问:
On the converse of the $n$th term test
A student asked a very insightful question in my Calculus class this morning.  I did not know how to answer him.  (Admittedly, I am not an analyst by trade:  once I passed my qualifiers I never looked back.)  I would like to know if anyone here can give a precise answer, and if that can be massaged into an answer understandable by someone in Calculus II.

The topic of the lecture was the $n$th term test (or "divergence test") for infinite series.  I presented it as:

Theorem: If $\sum_n a_n$ converges then $\lim_{n \to \infty} a_n=0$.

I proved this, then had them state the contrapositive:

Divergence Test:  If $\lim_{n \to \infty} a_n \neq 0$ then $\sum_n a_n$ diverges.

I then gave a litany of examples. To their credit, they never fell into the false-converse trap.  I have been harping on dogs/mammals/horses all semester so they are very good about avoiding that (if you are a dog then you are a mammal but the converse is false if you can find a horse).

My second example was the harmonic series, which of course has $\lim_{n \to \infty} a_n=0$ yet fails to converge.  Hence we definitely have horses in my theorem above (and they all spotted this).

Enter the sharp student.  He asked if there was an improvement on my first theorem so that the converse becomes true.  I had said earlier in my example that although $\lim_{n \to \infty} \frac{1}{n}=0$, it doesn't converge "fast enough" to $0$ to make the harmonic series converge.  The student asked for a measure of "fast enough" or at least a precise statement of this.  What he is fishing for is something like:

Improved Theorem:  If $\sum_n a_n$ converges then $\lim_{n \to \infty} a_n=0$ and (extra-nice condition on the speed of the convergence to $0$).

My instinct is there is no answer in terms of the terms $a_n$.  The only answer is that the terms must vanish quickly enough to make the sequence of partial sums convergent (which is an unsatisfying answer to his question).  If he only cared about $p$-series then I can be precise ($p > 1$), but he is asking about generic series whose terms vanish in the limit.

I hope my question is clear.  Am I correct that all this is much too subtle to have a nice, clean answer on the "rate" of convergence of the terms to $0$?

答: Note that if $a_n = \frac{1}{n^{1 + \epsilon}}$ where $\epsilon > 0$, then the series converges by the integral test (or p-series). Then you can argue via the limit comparison test, e.g. if the terms approach zero in a similar fashion (as $\frac{1}{n^{1 + \epsilon}}$, then the series converges as well.

Title: On the converse of the $n$th term test


问:
$\mu (A \bigtriangleup B ) = \mu(B) - \mu(A) + 2\mu(A \setminus B)$
How can I prove it?:
$\mu (A \bigtriangleup B ) = \mu(B) - \mu(A) + 2\mu(A \setminus B)$

I started with: $\mu (A \bigtriangleup B )=\mu((A \cup B)\setminus B) + \mu ((A \cup B) \setminus A) = 2\mu (A \cup B) - \mu(B) - \mu(A)=... $

Does anyone have an idea how to finish it?

答:
It is enough to prove that $\mu (A \bigtriangleup B )+\mu(A)  = \mu(B) + 2\mu(A \setminus B)$.

But we have that

$\mu (A \bigtriangleup B )+\mu(A)=\mu(A\setminus B)+\mu(B\setminus A)+\mu(A)=\mu(A\setminus B)+\mu(A\cup B)$,

and

$\mu(B) + 2\mu(A \setminus B)=\mu(A\cup B)+\mu(A \setminus B)$,

so they are the same.

ContentLicense: CC BY-SA 4.0


问:
prove $\text{abs}(a + b) + \text{abs}(a - b) = \text{max}(a, b) \cdot 2$
after playing around with absolute value for a little, i noticed that $|a + b| + |a - b|$ always appeared to be the larger of $a$ and $b$ multiplied by $2$, whenever both $a$ and $b$ were positive.

$$
|a + b| + |a - b| =
\begin{cases}
2a, & \text{if $a \ge b$} \\
2b, & \text{if $a < b$}
\end{cases}
$$

can this be proven to always be the case, and if so, how do i prove it?

答:
If $a\ge b$  then $|a-b|=a-b$ and $\max\{a,b\}=a$, and
$$ a+b+|a-b|=a+b+a-b=2a=2\max\{a,b\}.$$

If $a\le b$  then $|a-b|=-(a-b)=b-a$  and $\max\{a,b\}=b$, and
$$ a+b+|a-b|=a+b+b-a=2b=2\max\{a,b\}.$$

As this covers all cases for real $a,b$, we have
$$ a+b+|a-b|=2\max\{a,b\}$$
for all $a,b\in\Bbb R$.


问:
Is it coincidental that matrices are useful for solving systems of equations or is there a deeper (geometric) connection?
Let's say we have this simple system of equations:

$$-x + y = 1$$
$$ x + y = 3$$

They intersect at the point $(1,2)$ and this is the graph:

One way of going about and solving this system of equations is to convert it into matrix form:

$$\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix} * \begin{bmatrix}
x\\
y
\end{bmatrix} = \begin{bmatrix}
1\\
3
\end{bmatrix}$$

When my brain sees this matrix form, it has the following geometric meaning: If we apply the linear transformation encoded by $\begin{bmatrix}
-1 & 1\\
1 & 1
\end{bmatrix}$ to vector $\begin{bmatrix}
x\\
y
\end{bmatrix}$, then the output vector is $\begin{bmatrix}
1\\
3
\end{bmatrix}$.
We can then use Gaussian Elimination to determine $\begin{bmatrix}
x\\
y
\end{bmatrix}$.

My question: Is the matrix form of the system of equations simply a useful trick to algorithmically solve these systems of equations use Elementary Row Operations, etc. Or is there a deeper geometric connection between the intersection of two lines and the linear transformation of the vector that describes said intersection.

答: You are right: matrices can be thought of as representing linear transformations. An $(m \times n)$-matrix takes as inputs vectors in an $n$-dimensional vector space and outputs vectors in an $m$-dimensional vector space. The crucial observation here is that in order to completely determine a linear transformation $T: V \to V'$, it suffices to know what it does on a certain chosen basis $\{e_1, \dots, e_n\}$ of $V$, since for any other $v \in V$, writing $v = \sum_{i=1}^n v_i e_i$ means that we have to get $T(v) = \sum_{i=1}^n v_i T(e_i)$. Still the $T(e_i)$ are vectors, but we would like to express these using numbers as well. So we chooose another basis $\{e_1', \dots, e_m'\}$ of $V'$, so that we can write any vector $v' \in V'$ as a linear combination of the $e_i'$. In particular, there are certain numbers $T_{ij}$ such that we have $T(e_i) = \sum_{j = 1}^m T_ij e_j'$. We conveniently put all these $n \cdot m$ numbers in a rectangular shape because it makes calculations easier, and tadá: we get our usual matrix notation.


问:
Mistake proving that $3 = 0$
I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:

Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.

Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$

From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$

Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.

If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$

I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:

 He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?

答:
Consider the equation $x=x^2$. Its solutions are easy to find: they are $0$ and $1$, right?!

However, if $x$ is a solution of that equation, then $x=x^2$, and therefore $x=(x^2)^2=x^4$. But the equation $x=x^4$ has other solutions, besides $0$ and $1$, namely $-\frac12+\frac{\sqrt3}2i$.

The problem lies in the if-then clause of the first sentence of the previous paragraph: we are saying (correctly) that if $x$ is a solution of the given equation, then $x$ is also a solution of another equation. But there is no reason for you to think that the new equation only has the solutions of the original one!

That's what happens in your example: if $x$ is a solution of $x^2+x+1=0$, then $x$ is a solution of $x^2=\frac1x$, but you have no reason to suppose that every solution of $x^2=\frac1x$ is also a solution of $x^2+x+1=0$.

Title: Mistake proving that $3 = 0$


答:
Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.

From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.

To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.

Thus, the equation $(3)$ will be true when either $f(x)=0$, which are the solutions of the original problem, or when $\frac{1}{x}-1=0$, that is, $x=1$. That´s what he means when he says he is adding the solution $x=1$.

Title: Mistake proving that $3 = 0$

Body:
I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:

Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.

Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$

From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$

Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.

If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$

I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:

 He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?


答: The reasonning only shows that if $x$ satisfies $x^2 + x+1=0$ then it satisfies $x^3=1$. The fact is that you cannot "go back" to prove the reverse, since at step 3 you reuse the hypothesis that $x^2+x+1=0$.

Title: Mistake proving that $3 = 0$


问:
How do I convert from a semi-annual interest rate to montly?
On a practice exam I am given a $.06$ compounded semi-annually and need it converted to monthly. I assumed it would be $.005$ because it has annual percentage of $.06/12$ but my answer is slightly off so I wanted to have this verified

答:
The equation for semi-annually is $A=P(1+\frac{r}{2})^{2t} $ while for monthly it is $A=P(1+\frac{R}{12})^{12t}$ . Making these equations equal with r = 0.06 yields
$$P(1+\frac{0.06}{2})^{2t}=P(1+\frac{R}{12})^{12t}$$
If we solve the above equation we get R = 0.05926…


答:
If we have interest rate $i$ compounded in $n$ periods per year; the effective per annum interest is given by

$$(1+\frac{i}{n})^n$$

If we want to calculate some $i'$ associated with an $n'$ for the same per annum rate we must set the two to be equal:

$$(1+\frac{i}{n})^n=(1+\frac{i'}{n'})^{n'}$$

Given that in your case you know $n'$ that leaves 1 equation, 1 variable; easy to solve.

Title: How do I convert from a semi-annual interest rate to montly?

Body: On a practice exam I am given a $.06$ compounded semi-annually and need it converted to monthly. I assumed it would be $.005$ because it has annual percentage of $.06/12$ but my answer is slightly off so I wanted to have this verified


答:
You made a mistake When you said that if $$x^3=1$$

then $$x=1$$

In fact,

$$x^3=1\iff x^3-1=0\iff (x-1)(x^2+x+1)=0$$

$$\implies x=1 \text{ xor } x^2+x+1=0$$

Title: Mistake proving that $3 = 0$


问:
Is it possible that individually, pieces of evidence increase the chance of guilt but together they decrease the chance of guilt?
Let G be the event that a certain individual is guilty of a certain robbery. In gathering evidence, it is learned that an event $E_1$ occurred, and a little later it is also learned that another event $E_2$ also occurred.
Is it possible that individually, these pieces of evidence increase the chance of guilt (so $P(G|E1)$ $> P(G)$ and $P(G|E2) > P(G)$), but together they decrease the chance of guilt (so $P(G|E1, E2)$ $ < P(G)$)?

This is the example in the solution:

Suppose that the crime was
committed between 1 pm and 3 pm on a certain day. Let E1 be the event
that the suspect was at a nearby coffeeshop from 1 pm to 2 pm that
day, and let E2 be the event that the suspect was at the nearby
coffeeshop from 2 pm to 3 pm that day. Then P(G|E1) > P(G), P(G|E2) >
P(G) (assuming that being in the vicinity helps show that the suspect had the opportunity to commit the crime), yet
P(G|E1, E2) < P(G) (as being in the coffeeshop from 1 pm to 3 pm gives
the suspect an alibi for the full time).

I've been trying to think of another example but I can't think of any. Can someone provide another example? Moreover, in terms of general problem-solving heuristics, I find myself terrible at coming up with examples/counter-examples. My only strategy is to look for simple/extreme cases, but there's no simple/extreme cases for this question.

Thank you very much!

答: Here is an example based on "consumable" items. Suppose that a suspect has only enough money/resources to buy one expensive lockpick, and it can be shown by looking at his bank account that he only made one purchase of such high-level lockpick. Furthermore, this lock pick has come out so recently that you cannot find it anywhere- no one drops it, and no one lends it to others for free. This lock pick, being designed to pick the most difficult-to-overcome locks, disintegrates upon use. Suppose that it was found that a lock was picked at the scene of the crime that required a lock-pick of that caliber to get through. Later, it was found that another lock in the same building was picked too. Individually, the pieces of evidence show that the man could certainly have been the thief because he purchased a lock pick that could get through those locks. But together, he could not have been the one because he did not have two lock picks.</p>


问:
Probability of a certain roll with six dice, each with a single marked side
There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".

I calculate the probability of rolling three sets of two with 6 dice:

$$\frac{{6 \choose 3} {6 \choose 2} {4 \choose 2} {2 \choose 2}}{6^6} = \frac{1800}{6^6}$$

As there are ${6 \choose 3}$ ways to assign the numbers, ${6 \choose 2}$ ways to arrange the 1st pair of dice into the available positions, then ${4 \choose 2}$ for the 2nd pair and then ${2 \choose 2}$ for the 3rd pair.

An example of this kind of roll would be "112233" or "242554".

How can I determine how many of these rolls have 0 gold sides and how many have 1 gold side?

Finding how many have 2 gold sides is:

$${3 \choose 2} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 360$$

As there are ${3 \choose 2}$ ways to assign the gold sides, $\frac{2}{6}$ chance the dice with the gold A is an A, then $\frac{2}{5}$ chance the dice with the gold B is a B, and then $\frac{2}{4}$ chance the dice with the gold C is not a C (in the context "AABBCC", where A, B, and C are different numbers).

And similarly 3 gold sides:

$${3 \choose 3} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 120$$

But for 0-1 gold sides it's more complicated as the chance for the dice with the gold B depends on the position of the dice with the gold A, and so on for the gold C.

