问:
Probability of rolling x or lower twice in a row over multiple dice rolls?
If I'm rolling, for example, 6 twenty-sided dice, one at a time, and I "fail" if I roll a 10 or lower twice in a row, how do I calculate the chances of success/failure?

Example:

6d20, success on 11 or higher.

I roll the dice, results in order are (13, 7, 17, 18, 2, 11). This is a success because the two failures (7, 2) were not rolled one after the other.

If instead I roll (15, 4, 8) it's a failure and I stop rolling, because I rolled <11 twice in a row.

How do I calculate the chances of success/failure? Over the 6 rolls in the example, or any number of rolls?

EDIT: I am not very knowledgeable when it comes to math and was hoping for a relatively simple explanation. If it's possible.

答:
Note that the chances of rolling a 10 or lower are the same as rolling an 11 or higher. That means this is the same as flipping a coin, and losing if we flip two tails in a row. (Think of it as if you roll an 11 -20, its the same as a tails and otherwise heads)

We want to know the chances that we don't ever get two adjacent tails.

If our most recent flip is a tail that is not after another tails, then we call that scary because we are close to losing: one more tails and we lose.

If our most recent flip was heads, we call that easy, since it means we can't lose in a single flip.

And, if our most recent flip was a tails after another tails, we call that a lose.

Now, initially, there is a 50-50 chance (on your first flip) that you flip heads or tails, meaning we can either go to an easy state or a scary state.

From the easy state, there is a 50-50 chance of ending up at an easy state (by flipping a heads) or flipping to a scary state (by flipping tails).

From the scary state, there is a 50-50 chance of ending up at the easy state and the lose state.

Let $P(i)$ represent the probability of losing by flip $i,$ and $S(i)$ represent the probability that we end up at a scary state by flip $i,$ as well as $Z(i)$ representing being at an easy state by flip i.

Note that $P(i) = \frac{1}{2} S(i-1) + P(i-1)$ because you need to get the 50% chance of flipping a tails if you are at a scary step, and already losing in a previous step means you have already lost.

$Z(i) = \frac{1}{2} (Z(i-1) + P(i-1) + S(i-1)),$ since at any step, we have the probability of flipping a heads and reaching an easy state.

$S(i) = \frac{1}{2}P(i-1).$

$P(1) = 0, S(1) = \frac{1}{2}, Z(1) = \frac{1}{2}.$

We can then use these recurrences and initial values to obtain $P(6),$ which should be the probability that we lose before or by the 6th flip.

Sorry for the long winded answer!


问:
I can't understand the result of this limit
$$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$

I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4  (numerator)$$ and $$-n ^ 3(denominator)$$

and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb question, but I didn't understand why is $+\infty$ instead of $-\infty$.

答:
At the moment, we have
$$  \lim_{x \rightarrow \infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3} = \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}  $$
because the expression in the limit has no $x$s (and the limit is as $x$ goes somewhere).

Assuming what was intended was "$\lim_{n \rightarrow \infty}$", we compute, "factoring out the big term" in the numerator and denominator,
\begin{align*}
\lim_{n \rightarrow \infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
    &= \lim_{n \rightarrow \infty} \frac{n^4}{n^3} \cdot \frac{1-n^{-1}+n^{-4}}{n^{-5/2} + n^{-1}-1}  \\
    &=  \lim_{n \rightarrow \infty} n \cdot \frac{1-0+0}{0 + 0-1}  \\
    &=  -1 \cdot \lim_{n \rightarrow \infty} n  \\
    &= -\infty  \text{.}
\end{align*}

We can also guess at this in the graph of the limitand.

The expression in the limit does approach the simpler expression $-n$ as $n$ increases (after some excitement on $[0,2]$, which does not matter for the limit as $n \rightarrow \infty$ because $[0,2]$ is not near $\infty$).

ContentLicense: CC BY-SA 4.0


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
The vector $(1,-2)$ is also neither positive nor negative, yet there is value in having the operation $-(1,-2)=(-1,2)$.

Note: there is no natural meaning for "positive" for vectors, this is just an analogy.

ContentLicense: CC BY-SA 3.0

Title: What is $-i$ exactly?


问:
Compact 3-manifold with compressible boundary tori?
There is much to be said about 3-manifolds with boundary consisting of a possibly empty collection of incompressible tori.

I don't seem to know where to look to find much about 3-manifolds with compressible torus boundary.

If $M^3$ is compact, with infinite $\pi_1$, and with non-empty boundary, consisting only of compressible tori, a solid torus is an option, but what else can happen?

答: Assuming that your manifold is, say, compact, connected and orientable, whose boundary is a union of tori, then $M$ is homeomorphic to a connected sum of a manifold $M'$ whose boundary tori are all incompressible and some number of solid tori. This is a consequence of the Loop Theorem.


问:
Characterization of the image of the Laplacian
I'm trying to understand the set of functions  $\mathbb R^3 \rightarrow \mathbb R$ and how the Laplacian interacts with them. Recall that the laplacian $\Delta: (\mathbb R^3 \rightarrow \mathbb R)\rightarrow (\mathbb R^3 \rightarrow \mathbb R)$ is defined as

$$
\Delta(f) \equiv \frac{\partial^2 f}{\partial x^2} + \frac{\partial ^2 f}{\partial y^2} + \frac{\partial ^2 f}{\partial z^2}
$$

Now, is there some convenient characterization of the image of the Laplacian, the set:

$$
Im(\Delta) \equiv \{ \Delta(f) \mid  f:   \mathbb R^3 \rightarrow \mathbb R \}
$$

Searching for "image of Laplacian" is not useful since it gives me results about image processing that I don't care about!
Is there a nice characterization of the set $Im(\Delta)$?

答: Search for codomain instead. You may like Sec. 4 here.</p>


问:
What is the probability of an outcome in general, if you know its respective probabilities in a bunch of other situations?
Say there are choices A, B, C, D, and E, and I want to know the probability of A being selected, but I only know the probability of A when compared to only 2 other choices.

For example,
I know for a fact that given the choice between A, B, C : I have a P(A) = .2

A, B, D : P(A) = .3

A, B, E : P(A) = .8

A, C, D : P(A) = .6

A, C, E : P(A) = .43

A, D, E : P(A) = .61

Is there a way for me to find out the probability that A will chosen? In the example I listed all the possible combinations, but what if I was missing one, let's say I didn't have the (A,D,E) variant? How would that effect the probabilities? What if I had the probabilities for all the other 3 choice combinations that don't include A like (B,C,D), (B,C,E)... etc? Would those help me?

For the record, all the numbers I gave are made up, and so is this problem. I'm simply very curious.

答:
What you know in the first case is the probability of $A$ given $ A \cup B \cup D $. From Bayes' theorem you have

$$ P(A | A\cup B\cup C) = \frac{P(A)P(A\cup B\cup C | A)}{P(A\cup B\cup C)} $$
Since your choices are distinct this can be written as

$$  P(A | A\cup B\cup C) = \frac{P(A)}{P(A)+P(B)+P(C)} $$
or using the actual numbers

$$ 0.3 = \frac{P(A)}{P(A)+P(B)+P(C)} \Rightarrow $$
$$ P(A) = 0.3P(A) + 0.3P(B) + 0.3P(C) \Rightarrow $$
$$ 0.7P(A) - 0.3P(B) - 0.3P(C) = 0. $$

If you have four of these kind of relations, together with $ P(A) + \dots + P(E) = 1$, you have a 5x5 linear system of equations, if the numbers are good, you can solve it and find the probabilities.


Title: I can't understand the result of this limit

Body:
$$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$

I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4  (numerator)$$ and $$-n ^ 3(denominator)$$

and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb question, but I didn't understand why is $+\infty$ instead of $-\infty$.

Body:
\begin{align} \lim_{n\to\infty} \left(\frac{n^4-n^3+1}{-n^3+n^2+\sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \frac{n^3-n^2+\frac1n}{-n^3 + n^2 + \sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \left(-\ \frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}} \ .\ \frac{ \frac{1}{n^3} }{ \frac{1}{n^3} }\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \ \left(\frac{1-\frac{1}{n}+\frac{1}{n^4}}{1-\frac{1}{n} -\frac{\sqrt{n}}{n^3}} \ \right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n\right) \ .\ 1 \\
\\
=-\infty.
\end{align}


问:
Finding Concentric Circles from Points
I'm wondering if there is any elegant solution to the following problem. Consider $M$ points known to be on $N$ concentric circles. In addition, assume $M>N$. See the attached picture. Is there a known way of efficiently finding the center and fitting circles to these points? In the picture the points are on radial lines, but the question is for a general configuration of points on circles. As noted here the question does not have a unique solution in general. So, a refinement of this question would be, is there any $M>N$ for which the question has a unique solution? Also, is it possible to reformulate this as a least square problem (and use, for example SVD to solve)?

答:
Consider the following:

Note that there are an infinite number of centers that satisfy your condition.

In this, $M=N+1,$ and there is no unique answer.

Maybe a rephrasing of your question would be solveable?

Title: Finding Concentric Circles from Points


问:
Open and closed from the topologic space through the base
Be $X$ an infinite set and $x_0$ a point of $X$. Be $B = \{\{x\} : x \neq x_0\} \cup \{A \subseteq X : x_0 \in A$ and $X -A$ is finite$\}$.

How are the open and closed of this topology?
Knowing the basis I can find out who are the open of this topology? I just know they're unions of the base elements. But it's still a little confusing for me.

答:
I assume $B$ is a base (which was answered in your previous question Base of topologic space ).
The answer to this question provided by Kavi Rama Murthy in the comments is correct (but the $A$ there should be $U$). Here are the details.

If $x_0\not\in U$ then $U$ is open since $U=\cup\{\{x\}:x\in U\}$ and each $\{x\}$ is open, since $\{x\}\in B$.

If $x_0\in U$ and $X\setminus U$ is finite
then $U\in B$, hence $U$ is open.

Conversely, if $x_0\in U$ and $U$ is open then there is some $A\in B$ with $x_0\in A\subseteq U$.
Then $X\setminus A$ finite (as it is not possible that $A=\{x\}$ for some $x\not=x_0$).
Hence $X\setminus U$ is also finite since $X\setminus U\subseteq X\setminus A$.

Thus, the open sets are exactly those sets that either do not contain $x_0$, or if they contain $x_0$ then their complement must be finite.

Taking complements (of the open sets as described above), the closed sets are all sets that contain $x_0$, as well as all finite sets (regardless whether they contain $x_0$ or not).


问:
Aproximating rational with fraction with "smallest numerator and denumerator possible"
For example $0.795=\frac{159}{200}$. But is there a way to find fraction with smaller numerator and denumerator that will represent number $0.795xyz...$ i.e. it will approximate our given number?

I need algorithm or some procedure for this.

答:
This is solved with continued fractions.  Your specific example $0.795$ is the continued fraction $[0;1,3,1,7,5]$.  It has convergents (best approximations) $0,1,\frac{3}{4},\frac{4}{5},\frac{31}{39},\frac{159}{200}$.  </p>

<p>$\frac{4}{5}=0.8$ is pretty good, and $\frac{31}{39}=0.7948718\ldots$ is even better.</p>

Body:
This is solved with continued fractions.  Your specific example $0.795$ is the continued fraction $[0;1,3,1,7,5]$.  It has convergents (best approximations) $0,1,\frac{3}{4},\frac{4}{5},\frac{31}{39},\frac{159}{200}$.

$\frac{4}{5}=0.8$ is pretty good, and $\frac{31}{39}=0.7948718\ldots$ is even better.


问:
SubGroups of $S_4$ of order $8$ and $12$. Describe by hand
What's a "good" way of describing $S_4$ subgroups by hand? The following post gives a pretty detailed stategy but i can't figure out some tools that are used:

How to enumerate subgroups of each order of $S_4$ by hand

I think I'm not familiar with the use of the "Normalizer" that is used in this post, it may be useful to understand the strategy.

For subgroups of orders less than $8$, I can describe them mentally with intuition, but for subgroups of order $8$ and $12$, I realize I may not be using a good method as I can't describe them by hand.

More precisely, for the example of $A_4$ , the order $12$ subgroup of $S_4$, I understand the following reasoning:
Prime factorization of $12$ is $2^2*3$, so, we know that this subgroup of order $12$ should (if it exists, what I can't prove except by exposing $A_4$) contain a $2$-sylow or a $3$-sylow, more precisely a subgroup of order $2^2$ or $3$. Since this point, I can't really figure out how to move forward.

I also tried to find a proper post that explains it but I never found explanation about the use of the normalizer,

答:
To get a subgroup of order $12$, you might notice this is half of $24 = |S_4|$.  You may also remember that elements of $S_4$ are either odd or even (and there's half of each). This might lead you to ask whether the set of all odd permutations, or the set of all even permutations, is a subgroup. If one of these is a subgroup, then we win. Of course, if you remember the definition of an even permutation, you'll find that the even elements do form a subgroup. This is $A_4$.

