问:
Part/ whole relation in geometry and in set theory? Did set theory overthrow the axioms of euclidian geometry?
Euclid's Elements treat as axiomatic the proposition " the whole is greater than the part" .

Set theory is said to overthrow this so-called self evident truth, because  for some sets ( namely infinite sets) the part ( proper part) has the same size ( cardinality) as the whole.

What is exactly the impact of this set-theoretic result (1) on geometry and (2) on mathematics in general?

Is it actually the " blow" it is sometimes said to represent for mathematical rationality?

答:
These kind of problems usually arise from applying different definitions to the same situation, so I will try to make things as clear as I can.

When Euclid wrote "the whole is greater than the part", he had a point of view of mathematics very different from our point of view. All of ancient Greek mathematics, actually, was heavily focused on geometric problems; and, given the fact they had much less "mathematics" than us, their problems were, necessarily, simpler as well.

Euclid wanted mainly to, and here I am not criticizing his scope, measure and build geometric entities as precisely as possible (and the "as precisely as possible" was indeed the revolutionary part of his work). So, if you take any reasonable segment of a finite straight line, of course you see that what you obtained, the part, is less than what you started with, the whole.

Now, let us move to a slightly more modern view of mathematics, which includes, e.g., the real numbers. What is a reasonable segment of a finite straight line? Well, the closed interval
$$[0,1]=\{x\in\mathbb{R} : 0\leq x\leq 1\}$$
is, arguably, a finite straight line, and $[0,1]\setminus\{1/2\}$ is, also arguably, a reasonable segment of a finite straight line. And yet, you could not distinguish them with a ruler: the case here is that, for Euclid, this would not be a reasonable segment of $[0,1]$, given, among other things, he had not a clear concept of the real numbers, so you cannot look at this problem from his point of view and expect his axiomatic to work flawlessly.

You can then say "OK, but $[0,1]$ IS larger than $[0,1]\setminus\{1/2\}$, in the sense that the former is a proper subset of the latter", and then it becomes natural to look at things from Cantor's point of view. Cantor's work, or rather its later formalization, is exceptionally precise, and statements such as "$X$ is greater or equal to $Y$" have a very specific meaning.

In set theory, this would mean (ignoring Axiom-of-Choice-related problems) that there exists a surjective function (i.e., such that, for every $y\in Y$, there exists $x\in X$ satisfying $f(x)=y$) $f:X\rightarrow Y$; two sets $X$ and $Y$ are then of "equal sizes" whenever one is greater or equal to the other and vice-versa, what comes down to there existing a function $f:X\rightarrow Y$ which is both injective (that is, if $f(x_{1})=f(x_{2})$, then $x_{1}=x_{2}$) and surjective.

Surprise, surprise, in set theory, $[0,1]$ and $[0,1]\setminus\{1/2\}$ are both of the same size! But, as explained, this would probably not be a problem for Euclid: the ends, and means, of his axiomatic did not involve the treatment of pathological things as these. Another example, that may make things clearer, is Galileo's paradox: think of the natural numbers
$$\mathbb{N}=\{0, 1, 2, 3, 4, 5, 6, ...\}$$
and its subset of even numbers, $E=\{0, 2, 4, 6, ...\}$. To Euclid, to measure both of these sets had no meaning, they were not geometrical objects; to the mathematicians just before set theory, it was clear that $\mathbb{N}$ is larger than $E$, given it contains all elements of the latter, and so many more! But, in set theory, and using the definition found in set theory, both are of the same size: take $f:\mathbb{N}\rightarrow E$ such that $f(n)=2n$.

So, to answer your questions:

What is exactly the impact of this set-theoretic result (1) on
geometry and (2) on mathematics in general?

(1) Huge. Set theory has altered all areas of mathematics deeply, some (algebra, topology come to mind) more than other (geometry), but it did not "overthrown" it. Of course, we have indeed some problems when putting the two together (Banach-Tarski's paradox is an example, coming from confronting intuition with counter-intuitive tools such as the axiom of choice), but set theory actually helped solidify geometry as it stands today.
(2) More than huge. This specific result is a tiny part of set theory's many, sometimes apparently absurd, contributions to modern math, all steaming from Cantor's attempt to measure sets.

Is it actually the " blow" it is sometimes said to represent for
mathematical rationality?

No! As I explained, people that think this "destroys" our rationality are usually being too dramatic or applying different lines of thought to the same problem. Yes, set theory is sometimes counter-intuitive, but how really good is our intuition anyway? How many times have you seem an infinite collection of things and operated with it to gain that intuition?


问:
How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series?
I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.

I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:

Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.

https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454

答:
Firstly, in the linked StackOverflow question, the program does integer division at each step, so "n/2" in that context actually means the greatest integer less than or equal to $\frac{n}{2}$: more correctly, it should be written as $\left\lfloor \frac{n}{2} \right\rfloor$ (where $\left\lfloor x \right\rfloor$ is the floor function, e.g. $\left\lfloor \frac{7}{2} \right\rfloor = \left\lfloor 3.5 \right\rfloor = 3$).

Secondly, you missed a clause mentioned at the linked StackOverflow question: the correct statement is that if $n$ is a power of $2$, then $\left\lfloor n \right\rfloor + \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \dots$ (which in this case is the same as $n + \frac{n}2 + \frac{n}4 + \dots + 1$) is exactly $2n - 1$.

For this special case when $n$ is a power of $2$ (which is what it takes for all the numbers $\frac{n}2, \frac{n}4, \dots$ to be integers, all the way to $1$), this is easy to prove. When $n$ is a power of $2$, say $n = 2^k$, the sum is
$$2^k + \frac{2^k}{2} + \frac{2^k}{4} + \dots + 1
= 2^k + 2^{k-1} + 2^{k-2} + \dots + 1
= 2^{k+1} - 1
= 2n - 1
$$
For example, $16 + 8 + 4 + 2 + 1 = 31$.

This can be proved by induction, or it follows from the formula for the geometric series, which states that $$a + ar + ar^2 + \dots + ar^{m-1} = a\frac{1-r^m}{1-r}.$$

In this case, we have $a = n = 2^k$, $r = 1/2$, and $m$ (the number of terms) is $k+1$, so the left-hand side is the sum $n + \frac{n}{2} + \dots + 1$, and the right-hand side is $$n \frac{1 - (1/2)^{k+1}}{1 - (1/2)} = n\frac{2 - (1/2)^k}{1} = 2n - n/2^k = 2n - 1.$$

But more generally, when $n$ need not be a power of $2$, still we can upper-bound the sum by $n + \frac{n}{2} + \frac{n}{4} + \dots$ (all the way to infinitely many terms). The formula for the infinite geometric series (when $|r| < 1$) is
$$a + ar + ar^2 + \dots = a\frac{1}{1-r}.$$
Here with $a = n$ and $r = 1/2$, we have $$n + \frac{n}{2} + \dots = n\frac{1}{1-1/2} = 2n.$$
As our finite sum is an integer strictly less than this upper bound, we can say it's at most $2n - 1$.

ContentLicense: CC BY-SA 3.0


问:
Conditional Probability- With replacement
Say a bag contains 2 white balls and 3 black balls. 3 balls are drawn randomly with replacement.
a. Find the P(1st is white | 2nd is black)
b. Find the P(1st is white | at least 1 black ball is drawn)

For a. I found the P(W)=2/5 and the P(B)=3/5, then I multiplied those together to get 6/25. Then I used the formula for condition probability P(WB)/P(2nd is B)= (6/25)/(3/5)= 2/5, is this correct?

And then for b. I got stuck at P(at least 1 ball is B), I know I should do 1-P(only white balls are drawn) and to find that I thought about doing 5C2/5C3. Is this the correct way to find the P(at least 1 black ball is drawn?

答:
Your answer for (a) is correct.

For (b) the probability is as you suggest  $1-(\frac{2}{5})^3=\frac{117}{125}$.

On the top line you have the probability that the first ball is white and then the next two balls are not both white. This is $(\frac{2}{5})(1-(\frac{2}{5})^2)=\frac{42}{125}$.

So the conditional probability is $\frac{42}{117}.$

ContentLicense: CC BY-SA 4.0


答:
(a) You're right. Notice that the multiplication you did is justified because the events B and W are independent (because the balls are drawn with replacement), and the $P(B)$ cancels out in the end. In general, such a calculation shows that if A and B are independent events then $P(A|B)=P(A)$, which matches our intuition about what "independence" is.

(b) You're right that you want to find $1-P(A)$ where $A$ is the event that only white balls are drawn. Remember that we're drawing with replacement, so each draw is independent. Therefore, $P(A)=P(\text{all balls white})=P(\text{first ball white})P(\text{second ball white})P(\text{third ball white})=\frac23\frac23\frac23.$

Title: Conditional Probability- With replacement

Body:
Say a bag contains 2 white balls and 3 black balls. 3 balls are drawn randomly with replacement.
a. Find the P(1st is white | 2nd is black)
b. Find the P(1st is white | at least 1 black ball is drawn)

For a. I found the P(W)=2/5 and the P(B)=3/5, then I multiplied those together to get 6/25. Then I used the formula for condition probability P(WB)/P(2nd is B)= (6/25)/(3/5)= 2/5, is this correct?

And then for b. I got stuck at P(at least 1 ball is B), I know I should do 1-P(only white balls are drawn) and to find that I thought about doing 5C2/5C3. Is this the correct way to find the P(at least 1 black ball is drawn?


问:
If $H_1 \stackrel d{\hookrightarrow} H_2$, $X \subset H_1$ and $Y \subset H_2$ then $X \stackrel d{\hookrightarrow} Y$?
Let $H_1=(H_1, (\cdot, \cdot )_1)$ and $H_2=(H_2, (\cdot, \cdot )_2)$ be Hilbert spaces. Suppose that $H_1$ is continuously and densely embedded in $H_2$. Simbolically, $H_1 \stackrel  d{\hookrightarrow} H_2$. Let $X \subset H_1$ and $Y \subset H_2$ be closed subspaces such that $X \subset Y$, with $X,Y \neq \emptyset$. I know that
$$
X=(X, (\cdot, \cdot )_1) \quad \text{and} \quad Y=(Y, (\cdot, \cdot )_2) \tag{1}
$$
are Hilbert spaces.

Question. Is $X \stackrel  d{\hookrightarrow} Y$?

Its clear for me that, since $X \subset Y$ and due to $(1)$ that $X$ is continuously embedded  in $Y$. But I don't know (and I couldn't prove) that $X$ is dense in $Y$. I tried using the Hahn-Banach theorem, but without success.

答:
This doesn't seem true to me without more restrictions.
For example, take $X$ the zero vector space in $H_1$, then $X$ is not dense in $Y$.

Edit: What if we consider a separable Hilbert space $H_2$ with basis $\beta$, then let the space generated by $\beta$ be $H_1$.

Taking a non-trivial subspace $Y \subset H_2$ and take a sub-basis $\beta_Y$ for $Y$, then remove one or more vectors from the said basis $\beta_Y$ (by construction or by a unilateral shift). Take the space generated by this shrunken basis to be $X$, it is indeed a subspace of $H_1$, and is a subspace of $Y$, but it is not dense.


问:
Prove: If $a$ divides $b c$ and $1 = a r + c s$, then $a$ divides $b$, where $a,b,c,r$ are integers.
I'm getting started with proofs but I have no idea how to do this. I have gotten to $b c = a k$ where $k$ is some integer by divisibility. I'm not sure I can use division yet as they will no longer be integers. I have no idea how to prove it and am seeking helpful ideas.

答:
Hints

Suppose that  $1 = ar + cs$ and suppose that $a,c$ have a common factor $d$.  Can you prove that therefore $d$ must divide $1$?

Suppose that $a,c$ are relatively prime and that $a$ divides $(bc)$.
Can you prove that therefore $a$ divides $b$?

Have you been taught about the prime factorization theorem?

If $a,c$ are relatively prime, is it possible for any prime number to appear in both the prime factorization of $a$ and the prime factorization of $c$?

If $a$ divides $(bc)$ then how does the prime factorization of $a$ compare with the prime factorization of $(bc)$?  What conclusion can you then draw, under the assumption that the prime factorizations of $a$ and $c$ are completely disjoint?