答:
Actually I didn't get how you calculated the number of possibilities for $2,3$ gold sides (though I have the same answers) but anyways this is my method (we'll solve for $2,3$ gold sides and then go for $1,0$ gold sides):

First of all, let's partition our $6$ dices into $3$ pairs of dices. Clearly there are $\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$ ways to do that. Throughout this solution, the number of the two dices in each pair shall be equal and the number of each dice means the number of its golden side. Also we'll name pairs with a dice with gold side as a golden pair. (Trivially no golden pair has more than one dice with gold side)

Now for the case with $3$ gold sides, exactly one dice in each pair shall be golden. Hence, there are $2^3=8$ ways to choose the number of the dices of each pair. So there are $8*15=120$ ways for that.

For the case with $2$ gold sides, we'll first choose the pairs that shall have a golden side. There are $\binom{3}{2}=3$ choices for this. Then there are two choices for each of these pairs to choose which one of the dices of each of these pairs shall have golden side. So there are $3*2^2=12$ choices for this and finally for the third pair, there are $2$ possibilities for the number of the dices of this pair. (Excluding two numbers because it shouldn't be a golden pair and excluding another couple of numbers because the dices of this pair shall have different numbers than the number of first and second pairs) So finally there are $15\cdot12\cdot2=360$ choices in this case.

For the case for only one golden side, there are $3$ ways to choose the golden pair and $2$ choices to choose the dice with golden side. Now for the two other pairs, if the number of the second pair is the number of one of the dices in the third pair, then there are two choices for the number of the second pair and $3$ choices for the number of the third pair. Otherwise there is only $1$ choice for the number of the second pair, and $2$ choices for the number of the third pair. Therefore there are $8$ choices for the numbers of the two other pairs. So at last there are $15\cdot3\cdot2\cdot8=720$ possibilities for this case.

Finally for no golden sides, there are $4$ choices to choose the number on the dices of the first pair and WLOG assumer that this number is the number of one of the dices in the third pair. Then like the previous case, if the number of the second pair is in the third pair, then there is only $1$ possibility for the number of the second pair and $4$ choices for the number of the third pair. Otherwise there are $2$ possibilities for the number of the second pair and $3$ choices for the number of the third pair. Therefore there are $10$ different possibilities for choosing the number of the two other pairs. Hence there are $15\cdot4\cdot10=600$ different possibilities for this case.

P.S. You can verify the answer by noting that $120+360+720+600=1800$.


问:
Understanding how two sets are equal
I am trying to understand this formal statement,

$$\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \implies A = B)$$

What do the universal quantifiers $$\forall A \forall B$$ indicate? Do they mean that we are talking about all of the elements in the sets A and B? And thus, that the two sets are equal?

答: The meaning of the quantifiers is that in order for any two sets to be defined as equal to one another, the statement ($X \in A \Leftrightarrow X \in B$) must hold.

Title: Understanding how two sets are equal


问:
SLLN tosses of a balanced die
Consider a suquence of tosses of a balanced die. Let $X_k$ be the result of the $k-$th toss and $Y_n=\max_{k\le n} X_k$.

I would like to determine almost surely convergence

Find the convergence almost surely of $(Y_n)_{n\ge1}$.

If $Y_1=\max\{X_1\}$, $Y_2=\max\{X_1,X_2\},\ldots,Y_n=\max\{X_1,\ldots, X_n\}$

since $ X_1,\ldots,X_k $ are independent random variables with the same distribution, then
for some $j\in\{1,\ldots,n\}$, $Y_n=\max\{X_1,\ldots,X_n\}=X_j$, then the expected value
$$E(Y_n)=E(X_j)$$
but all the random variables have the same distribution, hence the same expected value $(E(X_j)=\frac{7}{2})$.

With that, we can use the SLLN but I'm not sure if that's correct.

答: $E(Y_n)$ has nothing to do with this.  Clearly if $X_k=6$ then $Y_n=6\ \forall n\geq k$ and then $Y_n\to6$.  Therefore, we just need to show $\Pr(X_n\neq6,\ n\geq1)=0$.  I leave that to you.

Title: SLLN tosses of a balanced die


问:
How to interpret a time varying probability in the context of failure rate analysis?
I am having trouble interpreting a time varying probability in the context of failure rate analysis.

For instance, suppose I define $X(t) \sim \mathcal{N}(\mu,\,\sigma^{2})$, a function $g(t)$ and then a time varying probability as $p(t) = P(X(t) < g(t))$. Now If I call the event $\{X(t) < g(t)\}$ the "failure" that can only happen once, what does the function $p(t) = P(X(t) < g(t))$ represent in this context?

EDIT 1: I am ultimately interested in the failure time distribution, which I suspect can be constructed out of the above function $p(t)$

答:
Probably you are confused about probability. $p(t) = \text{Pr}(X(t) < g(t))$ is the probability of failure with respect to time. And while it is true that it can only fail one time for one single object, the probability is realized over $\infty$ objects, or more realistically and practically, a finitely many number $N$ of objects, e.g. 100s or 1000s. Each object only fails once, but this probability density function of failure has to consider many objects.

Also, you don't need any simulation to verify this. You just have to straighten out the concept of probability.


问:
What books introduce Number Theory to a Grade 11 student?
I'm in Grade 11. I'm interested in elementary number theory and would like to study it. I'm not intending to enter any competitions.</p>

答:
I recommend Number Theory: Step by Step published in Dec. 2020 by Kuldeep Singh for two reasons. It provides solutions online to EVERY exercise. It uses color.

As Senior Lecturer at University of Hertfordshire, he's taught number theory and linear algebra since Sep 1992 to first and second year undergraduates. I'm not he, but I know of him.


问:
Without using calculus determine the MLE of $(\theta_1,\theta_2)$
If $(x_1,\cdots,x_n)$ is a sample from a $\mathcal{U}(\theta_1,\theta_2)$ (uniform) distribution with $$\Omega=\{(\theta_1,\theta_2)\in\mathbb{R}^2:\theta_1<\theta_2\}$$
determine the MLE of $(\theta_1,\theta_2)$ without using calculus.

OK, I know that, $$f(x|(\theta_1,\theta_2))=\frac{1}{(\theta_2-\theta_1)}$$ and

$$L(\theta_1,\theta_2)=\prod_{i=1}^n\frac{1}{(\theta_2-\theta_1)}=\frac{1}{(\theta_2-\theta_1)^n}$$

Now I need to maximize the $\theta_2$ and minimize the $\theta_1$ values in order to miximize the likelihood. I know how to find those value using
LL (Log likelihood function) and partial derivatives. But how to find them without calculus? I have no idea where to start.

答:
The MLE is when $\theta_2=x_{(n)},\theta_1=x_{(1)}$. This is roughly because $\frac 1 {(\theta_2-\theta_1)^n}\le\frac 1 {(X_{(n)}-X_{(1)})^n}$ for all possible other choices of thetas. To see this, consider that the difference between the max and min values is 5. If you choose the max to be theta_2, and the min to be theta_1, then the likelihood is 1/5^n. Anything beyond these "boundaries" of data and you will get a fraction "smaller" than 1/5, so the fraction to the power of n will be "smaller" than 1/5^n, and you would get a "smaller" likelihood. Intuitively, these data are the most likely when the uniform distribution has boundaries of the smallest data point and the largest data point.

You should put some indicator variables to be more correct, i.e.
for an individual data point, $f(x|\theta_1,\theta_2)=\frac{1}{\theta_2-\theta_1}1_{\theta_1\le x\le\theta_2}$


答:
Here is a really stupid version of proof that noone would fall for:

 Start with equation (1). Multiply both sides with $x - 1$ and simplify to get: $x^3 - 1 = 0$. Conclude that $x = 1$ is a solution and proceed as in the original.

The question is now: how is the stupid version equivalent to the original? In other words:

 How can substitution be equivalent to multiplication?

The other answers don't address this issue and I am also not so sure myself. I'd like to see answer to this, if I find one I will edit it in to this answer.

Title: Mistake proving that $3 = 0$


问:
$2^{q-1}\equiv 1\pmod{q}.$

The question is asking to show that $q$ must be prime given
$$ 2^{q-1}\equiv 1\pmod{q}. $$

答:
If $q=2n$ where $n$ is odd, $\phi(q)=\phi(2)\phi(n)=\phi(n)<n$

$q-1=2n-1$ will be $>n$ if $n>1$

So, for $n>1, \text{ord}_q2<q-1$

Else if $q= m\cdot n$ where $(m,n)=1$  where $m>n>2$

Using Euler's Totient Theorem, $2^{\phi(n)}\equiv 1\pmod n$ and $2^{\phi(m)}\equiv 1\pmod m$

$$\implies 2^{\text{lcm}(\phi(n),\phi(m))}\equiv1\pmod {lcm(m,n)}$$

and lcm$(m,n)=m\cdot n=q$

As $\phi(n),\phi(m)$are both even for $m,n>2, 2$ divides $(\phi(n),\phi(m))$

$\implies \text{lcm}(\phi(n),\phi(m))< \phi(n)\cdot \phi(m)=\phi(q)$

So, $\text{ord}_q2<\phi(q)\le q-1$

So, a necessary condition for $\text{ord}_q2=q-1$ is $q$ must be prime

Now we know if $\text{ord}_q2=d$ and $2^r\equiv1\pmod q, d$ must divide $r$

As $341=11\cdot31$

$2^2\equiv4,2^5\equiv-1\pmod {11}\implies \text{ord}_{11}2=10 $

and $2^5\equiv1\pmod {31}\implies \text{ord}_{31}2=5 $

$\implies \text{ord}_{(11\cdot31)}2=$lcm$(5,10)=10$ which successfully divides $340$ which eventually turns out to be $341-1$

ContentLicense: CC BY-SA 3.0

Title: $2^{q-1}\equiv 1\pmod{q}.$


答:
There is so much you are missing that it is difficult for me to know where to begin addressing your issues.

First of all, if your density $$f(x \mid \theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1}$$ is not actually a function of $x$, then you are saying that any value of $x$ is equally likely, even if it is not in the interval $[\theta_1, \theta_2]$.  Is that really what you want to say?

Second, do you know anything about sufficient statistics?  Trying to maximize a likelihood function without understanding data reduction and sufficiency is like trying to do calculus without knowing basic algebra.

Third, your likelihood function is nonsense.  It doesn't even depend on the sample, so what function of the sample will produce a maximum?

Title: Without using calculus determine the MLE of $(\theta_1,\theta_2)$

Body:
If $(x_1,\cdots,x_n)$ is a sample from a $\mathcal{U}(\theta_1,\theta_2)$ (uniform) distribution with $$\Omega=\{(\theta_1,\theta_2)\in\mathbb{R}^2:\theta_1<\theta_2\}$$
determine the MLE of $(\theta_1,\theta_2)$ without using calculus.

OK, I know that, $$f(x|(\theta_1,\theta_2))=\frac{1}{(\theta_2-\theta_1)}$$ and

$$L(\theta_1,\theta_2)=\prod_{i=1}^n\frac{1}{(\theta_2-\theta_1)}=\frac{1}{(\theta_2-\theta_1)^n}$$

Now I need to maximize the $\theta_2$ and minimize the $\theta_1$ values in order to miximize the likelihood. I know how to find those value using
LL (Log likelihood function) and partial derivatives. But how to find them without calculus? I have no idea where to start.


答: In an exercise where your task is solving equations, equations are not theorems: equations are conditions. The theorem is, if we want to see it in set-theoretical terms, that the set of the values satisfying the first condition is equal to the set of values satisfying the last one. It is certainly true (and your proof is as good as any) that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\{x\in\Bbb R\,:\, 3=0\}$. Does that prove that $3=0$? No, of course. It just proves that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\varnothing$. Notice, as a side remark, that at some point you use the fact that $\{x\in\Bbb R\,:\, x^3=1\}=\{x\in\Bbb R\,:\, x=1\}$ and that this, while crucial in your proof that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\{x\in\Bbb R\,:\, 3=0\}$, is no longer true if you substitute $x\in\Bbb R$ with $x\in\Bbb C$.

Title: Mistake proving that $3 = 0$


问:
How to calculate sin/cos/tan of a Quaternion?
I would like to learn about Quaternions. I've read this article: https://en.wikipedia.org/wiki/Quaternion

Most of the article was not hard to understand, except the (Exponential, logarithm, and power) part.

To calculate the exponential, I have to calculate sin/cos of the imaginary part, but that functions aren't defined on Quaternions (in the article).

答:
Given a quaternion q

To calculate exp(q) the trig functions of quaternions are not necessary.

But if you did want to find a trig function of q:

The first way is the Taylor series for said trig function. Even python can handle the first 15 digits in about 0.005 seconds for small values.

The second way is to use the polar form of the quaternion.
First, we should start with trig of complex numbers:

```<code>cos(xi)=cosh(x)

sin(xi)=i sinh(x)

tan(xi)=i tanh(x)
</code>```

so...

```<code>cos(a+bi) = cos(a) cosh(b) - i sin(a) sinh(b)
</code>```

Expanding this for quaternions:

```<code>cos(xj)=cosh(x)

sin(xj)=j sinh(x)

tan(xj)=j tanh(x)

cos(xk)= cosh(x)

sin(xk)=k sinh(x)

tan(xk)=k tanh(x)
</code>```

Polar form for quaternions uses three variables, which I will denote as r, p, and n.

q = a+bi+cd+dj = a+v = r(cos(p)+n sin(p))

r = abs(q)
p = arccos(a/r)
n = v/abs(v)

r is the distance from 0.
p is the angle from 0
n is the direction of the angle

This is can be expanded for any versor (a quaternion q with abs(q) = 1) n with no real part:

```<code>cos(xn) = cosh(x)

sin(xn) = n cosh(x)
</code>```

So the cosine can be gotten from the polar form of any quaternion:
From a python quaternion-processing code I wrote:

```<code>    def __cos__(q, /):
        '''return cos of self in radians'''
        r, p, n = q.__polar__()
        return cos(q.real)*cosh(r*sin(p))-n*sin(q.real)*sinh(r*sin(p))
</code>```

And in math-language:

```<code>a = r(cos(p)

cos(q)

= cos(r(cos(p)+n sin(p)))

= cos(a+n*sin(p))

= cos(a) cos(n r sin(p)) - sin(a) sin(n r sin(p))

= cos(a) cosh(r sin(p)) - n sin(a) sinh(r sin(p))
</code>```

Python can process this in 0.00015 seconds.

