

Is the author suggesting that we use critical points to find local maxes and mins? If so, wouldn't that mean the converse of Fermat's modified Theorem is true?

Is the fact that local maxes and mins occur at critical numbers a consequence of Fermat's modified Theorem or its converse?

I suspect the answer is no to the first question, and the fact that local maxes and mins occur at critical points is a consequence of Fermat's modified Theorem which implies that if a local max or min exists, then it will always occur at a critical value. Therefore, finding critical values is only finding potential spots where local maxes or mins might occur.

答:
The author states that if $f'(c)$ exists, then it must be $0$. Notice that in your examples, it did not exist.

To answer your question:

Is the author suggesting that we use critical points to find local maxes and mins?

He is.

If so, wouldn't that mean the converse of Fermat's modified Theorem is true?

No. As you mentioned earlier, there are critical points at which the maximum or minimum may not occur at.

Is the fact that local maxes and mins occur at critical numbers a consequence of Fermat's modified Theorem or its converse?

I will explain this to you in a different way. First, you must convince yourself that if $f$ is continuous on an interval $[a,b]$, then $f$ must attain a minimum and a maximum on $[a,b].$ There are three possible scenarios:

The extrema value occurs at some point $c \in (a,b)$ such that $f'(c) = 0$;

The extrema value occurs at some point $c \in (a,b)$ such that $f'(c)$ does not exist;

The extrema value occurs at $f(a)$ or $f(b)$. In other words, the endpoint(s).

We call all possible points $c$ in all three of the cases above critical points of $f$. Now, this is only tells us where the critical points are. This implies that if we want to find the maximum or minimum value of $f$ on some interval, then we still have to check all of those points.

Therefore, finding critical values is only finding potential spots where local maxes or mins might occur.

This is correct.


问:
Parametrization of probability in a limit for a random graph

I don't understand the part where the authot exchanges $p$ for $c/n$. This can only work if $c$ is a function of $n$ namely $pn$.

I hope somebody can tell me what the author is doing here and why he is allowed to first take $n\rightarrow\infty$ and then $c\rightarrow 0$ even though if $n\rightarrow \infty$ then automatically $c\rightarrow \infty$.

Can also someone tell me how the author comes to the conclusion that if $p=10^{-6}/n$ then $G(n,p)$ is unlikely to have a triangle and also how the second conclusion that if $p=10^6/n$ then $G(n,p)$ is likely to have a triangle  is even possible ? For example if $n<10^6$ then $p>1$ but this is not possible because $p$ is a probability ranging from 0 to 1.

答:
Lots to explain here. Hope the following helps.

First, the parametrization of $G(n,p)$ as $p=c/n$ is a standard parametrization, sometimes called the coarse parametrization if $c$ is a constant. It just means that we define $c=pn$ and we will study different regimes of $G(n,p)$ depending on the value of $c$. For instance, its common to look for properties of $G(n,p)$ depending on the three cases $c<1$, $c=1$ and $c>1$. In general $c$ will be a function of $n$ indeed, but studying the case where $c$ is a constant often helps us understand how the general case behaves.

It is not true that $c\to\infty$ automatically when $n\to\infty$. For instance you could have :

$p=\frac{\log n}{n}$ and $c(n)=\log n$, then $c\to\infty$ when $n\to\infty$

$p=1/ n^2$ and $c(n)=1/n$, then $c\to 0$ when $n\to\infty$

$p=2/n$, then $c$ is a constant.

But in here, the author does not assume the asymptotic of $c(n)$. When he states the two different cases ($c\to 0$ and $c\to\infty$), it is to present the following fact :

If you have $p=c/n$, with $c$ a constant, then you know that the limit probability of having no triangle is exactly $e^{-c^3/6}$. Therefore if $p=c/n$ where $c$ is a very small constant, then the limit probability of having no triangle will be very close to $1$. If $c$ is very large, then the limit probability of having no triangle will be very close to $0$.

The author does not value
$$\lim_{n\to\infty} \lim_{c\to } \mathbb{P}[X=0]$$
He says, assume $p=c/n$, then
$$\lim_{c\to 0} \lim_{n\to\infty} \mathbb{P}[X=0] = 1$$
and
$$\lim_{c\to \infty} \lim_{n\to\infty} \mathbb{P}[X=0] = 0$$

This is why the author say that if $p=10^{-6}/n$, then the probability of no triangle is almost 1 : $$\lim_{n\to\infty} \mathbb{P}[X=0]=e^\frac{-c^3}{6}\approx 1$$

This helps concluding on the other cases as follow. Assume that $p=n^{-0.9}$. You can take $c$ a constant as large as you want, for $n$ large enough you will have $p>c/n$. Therefore because we know the limit probability when $c$ is constant is $e^{-c^3/6}$ which tends to $0$ when $c\to\infty$, then we know $\lim_{n\to\infty} \mathbb{P}[X=0]=1$. (caveat : we need some details on the monotonicity of the probability to make this argument formal.)
You conclude on the other case $p=1/(n \ln n)$ in the same way with $p<c/n$ for $c$ as small as you want.

Regarding your last remark, it's true that $p$ is ill-defined if $n<10^6$, but we are only interested in asymptotic ($n\to\infty$), so you can assume " $n$ is large enough".


问:
Proving a piecewise function is discontinuous at a point

Give a $\delta,\epsilon$ proof that the following function,
$f(x) = \begin{cases} x,  & \text{if $x \geq 0$} \\ x+3, & \text{if $x
 < 0$} \end{cases}\quad \quad $

is discontinuous at $x=0$.

Attempt:

If $f(x)$ is not continuous at $x=0$, then $\exists \epsilon>0$ such that $\forall \delta > 0$, $\exists x$ such that $0<|x-0|<\delta$ and $|f(x)-f(0)|=|x+3| \geq \epsilon.$

$\quad$Choose $\epsilon=1$. We must show that for a given $\delta$, then $\exists x$ such that $|x+3|\geq1$.

$\quad \quad$Let $a=x+3,b=-x,a+b=3$. By the Triangle Inequality, we have:

$\begin{align}
\quad \quad \quad \quad|a+b|\leq|a|+|b|&\Rightarrow |3|\leq|x+3|+|-x| \\
&\Rightarrow 3-|-x|\leq|x+3|
\end{align}\\$

$\begin{align}
\quad \quad \text{Let } \delta\leq1 &\Rightarrow 0 < x <1\\
&\Rightarrow 3 < x+3 < 4
\end{align}\\$

$\rule{18cm}{.03cm}$

I am unsure as to how to continue with the proof. I believe that after manipulating our $3<x+3<4$, we can use it to create an inequality such that $\epsilon=1<3...\leq|x+3|.$ Which would lead us to the following result:

If $\delta\leq1$, every $x$ where $|x|<\delta$ has $|x+3|\geq \epsilon=1$.

If $\delta >1$, then $(x \bigr ||x|<1)$ is contained in the set of $x$, where $|x|<\delta$. Thus the set of points where $\delta>1$ contains points where $|x+3|\geq \epsilon=1$.

Therefore $f(x)$ is discontinuous at $x=0$.

答:
You cannot start the proof with 'If $f$ is not continuous at $0$' because that is what you want to prove.

Suppose $f$ is continuous at $0$.  Then there exists $\delta >0$ such that $|f(x)-f(0)| <1$ if $|x-0| <\delta$. Now take $x=-\frac  1n$ with $n >\frac 1{\delta}$. Then $|x| <\delta$ but $|f(x)-f(0)|=3-\frac 1 n >1$. This contradcition proves the result.


问:
Birthday problem (combinatorics), without using inverse solution
There is that classic question about how many people in a room is required so that at least one pair of people share a birthday, with > 50% probability, the answer is 23. The standard textbook solution is to solve it using:

$$\mathbb{P}\left(\text{At least one shared birthday}\right) = 1 - \mathbb{P}\left(\text{zero shared birthdays}\right)$$

Since the probability of zero shared birthdays is easier to calculate/derive. From what I know, this is calculated as the series:

$$\mathbb{P}\left(\text{zero shared birthdays}\right)=\frac{365-1}{365}\times\frac{365-2}{365}\times\ldots\times\frac{365-22}{365}$$

the above makes sense, because for each successive person in the group, they can't share any birthday with any previous person, so the number of available days diminishes by 1 each time.

However, I am really struggling to derive $\mathbb{P}\left(\text{At least one shared birthday}\right)$, without using the inverse. Some sort of recursion/series is the way to go, something like:

$$\mathbb{P}\left(\text{23 people share at least 1 birthday}\right)=x+\mathbb{P}\left(\text{22 people share at least 1 birthday}\right)$$

I believe the base cases are:
$$\mathbb{P}\left(\text{1 person share a birthday}\right)=0$$
$$\mathbb{P}\left(\text{2 person share a birthday}\right)=1/365$$

Can someone help me derive it, perhaps by spelling out the series?

答:
Indeed, just take $$p_1=0\quad \quad p_n=p_{n-1}+\frac {n-1}{365}\times (1-p_{n-1})$$

The logic being: in order to have at least one pair with $n$ people, you must either have a pair with the first $n-1$ of them, or you must have no pair with the first $n-1$ but the $n^{th}$ person must match one of the prior $n-1$ birthdays.  And of course these two events are mutually exclusive.


问:
Solution of $f'(t) = f(t)^2$ when $f: U \subset \mathbb{R} \to \mathbb{C}$

Let $f: U \subset \mathbb{R} \to \mathbb{C}$ be a continuously differentiable function (that is, its component functions are continuously differentiable) defined on a suitable open subset $U$ of $\mathbb{R}$, such that $f'(t) = f(t)^2, \forall t \in U$.

Can  we determine the solution to the above equation by "abstracting" of the real-valued case? By this I mean that if we say $f(t) = x(t) + iy(t)$, then $f' = f^2$ implies that
$$
x'(t) = x(t)^2 - y(t)^2, y'(t) = 2x(t)y(t), \forall t \in U,
$$
so by existence and uniqueness, we above system has a unique solution for each initial condition, so we can determine $f$ from the real ODEs.

However, can we determine the solution using only the complex system, i.e. can we say that
$$
f(t) = \frac{1}{c-t}, \forall t \in U
$$
for some $c \in \mathbb{C}$ and then determine $x$ and $y$ from this?

I do not think that the above is correct. When I plug the real system into Wolfram Alpha, the solution is completely different.

Edit

I was thinking about this question when I tried to "merge" the real and the complex case.

I know that if $f$ would have been defined on a "bigger" subset of $\mathbb{C}$ (that would not include only real numbers) and if it were holomorphic, then we could the existence and uniqueness and standard results from the complex case to get the solution.

答:
Perhaps, you didn’t check the computations carefully because with $$x(t)=\frac{a-t}{(a-t)^2+b^2}\,,~\,~\,~\,~ y(t)=\frac{-b} {(a-t)^2+b^2}\,$$ then we have
$$x’(t)=\frac{(a-t)^2-b^2}{((a-t)^2+b^2)^2}=x(t)^2-y(t)^2\,,$$
$$y’(t)=\frac{-2b(a-t)}{((a-t)^2+b^2)^2}=2x(t)y(t)\,,$$
which is in perfect agreement.


问:
Explain why the number of integer solutions is equal to the number of integer partitions of $n$
I have to explain how the number of integer solutions to $x_1 + 2x_2 + 3x_3 + \dots + nx_n = n$ with $x_i \ge 0$ for $i = 1,2,\dots,n $ is equal to the number of integer partition of $n$.

I made an attempt. Assume $x_1 = a_1, x_2 = a_2, \dots , x_n = a_n$ Then we have a partition of $n$ equal to $a_1 + 2a_2 + 3a_3 + \dots + na_n = n$

$$\Rightarrow  (1 + 1 + \cdots + 1) + (2 + 2 + \cdots + 2) + \cdots + (n + n + \cdots + n) = n$$

Got up this point, but I don't know what to do next. Can you help?

答:
HINT: A partition of $n$ is completely specified when we know the number of parts of each size. For instance, the partition $2+2+2+1+1$ of $8$ is completely specified by saying that it has $3$ parts of $2$ and $2$ parts of $1$. Interpret the numbers $x_1,\ldots,x_n$ as numbers of parts in some way. I’ve put a further hint in the spoiler-protected box below.

 $x_1+x_2+\ldots+x_n$ is the total number of parts in the partition corresponding to that solution to the equation.

