问:
Smallest $k$ Such that $13 + 4 \cdot k \cdot p^2$ is a Perfect Odd Square
Given a prime number $p$, I am looking to find the smallest positive integer $k$ such that the following equation $$13 + 4 \cdot k \cdot p^2$$ produces a perfect odd square. All variables are integers. For example, for the prime $43$, $k = 3$. For $p=103$ , it turns out that $k = 1391$. A computer program can solve this for small prime numbers. It is easy to prove that $k$ has to be odd too, which improves the search. But for larger primes, say $p>10^4$, the naive approach of incrementing $k$ untill a suitable value is found just takes a long time.

It is important to mention that not all primes have any solution at all. For those which do have a solution, I am interested in an efficient way for finding it.

Is there any other approach to tackle this? Perhaps one that relates to number theory? Or any other field really which may prove useful.

答:
There is one major optimization screaming at me here.

Check each square sequentially for whether or not it is the “odd square” that the formula equals. This will be faster because $n^2$ (for odd $n$) grows faster than the current linear formula dependent on $k$.

Of course you would start with the first square greater than $13 + 4p^2$ since any lower square is impossible.

This method will be faster when $\frac {n^2}{4p^2} > n - \sqrt{4p^2} = n - 2p$.

I do not know whether or not this equation ever works out to being true. However, for sufficiently large $p$ I strongly suspect that iterating through the squares will be faster.

One may note that my formula assumes that every multiple of $p^2$ needs to be as well as every $n^2$. This cancels out as I would divide both sides by $2$. Therefore it is irrelevant.

EDIT:

I thought about this a bit more. For sufficiently small $k$ iterating through squares will be slower (because the growth rate of sequential squares will be smaller than the growth of sequential multiples of $4p^2$). Once $k > 2p^2 - 1$ the growth of sequential squares outpaces the linear growth of your formula. Therefore you should add something in your code to start counting by squares once you reach $k = 2p^2 - 2$. The value of $n$ to start iterating squares would then be $n = 2p^2 - 1$.

This should be about as fast you can get (assuming $k$ exists) other than iterating through odd values of $k$ and $n$.


问:
Calculate center of mass using integration
How to calculate center of gravity having figure described by those equations with integration? This can and should be solved by numerical analysis.

$y < -x + 1\\ y > x^2 - 1\\ y < \sin(x)$

Firstly I found bounding points for integration:

$-x + 1 = \sin(x) \\ x_0 = -0.63673265$

$sin(x) = -x + 1 \\ x_1 = 0.51097343$

$-x + 1 = x^2 - 1 \\ x_2 = 1$

Then I've calculated integral of $g(x)$ from $x_0$ to $x_1$:

$\int_{x_0}^{x_2} |x^2 - 1| dx = 1.2173486603176453$

Then integral of $h(x)$:

$\int_{x_0}^{x_1} |\sin(x)| dx = 0.32334348721908174$

Then integral of $f(x)$:

$\int_{x_1}^{x_2} |-x + 1| dx = 0.11957349302898969$

So the area between is equal to
$$\int_{x_0}^{x_1} |\sin(x)| dx + \int_{x_1}^{x_2} |-x + 1| dx - \int_{x_0}^{x_2} |x^2 - 1| dx = -0.7744316800695739$$

Which is something wrong I guess.

答:
The lower border of the box around your graph is at about $y=-1.5$, but you're treating it like the $x$-axis.  Your first integral has little to do with your shaded area.  Better to

$$\int_{x_0}^{x_1}  \sin(x) - (x^2-1) \; dx + \int_{x_1}^{x_2}  -x+1 - (x^2-1) \; dx.$$

ContentLicense: CC BY-SA 4.0


问:
How to draw Root locus for two gains
In the book I'm reading, one of the exercises asks to draw the root locus for this closed-loop  system
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + K_2 s + K_1}
$$

I never encounter such a problem for two gains. I usually do it for a single gain where I can change the gain and observe how the poles of the closed-loop system move in the s-plane. Any suggestions for handling this?

答:
Because we want to calculate our transfer function in the complex plane modulus and phases are sufficient to trace it.

Let $$H : s \to \dfrac{1}{s^2+K_1s+K_2}$$

Setting $s=j \omega$

$$ H(jw)=\dfrac{1}{(K_2-\omega^2)+j\omega K_1}$$
which is by multiplying by conjugate quantity

$$ H(jw)=\dfrac{(K_2-\omega^2)-j\omega K_1}{(K_2-\omega^2)^2+(\omega K_1)^2}$$

Now for gain:

$$ |H(j\omega)|=\dfrac{1}{\sqrt{(K_2-\omega^2)^2+(\omega K_1)^2}}\\H(0)=1 \\ |H(j\omega)_{\omega \to \infty}|=0$$

Now for phase:

$$ Arg(H(j\omega))= \arctan\left(\dfrac{-\omega K_1}{K_2-\omega^2}\right) +\mathbb{1}_{\omega \geq \sqrt{K2}}(\omega)\pi$$

So here to gain many time it is fundamental to take time to study the phase of $H$.

Variations are as follow (phases must be continous)

$$ \text{On} \ [0,\sqrt{K2}] \ , \text{decreasing}\\
    \text{At} \ \sqrt{K_2} \  \text{we get a phase of} \ -\pi/2 \mod \pi\\\text{Then decreasing to } -\pi \mod \pi $$

So you will get this form :

The modulus at intersection with imaginary axis will be $\frac{1}{K_1\sqrt{K_2}}$.

So the diagramm consists in fixing $K_1$ and trace your diagramm in function and making it for different value of $K_1$ as it is made on the picture ( don't take in account the letter $z$)

Title: How to draw Root locus for two gains


问:
Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$
What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.

Body:
$12(8^x+27^x)=14(12^x+18^x)$

$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$

Divide by$12^x$

$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$

$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$

We put :$t=(\frac{3}{2}) ^x$

$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$

Multiply by $t$

$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$

We can see $-1$ is a solution of the equation

So:after division by $t+1$ we see

$ t^2 - \frac{13}{6}x+1=0$

$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $

So :

$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $

$\Rightarrow $ $x_1=1$

And

$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $

$\Rightarrow $ $x_2=-1$

Finally

$x=-1$ or $ x= 1$


问:
Using $\gcd(a,b)$ to find gcd of other values $\gcd(a^2,b)$ and $\gcd(a^3,b)$
If $\gcd(a,b) = p$, a prime, what are the possible values of $\gcd(a^2,b)$ and $\gcd(a^3,b)$?