Do the same for the rest of the trig functions to find them.

for cot, sec, and csc...

```<code>cot(xi) = coth(x)

sec(xi) = i*sech(x)

csc(ci) = i*csch(s)
</code>```

and for hyperbolic trig,

```<code>cosh(xi) = cos(x)

sinh(xi) = i*sin(x)

tanh(xi) = i*tan(x)

coth(xi) = cot(x)

sech(xi) = i*sec(x)

csch(ci) = i*csc(s)
</code>```

Body:
Given a quaternion q

To calculate exp(q) the trig functions of quaternions are not necessary.

But if you did want to find a trig function of q:

The first way is the Taylor series for said trig function. Even python can handle the first 15 digits in about 0.005 seconds for small values.

The second way is to use the polar form of the quaternion.
First, we should start with trig of complex numbers:

cos(xi)=cosh(x)

sin(xi)=i sinh(x)

tan(xi)=i tanh(x)

so...

cos(a+bi) = cos(a) cosh(b) - i sin(a) sinh(b)

Expanding this for quaternions:

cos(xj)=cosh(x)

sin(xj)=j sinh(x)

tan(xj)=j tanh(x)

cos(xk)= cosh(x)

sin(xk)=k sinh(x)

tan(xk)=k tanh(x)

Polar form for quaternions uses three variables, which I will denote as r, p, and n.

q = a+bi+cd+dj = a+v = r(cos(p)+n sin(p))

r = abs(q)
p = arccos(a/r)
n = v/abs(v)

r is the distance from 0.
p is the angle from 0
n is the direction of the angle

This is can be expanded for any versor (a quaternion q with abs(q) = 1) n with no real part:

cos(xn) = cosh(x)

sin(xn) = n cosh(x)

So the cosine can be gotten from the polar form of any quaternion:
From a python quaternion-processing code I wrote:

    def __cos__(q, /):
        '''return cos of self in radians'''
        r, p, n = q.__polar__()
        return cos(q.real)*cosh(r*sin(p))-n*sin(q.real)*sinh(r*sin(p))

And in math-language:

a = r(cos(p)

cos(q)

= cos(r(cos(p)+n sin(p)))

= cos(a+n*sin(p))

= cos(a) cos(n r sin(p)) - sin(a) sin(n r sin(p))

= cos(a) cosh(r sin(p)) - n sin(a) sinh(r sin(p))

Python can process this in 0.00015 seconds.

Do the same for the rest of the trig functions to find them.

for cot, sec, and csc...

cot(xi) = coth(x)

sec(xi) = i*sech(x)

csc(ci) = i*csch(s)

and for hyperbolic trig,

cosh(xi) = cos(x)

sinh(xi) = i*sin(x)

tanh(xi) = i*tan(x)

coth(xi) = cot(x)

sech(xi) = i*sec(x)

csch(ci) = i*csc(s)


问:
Relation involving entropy and dimension
Assume $n$ is a postive integer much larger than $\ell$. Let $\tau$ be a distribution over $\mathbb{F}_p^\ell$ satisfying $n\cdot\tau(x)\in\mathbb{N}$ for all $x\in \mathbb{F}_p^\ell$ (and $\sum_{x\in\mathbb{F}_p^{\ell}}\tau(x)=1$.)
A matrix $M\in \mathbb{F}^{n\times \ell}$ is called to be in $\mathcal{M}_{n,\tau}$ if the number of $x\in\mathbb{F}^\ell$ as rows of $M$ is same as $n\tau(x)$ for all $x\in\mathbb{F}^\ell$. On the other hand, the occurrence of rows of $M\in\mathbb{F}^{n\times \ell}$ divided by $n$ also defines a distribution over $\mathbb{F}_p^{\ell}$.

For any two matrices $M_1,M_2\in\mathcal{M}_{n,\tau}$, let $M$ be the $n\times (2\ell)$ matrix such that $M=[M_1,M_2]$, i.e., its first submatrix is $M_1$ and the second submatrix is $M_2$. Then $M$ defines some distribution $\mu$ over $\mathbb{F}_p^{2\ell}$.

I am wondering is it true that the entropy $H(\mu):=-\sum_{x\in supp(\mu)}\mu(x)\log\mu(x)\le H(v)$?

If not, let $d(\tau)=\dim(span(supp(\tau)))$. Is it true that
$$H(\mu)d(\tau)<H(\tau)d(\mu)$$
when $d(\mu)<2d(\tau)$?

答:
We can treat the first question  without reference to the vector space - you're effecive constructing a law on two random variables $(U,V)$ such that the marginal laws of $U$ and $V$ are both $\tau$ (up to some integrality conditions). Then $$ H(\tau) = H(U) \le H(U) + H(V|U) = H(\mu)\le H(U) + H(V) = 2H(\tau),$$ where for a random variable $X$, $H(X)$ is the entropy of its law. As long as $n\tau$ is favourible, the upper bound can be achieved - for example, if $\tau(x) \in \{0, 1/k\}$ for some $k$ that divides $n$.

For the second question, consider the case where $\tau$ is uniform on two linearly independent vectors $v_1,v_2$, and $n$ is divisible by $4$. Then by uniformly mixing up the $v_1$ rows with $v_2$ rows, we can generate a $\mu$ that is uniform on $\{(v_1, v_1), (v_1, v_2), (v_2, v_1), (v_2,v_2)\}.$ But the dimension of the space spanned by this is $3$, since $(v_1,v_2) = (v_1,v_1) + (v_2, v_2) - (v_2, v_1)$. In this case $H(\tau) = 1, H(\mu) = 2,$ $d(\tau) = 2, d(\mu) = 3,$ and $4 > 3$. This should be embeddable into quite rich $\tau$s.


问:
Multivariate Elliptic Integral
I want to find a "closed form" solution to the following integral:

$$I(a)=\int_{-1}^1\int_{-1}^1\int_{-1}^1\sqrt{\frac{a+ xy+yz+xz }{(1-x^2)(1-y^2)(1-z^2)}}\, \text{d}x\,\text{d}y\,\text{d}z$$

where $a>1$ to keep the integrand real valued.
I hope that the symmetry of the integrand might help.
I have tried integrating one variable after the other, but this involves Integrals of elliptic integrals that I am not familiar with:

Another form of the integral is:
$$I(a)=\int_{0}^\pi\int_0^\pi\int_0^\pi\sqrt{ a+ \cos(x)\cos(y)+\cos(y)\cos(z)+\cos(x)\cos(z) } \,\text{d}x\,\text{d}y\,\text{d}z$$

Evaluating the first integral, Mathematica gives:
$I(a)=\int_0^\pi\int_0^\pi 2\sqrt{ a+\cos(y)+\cos(y)\cos(z)+ \cos(z) }E\left[\frac{2\cos(y)+2\cos(z) }{ a+\cos(y)+\cos(y)\cos(z)+ \cos(z) }\right]  \,\text{d}y\,\text{d}z$

Where $E$ is the Elliptic Integral of second kind. I dont know how to proceed.
Is there maybe a nice coordinate system I could transform into? Maybe I have to try a completely different approach?
Any insight would be helpful.

Edit:

I have made some progress by using the FCC Lattice Green's function.
It is defined by:
$$G(a)=\frac{1}{\pi^3}\int_{0}^\pi\int_0^\pi\int_0^\pi\frac{1}{ 1 - \frac{a}{3}( \cos(x)\cos(y)+\cos(y)\cos(z)+\cos(x)\cos(z)) } \,\text{d}x\,\text{d}y\,\text{d}z$$

and has a known closed form solution in terms of Complete Elliptic Functions that can be found here (equation 2.20)

I will add more info soon.

答:
I was able to reduce the integral to a one-dimensional integral. I felt this was to long for an edit to my question.

I will consider a slightly different Integral than I originally stated:

$$J(a)=\frac{1}{\pi^3}\int_{0}^\pi\int_0^\pi\int_0^\pi\sqrt{ 1-\frac{a}{3} (\cos(x)\cos(y)+\cos(y)\cos(z)+\cos(x)\cos(z)) } \,\text{d}x\,\text{d}y\,\text{d}z$$
but it can easily be transformed into the original by considering $I(a) = \sqrt{a}J(-\frac{3}{a})$
I assume $\lvert a\rvert < 1$, but this restriction can probably be lifted to a $-3 < a < 1$.
Using the series expansion of the integrand and by exchanging integrals and summation we can rewrite J(a) as

$$J(a) = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} c_n (-a)^n $$
where the coefficients $c_n$ are defined by
$c_n = \frac{1}{\pi^3}\int_{0}^\pi\int_0^\pi\int_0^\pi \frac{1}{3^n}( \cos(x)\cos(y)+\cos(y)\cos(z)+\cos(x)\cos(z))^n$
Both functions $\sum_n^{\infty} \binom{\frac{1}{2}}{n}(-a)^nx^n = \sqrt{1-ax}$ and $\sum_n^{\infty} c_n x^n =W_1(x)$ are known.

$$W_1(x) =\frac{3\sqrt{3}}{(3+x)^{\frac{3}{2}}}(2-\sqrt{1-x})[ \frac{2}{\pi} K(\frac{1}{2}-\frac{2\sqrt{3}x}{(3+x)^{\frac{3}{2}}} -\frac{\sqrt{3}(3-x)\sqrt{1-x}}{(3+x)^{\frac{3}{2}}})]^2$$

is the Lattice Greens function of the FCC lattice. $K(k)$ is the complete Elliptic Integral of the first kind. A link to a paper with this closed form can be found here. The formula I used is Nr. (2.26).

The Integral is then the Hadamard product of both Series $W_1(x)$ and $\sqrt{1-ax}$ at x = 1:

$$J(a) = \frac{1}{2\pi}\int_0^{2\pi}\sqrt{1-ae^{-it}}W_1(e^{it}) \text{d}t$$

This can be rewritten as a contour integral:

$$J(a) = \frac{1}{2\pi i}\oint_{\lvert z \rvert=1}\frac{1}{z}\sqrt{1-\frac{a}{z}}W_1(z)$$

Now $W_1(z)$ has its Branch points at $-3,1, \infty$ so they wont bother us. But $\sqrt{1-\frac{a}{z}}$ has its branch points at $0$ and $a$. So we can modify the circular contour to a dogbone contour around the branch cut from 0 to a. With a circle around 0 with infinitesimal radius $\epsilon$, a circle around $a$ with infinitesimal radius $r$ and the back and forth integrals on the Real line which are complex conjugates of each other.

$$\oint_{\lvert z\rvert=1} = \oint_{\lvert z \rvert=\epsilon}  +\oint_{z = a + ry, \lvert y\rvert=1 } - 2i \text{Im}[\int_\epsilon^{a-r} ] $$

The Integral around $a$ just gives $0$ for $r\xrightarrow{}0$. The Integral around $0$ is trickier since it diverges for $\epsilon = 0$ . This can be solved by first integrating for infinitesimal $\epsilon$, then adding the integrals together, and then letting $\epsilon$ got to 0. This way all divergences cancel. In the neighbourhood of $z=0$ the Lattice Greens function is just $1$. Therefore the integral gives

$$\frac{1}{2\pi i} \oint_{\lvert z \rvert=\epsilon}\frac{1}{z}\sqrt{1-\frac{a}{z}}W_1(z)\text{d}z \approx \frac{1}{2\pi i} \oint_{\lvert z \rvert =\epsilon}\frac{a}{z^{3/2}} \text{d}z + \mathcal{O}(\epsilon)= \frac{2}{\pi}\sqrt{\frac{a}{\epsilon}}=  \frac{2}{\pi}+\frac{1}{\pi}\int_\epsilon^a\frac{\sqrt{a}}{z^\frac{3}{2}}\text{d}z$$

After adding this to the remaining integral we arrive at the 1D Integral form:

$$J(a)= \frac{2}{\pi}+\frac{1}{\pi}\int_0^a\frac{\sqrt{az}-\sqrt{az-z^2}W_1(z)}{z^2}\text{d}z = \frac{2}{\pi}+\frac{1}{\pi}\int_0^1\frac{\sqrt{ z}-\sqrt{ z-z^2}W_1(az)}{z^2}\text{d}z $$

I have checked the equality numerically for values of $-3<a<1$.
Unfortunately I am stuck here. This is quite a lot nicer than the original expression but sadly still far away from any final solution. Maybe after a nice substitution one can use contour integration again. But the Branch structure of $W_1(x)$ might cause a lot of Problems. Maybe someone can handle the Series expansion of the Integral. The closed form of $c_n$ can be found here in equation (13) and (17).

Edit:
An even nicer form of the resulting integral is:

$$J(a)=1-\frac{1}{\pi}\int_0^1\frac{\sqrt{ z-z^2}}{z^2}(W_1(az)-1)\text{d}z$$

If one applies the Euler substitution $z-> \frac{1}{1+t^2}$ one can get rid of the square root

$$J(a)=1-\frac{2}{\pi}\int_0^\infty \frac{t^2}{1+t^2}(W_1(\frac{a}{1+t^2})-1)\text{d}t$$

Title: Multivariate Elliptic Integral


问:
Evaluate the Beta(3,1) and Beta(91,11) probabilities
In here there is one diagram

And a statement:

After observing 90 positive reviews and 10 negative reviews, our posterior estimate on $θ_A$ has a $beta(91, 11)$ distribution. After observing 2 positive reviews, our posterior estimate on $θ_B$ has a $beta(3, 1)$ distribution. The probability that a sample from $θ_A$ is bigger than a sample from $θ_B$ is 0.713.