Added: Here is an example to illustrate more fully the correspondence between solutions in non-negative integers to $$1x_1+2x_2+3x_3+\ldots+nx_n$$ and partitions of $n$. Specifically, I’ll match each of the $7$ partitions of $5$ with the corresponding solution to $1x_1+2x_2+3x_3+4x_4+5x_5=5$.

$$\begin{align*}
&\color{red}5=\color{blue}5:\\
&1\cdot0+2\cdot 0+3\cdot0+4\cdot0+\color{blue}{5\cdot 1}=\color{red}5\\
&\color{blue}{x_5=1};x_1=x_2=x_3=x_4=0\\
&\\
&\color{red}5=\color{blue}{4+1}\\
&\color{blue}{1\cdot1}+2\cdot 0+3\cdot0+\color{blue}{4\cdot1}+5\cdot0=\color{red}5\\
&\color{blue}{x_4=1,x_1=1};x_2=x_3=x_5=0\\
&\\
&\color{red}5=\color{blue}{3+2}\\
&1\cdot0+\color{blue}{2\cdot1}+\color{blue}{3\cdot1}+4\cdot0+5\cdot0=\color{red}5\\
&\color{blue}{x_3=1,x_2=1};x_1=x_4=x_5=0\\
&\\
&\color{red}5=\color{blue}{3+1+1}\\
&\color{blue}{1\cdot2}+2\cdot0+\color{blue}{3\cdot1}+4\cdot0+5\cdot0=\color{red}5\\
&\color{blue}{x_3=1,x_1=2};x_2=x_4=x_5=0\\
&\\
&\color{red}5=\color{blue}{2+2+1}\\
&\color{blue}{1\cdot1}+\color{blue}{2\cdot2}+3\cdot0+4\cdot0+5\cdot0=\color{red}5\\
&\color{blue}{x_2=2,x_1=1};x_3=x_4=x_5=0\\
&\\
&\color{red}5=\color{blue}{2+1+1+1}\\
&\color{blue}{1\cdot3}+\color{blue}{2\cdot1}+3\cdot0+4\cdot0+5\cdot0=\color{red}5\\
&\color{blue}{x_2=1,x_1=3};x_3=x_4=x_5=0\\
&\\
&\color{red}5=\color{blue}{1+1+1+1+1}\\
&\color{blue}{1\cdot5}+2\cdot0+3\cdot0+4\cdot0+5\cdot0=\color{red}5\\
&\color{blue}{x_1=5};x_2=x_3=x_4=x_5=0
\end{align*}$$


问:
Where is the additional lambda while solving problems for an Exponential distribution?
I've been studying the problems for an exponential distribution and I have this somewhat basic question.

Known

The pdf in an exponential distribution where  $x \ge 0$ is given by:

$$ f(x) = \lambda e^{-\lambda x}  $$

Example Problem

Suppose that the amount of time one spends in a bank is exponentially distributed with mean ten minutes, that is, λ = 1/10 . What is the probability that a customer will spend more than fifteen minutes in the bank?

Example Solution

If X represents the amount of time that the customer spends in the bank, then the probability is just:

$$P\{X > 15\} = e^{-15\lambda} = e^{-3/2}$$

My question

Where is the other $\lambda$ from the pdf ? (i.e. why did we take the solution above to be $e^{-15\lambda}$ and not $\lambda e^{-15\lambda}$ ?)

答: $\lambda e^{-\lambda x}$ is the density function. The distribution function is given by $P(X \leq x) =1-e^{-\lambda x}$ which gives $P(X>x)=e^{-\lambda x}$.  $\lambda e^{-\lambda x}$ is the derivative of $1-e^{-\lambda x}$.


问:
Can every ordinal be reachable from $\aleph_0$, in ZF?
Define: $ x \text{ is reachable from } y \iff \\  \exists z (|z|\leq |y| \land \forall m \in z (|m| \leq |y|) \land |x|\leq|\bigcup (z)|)$

Given that ZF is consistent with only $\aleph_0$ being regular*, and so any ordinal supernumerous to $\aleph_0$ would have $\aleph_0$ as its confinality; then would this mean that ZF is consistent with saying that every ordinal is reachable from $\aleph_0$?

答:
(My previous answer was of course nonsense as Andreas Blass pointed out.)

In natural language your definition is: "$x$ is reachable from $y$ iff $x$ can be covered by $y$-many sets each of which has $y$-many elements." Of course in the absence of choice this is not the same as admitting a surjection from $y^2$.

So your question is asking whether it is consistent that every ordinal is a countable union of countable sets. In fact, already we can show in $\mathsf{ZF}$ that $\omega_2$ is not a countable union of countable sets (although per Gitik it is consistent that it has countable cofinality). See e.g. here.


问:
Prove by strong induction the Jacobstahl Series
I am just beginning to learn about strong induction, I can't figure out how to complete this homework question:

Jacobstahl numbers are defined as
$$J_1 = 1$$
$$J_2 = 1$$
$$J_n = J_{n-1} + 2J_{n-2}$$

Prove:
$$J_n = 2^{n-1}-J_{n-1}$$
for n >= 2.

Here is what I have done so far:

Assume:

$$J_k = 2^{k-1} + J_{k-1}$$
then
$$J_{k+1} = 2^{k+0} - J_{k+0}$$

Then from the definition:

$$J_{k+1} = 2^{k+0} - J_{k+0} = 2^{k-1}-J_{k-1}+2(2^{k-2}-J_{k-2})$$
$$2^{k+0} - J_{k+0} = 2^{k-1}-j_{k-1}+2^1*2^{k-2} - 2^1*J_{k-2}$$
$$= 2(2^{k-1})-J_{k-1}-2J_{k-2}$$
$$= 2^{k+0} - J_{k-1}-2J_{k-2}$$

So I have gotten part of the way there, by finding the $2^{k+0}$ term, but I'm not sure what to do with the rest of it.  I think I've probably gone wrong in setting up the assumption, but I can't work out how.

答:
Your assumption for strong induction should be that $J_k=2^{k-1}-J_{k-1}$ for $k=2,\ldots,n$; from that assumption you want to prove that $J_{n+1}=2^n-J_n$. If you then attempt the usual kind of argument, your calculation of $J_{n+1}$ will start like this:

$$\begin{align*}
J_{n+1}&=J_n+2J_{n-1}\\
&=\left(2^{n-1}-J_{n-1}\right)+2J_{n-1}\\
&=2^{n-1}+J_{n-1}\\
&=2^{n-1}+2^{n-2}-J_{n-2}\,.
\end{align*}$$

It’s pretty clear that if you continue in the most straightforward fashion, replacing $J_{n-2}$ by $2^{n-3}-J_{n-3}$, you’ll still have a $J_k$ term of some kind: you’re not obviously reaching what you want.

It actually is possible to get somewhere this way, but for a simple induction proof you’re better off looking at the desired relationship in a different way. Notice that $J_k=2^{k-1}-J_{k-1}$ is equivalent to $J_k+J_{k-1}=2^{k-1}$. Thus, we could just as well assume as our induction hypothesis that $J_k+J_{k-1}=2^{k-1}$ for $k=2,\ldots,n$ and try to prove that $J_{n+1}+J_n=2^n$. Then

$$J_{n+1}+J_n=J_n+2J_{n-1}+J_n\,;$$

now just simplify that and apply the induction hypothesis. (You’ll find that you don’t actually need strong induction here: ordinary induction is enough.)


问:
Why does separation of variable gives the general solution to a PDE
I was reading about physics and came across the method of using separation of variables to solve specific PDEs, but I can't figure out why the specific solutions give rise to the general solution (the book didn't give any explanation for all these).

The specific example in the book was the Laplace Equation in $2$ variables:
$$\frac {\partial^2 V}{\partial x^2}+\frac {\partial^2 V}{\partial y^2}=0$$
For the above example, separation of variable is essentially solving for the eigen-vectors of the operator $\frac {\partial^2 }{\partial x^2}$ and $\frac {\partial^2 }{\partial y^2}$, which are Hermitian and commutes with each other. I know that in the finite dimensional case, such operators are simultaneously diagonalizable, then solving for the eigen-vectors will give all the solution, but I'm not sure does this work for infinite dimension. I'm also not sure does this approach works in the general case, for other PDEs that can be solved by separation of variable.

All the other post I find on here are all explaining how or when separation of variable work, instead of why such techniques will give the general solutions.

Another side question is: What kind of classes will cover these topics? The only undergraduate class that seems relevant at my university is Linear Analysis, which doesn't cover this. The graduate PDE sequence have graduate Real Analysis sequence as pre-requisite, which I don't think I'll be able to take soon.

答:
There are several key ingredients I will briefly describe here. I won't go into too much detail as you've mentioned that you don't have a graduate real analysis background yet. But indeed a full description of the theory is a standard part of a graduate course in linear PDE. So I hope that answers your side question as well.

We start with a strongly elliptic linear operator (such as the Laplacian) and, along with some nice boundary condition, we restrict to some appropriate solution (Hilbert) space.

In that solution space, we can prove under fairly general conditions that the eigenvalues of the operator are countable and that eigenvectors (eigenfunctions) form an orthogonal basis for the solution space. This is the infinite-dimensional generalization of the diagonalizability result from regular matrix theory. The proof relies on the spectral theorem for compact operators. The key here is that, up to a shift, the inverse of a strongly elliptic operator is compact.

This demonstrates that if we can construct all the eigenvectors of the operator, the general solution can be written as a decomposition of these eigenvectors.

It remains to find the eigenvectors; in special cases (most famously, 2D Laplacian on a rectangle) this can be done via separation of variables. Therefore it remains to address "Why does separation of variables produce all eigenvectors?" To answer this question, we note that we proved that the eigenvectors form a complete basis. Next, we see that because of the specific symmetry of the Laplacian on the rectangle, using separation of variables reduces the problem to a pair of second-order equations in one-dimension; in this process we produce the eigenvectors of these one-dimensional operators, and then from the existing theory (in particular, Sturm-Liouville theory) we know that we have produced a set of functions that span the space. As we have produced a basis, no other eigenvectors are needed to form a general solution.


问:
Fliping 11 coins with $0$ and $1$ on their sides, probability of the event $\text{ there are more zeros than ones }$
you flip 11 coins wich has sides of $1$ and $0$, what is the probability of the event $$\text{ there are more zeros than ones }$$

my first attempt was just write $$6\cdot \frac{1}{2}$$ but first of all this is not resonable because it is bigger than 1. then I realised 2 things:

I dont understand how to determine my probability space

How do I considers another cases with there is more zeros than ones, for example I got 7 zeros and 4 ones.

I hope some 1 can help

答: The answer is of course $1/2$, because there are only two equally likely outcomes:  more $0$s or more $1$s.


答:
We need the area under of all of the curve to the right of $15$. The indefinite integral is $-e^{-\lambda x}$. The definite integral is $F(\infty) - F(x)$. Since $-e^{-\infty}=0$, this simplifies to gives us $e^{-\lambda x}$.

In fact, the reason we have the extra $\lambda$ in the PDF is to get it to integrate to $1$. Without the $\lambda$, the integral of the PDF from $0$ to $\infty$ would be $\frac 1 {\lambda}$. Since the total probability has to be $1$, we have to normalize the PDF.

Title: Where is the additional lambda while solving problems for an Exponential distribution?

Body:
I've been studying the problems for an exponential distribution and I have this somewhat basic question.

Known

The pdf in an exponential distribution where  $x \ge 0$ is given by:

$$ f(x) = \lambda e^{-\lambda x}  $$

Example Problem

Suppose that the amount of time one spends in a bank is exponentially distributed with mean ten minutes, that is, λ = 1/10 . What is the probability that a customer will spend more than fifteen minutes in the bank?

Example Solution

If X represents the amount of time that the customer spends in the bank, then the probability is just:

$$P\{X > 15\} = e^{-15\lambda} = e^{-3/2}$$

My question

Where is the other $\lambda$ from the pdf ? (i.e. why did we take the solution above to be $e^{-15\lambda}$ and not $\lambda e^{-15\lambda}$ ?)


问:
Are the following union the same as the intersection?
I solved this problem, however, then given solution was different then mine in the following way: For a set $D$ defined as $D:= \{x \in E: f(x) < 0\}$, I wrote that I could define the following set
$D_k := \{x \in E: f(x) < \frac{1}{k}\}$, hence, the following holds:

$$\bigcap_{k=1}^{\infty}D_k = D.$$

Is this not the case? I think, if anything, there might be a problem at $f(x)=0$?

答:
If $f(x)=0$ then $x \in\bigcap D_k$ but $x \notin D$.

$\bigcap D_k$ is $\{x \in E: f(x) \leq 0\}$.


问:
Linear Algebra: Rowspace, column space (Concept question)
freshman student here looking to clarify the concepts.

1.) Is the column space also the solution space of a matrix?

2.) Lets say a matrix is formed by a1,a2,a3,a4,a5.  I find the rref form of this matrix; turns out a1,a2,a3 are linearly independent.  Can it suffice to say that a1,a2,a3 are the basis for both solution space and column space of a matrix?  Can this tell me anything about row space?