问:
Prove that the number of positive divisors of a is given by the product $(e_1 + 1)(e_2 + 1)...(e_n + 1)$.
Let $a$ be a positive integer with factorization $a = p_1^{e_1} p_2^{e_2}...p_n^{e_n}$. Prove that the number of positive divisors of a is given by the product $(e_1 + 1)(e_2 + 1)...(e_n + 1)$.

I know that from the fundamental theorem of arithmetics, that if $a\geq2$, then $a$ can be factored as a product of prime numbers. However, I'm not sure how I can use this knowledge to answer this question. Any help is appreciated.

答:
A positive divisor is of the form $ p_1^{a_1} p_2^{a_2}...p_n^{a_n}$.

There are $e_i+1$ choices for each $a_i$ and so the number of choices is the product of these, as you have given.


问:
Listing subgroups of a group
I made a program to list all the subgroups of any group and I came up with satisfactory result for $\operatorname{Symmetric Group}[3]$ as

$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 2 \\
\end{array}
\right)\right]\right\}\right\}$

It excludes the whole set itself though it can be added seperately. But in case of $SymmetricGroup[4]$ I am getting following

$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
 3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
 2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 4 \\
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 4 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 4 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 2 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 2 & 4 & 3 \\
\end{array}
\right)\right]\right\}\right\}$

The matrix form shows double transposition. Can someone please check for me if I am getting appropriate results? I doubt I am!!

答:
I have the impression that you only list the cyclic subgroups.

For $S_3$, the full group $S_3$ ist missing as a subgroup (you are mentioning that in your question).

For $S_4$, several subgroups are missing. In total, there should be $30$ of them. $14$ of them are cyclic, which are exactly the ones you listed. To give you a concrete example, the famous Klein Four subgroup
$$\{\operatorname{id},(12)(34),(13)(24),(14)(23)\}$$
is not contained in your list.

Title: Listing subgroups of a group


问:
Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

答:
Not a proof of the statement but close.

Also, one should notice that the statement does not hold water for numbers in the form
$x=\frac{\pi*m}{k}$ where $m <=k$

The question remains if these are the only "bad" x's.

For the proof,
let $k_n$ be an increasing sequence of integers and $\sigma < \frac{1}{\sqrt2}$.

Let $E = \{x; \overline {\lim}_{n \to \infty} |\sin(k_{n}x)| < \sigma \}$.

Lastly, let $\chi_{E}$ be characteristic function of $E$.

Then $m(E)=0$.

To see this, note that on one hand
$$\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx$$ tends to $0$ because this is just a Fourier coefficient of integrable function.

At the same time since $\cos(2k_{n}x) = \frac { 1-2\sin^{2}k_nx}{2}$ one gets:
$$\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx = \frac{1}{2}m(E) - \int_{0}^{2\pi}\chi_{E} \sin^{2}(k_{n}x)dx $$, so regrouping and taking upper limits:
$$\frac{1}{2}m(E) <= \overline {\lim}_{n \to \infty}\int_{0}^{2\pi}\chi_{E} \cos(2k_{n}x) dx + \overline {\lim}_{n \to \infty}\int_{0}^{2\pi}\chi_{E} \sin^{2}(k_{n}x) dx $$
As remarked above the first term tends to $0$.

The second one can be estimated by passing with upper limit inside the integral (Fatou's Lemma) and using definition of $E$, to get $<= \sigma^{2}m(E)$.

//////////////////////

This comment relates to the reformulated version of the problem added by Michelle on 2/15/2021.
The new statement suggest much simpler solution.
One looks at unit vectors (complex numbers) on the unit circle and the angle between them.
If two vectors have angle between them (mod $2\pi$) , say $\alpha$, which is less then $\frac{\pi}{k}$ then their image under $f$ maps this angle into $k*\alpha$ (mod $2\pi$). Hence, for any two vectors $z_1, z_2$ if $z_1 != z_2$ there is an $n$ such that the angle between $z_1^n$ and $z_2^n$ is at least $\frac{\pi}{k}$. This gives proper estimates on $\sigma$ that is $\sigma >= |1-e^{i*\frac{\pi}{4}}|$


答: If $a=p_{1}^{e_1} \ldots p_{n}^{e_n}$ then $b$ is a divisor of $a$, if, and only if, $b=p_{1}^{f_1} \ldots p_{n}^{f_n}$ where $0\leq f_i\leq e_i$. The is just a combinatorics problem and using the multiplicative principle you get that the number of positive divisors of a is given by the product $(e_1+1)(e_2+1)...(e_n+1)$.

Body:
Let $a$ be a positive integer with factorization $a = p_1^{e_1} p_2^{e_2}...p_n^{e_n}$. Prove that the number of positive divisors of a is given by the product $(e_1 + 1)(e_2 + 1)...(e_n + 1)$.

I know that from the fundamental theorem of arithmetics, that if $a\geq2$, then $a$ can be factored as a product of prime numbers. However, I'm not sure how I can use this knowledge to answer this question. Any help is appreciated.


答:
Let us denote $B_1, B_2, B_3$ to represent the independent events of drawing a black ball in the indicated draw; and likewise their corresponding complements shall be $W_1, W_2, W_3$.

(a) No need to multiply then divide by the same thing.   It is simply that $\mathsf P(W_1\mid B_2)= \mathsf P(W_1)$ since these events are independent (vis "with replacement"), so the result is as you had: $\mathsf P(W_1\mid B_2)= 2/5$

$$\mathsf P(W_1\mid B_2)=\dfrac{\mathsf P(W_1)\,\require{cancel}\cancel{\mathsf P(B_2)}}{\cancel{\mathsf P(B_2)}}$$

(b) "At least one is black" is the event $B_1\cup B_2\cup B_3$, and this will be the complement of $W_1\cap W_2\cap W_3$, as you reasoned.  However, there will be no binomial coefficients involved in this evaluation, as the draws are made with replacement.

The definition of conditional probability, and the independence of the draws, leads to:

$$\mathsf P(W_1\mid B_1\cup B_2\cup B_3)=\dfrac{\mathsf P(W_1\cap( B_2\cup B_3) )}{\mathsf P(B_1\cup B_2\cup B_3)}=\dfrac{\mathsf P(W_1)\,(1-\mathsf P(W_2)\,\mathsf P(W_3))}{1-\mathsf P(W_1)\,\mathsf P(W_2)\,\mathsf P(W_3)}$$


问:
Probability density function of Z where Z=X/Y, X,Y are exp random variables with parameter is 1?
What I thought is, for any a that Z = a, if Y = y, then X should be ay. The probability X = ay and Y = y is $e^{-ay}e^{-y}$. Then take intergal $\int_{0}^{\infty } e^{-y}e^{-ay}dy=\frac{1}{a+1}$.

But it's not the right answer, right answer is $\frac{1}{(1+a)^{2}}$.

I use the other method that calculate $F_{Z}(z)$ first then derivative it and gets the right p.d.f. But I am confused where my thinking above is wrong...

答:
If $X$ and $Y$ are independent continuous random variables with densities $f_X$ and $f_Y$, then the density of the ratio $Z:= \frac XY$ is given by
$$
f_Z(z) = \int_{\mathbb R} |y|f_X(zy)f_Y(y)\ \mathsf dy.
$$
In the case where $X\sim\mathsf{Expo}(\lambda)$, $Y\sim\mathsf{Expo}(\mu)$ we have
\begin{align}
f_Z(z) &= \int_0^\infty y\lambda e^{-\lambda zy}\mu e^{-\mu y}\ \mathsf dy\\
&= \lambda\mu \int_0^\infty ye^{-(\lambda z+\mu)y}\ \mathsf dy\\
&= \frac{\lambda  \mu }{(\mu +\lambda  z)^2}.
\end{align}
If $\lambda=\mu=1$ this simplifies to
$$
f_Z(z) = \frac1{(1+z)^2}\cdot\mathsf 1_{(0,\infty)}(z).
$$


问:
Why is this an incorrect application of substitution theorem? (Buck)

Let $C=\int_{-1}^{1}[\sin 1 / x]^{2} d x$. Since the integrand is bounded and continuous on [-1,1] except at the point $x=0,$ the integral exists. Moreover, since the integrand is never negative, $C>0 .$ Put $u=1 / x$ so that $f(x)$ becomes $(\sin u)^{2}$ and $d x=-u^{-2} d u$. When $x=1, u=1,$ and when $x=-1, u=-1$
Thus,
$$
C=\int_{-1}^{1}(\sin u)^{2}\left(-u^{-2}\right) d u=-\int_{-1}^{1}\left[\frac{\sin u}{u}\right]^{2} d u
$$
The integrand in this integral is continuous on [-1,1] since the discontinuity at $u=0$ is removable; thus, the new integral exists, and is negative!

Is it because we are supposed to integrate over $[-1,1]$, but after performing the change in variables, the interval for $u$ should be $(-\infty,-1] \cup [1,\infty)$, since $u=1/x$, not $[-1,1]$, i.e. an improper integral?

Source:

Buck, Advanced Calculus, pg 183-184

答:
Yes. I think their point is a simple one. You cannot simply plug in the lower and upper limits of integration into $u=1/x$ and put them as the new limits. The new integral is

$$\lim_{a\rightarrow-\infty}\int_a^{-1}\left(\frac{\sin u}{u}\right)^2du+\lim_{b\rightarrow\infty}\int_1^{b}\left(\frac{\sin u}{u}\right)^2du$$

because $x\in[-1,0)$ means $u\in(-\infty,-1]$ and $x\in(0,\infty]$ means $u\in(1, \infty)$.


问:
Show how union is analogous to logical inclusive OR
The problem I am trying to solve is showing that a union between sets represents the same action as an inclusive OR. Here is a picture of the work I have:

This is my for a class activity and my professor says that it is okay, but that it would be more elegant to show with truth tables. I know what the truth table looks like for inclusive OR but I'm not sure how to show that relates to union because in my head, I think of them as the same. I'm probably overthinking this, but does anyone have an idea? Thanks!

答:
There are many approaches to understanding how inclusive disjunction (your inclusive "or", or $\vee$) is related to union, but let us look from your professor's point of view of truth tables.

Let $X$ and $Y$ be sets: when is $a$ in the union of $X$ and $Y$, $X\cup Y$? When it is in either $X$ (meaning, it is in $X$ and it is NOT in $Y$), or in $Y$ (again, meaning it is in $Y$ and it is NOT in $X$), or simultaneously in both. When $a$ is not in $X\cup Y$? When is, simultaneously, not in $X$ and not in $Y$.

So, we have four cases to consider:

$a$ is in $X\cup Y$ if $a$ is in $X$ and $a$ is not in $Y$;

$a$ is in $X\cup Y$ if $a$ is not in $X$ and $a$ is in $Y$;

$a$ is in $X\cup Y$ if $a$ is in $X$ and $a$ is in $Y$;

$a$ is not in $X\cup Y$ if $a$ is not in $X$ and $a$ is not in $Y$.

Now, the truth table for the inclusive disjunction is given by (where $T$ stands for true and $F$ for false)

$$\begin{array}{|c|c|c|}
\hline
 \vee & T & F \\ \hline
T & T & T \\ \hline
F & T & F \\ \hline
\end{array}$$

This means that, for example, when you have a (inclusive) disjuntion of a true statement (line started with $T$) and a false statement (column started with $F$), you have again a true statement (look at where the line started with $T$ meets the column started with $F$).

Well, if write down the content of your table, what you have is the following:

A (inclusive) disjunction is true if the first statement is true and the second statement is false;

A disjunction is true if the first statement is false and the second statement is true;

A disjunction is true if the first statement is true and the second statement is true;

A disjunction is false if the first statement is false and the second statement is false.

Finally, if in our previous list you replace

"$a$ is in $X\cup Y$" by "a disjunction is true";

"$a$ is not in $X\cup Y$" by "a disjunction is false";

"$a$ is in $X$" by "the first statement is true";

"$a$ is not in $X$" by "the first statement is false";

"$a$ is in $Y$" by "the second statement is true";

"$a$ is not in $Y$" by "the second statement is false",

you obtain precisely the list extracted from our table, and the equivalence between the two notions becomes more evident!