Now I am trying to understand how to get this 0.713?

What I know for Beta distribution the first moment (mean) is:

$E(X) = \Large \frac{a}{a + b}$

I can find the mean for the $\mu_{A} = \frac{91}{102} =0.892$ (red)

I can find the mean for the $\mu_{B} = \frac{3}{4} =0.75 $ (blue)

PDF:
$f(x)=\frac{1}{B(a, b)} x^{a-1}(1-x)^{b-1}, \quad x \in(0,1)$

Probability density function PDF(A):
$\left\{\begin{array}{ll} \frac{1}{B(91,11)} x^{90}(1-x)^{10} & 0<x<1 \\ 0 & \text { (otherwise) }\end{array}\right.$

Probability density function PDF(B):
$\left\{\begin{array}{ll}3 x^{2} & 0<x<1 \\ 0 & \text { (otherwise) }\end{array}\right.$

I can approximate the blue curve with the line $y=3x$, and the average $y$ value will be 1.5

The red curve I can possible approximate with Normal distribution, with the mean of 0.892, and variance 0.03056 because of

$ Beta(a, b)$ as $Normal \left(\frac{a}{a+b}, \sqrt{\frac{ab}{\left(a+b\right)^{2}\left(a+b+1\right)}}\right)$

Now the peak of the normal distribution should be $\frac{1}{\sigma\sqrt{2\pi}}$ is ~13. If I create a triangle from normal distribution with the height 13 and bottom $2*\sigma$, this means the average $y=6.5$.

Again I cannot find 0.713.

Any idea how John Cook found it?

答:
If $\theta_A$ has the PDF $\frac{101!}{90!10!} x^{90} (1-x)^{10}$ for $x \in [0,1]$, and $\theta_B$ has the PDF $3x^2$ for $x \in [0,1]$, and they're independent, then their joint PDF is
$$
     \frac{3 \cdot 101!}{90! 10!} x^{90} (1-x)^{10} y^2
$$
for $(x,y) \in [0,1]^2$. To find the probability $\theta_A > \theta_B$, integrate over the region where $x > y$:
$$
   \int_0^1 \int_0^x \frac{3 \cdot 101!}{90! 10!} x^{90} (1-x)^{10} y^2 \,dy\,dx = \frac{4991}{7004} \approx 0.713.
$$
We can also approximate by an easier integral: since $\theta_A$ has expected value $\frac{91}{102}$, and is pretty concentrated around that value, we could estimate $\Pr[\theta_A > \theta_B]$ as $\Pr[\frac{91}{102} > \theta_B] = \int_0^{91/102} 3x^2\,dx$. This gives us $(\frac{91}{102})^3 \approx 0.710$, which is pretty close.


答:
The variables $\theta_A$ and $\theta_B$ are the posterior estimates for the proportion of positive reviews.  If these are independent (which is assumed to be the case), then $$\Pr[\theta_A > \theta_B] = \int_{\theta = 0}^1 \Pr[\theta_A > \theta]f_{\theta_B}(\theta) \, d\theta = \int_{\theta = 0}^1 \int_{\psi = \theta}^1 f_{\theta_A}(\psi) f_{\theta_B}(\theta) \, d\psi \, d\theta,$$ where $f_{\theta_A}$ and $f_{\theta_B}$ are the respective posterior densities, which we are told are beta distributed.  Thus
$$\Pr[\theta_A > \theta_B] = \frac{\Gamma(91+11)}{\Gamma(91)\Gamma(11)}\frac{\Gamma(3+1)}{\Gamma(3)\Gamma(1)} \int_{\theta=0}^1 \int_{\psi = \theta}^1 \psi^{90} (1-\psi)^{10} \theta^2 \, d\psi \, d\theta.$$

Title: Evaluate the Beta(3,1) and Beta(91,11) probabilities

Body:
In here there is one diagram

And a statement:

After observing 90 positive reviews and 10 negative reviews, our posterior estimate on $θ_A$ has a $beta(91, 11)$ distribution. After observing 2 positive reviews, our posterior estimate on $θ_B$ has a $beta(3, 1)$ distribution. The probability that a sample from $θ_A$ is bigger than a sample from $θ_B$ is 0.713.

Now I am trying to understand how to get this 0.713?

What I know for Beta distribution the first moment (mean) is:

$E(X) = \Large \frac{a}{a + b}$

I can find the mean for the $\mu_{A} = \frac{91}{102} =0.892$ (red)

I can find the mean for the $\mu_{B} = \frac{3}{4} =0.75 $ (blue)

PDF:
$f(x)=\frac{1}{B(a, b)} x^{a-1}(1-x)^{b-1}, \quad x \in(0,1)$

Probability density function PDF(A):
$\left\{\begin{array}{ll} \frac{1}{B(91,11)} x^{90}(1-x)^{10} & 0<x<1 \\ 0 & \text { (otherwise) }\end{array}\right.$

Probability density function PDF(B):
$\left\{\begin{array}{ll}3 x^{2} & 0<x<1 \\ 0 & \text { (otherwise) }\end{array}\right.$

I can approximate the blue curve with the line $y=3x$, and the average $y$ value will be 1.5

The red curve I can possible approximate with Normal distribution, with the mean of 0.892, and variance 0.03056 because of

$ Beta(a, b)$ as $Normal \left(\frac{a}{a+b}, \sqrt{\frac{ab}{\left(a+b\right)^{2}\left(a+b+1\right)}}\right)$

Now the peak of the normal distribution should be $\frac{1}{\sigma\sqrt{2\pi}}$ is ~13. If I create a triangle from normal distribution with the height 13 and bottom $2*\sigma$, this means the average $y=6.5$.

Again I cannot find 0.713.

Any idea how John Cook found it?


问:
Find the quotient ring $T_{2}(\mathbb{Z_{4}})/I$.

Let $T_{2}(\mathbb{Z_{4}})= \left\{ \begin{bmatrix}a&b\\0&c\end{bmatrix}:a, b, c \in \mathbb{Z_{4}}\right\}$ be the ring of all upper triangular matrices on $\mathbb{Z_{4}}$.

Prove that $I =\left\{ \begin{bmatrix}0&b\\0&c\end{bmatrix}: b,c \in \mathbb{Z_{4}}\right\}$ is an ideal of $T_{2}(\mathbb{Z_{4}})$. Find the quotient ring $T_{2}(\mathbb{Z_{4}})/I$.

I already have proof that $I$ is an ideal of $T_{2}(\mathbb{Z_{4}})$. Could someone help me find the quotient ring?

答:
HINT:

Consider the following ring homomorphism $f: T_2 (\Bbb Z_4) \to \Bbb Z_4$ defined by
$$\begin{bmatrix}a&b\\0&c\end{bmatrix} \mapsto a$$
Show that this is an onto ring homomorphism and show that $I= \ker f$.

Then apply the first isomorphism theorem to get
$$T_2 (\Bbb Z_4) / I  \cong \Bbb Z_4$$


问:
Are there more possibilities for having twin primes occurring than cousin primes?
Let any prime be $X$

It started out with an experiment to find out if there are some patterns in primes with $X \mod 6 = 1$ Versus primes with $X \mod 6 = 5$.

I have then checked to see what are the multiples of $6$ before getting the remainders and obviously some duplicates came in due to cousin primes, such as in the case of $27$ being the multiple: $(27 * 6) + 1= 163$ and $(27 * 6)+5=167$

But here is the thing:

All these multiples shared a limited set of possible endings, they can only end in either: $1,2,3,6,7,8$ and always excluding possible endings in 0,4,5,9

Let the multiple be $Y$

Below is a short example of the first hundreds of $Y$'s generated from  $(6 * Y)+1 = prime$ and $(6 * Y)+5 = prime$ while sharing the same $Y$ (I have further tested the first 10000):

Why are the endings limited to a limited amount of possible digits, and does it imply that in a probabilistic fashion, there are more possibilities for twin primes to occur than cousin primes?

答:
If I understand your first query, you ask why $Y$ cannot end in the digits $0,4,5,9$

Looking at cases, if $Y$ ends in $0,5$, then $6Y$ ends in $0$ and $6Y+5$ ends in $5$, yielding a number which is divisible by $5$ and hence not prime, so it can't be part of a pair of cousin primes.

Similarly, if $Y$ ends in $4,9$, then $6Y$ ends in $4$ and $6Y+1$ ends in $5$, yielding a number which is divisible by $5$ and hence not prime, so it can't be part of a pair of cousin primes.

There is a similar situation regarding twin primes. Consider twin primes of the form $6Z-1,\ 6Z+1$. You will find that $Z$ cannot end in $1,4,6,9$ for similar reasons. The argument for $4,9$ is the same as for the case of $Y$.

If $Z$ ends in $1,6$, then $6Z$ ends in $6$ and $6Z-1$ ends in $5$, yielding a number which is divisible by $5$ and hence not prime, so it can't be part of a pair of twin primes.

In both instances, there are $4$ forbidden end digits, so I don't think this would form a basis to argue a difference in the density of twin vs cousin primes.


问:
Upper-bound on number of connected components, given maximal degree
If an undirected graph $G = (V, E)$ has bounded degree $\deg(v) \leq d$ for each $v \in V$, how can we bound from above the number of connected components $C(G)$ of $G$?

The following very rough proof sketch argues that
$$C(G) \leq |V| - \frac{|E|}{2\cdot d}.$$
Indeed, write $V^* \subset V$ for the number of vertices of positive degree; i.e., $V^* = \{ v \in V | \deg (v) > 0 \}$. We first argue that $C(G) \leq |V| - \frac{|V^*|}{2}$. Indeed, $|V^*|$ is identified under the reachability relation in at least a 2-1 manner (and more in general), and will decrease the number of connected components by $\frac{|V|}{2}$. Finally, $|V^*| \geq \frac{|E|}{d}$. The idea is that the $|E|$ total edges must be "distributed" among the elements of $V^*$; in the "worst" case, each vertex alone uses $d$ among those edges. (This doesn't really make sense because each edge belongs to two vertices, not one.)

Can we get a tighter bound? Can we get a more rigorous / elegant proof? Thanks.

答:
Claim. In every simple graph with at least one vertex and degree bounded by $d$, we have
$$ |V|-\frac{|E|}d\ge 1.$$

Proof. It is enough to prove the claim when $$d=\max\{\,\deg(v)\mid v\in V\,\}$$ as making $d$ larger makes the claim only weaker. Under this assumption, we have $$|E|\ge d$$ already from the edges incident with one vertex of maximal degree.

By the handshaking lemma (i.e., by counting the edge-vertex incidences in two ways), we have
$$ d|V|\ge 2|E|$$
and so
$$ |V|\ge\frac{2|E|}{d}.$$
This makes
$$ |V|-\frac{|E|}{d}\ge \frac{2|E|}{d}-\frac{|E|}d=\frac{|E|}d\ge1$$
as was to be shown. $\square$

The claim holds per connected component so that we immediatyly improve it to
$$ C(G)\le |V|-\frac{|E|}d.$$
Note that this inequality is sharp as witnessed by graphs where each component is $d$-regular (e.g., when $G$ consists of several disjoint copies of $K_{d+1}$).


问:
maximum of multivariable convex function
There is a statement that says that every real convex function of one variable on compact convex set attains maximum on its boundary. Is there analogous result for real convex function of more then one variable?</p>

答:
Suppose $f : U \to \Bbb R, U \subseteq \Bbb R^n$ is convex, and $U$ is compact and convex.

Let $p$ be any interior point of $U$ and $\ell$ be any line passing through $p$. Because $U$ is compact and convex, $\ell \cap U$ is a line segment, a compact convex set, and the endpoints are on the boundary of $U$. Further, the restriction $f|_\ell$ of $f$ to $\ell$ is convex, so by the result you've quoted, the maximum of $f|_\ell$ is at one of the endpoints $b$. Therefore $f(b) \ge f(p)$.

That is, for every interior point $p$ of $U$, there is at least one boundary point $b$ with $f(b) \ge f(p)$. Since $U$ is compact, $f$ must have a maximum, which by this result must be taken on by a point on the boundary.


问:
Find all $m \in \mathbb{N}$ such that $m^{2}-8m+8$ Is a perfect square
I think that this problem is easy, but i am not sure whether my solution is correct or not.

My solution:

Set: $m^{2}-8m+8 =x^2$, with $x \in \mathbb{N}$
$$m^{2}-x^{2}=8(m-1)$$
$$\implies 8\mid m^{2}-x^{2}$$
$$\implies 8=m-x$$
This last line is the critical one because i don’t have a proof of this statement.
$$8+x=m$$
If we plug this result to the original equation, we get:
$m=1$ and $m=7$.

答:
Your implication $8\mid m^2-x^2\implies 8=m-x$ does not necessarily hold. Consider, for instance $8\mid 7^2-3^2$, whereas $8\not= 7-3$...

Hint.

$$m^2-8m+8=x^2\iff (m-4)^2-8=x^2\iff (m-4+x)\cdot (m-4-x)=8$$What are the factors of $8$?


问:
Using $f'$ to prove that $f$ is one-to-one. (Verification)
I would like to verify if I am doing this right, and if I am on the right track. I feel like my proof is a bit short, so I would like some verification.

Problem: Let $f(x) = 4x^3 + 5x$. Use $f'$ to prove $f$ is one-to-one $\forall x\in\mathbb{R}$.