3.) What is the purpose of the row space?

4.) Is the basis for row space also the basis for the solution space in a square matrix?  Here on out it becomes increasingly difficult for me to visualize.

5.) Apart from dimension theorem, what other purpose does rank actually serve?  So I am familiar with the formulas regarding rank; rank <= min(n,m) or rank = min(dimension(rowspace),dimension(colspace)), but I am not actually sure I understand the implications of this.

6.) If dim(rowspace)<dim(colspace), can this tell me anything about the solution space or any implications (apart from implications on the rank)?

Sorry for the long-winded question.  Hope anyone can provide insight on anything above, for me and my friends.

答:
The column space is range or the image of the transformation.  This is analogous to the range of single variable functions.  i.e. the set of all possible outputs of $Ax.$

If a $5\times 5$ matrix has a rank of 3, then you can choose any set of 3 independent column vectors as a basis of the column space.

The row-space is a little bit less intuitive to me.  It can be thought a basis of set of all linear equations consistent with the matrix.  Or, you could think of it as the basis of the set of vectors that is orthogonal to the kernel of the matrix.

The rank of a matrix tells me how much information is getting compressed by this transformation.  If the matrix is full rank, it isn't getting compressed at all.  If it a rank 2 matrix, then the output maps onto a plane.

The rank nullity theorem says that the Dimension of the range plus the dimension of the kernel equals the dimension of the domain.

The dimension of the row space always equals the dimension of the column space which equals the rank of the matrix.

Does this help?


答:
By symmetry, the probability of having more heads than tails will be exactly equal to the probability of having more tails than heads, given the coin is fair.

You cannot have an equal number of heads than tails with an odd number of coins, so the required probability is simply $\frac 12$.

Slightly more formally, let the probability of more heads than tails be $p$. Then the probability of more tails than heads is also $p$. Finally, let the probability of equal heads and tails be $q$. You have that $p +p +q =1 \implies p = \frac{1-q}2$. Now note that $q=0$ for an odd number of coins (it would be non-zero for an even number). Hence, $p = \frac 12$.

Title: Fliping 11 coins with $0$ and $1$ on their sides, probability of the event $\text{ there are more zeros than ones }$

Body:
you flip 11 coins wich has sides of $1$ and $0$, what is the probability of the event $$\text{ there are more zeros than ones }$$

my first attempt was just write $$6\cdot \frac{1}{2}$$ but first of all this is not resonable because it is bigger than 1. then I realised 2 things:

I dont understand how to determine my probability space

How do I considers another cases with there is more zeros than ones, for example I got 7 zeros and 4 ones.

I hope some 1 can help


问:
The set of differences for a set of positive Lebesgue measure
Quite a while ago, I heard about a statement in measure theory, that goes as follows:

Let $A \subset \mathbb R^n$ be a Lebesgue-measurable set of positive measure. Then we follow that $A-A = \{ x-y \mid x,y\in A\}$ is a neighborhood of zero, i.e. contains an open ball around zero.

I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5):

Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.

The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence $\{x_n\}$ converging to zero such that none of the elements are contained in $A$, we might be able to define an ascending/descending chain $A_n$ such that the union/intersection is $A$ but the limit of its measures zero. I am in lack of ideas for the definition on those $A_n$.

I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help.

答: If you need to find information on the subject, the first proof of the fact that the set of differences contains a neighbourhood of the origin is (for Lebesgue measure on the line) due to Steinhaus.  There is a substantial collection of generalizations of the result.  </p>

ContentLicense: CC BY-SA 3.0


问:
Set Theory: show that function h is injective
If $h$ belongs to the set of functions from $X$ to $X$, why is $h$ injective if and only if, for any functions $f$ and $g$ in that set, if $h \circ f =  h \circ g$, then $f=g$?

I'm trying to understand this question about set theory. I'm studying Halmos's Naive Set Theory book. I'll post my progress later (if any :/)

So, h composite f means the domain of h equals the range of f. h composite g means the domain of h equals the range of g. then the range of f and g are the same.

答:
$h$ is injective iff $h(x)=h(y)$ implies $x=y$.

Suppose $h$ is injective. Then if $(h \circ f)(x) = h(f(x))= (h \circ g)(x) = h(g(x))$, it follows that $f(x)=g(x)$. Since $x$ is arbitrary, we have $f=g$.

Now suppose that $(h \circ f) = (h \circ g)$ implies $f=g$. Now suppose $h(x)=h(y)$. Let $f(t) = x, g(t) = y$ for all $t$ in the domain. Then $h(x)=h(y)$ implies $(h \circ f)(t) = h(x)= (h \circ g)(t) = h(y)$ for all $t$, or in otherwords, $(h \circ f) = (h \circ g)$. Hence $f=g$, and so $f(t)=g(t)$, or in other words, $x=y$. Hence $h$ is injectve.


答:
Basically, the situation when there are more 0s than 1s are the cases where you flipped 6 times 0s + 7 times + ...  until you hit 11 times.

The probability that you flipped $n$ zeros out of $11$ is:

$$ \binom{11}{n}  \left( \frac{1}{2} \right)^{11}$$
Because we choose $n$ coins out of $11$ and then calculating each probability that it hits a zero or a one: $ \overset{\text{0s}}{\frac{1}{2^n}} \cdot \overset{\text{1s}}{\frac{1}{2^{11-n}}}$

You need to calculate the sum:

$$ \sum_{n = 6}^{11}  \binom{11}{n}  \left( \frac{1}{2} \right)^{11}$$

You can also interpret this question how David has, because this question is a dichotomy (Either you get more 0s or more 1s)


问:
Extend Petersen's Theorem
This was a homework problem.

Petersen's Theorem: Every cubic, bridgeless graph contains a perfect matching.

Show that Petersen’s theorem (Theorem 8.11) can be extended somewhat by proving that if $G$ is a bridgeless graph, every vertex of which has degree $3$ or $5$ and such that $G$ has at most two vertices of degree $5$, then $G$ has a $1$-factor.

My idea was about splitting into $3$ cases, $0$ $5$-degree vertices, $1$ $5$-degree vertice and $2$ $5$-degree vertices. With $0$, it's obvious, but I can't figure out $1$ and $2$.

This is what I've done.

Case 1: $G$ has $0$ vertices of degree $5$.
Automatically proven by using thm 8.11 (every $3$-regular bridgeless graph contains a $1$-factor)

Case 2: $G$ has $1$ vertice of degree $5$
Consider $S$ is a subset of $V(G)$

Let $|S| = k$ and $|Ko(G - s)| = j$. ($Ko$ being the number of components of a graph)

$3  (k-1) + 5 \ge 3j$ or $3  (k-1) + 5 \ge 3(j-1) + 5$

$3  (k-1) + 5 \ge 3j \implies 3k - 3 + 5 \ge 3j \implies 3k + 2 \ge 3j \implies k + ⅔ \ge j$

or $3  (k-1) + 5 \ge 3j-3+5 \implies 3 k \ge j$

Case 3: G has 2 vertices of degree 5

$3  (k-2) + 5(2) \ge  3j \implies 3k - 6 + 10 \ge 3j \implies 3k + 4 \ge 3j \implies k + 4/3 \ge j$

or $3(k-2) + 5(2) \ge 3(j-1)+5 \implies 3k+4 \ge 3j+2 \implies 3k+2 \ge 3j$

or $3(k-2) + 5(2) \ge 3(j-2)+10 \implies 3k+4 \ge 3j+4 \implies k \ge j$

Where do I go from here? Is my setup completely wrong?

答: There are some typos/mistakes in the inequalities in the cases $1$ and $2$. After they are fixed you get in case $1$ that $k\geq j+2/3$. However since they are both integers, $k\geq j$, which is what you need to apply Tutte's theorem. In case $2$ you only get $k\geq j+4/3$, which means that either $k\geq j$ (which is good), or, potentially, that $k=j-1$, which would be bad. However, this is actually not possible: either there is at least one edge between the vertices if $S$ or there isn't. If there is then $3(k−2)+5(2)-2\geq 3j$ and we are back to $k\geq j+2/3$. But if there isn't then all $3k+4$ edges go out to $j=k+1$ components, with all of them getting odd number and at least 3 each - which is is not possible (you can check why). So in all cases $k\geq j$ and you can apply Tutte's theorem.

Title: Extend Petersen's Theorem


问:
$y'+a(t)y=f(t)$, $\lim_{t\to \infty} f(t)= 0 \implies \lim_{t\to \infty} y(t) = 0 $
I've come across the following ODE problem (1.2.19 Braun,4th edition, Differential Equations and Their Applications):

given $a(t)\geq a_{min} > 0$ and $\lim_{t\to \infty}f(t) = 0$, with $a$ and $f$ continuous on $(-\infty, \infty)$, every solution $y$ of $y'+a(t)y=f(t)$ vanishes as t tends to infinity.

The integrating factor for this ODE is $\mu (t) = e^{\int a(t)} \geq e^{a_{min} \int dt} = e^{a_{min}t}$.

Hence we have $$\int (\mu (t)y)'dt = \int \mu(t)f(t)dt\\ e^{\int a(t) dt}y(t)= \int e^{\int a(t)}f(t)dt \\ y(t) = e^{-\int a(t) dt} \int e^{\int a(t)}f(t)dt \\ \ . $$

This is where I get stuck.

I tried using the information on $f$ by rewriting it as a product of a bounded part $b(t)$ and an unbounded part $d(t) > 0$, with $d(t) \to \infty $ as $t \to \infty$: $$f(t) = \frac{b(t)}{d(t)}\ . $$

Now, I guess, I need to show that $e^{-\int a(t) dt} $ vanishes faster than $\int e^{\int a(t)}f(t)dt$ grows as $t\to \infty$. I don't see how.

答:
Let's let $Y=\lim_{t\to\infty}y(t)$. First, let's say that $Y$ is positive, or that $Y=\infty$. Then for some real $\delta_1$, and some fixed positive real $r$, whenever $t>\delta_1$, we have $r<y(t)$. (Which positive real $r$, you may ask? If $Y$ is positive and finite, then $\frac Y2$ works. If $Y$ is infinite, pick your favorite number.)

So then for $t>\delta_1$ we have $y'=f(t)-a(t)y<f(t)-a_{min}r$. Now, we are given that $f$ approaches $0$, so for some real $\delta_2$, whenever $t>\delta_2$ we have that $f(t)<\frac{a_{min}r}2$. So, whenever $t>\max{(\delta_1,\delta_2)}$, we have that $y'(t)<-\frac 12a_{min}r$.

The mean value theorem guarantees that if $\max(\delta_1,\delta_2)\leq t_1<t_2$ then $y(t_2)-y(t_1)\leq-\frac12a_{min}r(t_2-t_1)$. So, if $t_1>\max(\delta_1,\delta_2)$ then $y(t)\leq y(t_1)-\frac12a_{min}r(t-t_1)$, where the right hand side of that final inequality approaches $-\infty$, and so $Y=-\infty$.

So, if $Y$ is positive (or if the limit approaches positive infinity) then $Y=-\infty$. In much the same way, if $Y$ is negative (or negative infinity) then $Y=\infty$. Either way, we get a contradiction, and so the only conclusion left is that $Y=0$.


答:
In fact, the solution of the equation is
$$ y(t) = e^{-\int_0^t a(s) ds}\bigg[y_0+ \int_0^t e^{\int_0^s a(\tau)d\tau}f(s)ds\bigg]. $$
Since $\lim_{t\to\infty}f(t)=0$, for any $\epsilon>0$, there is $N_1>0$ such that for $t>N_1$, $|f(t)|<\frac12 a_{min}\epsilon.$
Thus for $t>N_1$,
\begin{eqnarray}
|y(t)| &\le& e^{-\int_0^t a(s) ds}\bigg[|y_0|+ \int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds+\int_{N_1}^t e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]\\
&\le& e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min} \epsilon\int_{N_1}^te^{-\int_s^t a(\tau) d\tau}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min}\epsilon \int_{N_1}^te^{-\int_s^t a_{min} d\tau}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12 a_{min}\epsilon \int_{N_1}^te^{-a_{min}(t-s)}\\
&\le&e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]+\frac12(1-e^{-a_{min}(t-{N_1})})\epsilon.
\end{eqnarray}
Since
$$\lim_{t\to\infty}e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]=0, $$
there is $N_2>0$ such that when $t>N_2$,
$$ e^{-a_{min}t}\bigg[|y_0|+\int_0^{N_1} e^{\int_0^s a(\tau)d\tau}|f(s)|ds\bigg]<\frac12\epsilon. $$
Let $N=\max\{N_1,N_2\}$ and then when $t>N$,
$$ |y(t)|<\frac12\epsilon+\frac12\epsilon=\epsilon. $$
So
$$ \lim_{t\to\infty}y(t)=0. $$

Title: $y'+a(t)y=f(t)$, $\lim_{t\to \infty} f(t)= 0 \implies \lim_{t\to \infty} y(t) = 0 $

Body:
I've come across the following ODE problem (1.2.19 Braun,4th edition, Differential Equations and Their Applications):

given $a(t)\geq a_{min} > 0$ and $\lim_{t\to \infty}f(t) = 0$, with $a$ and $f$ continuous on $(-\infty, \infty)$, every solution $y$ of $y'+a(t)y=f(t)$ vanishes as t tends to infinity.