问:
How is functional value defined?
"Suppose $\phi$ is a function of the coordinates $x^k$, and let denote $\bar{\phi}$ the functional value under a transformation to a new set of coordinates $\bar{x}^k$. Then $\phi$ is called a scalar or invariant with respect to the coordinate transformation if $\phi=\bar{\phi}$. A scalar or invariant is also called a tensor of rank zero."

Maybe this is obvious but the term "functional value" has not been defined anywhere in the book. I don't know what it means and therefore don't understand the definition.

The definition is from "Schaum's Outline of Vector Analysis", chapter 8.

答: A functional is a function that takes a vector as input and gives a number as output.  A functional value is the value of the functional when you plug in a vector.  They are saying that if you get the same number in different coordinate systems, it is a "tensor of rank zero."</p>

Title: How is functional value defined?


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
Here's a way to think about it: imagine initializing a "total damage counter" as $D_0=0$, and on the $n$th turn, when $d_n$ damage is done to the boss, increment it to $D_n=D_{n-1}+d_n$ and color the interval $(D_{n-1},D_n]$ according to which player attacked. If we allow this to continue forever, it will color the whole number line with a random pattern of three colors.

Notice that the player who kills the boss is the one whose color covers the point 1000. Since this point is far away from 0, we can think of it as being randomly aligned with our color pattern, so it's as though we're picking a random point on the number line and asking what color it is. This only depends on the ratio of paint colors that our pattern uses, and (by the law of large numbers) this is approximately just the ratio among the expected values of the players' attacks. So e.g. the probability that player A kills the boss is about $37.5/(37.5+50+45)$.


问:
Product Model Covariance
I am investigating a simple model related to the random walk (Just that you work with the product here).

What you do is setting $S_0=1$ and $S_{k}=u\cdot S_{k-1}$ with some probability $p$ and $S_{k}=d\cdot S_{k-1}$ with some probability $q=1-p$. Those choices are of course independent. Therefore $S_n=\prod _{k=1}^n X_k$ with $X_k \in \{u,d\}$

Now the question is $\operatorname{Cov}(S_n,S_l)=?$

Also I would like to know a way how to test some formula I found for this with mathematica if that exists. For example I generated two tables $A,B$ that simulate $S_n$ for $n=10$ (with $100.000$ values each). Now $\text{Variance}(A)$ gives me (almost) the same result in mathematica as $\text{Variance}(B)$ or $\text{Covariance}(A,A)$. But even plugging in $\text{Covariance}(A,B)$ I start getting really tiny values, is that correct?

Now I will provide some own thoughts I did about this problem:

Certainly you have $P(S_n=u^k d^{n-k})=\binom{n}{k} p^k (1-p)^{n-k}$.

With this you can easily see that

$E(S_n)=\sum _{k=0}^n u^k d^{n-k} P(S_n=u^k d^{n-k}) = (d q+p u)^n$

and simillary

$E(S_n^2)=(d^2 q+p u^2)^n$

This also provides the variance, but the covariance seems very tough (at least for me).

答:
Your computations are correct. The computation of the covariance is best approached by writing $S_{n+k}$ as $S_{n+k}=S_nT_k$ where $T_k$ is independent of $S_n$ and distributed like $S_k$. Hence, using the fact that $E(S_n)=m^n$, $E(S_k)=m^k$ and $E(S_n^2)=v^n$ with $m=pu+(1-p)d$ and $v=pu^2+(1-p)d^2$, one gets
$$
\mathrm{Cov}(S_n,S_{n+k})=E(S_n^2)E(S_k)-E(S_n)^2E(S_k)=(v^n-m^{2n})m^k.
$$</p>

Title: Product Model Covariance


问:
How to solve this simultaneous equation of $3$ variables.
I've stuck in this equation system. No clue how to start ?

$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$

Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.

答:
Assume that $(a,b,c)\ne \lambda(1,1,1)$ and put $x:=a+u$, $\>y:=b+v$, $\>z:=c+w$. Then $(1)$ and $(2)$ amount to the system
$$\eqalign{\phantom{a}u+\phantom{b}v+\phantom{c}w&=0\cr a u+ bv+cw&=0\cr}$$
with the solution
$$u=t(b-c),\quad v=t(c-a),\quad w=t(b-c)\qquad(t\in{\mathbb R})\ .$$
Plug this into the equation $(3)$:
$$a(a+u)^2+b(b+v)^2+c(c+w)^2=a^3+b^3+c^3$$
and obtain a quadratic equation for $t$ with one obvious solution $t=0$ and a second solution
$$t={2(a-b)(b-c)(c-a)\over ab(a+b)+bc(b+c)+ca(c+a)-6abc}\ .\tag{4}$$
Cases where the denominator in $(4)$ is zero have to be dealt with separately.</p>


问:
Integral equations solutions
In my research, I have reached a point where I need to find a function defined on the interval $[0,\pi]$ and satisfies the following properties:

It's an even function:

$f(-\theta) = f(\theta)$

It's normalised as:

$\int^\pi_0f(\theta)d\theta = \frac{\pi}{2}$

For any $\phi$ in the interval $[0,\pi]$,the following property is satisfied:

$\int^\pi_0f(\theta)f(\theta - \phi)d\theta = \frac{\pi}{2} \cos^2\frac{\phi}{2}$

I have tried many functions, but non of them worked. I would be very much in your debt if you help me find such a function.
Thank you in advance.

答:
Your first condition looks strange; let's shift the limits to $-\phi$ and $-\phi + \pi$. That way we get $f(\theta)$ inside.  Now, you want that integral to have a constant value regardless of the value of $\phi$.  But that can only be satisfied by a constant function. If you shift the interval $\delta \theta$ to the right, the integral can't change, which means the area added has to equal the area lost, so the heights have to be equal.  So the first condition requires $f(\theta - \phi) = 1/2$.
Given that constraint, the second requirement is impossible to satisfy.

Title: Integral equations solutions


问:
Unbiased estimator and getting estimate from estimator
I got a unbiased estimator but I don't know how to interpret it and adjust it to get estimate.

The original problem is to find out the unbiased estimator for $\lambda$ in Zero-truncated Poisson Distribution.

So, I am trying to find $T(X)$ such that $E(T(X)) = \lambda$.

Let, $X \sim ZTP(\lambda)$ then, the PMF of $X$ is $ \frac{\lambda^{k}}{k!(e^{\lambda}-1)}, \quad for \; k > 0$

\begin{align}
E(T(X)) = \sum_{x=1}^{\infty}T(x)\cdot\frac{\lambda^{x}}{x!(e^{\lambda}-1)}
\end{align}
By expanding the above and Taylor Series then,
\begin{align}
T(1)\frac{\lambda^{1}}{1!} + T(2)\frac{\lambda^{2}}{2!} + T(3)\frac{\lambda^{3}}{3!} + \cdots = \lambda(e^{\lambda}-1) = \lambda(\lambda + \frac{\lambda^{2}}{2!} + \frac{\lambda^{3}}{3!} + \cdots)
\end{align}

Two power series can be identical if their all coefficients match so,

\begin{align}
T(1) = 0, \quad T(2) = 2, \quad T(3) = 3, \quad \cdots
\end{align}

\begin{equation}
T(X) = \begin{cases}
0 & X = 1 \\
X & X > 1
\end{cases}
\end{equation}

Finally got unbiased estimator for $T(X)$ for $\lambda$.

And if I get some random samples, $X_{1}, X_{2}, X_{3}$ which are equivalent to $X_{1}=3, X_{2}=8, X_{3} = 1$ then how can I get the estimate from estimator $T(X)$?

Is it $3$? $8$? $1$? or maybe any value can be an estimator among those?

or maybe the mean of three values, which is equivalent to $4$ or maybe any other value?

To give some detail about this problem, actually this problem came from the online lecture I took.

And the lecturer used this example as a method to get an unbiased estimator by solving an equation.

So you don't need to consider whether the estimator is a proper value or not.

Btw, what makes me so confused is he does not make a clear notation between random variable such as $X$ and random sample $X_{i}$

so I got a trouble to distinguish them.

It seems like the lecturer used notation $T(X)$ instead of using $T(X_{i})$ but I am not sure.

Please help me!

答: Your method found an unbiased estimator $T$ for a single observation $X$.  So if you take the sample mean of the estimates for $n$ iid observations $T(X_1), \ldots, T(X_n)$, then this will also be unbiased, since by linearity of expectation, $$\operatorname{E}\left[\frac{T(X_1) + \cdots T(X_n)}{n}\right] = \frac{1}{n} \sum_{i=1}^n \operatorname{E}[T(X_i)] = \frac{1}{n} (n \lambda) = \lambda$$ because each $\operatorname{E}[T(X_i)] = \lambda$ by your construction.  So your unbiased estimator for a sample is $$T(X_1, \ldots, X_n) = \frac{1}{n}\left( \sum_{i=1}^n X_i - \mathbb 1(X_i = 1) \right),$$ in other words, take the sample total $\sum X_i$, subtract the number of observations in the sample that equal $1$, and then divide the result by $n$.  For example, if you observe the sample $$(x_1, \ldots, x_n) = (3, 5, 2, 7, 1, 1, 3, 2, 4, 1, 10),$$ with $n = 11$, we would compute the total $39$, subtract $3$ because three of the observations are $1$, and divide by $11$ to obtain the estimate $T = \frac{36}{11}$.


问:
Getting stuck when solving second order non-homogeneous linear differential equation
I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$

I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$

We have a non-distinct real root of multiplicity 2 which is -5.

The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$

We now find the candidate for the undermined coefficients method.

$F(x) = (3x+2)e^{-5x}$

$F'(x) = (-5Ax + A - 5B)e^{-5x}$

$F''(x) = (25Ax - 10A + 25B)e^{-5x}$

$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.

So our particular solution is $y_{p} = (Ax + B)e^{-5x}$ (Should the exponential be multiplied by C?)

My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...

Thank you.

答:
HINT

Here it is another way to approach such ODE.

I would start with noticing that

\begin{align*}
y'' + 10y' + 25 = (3x+2)e^{-5x} \Longleftrightarrow (ye^{5x})'' = 3x+2
\end{align*}

Can you take it from here?

EDIT

According to the integrating factor method, we deduce that

\begin{align*}
y'' + 10y' + 25 = (3x+2)e^{-5x} & \Longleftrightarrow (y' + 5y)' + 5(y' + 5) = (3x+2)e^{-5x}\\\\
& \Longleftrightarrow u' + 5u = (3x+2)e^{-5x}\\\\\
& \Longleftrightarrow u'e^{5x} + 5ue^{5x} = 3x + 2\\\\\
& \Longleftrightarrow (ue^{5x})' = 3x + 2\\\\
& \Longleftrightarrow (y'e^{5x} + 5ye^{5x})' = 3x+2\\\\
& \Longleftrightarrow (ye^{5x})'' = 3x + 2
\end{align*}

where $u = y' + 5y$.


问:
What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$? v2.0
Suppose $X$ and $Y$ are two independent non-negative random variables, such that $E[X] = E[Y] = 1$. What is $\sup_X \{ \inf_Y \{P(X > Y)\}\}$?

I managed to derive the following facts:

$\frac{1}{4} \leq \sup_X \{ \inf_Y \{P(X > Y)\}\} \leq \frac{1}{2}$.

One can get the upper bound simply by examining the case, when $X$ and $Y$ are i.i.d.

To get the lower bound we have to look at $X$, that takes values $0$ and $2$ with equal probability. In this case, for any $Y$ satisfying our condition, we have

$$P(X > Y) = P(X = 2)(1 - P(Y \geq 2)) = \frac{1}{2}(1 - P(Y \geq 2)) \geq \frac{1}{4}$$

by Markov inequality.

However, I do not know, whether any of those bounds is actually tight.

Here is a similar question for random variables without non-negativity condition (and with answer $0$).