Attempt: If we are going to be using $f'$ to show that $f$ is a one-to-one function $\forall x\in\mathbb{R}$, then one of the two conditions must be met.

If $f'(x) > 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.

If $f'(x) < 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.

The derivative of $f$ is simply: $f'(x) = 12x^2 + 5$. If $f'(x) = 0$, then there will be no solutions to $0 = 12x^2 + 5$. Therefore, there can only be one interval, which is $x\in(-\infty, \infty)$.

If we substitute any value of $x$ into $f'(x) = 12x^2 + 5$, then the value is positive, such that $f'(x) > 0$ which means that $f'$ is increasing on $x\in(-\infty, \infty)$.

And since the function is never decreasing, and is always going to be increasing, which proves that $f$ is a one-to-one function. $\square$

Let me know if this is correct, and if there are any suggestions that I can do to make this solution better. Thanks!

答: The proof is correct. However, it is not true that one of those two conditions must be met. For instance$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&x^3\end{array}$$is one-to-one, but you have $f'(0)=0$.


答:
There is a much easier way to do this, if you are interested in a solution:

 Note that $m^2-8m+16$ is a square $x^2=(m-4)^2$. Thus $m^2-8m+8 =x^2-8$ for $x=m-4$.

 However, also note the following: for all integers $y \ge 4$  and any other integer $z$, the following inequality holds: $|y^2-z^2| > 8$.

 So $x^2=(m-4)^2$ can't be larger than 9...Can you finish from here.

Title: Find all $m \in \mathbb{N}$ such that $m^{2}-8m+8$ Is a perfect square

Body:
I think that this problem is easy, but i am not sure whether my solution is correct or not.

My solution:

Set: $m^{2}-8m+8 =x^2$, with $x \in \mathbb{N}$
$$m^{2}-x^{2}=8(m-1)$$
$$\implies 8\mid m^{2}-x^{2}$$
$$\implies 8=m-x$$
This last line is the critical one because i don’t have a proof of this statement.
$$8+x=m$$
If we plug this result to the original equation, we get:
$m=1$ and $m=7$.


答:
You can significantly reduce your working by noting that $12 x^2 + 5$ cannot be less than zero (and cannot even go as low as that) because $x^2 \ge 0$ for all reals.

The rest is indeed correct.

Title: Using $f'$ to prove that $f$ is one-to-one. (Verification)

Body:
I would like to verify if I am doing this right, and if I am on the right track. I feel like my proof is a bit short, so I would like some verification.

Problem: Let $f(x) = 4x^3 + 5x$. Use $f'$ to prove $f$ is one-to-one $\forall x\in\mathbb{R}$.

Attempt: If we are going to be using $f'$ to show that $f$ is a one-to-one function $\forall x\in\mathbb{R}$, then one of the two conditions must be met.

If $f'(x) > 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.

If $f'(x) < 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.

The derivative of $f$ is simply: $f'(x) = 12x^2 + 5$. If $f'(x) = 0$, then there will be no solutions to $0 = 12x^2 + 5$. Therefore, there can only be one interval, which is $x\in(-\infty, \infty)$.

If we substitute any value of $x$ into $f'(x) = 12x^2 + 5$, then the value is positive, such that $f'(x) > 0$ which means that $f'$ is increasing on $x\in(-\infty, \infty)$.

And since the function is never decreasing, and is always going to be increasing, which proves that $f$ is a one-to-one function. $\square$

Let me know if this is correct, and if there are any suggestions that I can do to make this solution better. Thanks!


问:
Prove: If $\lim_{n\rightarrow\infty}|a_n| = 0$, then $\lim_{n\rightarrow\infty}a_n = 0$

Using the squeeze theorem, prove the following:

If $\lim_{n\rightarrow\infty}|a_n| = 0$, then $\lim_{n\rightarrow\infty}a_n = 0$.

Let $f(x) = |a_n|$. Let $g(x) = a_n$ such that $\forall x : g(x) \leq f(x)$. How do I continue this proof using the squeeze theorem? I want to construct a function $h(x)$ that lies in between those two functions.

答:
The accepted answer uses the squeeze Theorem.

Alternatively, note that
$$
|a_n-0|=||a_n|-0|\tag{1}
$$

Now if you unwrapping the definition of limits, the consequence is immediate by (1).


问:
Show there is a real number $c$, such that for any continuous $f$, $f'(x) =cx$ for some $x$ in $(0,1)$
Here is the full problem statement.
Show that there is a unique real number $c$ such that for any function $f$ that is differentiable on $[0, 1]$ and that takes the values $f (0) = 0$ and $f (1) = 1$ there is some point $x \in (0,1)$ (depending on the function $f\;$) such that $f'(x) = cx$.

I think the correct value of $c$ is $2$, by considering $f(x) = x^2$, and can also use that function to prove uniqueness of $c$. I’m having a hard time proving that any $f$ must have such a point. I think it will be an application of MVT, but would appreciate a hint.
Cheers!

答: Let $g(x)=f(x)-x$, $g(0)=g(1)=0$ implies there exists $x_0$ such that $g'(x_0)=0$ implies $f'(x_0)=1$.


答:
The fact that there is no sign change in the derivative is enough to prove. But I think you're missing the reasoning. What needs to exist to not be one-to-one? You need the curve to "turn around" at some point. And what causes a function to "turn around"? A local extrema. So in your case, you simply have no critical points, therefore no local extrema. In the example mentioned:

$$
f(x) = x^3
$$

Look at the derivative:

$$
f'(x) = 2x^2 \longrightarrow x = 0 \text{ is a critical point}
$$

But is there a sign change? There isn't--it's positive on both sides. So again, there are no local extrema therefore this function is also one-to-one.


问:
How to calculate the centered approximation of ${\frac{{df}}{{dt}} = - \frac{f}{\tau }}$?
In the textbook of An Introduction to Computational Methods in Hydrodynamics by Willy Benz. 2.2 Differential Equations.

I know how to derive the general formula above
$$
f\left( {{x_0} + \Delta x} \right) = f\left( {{x_0}} \right) + {\left( {\frac{{\partial f}}{{\partial x}}} \right)_{{x_0}}}\Delta x + \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_{{x_0}}}\Delta {x^2} + O\left( {\Delta {x^3}} \right)
$$
$$
f_{k + 1}^n = f_k^n + {\left( {\frac{{\partial f}}{{\partial x}}} \right)_k}\Delta x_{k + \frac{1}{2}}^n + \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_k}{\left( {\Delta x_{k + \frac{1}{2}}^n} \right)^2} + O\left( {\Delta {x^3}} \right)
$$

--
$$
f\left( {{x_0} - \Delta x} \right) = f\left( {{x_0}} \right) + {\left( {\frac{{\partial f}}{{\partial x}}} \right)_{{x_0}}}\left( { - \Delta x} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_{{x_0}}}{\left( { - \Delta x} \right)^2} + O\left( {{{\left( { - \Delta x} \right)}^3}} \right)
$$
$$
f_{k - 1}^n = f_k^n + {\left( {\frac{{\partial f}}{{\partial x}}} \right)_k}\Delta x_{k - \frac{1}{2}}^n + \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_k}{\left( {\Delta x_{k - \frac{1}{2}}^n} \right)^2} + O\left( {{{\left( {\Delta x} \right)}^3}} \right)
$$

--
$$
f_{k + 1}^n - f_{k - 1}^n = {\left( {\frac{{\partial f}}{{\partial x}}} \right)_k}\left( {\Delta x_{k + \frac{1}{2}}^n - \Delta x_{k - \frac{1}{2}}^n} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_k}\left( {{{\left( {\Delta x_{k + \frac{1}{2}}^n} \right)}^2} - {{\left( {\Delta x_{k - \frac{1}{2}}^n} \right)}^2}} \right) + O\left( {\Delta {x^3}} \right){\text{ }}
$$
$$
{\left( {\frac{{\partial f}}{{\partial x}}} \right)_k} = \frac{{f_{k + 1}^n - f_{k - 1}^n}}{{2 \cdot \Delta {x_k}}} - \frac{1}{2}{\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_k}\frac{{\left( {{{\left( {\Delta x_{k + \frac{1}{2}}^n} \right)}^2} - {{\left( {\Delta x_{k - \frac{1}{2}}^n} \right)}^2}} \right)}}{{2 \cdot \Delta {x_k}}} + O\left( {\Delta {x^2}} \right)
$$

But I am stuck below:

given ${\frac{{df}}{{dt}} =  - \frac{f}{\tau }}$ & $f(0)=1$

I know how to get these 2,

$$
\begin{array}{l}
\left[ {{\text{forward difference approximation}}} \right]\\
\frac{{\delta f}}{{\delta t}} \equiv \frac{{{f^{n + 1}} - {f^n}}}{{\Delta t}}\\
- \frac{1}{\tau } \cdot f\left( {{t_0}} \right) \approx \frac{{f\left( {{t_0} + \Delta t} \right) - f\left( {{t_0}} \right)}}{{\Delta t}}
\end{array}
$$
$$
\begin{array}{l}
\left[ {{\text{backward difference approximation}}} \right]\\
\frac{{\delta f}}{{\delta t}} \equiv \frac{{{f^n} - {f^{n - 1}}}}{{\Delta t}}\\
- \frac{1}{\tau } \cdot f\left( {{t_0}} \right) \approx \frac{{f\left( {{t_0}} \right) - f\left( {{t_0} - \Delta t} \right)}}{{\Delta t}}
\end{array}
$$

but how can you get the centered approximation of the given function as following?

and how can you get numerical solutions of it?

$$
\frac{{{f^{n + 1}} - {f^n}}}{{\Delta t}} =  - \frac{1}{{2\tau }}\left( {{f^{n + 1}} + {f^n}} \right)
$$
$$
\frac{{f\left( {{t_0} + \Delta t} \right) - f\left( {{t_0}} \right)}}{{\Delta t}} =  - \frac{1}{{2\tau }} \cdot \left( {f\left( {{t_0} + \Delta t} \right) + f\left( {{t_0}} \right)} \right)
$$

Following is the graph for a better visualization of the solutions at different time points.

Update 20210201_174816

$$
\begin{array}{l}
\frac{{{f^{n + 1}} - {f^n}}}{{\Delta t}} =  - \frac{1}{\tau } \cdot \frac{{{f^{n + 1}} + {f^n}}}{2}\\
\frac{{{f^{n + 1}} - {f^n}}}{{\Delta t}} =  - \frac{1}{{2\tau }}\left( {{f^{n + 1}} + {f^n}} \right)\\
{f^{n + 1}} - {f^n} =  - \frac{{\Delta t}}{{2\tau }}\left( {{f^{n + 1}} + {f^n}} \right)\\
set:\left\{ {\alpha  = \Delta t/\tau } \right.\\
{f^{n + 1}} - {f^n} =  - \frac{\alpha }{2}\left( {{f^{n + 1}} + {f^n}} \right)\\
\left( {1 + \frac{\alpha }{2}} \right){f^{n + 1}} = {f^n} \cdot \left( {1 - \frac{\alpha }{2}} \right)\\
{f^{n + 1}} = \frac{{\left( {1 - \frac{\alpha }{2}} \right)}}{{\left( {1 + \frac{\alpha }{2}} \right)}}{f^n}\\
given:\left\{ {\alpha  = \Delta t/\tau  = 0.2} \right.\\
{f^{n + 1}} = \frac{{\left( {1 - \frac{{0.2}}{2}} \right)}}{{\left( {1 + \frac{{0.2}}{2}} \right)}}{f^n}\\
{f^n} = {\left( {\frac{{\left( {1 - \frac{{0.2}}{2}} \right)}}{{\left( {1 + \frac{{0.2}}{2}} \right)}}} \right)^n} \cdot f\left( 0 \right) = {\left( {\frac{{\left( {2 - 0.2} \right)}}{{\left( {2 + 0.2} \right)}}} \right)^n} \cdot 1\\
{f^{2\tau /0.2\tau }} = {\left( {\frac{{\left( {2 - 0.2} \right)}}{{\left( {2 + 0.2} \right)}}} \right)^{10}} = 0.134430632749
\end{array}
$$

Above does not match with the textbook, which is 0.1407.

答:
The comment right below Eq. (2.2.4) is giving you a clue

Notice that (2.2.4) has been centered in a somewhat surprising way. Using a straightforward analog of (2.1.9) for estimating the time derivative of $f$ would lead to an intrinsically unstable and therefore useless numerical scheme.

Apparently that is further discussed in section 3, but I can give you an argument here.

How NOT to do it

If you stick to Eq. (2.1.9) you end up with

$$
\frac{f^{(n+1)} - f^{(n-1)}}{2\Delta t} = -\frac{f(\text{where should I evaluate this?})}{\tau} \tag{1}
$$

the problem is that if you decide evaluate the right hand side (RHS) at $t^{(n)}$ then you will two initial conditions to solve the iteration. A better approach is to evaluate in the average of the points of the grid

$$
\frac{f^{(n+1)} - f^{(n-1)}}{2\Delta t} = -\frac{f^{(n+1)} + f^{(n-1)}}{2\tau} \tag{2}
$$

which can be expressed as

$$
f^{(n+1)} = \frac{1 - \alpha}{1 + \alpha} f^{(n-1)} ~~~\alpha= \frac{\Delta t}{\tau} \tag{3}
$$

and the issue with this one is that you end up with parallel tracks for even/odd indices.