The integrating factor for this ODE is $\mu (t) = e^{\int a(t)} \geq e^{a_{min} \int dt} = e^{a_{min}t}$.

Hence we have $$\int (\mu (t)y)'dt = \int \mu(t)f(t)dt\\ e^{\int a(t) dt}y(t)= \int e^{\int a(t)}f(t)dt \\ y(t) = e^{-\int a(t) dt} \int e^{\int a(t)}f(t)dt \\ \ . $$

This is where I get stuck.

I tried using the information on $f$ by rewriting it as a product of a bounded part $b(t)$ and an unbounded part $d(t) > 0$, with $d(t) \to \infty $ as $t \to \infty$: $$f(t) = \frac{b(t)}{d(t)}\ . $$

Now, I guess, I need to show that $e^{-\int a(t) dt} $ vanishes faster than $\int e^{\int a(t)}f(t)dt$ grows as $t\to \infty$. I don't see how.


问:
Solving a cubical nonlinear matrix equation
I'm trying to solve an matrix optimization problem. The stationary equation drawn from the opt problem is to find the solution for $X$ of the following nonlinear matrix equation:

$AXX^TAX-AXB+C=0$

where $A$, $B$, $C$ are known. $A$ and $B$ are symmetric.

Thanks!

答:
We want to find a root of the matrix function $F(X)=0$ where
$$\eqalign{
 F &= AXX^TAX - AXB + C \\
}$$
Calculate the differential
$$\eqalign{
dF &= A\,dX\,X^TAX + AX\,dX^TAX + AXX^TA\,dX - A\,dX\,B \\
}$$
Then vectorize the matrices, and calculate the Jacobian
$$\eqalign{
 f &= {\rm vec}(F) \\
df &= {\rm vec}(dF) \\
  &= (X^TA^TX\otimes A)dx + (X^TA^T\otimes AX)P\,dx
   + (I\otimes AXX^TA)dx - (B^T\otimes A)dx \\
  &= \Big((X^TA^TX\otimes A) + (X^TA^T\otimes AX)P
   + (I\otimes AXX^TA) - (B^T\otimes A)\Big)dx \\
  &= J\,dx \\
}$$
where $P$ is the permutation matrix required to vectorize $X^T,\;$ i.e.
$${\rm vec}(X^T) = P\,{\rm vec}(X)\\$$

Now apply Newton's insight and solve for the $dx$ which produces $\,df=-f$
$$\eqalign{
J\,dx &= -f \qquad\implies\quad dx &= -J^{-1}f \\
}$$
which suggests the following iterative scheme
$$\eqalign{
F_k &= F(X_k) \\
f_k &= {\rm vec}(F_k)\\
x_k &= {\rm vec}(X_k) \\
J_k &= (X_k^TA^TX_k\otimes A) + (X_k^TA^T\otimes AX_k)P
     + (I\otimes AX_kX_k^TA) - (B^T\otimes A) \\
x_{k+1} &= x_k - J_k^{-1}f_k \\
X_{k+1} &= {\rm Reshape}(x_{k+1}) \\\\
}$$
This is just the basic algorithm. To improve its convergence you'll probably need to consider techniques like line-searches, trust-regions, etc.


问:
Basis of Kernel of Linear Transformation
Determine a basis of $ker(T)$ and $T(R^5)$ for a linear map $T : R^5 → R^3$ whose matrix (relative to the standard basis) equals

$\begin{bmatrix}1 & 0 & 0 & 5 & 9\\
0 & 1 & -1 & -3 & 2\\
0 &2 & -2 & -6 & 4
\end{bmatrix}$

To find the basis for kernel, I just let the matrix time a vector [v1,...,v5] s.t. it equals 0 and solve for the basis. For $T(R^5)$ basis, do I just reduce the matrix to rref and find the basis of the rref?

In addition, what does it mean "(relative to the standard basis)" in the question?

答:
You find the kernel if you evaluate the augmented matrix

\begin{pmatrix}
    1 & 0 & 0 & 5 & 9 &\bigm| & 0 \\
    0 & 1 & -1 & -3 & 2&\bigm| & 0 \\
    0 & 2 & -2 & -6 & 4 &\bigm| & 0
\end{pmatrix}

(this is equivalent to solving $Ax = 0$).

The remainining linearly independed row vectors generate the kernel.

Relative to the standard basis means that your matrix has been evaluated on the standard basis $e_1,e_2,...,e_5$ of $\mathbb{R}^5$. In other words: the columns of your matrix are the images of each standard basis vector

$$T(e_1) = \begin{pmatrix}
    1 \\ 0 \\ 0 \end{pmatrix}, T(e_2) = \begin{pmatrix}
    0\\ 1 \\ 2 \end{pmatrix},... $$

and so on and so forth.


问:
How to write an algorithm as an equation?
Alright, it's with trepidation that I ask this, because dear god do I feel stupid. A question asked:
Whenever we run this formula we double x, then subtract 3 from the result. x = 5. What's the result if we do this 4 times? What's the result if we do it 12 times?
5*2 = 10, 10 - 3 = 7. 7*2 = 14, 14-3 = 11 and so on.

At first I thought it would be simple: ((x*2)-3)*12; but that really doesn't come close to the correct value. Then I thought "aha, exponentiation!" but that's clearly not correct either.

Truth be told, I've never felt so bad at math as I do right now. Right now I just wonder: how would you write that? Where can I learn more on this? It's obvious that I'm going to need to refresh my math skills (and I want to learn more, because math is cool but it's been a while since I did any).

Additionally I feel a bit ashamed asking such a simple question here, because judging by the "similar questions" list I'm the village idiot. The village itself seem to consist of people who are amazingly good at math, so yeah -- it sucks to have to admit to being lost at something like this.

In fact, I'm so lost I don't even know what tags I should use.

Edit: I know the answer (I wrote a short for loop), I just want to know how I'd approach it mathematically instead.

Body:
For the particular function you have, we can figure out $f^{(n)}(x)$.

$f^{(1)}(x)=2x-3$

$f^{(2)}(x)=2(2x-3)-3=2^2x-(2+1)3$

$f^{(3)}(x)=2(2^2x-(2+1)3)-3 = 2^3x-(4+2+1)3$

$f^{(4)}(x)=2(2^3x-(4+2+1)3)-3 = 2^4x - (8+4+2+1)3$

The pattern is $f^{(n)}(x)=2^nx-(2^n-1)3$, which can be rearranged as $2^n(x-3)+3$.  To prove this I would use induction.


答:
Relative to the standard basis just means we don't have to use a change of basis matrix to compute the kernel.  So we get $\begin{bmatrix}1 & 0 & 0 & 5 & 9\\
0 & 1 & -1 & -3 & 2\\
0 &2 & -2 & -6 & 4
\end{bmatrix}\to\begin{bmatrix}1 & 0 & 0 & 5 & 9\\
0 & 1 & -1 & -3 & 2\\
0 &0 & 0 & 0 & 0
\end{bmatrix}$.  Now "back-substitute".  Get $b-c-3d+2e=0,a+5d+9e=0$.  So $\{(-9,-2,0,0,1),(-5,3,0,1,0),(0,1,1,0,0)\}$ is a basis for the kernel.

Title: Basis of Kernel of Linear Transformation

Body:
Determine a basis of $ker(T)$ and $T(R^5)$ for a linear map $T : R^5 → R^3$ whose matrix (relative to the standard basis) equals

$\begin{bmatrix}1 & 0 & 0 & 5 & 9\\
0 & 1 & -1 & -3 & 2\\
0 &2 & -2 & -6 & 4
\end{bmatrix}$

To find the basis for kernel, I just let the matrix time a vector [v1,...,v5] s.t. it equals 0 and solve for the basis. For $T(R^5)$ basis, do I just reduce the matrix to rref and find the basis of the rref?

In addition, what does it mean "(relative to the standard basis)" in the question?


答:
The 3 row is a multiple of the second row. We have 2 independent row vectors.  The dimension of the kernel will be 3.

$\begin{bmatrix} 0\\1\\1\\0\\0\end{bmatrix}$ is in the kernel.

We need two more.

Consider,
$\begin{bmatrix} -5\\u\\v\\1\\0\end{bmatrix},\begin{bmatrix} -9\\w\\x\\0\\1\end{bmatrix}$

We know that whatever the values of $u,v,w,x$ are, they will first row will multiply to $0.$  Find values that make $0$ when multiplied by the other rows.

Clearly these 3 vectors are independent.


答:
The answer by @Christopher is very complete and definitely better than what this answer will be. But I would like to make some comments on Separation of Variables.

Separation of Variables is a process of splitting a multi-dimensional problem, into several single dimensional problems. However, this relies on an inherent symmetry of the domain, which itself determines the coordinates which allow for separation of variables.

If the question is posed in a rectangle, then it is quite natural that the problem given in rectangular coordinates can be broken down into two one dimensional problems in each orthogonal dimension. If the problem is posed on a circle, then polar coordinates are required. However, if the problem is given on a completely arbitrary domain then it is unlikely that you could find a coordinate system that can reflect the symmetry of the domain and allow for separation of variables.

If you get deeper into Lie theory, one can describe a group theoretic method of determining the possible coordinate systems that allow for a given equation to be separable. However, I don't think I have a deep enough understanding on this to comment further.

Title: Why does separation of variable gives the general solution to a PDE

Body:
I was reading about physics and came across the method of using separation of variables to solve specific PDEs, but I can't figure out why the specific solutions give rise to the general solution (the book didn't give any explanation for all these).

The specific example in the book was the Laplace Equation in $2$ variables:
$$\frac {\partial^2 V}{\partial x^2}+\frac {\partial^2 V}{\partial y^2}=0$$
For the above example, separation of variable is essentially solving for the eigen-vectors of the operator $\frac {\partial^2 }{\partial x^2}$ and $\frac {\partial^2 }{\partial y^2}$, which are Hermitian and commutes with each other. I know that in the finite dimensional case, such operators are simultaneously diagonalizable, then solving for the eigen-vectors will give all the solution, but I'm not sure does this work for infinite dimension. I'm also not sure does this approach works in the general case, for other PDEs that can be solved by separation of variable.

All the other post I find on here are all explaining how or when separation of variable work, instead of why such techniques will give the general solutions.

Another side question is: What kind of classes will cover these topics? The only undergraduate class that seems relevant at my university is Linear Analysis, which doesn't cover this. The graduate PDE sequence have graduate Real Analysis sequence as pre-requisite, which I don't think I'll be able to take soon.


问:
Proving that V is open
Let $n ∈ \mathbb{N}$, let $a \in \mathbb{R^n}$, let $s,r \in R$ with $s < r$ and $V = \{x \in \mathbb{R^n} : s<||x-a||<r\}$. Prove that V is open.

I'm not too sure how to go about this one here, but for $V$ to be open, for every $a \in V$ there exists some $\epsilon >0$ such that $B_{\epsilon}(a) \subseteq V$.

From what I can see we just need to take some $r \in V$ and show that $B_{\epsilon}(r) \subseteq V$. If this is the case, to prove that this ball is a subset would I actually go about it by taking some element in the ball and finding that element in $V$ or am I just confused about the problem? My definitions are coming from real analysis because I am learning this information from a real analysis text as well.

Title: Proving that V is open

Body:
Take some $x_0\in V$. Then $s<\|x_0-a\|<r$. Now take $\epsilon=\min\{r-\|x_0-a\|,\|x_0-a\|-s\}>0$ and consider $B_\epsilon(x_0)=\{y\in\Bbb R^n:\|y-x_0\|<\epsilon\}$.

If $y\in B_\epsilon(x_0)$ then $\|y-a\|\le\|y-x_0\|+\|x_0-a\|<\epsilon+\|x_0-a\|\le r-\|x_0-a\|+\|x_0-a\|=r$; and $s=s-\|x_0-a\|+\|x_0-a\|\le-\epsilon+\|x_0-a\|<-\|y-x_0\|+\|x_0-a\|\le\|y-a\|$. Therefore $s<\|y-a\|<r$, so $y\in V$.