答:
Let $X$ be uniform in $[0,2]$. Then every non-negative variable $Y$ independent of $X$ satisfies $$P(X \le Y\, |\, Y) \le Y/2 \,,\; \; \; (*)$$
with equality iff $Y \le 2$. Next, recall the assumption $E(Y)=1$. Taking expectation in (*) using the tower property [1] gives
$$P(X \le Y) \le E(Y/2)= 1/2 \,,$$
with equality iff $Y \le 2 $ a.s.
Thus for this choice of $X$ and for every $Y \ge 0$  independent of $X$ that satisfies $E(Y)=1$, we have
$$P(X>Y)\ge 1/2 \,,$$
so
$$\sup_X \inf_Y P(X>Y)=1/2 \,,$$
where $X,Y$ are assumed to be independent, non-negative of mean 1.

[1] https://en.wikipedia.org/wiki/Law_of_total_expectation


答:
Since your root $-5$ is double, the right form of $y_p$ is
$$
Y_p=x^2(ax+b)e^{-5x}.
$$

Title: Getting stuck when solving second order non-homogeneous linear differential equation

Body:
I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$

I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$

We have a non-distinct real root of multiplicity 2 which is -5.

The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$

We now find the candidate for the undermined coefficients method.

$F(x) = (3x+2)e^{-5x}$

$F'(x) = (-5Ax + A - 5B)e^{-5x}$

$F''(x) = (25Ax - 10A + 25B)e^{-5x}$

$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.

So our particular solution is $y_{p} = (Ax + B)e^{-5x}$ (Should the exponential be multiplied by C?)

My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...

Thank you.


问:
Showing remainder of taylor polynomial goes to $0$
To show the remainder of the taylor polynomial for $e^x$ about $x=0$, goes to $0$, is all I need to do suggest there is some $c \in (0,x)$ such that $r_n(x)=\frac{e^cx^n}{n!}$, where $r_n$ is the remainder term of the taylor polynomial, then note that this approaches $0$ as $n \rightarrow \infty$?

答: This is essentially correct but observe that $c$ can depend on $n$. But for any fixed $x$ you can bound $e^c$ by $\max\{1,e^x\}$ and then take the limit of $|r_n(x)|$ as $n\to\infty$.


问:
How do you find inflection points from a graph?
Before you get upset that it's already answered somewhere on the internet, I thought so as well. In fact, I immediately googled it and went to Khan Academy, only to get 10x more confused. In the KA lesson, the inflection pts are the critical pts, but in their lesson quiz, it's the pt where the concavity changes. So which one is it?

KA Lesson Quiz

KA Lesson

答:
Because the graph in the Lesson is of the derivative, the leftmost inflection point is the leftmost point of where the slope of that graph is 0. Note that the derivative is positive to the left and right of the inflection point. This means the graph still has a positive slope. However it has changed from getting steeper to getting less steep, at this inflection point. You can imagine that the graph changes from "curving upward" to "curving downward."

For the quiz, the answers are A and C. For A, the second derivative changes from negative to positive, and for C, from positive to negative. The second derivative is a rate of change of the first derivative. So for A, the slope changes from getting more negative/less positive to getting more positive/getting less negative. For C, it is the opposite.


问:
What is the language recognized by the finite automaton?
I have no idea how to approach this?

答:
We are dealing with a deterministic finite automaton according to the provided picture. It is defined as a $5$-tuple $(Q, \Sigma, \delta, q_0, F)$.

The set of states: $Q=\{\{1,2\}, \{1,2,3\}, \{2,3\}\}$

The set of input symbols $\Sigma=\{a,b, \epsilon\}$

The transition function $\delta: Q \times \Sigma \to Q:(\{1,2\},a) \mapsto \{1,2,3\}$, $(\{1,2,3\},a) \mapsto \{1,2,3\}$, $(\{1,2,3\},b) \mapsto \{2,3\}$, $(\{2,3\},b) \mapsto \{2,3\}$, $(\{2,3\},a) \mapsto \{1,2\}$

The inital state $q_0:=\{1,2\}$

The set of accept states $F \subseteq Q$, which is $Q$

So you always start with $a$ or $\epsilon$ (notice that the empty word is accepted because the initial state is an accept state).

Example of accepted words: $aabbb$, $a$, $aba$

Example of unaccepted words: $b$, $bbabaaaaaba$

Can you go from here?

Body: I have no idea how to approach this?


问:
Find a geometric solution for a equilateral tringle within 3 circles
For the task of constructing an equilateral triangle with a vertex on each of three concentric circles (the subject of a previous question), I found a solution on: http://mathafou.free.fr/pbg_en/sol113.html, which is hard to understand.

Red triangle is the solution. Quoting from the reference:

With point P as center, draw circles (A), (B) and (C) with radii a,b,c. We can fix A anywhere on circle (A).

So in the graphic is A fixed on the bottom of the outter circle.
A circle is drawn with the radius AP and cuts at M.

The locus of C is then (C). B is deduced from C by rotating with center A and angle π/3.

There is no B in the graph. Just confusing.

The locus of B is then circle (C') image of (C) using this rotation. But it is also (B) from definition. Point B is then the intersection point of (B) and (C').

Why is the circle (C') with middle point drawn here?

Point C is then deduced by inverse rotation.

Ok. We have B1 and now we rotate back 60 degrees to have C.

There are two intersection points B1 and B2 giving the two triangles obtained from analytic solution.
Only one of them contains point P.

Yes, B1, but why? "We must have angle ABP < π/3"

Please, I need some help.

答:
Comment: Experimental approach

$r_1=40, r_2=55, r_3=70$, in this case the difference between radii is equal.

The locus of center of triangle ABC , Q is a circle center on O and radius OQ and we have:

$OQ\approx \frac{\sqrt{40+55+70}+\sqrt{40}+\sqrt{55}+\sqrt{70}}2\approx 17.48$

This value is close to $17.88$ shown in figure.

In this figure the difference of radii is not equal. $r_1=47, r_2=60, r_3=80$ and we have:

$OQ\approx\frac{\sqrt{47+60+80}+\sqrt{47}+\sqrt{60}+\sqrt{80}}2\approx 18.6$

This value is close to $20.8$ shown in figure. So for a rough drawing:

1- calculate OQ and draw a circle $r=OQ$ center on O.

2- take an arbitrary point on this circle (Q). this is the center of triangle. In a particular case one of altitudes of triangle crosses the center O. So draw a line connecting O and Q, this line is perpendicular bisector of BC.

3- The angles both QB and QC make with OQ is $60^o$.By drawing the rays of these angles you get vertices B and C on circles with radii $r_2$ and $r_1$ respectively. Vertex A is the intersection of OQ with third circle with radius $r_3$.

May be by analytic geometry we can find a more accurate formula for $OQ$ in terms of $r_1, r_2$ and $r_3$.


答: Keep in mind the picture in the one you called "KA Lesson" [second one you refer to] is already the derivative of the original function. So where that graph has a local min or max, the concavity of original function changes.</p>

Title: How do you find inflection points from a graph?

Body:
Before you get upset that it's already answered somewhere on the internet, I thought so as well. In fact, I immediately googled it and went to Khan Academy, only to get 10x more confused. In the KA lesson, the inflection pts are the critical pts, but in their lesson quiz, it's the pt where the concavity changes. So which one is it?

KA Lesson Quiz

KA Lesson


问:
When does Big-O notation apply to functions with arbitrary constants?
I am having trouble understanding if I can apply Big-O notation to the following circumstance.
Suppose that
$$Mxk_0=\sum_{j=1}^{M}xk_j, \hspace{0.5cm}\mbox{where $0<k_j<1$.}$$
Regard all $k_j$ as being arbitrary in their collective interval. This leads me to believe that $k_0<1$ must be a constant. Now, let $g(x)=xf(x)$. We have
$$\frac{g(x)}{k_0x}=\frac{1}{k_0}f(x).$$
Thus, $$\frac{1}{k_0}f(x)=O(f(x)).$$
Is this correct? I think it uses Big-O notation correctly, but I don't feel satisfied in justifying it because $k_0$ feels weirdly fickle.

答:
You might be worried that since $k_0$ can be arbitrarily close to $0$, the factor $\frac1{k_0}$ can be arbitrarily large. It's still valid to carry out $O$-notation reasoning treating $\frac1{k_0}$ as a constant, though. The point is that even if $\frac1{k_0}$ is really large, as long as we hold it constant, eventually $x$ will get big enough (or small enough, depending on which version of $O$ we're using) that it doesn't matter. One can formalize this argument and prove it from the definition of $O$-notation.

Still, it's often useful/informative to keep track of which parameters the constant factor hidden in the $O$ expression depends on. E.g. a paper might present its result as "$g=O(f)$, with a constant factor that does not depend on $\epsilon$", which is a stronger statement than just "$g=O(f)$".


问:
Calculating Conditional Probability $P(X<3|X+Y) $
Suppose that $ X,Y$ are independent unit $\text{Exp}(1)$ rv's find  $P(X<3|X+Y) $.

I am hoping to solve this problem, but I am actually not sure how to set up things correctly. I think we need to find the Density of $Z=X+Y$ and we can do something with that, but I am uncertain.

Any advice would be appreciated.

答:
Tip: The sum of two independent exponential distributions has this pdf.

$$\begin{align}f_{\small X+Y}(z) &=\frac{\mathrm d~~}{\mathrm d z}\int_0^z\int_0^{z-x}\mathrm e^{-(x+y)}\mathrm d y\mathrm dx \\[1ex]&= z\mathrm e^{-z}\mathbf 1_{0\leqslant z}\end{align}$$

Which leads to the conditional distribution$$f_{\small X\mid X+Y}(x\mid z) = \dfrac{f_{\small X}(x)\,f_{\small Y}(z-x)}{f_{\small X+Y}(z)}\\\vdots$$


问:
How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series?
I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.

I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:

Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.

https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454

答:
The formula provided on the wikipedia page works if you know how many terms you are adding. For example, if you want to calculate the sum $1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \frac{1}{2^9}$, you know you are adding 10 terms (so $n = 10$). Also, $a = 1$ (the number you begin with) and $r = \frac{1}{2}$ (the common ratio). Now you could use the provided formula to calculate the sum:

\begin{align*}
a \frac{1-r^n}{1-r} &= 1 \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}}\\
&= \frac{1 - \frac{1}{2^{10}}}{\frac{1}{2}}\\
&= 2(1 - \frac{1}{2^{10}})\\
&= 2 - \frac{2}{2^{10}}\\
&= 2 - \frac{1}{2^9}
\end{align*}

You are talking abouth the sum $a + \frac{a}{2} + \frac{a}{4} + \dots + 1$, so the last term is 1, instead of $ar^{n-1}$. Note that $r = \frac{1}{2}$ in your case, as each term is twice as small as the term before. The formula could still solve this problem, but you will have to set $a\left(\frac{1}{2}\right)^{n-1}$ equal to 1:

\begin{align*}
a\left(\frac{1}{2}\right)^{n-1} &= 1\\
a &= \frac{1}{\left(\frac{1}{2}\right)^{n-1}} = \frac{1}{\frac{1}{2^{n-1}}} = 2^{n-1}
\end{align*}

So, your formula will only work when $a$ is a power of two. For example, if $a = 4$:

$$
4 + 2 + 1 = 7 = 8 - 1 = 2*4 - 1
$$

So for $a=4$, your formula works. Let's try $a = 16$:

$$
16 + 8 + 4 + 2 + 1 = 31 = 32 - 1 = 2 * 16 - 1
$$

Again, your formula works. How would we prove this? The formula you found on wikipedia has a $r^n$ term that we would like to get rid of:

\begin{align*}
a\left(\frac{1}{2}\right)^{n-1} &= 1\\
\left(\frac{1}{2}\right)^{n-1} &= \frac{1}{a}\\
\left(\frac{1}{2}\right)^n &= \frac{1}{a} * \frac{1}{2} = \frac{1}{2a}
\end{align*}

So, $r = \frac{1}{2}$ and $r^n = \left(\frac{1}{2}\right)^n = \frac{1}{2a}$. Now, let's plug that in:

\begin{align*}
a \frac{1-r^n}{1-r} &= a \frac{1 - \frac{1}{2a}}{1 - \frac{1}{2}}\\
&= a \frac{1 - \frac{1}{2a}}{\frac{1}{2}}\\
&= 2a(1 - \frac{1}{2a})\\
&= 2a - \frac{2a}{2a}\\
&= 2a - 1
\end{align*}

So there you have it: for every $a = 2^k$ (with $k \in \mathbb{N}$):

$$
a + \frac{a}{2} + \frac{a}{4} + \dots + 1 = 2a - 1
$$


答:
Inflection points are places where the concavity changes. Inflection points only occur at critical points, but not all critical points are inflection points. For example, all cookies are desserts, but not all desserts are cookies. So, first you need to find the critical points of $f(x)$ where the concavity COULD change. $f(x)$ is concave-down if $f''(x)<0$ and concave up when $f''(x)>0$, so the concavity changes when you cross from positive to negative, which MIGHT happen when $f''(x)=0$, or at an area in the graph where $f''(x)$ is undefined. After you find these critical points, you need to make sure the concavity actually does change by testing points close to the left and points close to the right of the critical point to make sure the sign changes. Then, that's when you find the inflection points.