Through examples I've been able to find the answer, but I don't know how to come up with a proof.

Update: I think gcd of either is $p^2$, but again this is just by working examples.

I'm not familiar with much number theory. I've just started the second section of my textbook, so all I know is just divisibility and primes.

答:
Suppose $\gcd(a,b)=p$ is prime.  Then both $a$ and $b$ are divisible by $p$ at least once, and at least one of the two is divisible by $p$ only once i.e. by $p$ but not by $p^2$.  So suppose $a=pn$ where $n$ is not divisible by $p$, and $b=p^{50}m$.  Then $\gcd(a^2,b)=p^2$ and $\gcd(a^3,b)=p^3$, etc.

But not suppose $a=pn$ and $b=pm$ where $n,m$ are not divisible by $p$.  Then $\gcd(a^2,b)=\gcd(a^3,b)=p$.

Also if $a=pn$ where $n$ is not divisible by $p$, then it's possible that $\gcd(a^2,b)=\gcd(a^3,b)=p^2$.  You should be able to figure out what would have to be true of the factorization of $b$ for that to happen.

ContentLicense: CC BY-SA 3.0


问:
why use the root to calculate the norm of vector?
I understand that the norm comes from the Pythagorean theorem: $$a^2+b^2=c^2.$$
So why is the vector length not $c^2$?

Why use root like that $c=\sqrt{a^2+b^2}$?

Thank you.

答:
By definition, a norm must be absolutely homogeneous (or absolutely scalable) that is

$$\|\lambda \vec v\|= \lambda \|\vec v\|$$

According to the physical interpretation of length, we have that the square root preserves the dimension.

The main point is that the norm of a vector is defined in such way it is "comparable" with the components of the vector. For this reason, in the case of euclidean norm we take the square roots. For the case $(2,0)$, using square root we have that $\|(2,0)+(2,0)\|=4$, which corresponds to the "natural" definition of length, otherwise we would obtain $16$.


问:
Can set in a metric space be open and have all of its points isolated?
I have been going through some questions in my real analysis book and I came to this one.
It says the answer is no, but I think that the set of Natural numbers in the metric space of Natural numbers with |x-y| metric is both open (its the whole space) and all of its points are isolated.</p>

答:
Yes.  Consider any discrete metric space.  That's $d(x,y)=\begin{cases}0,x=y\\1,x\ne y\end{cases}$.

All the sets are then counterexamples.


答:
You do not have enough information to answer unambiguously. The most we can say is that $\mathrm{gcd}(a^2,b)$ is either $p$ or $p^2$, and $\mathrm{gcd}(a^3,b)$ is one of $p$, $p^2$, and $p^3$.

For suppose $\mathrm{gcd}(a,b)=p$. Then there are integers $x,y$ such that $ax+by=p$, so (squaring this expression) $a^2x^2+b(2axy+by^2)=p^2$, showing that a linear combination of $a^2,b$ with integer coefficients is $p^2$, which implies that $\mathrm{gcd}(a^2,b)$ divides $p^2$. Since $\mathrm{gcd}(a,b)$ divides any combination of $a,b$ and therefore of $a^2,b$, then $p$ divides $\mathrm{gcd}(a^2,b)$.

Now: $\gcd(p,p^2)=p$ and $\gcd(p^2,p^2)=p^2$, so certainly $p^2$ is possible. On the other hand, $\gcd(p,p)=p$ and $\gcd(p^2,p)=p$, so $p$ is possible as well.

A similar analysis, with similar examples, gives the result for $\mathrm{gcd}(a^3,b)$.

Title: Using $\gcd(a,b)$ to find gcd of other values $\gcd(a^2,b)$ and $\gcd(a^3,b)$

Body:
If $\gcd(a,b) = p$, a prime, what are the possible values of $\gcd(a^2,b)$ and $\gcd(a^3,b)$?

Through examples I've been able to find the answer, but I don't know how to come up with a proof.

Update: I think gcd of either is $p^2$, but again this is just by working examples.

I'm not familiar with much number theory. I've just started the second section of my textbook, so all I know is just divisibility and primes.


问:
Propositional calculus help proving theorem
I am asked to prove:

(Y => Z) ∧ X ≡ (Y ∧ X) => (Z ∧ X) ≡ X

What I have done:

(¬Y V Z) ∧ X

(X ∧ ¬Y) V (X ∧ Z)

I don't know where to go from there. I would sincerely appreciate any help.

答:
To prove: (Y => Z) ∧ X ≡ (Y ∧ X) => (Z ∧ X) ≡ X

step one using double negation

(Y => Z) ∧ X ≡ ¬(¬((Y => Z) ∧ X))

step two using De Morgan's laws

(Y => Z) ∧ X ≡ ¬(¬(Y => Z) V ¬X )

step three using the negation of an implication

(Y => Z) ∧ X ≡ ¬((Y ∧ ¬Z) V ¬X )

step four using De Morgan's laws again

(Y => Z) ∧ X ≡ ¬(Y ∧ ¬Z) ∧ X)

step five using De Morgan's laws again

(Y => Z) ∧ X ≡ (¬Y V Z) ∧ X :(1)

Now:

(Y => Z) ∧ X ≡ (Y ∧ X) => (¬Y V Z) ∧ X ≡ Y ∧ X => (¬Y ∧ X) V (Z ∧ X)  ≡ Y ∧ X

This means that  Y ∧ [(¬Y ∧ X) V (Z ∧ X)]  ≡ Y ∧ (Y ∧ X)

=> (Y ∧ (¬Y ∧ X)) V (Y ∧ (Z ∧ X))  ≡ Y ∧ X =>  (Y ∧ (Z ∧ X))  ≡ Y ∧ X

=>  Z ∧ X  ≡ X Proved!