As for subgroups of order $8$, you might first realize that once you've found one subgroup of order $8$, you're done. This is a Sylow 2-subgroup, and so every subgroup of order $8$ is conjugate (thus isomorphic). So you might start thinking about what groups have order $8$ (there's only so many) and try to find one which is a subgroup of $S_4$.

After some thought, you'll probably land on the dihedral group $D_8$, which measures the symmetries of a square. Since a square has $4$ vertices, this feels plausible as something to do with $S_4$.

Since $D_8$ is defined geometrically, let's try to reason about it geometrically and show that it really is a subgroup of $S_4$, and then we'll be done.

Notice every element of $D_8$ defines a permutation of the vertices. For instance, the "rotate clockwise" action is the element $(4, 1, 2, 3)$ and the "flip along a vertical axis" element is $(4, 3)(1, 2)$. So then the permutations which arise as symmetries of this square form a subgroup isomorphic to $D_8$.

As an aside, this shows why these groups are all conjugate! Notice the only difference between the subgroups you get is the initial labeling of the vertices. So by relabeling the vertices (which corresponds to conjugation), you can move from one subgroup to another. You might want to think of this as a kind of "change of basis" for these symmetries.

I hope this helps ^_^


问:
Prove that a simple path in a graph exists, if a path exists
I learned that a simple path is a path p = $v_0,...,v_m$ with each $v_i \neq v_{i+1}$ with i $\in [m]$ and its a closed path/cycle, if $v_0 = v_m$, so the last node of the path is the same as the beginning.

Now I came across the statement: "In a simple graph exists a simple path between the nodes u and v, if there is a path between u und v."

But that path could also be a loop, if u = v, which is per definition not a simple path.

So why is this statement true?

答:
If $u=v$, then the sequence "$u$" is a simple $u,v$-path with length $m=0$.

(Note that usually, what you are calling a "path" is called a "walk", and what you are calling a "simple path" is just called a "path"; a closed walk with no repeated vertices except the first and last is called a "cycle".)


问:
Proof of strong Markov property in Oksendal
I am reading Oksendal's SDE book on the strong Markov property for Ito diffusions. There is a step in the proof that I cannot follow. Here is some notation:

$\mathcal{F}_t$ is the sigma-field generated by the Brownian motion $B_r, r \le t$. $E^{x}[\cdot]$ denotes the expected value assuming that the diffusion process $X_t$ satisfies $X_0=x$.

Let $f_1,f_2$ be bounded Borel functions and $\tau$ be a $\mathcal{F}_t$-stopping time. Let $0 \le h_1 \le h_2$.

Then $E^{x}[f_1(X_{\tau + h_1}) f_2(X_{\tau+h_2}) \, | \, \mathcal{F}_{\tau}] = E^{X_{\tau}}[f_1(X_{h_1}) f_2(X_{h_2})]$.

The last line of the proof goes as follows:
$E^{X_{\tau}}[f_1(X_{h_1}) E^{x}[f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}]] = E^{X_{\tau}}[f_1(X_{h_1}) f_2(X_{h_2})]$.

My question is: it seems that the proof is saying that $ E^{x}[f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}] = f_2(X_{h_2})$. This can happen if $X_{h_2}$ is $\mathcal{F}_{h_1}$ measurable which I don't understand since $h_2 \ge h_1$. Am I missing something? I would appreciate some clarification.

答:
Note that $E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]$ does not depend on $x$. This is apparent from the previous line of Øksendal's derivations or, equivalently, from the ordinary Markov property. (It equals
$E^{X_{h_1}}[f_2(X_{h_2-h_1})]$, where $E^y[f_2(X_{h_2-h_1})] = E[f_2(X^y_{h_2-h_1})]$.)

Thus for any $z\in\mathbb{R}$, we find that
\begin{align*}
E^z\!\left[f_1(X_{h_1})E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right]
&=
E^z\!\left[f_1(X_{h_1})E^z[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right] \\
&=
E^z\!\left[E^z[f_1(X_{h_1})f_2(X_{h_2}) \, | \, \mathcal{F}_{h_1}]\right]
=
E^z[f_1(X_{h_1})f_2(X_{h_2})],
\end{align*}
where the last equation follows from the law of iterated expectations. Consequently,
$$
E^{X_{\tau}}\!\left[f_1(X_{h_1})E^x[f_2(X_{h_2}) \, |\, \mathcal{F}_{h_1}]\right]
=
E^{X_{\tau}}[f_1(X_{h_1})f_2(X_{h_2})],
$$
as desired.

I am very happy about your question; it might at first seem to simply be resolved by the law of iterated expectations, but as my answer attempts to highlight, there is a bit more to it. (Good understanding of the ordinary Markov property is a must.)


问:
For $P(x)$ a degree-$n$ polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,\ldots,n$, find $P(n+1)$.
The problem, from USAMO $1975$, reads:

If $P(x)$ denotes a polynomial of degree $n$ such that $P(k) = \frac{k}{k+1}$ for $k = 0,1,2, \ldots, n$. Find $P(n+1)$.

My attempt:

We can represent $p(x)$ as another polynomial which would satisfy all the given conditions. This polynomial would be:

$P(x)$ = $x(x-1)(x-2).....(x-n) + \frac{x}{x+1}$

In other words,

$\frac{P(x)(x+1)-x}{(x+1)}$ = $x(x-1)(x-2)....(x-n)$, or:

${P(x)(x + 1)-(x)}$ = $x(x^2$ $- 1)(x - 2)......(x-n)$

now, simply putting $x = n + 1$, we have

${P(n+1)(n+2)-(n+1)} = \frac{(n + 1)(n^2+ 2n)(n - 1)!}{2}$           (after some simplification)

thus, our final expression would look something like:

$P(n + 1) = \frac{(n + 1)(n^2+ 2n)(n - 1)!+2(n+1)}{2(n + 2)}$

which is not the final answer.

Body: So $$Q(x) = (x+1)P(x)-x$$ is of degree $n+1$ and has zeroes $0,1,2,...,n$, thus we can write it $$Q(x) = ax(x-1)(x-2)\cdots (x-n)$$ so $$(x+1)P(x) =ax(x-1)(x-2)\cdots (x-n)+x$$ which is valid for all $x$ and in particulary for $x=-1$: $$0 =a(-1)^{n+1}(n+1)!-1$$ and for $x=n+1$ we get: $$(n+2)P(n+1) =a(n+1)!+n+1$$ so $$P(n+1)= {(-1)^{n+1} +n+1\over n+2}$$


问:
convergence of series with absolute value
prove or show false:

if $\sum_{n=1}^{\infty}\left |a_{n}  \right |$ converges, then $\sum_{n=1}^{\infty}\frac{n+1}{n}a_{n}$ converges as well.

Thank you very much in advance,

Yaron.

答:
Just notice that,

if $\sum_{n\geq 1}|a_n|<\infty$ , then  $\sum_{n\geq 1}a_n<\infty$. Also, if $\sum_{n\geq 1}a_n$ converges, then $\sum_{n\geq 1}\frac{a_n}{n}$ converges too by Dirichlet's test, so we get

$$ \sum_{n \geq1}^{\infty}\frac{n+1}{n}a_n = \sum_{n \geq 1}a_n + \sum_{n \geq 1}\frac{a_n}{n}< \infty.$$


问:
Solutions to this non-linear ODE
any clue for the equation $y' = \lfloor{y}\rfloor$ ? Thanks !
I've tried basic tricks but none worked out.

答:
Of course there are constant solutions $y = c$ for $0 \le c < 1$, and solutions $y = at + b$ where $a$ is a nonzero integer, defined for $a < at + b < a+1$.
You can't go past $at+b = a$ or $a+1$ unless you allow weak solutions.

Title: Solutions to this non-linear ODE


问:
Continuity at a point and empty interior
I am trying to solve the following problem:

Let $f:[a, b] \rightarrow \mathbb{R}$ be Riemann integrable. Suppose that $f$ is such that for any $c \in [a,b]$, $f$ continuous at $c$ implies $f(c)=0$. I need to prove that then the set $$
X=\{x \in[a, b] ; f(x) \neq 0\}
$$ has empty interior (notation: $\operatorname{int}(x)=\varnothing$).

What have i tried so far?

Since f is continuos at $c \in [a, b]$ for every $\epsilon > 0$ there exists $\delta > 0$ such that $\forall$ $\bar{x} \in [a,b]$ and $|\bar{x} - c|<\delta$ then $|f(\bar{x})-f(c)|<\varepsilon$

Since $f(c) = 0$ we have that $|f(\bar{x})|<\varepsilon$

Now i am stuck. Since i am trying to prove that $X$ has empty interior, i should come to the conclusion that there does not exist $\bar{\epsilon} > 0$ such that $$
(x-\bar{\epsilon}, x+\bar{\epsilon}) \subset X
$$ but i just cannot see the connection between what i have already done and the conclusion i should reach.

Can someone help?

Thanks in advance, Lucas

答:
Hints for a proof from scratch, using your idea:

$1).\ $let $D_{n}=\{x:\omega(f,x)\ge 1/n\}$ and note that the points of discontinuity of $f$ are given by $\bigcup D_n.$

$2).\ f\ $ is integrable so we may take a partition $I_j$ such that $\sum \omega(f,I_j)|I_j|<\epsilon/n$ and let $\Delta$ be the set of indices $j$ such that $D_{1/n}$ has non-empty intersection with the interior of the $I_j.$

$3).\ $ Conclude that $D_{1/n}$ is contained in $\bigcup_{j\in \Delta} I^\circ_j$ plus possibly the endpoints of these intervals.

$4).\ $ Conclude that $\sum_{j\in \Delta}|I_j|<\epsilon$ and therefore that $D_{1/n}$ can be covered by finitely many $\textit{open}\ $intervals the sum of whose lengths is less than $2\epsilon$ (the factor of $2$ enters because you have to consider the endpoints of the $I_j$, which is no problem since the number of these is finite).

$5).\ $ Conclude now that as $\epsilon$ is arbitrary, $D_{1/n}$ and hence $D$ cannot contain an open interval-


问:
Bellmann Equation loss function optimal Q-Value
I am currently working on reinfocement learning and there is this Bellman Equation which I need, so I can minimize the loss-Function calculated by my neural Network. When we calculate the loss, we compare the Q-Value that is generated by my neural Network q(s,a) and subtract that from the optimal Q-Value q*(s,a). I dont understand the difference between q* and q, because if we already have the optimal q-Value, then why do I even bother to compute q(s,a)? Or in Q-Learning, where I can look in my Q-Table to get the maxarg(q(s´,a)) to update my table. I dont understand the difference between those two because right now the way I get my q(s,a) is the same as q*(s,a). Help is really appreciated, I googled the whole weekend and couldnt find any solution.