Let's try an example $f(x) = x^4$. We get $f'(x) = 4x^3$, so $f''(x)=12x^2$. So, it looks like $f''(x)$ $=0$ when $x=0$, and $f''(x)$ is defined everywhere, so we don't look for undefined regions. Now, we need to make sure this critical point actually is an inflection point. So, we test near the critical point on the left $f''(-1) = 12(-1)^2=12$. This is positive, so graph concave up left of $x=0$. Now we test on the right $f''(1)=12(1)^2=12$. This is positive, so the graph is concave up on the right of $x=0$ as well. Since the graph is concave up to the left and right of $x=0$, the concavity doesn't change at $x=0$, thus $x=0$ is not an inflection point. So, there are no inflection points for $x^4$.

Let's try $f(x) = x^3$. We get $f''(x) =6x$. So, it looks like $f''(x)$ $=0$ when $x=0$, and $f''(x)$ is defined everywhere, so we don't look for undefined regions. Now, we need to make sure this critical point actually is an inflection point. So, we test near the critical point on the left $f''(-1) = 6(-1)=-6$. This is negative, so graph concave down left of $x=0$. Now we test on the right $f''(1)=6(1)=6$. This is positive, so the graph is concave up on the right of $x=0$ as well. Since the graph is concave down to the left and concave up to the right of $x=0$, the concavity changes at $x=0$, thus $x=0$ is an inflection point. So, there is an inflection point for $x^3$.


问:
Ideal of height $>1$ containing no non-zero prime ideals.
Edit: Badam Baplan gives a remarkable example below, which answers the question completely. I would still appreciate any other examples that others may come up with; in particular, is an example still possible if we ask for $R$ to be Noetherian?

Let $R$ be a (commutative, unital) ring and $I<R$ an ideal. Recall that we define $$\operatorname{ht}I=\min\{\operatorname{ht}P:P\in\operatorname{Spec}R,P\geqslant I\},$$ where $\operatorname{ht}P$ is defined in the usual way for a prime ideal $P$. Is there a nice example of an integral domain $R$ and an ideal $I$ such that (i) $I$ contains no non-zero prime ideals, and (ii) $\operatorname{ht}I>1$? Note: by Krull's height theorem, if $R$ is Noetherian then $I$ cannot be principal.

If we drop the condition that $R$ is an integral domain, then we can get examples where $\operatorname{ht}I$ is arbitrarily large. For instance, let $F$ be your favorite field, and define $A$ to be the polynomial ring $F[x_1,\dots,x_n]$. Choose any non-zero ideal $M\leqslant A$, with some $o\in M\setminus\{0\}$ of minimal degree. (For later, note that $o\notin x_1M+\dots+x_nM$; call this fact $(\star)$.) Now, consider the ring structure on the product $R:=A\times M$ defined by

$(a,m)+(b,n)=(a+b,m+n)$

$(a,m)\cdot(b,n)=(ab,an+bm)$

for all $a,b\in A$ and $m,n\in M$. Note that every element $(0,m)$ is nilpotent, so any prime ideal of $R$ must contain $0\times M$. Conversely, if $P$ is a prime ideal of $A$, then $P\times M$ is a prime ideal of $R$. With this in mind, consider the ideal $I<R$ generated by the elements $(x_i,0)$ for $1\leqslant i\leqslant n$. First note that $I$ contains no prime ideals; indeed, one can explicitly compute $I$ to be the product $$(x_1A+\dots+x_nA)\times(x_1M+\dots +x_nM),$$ where both direct factors are considered as ideals of $A$. In particular, by fact $(\star)$, we have $(0,o)\notin I$, and so (by the remark above, since $(0,o)$ is nilpotent), $I$ cannot contain any prime ideal.

On the other hand, consider the prime ideal $P:=\langle x_1,\dots,x_n\rangle\times M>I$. This is the unique minimal prime lying above $I$, since $P=I+(0\times M)$ and every prime must contain $0\times M$. However, we have $\operatorname{ht}P=n$, since $$0\times M<\langle x_1\rangle\times M<\langle x_1,x_2\rangle\times M<\dots<P$$ is a strictly ascending chain of primes below $P$. Thus, since $P$ is the unique minimal prime above $I$, we have $\operatorname{ht}I=n$.

One can extend this example in the natural way to find an example where $\operatorname{ht}I=\infty$. So this resolves the question completely if we do not require $R$ to be an integral domain. However, I'm struggling to find an example where $R$ is an integral domain. Can anyone think of a nice example?

I tried attempting to consider the example above as a quotient of an appropriate polynomial ring, but didn't make any progress. My thinking was that we could perhaps find a desirable prime lying below the kernel of the projection map, and then quotient by that to find a domain with some similar behavior. However, this doesn't seem to work out very nicely.

As a smaller follow-up question, what is the geometric significance (if any) of such a ring?

答:
One of my favorite examples of this kind of wonky behavior originates from Krull's (false) conjecture that every completely integrally closed domain is the intersection of its one-dimensional valuation overrings.  Nakayama disproved this conjecture in On Krull's conjecture concerning completely integrally closed integrity domains, I and II.   He did this by constructing an integral domain $D$ which possesses the following properties:

$D$ is completely integrally closed

$D$ is Bezout

Every nonzero prime of $D$ has infinite height

So for Nakayama's purposes every valuation overring of $D$ is a localization of $D$ at a prime ideal and thus has infinite dimension.

For the purposes of your question, every nonzero principal ideal of $D$ has infinite height, but a principal ideal of $D$ never contains a prime ideal.

Indeed, for any Archimedean domain (i.e. $\bigcap d^nD = 0$ for all $d \in D$) it is is impossible for a prime to be properly contained in a principal ideal.  Consequently a principal prime in an Archimedean domain necessarily has height $1$.  Hence in an Archimedean domain with no finite height nonzero primes, it is impossible for a nonzero prime to be contained in a principal ideal.

(Updated to mention low-dimensional Noetherian examples)
In the 1970's there was focused interest in the possible poset structure of the prime ideals of a Noetherian ring.  One of the motivating questions was a specialization of your own:  do finite intersections of maximal ideals contain prime ideals?

Among the advances in this period was a construction of Heitmann, which takes in a number of compatible 'special' Noetherian domains $D_\alpha$ with prime posets $P_\alpha$ and spits out a Noetherian domain having (up to order isomorphism) as its prime poset $\coprod P_\alpha$, the disjoint union of the $P_\alpha$ glued together at $(0)$.

This construction gives you a way to find $n$-dimensional Noetherian domain counterexamples to your question.  For example, you can apply it to a couple of $n$-dimensional rings $D_1, D_2$.  Pick a maximal ideal $M_i$ coming from the $P_i$ component.  Clearly $M_1 \cap M_2$ is height $n$.  But $M_1 \cap M_2$ cannot contain a nonzero prime, by disjointness of the $P_i$.

You can check out Heitmann's paper Prime ideal posets in Noetherian rings for details of the construction.


问:
In $a - (a + 2)=-2$, how did the $2$ become $-2$?
I'm new to math. I would like to ask for you help to explain solving the following:

$$a - (a + 2)= -2$$

I didn't understand how the $2$ became $-2$.

Is there a good reference to learn these step by step?

Thanks

答:
Perhaps think of '$a$' as some other number like $5$.

$$5-(5+2)=-2$$

$$5-(7)=-2$$

$$-2=-2$$
Or think of using $10$ in place of '$a$'
$$10-(10+2)=-2$$
$$10-(12)=-2$$
$$-2=-2$$

In fact it does not matter what number is used in place of '$a$' the statement given is always true: if you subtract 'some number' minus 'two more than that number' you will get two less than you start with.


答:
I give you $a$ sweets

I give your brother $b$ sweets.

How much more sweets do you have compared to your brother? Well, $a-b$.

Now you found out that your brother has $a+2$ sweets. Well, it is $a-(a+2)$.  You actually have $2$ sweets less than your brother, not more. So $a-(a+2)=-2$.

By distributive law, we have $c(d+e)=cd+ce$.

$$-(a+2)=-a-2$$

Hence $$a-(a+2)= a+ (-1)(a+2)=a+(-a-2) = (a-a) -2=-2$$

where we use distributive law in the second equation  and associativity of addition in the next equation.

Title: In $a - (a + 2)=-2$, how did the $2$ become $-2$?

Body:
I'm new to math. I would like to ask for you help to explain solving the following:

$$a - (a + 2)= -2$$

I didn't understand how the $2$ became $-2$.

Is there a good reference to learn these step by step?

Thanks


问:
Average Rate of Change of a Quadratic equation but with variables
I received a question that gives the standard form of a quadratic: $y=ax^2+bx+c$ and I am to prove that within the interval $r \le x \le s$, the average rate of change is: $a(s+r) + b$.

I was thinking of using this equation: $\frac{f(s)-f(r)}{s-r}$ and plug in $r$ and $s$ into the original equation to find $f(r)$ and $f(s)$. However, my math does not seem to be working out as I am not getting the correct AROC. Does anyone know the correct way to go about solving this problem?

Thank you!

答:
Your equation is correct, with AROC you'll usually find a factor of $(s-r)$ in the numerator that will cancel with the denominator to get to the result you're looking for (and this becomes a very useful algebraic trick when you get to instantaneous rates of change):

$$ \text{AROC} = \frac{f(s)-f(r)}{s-r} $$
$$ = \frac{(as^2+bs+c)-(ar^2+br+c)}{s-r} $$
$$ = \frac{as^2+bs-ar^2-br}{s-r} $$
$$ = \frac{a(s^2-r^2)+b(s-r)}{s-r} $$
$$ = \frac{a(s-r)(s+r)+b(s-r)}{s-r} $$
$$ = \frac{(s-r)\left(a(s+r)+b\right)}{s-r} $$
$$ = a(s+r)+b, s\neq r .$$


答: Depending on the class, you may be being asked to prove that somewhere on the interval the average rate of change is the expression you gave. In a calculus class, you may be talking about the mean value theorem...proving that the average rateof change is equal to the average rate of change.

Title: Average Rate of Change of a Quadratic equation but with variables

Body:
I received a question that gives the standard form of a quadratic: $y=ax^2+bx+c$ and I am to prove that within the interval $r \le x \le s$, the average rate of change is: $a(s+r) + b$.

I was thinking of using this equation: $\frac{f(s)-f(r)}{s-r}$ and plug in $r$ and $s$ into the original equation to find $f(r)$ and $f(s)$. However, my math does not seem to be working out as I am not getting the correct AROC. Does anyone know the correct way to go about solving this problem?

Thank you!


问:
Why, given a linear transformation $L:V\rightarrow W$ and basis $B$ for $V$, when $L(B)=\dim(W)$ then $L(B)$ is a basis for $W$?
I am trying to understand a proof for the following proposition: given a linear transformation $L:V\rightarrow W$ between finite vector spaces $V$ and $W$, $\dim(V)=\dim(W)=n$, if $L$ is injective then it is an isomorphism. The proof goes like this: let $B$ be a basis for $V$. Since $B$ is lin. indep. and $L$ is injective, it takes $B$ to a lin. indep. set, i.e. $L(B)$ is lin. indep. Then since $|L(B)|=n=\dim(W)$, $L(B)$ is a basis for $W$ and $L$ is surjective.

I am not sure how the last statement is formed. In particular, how does $|L(B)|=\dim(W)$ imply that $L(B)$ is a basis for $W$? As for the finishing statement am I right that $L(B)$ being a basis for $W$ means the image of $L$ is the same as its codomain, which implies surjectivity?