Source: https://www.tu-ilmenau.de/fileadmin/media/num/neundorf/Dokumente/Lehre/hm/Aussagen_Wertetafel.pdf


答:
You don't want the absolute values in the integrals when you're finding area. It's easier to explain why using two functions instead of three, so here's a Desmos graph of just $g(x)$ and $h(x)$:

Now, let's think about $\int_{x_0}^{x_2}g(x)dx$ and $\int_{x_0}^{x_2}h(x)dx$. Let's color the areas that the integrals measure: For the integral of $g$, let's color the positive area blue and the negative area red. For the integral of $h$, let's color the positive area green and the negative area orange:

Now, the region that we want to measure is the red area that is not orange, as well as the green area that is not blue. In order to properly capture the red area, we need to make sure we are subtracting the negative area in $\int g$ (giving a positive value) while adding the negative area in $\int h$ (which will eliminate the contribution of the orange region). Then we need to add the positive area in $\int h$ in order to get the green area while subtracting the positive area in $\int g$ to get rid of the blue area. So that means that in order to find the area between $g$ and $h$, we need to evaluate $\int_{x_0}^{x_2}h(x)-g(x)dx$.

Going back to the original issue, the area in between is $A=\int_{x_0}^{x_1}\sin x\,dx+\int_{x_1}^{x_2}-x+1\, dx-\int_{x_0}^{x_2}x^2-1\, dx$.

Then, to find the $x$-coordinate, you need to multiply the integrands by $x$:

$$x_{\text{CM}}=\frac 1A\cdot\left(\int_{x_0}^{x_1}x\sin x\,dx+\int_{x_1}^{x_2}x(-x+1)\, dx-\int_{x_0}^{x_2}x(x^2-1)\, dx\right)$$

The $y$-coordinate is similar but more difficult; with inverse functions it's essentially the same process.

Title: Calculate center of mass using integration

Body:
How to calculate center of gravity having figure described by those equations with integration? This can and should be solved by numerical analysis.

$y < -x + 1\\ y > x^2 - 1\\ y < \sin(x)$

Firstly I found bounding points for integration:

$-x + 1 = \sin(x) \\ x_0 = -0.63673265$

$sin(x) = -x + 1 \\ x_1 = 0.51097343$

$-x + 1 = x^2 - 1 \\ x_2 = 1$

Then I've calculated integral of $g(x)$ from $x_0$ to $x_1$:

$\int_{x_0}^{x_2} |x^2 - 1| dx = 1.2173486603176453$

Then integral of $h(x)$:

$\int_{x_0}^{x_1} |\sin(x)| dx = 0.32334348721908174$

Then integral of $f(x)$:

$\int_{x_1}^{x_2} |-x + 1| dx = 0.11957349302898969$

So the area between is equal to
$$\int_{x_0}^{x_1} |\sin(x)| dx + \int_{x_1}^{x_2} |-x + 1| dx - \int_{x_0}^{x_2} |x^2 - 1| dx = -0.7744316800695739$$

Which is something wrong I guess.


问:
Proving a statement
The question is

Prove that if $A\subset B$ and $B\subset C$, then $A\subset C$.

Now the logical equivalence should be

If $p\rightarrow q$ $ \land q \rightarrow r$, then $p\rightarrow r$.

Now I know that the statement "If $A\subset B$ and $B\subset C$, then $A\subset C$." is true.

However, the truth tables for $p\rightarrow q$ $ \land q \rightarrow r$ and $p\rightarrow r$ are not the same.

Am I doing something wrong here?

答: Is there any particular reason you do not want to do a “normal” set theory proof? For example, you could start by letting $x \in A \subset B.$ Then, this implies that $x \in B$ and $x \in A.$ Since $x \in B$, and $B \subset C,$ then that implies $A \subset C.$

Title: Proving a statement


答:
hint

The truth tables are used when we deal with propositions. With sets, we speak set-Theory language.

Let us prove that
$$(A\subset B \wedge B\subset C)\implies  A\subset C$$

we assume the premisses are true and prove so is the conclusion .

Let $ x\in A$.

$$x\in A \implies x\in B \;because\; A\subset B$$

$$x\in B \implies x\in C \;since\; B\subset C$$

thus, by transitivity,
$$x\in A\implies x\in C$$

So,

$$A\subset C$$

Title: Proving a statement

Body:
The question is

Prove that if $A\subset B$ and $B\subset C$, then $A\subset C$.

Now the logical equivalence should be

If $p\rightarrow q$ $ \land q \rightarrow r$, then $p\rightarrow r$.

Now I know that the statement "If $A\subset B$ and $B\subset C$, then $A\subset C$." is true.

However, the truth tables for $p\rightarrow q$ $ \land q \rightarrow r$ and $p\rightarrow r$ are not the same.

Am I doing something wrong here?


问:
Extending the action $S_5$ on $2$-subsets of $\{1,\cdots,5\}$ to an action of $S_6$.
The symmetric group $S_5$ acts on the set $\binom{5}{2}$ of ten $2$-subsets of $[5]=\{1,\cdots,5\}$. In The Finite Simple Groups (Wilson), problem 2.21 asks the reader to extend the group action $S_5\to S_{10}$ to an action $S_6\to S_{10}$.

There's probably a "hands-on" way to do this by writing down explicit cycle notations and relations and such, but is there a conceptual reason for this? Is this explained by some exceptional object, maybe like ${\rm Out}\,S_6$ somehow?

(For comparison, problem 2.24 asks to show things about a group generated by a couple of given permutations in $S_8$, without mentioning it's just ${\rm PSL}_2(\Bbb F_7)$ acting on the projective line $\Bbb F_7\Bbb P^1$, so hiding the story behind a problem seems like something the text would do.)

答:
An $S_6 \to S_{10}$ homomorphism shows up in the exceptional group $M_{24}$ on 24 points. Take an octad $O$ and dodecad $D$ which intersect in 6 points.  (The details of the construction are in a later chapter of Wilson's book.)  This partitions the 24 points as follows

(a) 6 points which lie in both $O$ and $D$

(b) 6 points which lie in $D$ but not $O$

(c) 2 points which lie in $O$ but not $D$

(d) 10 points which lie in neither $D$ nor $O$

The subgroup of $M_{24}$ which simultaneously stabilizes $D$ and $O$ is isomorphic to $S_6$, and maps each of the subsets (a) through (d) to itself.  In particular, for (d), we get a homomorphism $S_6 \to S_{10}$.