We conclude $B_\epsilon(x_0)\subseteq V$, so $V$ is open.


问:
MaxMin of Sum of Fractionals Optimization Problem
Let $\mathbf{x}^{(1)},\ldots,\mathbf{x}^{(L)}$ be $L$ vectors of $N$ variables. Then, how can I solve the following fractional optimization problem?
\begin{align}
\max_{\mathbf{x}^{(1)},\ldots,\mathbf{x}^{(L)}}&\min_n\quad \sum_{\ell} \frac{x_n^{(\ell)}}{\alpha+\sum_{m}\beta_m^{(n,\ell)}x_m^{(\ell)}}\\
\text{subject to}&\quad\sum_{\ell=1}\mathbf{A}^{(\ell)}\mathbf{x}^{(\ell)}\leq \mathbf{1},
\end{align}
where $\alpha$ and $\beta_m$'s are non-negative constants. Moreover, $\mathbf{A}^{(\ell)}$'s are $M\times N$ matrices with non-negative entries. Inequlity is entry-wise and $\mathbf{1}$ is all-one vector.

答:
As I explained before, the problem is difficult. Here is an approach that will give an approximate solution (feasible but not necessarily optimal). The difficulty in this approach is the distribution of the budget constraints.

For each $\ell$, you first estimate a Pareto frontier of $N$ objectives:
\begin{align}
\max_{\mathbf{x}^{(\ell)}}& \frac{x_n^{(\ell)}}{\alpha+\sum_{m}\beta_m^{(n,\ell)}x_m^{(\ell)}}\\
\text{subject to}&\quad\mathbf{A}^{(\ell)}\mathbf{x}^{(\ell)}\leq \mathbf{b}^{(\ell)}.
\end{align}
Some Pareto optimization techniques use a weighted sum of the $N$ objectives, which does not simplify the problem at all. You should use a method where you optimize one objective while constraining the other objectives. The right hand side in the constraints should be picked in a way such that $\sum_\ell \mathbf{b}^{(\ell)} = \mathbf{1}$, for example via $\mathbf{b}^{(\ell)} = \mathbf{1}/L$. The result of this step is a large set of Pareto-optimal solutions (say $(x^{(\ell,k)})_{k\in K}$).

The next step is to pick one $x^{(\ell)}$ for each $\ell$ that optimizes the original objective. Precompute $p_{\ell n k} := (x_n^{(\ell,k)}) / (\alpha+\sum_{m}\beta_m^{(n,\ell)}x_m^{(\ell,k)})$. The following mixed integer linear optimization problem picks the optimal $x$ within the set of Pareto optimal solutions:
$$\max_y \min_n \{ \sum_{\ell,k} p_{\ell nk}y_{\ell k} : \sum_k y_{\ell k}=1 \; \forall \ell, \; y_{\ell k} \in\{0,1\} \}, $$
which you can solve as:
$$\max_{y,t} \{ t : t \leq \sum_{\ell,k} p_{\ell nk}y_{\ell k} \; \forall n, \; \sum_k y_{\ell k}=1 \; \forall \ell, \; y_{\ell k} \in\{0,1\} \}. $$


问:
Solution of $f'(t) = f(t)^2$ when $f: U \subset \mathbb{R} \to \mathbb{C}$

Let $f: U \subset \mathbb{R} \to \mathbb{C}$ be a continuously differentiable function (that is, its component functions are continuously differentiable) defined on a suitable open subset $U$ of $\mathbb{R}$, such that $f'(t) = f(t)^2, \forall t \in U$.

Can  we determine the solution to the above equation by "abstracting" of the real-valued case? By this I mean that if we say $f(t) = x(t) + iy(t)$, then $f' = f^2$ implies that
$$
x'(t) = x(t)^2 - y(t)^2, y'(t) = 2x(t)y(t), \forall t \in U,
$$
so by existence and uniqueness, we above system has a unique solution for each initial condition, so we can determine $f$ from the real ODEs.

However, can we determine the solution using only the complex system, i.e. can we say that
$$
f(t) = \frac{1}{c-t}, \forall t \in U
$$
for some $c \in \mathbb{C}$ and then determine $x$ and $y$ from this?

I do not think that the above is correct. When I plug the real system into Wolfram Alpha, the solution is completely different.

Edit

I was thinking about this question when I tried to "merge" the real and the complex case.

I know that if $f$ would have been defined on a "bigger" subset of $\mathbb{C}$ (that would not include only real numbers) and if it were holomorphic, then we could the existence and uniqueness and standard results from the complex case to get the solution.

答:
If $f(t) = F(t,c)$ is an analytic function of $(t,c)$ in some (open, connected) domain $D \subseteq \mathbb C^2$ containing the product of real intervals $(a,b) \times (c_0, c_1)$ and satisfies the differential equation
$f'(t) = G(f(t),t)$ there for real $t$ and $c$, where $G$ is an analytic function of two variables, then  by analytic continuation it satisfies the differential equation in all of $D$.  Thus a general solution that works for real $c$ will also work for complex $c$.  In particular this applies to your differential equation: $f' = f^2$: the general solution $F(t,c) = 1/(c-t)$ is always a solution to the differential equation on $\{(t,c): t \ne c\}$.


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答:
For a real number $x$,
$$\lfloor x\rfloor=\max\{n\in\mathbb{Z}\mid n\leq x\}.$$
I'd like to add though, that "$\lfloor x\rfloor$" is mathematical notation, just as much as the right side of the above equation is; the right side might use more "basic" constructions, but you can then ask about
$$\max,\qquad\in,\qquad \mathbb{Z},\qquad {}\mathbin{\mid}{},\qquad \leq$$
and so on. At some point you just have to start writing notation and explaining it in words and hope your readers understand. So I disagree with your phrasing of the question.</p>


答: The function $f:\Bbb R^n \to \Bbb R$ given by $f(x) = \|x-a\|$ is continuous and $V = f^{-1}\big(]s,r[\big)$. So $V$ is the continuous inverse image of an open set, hence open.

Title: Proving that V is open

Body:
Let $n ∈ \mathbb{N}$, let $a \in \mathbb{R^n}$, let $s,r \in R$ with $s < r$ and $V = \{x \in \mathbb{R^n} : s<||x-a||<r\}$. Prove that V is open.

I'm not too sure how to go about this one here, but for $V$ to be open, for every $a \in V$ there exists some $\epsilon >0$ such that $B_{\epsilon}(a) \subseteq V$.

From what I can see we just need to take some $r \in V$ and show that $B_{\epsilon}(r) \subseteq V$. If this is the case, to prove that this ball is a subset would I actually go about it by taking some element in the ball and finding that element in $V$ or am I just confused about the problem? My definitions are coming from real analysis because I am learning this information from a real analysis text as well.


问:
is it possible to fit 6n(n-1) + 1 = 37 into the quadratic formula?
This question results from my original question here.

I'm trying to find x for ... 6x(x-1) + 1 = 37 ... the answer is 3 but I need to know how to get to that answer. I've been advised to use the quadratic formula to do this and to use that I need something like:

```<code>ax2+bx+c=0
</code>```

Here's my steps so far ...

```<code>6x(x-1) + 1 = 37
6x(x-1) + 1 - 37 = 0
6x(x-1) - 36 = 0
</code>```

but now I'm stuck. Can I use the quadratic formula to solve this? What are the next steps to arrange this into the format ax2+bx+c=0? If this is not possible, how would I solve this? Many thanks

答: $$6x(x-1) - 36 = 0 \Rightarrow 6x^2 - 6x - 36 = 0.$$ The quadratic formula states that $$x_i = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ are the solutions of $ax^2 + bx + c = 0.$ Therefore, $$x_i = \frac{-(-6) \pm \sqrt{36 - 4 \cdot 6 \cdot (-36)}}{2\cdot6}.$$ Alternatively, we can divide our original equation by $6$ to make our computations easier. That is, solving $x^2 - x - 6 = 0$ will give the same solutions as the original equation. Can you finish from here?


问:
Can a function that is undefined be lebesgue integrable?
Let $a \in \mathbb{R}$ be such that $a>0$.

If I have a function $f:[-a,a] \rightarrow\mathbb{R}$ that is defined by $f(x) = \frac{1}{g(x)}$ and if $g(c)=0$ for some $c\in (-a,a)$.

Is it safe to conclude that $f$ is not lebesgue integrable over $[-a,a]$ with respect to the lebesgue measure?

I would say $f$ is not integrable over $[-a,a]$ as it is undefined at $c$.

However, if we ignore $c$ in the domain then $f$ is integrable and so can we say $f$ is integrable as it is only “not integrable” on a null set?

答:
There is no functions that are undefined at some point, that is, if $f(x):=\frac1{g(x)}$ and $g(x)=0$ for some $x$ then $f$ is not well-defined. However if you change the values of $f$ to something else, say zero, when $g$ is zero, then this new function is well-defined.

In Lebesgue integration theory we usually consider, instead of individual functions, equivalent classes of measurable functions. A class of measurable functions is a set of functions such the set of points where any two functions of this class are different is a set of zero measure, that is, if $\mathcal{L}_0(\mathbb{R})$ represent the set of Lebesgue measurable functions from $\mathbb{R}$ to $\mathbb{C}$, then for any $f\in \mathcal{L}_0(\mathbb{R})$ it equivalence class is defined by

$$
[f]:=\{h\in \mathcal{L}_0(\mathbb{R}): f(x)=h(x) \text{ almost everywhere }\}
$$

Almost everywhere means that the set $\{x\in \mathbb{R}: f(x)\neq h(x)\}$ have zero measure. In this sense if $g$ is a measurable function and the set $\{x\in \mathbb{R}: g(x)=0\}$ have measure zero, then in the equivalent class of $g$, namely $[g]$, there is some function $g_1$ such that $g_1(x)=1$ whenever $g(x)=0$, therefore the equivalent class of $\frac1{g_1}$ is well-defined. So, in your example, we can consider $f$ as a representative of the class $\left[\frac1{g_1}\right]$.


答:
Note that $x(x-1)=x^{2}-x$ by the Distributive Property of Multiplication. Then, we have:

$6x(x-1)+1=37$

$6(x^{2}-x) - 36 = 0$

$x^{2}-x-6 = 0$

Can you find the solutions now? (Hint: Factoring works well.)

If not:

Because $x^{2}-x-6 = (x-3)(x+2)$, $x^{2}-x-6=0$ has the solutions $\boxed{x=-2,3.}$

Title: is it possible to fit 6n(n-1) + 1 = 37 into the quadratic formula?

Body:
This question results from my original question here.

I'm trying to find x for ... 6x(x-1) + 1 = 37 ... the answer is 3 but I need to know how to get to that answer. I've been advised to use the quadratic formula to do this and to use that I need something like:

```<code>ax2+bx+c=0
</code>```

Here's my steps so far ...

```<code>6x(x-1) + 1 = 37
6x(x-1) + 1 - 37 = 0
6x(x-1) - 36 = 0
</code>```

but now I'm stuck. Can I use the quadratic formula to solve this? What are the next steps to arrange this into the format ax2+bx+c=0? If this is not possible, how would I solve this? Many thanks


问:
For every integer a, b, c, if 3c is not divisible by a, then b is not divisible by a or 3c is not divisible by b
I can't seem to figure this proof out. How would I prove this by contradiction and contraposition?

I tried doing it by saying $3c=bk=(ak)k=a(kk)$
since $b=ak$, $3c=bk$ for some interger $k$

$3c=a(kk)$ contradicts "3c is not divisible by a"
contraposition would follow suit since we prove that $3c=a(kk)$

What am I doing wrong?

答:
First figure out how to say this by contrapositive.

The statement is of the form  $\lnot P \to \lnot Q \lor \lnot R$.

Now $\lnot Q \lor \lnot R \iff \lnot (Q \land R)$ so the statment that you are trying to prove is equivalent to

$\lnot P \to \lnot(Q \land R)$ and that is equivalent to the contrapositive:

$(Q\land R) \to P$.

So it is equivalent to prove the following:

If $b$ is divisible by $a$ and $3c$ is divisible by $b$ then $3c$ is divisible by $a$.

That is almost too easy to deserve discussion:

Pf:  $a|b$ so there exist an integer $k$ so that $b = ka$.  And $b|3c$ so there exists an integer $j$ so that $3c = jb$.  So $3c =j(ka) = (jk)a$.  $jk$ is an integer so $a|3c$.  QED.

To show that is equivalent to the title statement:

Pf:  Note:  If $a|b$ and $b|3c$ then we would have $a|3c$.  But we were given that $a \not \mid 3c$. So we can't have both $a|b$ and $b|3c$.  So we must have either $a\not \mid b$ or $b\not \mid 3c$.