答:
1st Question:

Given that $\dim(V)=\dim(W)=n$.
Let $B=\{b_1,b_2,\ldots, b_n\}$ be a basis of $V$.
To show $L(B)$ is a basis for $W$.

We first show that the set $L(B)=\{L(b_1),L(b_2),\dots, L(b_n)\}$ is linearly independent.

Let $c_1L(b_1)+c_2L(b_2)+\cdots+c_nL(b_n)=0 $

$\implies L(c_1b_1)+L(c_2b_2)+\cdots +L(c_nb_n)=0$

$\implies L(c_1b_1+c_2b_2+\cdots +c_nb_n)=0=L(0)$

Since $L$ is injective so $c_1b_1+c_2b_2+\cdots +c_nb_n=0$.

Since $B$ is a basis of $V$ and $b_i$'s are linearly independent, we have $c_i=0 \forall i.$

2nd Question:

Now $L(B)\subset W$ is a linearly independent set of $W$ having $n$ elements. We show that $L(B)$ spans $W$, i.e. every element of $W$ is a linear combination of members of $L(B)$.

Let $w\in W$. If $w\notin \text{span}(L(B))$, then $\{w,L(b_1),L(b_2),\dots, L(b_n)\}$ is a linearly independent set of $W$,  which is false as $W$ can have atmost $n$ linearly independent elements since $\dim(W)=n$.


答:
The minus is distributive, like multiplying the amount in parentheses by $(-1).\space$ Expanding, we get.

$a-(a+2)=a-a-2\quad =0-2=-2$


问:
An uncountable Hausdorff space such that every countable subset is closed is discrete
Let $X$ be a topological space. Is the following claim true: If $X$ is uncountable, Hausdorff, and every countable subset of $X$ is closed, then $X$ is discrete.

This claim is false without the Hausdorff assumption (If every countable subset of a space is closed, is the topology discrete?), but the example given in the link does not work since it is not Hausdorff. I think my claim is not true (but not sure) and I cannot found a counterexample. Any hints?

答:
Let $X=\Bbb R$; every point of $\Bbb R\setminus\{0\}$ is isolated, and open nbhds of $0$ are co-countable. Thus, the topology on $X$ is

$$\wp(\Bbb R\setminus\{0\})\cup\{\Bbb R\setminus C:C\subseteq\Bbb R\setminus\{0\}\text{ is countable}\}\,.$$

Every countable set is closed, and $X$ is Hausdorff, but $0$ is a limit point of $X\setminus\{0\}$.


问:
What is the purpose of a time variable in a sinus function?
The input of a system is a sine wave of frequency $1/2$ radians per second. The teacher has written it this way, but I do not understand what the $t$ is supposed to mean.

$$\begin{align}\text{Input} &= \sin\left(\frac12t\right)\\
\omega&=\frac12\text{radians}/\Bbb s\propto\text{frequency}\end{align}$$

There is also that unidentified symbol that looks like alpha($\alpha$) before the word frequency on the second line. What is this symbol and what does it mean??

Thank you!

答: The alpha-looking symbol means "is proportional to", as stated in the comments. You can get it in mathjax using \propto, shown here: $\propto$. So if $a\propto b$, then we know that $a=kb$ for some $k\in\Bbb Z$ and $k\ne0$.


答:
It would be helpful to think of an automaton as some kind of "Snakes and Ladders Game". The circles which represent states correspond to the squares in the board game. The arrows represent movement from one square to another. The label on the arrow specifies which input ("dice throw") leads to that move. In board game there is a designated square reaching which is the goal. Here some states are marked specially to designate final state(s).

The language recognized by an  automaton is the sequence of dice throws (inputs) which start from a specially marked initial state and lead to the final state.

The problem you stated was "I have  no idea ...". Now  I have given you an Idea that will help you solve this kind of problems.

Title: What is the language recognized by the finite automaton?

Body: I have no idea how to approach this?


问:
Let $f$ be a curve reparametrized by arc length. Show that the vector $f''(s)$ is orthogonal to the vector $f'(s)$, for all $s∈I$.
Let $f:I⊆\mathbb{R}→\mathbb{R^n}$ be a curve reparametrized by arc length. Show that the vector $f''(s)$ is orthogonal to the vector $f'(s)$, for all $s∈I$.

$(*)$ Definition: $\varphi=\phi^{-1}:[0, \ell(f)] \rightarrow [a,b]$ is such that $\overline{f}=f \circ \phi$ is a reparametrization of the curve by arc length. That is $|\overline{f'}(s)|=1$, for all $s \in [0, \ell(f)]$.

I have no idea what to do here please someone who can help me

答: The intuition is: Imagine you're following some trajectory $f(t)$. We can split your acceleration vector $f''(t)$ into a component parallel to your direction of motion and a component perpendicular to it (i.e. write $f''(t)=a(t)f'(t)+b(t)$ where $a$ is real-valued and $b(t)\cdot f'(t)=0$). Now whenever $a(t)>0$, your speed is increasing, and whenever $a(t)<0$, your speed is decreasing. But if $f$ is arc-length-parametrized, your speed is constant, so $a(t)=0$ everywhere and $f''(t)=b(t)$ is just the perpendicular part (which describes how you're turning).


问:
What is wrong with my probability formula?

My thought process was this, if they attack they have a 25% chance of getting 3 points, 25% chance of getting 1 point and 50% chance of getting 0 points

Now on the other hand if they go defensive, theres a 15% chance of them getting 3 points, 50% chance of them getting 1 point and 35% chance of them getting 0 points.

This seems obvious to me that the obvious route would be defensive, yielding more points with a lower probability of actually loosing, which gets you 0 points. it simply asks for the higher number of expected points, so I would go defensive. If I am wrong, please correct me, thank you in advance!

答:
If you calculate the expected value of each strategy, you can see which one will, on average, win out over a long series of games.

The expected value of a random variable $X$ is defined as $E[X] = \sum_i {x_i*P[X=x_i]}$. Essentially, this is the weighted average over all possible outcomes, weighted by the probabilities.

In this case, the expected value of any strategy is:

$$ E[\text{Strategy}] = 3*P[\text{Win}]+1*P[\text{Draw}]+0*P[\text{Loss}]$$

If the strategy is "attacking," then we have

$$ E[\text{Attack}] = 3*0.25 + 1*0.25 + 0*0.50 = 1 . $$

Similarly, if the strategy is "defensive," then we have

$$ E[\text{Defense}] = 3*0.15 + 1*0.50 + 0*0.35 = 0.95 . $$

So, it looks like attacking is a slightly better strategy.

Of course, this is not meant to explain what happens in any one match, but what would happen on average if you were able to play the same match several times under the exact same conditions.

You could expect to win, on average, 1 point per match if you employed only the attacking strategy, and 0.95 points per match if you employed only the defensive strategy.


问:
Shortcut in calculating examples of elements of a given order?
My question is:

Find all possible orders of elements of the group of units $G_{31}$. Give an example of an elememt of each possible order.

I did the question, but I felt I did it a long way. As $31$ is prime, elements of $G_{31} = \{1, 2, \cdots 30\}$. Possible orders are divisiors of $\varphi(31) = 30 = 1,2,3,5,6,10,15,30$. So, to get an example of each element, I litterally went through all elements checking their orders. Like, I didn't have to go through ALL of them because I noticed after a while if I was going to get the order I wanted or if I wasn't, but it still took a while calculating all of this.

Is there a shortcut to finding all this? So, I think I'm asking, is there a way to solve

$$a^{10} \equiv 1 \mod 31,$$
$$a^{15} \equiv 1 \mod 31$$
$$\mathrm{etc}$$
without trial and error?

答: Since $31$ is prime, the ring $\mathbb Z/31\mathbb Z$ is a field and henc its multiplicative group $G_{31}$ is cyclic. Therefore, you can find an element $g$ of order $30$. Then the elements of order $d|30$ are precisely the $g^k$ with $\gcd(k,30)=\frac{30}d$.</p>


问:
$X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?
Let $(\Omega, \mathcal{F}, P)$ be a probability space,  $\mathcal{G}\subset \mathcal{F}$ be a sigma algebra, $X, Y : \Omega \to \mathbb{R}^n$ be random vectors, and $f$ be a measurable function on $\mathbb{R}^{2n}$.
Suppose that $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, and $Y$ is $\mathcal{G}$-measurable. Is it true that $f(X,Y)$ is independent of $\mathcal{G}$?

Motivation: This question arises from a randomized algorithm. In this algorithm, each iteration receives some randomness that is independent of the previous iterations. In my question, $\mathcal{G}$ indeed represents the randomness of the algorithm up to iteration $k$, $Y$ is a vector generated by iteration $k$, $X$ is some new randomness injected into the algorithm at iteration $k+1$, and $f(X,Y)$ is a quantity computed from $X$ and $Y$. It turns out that $f(X,Y)$ is statistically independent of $Y$. I would like to prove that $f(X,Y)$ is independent of the first $k$ iterations, which will be interesting and convenient.

Any comments or criticism will be appreciated. Thank you.

A related question: Uniform distribution on the unit sphere rotated by a random orthogonal matrix.

答:
Yes, $f(X,Y)$ is independent of $\mathcal G$. Let $\varphi$ be a bounded Borel function and $G$ a bounded $\mathcal G$-measurable function. Let $F=f(X,Y)$. We need to show that
$$
E[\varphi(F)\cdot G] =E[\varphi(F)]\cdot E[G].
$$
Because $Y$ is $\mathcal G$-measurable and $X$ is independent of $\mathcal G$, you have
$$
E[\varphi(F)\cdot G] = \int E[\varphi(f(x,Y))\cdot G]\mu(dx),
$$
where $\mu$ is the distribution of $X$.
But, for each fixed $x$,
$$
E[\varphi(f(x,Y))\cdot G]=E[\varphi(f(x,Y))\cdot E[G\mid Y]],
$$
and so
$$
\eqalign{
E[\varphi(F)\cdot G]
&=\int E[\varphi(f(x,Y))\cdot E[G\mid Y]]\mu(dx)\cr
&=  E\left[\varphi(f(X,Y))\cdot E[G\mid Y]\right] \cr
&=  E[\varphi(f(X,Y))]\cdot E[E[G\mid Y]] \cr
&=  E[\varphi(F)]\cdot E[E[G\mid Y]] \cr
&=  E[\varphi(F)]\cdot E[G]. \cr
}
$$
(The third in this string of equalities follows because $F$ is independent of $Y$ and therefore independent of $E[G\mid Y]$.)


问:
Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$

In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$

(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)

Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.

I was referred by a friend to the identities
$$\begin{align}
a+b+c &= 2s \tag{1} \\[4pt]
ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt]
abc &= 4Rrs \tag{3}
\end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.

Any other idea?

答:
The area of the triangle can be expressed as $ \frac12r(a+b+c)= \frac{abc}{4R}$. Then

\begin{align}
\frac rR =\frac12 \frac{\frac{abc}{R^3}}{\frac{a+b+c}R}
=\frac{2\sin A \sin B \sin C}{\sin A +\sin B +\sin C}\tag1
\end{align}
Note
$$\sin A +\sin B +\sin C
= 2\cos\frac A2\sin\frac A2 + 2\sin\frac {B+C}2\cos\frac{B-C}2\\
= 2 \cos\frac A2 (\cos\frac {B+C}2+\cos\frac{B-C}2) = 4 \cos\frac A2 \cos\frac B2 \cos\frac C2
$$

Substitute into (1)
$$\frac rR =  4 \sin\frac A2 \sin\frac B2 \sin\frac C2\tag2
$$
Similarly
$$\cos A +\cos B +\cos C
= 1-2\sin^2\frac A2 +2\cos\frac {B+C}2\cos\frac{B-C}2\\
=1-  2 \sin\frac A2 (\cos\frac {B+C}2-\cos\frac{B-C}2)=1+ 4 \sin\frac A2 \sin\frac B2 \sin\frac C2
$$
Substitute into (2)

$$\cos A+\cos B+\cos C=1+\frac rR$$


问:
Poisson Process - Joint Distribution and estimation of $p_i$
Two copy editors read a $300$-page manuscript. The first found $100$ typos, the second found $120$, and their lists contain $80$ errors in common. Suppose that the author's typos follow a Poisson process with some unknown rate $\lambda$ per page, while the two copy editors catch errors with unknown probabilities of success $p_1$ and $p_2$. Let $X_0$ be the number of typos that neither found. Let $X_1$ and $X_2$ be the number of typos found only by $1$ or only by $2$, and let $X_3$ be the number of typos found by both.