It is not clear that this is an extension of the given $S_5 \to S_{10}$ in the problem, though.

An outer automorphism of $S_6$ also shows up here by considering (a) and (b).


问:
Probability desiderata: Why is AB|E not a function just of A|E and B|E?
So I've been reading Jaynes(2003) and Tribus (1969). In their exposition of the reasoning for the plausibility of events they say that the plausibility of $ AB|E$ is not a function of $ A|E$ and $B|E$. Only Tribus offers a mathematically worked out argument as to why. He does so in this section of the book: Take a look here.

Some context as to what each symbol means:

$A, B, E$ are all letters representing a statement.

$b$ is the negation of $B$

Two statements written together will be taken to mean both statements
are true. (Conjunction).

$u, y, i, v, x$ are all real numbers.

$u = [AB|E], y = [B|AE], v = [A|E], x = [B|E]$

$i$ represents that a statement is impossible

$F_2(v, x)$ is just a function of $v$ and $x$

What I find confusing here is that the demonstration begins y saying that $A$ is equal to NOT $B$ but later on $A$ and $B$ are said to be independent, which is contradictory since determining the value of $A$ will determine the value of $B$ and vice-versa. Hence I don't see why $[AB|E']=i$ if $A$ and $B$ are independent.

Thank all of you for your help!

P.S.: We are not talking about probabilities just yet only what the authors call "credibilities" this will eventually evolve in to probability theory but in this context it does not apply.

答:
After thinking about it for a while I think this is a more clear reasoning as to why this function is inappropriate. I am not 100% confident this is the correct explanation so if anyone has any comments that would help improve this explanation please post them below. Anyway here it is:

if $E = ``A = b"$ then $F_2 = i, \forall i$

Take $E' = ``A \: and \: B \: are \: independent"$ (also for clarity denote $A'$ and $B'$ as statements given $E'$)

For some combination of $A, A', B, B', E, E'$ make it so that at least in one case $[A|E]=[A'|E']=v' \land [B|E]=[B'|E']=x'$

Input these values in $F_2$

Since $F_2 ([A|E], [B|E]) = i = F_2 (v', x')$

Using $F_2$ under $E'$ we get:

$i = F_2 (v', x') = F_2 ([A'|E'], [B'|E'])$ which makes no sense, since $A'$ and $B'$ are independent and so the credibility that they both happen must not be impossible. This would mean $i=F_2 ([A'|E'], [B'|E'])\neq i$ which is a contradiction.

One last comment, this shows that $F_2$ is an inadequate function to calculate the plausibility of $AB|E$. However, there is no proof to my knowledge, that shows that there must be a function by which the plausibility of $AB|E$ can be calculated. Hence, I find this proof by contradiction somewhat flawed as it is not clear that there must be an alternative way of calculating the plausibility of $AB|E$. Once again if anyone knows how to solve this problem your comments are highly welcomed.

I leave this here in hopes someone finds it useful in the future.


问:
Var(X_2months) = 2 Var(X_1month) with independent random variables?
My instructor insists that
$$\mathrm{Var}(X_{2\text{months}}) = 2 \mathrm{Var}(X_{1\text{month}}) $$
with independent random variables $X_{1\text{month}}$ and $X_{2\text{months}}$, but generally with a random variable $Y$
$$\mathrm{Var}(2Y) = 2^2\mathrm{Var}(Y).$$
So
$$\mathrm{Var}(X_{2\text{months}}) = \mathrm{Var}(2X_{1\text{month}}) = 2^2 \mathrm{Var}(X_{1\text{month}}).$$
Which one is right and why?</p>

答:
From wikipedia, I found a proper answer that solved the problem.

Let the sum of random variables be $Y = \sum_{i=1}^{N} X_{i}$.

$Var(Y) = \sum_{i=1}^{N} Var(X_{i}) + 2 \sum_{i<j} Cov(X_{i}, X_{j})$

or

$Var(Y) = \sum_{i=1}^{N} \sum_{j=1}^{N} Cov(X_{i} , X_{j})$


答:
Recall that integers can be factored uniquely into a product of primes, so we can write $a=p_1^{d_1}\cdots p_m^{d_m}$ and $b=q_1^{e_1}\cdots q_n^{e_n}$ where the $p_i$ and $q_i$ are prime. By relabeling, we can assume that $p_1=q_1,\ldots,p_k=q_k$ and the remaining prime factors are distinct ($k$ might be $0$). One very nice consequence of unique factorization is
$$\gcd(a,b)=p_1^{\min(d_1,e_1)}\cdots p_k^{\min(d_k,e_k)}.$$
Since $\gcd(a,b)=p$, we know that $k=1$, $p_1=p$ and $\min(d_1,e_1)=1$. Note that the factorizations of $a^2$ and $a^3$ are $p_1^{2d_1}\cdots p_m^{2d_m}$ and $p_1^{3d_1}\cdots p_m^{3d_m}$ respectively, so we have
$$\gcd(a^2,b)=p^{\min(2d_1,e_1)} \text{ and } \gcd(a^3,b)=p^{\min(3d_1,e_1)}.$$
Note that if $e_1$ is sufficiently large then these are $p^2$ and $p^3$ respectively, but if $e_1=1$ both are $p$ and if $e_1=2$ then both are $p^2$. Thus the possibilities are $p,p^2$ and $p^3$.</p>


问:
Decomposing a general complete seperable metric space into a meager set and a null set
Suppose $X$ is Polish, i.e, a seperable and completely metrizable space. And let $\mu$ be a Borel probability measure on $X$. Furthermore suppose $\mu[\{x\}] = 0$ for every isolated point $x \in X$. Then $X = M \cup N$ where $M$ is meager and $N$ is measure $0$.

This feels simple but I'm having a brain fart.

Firstly, thanks to separability there can only be countably many isolated points. All we really need then is the ability to take neighborhoods of arbitrarily small measure around any non-isolated point. Then the result follows fairly easily by reproducing in $X$ the proof that $\mathbb{R}$ is a union of  meager and a null set.