Here is an image that maybe underlines my problem

答:
Okay I just realized my mistake when I programmed it. Its actually quite obvious. The q-value of q(s,a) should be equal to the calculation we do with the Bellman Equation.

```<code>q_optimal = reward + discount_rate * np.max(q_table[new_state, :])
q_now = q_table[state, action]
loss = q_optimal - q_now
</code>```

As you can see, q_optimal is calculated differently than q_now, so there are definetely not the same, I didnt get that at first.

Body:
Okay I just realized my mistake when I programmed it. Its actually quite obvious. The q-value of q(s,a) should be equal to the calculation we do with the Bellman Equation.

q_optimal = reward + discount_rate * np.max(q_table[new_state, :])
q_now = q_table[state, action]
loss = q_optimal - q_now

As you can see, q_optimal is calculated differently than q_now, so there are definetely not the same, I didnt get that at first.


问:
Distribution of $X_t=\mathbb{E}(B_1\mathbb{I}_{|B_1|\geq1}-B_1\mathbb{I}_{|B_1|<1}|\mathcal{F}_t)$?
I am asked to consider the process

$$X_t=\mathbb{E}(X|\mathcal{F}_t),\quad X=B_1\mathbb{I}_{|B_1|\geq1}-B_1\mathbb{I}_{|B_1|<1}=\begin{cases}B_1&\text{if }|B_1|\geq1\\-B_1&\text{otherwise}\end{cases}$$

for $t\in[0,1]$, where $\{B_t\}$ is a standard BM and $\mathcal{F}_t=\sigma(B_s:s\in[0,t])$. I am tasked to obtain the following:

i) Find the distribution of $X$ and $X_t$, and prove $X_t$ is a martingale;

ii) Show that $X_t=B_t-2f(B_t,\sqrt{1-t})$ where $f(x,s)=\mathbb{E}\left[(x+sZ)\mathbb{I}_{|x+sZ|<1}\right]$ and $Z\sim\mathcal{N}(0,1)$, and consequently derive a PDE with an initial condition based on $s$.

For i), the comments and current answer have explained why it is a martingale. I also now know that for ii), we have to use Ito’s formula and make the drift term disappear.

Any guidance is greatly appreciated!

答:
This is not a full answer but it may be a hint to further solution.

$X_t$ is Levy martingale hence it's martingale.

Put $\xi = B_t$ and $\eta = B_1 - B_t$.

Thus $X_t = E( B_1 - 2 B_1 I_{|B_1| < 1} | \mathcal{F}_t) = E(h(\xi,
\eta) | \mathcal{F}_t) $
where $h(a,b) = a + b- 2(a +b) I_{|a + b| < 1}$.

We know that $\xi$ is $\mathcal{F}_t$-measurable. As $\eta$ and $\mathcal{F}_t$ are independent we have

$$X_t = E(h(\xi, \eta) | \mathcal{F}_t) = g(\xi),$$
where $g(x) =  E(h(\xi, \eta) | \xi = x)$.

Further, $g(x) =  E(h(\xi, \eta) | \xi = x) =  E(h(x, \eta) | \xi = x) = Eh(x, \eta) $ by independence. Put $\eta = \sqrt{1-t} \zeta$. As $\eta \sim N(0, 1-t)$ we have $\zeta \sim N(0,1)$. Thus
$$g(x) = Eh(x, \eta) = E( x + \eta- 2(x +\eta) I_{|x + \eta| < 1} )= x - 2E(x +\eta) I_{|x + \eta| < 1} $$
$$= x -  2E(x +\sqrt{1-t} \zeta) I_{|x +\sqrt{1-t} \zeta| < 1} = x - 2f(x, \sqrt{1-t})$$
and
$$X_t = g(\xi) =  \xi - 2f(\xi, \sqrt{1-t}) = B_t - 2f(B_t, \sqrt{1-t}) .$$

We proved that $X_t = H(B_t, t)$. Now we may try to find the distribution of $X_t$ or use Ito formula, but a function $f(x,s)$ doesn't look good.


问:
Why does a cumulative distribution have the quality that $P(X<b)=\lim_{n\to\infty}\left[P\left(X\leqslant b-\frac1n\right)\right]$?
I came across the following passage in Ross' "First Course in Probability":

If we want to compute the probability that X is strictly less than b , we can apply the continuity property to obtain:
$$P\bigl(X<b\bigr)=\lim_{n\to\infty}\left[P\left(X\leqslant b-\frac1n\right)\right].$$

The continuity Ross is referring to, is the right continuity of the cumulative function but the property he mentions seems to me to be a property of a left continuity since the sequence $b-\dfrac1n$, where $n$ goes to infinity, is an increasing sequence that converges to $b$ from left to right.
Will be grateful for any enlightening remarks on this.

答:
This is a consequence of continuity of measures. (As pointed out in a different answer, please consult Section 2.6 in Ross'.)

Continuity from below

Let $\mu$ be a measure on $(\mathcal{X},\mathcal{E})$, and let $E,E_1,E_2,\ldots$ be sets in $\mathcal{E}$. If $E_n \nearrow E$, then $\mu(E_n) \nearrow \mu(E)$.

Application

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a background probability space, and let $X : (\Omega,\mathcal{F}) \to (\mathbb{R},\mathbb{B}(\mathbb{R}))$ be some random variable. Denote by $X(\mathbb{P})$ the pushforward measure on $(\mathbb{R},\mathbb{B}(\mathbb{R}))$ given by $X(\mathbb{P})(A) = P(X \in A)$ for $A\in\mathbb{B}(\mathbb{R})$.

Let $b\in\mathbb{R}$, let $B = (-\infty,b)$, and let $B_n = (-\infty,b-1/n]$ for $n\in\mathbb{N}$. Clearly, $B, B_1, B_2, \ldots$ are subsets of $\mathbb{B}(\mathbb{R})$. Note that $B_n \nearrow B$. Then according to the continuity from below of the probability measure $X(\mathbb{P})$, we find that
$$
X(\mathbb{P})(B_n) \nearrow X(\mathbb{P})(B).
$$
Rewriting according to the definition of the pushforward measure yields
\begin{align*}
\mathbb{P}(X \leq b-1/n) \nearrow \mathbb{P}(X < b)
\end{align*}
as desired.

Direct proof from axioms

Recall Kolmogorov's axioms: $\mathbb{P}$ is non-negative, $\mathbb{P}(\Omega)=1$, and if $E_1,E_2,\ldots$ is a sequence of mutually exclusive events, then
\begin{align*}
\mathbb{P}\!\left(\cup_{n=1}^\infty E_n\right) = \sum_{n=1}^\infty \mathbb{P}(E_n).
\end{align*}
This last axiom is also known as subadditivity.

Let $F_n = (X \leq b-1/n)$ for $n\in\mathbb{N}$. Note that $\cup_{n=1}^\infty F_n = (X < b)$. Define $E_n$ recursively by $E_1 = F_1$, $E_2 = F_2 \setminus F_1$, $E_3 = F_3 \setminus F_2$, and so on. For example, $F_2 = (b-1 < X \leq b-1/2)$. Note that $E_1,E_2,\ldots$ is a sequence of mutually exclusive events, and that $\cup_{n=1}^\infty E_n = \cup_{n=1}^\infty F_n = (X < b)$. Then by twice applying first countable subadditivity and then finite subadditivity (i.e. Kolmogorov's last axiom),
\begin{align*}
\mathbb{P}(X < b) &= \mathbb{P}\!\left(\cup_{n=1}^\infty E_n\right) \\
&= \sum_{n=1}^\infty \mathbb{P}(E_n) \\
&= \lim_{N\to\infty} \sum_{n=1}^N \mathbb{P}(E_n) \\
&= \lim_{N\to\infty} \mathbb{P}(\cup_{n=1}^N E_n) \\
&= \lim_{N\to\infty} \mathbb{P}(F_N) = \lim_{N \to \infty}\mathbb{P}(X \leq b - 1/N).
\end{align*}


问:
Exercises in FOL
From Section 4.2 of "An introduction to mathematical logic" By Richard Hodel.

I would appreciate comments on how to proceed with exercise $(3)$ and whether $(4)$ is correct:

Express these syllogisms in the language of FOL and in each case decide whether the syllogism is valid or invalid.

$ (3)$ $(i)$ No wheelbarrows are comfortable vehicles. $(ii)$ No uncomfortable vehicles are popular. $\therefore$ No wheelbarrows are popular.

soln.

$(i)$  $\forall x[W(x) \to \lnot C(x)]$ $\equiv \forall x[\lnot W(x) \lor \lnot C(x)]$

$(ii) \forall x [\lnot C(x) \to \lnot P(x)]\equiv \forall x (C(x) \lor \lnot P(x))$

$\therefore \forall x [W(x) \to \lnot P(x)]\equiv \forall x (\lnot W(x) \lor \lnot P(x))$

I'm stuck with drawing conclusion now:

Since in $(i), (ii)$ there are $\lnot C(x)$ and $C(x)$ respectively, we are left with $\lnot W(x)$ and $\lnot P(x)$ from $(i)$ and $(ii)$. So we have $\lnot W(x) \land \lnot P(x)$. But we need $\lnot W(x) \lor \lnot P(x)$.

$(4)$ $(i)$ Some healthy people are fat. $(ii)$ No unealthy people are strong. $\therefore$ Some fat people are not strong.

$(i)$ $\exists x(H(x) \land F(x))$

$(ii) \forall x (\lnot H(x) \to \lnot S(x))\equiv \forall x( H(x) \lor \lnot S(x)).$

$\therefore \exists x(F(x) \land \lnot S(x)).$

From $(i)$ I know that $(H(x) \land F(x))$ is true, so both relations here must be true for particular $x$.

For $(ii)$ there is $H(x)$ as well, but for all $x$. Since here logical operation is $\lor$, $\lnot S(x)$ can be either $T$ or $F$ for $(ii)$ to be true.

Thus putting $(i),(ii)$ together: $\exists x[(H(x) \land F(x)) \land \lnot S(x)] \lor [(H(x)\land F(x)) \land S(x)].$

Therefore initial syllogism $\exists x(F(x) \land S(x))$ is not valid.

答:
In $(3)$ you actually do get the desired conclusion. I don’t know exactly what formal derivational rules are available to you, so I’ll reason informally. Let $x$ be arbitrary, and suppose that $W(x)$. Then (i) tells you that $\neg C(x)$, at which point (ii) tells you that $\neg P(x)$. In other words, $\forall x\big(W(x)\to\neg P(x)\big)$. In both steps I’m using the tautology $\big((p\lor q)\land\neg p\big)\to q$.

In $(4)$ you don’t know that $H(x)\land F(x)$ is true: you know that there is at least one entity $c$ such that $H(c)\land F(c)$ is true. I don’t know whether you’re permitted to use this approach, but I’d show that the syllogism is invalid by producing a counterexample. Suppose that there is exactly one person, $c$, in the universe of discourse, and that $F(c)$, $H(c)$, and $S(c)$ are all true. Clearly (i) is satisfied, since $H(c)\land F(c)$ is true. (ii) is vacuously satisfied, since $\neg H(x)$ is false for all $x$ (in the universe of discourse). Finally, $F(c)\land\neg S(c)$ is false, so there is no $x$ (in the universe of discourse) such that $F(x)\land\neg S(x)$ is true. Thus, it is consistent that (i) and (ii) hold, but the conclusion does not.

Title: Exercises in FOL


问:
Infinite Series question

The first one, the effective resistance is $2R$, then $5R/3$ then $13R/8$ etc....
My job is to find the pattern/equation so I can find the total resistance when $20$ resistors are connected. Of course I can do the long way by using this formula I created
$R_n = (R_{n-1} \times R)/(R_{n-1} +R) +R$. Can you guys help me find an equation where I can just input the value of $n$ and simply get an answer. Thanks!

The second question is: What would I get if I had an infinite number of resistance.

Help is very much appreciated :)

答:
The second question is easier. Suppose that the resistance for infinite number of elementary parts is equal to $R_{\infty}$. Then it should not change if you add one more elementary part. This gives the equation
$$R_{\infty}=\frac{RR_{\infty}}{R+R_{\infty}}+R,$$
with the solution $\displaystyle R_{\infty}=\frac{1+\sqrt{5}}{2}R$.

Now concerning the first question, you should first check that the initial recurrence relation
$$R_{n+1}=\frac{RR_n}{R+R_n}+R,$$
being rewritten in terms of the variable
$$u_n=\frac{1}{R_n-R_{\infty}}+\frac{R+R_{\infty}}{R_{\infty}(2R+R_{\infty})},\tag{1}$$
is equivalent to
$$u_{n+1}=\frac{(R+R_{\infty})^2}{R^2}u_n.$$
(this is just a rephrasing of approach of Did in simpler terms). This has the obvious solution
$$u_{N}=\left(\frac{R+R_{\infty}}{R}\right)^{2(N-1)}u_1.\tag{2}$$
Now from (1), (2) and $R_1=2R$ one obtains a general formula for $R_N$.

Title: Infinite Series question


问:
$\frac{1}{u_n}W_n-v_n\Rightarrow W$ implies $u_n \to +\infty$
Suppose $(X_n)_n$ are i.i.d. random variables and let $W_n = \sum_{k=1}^n X_k$. Assume that there exist $u_n>0 , v_n \in \mathbb{R}$ such that

$$\frac{1}{u_n}W_n-v_n\Rightarrow W$$

where $W$ is not degenerate. Show that

$$u_n\to \infty , \frac{u_n}{u_{n+1}}\to 1$$

What happens to $u_n$ if $W$ is degenerate?

Hint: You may need to consider ${u_{2n}}/{u_n}$.

Is the following attempt, for the first part, true?

In order to remove $v_n,$ we can consider $\frac{1}{u_n}\sum_{k=1}^n(X_{2k+1}-X_{2k})$ which converges in distribution to a non-degenerate random variable $Y.$ So we can suppose that $\frac{1}{u_n}W_n$ converges in distribution to $W$.

If $W$ is non-degenerate then there exist $x \in \mathbb{R};|\phi_W(x)|<1,$ since $\frac{1}{u_n}W_n-v_n\Rightarrow W$ then there exist $k \in \mathbb{N};|\phi_{X_1}(\frac{x}{u_k})|<1$ which means that $X_1$ is not degenerate, if $(u_n)_n$ is bounded from above then there exist a subsequence $(u_{k_n})$ such that $W_{k_n}$ converges in distribution.
Let $(u_{q_n})_n$ be an arbitrary subsequence, since $X_1$ is not-degenerate then $\sum_{l=1}^{q_n}X_l=W_{q_n}$ doesn't converges in distribution, so $u_{q_n}$ is not bounded from above and we can extract a subsequence from $u_{q_n}$ diverging to $+\infty.$