HOW TO DO IT

Use forward difference on the derivative, and evaluate the RHS on the average of the grid

$$
\underbrace{\frac{f^{(n+1)} - f^{(n)}}{\Delta t}}_{\text{forward}} = -\frac{1}{\tau}\underbrace{\frac{f^{(n+1)} + f^{(n)}}{2}}_{\text{average point}} \tag{4}
$$

Rearranging a bit

$$
f^{(n+1)} = \frac{2 - \alpha}{2 + \alpha} f^{(n)} \tag{5}
$$

whose solution is

$$
f^{(n)} = \left(\frac{2 - \alpha}{2 + \alpha}  \right)^n f^{(0)} \tag{6}
$$

EDIT 20210203

These are the solutions for all the methods

\begin{eqnarray}
f_{\rm forward}^{(n)} &=& (1 - \alpha)^n f^{(0)} \\
f_{\rm backward}^{(n)} &=& \left(\frac{1}{1 + \alpha}\right)^n f^{(0)} \\
f_{\rm centered}^{(n)} &=& \left(\frac{2 - \alpha}{2 + \alpha}\right)^n f^{(0)}
\end{eqnarray}

and in graphical form

You can see that lines are in a different order in the picture you posted ... I'm leaning towards a problem with the book


问:
Distribution of the minimum of discrete Uniform R.V.s

What is the cumulative distribution function of the minimum of $n$ discrete uniform i.i.d  R.V.s?
In particular when $X_i\sim\text{Unif}\,\{1,6\}$.

For $n=2$ the problem becomes quite simple as listing all possible choices is not too  bad. This becomes fairly hard when $n$ is large. Are there any closed forms?

答:
Let $Y=min\{X_1,...,X_n\}$. The cdf of Y is $G(y)=P(Y\le y)$. But the probability that the minimum is less than y is 1 minus the probability that it is greater than y, $1-P(Y>y)$. Since the minimum is at least y, everything in the sample is at least y, i.e. $1-P(X_1>y)...P(X_n>y)$. But since the $X_i$ are discrete iid on $Unif[1,6]$, this is just $1-[P(X>y)]^n$. Again doing the trick that $P(A)=1-P(A^c)$, this is equivalent to $1-[1-P(X\le y)]^n=1-[1-F_X(y)]^n$.

The actual cdf is $G(Y)=\begin{cases}0&y<1\\1-(\frac 5 6)^n&1\le y<2\\1-(\frac 2 3)^n&2\le y<3\\1-(\frac 1 2)^n&3\le y<4\\1-(\frac 1 3)^n&4\le y<5\\1-(\frac 1 6)^n&5\le y<6\\1&y\ge6\\\end{cases}$


问:
Finding a matrix with the following property
I have one $n \times n$ symmetric matrix $B$. Let $p$ be a scalar, I want to multiply the diagonal elements of $B$ by $p$. Let now $C$ denote the resultant matrix of the process described. Is there some matrix $A$, such that $AB = C$?</p>

答: Assuming $B$ is nonsingular, $A = C B^{-1}$.  Or did you want an $A$ that doesn't depend on $B$ and $C$?  That won't exist.</p>


答:
let's tryna see wha we get
$x^2$ is important function
the $f(x) -x^2$ is zero at the extremes
So derivative zero is what u want

Title: Show there is a real number $c$, such that for any continuous $f$, $f'(x) =cx$ for some $x$ in $(0,1)$

Body:
Here is the full problem statement.
Show that there is a unique real number $c$ such that for any function $f$ that is differentiable on $[0, 1]$ and that takes the values $f (0) = 0$ and $f (1) = 1$ there is some point $x \in (0,1)$ (depending on the function $f\;$) such that $f'(x) = cx$.

I think the correct value of $c$ is $2$, by considering $f(x) = x^2$, and can also use that function to prove uniqueness of $c$. I’m having a hard time proving that any $f$ must have such a point. I think it will be an application of MVT, but would appreciate a hint.
Cheers!


问:
Probability density function and Radon-Nikodym derivative
Consider a probability space $(\Omega,\Sigma,P)$.

We say that a real random variable $X\colon \Omega \to \mathbb{R}$ is absolutely continuous when there exists a function $f_X\colon \mathbb{R} \to [0,+\infty)$ such that $\mu_X((-\infty,x])=\int_{-\infty}^x f_X(t)\,\text{d}t$ for all $x \in \mathbb{R}$, where $\mu_X\colon \mathcal{B}(\mathbb{R}) \to [0,1] \mid \mu_X(A)=P(X \in A)$, namely $\mu_X$ is the Borel pushforward measure (or image measure) of $X$. $f_X$
is said to be the probability density function (PDF) of $X$.

$ $

Now consider Radon-Nikodym derivative:

Consider the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu_X)$. The Radon-Nikodym derivative $\frac{\text{d}\mu_X}{\text{d}\lambda}$ is a measurable function $f\colon \mathbb{R} \to [0,+\infty)$ such that for all $A \in \mathcal{B}(\mathbb{R}) \quad \mu_X(A)=\int_A f\,\text{d}\lambda$ where $\lambda$ is the Lebesgue measure.

$ $

So my question: is it true that $f_X=f$?

I have foound on wikipedia that: "The probability density function of a random variable is the Radon–Nikodym derivative of the induced measure with respect to some base measure (usually the Lebesgue measure for continuous random variables)". So I suppose that the answer to my question is "yes".

$ $

We know that $\forall \,x \in \mathbb{R} \quad \mu_X((-\infty,x])=\int_{-\infty}^x f_X(t)\,\text{d}t=\int_{(-\infty,x]} f_X\,\text{d}\lambda$, where obviously $(-\infty,x] \in \mathcal{B}(\mathbb{R})$. In order to show that $f_X=f$ we should prove that the above relation holds for every $A \in \mathcal{B}(\mathbb{R})$.

Thank you for your time!

答:
The answer to your question is yes.

If $f:\mathbb R\to\mathbb R$ is a nonnegative measurable map such that $\int_{\mathbb R}f(x)\,dx=1$, then there exists a unique probability measure $\mu$ on $\mathcal B(\mathbb R)$ such that for all $x\in\mathbb R$, $\mu((-\infty,x])=\int_{-\infty}^xf(t)\,dt$. This is the very well known fact that the cumulative distribution function characterises a probability measure. So this measure $\mu$ is the one defined for all $A\in\mathcal B(\mathbb R)$ by $\mu(A)=\int_Af(x)\,dx$. We deduce that a real-valued random variable $X$ is absolutely continuous iff its distribution $\mu_X$ is absolutely continuous with respect to the Lebesgue measure $\lambda$, in which case $X$ admits the probability density function $f_X=\frac{d\mu_X}{d\lambda}$.

You could also check that if $X:\Omega\to\mathbb N$ is a discrete random variable, then the distribution $\mu_X$ of $X$ admits a density with respect to the counting measure $\nu=\sum_{n\in\mathbb N}\delta_n$. More precisely, the density $\frac{d\mu_X}{d\nu}=f_X:\mathbb N\to\mathbb R$ is defined for all $n\in\mathbb N$ by $f(n)=\mathbb P(X=n)$. Indeed we have for any $A\subset\mathbb N$ and measurable bounded map $h:\mathbb N\to\mathbb R$ that
$$
\mathbb E[h(X)]=\sum_{n\in\mathbb N}h(n)\mathbb P(X=n)=\int_{\mathbb N}h(n)f_X(n)\,\nu(dn).
$$


问:
Norm of the right shift operator in $B(H)$
Let $H$ be an infinite dimensional separable complex Hilbert space.

Let $T \in B(H)$ be a unilateral weighted shift,

i.e. there exists an orthonormal basis $(e_n)_{n\in \mathbb{N}}$ for $H$ and a sequence $(w_n)_{n\in \mathbb{N}} \in \ell^{\infty}$ such that $Te_n= w_n e_{n+1}$.

Q: Write $\|T \|$ in terms of the sequence of weights $(w_n)_{n\in \mathbb{N}}$ for $T$.

Attempt:

So we have $\|T e_n\|= \| w_n e_{n+1} \|$.

$\|Te_n \|\leq | w_n| \| e_{n+1} \|$

i.e. $\|T \|\leq | w_n| $

On the other hand,

$\|T \|= \sup\{|Tx| : \| x\|=1 \} \geq |Te_n| = |w_n e_{n+1}| = |w_n| $

Thus, $\|T\| = |w_n|$.

Could you let me know if my work is correct?

Thank you!

答: You are right about your second inequality $||T|| \geq |w_n|$, but don't forget that it's true for all $n \in \mathbb{N}$. In the first part of your proof you have a logic mistake -- the equality $||T e_n|| = |w_n|||e_{n+1}||$ (and it's indeed an equality) implies only that $||T|| \geq |w_n|$ :) yes, again, because the norm of an operator is such number that the inequality $||Tx|| \leq ||T||||x||$ holds for all $x \in H$. Now that you now the answer from a comment, you should be able to prove that the last property of the norm is satisfied for it. Combine this with your second inequality which is true for all $n \in \mathbb{N}$ and you'll get the result.


问:
Kernel preshef is a (pre)sheaf
I'm not the first one asking for clarifications on the subject.
But my question is slightly different, and possibly easier to answer.

Suppose $\varphi\colon \mathcal{F}\to \mathcal{G}$ be a morphism of sheaves on a topological space $X$, so that $V\subseteq U$ (inclusion of open subsets) implies $\varphi(V)\circ \rho^\mathcal{F}_{UV}=\rho^\mathcal{G}_{UV}\circ \varphi(U)$, where  $\rho^\mathcal{F}_{UV}\colon \mathcal{F}(U)\to\mathcal{F}(V)$, $\rho^\mathcal{G}_{UV}\colon \mathcal{G}(U)\to\mathcal{G}(V)$ restriction maps.
The kernel presheaf of $\varphi$ works as follows: $U\mapsto \mathrm{Ker}(\varphi(U))\subseteq \mathcal{F}(U)$.
This means that $V\subseteq U$ implies $\rho^\mathcal{F}_{UV}\colon \mathrm{Ker}(\varphi(U))\to \mathrm{Ker}(\varphi(V))$, restriction to the kernel of the above restriction map.

So I would like to prove that $\rho^\mathcal{F}_{UV}(\mathrm{Ker}(\varphi(U)))=\mathrm{Ker}(\varphi(V))$.
I easily proved $(\subseteq)$, but I have no idea for the converse $(\supseteq)$.
Can someone help me, please?

答: it's not true unless $\mathcal{F}$ is special since it's not you are just free not to verify

Title: Kernel preshef is a (pre)sheaf


问:
Residue of $f(z) = \frac{2z+3e^z}{(e^z-1)^2}$ at $z=0$
I tried using finding $\displaystyle{\lim_{z \to 0}\frac{2z+3e^z}{(e^z-1)^2}}$ and $\displaystyle{\lim_{z \to 0}\frac{d}{dz} z^2\frac{2z+3e^z}{(e^z-1)^2}}$,

But neither of the limits seem to work out to a value. Is there any better way to find the residue?

答: Let$$f(z)=\begin{cases}\left(\frac{e^z-1}z\right)^2&\text{ if }z\ne0\\1&\text{ if }z=0.\end{cases}$$Then $f$ is analytic in the neighborhood of $0$ and the function whose residue you want to compute is $\frac{2z+3e^z}{z^2f(z)}$. If you define$$g(z)=\frac{2z+3e^z}{f(z)},$$then $g$ is an analytic function in the neighborhood of $0$ and so\begin{align}\operatorname{res}_{z=0}\left(\frac{2z+3e^z}{(e^z-1)^2}\right)&=\operatorname{res}_{z=0}\left(\frac{g(z)}{z^2}\right)\\&=g'(0)\\&=\frac{5f(0)-3f'(0)}{f^2(0)}\\&=2,\end{align}since\begin{align}f(z)&=\left(1+\frac z2+\frac{z^2}{3!}+\cdots\right)^2\\&=1+z+\frac7{12}z^2+\cdots,\end{align}which implies that $f(0)=f'(0)=1$.


问:
using diagonals of the a trapezoid inside a triangle you can find the midpoint of the bases
In the following triangle, AB is parallel to DE. F is an intersection point of the segments BD, AE and CG. Is it true that AG = GB?

答: Yes, use Ceva Theorem: $${AG\over GB}\cdot {BE\over EC}\cdot {CD\over DA}=1$$ and Thales theorem: $${CD\over DA} = {EC\over BE} $$ and thus a conclusion.


问:
Inconvenience about definition of the 1 to 1 function
I read a from my discrete mathematic book (Discrete Mathematics and Its Application),
 Definition of the 1 to 1 function is

$( ∀a,∀b( f(a) = f (b) → a = b)$
 the problem of this statement is

if $f(a) \neq\ f(b)$   and $a = b$ it means that same numbers have different images and

it violates the definition of the function.

I think my definition would be more correct. which is

$f(A):$ $f(a) = f(b)$  $ \iff  a = b $

Can you explain to me why I am correct or incorrect?

答:
Your alternate definition is correct.

The usual definition is (as you said)
$$
\forall a, \forall b (f(a)=f(b) \Rightarrow a=b)
\tag1$$
The converse
$$
\forall a, \forall b (f(a)=f(b) \Leftarrow a=b)
\tag2$$
is true for every function.

Your alternative definition
$$
\forall a, \forall b (f(a)=f(b) \Leftrightarrow a=b)
\tag3$$
is true if and only if both $(1)$ and $(2)$ are true.  If we know in advance that $f$ is a function, this means that $(1)$ and $(3)$ are either both true or both false.

Both definitions are correct.  So we cannot say one is "more correct" than the other.  If we know in advance that $f$ is a function, then $(1)$ is easier to prove than $(3)$, which may explain why $(1)$ is usually given as the definition.