However, I'm not sure if this is true and I don't have many ideas otherwise. I think I'm overlooking something simple.

答:
$\newcommand{\N}{{\mathbb{N}}}$
Lemma 1.  The set
$$
  D = \big\{x\in  X: \mu (\{x\}) = 0 \big\}
  $$
is dense in $X$.

Proof.  First of all notice that, for all finite subsets $F\subseteq X$, one has that
$$
  \sum_{x\in F} \mu (\{x\}) \leq  \mu (X)=1.
  $$
This implies that the (possibly uncountably indexed) series
$$
  \sum_{x\in X} \mu (\{x\})
  $$
converges, so it has at most countably many nonzero terms, that is, $X\setminus D$ is countable.

Now suppose by contradiction that some point $a\in X$ is not in the closure of $D$.  Then there exists some $r>0$ such that
the ball  $B_r(a)$ is contained in $X\setminus D$.  Needless to say, $ B_r(a)$  is therefore countable.

We next wish to find some positive number $s<r$ such that the  sphere
$$
  S_r(a) = \{x\in  X: d(x, a)=s\}
  $$
becomes  empty.  This is in fact easy  since the interval $(0,r)$ is uncountable and $B_r(a)$ is countable.

Once such an $s$ is fixed, we deduce that the open ball of radius $s$ coincides with the corresponding closed ball, so
$B_s(a)$ is both open and closed.  Also  $B_s(a)\subseteq X\setminus D$.

Since $X$ is complete and $ B_s(a)$  is closed, we deduce that $ B_s(a)$  is also complete.  By  Baire's Theorem
(https://en.wikipedia.org/wiki/Baire_category_theorem) every complete countable metric space admits an isolated point.  It then follows that $B_s(a)$ contains at least one isolated point $x_0$, and because $B_s(a)$ is open as well, we have that
$x_0$ is also isolated relative to $X$.  By assumption $\mu (\{x_0\}) = 0$, but this contradicts the fact that $B_s(a)\subseteq X\setminus D$.
QED

Lemma 2.
Given any point $d$ in $D$, and any $\varepsilon >0$, there exists an open set $U$ containing $d$,   such that $\mu (U)<\varepsilon $.

Proof.
One has that
$$
  \{d\} = \bigcap_{n\in \mathbb N} B_{1/n}(d),
  $$
so
$$
  0 = \mu (\{d\}) = \lim_{n\to \infty } \mu \big (B_{1/n}(d)\big ),
  $$
and then  one can  find some $n$ such that  $\mu \big (B_{1/n}(d)\big )$ is as small as  needed.  QED

This said, let us proceed to the main argument.

It is well known that any subspace  of a separable metric space is also separable.  Therefore  $D$ is separable and we
may then find a countable subset
$$
  D_0 = \{d_k:k\in \N\}\subseteq D
  $$
whose closure contains $D$.   Since $D$ is dense, we deduce that the
closure of $D_0$ also contains $X$.  In other words,  $D_0$ is dense in $X$ as well.

For each pair $(k, n)$ of natural numbers,  use Lemma 2 to choose some open set $U_{k,n}$ cointaining $d_k$, and such that
$$
  \mu (U_{k,n})< {1\over 2^kn},
  $$
and define
$$
  V_n= \bigcup_{k\in {\bf N}} U_{k,n}.
  $$
We then have that
$$
  \mu (V_n)\leq  \sum_{k\in {\bf N}} \mu (U_{k,n}) \leq  \sum_{k\in {\bf N}} {1\over 2^kn} = {1\over n}.
  $$

Oberve that  $V_n$ is  open and also  dense, because it contains $D_0$,  so
$$
  N:= \bigcap_{n\in {\bf N}} V_n
  $$
is a dense $G_\delta $, by Baire, and clearly $\mu (N)=0$.  The complemrnt
$M:= X\setminus N$
is therefore a meagre set and we evidently have that $X=M\cup N$.

PS: After initially misreading the hypothesis, I hope this is now correct.


问:
Example 11.27 in Lee's Introduction to Smooth Manifolds
In Lee's "Introduction to Smooth Manifolds" there is the following quoted piece of text:

Example 11.27: Let $F : \mathbb{R}^3 \to \mathbb{R}^2$ be the map given by $$(u, v) = F(x, y, z) = (x^2y, y\sin z)$$ and let $\omega \in \mathfrak{X}^*(\mathbb{R}^2) $  be the covector field $$\omega = u dv + v du$$...

Now my interpretation of this is that the author is defining the function $u = \pi_1 \circ F : \mathbb{R}^3 \to \mathbb{R}$ where $\pi_1 : \mathbb{R}^2 \to \mathbb{R}$ is the standard projection function onto the first coordinate. And similar the author is defining $v = \pi_2 \circ F : \mathbb{R}^3 \to \mathbb{R}$. So essentially the author is saying that $u$ and $v$ are the first and second component functions of $F$ respectively in my interpretation.

But if my interpretation is correct then $du$ and $dv$ are covector fields on $\mathbb{R}^3$ and not on $\mathbb{R}^2$. What is my mistake here and what is the correct interpretation of this piece of text?

答:
A map $F:\Bbb{R}^3\to \Bbb{R}^2$ is specified by the data of two real valued functions depending (in this case smoothly) on the three coordinates on the domain. Write the coordinates on $\Bbb{R}^3$ as $(x,y,z)$ and write those of $\Bbb{R}^2$ as $(u,v)$. The first component function is $F^1=\pi_1\circ F$ and the second component is $F^2=\pi_2\circ F$ as in your notation. Sometimes the dependence of the pair $(u,v)$ on the coordinates $(x,y,z)$ introduced by the mapping is recorded as above:
$$ (u,v)=(x^2y,y\sin z)=(F^1(x,y,z),F^2(x,y,z)),$$
i.e. $u(x,y,z)=x^2y$ and $v(x,y,z)=y\sin z$. This is often how such transformations are written in calculus. I find this notation to be a bit ambiguous or confusing, but I think it is standard.

In any case, $u$ and $v$ are coordinate functions on $\Bbb{R}^2$ (the codomain) and $vdu+udv$ is indeed a differential $1-$form on $\Bbb{R}^2$.