That's it.  We're done.  Let's go home and eat lunch.


答:
$$\lfloor x \rfloor = \text{supremum} \{n \in \mathbb{Z} : n \leq x\}$$

For a subset $S \subset X$, where we have an order $\leq$ on $X$, a supremum or least upper bound of $S$ is an element $M \in X$ such that
$$x \leq M, \,\,\,\,\, \forall x \in S$$
and for any $m \in X$ such that $x \leq m$ for all $x \in S$, we have $M \leq m$.

Title: How to represent the floor function using mathematical notation?

Body:
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?


问:
Local ring dominated by discrete valuation ring, Hartshorne Thm. 6.9, Chap. I
I have a question about a statement in the proof of Theorem 6.9 of Hartshorne's Algebraic Geometry. The setup is this:
$Y$ is a projective curve and $C$ is an abstract nonsingular curve (although i think that it can be seen as a projecitve nonsingular curve for the matter of this question). Furthermore $\varphi \colon C \to Y$ is dominant morphism. Let $K$ be the funciton field of $C$ and $K'$ the function filed of $Y$.

Then he claims: For any $Q\in Y$ the local ring $\mathcal{O}_Q$ is dominated by some discrete valuation ring of $R$ of $K/k$, ($k$ being the algebraically closesd ground field). Take for example a localization of the integral closure of $\mathcal{O}_Q$ at a maximal ideal.

But i fail to see how this yields a discrete valuation ring, at least by means of simple arguments.

My thoughts so far: We can embed $\mathcal{O}_Q$ in $K$ with the induced field homomorphism $\varphi^* \colon K' \to K$. $\mathcal{O}_Q $ is a local noetherian domain of dimension $1$ (not necessarily integerally closed), so the integral closure  in $K$ is an integrally closed domain of dimension $1$. So if we were to show that it was noetherian, its localization at a prime ideal would be a discrete valuation ring.
Is there an easy way to see that the integral closure is noetherian?  I believe the Krull–Akizuki theorem tells us that the integral closure is indeed noetherian again; but this theorem is not mentioned anywhere in the book previously. Am I missing something completely? Or is there an alternative way to see that $\mathcal{O}_Q$ is dominated by a discrete valuation ring of $K/k$?

Any help is appreciated.

答: Instead of using the theorem of Krull-Akizuki one could also invoke the following result: the integral closure of an integral finitely generated algebra $A$ over a field $k$ in a finite extension of the fraction field of $A$ is a finitely generated $k$-algebra too. In particular it is noetherian. This result is mentioned although not proved in Hartshorne as Theorem 3.9 A.


答: It is clear (you can prove it easily) that the set $G = \{ x \in \mathbb{R}^n: ||x-a|| <r \}$ is open set in $\mathbb{R}^n$. And it is clear that the set $F = \{ x \in \mathbb{R}^n: ||x-a|| \le s \}$ is closed set in $\mathbb{R}^n$. Thus, the difference between an open set and closed set ($V = G-F$) is open.

Title: Proving that V is open


问:
Construct a morphism from a nonsigular projective curve onto P1 from a rational function, and its quasi-finiteness (II)
I'm currently doing exercise I.6.2 in Hartshorne's Algebraic Geometry. I've tried to prove this and get a so-called "proof" which full of gaps and uncertainty.

The exercise goes like this:

Let $Y$ be a nonsingular projective curve. Show that:
(1) every nonconstant rational function $f$ defined on $Y$ defines a surjective morphism $\phi: Y \rightarrow \mathbb{P}^1$,
(2) and that for every $P \in \mathbb{P}^1$, $\phi^{-1}(P)$ is a finite set of points.

Questions on (1) has already been posted in the question: Construct a morphism from a nonsigular projective curve onto $\mathbb{P}^1$ from a rational function, and its quasi-finiteness (I)

In this post, I'm going to raise some questions on the proof of (2). This proof is adapted from an online solution manual of Hartshorne. Here $k$ is the base algebraically closed field.

"Proof" of (2): Since $\phi$ is dominant, it incudes an inclusion $K(Y) \rightarrow K(\mathbb{P}^1)$. Since both fields are finitely generated extension fields of transcendence degree 1 of $k$, $K(\mathbb{P}^1)$ must be a finite algebraic extension of $K(Y)$. To show that $\phi$ is quasi-finite (i.e. $\phi(P)$ is a finite set), look at any open affine set $V$ in $P$. Its coordinate ring is $A(V)$, and by "Finiteness of Integral Closure" (i.e. Thm (I.3.9) of Hartshorne), $A(V)$ is a finite $A(V)$-module. The corresponding affine set to $A(V)$ is isomorphic to an open subset $U$ of $Y$. Clearly $U = \phi^{-1}(V)$, and thus $\phi$ is a quasi-finite morphism, as desired.

I feel quite confused following the above proof. Here are my questions:

Question 1: Is the induced inclusion map obtained by [Hartshorne's Corollary I.6.12]? (I will list the corollary below.) If so, it seems that the inclusion obtained should be $K(\mathbb{P}^1) \rightarrow K(Y)$ instead of claimed in the "proof". Am I getting something wrong when reading Hartshorne.

Question 2: I'm not quite familiar with the concept and properties of quasi-finite maps. I have only read the first chapter of Shafarevich's Algebraic Geometry Volume 1 besides Chapter 1 of Hartshorne. A similar concept is a finite map, which is defined as:

Defn: Let $f:X \rightarrow Y$ be a morphism between affine varieties. We call $f$ a finite morphism, if the coordinate ring $A(X)$ is integral over $A(Y)$.

And such a morphsim is called finite, probably because the following property holds:

FACT 1: Using the same notations as Defn, if $f$ is a finite map, then any point $y \in Y$ has at most a finite number of inverse images.

And may be in the proof above, we also use the fact that finiteness is a local property, namely,

FACT 2[Shafarevich Vol.1 Thm 1.13]: If $F: X \rightarrow Y$ is a morphism of affine variety, and every point $x \in Y$ has an affine neighborhood $x \in U$, such that $V=f^{-1}(U)$ is affine and $f: V \rightarrow U$ is finite, then $f$ itself is finite.

HOWEVER, the above seemingly satisfactory facts cannot apply here, since here we are talking about PROJECTIVE curves and PROJECTIVE line $\mathbb{P}^1$. So I'm asking whether such demanding results hold in projective case? (Is "quasi-finiteness" a corresponding notion in projective case to "finiteness" in affine case?) I'm hoping to get some reference on that. (I have looked up Milne's Algebraic Geometry notes, yet the materials there seems inadequate to solve the above exercise.)

OR, I'm trying to convert this exercise into an affine version. For example, fix a point $P$ and take an open affine chart containing it and so on.... Yet I feel it quite challenging to me to fill up all the gaps and details. If such ideas works, could anyone help me clean up the details?

IF the ideas above does not work, could anyone help me to clear up the details in the original proof listed above?

Question 3: It seems that the proof above (after filling up the details) is quite complicated and involves so many concepts and tools not mentioned in Hartshorne. Does Hartshorne really hope us to give such a proof? Is there any simple ones that Hartshorne is demanding?

Thank you all in advance! :)

P.S. [Hartshorne's Corollary I.6.12]:

The following three categories are equivalent:
(1) Nonsigular projective curves, with dominant morphisms;
(2) Quasi-projective curves, and dominant rational maps;
(3) Function fields of dimension one over $k$, and $k$-homomorphisms.

答:
Question 1: Yes. You are right. The arrow should be reversed. And the whole "proof" is just nonsense.

To prove this exercise, you only need to do a dimension count.

An "algebraic" proof: Take affine chart $V \in \mathbb{P}^1$ and $U := \varphi^{-1}(V) \in Y$. We have an inclusion $A(V) \to A(U)$. We need to prove $A(U)$ is integral over $A(V)$. And then the theorem you mentioned would tell us that the fiber is finite.

Now because $A(V)$ and $A(U)$ are both (Krull) dimension $1$ ($U$, $V$ are nonempty open sets), by Noether normalization lemma (the wiki of the lemma explained this fact well, we need the refined version https://en.wikipedia.org/wiki/Noether_normalization_lemma), $A(U)$ is indeed integral over $A(V)$.

However, Hartshorne did not include Noether normalization lemma in his book, so I guess the proof in his mind should be the following "geometrical" one:

Because the preimage of a closed set is again a closed set. $\varphi^{-1}(P)$ must be closed. However, it cannot be the empty set or the whole $Y$, so it must be a set consists of finite many points.


问:
Linear Algebra: Rowspace, column space (Concept question)
freshman student here looking to clarify the concepts.

1.) Is the column space also the solution space of a matrix?

2.) Lets say a matrix is formed by a1,a2,a3,a4,a5.  I find the rref form of this matrix; turns out a1,a2,a3 are linearly independent.  Can it suffice to say that a1,a2,a3 are the basis for both solution space and column space of a matrix?  Can this tell me anything about row space?

3.) What is the purpose of the row space?

4.) Is the basis for row space also the basis for the solution space in a square matrix?  Here on out it becomes increasingly difficult for me to visualize.

5.) Apart from dimension theorem, what other purpose does rank actually serve?  So I am familiar with the formulas regarding rank; rank <= min(n,m) or rank = min(dimension(rowspace),dimension(colspace)), but I am not actually sure I understand the implications of this.

6.) If dim(rowspace)<dim(colspace), can this tell me anything about the solution space or any implications (apart from implications on the rank)?

Sorry for the long-winded question.  Hope anyone can provide insight on anything above, for me and my friends.

答:
Let's go through these one by one.

Yes. The row space of a matrix $A$ can be thought of as all of the vectors $\vec b$ such that $A\vec x=\vec b$ has a solution. This is because multiplying a matrix by a vector essentially takes all the column vectors of the matrix and adds them up in a linear combination. The column space is just all the linear combinations of the columns, so the column space and the solution space are identical.

Yes, as long as $a_4$ and $a_5$ are not also independent. In general, in a matrix of rank $n$, any $n$ linearly independent column vectors will be a basis for the column space.

This is a bit of an open-ended question, but there's a few important things to know about the row space: first, the row space and the column space have the same dimension, $\text{rank }A$. This is not at all obvious without recognizing the second fact: every vector in $\text{Row }A$ is orthogonal to every vector in $\text{Nul }A$. (If you haven't learned that in class yet, take ten to twenty minutes playing with matrices of different sizes, and see if you can figure out why that's true.) It's also important to realize that the row space of $A$ is the column space of $A^T$, which means that a matrix and its transpose have the same rank.

No. As an extreme example, consider $\begin{bmatrix}1&0&0\\1&0&0\\1&0&0\end{bmatrix}$. The row space is spanned by $\begin{bmatrix}1\\0\\0\end{bmatrix}$ while the column space is spanned by $\begin{bmatrix}1\\1\\1\end{bmatrix}$. However, there's a few things you should notice about these spaces: first, they have the same dimension. Second, if we perform Gauss-Jordan elimination on our matrix, the column space will change, but the row space (and therefore the null space, by what we talked about in #3) does not. This is true for all matrices, not just square ones. (Note that, in this particular case, the row space and the column space are identical after Gauss-Jordan. This is not true in general.)

Rank is crucially important because of the rank theorem, as well as because the rank is equal to the dimension of the column space and the dimension of the row space. Finding the rank allows you to determine invertibility, and after you get a feel for rank as "the dimension of the range of $A$" you can easily see why inequalities like $\text{rank }(AB)\leq\min(\text{rank }A,\text{rank }B)$ are true. From there, you can look at things like matrices $A$ where $A^n$ eventually becomes the zero matrix, and you can analyze how rank affects transformations. There's also the ever-important Rank Theorem, that in an $m\times n$ matrix $A$, $\text{rank }A+\text{dim Nul }A=n$.

The dimension of your row space and the dimension of your column space have to be equal. Both of them are equal to the rank. If you get different dimensions, go back and check your work. I'll leave you with, once again, the fundamental property of rank:

$$\text{rank }A=\text{dim Col }A=\text{dim Row }A$$


问:
Order of convergence of Fixed Point Iteration
I have to determine the order of convergence of this iteration specification but I don't know how to do this in $\mathbb{R}^n$. (So far, I've only done this in $\mathbb{R}$).
Any help is highly appreciated (and sorry for my bad english).