($a$) Find the joint distribution of ($X_0;X_1;X_2;X_3$), expressed in $\lambda$, $p_1, p_2$.

($b$) Use the answer to ($a$) to find an estimates of $p_1, p_2$ and then the number of undiscovered typos. (Hint: Let $N(s)$ be the Poisson process, with Ti being the "time" when the i-th typo occurs, and the "time" variable is the page number.)

Here's the thing...

Well, I've got that ($a$) is, given that each error is found by neither editor, $1$ only, $2$ only, and both editors with probabilities $(1-p_1)(1-p_2),p_1(1-p_2),(1-p_1)p_2$ and $p_1p_2$, respectively, and assuming that $x_0+x_1+x_2+x_3=x$,

$\Pr[(X_0,X_1,X_2,X_3)=(x_0,x_1,x_2,x_3)]=\Pr[N=x]f(x_0,x_1,x_2,x_3)$

where $\Pr[N=x]=(300\lambda)^x\exp(-300\lambda)/x!$

and

$f(x_0,x_1,x_2,x_3) ={x\choose x_0,x_1,x_2,x_3}p_1^{x_1+x_3}p_2^{x_2+x_3}(1-p_1)^{x_0+x_2}(1-p_2)^{x_0+x_1}$

is the mass function of the multinomial distribution.

But, i'm not sure how to obtain the values for (b). Can someone help me?
Searching, i've got the following answer:

For (b), we are given that $X_1=20,X_2=40,X_3=80$. It is reasonable to guess that $p_1p_2=2(1-p_1)p_2=4p_1(1-p_2)$, which gives $p_2=4/5$, $p_1=2/3$. Then, $(1−p_1)(1−p_2)=1/15$, so a reasonable estimate for $X_0$ is $80/8=10$.

But why?

答: To start, $p_1p_2 = 80/\lambda$ and $(1-p_1)p_2 = 40/\lambda.$ Thus, $p_1p_2 = 2(1-p_1)p_2,$ Etc. Then later, $p_1p_2 = 8/15 = 80/\lambda$ implies $\lambda = 150.$ Etc. On average, there must be a typo on every two pages. (Sounds like my Answers on this site.)


问:
Equation $f(x,y) f(y,z) = f(x,z)$
How to solve the functional equation  $f(x,y) f(y,z) = f(x,z)$?</p>

答:
Set $g(x)=f(x,0)$ and $h(z)=f(0,z)$. Then, we have $f(x,z)=f(x,0)f(0,z)=g(x)h(z)$ for all $x$ and $z$. Apply this to the original equation to obtain $g(x)h(y)g(y)h(z)=g(x)h(z)$.

There are three possibilities now:

Either $g(x)=0$ for all $x$ (and thus $f(x,z)=0$ for all $x$ and $z$).

Or $h(z)=0$ for all $z$ (and thus $f(x,z)=0$ for all $x$ and $z$ again).

Or $g(x)\not=0$ and $h(z)\not=0$ for some $x$ and $z$. Then, we can cancel those two out to obtain $h(y)g(y)=1$ for all $y$. Thus, $h(y)=\frac{1}{g(y)}$ and we get $f(x,z)=\frac{g(x)}{g(y)}$. This can be easily checked to be a solution for any suitable $g(x)$.

It's easy to check there are two solutions: $f(x,y)=0$ for all $x,y$ and $f(x,y)=\frac{g(x)}{g(y)}$ with $g(x)$ being any function which is never equal to zero.

Title: Equation $f(x,y) f(y,z) = f(x,z)$


问:
Find absolute maximum and minimum
Find the absolute maximum and absolute minimum of the function $f(x,y)=xy-5y-25x+125$ on the region above $y=x^2$ and on or below $y=27$.

My solution:

Critical point: $(5, 25)$

$f$ has a saddle point at $(5, 25)$ so it can't be used to find absolute min/max (correct me if i'm wrong).

Boundary points: $(-5, 0), (5, 0), (0,27)$

$f$ has absolute max of $250$ at $(-5, 0)$

$f$ has absolute min of $-10$ at $(0, 27)$

These are my answers, but the online system (homework portal) doesn't accept them so they may be wrong. Any ideas?

答:
It is true that $ \ (5,25) \ $ constitutes a saddle point for the function, and so would not be an extremum over $ \ \mathbb{R} \ $ , but its role changes in the specified parabolic region.  Having ascertained that there is no critical point in the interior, we next need to check the boundary.

Along the line $ \ y \ = \ 27 \ $ , the function becomes

$$ f(x,y) \ = \ xy \ - \ 5y \ - \ 25x \ + \ 125 \ \longrightarrow \ f(x) \ = \ 27x \ - \ 135 \ - \ 25x \ + \ 125 \ = \ 2x \  - \ 10 \ \ . $$

Since this linear function has no relative extrema, we will evaluate it for now at the endpoints of this portion of the region boundary (noted by Rory Daulton)

$$ f(-3 \sqrt{3}) \  = \ 2(-3 \sqrt{3}) \  - \ 10 \  = \ -6 \sqrt{3} \  - \ 10 \ \approx \ -20.39 \ \ ,  $$
$$ f(+3 \sqrt{3}) \    = \ 6 \sqrt{3} \  - \ 10 \ \approx \ +0.39 \ \ .  $$

It isn't entirely clear whether the original problem intended to include the parabolic lower boundary, but we will look at our function on $ \ y \ = \ x^2 \ $ as well.  (I suspect it should be included, though: otherwise, the endpoints we  just worked with would not be part of the boundary, meaning there would be no absolute extrema for the region, as we shall see.)

On the parabola, the function is

$$  \longrightarrow \ f(x) \ = \ x · x^2 \ - \ 5 · x^2 \ - \ 25x \ + \ 125 \ = \ x^3 \ - \ 5 x^2 \ - \ 25x \ + \ 125 $$ $$ = \ (x - 5)^2 \ · \ ( x+5 ) \ \ . $$

The critical points on this portion of the boundary are found from

$$   f'(x) \  = \ 3x^2 \ - \ 10 x \ - \ 25\ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{10 \ \pm \ \sqrt{400}}{6} \ = \ -\frac{5}{3} \ \ , \ \ 5 , $$

producing the function values

$$   f(5) \ =  \ (5 - 5)^2 \ · \ ( 5+5 ) \ = \ 0 \ \ , \ \ f(-\frac{5}{3}) \ =  \ (-\frac{20}{3})^2 \ · \ \frac{10}{3} \ = \ \frac{4000}{27} \ \approx \ +148.1 \ \ . $$

So the saddle point at $ \ (5,25) \ $ found in your original analysis is a relative minimum on the parabolic section of the boundary.  However, the absolute extrema for the closed parabolic region are

maximum:   $ \ f(x,y) \ \ = \ \ \frac{4000}{27} \ $ at $ \ \left( \ -\frac{5}{3} \ , \ \frac{25}{9} \ \right) $

minimum:   $ \ f(x,y) \ \ = \ \ -6 \sqrt{3} \  - \ 10 \ $ at $ \ \left( \ -3 \sqrt{3} \ , \ 27 \ \right)  $

Title: Find absolute maximum and minimum


问:
Proof-check: Any subset of $1,...,2n$ with order $\geq n+1 $ contains some element which divides another, assuming $n>1$.
I just want to confirm my proof is error free, since it seems quite different from others I've come across.

Consider $\{\sigma,...,2n\}$ for $\sigma \in \{1,...,n\}$. Obviously we have:

$n \mid 2n$ , $n-1 \mid 2n-2$ , . . . , $\sigma \mid 2\sigma$

For all $x,y\in \{\sigma,...,n\}$ we know $2x \ne 2y$ unless $x=y$. Thus if $2x \notin \{\sigma,...,n\}$ then at least one novel member of every such pair $\{x,2x\}$ must be deleted from $\{\sigma,...,2n\}$ to prevent having an entry and its double in $\{\sigma,...,2n\}$. If instead $2x\in \{\sigma,...,n\}$ then we must delete at least two members (at least one novel) of every such triple $\{x,2x,4x\}$ from $\{\sigma,...,2n\}$ to prevent having an entry and its double and / or quadruple in $\{\sigma,...,2n\}$.

This means we must remove at least one novel element $\textit{for}$ each member of $\{\sigma,...,n\}$, which amounts to removing at least $n-\sigma+1$ elements from $\{\sigma,...,2n\}$ thus leaving us with less than $n+1$ elements if we are to have no elements which divide each other.

答:
This proof follows a generally bad pattern that can be described as "Let me think of the worst thing that could happen, and then show that it doesn't happen." Such proofs have two perils:

You may think of a bad thing, but fail to think of the worst, and because the proof doesn't contain a proof that your situation is the worst, no one (especially you) ever notices it.

The second-worst thing might also provide a counterexample if you're not careful.

If you want to show that for every subset $S$ of size $n_+1$, some property holds, your proof should almost certainly start with "Let $S = \{x_0, \ldots, x_n\}$ be a subset, where the $x_i$ are all distinct, so that $S$ has $n+1$ elements. We now show that ..."

Saying "Let $S$ be the subset containing the top $n+1$ elements" makes the assumption that this is the "worst case", and you're headed down a very bad path.


问:
Simplifying this equation to get sum of squares
$$\frac{1}{n\sum x_{i}^2-{(\sum x_{i}})^2}$$

I have this equation above. I am trying to simplify it such that I can get: $\dfrac{1}{nSS_{x}}$

Where $SS_{x}$ is the sum of squares of $x$. Any pointers on how to achieve this?

答:
It's easier to work backwards, and only compare the denominators of your fractions. The sum of squared deviations of $x$ is

$$ SS_x = \sum_i {(x_i-\bar{x})^2} $$

$$ \Rightarrow nSS_x = n\sum_i {(x_i-\bar{x})^2} $$

$$ = n\sum_i \left({x_i^2 - 2x_i\bar{x} + \bar{x}^2}\right) $$

$$ = n\sum_i {x_i^2} - n\sum_i {2x_i\bar{x}} + n\sum_i {\bar{x}^2} . $$

Since $\bar{x}$ and $\bar{x}^2$ are constants, we can factor them out of the sums:

$$ = n\sum_i {x_i^2} - 2n\bar{x}\sum_i {x_i} + n^2\bar{x}^2 , $$

Note, the extra factor of $n$ in the third term comes from the fact that $\sum_i {1} = n$. In the second term and third terms, we can substitute $n\bar{x} = \sum_i x_i$:

$$ = n\sum_i {x_i^2} - 2 \left( \sum_i x_i \right)^2 + \left( \sum_i x_i \right)^2 $$

$$ = n\sum_i {x_i^2} - \left( \sum_i x_i \right)^2 . $$


问:
Understanding selecting without replacement when a previous selection does not matter
This is likely a question with answers everywhere, but I am unsure how to phrase it so I find answers.

Lets say you have $100$ hammers where $23$ are defective.

Example 1: If you select two, one at a time, without replacement, what is the probability the second one is defective?

A1: The way I would naturally answer it is:

$\frac{23}{100}\times\frac{22}{99}+\frac{77}{100}\times\frac{23}{99} = \frac{23}{100}$.

This is the same probability that the first hammer chosen being defective.

Example 2: If you select $3$, one at a time, without replacement, what is the probability the first is defective but the last is not?

A1: Once again the way I would naturally answer it is:

$\frac{23}{100}\times\frac{22}{99}\times\frac{77}{98}+\frac{23}{100}\times\frac{77}{99}\times\frac{76}{98} = \frac{161}{900} = \frac{23}{100}\times\frac{77}{99}$.

This is the same probability that the first hammer chosen is defective, and the second not.