问:
Compactness of the whole space by compactness of subspaces.
Take double arrow space $[0,1]\times \{1,2\}$

Where topology is dictionary topology defined by linear order as $(a,b)<(c,d)$ iff $a<c$; or ;$a=c \;\&\; b<d$

I want to show that since $[0,1]$ and $\{1,2\}$ are compact sets in linear order topology themselves, then their product is compact (because finite product of compact sets is compact).

**Question:**First of all "because finite product of compact sets is compact" part is confusing me because we need to talk about some common topology.

And to show the above space is compact can we do some kind of trick and how to become sure the above idea is wrong?

答:
The topology on Double Arrow $X$ is more "subtle" than the product topology on $[0,1]\times \{1,2\}$. The topology is induced by the order $<$ that you defined, so e.g. a set like $[ (0,1), (0,2)\rangle$ is open as $(0,1)$ is clearly the minimum of $X$ under $<$ and $(0,1) < (0,2)$, but note there is no point inbetween them so that this basic open set of the order topology reduces to $\{(0,1)\}$. E.g. a basic neighbourhood of $(0,2)$ is an open interval $\langle a,b\rangle$ where $a,b \in X$ with $a < (0,2)$ and $(0,2) < b$. This forces $a=(0,1)$ here and e.g. for $b$ of the form $(c,1) \in X$, with $0< c \le 1$, we get that this interval then becomes the set $\langle 0,c \rangle \times \{1\} \cup [0,c\rangle \times \{2\}$ (check this!), where the first coordinate sets are intervals in the usual order of $[0,1]$, of course. Try to visualise some more intervals with endpoints of the form $(c,2)$ as well.

It turns out that there is no inclusion relation between the product topology and  this order topology, so we cannot say anything about the compactness of $(X,<)$ from the compactness of $[0,1] \times \{1,2\}$ as a product.

It does turn out that the order topology too is compact. It's also separable, but not second countable. The example has the Sorgenfrey line as a subspace, e.g. It's $T_5$ (completely/hereditarily normal and Hausdorff) as any ordered space is. The proof that double arrow is compact is easily found online; also on this site.


问:
On a problem of connectedness in $\mathbb{R}^{n}$
I'm trying to solve a problem from the book "Foundations of real and abstract analysis" by Douglas Bridges. I'm asking here because I think a have a solution to the excersice but I'm not quite sure about it. Here is the problem:

Let $S$ be a connected subset of the Euclidean space $\mathbb{R}^{n}$ . Prove that for each $r>0$ the set

$$T=\{x\in\mathbb{R}^{n}: \rho(x,S)\leq r\}$$
is also connected.

Here, we define $\rho(x,S)=\inf\{\rho(x,s):s\in S\}$ is the distance from the point x to the set S.

Here is my attempt: Suppose, for the sake of getting a contradiction, that there is $r>0$ for which $S$ is not connected, then there are disjoint open sets $C,D\subset\mathbb{R}^{n}$ with $S=(S\cap C)\cup( S\cap D)$ (everything is with the subspace topology) and a continuous function of $S$ onto $\{0,1\}$ given by:

$$f(x)=\left\{
\begin{array}{cl}
1&\mbox{if } x\in S\cap C\\
0&\mbox{if } x\in S\cap D
\end{array}\right.$$

Note that $S\subset A$ (the set $A$ is like a band around the set $S$ with distance $r$). Also, the restriction $f|_{S}$ is continous with respect to the subspace topology and

$$f|_{S}(x)=\left\{
\begin{array}{cl}
1&\mbox{if } x\in T\cap S\cap C\\
0&\mbox{if } x\in T\cap S\cap D
\end{array}\right.$$
is a continous function of $S$ onto $\{0,1\}$, but this is a contradiction since $S$ is connected, hence $A$ is connected.

Is my argument correct?

Thanks

答:
Suppose that $T=\{x: d(x,S)<r\}$ is not connected and consider a surjective continuous function $f:T\rightarrow \{0,1\}$. Let $x,y\in T$ such that $f(x)=0, f(y)=1$.

There exists $a\in S$ such that $\|x-a\|<r$, consider $a_t= a+t(x-a), t\leq 1$, $\|a_t-a\|<r$ implies that $a_t\in T$, we have $f(a_t)=f(a)=f(x)=0=f(a)$ since $\{a+t(x-a)\}$ is connected.

Similarly, let $b\in S$ with $d(b,y)<r$, write $b_t=b+t(y-b)$, $b_t\in T$ and $f(b_t)=f(b)=1$.

Since the restriction of $f$ to $S$ is continuous and $S$ is connected, it is constant contradiction since $f(a)=0$ and $f(b)=1$.


问:
Growth limit of $\log(n^2+1)$
While following up a solution to find the growth limit of $(n!+2^n)(n^3+\log(n^2+1))$:

$$(n!+2^n)(n^3+\log(n^2+1))\le n^3n!+n^32^n+n!\log(n^2+1)+2^n\log(n^2+1)$$

when $n\gt3$, we have $\log(n^2+1)\le n$ and $2^n \le n!$,

$$(n!+2^n)(n^3+\log(n^2+1))\le n^3n!+n^3n!+n!(n^2+1)+n!(n^2+1)$$

Without moving on further, I think we should have based on $\log(n^2+1)\le n$ a $n$ in place of $\log(n^2+1)$, but we have $n^2+1$ in place of $\log(n^2+1)$, do you know the reason or do you think this is right?

$$(n!+2^n)(n^3+\log(n^2+1))\le n^3n!+n^3n!+n!(n^2+1)+n!(n^2+1).$$

答:
Obviously it's wrong to say that $O(\log(n^2+1)) = O(n^2+1),\ $ because $O(\log(a)) \neq O(a).$

Furthermore, $O(\log(n^2+1)) \neq O(n).\ $ In fact, $\ O(\log(n^2+1)) = O(\log(n))\ $ because for large enough $n:\ 2\log(n) = \log(n^2) < \log(n^2+1) < 3\log(n)$.

But I think the second sentence of Gary's comment cuts to the heart of your question:

"You can bound the $\log$ by $n$ if you like, but it does not change the fact that the whole thing is $O(n^3n!)=O((n+3)!)$."

Title: Growth limit of $\log(n^2+1)$


问:
relation of medians of two way partition
Given a set of finite real numbers $X \subset \mathbb{R}$, one can obtain the median using $M: X \to \mathbb{R}$.

And $X$ can be partitioned into two subsets $A_i, B_i \subset X, A_i \cup B_i = X , A_i \cap B_i = \emptyset, A_i \neq \emptyset, B_i \neq \emptyset$, without losing generulaity we arrange the subsets such that $M(A_i) \leq M(B_i)$.

For any two different partition $(A_1, B_1)$ and $(A_2, B_2)$,
does such two partitions exists such that $M(A_1) \lt M(A_2)$ and $M(B_1) \lt M(B_2)$ ?

Body:
$X = \{ 1, 3, 4, 5, 10\}$

$A_1 = \{1,5\}$, $M(A_1)=3$

$B_1=\{3,4,10\}$, $M(B_1)=4$

$A_2=\{1,3,4,5\}$, $M(A_2)=3.5$

$B_2=\{10\}$, $M(B_2)=10$

satisfy these inequalities.


问:
Appending basis elements to a basis in topology.
I know in linear algebra, once we have a set of basis vectors, adding any vector to the basis causes the set to no longer be a basis, because of the loss of linear independence. However in topology, once there is a basis for a topology, can we always append another open set to the basis, and have it still be a basis?

For example the collection of open intervals $(a,b), a<b, a,b \in \mathbb{R}$ is a basis for the standard topology on $\mathbb{R}$. However,if we add any open set to this collection, does this new subset of elements from the topology still form a basis?

Would the collection of all open intervals as well as $\mathbb{R}$ itself be a basis?

Would the collection of all open intervals as well as all open rays still be a basis?

What about the collection of open intervals, along with the empty set?

I would presume attaching open sets to any basis does not disturb the collection from being a basis, because of the definition of a basis for a specific topology.

答:
Yes, we can add any number of open sets to a base for $(X,\tau)$ and we still have a base; we don't have to use these extra sets, nor is there any notion of independence: it's actually rather special when we can even find a base that is minimal, i.e. where we cannot leave out any set and still have a base (this happens for finite spaces and discerte spaces e.g.) but usually there will be more than one way to write an open set as a union of basic open sets,a nd that's never an issue. So the analogy with vector space bases is quite weak.

All the examples you mention are indeed other bases for $\Bbb R$. Note that bases don't often include the empty set as it does not contribute to any union, and we can always leave it out anyway (or add it too, as I said).


问:
Why defining homotopy on functions instead of spaces?
Apparently, the definition of homotopy formalizes the idea of continuous transformation between two things.

(*) Let's take this motivating example: I have $ S,S' \subset \mathbb{R}^3 $ two surfaces in space. If I wanted to define a continuous transformation between them I would request for a continuous function $\Gamma:S\times [0,1] \rightarrow \mathbb{R}^3$ to exist such that $\Gamma(\cdot,0)=id|_S $ and $Im(\Gamma(\cdot,1))=S' $.

Instead, the definition in use is not between spaces ($S$ and $S'$) but between functions. I think my definition (*) would be a special case of the definition in use.
In fact, by posing $f:=\Gamma(\cdot,0)=id|_S$ and $f':=\Gamma(\cdot,1)=id|_{S'}$ I have that $f\simeq f'$.

Can someone clarify what's the limitations (or what's wrong with) my definition (*)?

答:
I think this is a very nice question, and your definition (*) is very close to the definition of an isotopy. You want to also require that $\Gamma(S,t)$ is an embedding for all $t$ so that you can't collapse everything to a point (as was pointed out in the comments). Otherwise, any two maps would be equivalent.

I think it's useful as well to compare isotopy to the notion of homotopy equivalence. As an example the circle is homotopy equivalent to itself via the identity map, but it has many distinct (up to isotopy) embeddings in $\mathbb{R}^3$. Specifically, the unknot and trefoil can be shown to be non-isotopic using various knot invariants.


问:
Limit of $\log(n!)/\log((n+1)!)$?
I want to show that $$\lim_{n\to\infty}\frac{\log(n!)}{\log((n+1)!)}=1.$$
Obviously, $$\frac{\log(n!)}{\log((n+1)!)}=\frac{\log(n!)}{\log(n+1)+\log(n!)}\leq1.$$
But I get stuck with the other estimate. It looks very similar to other problems prior to this, but I cannot derive a converging lower bound. I am open to fundamental proofs and proofs using Stirling's formula.

答: How about $$\displaystyle\lim_{n \to \infty} \frac{\log(n!)}{\log((n + 1)!)} = \lim_{n \to \infty} \frac{\log(n!)}{\log(n + 1) + \log(n!)} = \lim_{n \to \infty} \left( \displaystyle 1 - \frac{\log(n + 1)}{\log(n + 1) + \log(n!)} \right) = 1$$

Title: Limit of $\log(n!)/\log((n+1)!)$?


问:
Topics to Study and Books to Read?
I'm an undergraduate studying materials science and engineering with a concentration in polymer science. I would like to go to graduate school and focus on theory and computation of synthetic polymer and biopolymer systems. So I'm planning of studying things such as Hamiltonian mechanics, statistical mechanics, molecular dynamics and monte carlo simulations.

I feel like college doesn't teach math that good. I want to study more topics and go over topics I studied in classes before and hopefully build my mathematical intuition. The areas I was thinking about were: single and multivariable calculus, linear algebra, ODEs and PDEs and statistics and probability. I also saw that some researchers use topology and graph theory to study protein structures. Are there any other areas I should include?

Besides what area to study, my other question is which books to study in a particular area? I took Courant's book "Introduction to Calculus and Analysis" out of the library today.  I'm trying to find the "best" or most recommended books because I would like to go through all the topics this summer, thus I don't have a ton of time. Good thing is a lot of it will be things I already know but just analyzed in more detail.

Would I be better off reading textbooks or doing MIT OpenCourseWare courses? I appreciate any responses or recommendations.

答:
Do read Apostol's calculus books , both volumes . They are probably one of the best calculus book available . They include almost all topics in details and with rigour and applications , you've listed . Just go for those 2 books by TMA , you won't need any other math book . For graph theory , you can pick up Diestel , it is available for free on internet .

And you can never know all the math , you'll require for these topics. Just read these books and work out their exercises . And then if you need to learn anything else , then you will have to pick it up along the way from internet or buy another book . I don't think , any one can give you a complete list of all the math books you should read . However , I am sure after Apostol's books , you are good to go . They will teach linear algebra, calculus,vector calculus and differential equations and probability and statistics as well , also you get an introduction to modern mathematics .

You will have to balance between both MIT OCW and books , like there is a separate OCW course called advanced calculus available that specifically uses apostol's books . You can do that , Nothing is ever learnt in all depth from just  a single source .

Title: Topics to Study and Books to Read?


Title: Limit of $\log(n!)/\log((n+1)!)$?

Body:
I want to show that $$\lim_{n\to\infty}\frac{\log(n!)}{\log((n+1)!)}=1.$$
Obviously, $$\frac{\log(n!)}{\log((n+1)!)}=\frac{\log(n!)}{\log(n+1)+\log(n!)}\leq1.$$
But I get stuck with the other estimate. It looks very similar to other problems prior to this, but I cannot derive a converging lower bound. I am open to fundamental proofs and proofs using Stirling's formula.

Body: Hint $\log(n!)=\log(\frac{(n+1)!}{n+1})=\log((n+1)!)-\log(n+1)$


答:
HINT add and subtract $\log(n+1)$ to the numerator, to get something like:
$$\frac{\log(n!)+\log(n+1)-\log(n+1)}{\log((n+1)!)}=1-\frac{\log(n+1)}{\log((n+1)!)}$$

Title: Limit of $\log(n!)/\log((n+1)!)$?