$x^{(k+1)} = \begin{pmatrix}
0 & -1/2 & -1/4\\
-1/2 & 0 & 0\\
0 & -1/2 & 0
\end {pmatrix}x^{(k)}+\begin{pmatrix} 3/4 \\ -2 \\ -1/2 \end{pmatrix}$
for $k \geq 0$

$x^{(k)} = \begin{pmatrix} x_1^{(k)} \\ x_2^{(k)} \\ x_3^{(k)} \end{pmatrix}$

答:
Your iteration is special case of the stationary iteration $$x_{n+1} = Gx_n + f$$ which can occasionally be used to solve the linear system $$x = Gx +f.$$ The initial guess $x_0$ must be selected by the user, but $x_0 = 0$ is a perfectly acceptable choice. If $\|G\|_2<1$, then $I-G$ is nonsingular, and the sequence $\{x_n\}_{n=0}^\infty$ is convergent and $$x_n \rightarrow x = (I-G)^{-1} f, \quad n \rightarrow \infty, \quad n \in \mathbb{N}$$
regardless of the choice of $x_0$. If $x_0 = 0$, then by induction on $n$ you can establish that $$x_n = \sum_{j=0}^{n-1} G^j f.$$ It follows, that $$\|x-x_n\|_2 \leq \left \| \sum_{j=n}^\infty G^j f \right\|_2 \leq \|G\|_2^n\|x\|_2 = \epsilon_n.$$ It follows that $x_n \rightarrow x$ at least linearly as $n \rightarrow \infty$ and $n \in \mathbb{N}$ because
$$ \frac{\epsilon_{n+1}}{\epsilon_n} = \frac{ \|G\|_2^{n+1} }{\|G\|_2^{n}} = \|G\|_2 \rightarrow \|G\|_2 < 1.$$
You are free to replace the $2$-norm with any other norm induced by a vector norm. In particular, you can use the $\infty$-norm for which your particular matrix $G$ satisfies $\|G\|_\infty = \frac{3}{4}$.

In any case, the order of convergence is at least $p=1$.


问:
Where I am getting the expected value calculation wrong?
Question 1 ) Locking. You have just bought an old bicycle for 80 dollars. You can either buy an unbreakable Kryptonite lock for 20 dollars or a cheap chain and lock for 8 dollars. You reckon no one would bother stealing your bike if it is locked with Kryptonite-too much work. But there is a 10 percent chance of the cheap chain being snipped, and your bike being stolen. If the bike is stolen, you won't bother to replace it. If we have listed all the relevant values, and you act so as to maximize expectation, which lock do you buy?

Planning ahead. You decided to buy a new bike after your first bike was stolen. Would it make any difference to your decision in question 1, if you already planned to buy another 80 dollars bike if your first one was stolen? Suppose you really work this out and see that if you bought a second bike you would still, buy a cheap lock for 8 dollars, and not a Kryptonite lock for 20 dollars.

My calculation is for the expected value of buying cheap lock the second time:

Exp(Cheap Lock) = Pr(stolen twice =0.01)(−80−8)+ Pr(not stolen =0.1)(−8) + Pr(not stolen =0.9)(−80) =$−80.8

The answer is -$17.6

答:
P(never stolen) = 0.90 cost $\$8$ EV $-7.2$

P(stolen once) = 0.09 cost $8 + 80 + 8 = \$96$ EV $-8.64$

P(stolen twice) = 0.009 cost $8 + 88 + 88 = \$184$ EV $-1.656$

We could stop here and say that there are only so many times a person lets his bike get stolen without changing his assumptions.  Maybe he buys the fancy lock if his bike is stolen twice.  In which case we would say:

P(stolen twice) = 0.009 cost $8 + 88 + 100 = \$196$ EV $-1.764$

Summing these we get $\$17.6$ as the expected loss.

Or we could keep doing the same old.... and we get an infinite sereis.

P(stolen n times) = $0.9\times 0.1^n$ cost $8 + n88$, etc.

$8 + (88)(0.90)\sum_\limits {i=0}^{\infty} i(0.1^i)\\
8 + \frac {88}{9} = 17.78$


问:
False duality of diagonalizable/constant group scheme?
The following is an introduction to constant group scheme and diagonalizable group scheme.

Let $M$ be an arbitrary abstract group and $S$ be a scheme. Let $M_S=\coprod_{i\in M}S_i$ where $S_i=S$,which can be constructed via glueing data of a scheme see tag 01JA. Then $M_S$ is called the constant group scheme over $S$ associated to $M$. And for any $S$-scheme $T$, $M_S(T)=\{\text{locally constant functions }f:|T|\to M\}$, where $|T|$ has Zariski topology and $M$ has discrete topology, and locally constant function is defined to be $\forall t\in T,\exists U\ni t\text{ open}$ s.t. $f(U)={f(t)}$. see tag 03YW

The dual of a group scheme $G$ over $S$ is a scheme representing the functor $T/S\mapsto \mathrm{Hom}_{\mathrm{Gr}-T}(G_T,\mathbb{G}_m)$. For a constant group scheme $M_S$, its dual is denoted as $D(M)$ or $D_S(M)$ and they are precisely the diagonalizable group schemes. In particular, it exists and $\DeclareMathOperator{\Spec}{Spec}D_S(M)\cong \Spec \mathbb{Z}[M]\times_{\Spec \mathbb{Z}}S$ where $\mathbb{Z}[M]$ is the group ring $\bigoplus_{i\in M}\mathbb{Z}_i$ with an obvious ring structure. Note that $D_S(M)$ is always commutative no matter whether $M$ is commutative.

It's known in a few reference that if $M$ is commutative then $D_SD_S(M)=M_S$ i.e. $M_S$ is reflexive. (e.g. SGA3,Expose VIII,Page 3,Theoreme 1.2 or [Groupe Algeacutebriques by Demazure and Gabriel, Chap II,§2,2.11 Dualite des groupes diagonalisables]). But none of the proof seems convincing. And I seem to find a counterexample in some sense.

From my understanding of the corresponding between $\DeclareMathOperator{\itSpec}{\mathit{Spec}}D_SD_S(M)(T)=\mathrm{Hom}_{\mathrm{Gr}-T}(\itSpec_{\mathcal{O}_T}\mathcal{O}_T[M],\mathbb{G}_m)$ and $M_S(T)$. A element in $\mathrm{Hom}_{\mathrm{Gr}-T}(\itSpec_{\mathcal{O}_T}\mathcal{O}_T[M],\mathbb{G}_m)$ corresponds to a map of $\mathcal{O}_T$-algebras $f:\mathcal{O}_T[t,t^{-1}]\to \mathcal{O}_T[M]=\bigoplus_{i\in M}\mathcal{O}_T$ s.t. $\epsilon f(t)=1$ and $\Delta(f(t))=f(t)\otimes f(t)$.

If we write $f(t)=\sum_{i\in M}a_i\cdot e_i\in \bigoplus_{i\in M}\mathcal{O}_T(T)$ then we have $a_i \cdot a_j=\delta_{ij}a_i$ and $1=\sum_i a_i$ (so each $a_i$ is an idempotent element in $\mathcal{O}_T(T)$). Clearly $\forall s\in T,1|_s=\sum_i a_i|_s$ so at least one of $a_i$ is supported on $s$. And $T_{s_i} \cap T_{s_j}=T_{s_is_j}=T_0=\emptyset$. So $(T_{s_i})_{i\in M}$ forms a disjoint open cover of $T$ which can be identified with a locally constant function from $T$ to $M$. But the only problem is, $\sum_{i\in M}a_i\cdot e_i\in \bigoplus_{i\in M}\mathcal{O}_T(T)$  has only finite terms, which means the induced locally constant map has only a finite image. And I believe that there exists locally constant functions $T\to M$ with infinite image, which cannot be induced from $\mathcal{O}_T[t,t^{-1}]\to \mathcal{O}_T[M]=\bigoplus_{i\in M}\mathcal{O}_T$.

So which part of my arugments is wrong? Or do we need some extra conditions on $M$ (like finite or finitely generated) or the base scheme $S$ (like locally Noetherian, locally connected or connected)? Or the duality only holds in some topology like fppf/etale topology?

答:
I am going to give an identification of $DD(M)(S)$ and $M_S(S)$.
Using the identification $D(M)=\mathrm{Spec}\mathcal{O_S}[M]$ and $\mathbb{G}_m=D(\mathbb{Z})$, we have
$$DD(M)(S)=\mathrm{Hom}_{\mathcal{O}_S-\mathrm{Hopf-alg}}(\bigoplus_{i\in \mathbb{Z}}e_i\mathcal{O}_S,\bigoplus_{i\in M}e_i\mathcal{O}_S)$$
i.e. every element is determined by a map $\phi$ of $\mathcal{O}_S$-modules s.t. the following commutative diagram holds
$$\require{AMScd}\begin{CD}
\bigoplus_{i\in \mathbb{Z}}e_i\mathcal{O}_S     @>\phi>>  \bigoplus_{i\in M}e_i\mathcal{O}_S\\
@VV{e_i\mapsto e_i\otimes e_i}V        @VV{e_i\mapsto e_i\otimes e_i}V\\
\bigoplus_{j,k\in \mathbb{Z}}e_{j}\mathcal{O}_S\otimes e_k\mathcal{O}_S     @>\phi\otimes \phi>>  \bigoplus_{a,b\in M}e_a\mathcal{O}_S\otimes e_b \mathcal{O}_S
\end{CD}$$
It's equivalent to the data $\mu :\mathcal O_S \cong e_1\mathcal{O}_S\stackrel{\phi}{\to}\bigoplus_{i\in M}e_i\mathcal{O}_S$ (written $\mu=\sum_i e_i \mu_i$ where $\mu_i:\mathcal{O}_S\stackrel{\mu}{\to}\bigoplus_{j\in M}e_j\mathcal{O}_S\to e_i\mathcal{O}_S\cong \mathcal{O}_S$) s.t. the diagram commutes
$$\require{AMScd}\begin{CD}
\mathcal{O}_S     @>\mu>>  \bigoplus_{i\in M}e_i\mathcal{O}_S\\
@VV{\mathrm{id}}V        @VV{e_i\mapsto e_i\otimes e_i}V\\
\mathcal{O}_S     @>{\mu\otimes\mu}>>  \bigoplus_{a,b\in M}e_a\mathcal{O}_S\otimes e_b \mathcal{O}_S
\end{CD}$$
and $\sum_i \mu_i=\mathrm{id}$.

It's equivalent to the condition that $(\mathcal{O}_S\otimes \mathcal{O}_S\stackrel{\Delta}{\to}\mathcal{O}_S)\circ(\mu_a \otimes \mu_b)=\delta_{ab}\mu_a$ and $\sum_i \mu_i=\mathrm{id}$.

Over any affine open, we have $(\mathcal{O}_S\otimes \mathcal{O}_S\stackrel{\Delta}{\to}\mathcal{O}_S)\circ(\mu_a \otimes \mu_b)=\mu_a \circ \mu_b$. So it's equivalent to the condition that $\mu_a \circ \mu_b =\delta_{ab}\mu_a$ and $\sum_i \mu_i =\mathrm{id}$.

Claim: It's equivalent to the data {$(U_i)_{i\in M}$ is a disjoint open cover of $S$}.

Proof of the claim. Given a disjoint open cover $(U_i)_{i\in M}$ of $S$, then each $U_i^c=\bigcup_{j\neq i}U_j$ is open, there exist an unique element $c_i\in \mathcal{O}_S(S)$ s.t. $c_i|_{U_i}=1$ and $c_i|_{U_i^c}=0$ using axioms of sheaf. With each $c_i$ we can associate $\mu_i:\mathcal{O}_S\to \mathcal{O}_S,u\mapsto c_i u$. It's not hard to see that $\mu_a \circ \mu_b=\delta_{ab}\mu_a$. As $\sum_i c_i=1$, we have $\sum_i \mu_i=\mathrm{id}$.

Reversely for each $s\in S$, $\mu_{i,s}:\mathcal{O}_{S,s}\to\mathcal{O}_{S,s}$ is completed determined by $c_{i,s}=\mu_{i,s}(1)$. We know that $c_{i,s}c_{j,s}=\delta_{ij}c_{i,s}$ and $\sum_{i}c_{i,s}=1$. So each $c_{i,s}$ is idempotent ($x^2=x$), but the only idempotent elements in a local ring is $0$ and $1$. So exactly one of $(c_{i,s})_{i\in M}$ is 1.

For any $i$, denote $U_{i}$ as the subset consisting of $s$ s.t. $\mu_{i,s}=\mathrm{id}$. Then $(U_{i})_{i\in M}$ is disjoint. If $s\in U_i$, then $\mu_{i,s}=\mathrm{id}$, clearly it extends to an open neighborhood of $s$, see tag 01CP. Thus $(U_{i})_{i\in M}$ is disjoint open cover of $S$. The result follows.$\square$

Clearly the data {$(U_i)_{i\in M}$ is a disjoint open cover of $S$} is equivalent to the data {locally constant function $S\to M$}. The result follows.