问:
Why can we extract the first term during integral?
I was reading a manuscript that is relevant to my thesis (essentially a proof for some tail bound) and encountered a formula that looks like this

$f'(z) = f'(0) + \int_0^z f''(w)dw$

for every $z \in [0, x]$.

Since my math background is pretty shaky, I don't quite understand why we are justified to do the step above. Can someone maybe explain this or point me to some useful resources? Many thanks.

答: Let $g(x)=f'(x)$  The formula is simply $g(z)-g(0)=\int_0^zg'(w)dw$, which is the general formula for the definite integral.


问:
using Lagrange multipliers to determine shortest distance between a point and straight line

Find the shortest distance between $(0,2,0)$ and the straight line
$x=y=z$ .

Let the required point on the straight line be $(x,y,z)$ . Now we set up the Lagrangian as $$L=x^2+(y-2)^2+z^2+\lambda(x-y)+\mu(y-z)+\nu(z-x)$$ where $\lambda,\mu,\nu$ are Lagrange multipliers. The extremizing equations are $$\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y}=\frac{\partial L}{\partial z}=0$$ which gives $$2x+\lambda-\nu=2(y-2)+\mu-\lambda=2z-\mu+\nu=0$$ Solving for $x,y,z$ and substituting in $x=y=z$ we get $$\begin{pmatrix}2&-1&-1 \\ 1&-2&1 \\ 1&1&-2\end{pmatrix}\begin{pmatrix}\lambda\\ \mu \\ \nu\end{pmatrix}=\begin{pmatrix}-4\\ -4 \\ 0\end{pmatrix}$$ But I keep getting the coefficient matrix as singular matrix, so it doesn't give us the unique value of Lagrange multipliers which help to minimize the distance required.

What is possibly going wrong there? Any help is appreciated.

答:
The distance of the point $(0,2,0)$ from the given line $(t,t,t)$ is given by minimizing the function $f(t)=\sqrt{t^2+(t-2)^2+t^2}; \; f(t) = \sqrt{3 t^2-4 t+4}$
$$f'(t)=\frac{6 t-4}{2 \sqrt{3 t^2-4 t+4}}$$
$$f'(t)=0\to t=\frac{2}{3}$$
And we have $$f\left(\frac{2}{3}\right)=2 \sqrt{\frac{2}{3}}$$


答:
You should not be asking this question.

In

$$a^2+b^2=c^2,$$

$a,b$ are the lengths of the sides of the right angle, and I am sure that you don't question that, by thinking that the sides are $a^2$ and $b^2$. So why should wonder that the third side be $c^2$ and not $c$ ???

Title: why use the root to calculate the norm of vector?

Body:
I understand that the norm comes from the Pythagorean theorem: $$a^2+b^2=c^2.$$
So why is the vector length not $c^2$?

Why use root like that $c=\sqrt{a^2+b^2}$?

Thank you.


问:
Evaluating Confidence Interval for difference of sample mean of two random variables
I have two normally distributed populations X and Y that have the same variance and are independent. I took two samples from X and Y and got the following:

For $X:\; n = 10,\; \sum_{i=1}^{10} X_i = 78,\;
\sum_{i=1}^{10} = 634.$

For x

For $Y:\; n = 8,\; \sum_{i=1}^8 Y_i = 36,\;
\sum_{i=1}^6 = 184.$

For Y

Now I need to build a confidence interval with 98% probability for the difference of the means.

I understand that because the sample is not large, I will need to use t-student distribution. My question is how to find the standard deviation to use in the confidence interval formula considering I only have the sum of $X_i$ and the sum of the $X_iˆ2.$

Shall I use that $Var(X) = E(X^2) - E(X)^2?$

答:
I will get you started on this by showing you how to compute the required sample standard deviations, pointing you to the formulas for the CI, and showing you the CI I got from software.

Formulas for sample variance. The sample variance for the $X_i$ can be obtained from
the following formula:

$$S^2_x = \frac{\sum_{i=1}^n(X_i-\bar X)^2}{n-1}
=\frac{\sum_{i=1}^n X_i^2 - \frac 1n(\sum_{i=1}^n X_i)^2}{n-1}\\
=\frac{\sum_{i=1}^n X_i^2 - n\bar X^2}{n-1}.$$
Take the square root of the sample variance to get the sample standard deviation.

Sample means and standard deviations for your data. Using R as a calculator: The sample mean (average) of the first sample is $\bar X = 7.8$

```<code>a.x = 78/10;  a.x
[1] 7.8
</code>```

and (using the formula above) the sample standard deviation of the first sample is $S_x = 1.6865.$

```<code>s.x = sqrt((634-(78)^2/10)/(10-1));  s.x
[1] 1.686548

sqrt((634-10*(7.8)^2)/(10-1))
[1] 1.686548   # alternate method
</code>```

Similarly, for the second sample: $\bar Y = 4.5$ and
$S_y = 1.7728.$

```<code>a.y = 36/8;  a.y
[1] 4.5

s.y = sqrt((184-(36)^2/8)/(8-1));  s.y
[1] 1.772811
</code>```

Finding the CI for the difference in population means. Now you are ready to find the formulas for the pooled
estimate of the common population standard deviation
and the formula for a confidence interval of the
difference $\mu_x - \mu_y$ in population means.

Instead of doing these computations here, I put sample sizes, means, and standard deviations into a recent release of Minitab in order to get the 95% CI for the difference
in population means from Minitab's pooled t test procedure.
Results below:

```<code>Two-Sample T-Test and CI

Sample   N  Mean  StDev  SE Mean
1       10  7.80   1.69     0.53
2        8  4.50   1.77     0.63

Difference = μ (1) - μ (2)
Estimate for difference:  3.300
95% CI for difference:  (1.566, 5.034)
T-Test of difference = 0 (vs ≠):
  T-Value = 4.03  P-Value = 0.001  DF = 16
Both use Pooled StDev = 1.7248
</code>```

The difference in sample means is $\bar X = \bar Y$ $= 7.80 - 4.50$ $= 3.30,$ which is the center of your 95% CI for the difference in population means $(1.566, 5.034),$ which also uses the
pooled estimate $S_p = 1.7248$ of the common
population standard deviation. (The pooled method
assumes that the two populations have the same
standard deviation.)