In case $W$ is degenerate, is it possible to know the behavior of $u_n$?

答:
Suppose $\frac {W_n} {c_n} $ converges in distribution to $W$ and $(c_n)$ does not tend to $\infty$. Then there is a subsequence $c_{n_k}$ converging to some real number $c$. This implies that $W_{n_k}$ converges in distribution to $cW$. Hence $(\phi(t))^{n_k} \to Ee^{itcW}$ where $\phi$ is the characteristic function of $X_i$'s. Since $Ee^{itcW}$ does not vanish for $t$ near $0$ it follows that $|\phi (t)|=1$ for al $t$ near $0$. This implies that $X_i$'s are a.s  constants. In this case $c_n \sim nc$.

If $W$ is allowed to be degenerate, take $X_n=0$ for all $n$. Obviously nothing can be said about $c_n$ in this case.


问:
Automorphism group of complex manifold
The (holomorphic) automorphism group  of the unit disc is $SU(1,1)$ which is not a complex Lie group!

So why (intuitively) the (holomorphic) automorphism group of a complex manifold  wouldn't have
to be a complex Lie group even though if it is finite dimensional?

答:
First of all, the automorphism group of a complex manifold $M$ is not always finite dimensional, the standard example is ${\mathbb C}^2$, its automorphism group contains the infinite-dimensional subgroups of "shears"
$$
(z,w)\mapsto (z, h(z)+w),
$$
where $h$ is an arbitrary entire function on ${\mathbb C}$. However, if $M$ is compact then indeed, $Aut(M)$ is naturally a real Lie group. Typically, though, the automorphism group is trivial.

Suppose you impose further assumptions on $M$, say, that it is a simply-connected Hermitian symmetric space (HSS). Two extreme cases of HSS's are:

Hermitian symmetric spaces of noncompact type. For instance, the open unit ball in ${\mathbb C}^n$.

Compact HSS. For instance, complex projective spaces.

As it turns out, in the second case, the automorphism group is naturally a complex Lie group, meaning that it has the structure of a complex Lie group $G$ such that the action map
$$
G\times M\to M
$$
is holomorphic. Briefly, the space of holomorphic vector fields on $M$ is finite-dimensional (since $M$ is compact) and is naturally a complex vector space (that much is indeed clear and is true regardless of compactness, etc.). Each holomorphic vector field is integrable (again, by compactness), hence, you conclude that the identity component of $G$ is naturally a complex Lie group.

On the other hand, in the former case, the automorphism group is never naturally a complex Lie group. Here is a reason (hopefully, intuitive enough). As it turns out, each HSS of noncompact type is biholomorphic to a bounded domain $D$ in ${\mathbb C}^n$ and the stabilizer of a point $x\in D$ is a compact connected nontrivial Lie subgroup $K< G$. Now, if $G$ were to admit a complex structure such that its action on $D$ is holomorphic, then $K$ would be also a complex Lie subgroup of $G$. But then, for each point $y\in D$ the $K$-orbit

$$
Ky=\{k(y): k\in K\}
$$
would be a compact complex submanifold of $D$, hence, of ${\mathbb C}^n$. Thus, $Ky$ would be a single point (namely, $y$). It would then follow that $K$ is a trivial subgroup, which is a contradiction.

Title: Automorphism group of complex manifold


问:
Getting Standard Deviation of Population from s. d. of individual samples. Given mean and standard deviation of samples, with no access to data.
I know mean and standard deviation of individual samples (let say $10$ data points each, and no access to the underlying data); could I get the standard deviation of the population (let say $100$ samples) from the standard deviation of the individual samples? Which is the relationship?

EDIT

Thanks to BruceET's Answer. I've applied approach-$2$ to our data, and I'm just a little puzzled on results I get when calculating $S_{all}$. I would expect that $S_{all}$ is greater than the average of the individual $S_{n}$ (since there is a spread as well of the mean of subsets); on the contrary I get an $S_{all}$ that is smaller. How this should be understood?

Processed Data

答:
Two approaches, depending on your application:

(1) If your purpose is to use SDs of several samples from
the same population to get a 'pooled' estimate of the SD of the population, then you can expand the idea of a 'pooled' variance used in a two-sample t test.

Suppose you have three random samples of sizes $n_1, n_2, n_3$ from the same
population with respective sample standard deviations $S_1, S_2, S_3.$ Then the pooled estimate of the population variance $\sigma^2$ is

$$S_p^2 = \frac{(n_1-1)S_1^2+(n_2-1)S_2^2+(n3-1)S_3^2}{n_1 +n_2+n_3-3}.$$

Take the square root to get the pooled estimate $S_p$ of $\sigma.$ The formula expands in the obvious way to pool more than three samples.

(2) If you need to find the sample standard deviation of the combined sample, then method (1) won't work. Because each sub-sample has its own mean, $S_p$ above
will not be exactly the same as the sample standard
deviation of all $n_1+n_2+n_3$ observations taken as a
grand sample.

To get the value of $S_{all},$ is a bit more tedious,
but it is possible. I will outline the procedure,
which is based on the right-hand side of the formula
below. [If you need to combine many samples, then you may want to write a brief computer program to implement this approach.]

$$S^2= \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)^2 \\
= \frac{1}{n-1}\left(\sum_{i=1}^n X_i^2 - \frac 1n\left(\sum_{i=1}^n X_i\right)^2\right) \\
= \frac{1}{n-1}\left(Q - \frac1nT^2\right)\!.$$

a.  For the first sample, use $n_1$ and $\bar X_1$ to get
$T_1 = n_1\bar X_1.$ Similarly, from the other samples get
$T_2$ and $T_3$ and then find $T = T_1+T_2+T_3,$ the grand total for the combined sample.

b. For the first sample, use $n_1, T_1,$ and $S_1^2$ to solve
for $Q_1,$ the sum of squares for the first sample.
Similarly, from the other samples get $Q_2$ and $Q_3$
and then find $Q = Q_1+Q_2+Q_3$ for the combined sample

c. Finally, use the displayed equation once again (with $n = n_1+n_2+n_3,\,$ $Q,$ and $T)$ to get
$S_{all}^2$ for the combined sample. Take the square
root to get $S_{all}.$


问:
Solving Equation of form: *num* = $x^3 + ax^2 + bx + c$.
The solution to the following:
$$
45113 = x^3 + ax^2 + bx + c
$$
is:
$$
45113 = 31^3 + 15 · 31^2 + 29 · 31 + 8
$$

My question is how does one go about solving this set of equation, I'd
like to write a program that solves this but need to know what kind of equation I'm looking at and what steps are involved at getting to the
solution. I'm familiar with Linear Diophantine and Chinese Remainder Theorem solving but not sure where, to begin with, this one. I just need to find a guide or know how to step by step solve this equation, but google searches aren't coming up with anything.

Any papers or guides on how to do this would be helpful, thank you.

Ok, here is how to solve, choose a value below the cube root of 45113. Then perform the following steps:

If there are restrictions on x apply them in selecting the integer x. For example select a positive integer less than the cube root of 45113, ie ≤ 35, for x.

Subtract x³ from 45113

Divide the result by x². The integer result is the coefficient “a” .

Subtract ax² from the result of step 1.

Divide this amount by x. The integer amount is the coefficient “b” .

Subtract bx from the result of step 3. The new result is “c” .
This will work, no matter the integer value of x.

答:
You have some degrees of freedom here. Assuming you want a solution or some solutions and not all solutions, I would do something like this based on Vieta's formulas

$$\begin{align*} 0 &= x^3+ax^2+bx+(c-n) \\
\\
&= x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_2r_3+r_1r_3)x-r_1r_2r_3
\end{align*}$$

Choose a $c$. $ (n-c)=r_1r_2r_3$ is the product of the roots of the cubic equation.

Factor an integer factor out of $n-c$, and call that root $r_1$.

Choose the 2 other roots, $r_2$ and $r_3$, such that $r_2r_3 =\dfrac{n-c}{r_1}$.  They need not be integers, and they can be complex conjugates if you want.  They should sum to an integer, and things are easier to manage if the real parts of their individual products with $r_1$ are also integers.

Set the values of $a = -(r_1+r_2+r_3)$, and $b=(r_1r_2+r_2r_3+r_1r_3)$ from Vieta's formulas.

Now you should have integer $c, r_1, a, b$ so you can write

$$n=r_1^3+ar_1^2+br_1+c$$


问:
Why is the volume of a sphere $\frac{4}{3}\pi r^3$?
I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?

答:
I wanted to post this as a comment to the answer by Justin L. but it is my first posting here so I couldn't. Archimedes was so proud of his proof that the volume of a sphere is two-thirds of the volume of a cylinder of the same height and diameter that he asked that a sculpture illustrating this would be placed at his tomb. The tomb was later found by the Roman orator Cicero who describes it as follows:

"I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder."

The American painter Benjamin West imagined the scene in his 1797 painting "Cicero Discovering the Tomb of Archimedes".

LastEditorDisplayName: anonymous


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
You must take care, even though $2^2 = (-2)^2 = 4$ this doesn't mean that $2 = -2$, indeed the problem here (thinking about reals for a while), is that if you consider negative numbers, then the function $f: \mathbb{R}\to\mathbb{R}$ given by $f(x) = x^2$ is not injective.

Since you have created $i$ to be $i=\sqrt{-1}$, then you are obviously allowed to create a new number $-i = -\sqrt{-1}$. Also, if you think on the function $f: \mathbb{C} \to \mathbb{R}^2$ given by $f(a+bi) = (a,b)$, this function identifies $\mathbb{C}$ with $\mathbb{R}^2$, and in this sense, a complex number is a point in the plane. Note that we have $f(i) = (0,1)$ and that $f(-i) = (0,-1)$, so there are two things to note: every number that can be written $ai$ for some $a \in \mathbb{R}$ can be identified with a point in the $y$ axis, and introducing $-i$ we have not just the upper $y$ axis, but the complete one.

EDIT: Because of the comments I thought it would be good to add this edit. Above I've provided a geometrical view of what happens when we introduce $-i$, now I'm just going to say a little about the algebraic point of view. Since every complex number $z \in \mathbb{C}$ is $z=a+ib$ for some $a,b \in \mathbb{R}$ if we do not allow things like $-i$ we would have some issues. For instance, consider quadratic equations with complex conjugate roots: one of them would be a well defined element of $\mathbb{C}$, but the other one wouldn't! Also, it wouldn't be possible to present for each $z$ some $-z$ with the nice property $z + (-z) = 0$ and it turns out that $\mathbb{C}$ wouldn't be a field. Also, if you use the standard operations of sum between complexes and you have $z = a+ib$ and you want the additive inverse  $-z$ you can prove rather easily that you must have $-z = (-a) + i(-b)$, so that $-i$ would just really be $i(-1)$, so in practice, the element $-i$ (and any other negative complex in general) appears naturally when we define the complexes as we do and when we introduce the operations as we do.

Title: What is $-i$ exactly?


问:
Field extension finitely generated by algebraic elements
I'm trying to prove the following:

Let $K/F$ be a field extension with fields $F\subseteq K$. If $K$ is finitely generated by elements that are algebraic over $F$, i.e. $K=F(a_1,\ldots , a_k)$ for $a_i \in K$ algebraic over $F$, then $K/F$ is a finite field extension.

In the lectures of my professor a proof of this is given, but very short and compact. I have problem with the following conclusion:
We assume $F(a_1, \ldots , a_k)=K$ for algebraic elements $a_i\in K$ over $F$. Then define $F_0:=F$ and $F_{i+1}:=F_i(a_{i+1})$. This gives us a tower of fields $F=F_0 \subseteq F_1 \subseteq \ldots \subseteq K=F_k$. So far everything is understandable, but now the part that I don't understand:

If we define $n_i:=[F(a_i):F] \in \mathbb{N}$, then it follows that $[F_{i+1}:F_i]\leq n_{i+1}$. Why does this follow?

答:
This depends on a theorem you’ve  probably seen. He’s claiming that
$[F_i(a_{i+1}):F_i]\le[F(a_{i+1}):F\,]$. But this follows from the theorem that if $b$ is algebraic over a field $k$, and $k'$ is any extension of $k$, then $[k'(b):k']\le[k(b):k]$. Very likely, you immediately see the proof of this last.


问:
Sequences of RVs have equal finite-dimensional distribution, then the "supremum" processes have equal finite-dimensional distributions??
Say $(X_{1},...,X_{n})$ and $(Y_{1},...,Y_{n})$ have the same finite dimensional distributions, can I say that for any $A\in \mathcal{B}(\mathbb R)$,

$$ P(\overline{X_{n}}\in A)=P(\overline{Y_{n}}\in A)$$ where $$\overline{X_{n}}:= \sup\limits_{j=1,...,n}X_{j}\text{  and   }\;\overline{Y_{n}}:= \sup\limits_{j=1,...,n}Y_{j}$$.

My idea: Consider the generating set of $\mathcal{B}(\mathbb R)$, namely, $\{ (-\infty,x]: x \in \mathbb R\}$. Then for $x\in \mathbb R$, we have:

$$ P(\overline{X_{n}}\in (-\infty,x])=P(\cap_{j\in [n]}\{{X_{j}}\in (-\infty,x]\})=P(\cap_{j\in [n]}\{{Y_{j}}\in (-\infty,x]\})=P(\overline{Y_{n}}\in (-\infty,x]).$$

We then use the $\pi-\lambda$ theorem to extend the result to the entire sigma algebra $\mathcal{B}(\mathbb R)$. Is this approach correct?

Additional question: Can I indeed say that the vectors $(\overline{X_{1}},...,\overline{X_{n}})$ and $(\overline{Y_{1}},...,\overline{Y_{n}})$ have the same finite-dimensional distributions?

Thanks for the help.

答:
You are right.
There is

lemma: if $\mathcal{\pi}$ is $\pi$-sistem of subsets of $\Omega$ then two measures are equal on $\mathcal{\pi}$ iff they are equal on $\sigma(\mathcal{\pi})$.
To prove this lemma we may use $\lambda$-systems.

Hence two probability measures on $\mathbb{R}$ with Borel sigma-algebra are equal iff they are equal on $(-\infty, x]$ for every $x$.

Hence the first statement is true.

Notice that instead of $\sup$ it's more natural to work with $\max$.

As $Z_1 = (\overline{X}_1, \overline{X}_2, \ldots, \overline{X}_n ) = h(X_1, \ldots, X_n)$ and $Z_2 = (\overline{Y}_1, \overline{Y}_2, \ldots, \overline{Y}_n ) = h(Y_1, \ldots, Y_n)$ it follows that
$$P(Z_1 \in B) = P(h(X_1, \ldots, X_n) \in B) = P( (X_1, \ldots, X_n) \in h^{-1}B )$$
$$ =P( (Y_1, \ldots, Y_n) \in h^{-1}B ) =  P(h(Y_1, \ldots, Y_n) \in B) = P(Z_2 \in B) $$
for all Borel $B$.

It follows that second statement is true.


问:
Logical propositions and predicates "quantifiers"
In logic, quantifiers $\forall$ and $\exists$ give mathematical objects. Is there any equivalent "quantifier" giving a logical proposition or predicate?

For example, if I denote the "for all predicate P" by $\square P$, I could write:
$$\forall x,\forall y,[x = y \Leftrightarrow (\square P,P(x)\Leftrightarrow P(y))]$$
My question is then: is there any standard notation for $\square$? Is this concept even useful? Can we consider propositions and predicates as a specific type of object?

Thanks in advance.

答:
Yes. The ability to quantify over predicates is the defining feature of second order logic. You can then ask if you can quantify over predicates that expect predicates and so forth, and these give rise to other higher order logics in an analogous way.

As an example in second order logic, you can say:

$$\forall P . \forall x . P(x) \lor \lnot P(x)$$

Here we are quantifying first over all predicates $P$, then over all elements $x$, then asserting the law of the excluded middle holds in every case.

These logics are fairly well studied, but they're substantially less well behaved than first order logic. In particular things like the compactness theorem fail because there's no good notion of proof theory for higher order logics.

A big part of the reason we use set theory as our foundations is because we can imitate higher order logic inside first order logic. Since everything is a set in ZFC, we can "quantify over propositions" by looking at the set of propositions (which is itself an element of our theory). This is a big part of where set theory gets its power, and I seem to recall a prominent set theorist saying "any higher order logic is set theory in sheep's clothing", but I'm forgetting where I saw it. I'm sure a quick google will bring it up if you're interested.

I hope this helps ^_^


问:
Integration is zero implies $g$ is continuous
Let $g$ be a monotone function such that
$$ \int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y)=0$$
where $f(x,y)= 1$ if $x-y \in \mathbb{Z}$ otherwise it is $0$.

The integration is w.r.t. Riemann-Stieltjes sense. How can we show that $g$ is continuous?

My try:

If $g$ is not continuous at $c$, $g$ can have only jump discontinuity at $c$ . I am not being able to use this fact.
Any help or hint will be appreciated. Thanks in advance.

Update:
If $f$ is not continuous at a point $c \in (0,1)$.
Let us choose $\epsilon>0$ such that, $(c-\epsilon, c+ \epsilon) \subset (0,1)$.
Then,
$$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq \int_0^1 \int_0^1 f(x,x) \,dg(x) \,dg(x)=
 \int_0^1 \int_0^1  \,dg(x) \,dg(x) = [  \int_0^1  \,dg(x) ]^2 \geq [  \int_{c -\epsilon}^{c+ \epsilon}  \,dg(x) ]^2 = [g(c+\epsilon)- g(c-\epsilon)]^2 >0$$, which is a contradiction. Thus $g$ is continuous at $c\in (0,1)$.

If $g$ is not continuous at $0$, then
$$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq [g(\epsilon)- g(0)]^2 >0$$, a contradiction.

If $g$ is not continuous at $1$, then
$$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq [ g(1)-
g(1-\epsilon)]^2 >0$$, a contradiction.

Thus, $g$ is continuous on [0,1].

I want  to justify the step
$$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq \int_0^1 \int_0^1 f(x,x) \,dg(x) \,dg(x)$$ with proof and hope the other parts are correct.

Any help or hint will be appreciated. Thanks in advance.

答:
If $g$ is not continuous at some $0< c\le 1$ then, without loss of generality, $g(x)=I(x-c)$ so that $g$ has one and only one unit jump at $x=c.$

Now, $f=0$ except on the corners of the unit square and the diagonal, where $f=1.$ Then since $f(c,c)=1$, we see that $\int_0^1f(x,c)dI(x-c)$ does not exist, because $f$ has a discontinuity at the point where $g$ has a jump.


问:
need help solving - system of equations
i was writing a model paper for a olympiad when i encountered this question:
i thought of using cramer's rule or just proceed with matrix inversion method but i am stuck trying to figure it out. is there any possible to find the answer?

Consider a system of equations:

$A_1 x+B_1y+C_1 = L$

$A_2 x+B_2y+C_2 = M$

$A_3x+B_3y+C_3 = N$

given L+M+N = 0 what kind of solutions will the above system of equations have ?

答:
If $L+M+N = 0$ then we would have $(A_1+A_2+A_3)x + (B_1+B_2 + B_3)y + (C_1 + C_2 + C_3) = 0 $ by adding all your equations. This is the equation for the line
$$ y = -\frac{C_1 + C_2 + C_3}{B_1+B_2 + B_3} - \frac{A_1+A_2+A_3}{B_1+B_2 + B_3}x.$$
So any solution to your equations will be on this line.</p>

Title: need help solving - system of equations


问:
Solve Compound Inequality Without Separation With Variables on Outside
I have the following compound inequality:

$3y-5 < -2y < 2+y$

I know I can separate them into two different equations, and go from there. I have been taught how to solve them without separation when the variable is in the center. But I have not been taught how to do this when there are variables on the outside like this one? How do you I "put the y variable back into the center?

The way I see it, I  can't do it. I can -y from both sides and add it to the center, but still leaves a 2y on the left. I can -3y and add it to the center, but that leaves a -2y on the right. The only way to solve it is through separation. Or am I wrong?

答:
Compound inequalities are just single inequalities that must be simultaneously true.  In other words, $$3y - 5 < -2y < 2 + y$$ is just a more compact way of saying $$3y - 5 < -2y \quad {\it and} \quad -2y < 2 + y.$$  Then you solve the individual inequalities:

$$3y - 5 < -2y$$ implies $$5y < 5$$ which is equivalent to $$y < 1.$$  The second inequality is $$3y > -2$$ or $$y > -\frac{2}{3}.$$  Now you put them back together; we must simultaneously have $y < 1$ and $y > -\frac{2}{3}$, or $$-\frac{2}{3} < y < 1.$$

I believe this process is what you call "separation."  I do not advise trying any other shortcut method that involves working both inequalities at the same time.  It's not that it can't be done, it's that it's potentially confusing and you may make a mistake, especially as the inequalities become more complicated.  Even in this case, you can see that there is no way to algebraically add or subtract any multiple of $y$ to both inequalities simultaneously in order to eliminate $y$ from the extremes.  This is because the the LHS inequality $3y - 5 < -2y$ leads to the RHS inequality $y < 1$, and vice versa.  The "ends" become the "middle."


问:
How to compute probability of event occurring with multiple CDFs
Suppose there are two independent processes each resulting in the same event occurring and we are at time t=0.

The first process says with uniform probability the event will occur between the 4sec and 12sec mark ie linear CDF with expectation at t=8sec.

The second process says the event will occur with 100% probability at t=10sec ie step function CDF.

What time will the event most likely occur?

答:
Let $X$ denote the time of occurrence of the event of interest.

If we have equal confidence in both assessments, then by the law of total expectation,
\begin{align*}
\mathbb{E}[X] = 0.5 \cdot 8 + 0.5 \cdot 10 = 9.
\end{align*}

But the question asks "What time will the event most likely occur?" Note that $\mathbb{P}(X=9) = 0$, since the cumulative distribution function of $X$ is continuous around $t=9$. Actually, the cumulative distribution function of $X$ is continuous on $(-\infty,10) \cup (10,\infty)$, that is in every point except $t=10$. Thus a better answer may be $t=10$, since $\mathbb{P}(X = 10) = 0.5$ while $\mathbb{P}(X=t) = 0$ for all $t\neq 10$.


问:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$.
I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.

I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.

答: Since $\frac{x}{10}, \frac{y}{10}, \frac{z}{10}\in[0; 1]$, we have that $\left(\frac{x}{10}\right)^2\leqslant\frac{x}{10}, \left(\frac{y}{10}\right)^2\leqslant\frac{y}{10}, \left(\frac{z}{10}\right)^2\leqslant\frac{z}{10}$. Thus \begin{align*}2x^2+y^2+z^2&=100\cdot \left(2\cdot\left(\frac{x}{10}\right)^2+\left(\frac{y}{10}\right)^2+\left(\frac{z}{10}\right)^2\right)\\&\leqslant 100\cdot \left(2\cdot \frac{x}{10}+\frac{y}{10} +\frac{z}{10}\right)\\&=100\cdot \left(\frac{x}{10}+1\right)\\&\leqslant 100\cdot (1+1)=200\end{align*} With equality if and only if $x=10, y=0, z=0.$


答:
If $a$ is algebraic over $F$ then $[F(a)\colon F]$ equals the smallest degree of  a polynomial  with coefficients in $F$ having $a$ as a root. Clearly, if $F'\supset F$, this degree only decreases ( there may be some new polynomials with coefficients in $F'$ having $a$ as a root). That's why $[F'(a)\colon F]\le [F(a)\colon F]$.

Notice that if $a$ is algebraic over $F$ then $F(a) = F[a]$, so by induction
$F(a_1, \ldots, a_k) = F[a_1, \ldots, a_k]$. Now, you can see a system of generators of $F[a_1, \ldots, a_k]$ (as an $F$ vector space) of cardinality $n_1\cdots n_k$.

Title: Field extension finitely generated by algebraic elements

Body:
I'm trying to prove the following:

Let $K/F$ be a field extension with fields $F\subseteq K$. If $K$ is finitely generated by elements that are algebraic over $F$, i.e. $K=F(a_1,\ldots , a_k)$ for $a_i \in K$ algebraic over $F$, then $K/F$ is a finite field extension.

In the lectures of my professor a proof of this is given, but very short and compact. I have problem with the following conclusion:
We assume $F(a_1, \ldots , a_k)=K$ for algebraic elements $a_i\in K$ over $F$. Then define $F_0:=F$ and $F_{i+1}:=F_i(a_{i+1})$. This gives us a tower of fields $F=F_0 \subseteq F_1 \subseteq \ldots \subseteq K=F_k$. So far everything is understandable, but now the part that I don't understand:

If we define $n_i:=[F(a_i):F] \in \mathbb{N}$, then it follows that $[F_{i+1}:F_i]\leq n_{i+1}$. Why does this follow?


问:
Fill the gaps in Eulers proof of Fermats last theorem for n=3
I wanted to read Euler's proof of Fermat's last theorem for n=3 but the versions I have found all seem to have a gap.  The key missing result is
$$ p^2 + 3q^2 = c^3=> p = r(r^2 + 3s^2),  q = 3r(r^2 - s^2) $$

for some r,s assuming p,q relatively prime and c odd.

Which is equivalent to the lemma below from a book by Edwards

Let p and q be relatively prime numbers such that pp +3qq is a cube. Then there exist integers a and b such that $p+q\sqrt{-3} = (a+b\sqrt{−3})^3$

It is clear from many of the proofs that if $p^2+3q^2 = c^3$ then $r^2+3s^2 = c$ for some r,s

From there it follows that

$$p^2 + 3q^2 = [r(r^2 + 3s^2)]^2 + 3*[3r(r^2 - s^2)]^2$$

But this result isn't sufficient to prove the necessary lemma.

Dan Harmon's blog, gets this wrong, but apparently the error Euler made was similar, so he is in good company.

This proof acknowledges the gap, and provides a reference to a proof of the required lemma, but I don't have access to the book
Edwards, Harold M.“Fermat’s Last Theorem: A Genetic Introduction
to Algebraic Number Theory.” Springer-Verlag. New York. 1977.

Euler wrote a paper in latin that is said to fix the problem with his first proof, but I can't see that it does this.  I can't read latin though, so maybe google translate misled me

I found a copy of Euler's proof in German(I think) on p159 here,Bergman but I can't read it.

答:
Below are excerpts from H. M. Edwards, giving proof of the Lemma needed to close the gap in Euler's published proof of Fermat's Last Theorem for the case $n=3$.

"Lemma. Let $a$ and $b$ be relatively prime numbers such that $a^2+3b^2$ is a cube.  Then there exist integers $p$ and $q$ such that $a+b\sqrt{-3}=(p+q\sqrt {-3})^3$.

Proof.  Let $a^2+3b^2=P_1P_2...P_n$ be a factorization into $4$'s and odd primes as in (4). If this factorization contains exactly $k$ factors of $4$ then $2^{2k}$ is the largest power of $2$ which divides $a^2+3b^2$ and, since $a^2+3b^2$ is a cube, it follows that $2k$ and hence $k$ are multiples of $3$. Moreover, any odd prime $P$ in the factorization must occur with a multiplicity which is a multiple of $3$. Thus $n$ is divisible by $3$ and the factors $P_1P_2...P_n$ can be arranged in such a way that $P_{3k+1}=P_{3k+2}=P_{3k+3}$. It follows that in the factorization of $a+b\sqrt{-3}$ given by (3) the factors corresponding to each group of three $P$'s are identical because the only choice is the choice of sign
$p^+_-q\sqrt{-3}$ and both signs cannot occur. Taking one factor from each group of three and multiplying them together then gives a number $(c+d\sqrt{-3})^3$ such that $a+b\sqrt{-3}={^+_-}(c+d\sqrt{-3})^3$. Since $-(c+d\sqrt{-3})^3=(-c-d\sqrt{-s})^3$ the desired result follows."

The Lemma evidently rests on some preliminaries since Edwards refers to results (3) and (4) in the course of his proof:

(3) "Let $a$ and $b$ be relatively prime. Then $a+b\sqrt{-3}$ can be written in the form$$a+b\sqrt{-3}={^+_-}(p_1{^+_-}q_1\sqrt{-3})(p_2{^+_-}
q_2\sqrt{-3})\cdot\cdot\cdot(p_n{^+_-}{q_n}\sqrt{-3})$$where the $p$'s and $q$'s are positive integers and ${p_i}^2+3q_i^2$ is either $4$ or an odd prime

If $a^2+3b^2$ is even then it is divisible by $4$. If $a^2+3b^2$ is not $1$ then it has factor $P$ equal to either $4$ or an odd prime and either (1) or (2) gives $a+b\sqrt{-3}=(p^+_-q\sqrt{-3})(u+v\sqrt{-3})$ where $p^2+3q^2=P$. Then $u$ and $v$ are relatively prime and the problem of taking a factor $p^+_-q\sqrt{-3}$ out of $u+v\sqrt{-3}$ is the same as that of taking one out of $a+b\sqrt{-3}$ except that $u^2+3v^2=(a^2+3b^2)/P$ is smaller than $a^2+3b^2$. Iterating this process must eventually lead to a stage $a+b\sqrt{-3}=(p_1{^+_-}{q_1}\sqrt{-3})\cdot\cdot\cdot(p_n{^+_-}{q_n}\sqrt{-3})(u+v\sqrt{-3})$ where $u^2+3v^2=1$. Then $u={^+_-1}$, $v=0$, $u+v\sqrt{-3}= {^+_-1}$ and the factorization is complete."

(4) "Let $a$ and $b$ be relatively prime. Then the factors in the above factorization of $a+b\sqrt{-3}$ are completely determined, except for the choice of signs as indicated, by the fact that $({p_1}^2+3{q_1}^2)({p_2}^2+3{q_2})^2\cdot\cdot\cdot({p_n}^2+3{q_n}^2)=a^2+3b^2$ is a factorization of $a^2+3b^2$ into odd primes and $4$'s. Moreover, if the factor $p+q\sqrt{-3}$ occurs then the factor $p-q\sqrt{-3}$ does not, and conversely. The thing to be proved in the first statement is that $p^2+3q^2=P$ determines $p$ and $q$, up to sign, if $p$ is $4$ or an odd prime. This is clear in the case $P=4$. If $P$ is an odd prime and if $a^2+3b^2$ were another representation of it then, by (2),$$a+b\sqrt{-3}=(p^+_-q\sqrt{-3})(u+v\sqrt{-3})$$and $P=P(u^2+3v^2)$, that is, $u^2+3v^2=1$, $u={^+_-1}$, $v=0$, $a+b\sqrt{-3}={^+_-}(p^+_-q\sqrt{-3})$ as was to be shown. The second statement is simply the observation that $p+q\sqrt{-3}$ and $p-q\sqrt{-3}$ would combine to give a factor $p^2+3q^2$, which is impossible if $a$ and $b$ are relatively prime."

Since (3) and (4) above make reference to (1) and (2), this lengthy citation of Edwards is not quite complete. I omit the proofs for now, but in (1) and (2) Edwards shows:

(1) "If $a$ and $b$ are relatively prime and if $a^2+3b^2$ is even then $a+b\sqrt{-3}$ can be written in the form$$a+b\sqrt{-3}=(1^+_-\sqrt{-3})(u+v\sqrt{-3})$$where the sign is appropriately chosen and where $u$ and $v$ are integers."

(2) "If $a$ and $b$ are relatively prime and if $a^2+3b^2$ is divisible by the odd prime $P$ then $P$ can be written in the form $P=p^2+3q^2$ with $p$ and $q$ positive integers and $a+b\sqrt{-3}$ can be written in the form $a+b\sqrt{-3}=(p^+_-q\sqrt{-3})(u+v\sqrt{-3})$ where the sign is appropriately chosen and where $u$ and $v$ are integers."

I hope this is of some use in filling the gap in question.


问:
If $A$ is a commutative Noetherian ring, then $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set.

Let $A$ be a commutative Noetherian ring (with unity) and $k\in \mathbb{N}$. Prove that $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set.

I'm trying to prove it. We know that $A/\mathfrak{p}$ is a domain. Since $|A/\mathfrak{p}|\leq k$, it follows that $A/\mathfrak{p}$ is a finite domain. Therefore $A/\mathfrak{p}$ is a field. Thus $\mathfrak{p}\in \operatorname{Max}(A)$, i.e., $D\subseteq\operatorname{Max(A)}$...

答:
Assuming $A$ has infinitely many prime ideals of norm $\le k$, there is some $p^d$ such that $A$ has infinitely many maximal ideals of norm $p^d$. Let $$A_0 = A/(p,\{ a^{p^d}-a,a\in A\})$$ It has infinitely many maximal ideals all of norm $p^e, e| d$. By induction for $n=0,1,2,\ldots$,

Take some $a_n\ne 0 \in A_n$ which is in infinitely many maximal ideals and let $A_{n+1}=A_n/(a_n)$.

$A_{n+1}$ has again infinitely many maximal ideals, all of norm $p^e,e|d$.

The sequence $A_n$ of quotient rings implies that $A$ wasn't Noetherian.

If such an $a_n\in A_n$ as above didn't exist then enumerate the maximal ideals $m_j$ of $A_n$, take some $b_n\ne 0\in A_n$, if $b_n=0\bmod m_j$ for infinitely many $j$ then we can take $a_n=b_n$, otherwise $b_n^{p^d-1}-1=0\bmod m_j$ for infinitely many $j$, that $a_n$ doesn't exist means that $b_n^{p^d-1}-1=0$ so $b_n\in A_n^\times$, thus $A_n$ is in fact a field, a contradiction.


问:
Convergence to O(1)
Could someone please explain $\sum\limits_{i=0}^{\infty}(i+1)\left(\dfrac{1}{2}\right)^i= 1 + 2\left(\dfrac{1}{2}\right)^1 + 3\left(\dfrac{1}{2}\right)^2 + \ldots$ converges to $O(1)$ as per my algorithmns lecturer. Thank you

答:
This is an Arithmetico–geometric sequence. We have
$$
{\displaystyle \sum _{k=1}^{\infty }{\color {blue}k}{\color {green}r^{k}}={\frac {r}{(1-r)^{2}}},\quad \mathrm {for\ } 0<r<1}
$$
Can you continue from here?

Title: Convergence to O(1)


答:
Recall a sequence is $O(1)$ if and only if it is bounded. So asking for a sum to be $O(1)$ is asking for its sequence of partial sums to be bounded. Since every summand is positive, this is asking that the sum is convergent.

So saying this sum is $O(1)$ is asking for it to be convergent. Of course, we can see this sum converges since (as noted in the comment), your sum is

$$\frac{1}{(1-\frac{1}{2})^2} = 4 = O(1)$$

Said another way:

$$
S_n =
\sum_{i < n} (i+1) \left ( \frac{1}{2} \right )^i \leq
\sum_{i < \infty} (i+1) \left ( \frac{1}{2} \right )^i =
4 = O(1)
$$

So the sequence of partial sums is $O(1)$.

I hope this helps ^_^

Title: Convergence to O(1)

Body: Could someone please explain $\sum\limits_{i=0}^{\infty}(i+1)\left(\dfrac{1}{2}\right)^i= 1 + 2\left(\dfrac{1}{2}\right)^1 + 3\left(\dfrac{1}{2}\right)^2 + \ldots$ converges to $O(1)$ as per my algorithmns lecturer. Thank you


问:
real numbers a vector space over rational numbers?
Let $V$ be set of real numbers and $K$ the field of rational numbers.

Is $V$ a vector space over $K$, with ordinary addition of real numbers
and multiplication by rational numbers?

答: Yes, $V$ is a vector space over $K$, but it is of infinite dimension over $K$.  Choosing a basis for $V$ over $K$ is a non-trivial matter that requires non-constructive foundational material such as the axiom of choice, see http://en.wikipedia.org/wiki/Axiom_of_choice</p>


问:
Why is the highest weight module in $\mathcal{O}$ category?
The $\mathcal{O}$ category is defined as follows:

Let $\mathfrak{g}$ be a Kac-Moody algebra and $V$ a
$\mathfrak{g}$-module. $V$ is an object in $\mathcal{O}$ if

$V$ has decomposition $V=\bigoplus_{\lambda\in \mathfrak{h}^*}V_\lambda$ with $V_\lambda=\{v\in V\ |\ hv=\lambda(h)v\ for\ all\  h\in\mathfrak{h}\}$

$\dim V_\lambda<\infty$

There is a finite set $\lambda_1,\cdots,\lambda_k\subset\mathfrak{h}^*$ st. each $\lambda$
with $V_\lambda\neq 0$ satisfies $\lambda\leq \lambda_i$ for some $i$

A highest weight module is defined as follows

A $\mathfrak{g}$-module $V$ is a highest weight module with highest
weight $\Lambda\in\mathfrak{h}^*$ if there exists a non-zero element
$v_\Lambda\in V$ st.

$x(v_\Lambda)=0$ for all $x\in\mathfrak{n}_+$.

$h(v_\Lambda)=\Lambda(h)v_\Lambda$ for all $h\in\mathfrak{h}$.

$U(\mathfrak{g})(v_\Lambda)=V$.

Kac states that $V$ is in the $\mathcal{O}$ category since $V=\bigoplus_{\lambda\leq \Lambda}V_\lambda$, $V_\Lambda=\mathbb{C}v_\Lambda$ and $\dim V_\lambda<\infty$, however I'm having trouble seeing why these holds.
I've managed to show that $U(\mathfrak{n}_-)(v_\Lambda)=V$ by the Poincare Birkhoff Witt theorem.

So the second condition of a highest weight module gives a weight space $V_\Lambda$ of $V$, however why does this imply that we have a decomposition $V=\bigoplus_{\lambda\leq \Lambda}V_\lambda$? How do we know all the other weight space $V_\lambda$ exist and why are these finite dimensional?

The third criteria for something to be in the $\mathcal{O}$ category is that we have a finite set, which bounds the roots. I assume that the finite set is just $\Lambda$, however how do we know that $\Lambda$ is in fact the biggest root?

Since we know that $U(\mathfrak{n}_-)(v_\Lambda)=V$, then by PBW we would have to show that
$$h(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)=(\Lambda-\sum k_j\alpha_j)(h)(f_{i_1}^{k_1}\cdots f_{i_d}^{k_d}v_\Lambda)$$
This would imply that $v_\Lambda$ is the highest root. I've tried shown in by induction on $k_1+\cdots+k_d$, however I'm having trouble with he inductive step.

答:
Let $V$ be a Kac-Moody algebra, $V$ a representation. For $\lambda \in \mathfrak{h}^*$, let $V_\lambda$ be the $\lambda$ weight space. Then we have:

For all $h \in \mathfrak{h}$, $h(V_\lambda) \subseteq V_\lambda$.

For all $i$, $e_i(V_\lambda) \subseteq V_{\lambda + \alpha_i}$.

For all $i$, $f_i(V_\lambda) \subseteq V_{\lambda - \alpha_i}$.

(1) follows from the definition of a weight space. (2) and (3) are easy to prove and I'll give you a hint: rearrange the commutation relation $[h, f_i] = -\alpha_i(h)f_i$ inside $U(\mathfrak{g})$ to read $h f_i = f_i(h - \alpha_i(h))$, then try to compute $h_i f v$ for some $v \in V_{\lambda}$.

Now, suppose that $(V, v_{\Lambda})$ is a highest-weight module. By this we mean that: $0 \neq v_{\Lambda} \in V_{\Lambda}$, $e_i v_{\Lambda} = 0$ for all $i$, and $V$ is generated by $v_{\Lambda}$. We want to show that $V$ is a weight module, with finite dimensional weight spaces, with all weights bounded above by $\Lambda$.

By the PBW theorem, $V = \sum \mathbb{C} f_{i_1}^{k_d} \cdots f_{i_d}^{k_d} v_{\Lambda}$, where the sum is over all possible choices of $d$, $i_1, \ldots, i_d$, and $k_1, \ldots, k_d$. Looking at point $3$ above, we can see that every such spanning vector is in a weight space, for example $f_1^2 f_3 v_{\Lambda} \in V_{\Lambda - 2 \alpha_1 - \alpha_3}$. Since $V$ is spanned by weight vectors, it is a weight representation.

We can already see that every weight of $V$ is bounded above by $\Lambda$, since $\lambda \geq \lambda - \alpha_i$ for all $\lambda$ and $\alpha_i$. In particular, if $V_{\lambda} \neq 0$ then $\lambda \leq \Lambda$.

As for finite-dimensionality, this comes from the fact that the simple roots are linearly independent, and there are only so many ways of writing a sequence $f_{i_1} \ldots f_{i_d} v_{\Lambda}$ so that the resulting vector is in the correct weight space. Take for example $\lambda = \Lambda - 2 \alpha_1 - \alpha_3$ Then $\lambda - \Lambda = 2 \alpha_1 + \alpha_3$ is an element of the root lattice, with "height" 3. There are only three sequences of $f_i$ that lead to vectors with this weight, namely $f_1^2 f_3 v_{\Lambda}$, $f_1 f_3 f_1 v_{\Lambda}$, and $f_3 f_1^2 v_{\Lambda}$. Therefore $\dim V_{\Lambda - 2 \alpha_1 - \alpha_3} \leq 3$. These inequalities do not in general give good bounds on the dimensions of weight spaces, but all the bounds they give show that the weight spaces are finite-dimensional, which is all that we need.


问:
Proof that a set is a subring of the real numbers ring

Is the set $\mathit{A}=\{x-y\sqrt[3]{2}\mid x, y\in \mathbb{Q}\}$ a subring of the ring $\mathbb{R}$?

Hi SE community, I have tried to proof that $0\in\mathit{A}$ so, $0 = x-y\sqrt[3]{2}\;$. Obviously I got a contradiction since $\sqrt[3]{2}\;$ is irrational and $x$ and $m$ rationals. I concluded that $A$ is not a subring since $0$ is not an element of it.

I am sure that $A$ is not a subring of $\mathbb{R}$ because I have checked that it is not closed under multiplication. But then I found that since  $(3)\;a+(-a) =0 \Rightarrow(x_1+x_1)+(y_1+y_1)\sqrt[3]{2}\;$, and $0 \in \mathbb{Q}, \; 0 \in A$. Then $A$ has to be a subring of $\mathbb{R}$. This last statement demolished my first proof.

I think that there is problem with statement $(3)$, I still thinking $A$ is not a subring. Please let me know why is wrong the last statement.

答: First note that $0 \in \mathbb{Q}$, so taking $x=y=0$ we have that $0-0\cdot \sqrt[3]{2} = 0 \in A$. I think your confusion comes from the fact that $\sqrt[3]{2}$ is irrational, as you noted, so any nonzero rational multiple of it is also irrational and any sum of it with a rational will be irrational as well. The thing to note is that this is only true for nonzero rational multiples, so we have no contradiction with writing a rational number $0$ as the difference between a rational and a $y\sqrt[3]{2}$ because in this case $y=0$.


问:
Tranversal paths between two points
I have been studying differential topology from Hirsch, and sometimes in proofs he takes two points $x,y\in M$ a path between them and then just says that we can assume that this is transversal to a certain submanifold that we are interested in. Now I have tried to prove this myself but I don't see why this is true.

One attempt well let's take a path $\gamma $ between them such that $\gamma(0)=x$ and $\gamma(1)=y$, then using the fact  that we  have that transversal maps are dense we can approximate this map by transversal maps $h_n:[0,1]\rightarrow M$, but here I don't think I have any control over $h_n(0)$ and $h_n(1)$ and so this could be not $x$ and $y$.

Second attempt, we show that $\{\gamma \in C^r([0,1],M) : \gamma(0)=x, \gamma(1)=y\}$ is open in the strong topology, however this doesn't seem to be very true to me , since when we consider neighborhoods we lose control over the initial and final point again.

Another idea would be to consider the path an embedding which is possible, and then use the fact that $C_S^r([0,1],\{0,1\},\gamma([0,1]),\{x,y\})$ is dense, but here we would need the fact that the manifold that we want the map to be transverse too to be closed.

So I am out of ideas , and would appreciate some help with this. Thanks in advance.

答:
Here is a useful corollary to the parametric transversality theorem (c.f. Guilleman and Pollack P.73):

Corollary. If, for $f: X\to Y$, the boundary map $\partial f: \partial X\to Y$ is transversal to $Z$, then there exists a map $g: X \to Y$ homotopic to $f$ such that $\partial g = \partial f$ and $g\pitchfork Z$.

Here $\partial f = f|_{\partial X}$. The full proof is a bit lengthy, so I won't write it out here, but the basic idea is to modify a homotopy $F$ of $f$ by a bump function which vanishes on neighborhoods of the boundary points. I don't have a copy of Hirsch unfortunately, but I imagine a similar result is discussed in his book.

Anyway, taking $X=[0,1]$ and $f$ to be some path with $f(0)=x$ and $f(1)=y$, then so long as $\partial f:\{0,1\}\to Y$ is transverse to $Z$ the result will follow. This is trivially true if $x,y\notin Z$ (which is the scenario in which this result would usually be applied).

Title: Tranversal paths between two points


问:
Is there a positive series of rational numbers that converge to $\pi$?
I'm looking for a series of rational numbers that approach $\pi$ that are all non-negative. In addition, I also want this series to have a general rule. So something like $\dfrac31+\dfrac1{10}+\dfrac4{100}+\ldots$ is not gonna cut it.

Does such a series exist?

答:
We know what the Taylor series for $\tan^{-1}(x)$ is the following.

$$ \tan^{-1}(x)=\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} $$

We also know that $\dfrac{\pi}4 = \tan^{-1}(1)$. So:

$$ \pi = \sum_{n=0}^{\infty} (-1)^n \dfrac{4}{2n+1} $$

So you can just take the sequence of rational numbers that converge to $\pi$ as:

$$ a_N = \sum_{n=0}^{N} (-1)^n \dfrac{4}{2n+1} $$

This follows the same comment mentioned by @JMoravitz.


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
In addition to another answer, I know a theorem in which we can extend the order of $\mathbb R$ to $\mathbb C$ such that two following are satisfied:

$P_1$: For all $a\in\mathbb R$ we have only one of $a>0,~a=0,-a>0$.

$P_2$: $a>0,b>0 \to a+b>0$.

In fact, if we let $\alpha=a+ib$ we can have $\alpha>0$ if and only if $a>0$ or $a=0,~b>0$. So for all $\alpha=a+ib$ we have just one of the following facts: $$a>0,a=0,-a>0$$ or $$a>0\to \alpha>0$$ or $$-a>0\to-\alpha>0$$ or if $a=0$ then $$b>0\to \alpha>0\\\\ b=0\to \alpha=0\\\\ -b>0\to-\alpha>0$$. Therefore, for any arbitrary $\alpha=a+ib\in\mathbb C$ we have just one of $$\alpha>0,~ \alpha=0,~-\alpha>0$$
I think this make a bit clear what we need here.

Title: What is $-i$ exactly?


答:
Some comments:

Both Tavish and Calvin Lin's solutions are good and are not just lucky. Let's break it down:

Lemma: If $a\ge 0, b\ge 0$ and $a+b$ is fixed, then $a^2+b^2$ obtains its maximum when either $a=0$ or $b=0$.

This can be proved either via property of parabola, or just algebraically. But it's easy to write it simply as $a^2+b^2 \le (a+b)^2$, and I believe nobody is against this.

Now if we fix $y+z$ we have $$2x^2 + (y^2 + z^2) \le 2x^2+ (y+z)^2 = 2x^2+ (10-x)^2 \tag 1$$

Next we apply the lemma on $x$ and $10-x$:
$$x^2+\left( x^2+(10-x)^2 \right)\le x^2 + (x+10-x)^2 = x^2 + 100 \tag 2$$
$$\le 100+100.\blacksquare$$

You can think of Calvin Lin's way as combining these two steps (or a generalized version of the lemma with more than two variables):
$$2x^2+y^2+z^2 = x^2 + (x^2+y^2+z^2) \le x^2 + (x+y+z)^2 = x^2 + 10^2 \le 200$$

Tavish's way is to apply the parabola property twice.

It's best understood from the picture below:

Title: Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$.

Body:
I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.

I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.


问:
which number corresponds to the right endpoint of line segment [0,1)?
I hold two basic opinion that

(1) Every line segment has two endpoints.

(2) The endpoint of a line segment is part of it.

So as for the line segment corresponding to [0,1],  if we remove its right endpoint,  which number corresponds to the right endpoint of the new line segment?  I don't think the number is still 1,  because we have already removed the point corresponding to 1,  thus created this new line segment, the point 1 is not part of the new line segment, so according to my opinion (2), I don't think the number is 1.

Update: A bounded continuous segment in a line is my definition of "line segment"

Update: I think I solved my question , see answer below

答: Your second "basic opinion" is simply untrue. The number is $1$.


问:
Proof of relative complement identity
I often see this identity for set theory,

$A-B=A\cap B^c$

It is easy to believe this identity by considering the venn diagram for a relative complement,

If we consider $B^c$ informally as the space not inside $B$ then the space of set $A$ not intersectiong with set $B$ would be included in $B^c$. So logically, $A\cap B^c$ would be the shaded area, which is the same space as $A-B$.

But how would one go about proving this identity purely from set theory and symbolic logic?

答:
Let $A = \{x\in U\mid p(x)\,\text{is true}\}$ and $B = \{x\in U\mid q(x)\,\text{is true}\}$, where $U$ is the universe of discourse.

Thus we have that
\begin{align*}
A - B & = \{x\in U\mid (x\in A)\wedge(x\not\in B)\}\\\\
& = \{x\in U \mid x\in A\}\cap\{x\in U \mid x\not\in B\} = A\cap B^{c}
\end{align*}

where we have used the definitions of difference, complement and intersection.

More precisely, given $X\subseteq U$ and $Y\subseteq U$, one has that
\begin{align*}
X^{c} & := \{x\in U \mid x\not\in X\}\\\\
X\cap Y & := \{x\in U \mid (x\in X)\wedge(x\in Y)\}
\end{align*}

Hopefully this helps!

Title: Proof of relative complement identity


问:
Abel's Test in Complex Analysis
I'm reading Complex Analysis by Freitag and Busam (2nd edition), and in exercise 14 for section I.2, they want us to prove the following variant of Abel's Test:

Given two sequences of complex numbers $(a_n)$ and $(b_n)$, if $$\sum_{n=0}^{\infty}a_n(b_n-b_{n+1})$$converges and $$(a_nb_{n+1})$$converges, then show $$\sum_{n=0}^{\infty}a_nb_n$$converges.

(I've rephrased it slightly because the actual problem makes use of the previous exercise (13) which was Abel's summation by parts, but all the given assumptions are there)  My question is this: Is this actually correct?  It seems phrased differently from what I find here, as there is no mention of monotone.  Furthermore, I think the following is a counterexample: $$a_n=1, b_n=1+\frac{1}{n+1}$$for all $n$.

I couldn't find an errata for this text.  If I misunderstood something, I hope someone here can point it out.  Thank you.

Sidenote: This is not a homework problem.  I'm just trying to teach myself complex analysis since my master's program never offered it (I'm slightly bitter).