问:
Can every ordinal be reachable from $\aleph_0$, in ZF?
Define: $ x \text{ is reachable from } y \iff \\  \exists z (|z|\leq |y| \land \forall m \in z (|m| \leq |y|) \land |x|\leq|\bigcup (z)|)$

Given that ZF is consistent with only $\aleph_0$ being regular*, and so any ordinal supernumerous to $\aleph_0$ would have $\aleph_0$ as its confinality; then would this mean that ZF is consistent with saying that every ordinal is reachable from $\aleph_0$?

答:
It's provable in ZF (without choice) that the ordinal $\omega_2$ is not the union of countably many countable sets. Proof: Suppose the contrary: $\omega_2=\bigcup_{n\in\omega}A_n$ with each $A_n$ countable. Without loss of generality, the $A_n$'s are disjoint (just replace each $A_n$ with $A_n-\bigcup_{k<n}A_k$).  For each $n$, let $f_n$ be the unique order-isomorphism from $A_n$ to its order-type $\alpha_n$, which, being a countable ordinal, is $<\omega_1$. Then you can map $\omega_2$ one-to-one into $\omega_1\cdot\omega$ by sending each $x\in A_n$ to $(\omega_1\cdot n)+f_n(x)$. Since $\omega_1\cdot\omega<\omega_2$, this is a contradiction.

If I remember correctly, this result is due to Tom Jech.

Edit: The result is indeed due to Jech:

MR0644752 Jech, Thomas

On hereditarily countable sets.

J. Symbolic Logic 47 (1982), no. 1, 43–47


答:
There are number of issues here. First, this is problem I.6.4, not I.6.2, but that's relatively minor. Next, this "proof" kind of stinks: there's an immediate error in the first line, in that the inclusion of function fields ought to the other way, $K(\Bbb P^1)\to K(Y)$, and it only goes downhill from there.

Here's a better approach: Consider the map of function fields $k(t)\to k(Y)$ by $t\mapsto f$.
This is a $k$-homomorphism between two function fields of dimension one, so by corollary I.6.12 it gives us a dominant (in particular, nonconstant) morphism $\varphi:Y\to \Bbb P^1$ which is given by $y\mapsto [1:f(y)]$ where $f(y)\neq \infty$.
If $\varphi(Y)$ misses a point, then by taking the copy of $\Bbb A^1$ which is the complement of that point, we can find a nonconstant regular function on $\Bbb A^1$ which pulls back to a nonconstant regular function on $Y$.
This is impossible by theorem I.3.4, so $\varphi(Y)=\Bbb P^1$ and $\varphi$ is surjective.

Note that since any point $P$ in $\Bbb P^1$ is closed, $\varphi^{-1}(P)$ is a closed set since morphisms are continuous.
On the other hand, the fiber over any point must be a proper subset of $Y$, since $\varphi$ is nonconstant.
Since all proper closed subsets of a curve are finite (exercise I.4.8, for instance), we are done.

Title: Construct a morphism from a nonsigular projective curve onto P1 from a rational function, and its quasi-finiteness (II)

Body:
I'm currently doing exercise I.6.2 in Hartshorne's Algebraic Geometry. I've tried to prove this and get a so-called "proof" which full of gaps and uncertainty.

The exercise goes like this:

Let $Y$ be a nonsingular projective curve. Show that:
(1) every nonconstant rational function $f$ defined on $Y$ defines a surjective morphism $\phi: Y \rightarrow \mathbb{P}^1$,
(2) and that for every $P \in \mathbb{P}^1$, $\phi^{-1}(P)$ is a finite set of points.

Questions on (1) has already been posted in the question: Construct a morphism from a nonsigular projective curve onto $\mathbb{P}^1$ from a rational function, and its quasi-finiteness (I)

In this post, I'm going to raise some questions on the proof of (2). This proof is adapted from an online solution manual of Hartshorne. Here $k$ is the base algebraically closed field.

"Proof" of (2): Since $\phi$ is dominant, it incudes an inclusion $K(Y) \rightarrow K(\mathbb{P}^1)$. Since both fields are finitely generated extension fields of transcendence degree 1 of $k$, $K(\mathbb{P}^1)$ must be a finite algebraic extension of $K(Y)$. To show that $\phi$ is quasi-finite (i.e. $\phi(P)$ is a finite set), look at any open affine set $V$ in $P$. Its coordinate ring is $A(V)$, and by "Finiteness of Integral Closure" (i.e. Thm (I.3.9) of Hartshorne), $A(V)$ is a finite $A(V)$-module. The corresponding affine set to $A(V)$ is isomorphic to an open subset $U$ of $Y$. Clearly $U = \phi^{-1}(V)$, and thus $\phi$ is a quasi-finite morphism, as desired.

I feel quite confused following the above proof. Here are my questions:

Question 1: Is the induced inclusion map obtained by [Hartshorne's Corollary I.6.12]? (I will list the corollary below.) If so, it seems that the inclusion obtained should be $K(\mathbb{P}^1) \rightarrow K(Y)$ instead of claimed in the "proof". Am I getting something wrong when reading Hartshorne.

Question 2: I'm not quite familiar with the concept and properties of quasi-finite maps. I have only read the first chapter of Shafarevich's Algebraic Geometry Volume 1 besides Chapter 1 of Hartshorne. A similar concept is a finite map, which is defined as:

Defn: Let $f:X \rightarrow Y$ be a morphism between affine varieties. We call $f$ a finite morphism, if the coordinate ring $A(X)$ is integral over $A(Y)$.

And such a morphsim is called finite, probably because the following property holds:

FACT 1: Using the same notations as Defn, if $f$ is a finite map, then any point $y \in Y$ has at most a finite number of inverse images.

And may be in the proof above, we also use the fact that finiteness is a local property, namely,

FACT 2[Shafarevich Vol.1 Thm 1.13]: If $F: X \rightarrow Y$ is a morphism of affine variety, and every point $x \in Y$ has an affine neighborhood $x \in U$, such that $V=f^{-1}(U)$ is affine and $f: V \rightarrow U$ is finite, then $f$ itself is finite.

HOWEVER, the above seemingly satisfactory facts cannot apply here, since here we are talking about PROJECTIVE curves and PROJECTIVE line $\mathbb{P}^1$. So I'm asking whether such demanding results hold in projective case? (Is "quasi-finiteness" a corresponding notion in projective case to "finiteness" in affine case?) I'm hoping to get some reference on that. (I have looked up Milne's Algebraic Geometry notes, yet the materials there seems inadequate to solve the above exercise.)

OR, I'm trying to convert this exercise into an affine version. For example, fix a point $P$ and take an open affine chart containing it and so on.... Yet I feel it quite challenging to me to fill up all the gaps and details. If such ideas works, could anyone help me clean up the details?

IF the ideas above does not work, could anyone help me to clear up the details in the original proof listed above?

Question 3: It seems that the proof above (after filling up the details) is quite complicated and involves so many concepts and tools not mentioned in Hartshorne. Does Hartshorne really hope us to give such a proof? Is there any simple ones that Hartshorne is demanding?

Thank you all in advance! :)

P.S. [Hartshorne's Corollary I.6.12]:

The following three categories are equivalent:
(1) Nonsigular projective curves, with dominant morphisms;
(2) Quasi-projective curves, and dominant rational maps;
(3) Function fields of dimension one over $k$, and $k$-homomorphisms.


问:
Congruence of invertible skew symmetric matrices
I am asking for some hints to solve this excercise. Given an invertible skew symmetric matrix $A$, then show that there are invertible matrices $ R, R^T$ such that $R^T A R = \begin{pmatrix} 0 & Id \\ -Id & 0 \end{pmatrix}$, meaning that this is a block matrix that has the identity matrix in two of the four blocks and the lower one with a negative sign.

I am completely stuck!

答:
The key fact is that every real and normal matrix is orthogonally similar to its real Jordan form. However, as real Jordan forms are seldom taught in universities, we may need to start from scratch here.

As $A$ is skew symmetric and invertible, its eigenvalues are conjugate pairs of nonzero pure imaginary numbers $\pm b_1i,\,\pm b_2i,\,\ldots,\pm b_ni$ (where the size of $A$ is $2n\times2n$).

Furthermore, as $A$ is skew symmetric, it is a normal matrix. Hence it possesses an orthonormal eigenbasis. (In some lecture notes, this fact is sometimes stated as "if $A$ is normal, $A=UDU^\ast$ for some unitary matrix $U$ and diagonal matrix $D$".) Let $u_1+iv_1,\,u_2+iv_2,\,\ldots,\,u_n+iv_n$ be $n$ orthogonal eigenvectors corresponding the eigenvalues $b_1i,\,b_2i,\,\ldots,\,b_ni$.

Note that $\left\{\frac{u_1}{\|u_1\|},\frac{u_2}{\|u_2\|},\ldots,\frac{u_n}{\|u_n\|},\frac{v_1}{\|v_1\|},\frac{v_2}{\|v_2\|},\ldots,\frac{v_n}{\|v_n\|}\right\}$ is an orthonormal basis of $\mathbb{R}^{2n}$ (why?). So, these $2n$ vectors form a real orthogonal matrix $Q$ and $A=Q\pmatrix{0&B\\ -B&0}Q^T$ (again, why?), where $B=\operatorname{diag}(b_1,b_2,\ldots,b_n)$. Now the rest should be straightforward.


问:
What is the reduced suspension of $I=[0,1]$?
How to visualise the reduced suspension of $I=[0,1]?$

Here is an answer, Smash product of $S^1$ with the interval $I$
but I don't get it. So, it will be helpful for me if you can give a pictorial view.Thanks

答:
We claim that, $\Sigma (I,0)\cong D^2/A$ where, $A=\{(r,0):r\in [0,1]\}$
Consider the map,

$F:I\times I\to D^2/A$ given by,

$F(t,r)=[re^{2πit}]$ then $F$ is continuous.

Next, by universal mapping property and noting that , $I\times I $ is compact and $D^2/A$ is Hausdorff, we conclude that,
$I×I/\left(\{0,1\}×I\bigcup I×\{0\}\right)\cong D^2/A$ and thus our claim is proved.


问:
Proving the sequence converges
I need to prove that $x_n = \frac{3^n}{n!}$ converges and to find the limit. I am wondering if I can use the squeeze theorem for this problem. If so, how would I structure my proof?

答:
Let $ e^x = \sum_{n=0}^{\infty} a_n x^n $, Sinc \frac{d}{dx} (e^x) = e^x, then

$$\frac{d}{dx} (\sum_{n=0}^{\infty} a_n x^n ) = \sum_{n=0}^{\infty} a_n x^n .$$

And

$$\sum_{n=1}^{\infty} n a_n x^{n-1}  = \sum_{n=0}^{\infty} a_n x^n, $$

equaling the coefficient of the x that has the same power, we get

$$a_{n} = \frac{1}{n} = a_{n-1},$$

so,

$$a_n = \frac{1}{n} a_{n-1} = \frac{1}{n} \frac{1}{n-1} a_{n-2}=....= \frac{1}{n!}a_0.$$

Since $e^0 = 1$ then we have $a_0 =1$. Thus $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, you need just to replace $x$ by 3.

Title: Proving the sequence converges


问:
Is 510511 the smallest Euclid number $ E_n$ that divides by the next prime $ p_{n+1}$?
That is $ 2.3.5.7.11.13.17+1 = 19.97.277$.
So I looking for examples of the form;
$ 2.3.5…...p_n +1 = p_{n+1}.m$.
Looking at a table of explicit factorizations, from https://oeis.org/A038507/a038507.txt
at ‘The online Encyclopedia of Integer Sequences’, there are no more examples on the list, which covers the first 60 primes.

See https://en.wikipedia.org/wiki/Euclid_number , and

https://oeis.org/A006862

I tried to use Wilson’s Theorem but couldn’t get this to work.

答:
$\prod_{i<1} p_i + 1 = 2$ is divisible by $p_1 = 2$ (an empty product is $1$ by convention).

$\prod_{i<2} p_i + 1 = 3 $ is divisible by $p_2 = 3$

If you don't count those, the first example is yours:

$\prod_{i<8} p_i + 1$ is divisible by $p_8$

The next examples are $p_{233}$ and $p_{431}$.

EDIT: There are no more such $p_n$ for $n \le 10000$.

Heuristically, $\prod_{i<n} p_i + 1$ has probability approximately $1/p_n$ of being divisible by $p_n$.  Since the series $\sum_{n} 1/p_n$ diverges, we should expect there to be infinitely many such $p_n$.  However, this is not a proof. Moreover, the series diverges very slowly, so these $p_n$ are likely to be very sparse.  The next $n$ could easily be in the billions.

EDIT: Sequence A338543 is now in OEIS.