答:
What you try to prove is equivalent to:

$\lim_{n\to\infty}\frac{\log((n+1)!)}{\log(n!)}=1.$

Substracting $1$ at both sides, that´s the same as:

$\lim_{n\to\infty}\frac{\log(n+1)}{\log(n!)}=0$.

But for a given $k>0$, there is some $N$ such that for any $n>N$, you have $(n+1)^k<n!$. For example, you can pick $N=3k$, and then for $n>N$, $(n+1)^k<(2(n-k))^k<2^k(n-k)^k<2\cdot3\dots\cdot k\cdot (n-k)\cdot\dots\cdot (n-1)n<n!$.

So for every $k$ there is $N$ such that for each $n>N$, $\frac{\log(n+1)}{\log(n!)}<\frac{1}{k}$. Thus, $\frac{\log(n+1)}{\log(n!)}$ converges to $0$, as we were trying to prove.

Title: Limit of $\log(n!)/\log((n+1)!)$?


问:
Digit Sums: A Math Project
I'm doing a Math project on Digits Sums, and I don't really want to try to figure out all the 4-digit numbers that add up to 2, 3, 4, 5, and 6. That's partly because it would be hard, and also, because I don't want to get anything wrong, or it would throw off my entire project... Can anyone help me?</p>

答:
Let me make sure I understand the question. Is this what you mean by "all the 4-digit numbers that add up to 6"?

6000;
1005 1050 1500 5001 5010 5100;
2004 2040 2400 4002 4020 4200;
1014 1041 1104 1140 1401 1410 4011 4101 4110;
3003 3030 3300;
1023 1032 1203 1230 1302 1320 2013 2031 2103 2130 2301 2310 3012 3021 3102 3120 3201 3210;
1113 1131 1311 3111;
2022 2202 2220;
1122 1212 1221 2112 2121 2211

That was a tedious couple of minutes, but I wouldn't go so far as to call it "sheer torture". Did I miss any?

Edit: OOPS! I don't know if I missed any, but towards the end I forgot that they had to be 4-digit numbers! Edited out the 5- and 6-digit numbers at the end.

Title: Digit Sums: A Math Project


问:
determine analytic function $f$ if $|f|=(x^2+y^2)e^x$
I'm self-studying complex analysis. Could someone please help me get this problem started?

I know an analytic function will satisfy the Cauchy-Riemann equations, but I only know the modulus of the function. I have no idea how to proceed

Body: Hint: $x^2 + y^2 = |z^2|$, and $e^{x} = |e^z|$, where $z = x+iy$.


问:
covariance of two random variables when one is related to the other

Let $X$ be a six sided die and $B$ is a Bernoulli random variable with
mean $p$. If $B=0$, then $Y=X-1$. If $B=1$, then $Y=X+1$. What is the
covariance of $X$ and $Y$?

I know $cov(X,Y) = E[XY] - E[X]E[Y]$ but I don't understand how to compute $E[XY]$ in this case. I tried to use iterated expectations $E[XY] = E[Y E[X \mid Y] ]$ but I can't see where that is going.

I can find the mean, variance of $X$ and $Y$. Any help will be much appreciated.

答:
$$E[XY] = E[E[XY \mid B]] = p E[XY \mid B=1] + (1-p) E[XY \mid B=0]$$

When $B=1$, you know $Y=X-1$, so $E[XY \mid B=1] = E[X(X-1) \mid B=1]$. If $X$ and $B$ are independent (I guess this needs to be assumed), then this equals $E[X(X-1)]$. The other term $E[XY \mid B=0]$ can be handled similarly.

In the end, we'd have $$E[XY]=pE[X(X+1)] + (1-p)E[X(X-1)] = E[X^2] - (1-2p) E[X]$$ and $E[Y] = E[E[Y \mid B]] = pE[X+1] + (1-p)E[X-1] = E[X] - (1-2p)$
and thus
$$\text{Cov}(X,Y) = E[X^2] - (1-2p)E[X] - E[X](E[X] -( 1-2p)) = \text{Var}(X)$$
as Matthew Pilling obtained with his much slicker approach.


问:
Calculate $f\left(x\right)=cxe^{-\frac{x^{2}}{2}}$ with $x>0$ (a) find c, (b)the mean of X, (c) the variance of X
Calculate $f\left(x\right)=cxe^{-\frac{x^{2}}{2}}$ with
$x>0$ (a) find c, (b)the mean of X, (c) the variance of X

(a)$$E\left[X\right]=\int_{0}^{+\infty}x\cdot cxe^{-\frac{x^{2}}{2}}dx=c\int_{0}^{+\infty}x^{2}\cdot e^{-\frac{x^{2}}{2}}dx$$
Integrating by parts
$$\begin{array}{cc}
u=x^{2} & du=2xdx\\
dv=e^{-\frac{x^{2}}{2}}dx & v=-e^{-\frac{x^{2}}{2}}
\end{array}$$
$$E\left[X\right]=c\left(\left(x^{2}\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+2\int_{0}^{\infty}xe^{-\frac{x^{2}}{2}}dx\right)$$
Integrating by parts again
$$\begin{array}{cc}
u=x & du=dx\\
dv=e^{-\frac{x^{2}}{2}}dx & v=-e^{-\frac{x^{2}}{2}}
\end{array}$$
$$E\left[X\right]=c\left(\left(x^{2}\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+2\left[\left(x\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}dx\right]\right)$$

Still not sure if this is the right procedure.
(b) Missing
(c) Missing

Body:
(a)

$f(x)$ is a valid pdf.

Therefore $$\int_0^\infty f(x) dx = c\int_0^\infty xe^{-\frac{x^2}{2}} dx = 1$$

Since  $$\int_0^\infty xe^{-\frac{x^2}{2}} dx = 1$$ therefore $c = 1$

(b)

$$E(X) = \int_0^\infty xf(x) dx = \int_0^\infty x^2 e^{-\frac{x^2}{2}} dx = \sqrt{\frac{\pi}{2}}$$

(c)
$$Var(X) = E(X^2) - (E(X))^2$$

$$E(X^2) = \int_0^\infty x^2 f(x) dx = \int_0^\infty x^3 e^{-\frac{x^2}{2}} dx = 2$$

Hence
$$Var(X) = 2 - \frac{\pi}{2} = \frac{4 - \pi}{2}$$


问:
Irregularity of $a^ib^jc^k∣\text{ if }i=1\text{ then }j=k$ using pumping lemma
This question has been asked here several times. However none of the examples I saw actually tried to prove it's irregularity using the standard pumping lemma.

I know how to prove irregularity using the specific case of $a^1b^pc^p$ however I'm struggling to build the pumping lemma for the whole language $a^ib^jc^k$
I know this won't prove the language to be regular, but I'm more interested in how to deal with lemmas with multiple variables on the exponents.
Let's suppose I've divided the string into the following:

$ x = \epsilon$

$ y = b^\beta$

$ z = b^{p-\beta}c^m$

Let $n = 2$. I get

$xy^2z= b^{p-\beta}c^n$

But this doesn't seem like I can find a pumping length for this language as it has too many variables. Is there a way to make this pumping lemma work? (I know it can't be used to prove the language regular, but I'm trying to find at least one pumping length which works for this language)

答:
The pumping lemma states that if $L$ is regular, there is a constant $N$ so that for all $\sigma \in L$ with $\lvert \sigma \rvert \ge N$ there is a division $\sigma = \alpha \beta \gamma$ with $\lvert \alpha \beta \rvert \le N$ and $\beta \ne \varepsilon$ for all $k$ it is $\alpha \beta^k \gamma \in L$.

Sadly, in this case if you pick e.g. $\sigma = a b^N c^N$, you can always pick $\alpha = \varepsilon$, $\beta = a$ and $\gamma = b^N c^N$, and everything works out beautifully. No contradiction to be had.

If you check out the proof of the pumping lemma, it is clear that you can prove variants that for example state $\lvert \beta \gamma \rvert \le N$, and $\beta$ can be pumped. Use that one to get a contradiction.

(Nice question, by the way.)


问:
Which Stats Text Is More Math-Intensive?
I have tried many stats textbooks and none of them really works for me.  The issue is that there are too many formulas and too little proofs or derivations.  Some of these formulas are really technical (ugly) and you can't get the feel on what's happening without seeing the derivation.

To give a flavor of what I'm looking for, I will point to Chapter 4 in Wasserman's All of Statistics.  In it, he derived the inequalities given in the chapter.  So that's good; you know where the numbers came from.

答: I will first recommend to you the textbook from Casella, Statistical Inference which is a very classical and outstanding textbook in graduate-level statistics. Since it almost covers all general materials about mathematical statistics, it provides lots of beautiful math proof inside. Secondly, I will recommend the book by Keener, Theoretical Statistics: Topics for a Core Course, which is based on the measure theory so it will much deeper than the first one and is involved lots of math.


问:
Given a circle $\{(x,y) \in \mathbb{R^2} : x^2 + y^2 = 1\}$, show that taking away points from the circumference gives us a disconnected space.
Given a circle $C = \{(x,y) \in \mathbb{R^2} : x^2 + y^2 = 1\}$, how do I show that that taking away points from the circumference gives us a disconnected space?

If we take away one point, it seems to still remain connected. If more than one point is scraped off, say $a, b \in C$ such that $a \neq b$, then the parts of the two sectors that form the subsets of $C$ will likely form the two disjoint non empty open sets. But how exactly are they open? How is open-ness defined for this case? We are probably not using the Euclidean metric, although we can. Maybe we can define a metric $d(r,s)$ that considers the distance from $r$ to $s$ in the clockwise direction. But in general, how do we know that it is open for any metric we consider?

答:
The circle minus a single point is homeomorphic to an open interval $(0,1)$ in $\Bbb R$ (or equivalently, $\Bbb R$ itself). So removing $n\ge 2$ points from the circle is the same topologically as removing $n-1$ points from $\Bbb R$, which gives us a space with $n$ components (open intervals and open segments), thus disconnected.

To see the first homeomorphism, note that e.g. $C\setminus\{(0,1)\}$ is homeomorphic to $(0,1)\subseteq \Bbb R$ via $t \to (\cos 2\pi t,\sin 2\pi t) \in C$ etc.


问:
Find all units, zero divisors, nilpotent elements and idempotent elements in $\Bbb Z[x]/(x^2-1)$
I have no clue on how to do unit elements (inversible), zero divisors (might not exist),  I know only for stuff like $\Bbb Z_n$, and I guessed that nilpotents you get from $(ax+b)^n=0$ and we have $a=0, b=0$ so only $0$ is nilpotent (like $b^n=0$ so b=0 and $a^nx^n=0$ so $a=0$ as well) and for idempotents, you just solve $(ax+b)^2=ax+b$ and you find $a,b$ which have to be something like $0$ and $1$?

Is this true for any case, like what has my function, $x^2-1$ have to do with any of the nilpotent/idempotent stuff?

Maybe I need like $x=\pm 1$, from the function and I put in the equation for idempotents that $x^2=1$? So this is literally the only information I have in all seminars, tutorials etc. I can't find anything related to this stuff, also I'm very bad at math, simple stuff is good.

答:
You just need to do the same, but instead of $(ax+b)^n$ is zero, you have $(ax+b)^n$ is a multiple of $x^2-1$ (that's an example for nilpotent elements). A polynomial $f$ is a multiple of $x^2-1$ iff $f(1)=f(-1)=0$. Using that, you can easily observe that we have no non-zero nilpotents: $(a+b)^n=0$ and $(-a+b)^n=0$ means $a=b=0$.

For other cases, you'll get similar equations, but you must remember that $x^2=1$ in your ring, so all elements can be reduced to linear functions. For example, in this ring

$$ (ax+b)^2 = a^2x^2 + 2abx + b^2 = 2abx + a^2+b^2, $$

so your idempotents have to satisfy $a=2ab$ and $b=a^2+b^2$. I leave details of the rest of the problem for you.


问:
Variables of Calculus
I've been learning calculus for a week and I recently felt overwelmed about the amount of variables we use on derivatives and integrals. Can someone give me an overview about what the variables $x$, $d$, $dt$ and $h$ means? And if I've missed some very important ones for derivatives or intergrals please don't hesitate to point them out to me in the answers.

Thanks,
Alex

答:
That's a pretty broad question, but here's a general answer:

You can usually think of $x$ as the "input" variable. Often you will see functions like
$$f(x) = x^2+1.$$
Here $x$ is the input, and $f(x)$ is the output. The choice of "$x$" is just a convention, though; you could just as well write
$$f(q) = q^2+1.$$
In either of these cases, the relationship is that "the output is equal to the input squared plus one".

$d$ by itself doesn't mean anything. If you see something like $dy/dx$, this indicates the (instantaneous) rate of change of $y$ with respect to $x$. This can be approximated by $\Delta y / \Delta x$, where $\Delta x$ and $\Delta y$ are the change, or difference (hence the letter $d$) in $x$ and $y$, respectively.

$dt$: this is the same as for $dx$ and $dy$ (see the previous note) but with respect to $t$ as opposed to $x$ or $y$.

$h$ is often used to represent a small increase in some variable. In that sense, you could just as well write it $\Delta x$.

Here's an example that combines some of these elements: if $y = x^2 + 1$, then
$$\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{\left((x+\Delta x)^2+1\right) - (x^2+1)}{\Delta x},$$
or we could write exactly the same thing using $h$ instead of $\Delta x$:
$$\frac{dy}{dx} = \lim_{h \to 0}\frac{\left((x+h)^2+1\right) - (x^2+1)}h$$

Title: Variables of Calculus