What I am getting at is the probabilites were the same when ignoring the "does not matter" choice. Assuming I am not mistaking something, how does this make sense? I can't seem to wrap my head around it intuitively or mathmatically. I would expect that because it is without replacement, even a "does not matter" changes the probability by reducing the total hammer count.

答:
answered in comments by David K

A probability interview questions on pen type

related to not having information on the unknown makes no changes to the probability


问:
A question about ideals, maximal ideals, and UFD's.
I'm working on the following problem in preparation for an exam.

Let $f(x) = x^3 + 2 \in \mathbb{Z}[x]$. Let $\alpha$ be any complex number satisfying $f(x)$. Let $K = \mathbb{Q}(\alpha)$ and suppose $R$ is the collection of element in $K$ with integer coefficients (with basis $1, \alpha$, and $\alpha^2$).

Next we are asked to show the following two equalities:

For every non-zero ideal $I \subseteq \mathbb{Z}$, show that $IR \cap \mathbb{Z} = I$.

Show that if $J \subseteq R$ is a nonzero ideal , then $J \cap \mathbb{Z} \neq \emptyset$.

Show that every nonzero prime ideal in $R$ is maximal.

Here is my work so far: It is clear that $IR$ contains $I$ and $\mathbb{Z}$ contains $I$. Thus $IR \cap \mathbb{Z} \supseteq I$. Now suppose $x \in IR \cap \mathbb{Z}$. Then $x = ir$ for some $i \in I$ and $r \in R$. Also, $ir \in \mathbb{Z}$. Now I am not sure how to proceed. I want to show that $x \in I$.

I know that $\mathbb{Z}$ is a UFD, and I believe that saying $x$ has a unique factorization in $\mathbb{Z}$ should help. In fact, since $x \in \mathbb{Z}$ and $I \subseteq \mathbb{Z}$ then $ir \in \mathbb{Z}$ says that $r$ is also in $\mathbb{Z}$. But then $x$ is a multiple of an element in $I$, and so $x \in I$. (is this correct?)

Unless I am mistaken, $0 \in J$ for every nonempty $J\subseteq R$.

I'm not sure where to start on part 3. I know what prime ideals are, and what maximal ideals are.

If someone has a better suggestion for the title, please let me know. I wasn't sure how to paraphrase this question.

答:
For what it's worth, I have an alternate approach.

1.) Considering that $\mathbb Z$ is a PID and $I$ is a nonzero ideal of $\mathbb Z,$ there exists a nonzero integer $n$ such that $I = n \mathbb Z.$ Consequently, every element of $IR \cap \mathbb Z$ is an integer $r$ of the form $r = nk(a + b \alpha + c \alpha^2),$ where $a, b, c,$ and $k$ are also integers. By viewing $R$ as a $\mathbb Q$-vector space with a $\mathbb Q$-basis of $\{1, \alpha, \alpha^2\},$ it follows by the uniqueness of a vector in $R$ that $b = c = 0,$ and we conclude that $r = n(ka)$ is in $I = n \mathbb Z.$ QED.

2.) Clearly, the intersection of a nonzero ideal $J$ of $R$ and $\mathbb Z$ is nonempty: by definition, $J$ contains $0,$ hence we have that $0 \mathbb Z \subseteq J \cap \mathbb Z.$ Unfortunately, there was a typo in the question; the actual problem statement asks us to prove that every nonzero ideal $J$ of $R$ satisfies $J \cap \mathbb Z \neq 0 \mathbb Z.$ Consider a nonzero ideal $J$ of $R.$ Observe that $R \cong \mathbb Z[x] / (x^3 + 2),$ hence $R$ is Noetherian. Consequently, the ideal $J$ of $R$ has a primary decomposition $J = Q_1 \cap \cdots \cap Q_n$ by nonzero primary ideals $Q_i$ such that $\sqrt{Q_i} = P_i$ is a nonzero prime ideal. Considering that radicals distribute over intersections, we have that $\sqrt J = P_1 \cap \cdots \cap P_n.$ By definition of the radical, we have that $(\sqrt{J})^k \subseteq J$ for some integer $k \gg 0,$ hence we have that $$J \supseteq (P_1 \cap \cdots \cap P_n)^k \supseteq (P_1 \cdots P_n)^k = P_1^k \cdots P_n^k.$$ Considering that $R$ is an integral extension of $\mathbb Z$ (because $1, \alpha,$ and $\alpha^2$ are integral over $\mathbb Z$), it follows that $\dim R = \dim \mathbb Z = 1,$ hence every nonzero prime ideal of $R$ is maximal. Contracting any nonzero prime ideal $P_i$ of $R$ to $\mathbb Z$ gives a nonzero maximal ideal $P_i \cap \mathbb Z = p_i \mathbb Z$ for some prime $p_i.$ Ultimately, we find that $J \cap \mathbb Z \supseteq P_1^k \cdots P_n^k \cap \mathbb Z \supseteq (P_1 \cap \mathbb Z)^k \cdots (P_n \cap \mathbb Z)^k = p_1^k \cdots p_n^k \mathbb Z.$ QED.

3.) Once we know that every nonzero ideal $J$ of $R$ satisfies $J \cap \mathbb Z \supseteq 0 \mathbb Z,$ it follows that $J \cap \mathbb Z = n \mathbb Z$ for some nonzero integer $n.$ Given that $J$ is a prime ideal of $R,$ we must have that one of the prime factors of $n$ is an element of $J.$ Call that prime factor $p.$  Observe that the action $(k + p \mathbb Z) \cdot (r + J) = kr + J$ is well-defined: indeed, if we have that $k + p \mathbb Z = \ell + p \mathbb Z,$ then $k - \ell$ is in $p \mathbb Z$ so that $k - \ell$ is in $J$ and $kr + J = \ell r + J.$ Consequently, we may view $R / J$ as a $\mathbb Z / p \mathbb Z$-module. But as $\mathbb Z / p \mathbb Z$ is a field, the integral domain $R / J$ is a finite-dimensional $\mathbb Z / p \mathbb Z$-vector space and hence must be a field. We conclude that $J$ is maximal. QED.


问:
Listing subgroups of a group
I made a program to list all the subgroups of any group and I came up with satisfactory result for $\operatorname{Symmetric Group}[3]$ as

$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 2 \\
\end{array}
\right)\right]\right\}\right\}$

It excludes the whole set itself though it can be added seperately. But in case of $SymmetricGroup[4]$ I am getting following

$\left\{\{\text{Cycles}[\{\}]\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 2 \\
 3 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 3 \\
 2 & 4 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{cc}
 1 & 4 \\
 2 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 3 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 2 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 4 & 2 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 1 & 4 & 3 \\
\end{array}
\right)\right]\right\},\left\{\text{Cycles}[\{\}],\text{Cycles}\left[\left(
\begin{array}{ccc}
 2 & 3 & 4 \\
\end{array}
\right)\right],\text{Cycles}\left[\left(
\begin{array}{ccc}
 2 & 4 & 3 \\
\end{array}
\right)\right]\right\}\right\}$

The matrix form shows double transposition. Can someone please check for me if I am getting appropriate results? I doubt I am!!

答:
By using GAP, you can make them all so easier and visible.</p>

```<code>  gap> LoadPackage("sonata");;
  gap> s4:=SymmetricGroup(IsPermGroup,4);;
  gap> g:=Subgroups(s4);;
  gap> List([1..Size(g)],k->g[k]);
</code>```

Title: Listing subgroups of a group

Body:
By using GAP, you can make them all so easier and visible.

  gap> LoadPackage("sonata");;
  gap> s4:=SymmetricGroup(IsPermGroup,4);;
  gap> g:=Subgroups(s4);;
  gap> List([1..Size(g)],k->g[k]);


问:
(Wrong site) Can't find mistake in resolution of second order non-homogenous linear differential equation
I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$

I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$

We have a non-distinct real root of multiplicity 2 which is -5.

The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$

We now find the right candidate for the undermined coefficients method.

$F(x) = (3x+2)e^{-5x}$

$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.

So our particular solution is $y_{p} = (Ax + B)e^{-5x}$

My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...

Thank you.

答:
You could try finding the solution of this ODE with Mathematica:</p>
```<code>eqn = y''[x] + 10 y'[x] + 25 y[x] == (3 x + 2) Exp[-5 x]
y[x] /. First@DSolve[eqn, y, x]
% // Factor
(* 1/2 \[ExponentialE]^(-5 x) (2 x^2+x^3+2 Subscript[\[ConstantC], \
1]+2 x Subscript[\[ConstantC], 2]) *)
</code>```

Body:
You could try finding the solution of this ODE with Mathematica:

eqn = y''[x] + 10 y'[x] + 25 y[x] == (3 x + 2) Exp[-5 x]
y[x] /. First@DSolve[eqn, y, x]
% // Factor
(* 1/2 \[ExponentialE]^(-5 x) (2 x^2+x^3+2 Subscript[\[ConstantC], \
1]+2 x Subscript[\[ConstantC], 2]) *)

OwnerDisplayName: Ferca


问:
number of matrices with a single nullspace point
Suppose I fix a vector $v\in \mathbb{R}^n$. I want to count the number of maximal rank (in this case $n-1$) linearly independent matrices $A_1,\ldots,A_m$ for which  $v$ lies in the nullspace of $all$ the $A_i$s, i.e., $A_iv=0$. What is the largest $m$ for which these set of matrices can exist?

答:
If $Av=0, Bv=0$ then clearly $(A+B)v = Av+Bv=0$ and for any
constant $k$,   $(kA)v = 0$. This shows that the number of such matrices is infinite. Better question would be: How many Linearly independent matrices $A$ exist such that $Av=0$ for a fixed vector $v$?


问:
the area principle of biholomorphic mapping
If $f=\frac 1z +\sum_{n=0}^\infty a_nz^n $ is a biholomorphic mapping on $B(0,1)\setminus\{0\}$,then prove that:
$$\sum_{n=1}^\infty n|a_n|^2\le1$$
I have known that when $D=B(0,R)\setminus\overline{B(0,r)},f:D\to G$ with f biholomorphic,the area of G is:
$$\pi \sum_{n=-\infty}^\infty n|a_n|^2(R^{2n}-r^{2n})$$
and I thought it would be easy if only I should exchange r with 0,but that turned out to be not so easy.T_T

答:
Considering the area of the image of $f$ is not helpful, because the area is infinite. The right idea is to look at the area of the complement of the image.

For $0<r<1$, the area bounded by the Jordan curve $f(|z|=r)$ is
$$-\frac{1}{2i} \int_{|z|=r} \overline{f(z)}f'(z)\,dz = \pi \sum_{n=-1}^\infty n|a_n|^2 r^{2n-1} $$
where I include $a_{-1}=1$ in the series.
Since this quantity expresses the area of  something, it is nonnegative. Let $r\to1$ and draw the desired conclusion.


问:
Direct product of simple groups case $G_1 \cong G_2$
I have the following exercise:

Find all normal non-trivial subgroups of $G_1 \times G_2$.  I managed to see that using projections, the normal subgroups are isomorphic to $G_1$ or $G_2$. But the professor say that we need to separate the case when $G_1\cong G_2$. But I don't understand why. Can you help me with this?

答:
I managed to see that using projections, the normal subgroups are isomorphic to $G_1$ or $G_2$.

This is true if we forget about the trivial subgroup and the entire group $G_1 \times G_2$. However, you're asked to actually find all of the normal subgroups! They are not always just $G_1 \times 1$ and $1 \times G_2$. For example, $\mathbb{Z}/2$ is simple, and $(\mathbb{Z}/2) \times (\mathbb{Z}/2)$ has three distinct normal subgroups isomorphic to $\mathbb{Z}/2$.


问:
How to get a reflection vector?
I'm doing a raytracing exercise. I have a vector representing the normal of a surface at an intersection point, and a vector of the ray to the surface. How can I determine what the reflection will be?

In the below image, I have d and n. How can I get r?

Thanks.

答:
In case you want to rotate about Y axis you can use the following instead. This is mostly useful for computer graphics applications. Note that $d$ is assumed to be pointing outward in the equation below (i.e. ignore the direction of $d$ in the picture below) and $n$ needs to be normalized:

$$r = 2 (d \cdot n) n - d$$

ContentLicense: CC BY-SA 2.5

Title: How to get a reflection vector?