Note:  The methods above are correct, but please proofread my typing, data entry, and computations.

Body:
I will get you started on this by showing you how to compute the required sample standard deviations, pointing you to the formulas for the CI, and showing you the CI I got from software.

Formulas for sample variance. The sample variance for the $X_i$ can be obtained from
the following formula:

$$S^2_x = \frac{\sum_{i=1}^n(X_i-\bar X)^2}{n-1}
=\frac{\sum_{i=1}^n X_i^2 - \frac 1n(\sum_{i=1}^n X_i)^2}{n-1}\\
=\frac{\sum_{i=1}^n X_i^2 - n\bar X^2}{n-1}.$$
Take the square root of the sample variance to get the sample standard deviation.

Sample means and standard deviations for your data. Using R as a calculator: The sample mean (average) of the first sample is $\bar X = 7.8$

a.x = 78/10;  a.x
[1] 7.8

and (using the formula above) the sample standard deviation of the first sample is $S_x = 1.6865.$

s.x = sqrt((634-(78)^2/10)/(10-1));  s.x
[1] 1.686548

sqrt((634-10*(7.8)^2)/(10-1))
[1] 1.686548   # alternate method

Similarly, for the second sample: $\bar Y = 4.5$ and
$S_y = 1.7728.$

a.y = 36/8;  a.y
[1] 4.5

s.y = sqrt((184-(36)^2/8)/(8-1));  s.y
[1] 1.772811

Finding the CI for the difference in population means. Now you are ready to find the formulas for the pooled
estimate of the common population standard deviation
and the formula for a confidence interval of the
difference $\mu_x - \mu_y$ in population means.

Instead of doing these computations here, I put sample sizes, means, and standard deviations into a recent release of Minitab in order to get the 95% CI for the difference
in population means from Minitab's pooled t test procedure.
Results below:

Two-Sample T-Test and CI

Sample   N  Mean  StDev  SE Mean
1       10  7.80   1.69     0.53
2        8  4.50   1.77     0.63

Difference = μ (1) - μ (2)
Estimate for difference:  3.300
95% CI for difference:  (1.566, 5.034)
T-Test of difference = 0 (vs ≠):
  T-Value = 4.03  P-Value = 0.001  DF = 16
Both use Pooled StDev = 1.7248

The difference in sample means is $\bar X = \bar Y$ $= 7.80 - 4.50$ $= 3.30,$ which is the center of your 95% CI for the difference in population means $(1.566, 5.034),$ which also uses the
pooled estimate $S_p = 1.7248$ of the common
population standard deviation. (The pooled method
assumes that the two populations have the same
standard deviation.)

Note:  The methods above are correct, but please proofread my typing, data entry, and computations.


问:
radical of sum of two ideals
$I$ and $J$ are ideals in $k[x_1,\cdots,x_n]$.

Show that $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$.

I have no idea how to prove it. Can someone help?

答:
You don't have to argue with elements. $\sqrt{I+J}$ is the smallest radical ideal which contains $I+J$, i.e. which contains $I$ and $J$. On the other side, $\sqrt{\sqrt{I}+\sqrt{J}}$ is the smallest radical ideal which contains $\sqrt{I}+\sqrt{J}$, i.e. which contains $\sqrt{I}$ and $\sqrt{J}$, i.e. which contains $I$ and $J$. Thus both radical ideals are equal.

In general, for every closure operator $C$ on a partial order, we have $C(A \vee B) = C(C(A) \vee C(B))$ for all elements $A,B$. This is purely formal. In the example above, we have the partial order of ideals and $C(A)=\sqrt{A}$. You can also take the partial order of subsets of a topological space and $C(A)=\overline{A}$. But actually here the union of two closed sets is already closed, so that the right hand side simplifies. The corresponding statement for radical ideals does not hold and this has actually an algebro-geometric interpretation. See here. Also note that even more generally $C(\vee_i A_i)=C(\vee_i C(A_i))$. In the case of subsets of a topological space the right hand side doesn't simplify.

Title: radical of sum of two ideals


问:
What are post-composition and pre-composition in category theory?
What are post-composition and pre-composition of morphisms in category theory?

The context of my question is to understand Lemma 1.2.3 in Riehl's book.

Lemma 1.2.3. The following are equivalent:

(i) $f: x \rightarrow y$ is an isomorphism in $C$.

(ii) For all objects $c \in C$, post-composition with $f$ defines a bijection $$f_{\ast} : C(c, x) \rightarrow C(c, y).$$

(iii) For all objects $c \in C$, pre-composition with $f$ defines a bijection $$f^{\ast} : C(y, c) \rightarrow C(x, c).$$

答:
In either case, it's about composing morphisms with $f$.

(ii) Take an element of $C(c,x)$, which is to say, a morphism $g:c\to x$. Then compose this with $f$ to get $$f\circ g:c\to y$$which is an element of $C(c,y)$. In this case, we first follow $g$ from $c$ to $x$, then afterwards we follow $f$ from $x$ to $y$. This means we are post-composing with $f$.

(iii) Take an element of $C(y, c)$, which is to say, a morphism $h:y\to c$. Then compose this with $f$ to get $$h\circ f:x\to c$$which is an element of $C(x,c)$. In this case, we first follow $f$ from $x$ to $y$, then afterwards we follow $h$ from $y$ to $c$. This means we are pre-composing with $f$.

Note that in $f\circ g$, we say $f$ is post-composed, even though we write $f$ in front of $g$. This might seem strange, but it is a natural result of our convention of writing functions to the left of their argument (we say $f(x)$ rather than $(x)f$).


问:
Stein and Shakarchi complex analysis exercise 8 chapter 3
For the following problem:

I considered the function $f(z)=\frac{1}{(1+z^2)^{n+1}}$ which is holomorphic except for poles at $ i$ and $-i$.
So I considered a semi-circular contour,

I have the residue of $\int_{\gamma_{R}} f(z) dz = \frac{\pi}{(2i)^n}$

If $C_R$ is the large semicircle of radius $R$, we have that $|\int_{C_R} f(z) dz| \leq \frac{M}{R^{2n+1}}$, so $R \to \infty$, the integral goes to zero.