答: Thanks to @Conrad for verifying my counterexample and providing a likely edit to the text.  They suggested more reasonable assumptions would be $\sum A_n(b_n-b_{n+1})$ and $(A_nb_{n+1})$ converge, where $$A_n=\sum_{k=0}^na_k.$$With these assumptions, the result follows immediately from the Cauchy Criteria for convergent sums and the identity $$\sum_{k=m}^n a_kb_k=A_nb_{n+1}-A_{m-1}b_m-\sum_{k=m}^nA_k(b_{k+1}-b_k)$$i.e. summation by parts, which was proven in the previous exercise.

Title: Abel's Test in Complex Analysis


问:
Show that there is exactly one element in $A$ congruent to $1$ modulo $p$ as follows.
Let $p$ be a prime. Let $A = \{a,2a,3a,\ldots,(p-1)a \}$, where $\gcd(a,p)=1$. Show that there is exactly one element in $A$ congruent to $1$ modulo $p$.

My attempt:
It is suffices to show that for each $a \in \{1,2,3,\ldots,p-1\}$, there is exist exactly one $b \in \{1,2,3,\ldots,p-1\}$ such that $ab \equiv 1 \pmod{p}$.
Suppose there is exist an element in $A$, say $c$, such that $ab \equiv ac \equiv 1 \pmod{p}$. Then,
\begin{equation*}
b \equiv 1\cdot b \equiv (ac)b \equiv (ab)c \equiv 1\cdot c \equiv c \pmod{p}
\end{equation*}
Hence, $b$ is unique and we're done.

Am i true? If not, how to show it?

答:
Yes, you’re essentially correct. On the one hand, if there are two such solutions, say $m,n$ and
$$ma\equiv na\equiv 1\pmod p\,$$
then you’d have that that $m\equiv n\pmod p$, which forces $m=n$. On the other hand, because $\gcd(a,p)=1$, then $a$ has an inverse modulo $p$, say $b$, hence $ab\equiv 1\pmod p$.


问:
How do I prove this sum of p-th powers identity?
So, I was playing around with the sum of p-th powers and I managed to find the following identity:

$$
S_{p+1} \left(n\right)=n\;S_p \left(n\right)-\sum_{i=1}^n S_p \left(i-1\right)
$$

where

$$
S_p \left(n\right)=1^p +2^p +3^p +\ldotp \ldotp \ldotp +n^p
$$

and

$$
S_{p+1} \left(n\right)=1^{p+1} +2^{p+1} +3^{p+1} +\ldotp \ldotp \ldotp +n^{p+1}
$$
I've scripted a code in MATLAB to generate all formulas up to a certain power and this recurrence relation seems to produce the correct formulas for the sum of p-th powers starting with $ S_0 \left(n\right) = n $, so it seems to be true at least for the cases I tested.

Now, I'm not sure if such a recurrence relation already exists and has been proven, either way I am trying to find a way to prove or maybe disprove that recursive formula.

I was thinking about using induction to prove that recurrence relation:
With the base case, $ S_1 \left(n\right)$ knowing $ S_0 \left(n\right) = n $
$$
S_1 \left(n\right)=n{\;S}_0 \left(n\right)-\sum_{i=1}^n S_0 \left(i-1\right)
$$
$$
S_1 \left(n\right)=n\left(n\right)-\sum_{i=1}^n(i-1)
$$
$$
S_1 \left(n\right)=n^2 -\sum_{i=1}^n i+\sum_{i=1}^n 1=n^2 +n-\sum_{i=1}^n i=n\left(n+1\right)-S_1 \left(n\right)
$$
$$
{2S}_1 \left(n\right)=n\left(n+1\right)
$$
$$
S_1 \left(n\right)=\frac{n\left(n+1\right)}{2}
$$

Now, I'm not quite sure how to approach the inductive step, can anyone guide me as to how to prove/disprove this recurrence?

答:
For the original identity, its' just algebraic manipulation.

Hint: Show that

$$LHS =  \sum (n-k) \times (n-k)^p = \sum n \times (n-k)^p - \sum k (n-k)^p = RHS$$

In terms of how to prove that $S_1(n) = \frac{ n(n+1) } { 2}$, you can do so by (another) straightforward induction.


答: Your opinions (1) and (2) are inconsistent with your (implied) opinion (3), that $[0,1)$ is a line segment. If you truly believe all three of these statements, then it follows, as night follows day, that you believe that $2+2=7$, that the Moon is made of green cheese, and that you are the Pope. I would recommend that you carefully consider your three beliefs, and see which one(s) you are really committed to, and which one(s) you are willing to abandon, in the interests of not being in a state of self-contradiction.

Title: which number corresponds to the right endpoint of line segment [0,1)?

Body:
I hold two basic opinion that

(1) Every line segment has two endpoints.

(2) The endpoint of a line segment is part of it.

So as for the line segment corresponding to [0,1],  if we remove its right endpoint,  which number corresponds to the right endpoint of the new line segment?  I don't think the number is still 1,  because we have already removed the point corresponding to 1,  thus created this new line segment, the point 1 is not part of the new line segment, so according to my opinion (2), I don't think the number is 1.

Update: A bounded continuous segment in a line is my definition of "line segment"

Update: I think I solved my question , see answer below


问:
Solve Compound Inequality Without Separation With Variables on Outside
I have the following compound inequality:

$3y-5 < -2y < 2+y$

I know I can separate them into two different equations, and go from there. I have been taught how to solve them without separation when the variable is in the center. But I have not been taught how to do this when there are variables on the outside like this one? How do you I "put the y variable back into the center?

The way I see it, I  can't do it. I can -y from both sides and add it to the center, but still leaves a 2y on the left. I can -3y and add it to the center, but that leaves a -2y on the right. The only way to solve it is through separation. Or am I wrong?

答:
$$3y-5 < -2y < 2+y\implies 2y-5 \lt -3y \lt 2$$

$$2y-5 \lt -3y\implies 5y\lt 5\implies y \lt 1
\quad \land \quad
-3y \lt 2\implies -2 \lt 3y\implies -\frac{2}{3} \lt y$$

$$\therefore -\frac{2}{3} \lt y \lt 1$$


答:
First , it is helpful to  inform me to check whether [0,1) has a right endpoint after removing .

Second , just as @ShubhamJohri informed that

A line segment may not contain its end points -- it only needs to
contain all points between them. Read here

so I misunderstood the definition of a line segment.


答: Comparing $\mathbb{C}$ with vectors in $\mathbb{R}^2$ the $i$ can be viewed as describing a direction in 2D space (vertical, if you want), and it gets its magnitude and orientation (i.e. sign) from the coordinate that is a real number. It turns out that defining $i$ as the $\sqrt{-1}$ (which doesn't exist in $\mathbb{R}$), we get a field extension of $\mathbb{R}$ that is algebraically very useful and has the "geometrical properties" of the usual $\mathbb{R}^2$ vectorspace.</p>

Title: What is $-i$ exactly?

Body:
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?


答:
Not a subring as you describe it, as you do not include the square of $\sqrt[3] 2.$ The word for what is missing is closure: your set is not closed under multiplication.

If you include all
$$ x + y \sqrt[3] 2 + z \sqrt[3] 4 $$
with rational $x,y,z$ you get a ring, with norm
$$ x^3 + 2y^2 + 4 z^3 -6xyz  $$
As any nonzero element has a multiplicative inverse, it is a field.

Note:
$$ \left( x + y \sqrt[3] 2 + z \sqrt[3] 4 \right) \left( \; (x^2 - 2yz) + (2z^2 - xy) \sqrt[3] 2 + (y^2-zx) \sqrt[3] 4 \right) =  x^3 + 2y^2 + 4 z^3 -6xyz$$
so
$$ \left( x + y \sqrt[3] 2 + z \sqrt[3] 4 \right) \frac{ \left( \; (x^2 - 2yz) + (2z^2 - xy) \sqrt[3] 2 + (y^2-zx) \sqrt[3] 4 \right)}{   x^3 + 2y^2 + 4 z^3 -6xyz} = 1$$

Title: Proof that a set is a subring of the real numbers ring

Body:
Is the set $\mathit{A}=\{x-y\sqrt[3]{2}\mid x, y\in \mathbb{Q}\}$ a subring of the ring $\mathbb{R}$?

Hi SE community, I have tried to proof that $0\in\mathit{A}$ so, $0 = x-y\sqrt[3]{2}\;$. Obviously I got a contradiction since $\sqrt[3]{2}\;$ is irrational and $x$ and $m$ rationals. I concluded that $A$ is not a subring since $0$ is not an element of it.

I am sure that $A$ is not a subring of $\mathbb{R}$ because I have checked that it is not closed under multiplication. But then I found that since  $(3)\;a+(-a) =0 \Rightarrow(x_1+x_1)+(y_1+y_1)\sqrt[3]{2}\;$, and $0 \in \mathbb{Q}, \; 0 \in A$. Then $A$ has to be a subring of $\mathbb{R}$. This last statement demolished my first proof.

I think that there is problem with statement $(3)$, I still thinking $A$ is not a subring. Please let me know why is wrong the last statement.


答: There strictly speaking is no endpoint of a line_segment or ray at an open boundary. That said, the real number corresponding to an end of a line_segment defined by and terminating at a finite value {be it [included or excluded)} is simply that number, i.e. 1.


问:
Reference for statistical inference concerning these topics
I am looking for book recommendations /lecture notes for self studying Inference. Following are the main topics of concern:

Parametric models, parameters, random sample and its likelihood,
statistic and its sampling distributions, problems of inference.
Examples from standard discrete and continuous models such as
Bernoulli, Binomial, Poisson, Negative Binomial, Normal, Exponential,
Gamma, Weibull, Pareto etc. Concept of sufficiency, minimal
sufficiency, Neyman factorization criterion, Fisher information,
exponential families. Maximum likelihood estimators, method of moment
estimators, percentile estimators, least squares estimators, minimum
mean squares estimators, uniformly minimum variance unbiased
estimators, Rao Blackwell theorem, Cramer Rao lower bond, different
examples. Statistical Hypotheses simple and composite, statistical
tests, critical regions, Type I and Type II errors, size and power of
a test, Neyman Pearson lemma and its different applications. Most
powerful test, uniformly most powerful test, unbiased test and
uniformly most unbiased test. Likelihood ratio test. Interval
estimation, confidence intervals, construction of confidence
intervals, shortest expected length confidence interval, most accurate
one sided confidence interval and its relation to UMP test.

Kindly suggest some good texts or lecture notes to help me break ice into this topic.

答:
I would recommend Probability and Statistical Inference by Hogg, Tenis, and Zimmerman (10th edition specifically!!).
I am not sponsored, but this book helped me a lot when I was trying to self study. Steps were easy to follow.

Chapter 1 introduces the the properties of the probability, counting rues, independent/dependent/mutually exclusive events, Bayes etc.

Then the Chapter 2 introduces all the Discrete distributions and the properties of them like you mentioned - Bernoulli, Binomial, Hypergeometric, Negative Binomial, and Poison distributions.

Chapter 3 introduces all the continuous distributions like Beta, Gamma, Exponential, Chi-square, Normal and Uniform distributions.

Chapter 5 introduces transformations and functions of random variables, moment generating function, central limit theorem, sampling distribution, Normal distribution, Z-distribution, and derivation of student t-distribution chebyshe's inequality, and more.

Chapter 6 introduces descriptive statistics(sample mean, sample variance, relative frequency, empirical rules, etc), sample percentile, quartiles, five-number summaries, Min, Max, Range, IQR, Box plot, order statistics, Maximum likelihood and estimators, method of moments estimation, biased/unbiased estimators,assymptotic distributions of maximum likelihood estimators, sufficient statistics, simple regression, and Bayesian estimation and more.

Chapter 7 basically talks about confidence interval for various distributions that we talked about chapter 2&3, sample size, and resampling methods.

Chapter 8 introduces definition of hypothesis and different types of statistical tests - test of equality of two means, test of equal variance, test of equal proportions and many more. Also talks about distribution-free hypothesis tests(nonparametric methods) - Wilcoxon's signed rank test and Wilcoxon's rank sum test.It also talks about critical region, type 1 error, type 2 error, significance level, p-value, power of a statistical test. When and how to reject the null hypothesis.

I haven yet studied chapter 9, but I hope you may find this book covers some of keywords that you mentions above and more.

Thank.