Therefore now we have $\int_{-\infty}^{\infty} f(x) dx = \frac{\pi}{(2i)^n}$ which is not what should be proved. I would appreciate if someone can point where I am going wrong.

Based on the feedback below, I have the residue as :
$2 \pi i(\lim_{z \to i} \frac {1}{(n!)} (\frac{d}{dz})^{n} (z-i)^n f(z))$ =
$2 \pi i(\frac {1}{(n!)} \frac {(-1)^n(n+1)(n+2)....(2n)}{(2i)^{n+1}})$
=$\frac{i^n(n+1)(n+2)......(2n)}{2.4.6......2n} . \pi$

But I dont have the final answer.
I would appreciate some feedback on this.

答:
Consider $\oint_{C_R}\frac{1}{(1+z^2)^{n+1}}dz=\oint_{C_R}\frac{\frac{1}{(z+i)^{n+1}}}{(z-i)^{n+1}}dz$. Write $f(z)=\frac{1}{(z+i)^{n+1}}$. Then $f'(z)=-\frac{n+1}{(z+i)^{n+2}},\, f''(z)=(-1)^2\frac{(n+1)(n+2)}{(z+i)^{n+3}},\dots,f^{(n)}(z)=(-1)^{n}\frac{(n+1)(n+2)\cdots(2n)}{(z+i)^{2n+1}}$.

Since $f$ is holomorphic at $z=i$, expand it in a Taylor series. The only meaningful term is at the $nth$ derivative. All other terms in the Taylor series integrate to zero due to $\frac{f(z)}{(z-i)^{n+1}}$ having an anti derivative.


问:
Calculate center of mass using integration
How to calculate center of gravity having figure described by those equations with integration? This can and should be solved by numerical analysis.

$y < -x + 1\\ y > x^2 - 1\\ y < \sin(x)$

Firstly I found bounding points for integration:

$-x + 1 = \sin(x) \\ x_0 = -0.63673265$

$sin(x) = -x + 1 \\ x_1 = 0.51097343$

$-x + 1 = x^2 - 1 \\ x_2 = 1$

Then I've calculated integral of $g(x)$ from $x_0$ to $x_1$:

$\int_{x_0}^{x_2} |x^2 - 1| dx = 1.2173486603176453$

Then integral of $h(x)$:

$\int_{x_0}^{x_1} |\sin(x)| dx = 0.32334348721908174$

Then integral of $f(x)$:

$\int_{x_1}^{x_2} |-x + 1| dx = 0.11957349302898969$

So the area between is equal to
$$\int_{x_0}^{x_1} |\sin(x)| dx + \int_{x_1}^{x_2} |-x + 1| dx - \int_{x_0}^{x_2} |x^2 - 1| dx = -0.7744316800695739$$

Which is something wrong I guess.

答:
$A = \displaystyle \int_{x_0}^{x_2} (y_u - y_l) dx$ where $y_u$ is the upper bound of $y$ and $y_l$ is the lower bound of $y$. Even when $y$ is negative, please note that this difference will be positive.

$A = \displaystyle \int_{x_0}^{x_1} (\sin x - (x^2-1))dx + \int_{x_1}^{x_2} ((1 - x) - (x^2-1))  dx $

$A = \displaystyle \int_{x_0}^{x_1} \sin x \, dx + \int_{x_1}^{x_2} (1 - x) \, dx - \int_{x_0}^{x_2}(x^2-1) \, dx = -0.0682262 + 0.119573 - (-1.21735)$

$A \approx 1.2686968$.


问:
Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$.

Let $G$ be a finite abelian group, and let $n$ divide $|G|$.  Let $m$ be the number of solutions of $x^n=1$.  Prove that $n\mid m$.

My attempt

It's tempting to find a way to use Lagrange's theorem.  Maybe something here is a subgroup of something else?  We can fix $n$ and take the subgroup of $G$ of all elements which solve $x^n=1$.  Proof that this is a subgroup:  Inverses of solutions are always solutions.  Because the group is abelian, products of solutions are solutions.  QED.

Great, so it's a subgroup, so $m$ divides the order of $G$.  So does $n$.  I'm not sure that this really got me anywhere.  It'd be nice if there were some relevant subgroup of order $n$.

Being finite and abelian then it has a representation as $G\cong C_{p_1^{n_1}}\times\dots\times C_{p_k^{n_k}}$, a product of cyclic groups of prime power order.  The solutions are exactly the product of solutions "in each factor", i.e. solutions of the form $\langle e, \dots, e, x, e, \dots, e\rangle$ where $x\in C_{p_i^{k_i}}$ for some $i$.  So perhaps something comes from thinking about the number of solutions to $x^n=1$ where $x$ is taken from $C_{p_i^{k_i}}$.

Again this is a subgroup so the number of solutions divides $p_i^{k_i}$, and $p_i^{k_i}$ divides $|G|$. And $n$ divides the order of $G$.  But at this point I'm not sure whether I'm on a productive path, since these facts don't seem to be enough to show that $n|m$.

In fact the more that I think about how $n$ is so-to-speak missing factors from $|G|$ the more I think that finding numbers which divide $|G|$ just isn't a productive path.

答: Let $p$ be a prime and assume $p^k\|n$. If one of the factors in the product representation of $G$ is $C_{p^r}$ for some $r\ge k$, then this contains a subgroup isomorphic to $C_{p^k}$, which consists of solutions to $x^n=1$. If no such factor exists, then there exist several factors $C_{p^r}$ with $r<k$, which are completely solutions to $x^n=1$. But their product must have cardinality at least $p^k$. So at any rate, $G$ has a subgroup of order $p^k$ of solutions of $x^n=1$. Combining all relevant primes, we obtain a subgroup of order  $n$ or - as we may have ignored a few factors - a multiple of $n$.


答:
We need $10$ of something related to a set of size $6$ and need to build that from the $2$-subsets from a set of size $5$.

There are exactly $10$ different ways to partition a set of size $6$ into two parts of size $3$ each. We associate each of these $10$ partitions with a specific $2$-subset by choosing the half of the partition which contains $6$ and selecting the other two points.

For example,

$\{1,2\}$ becomes $\{\{1,2,6\},\{3,4,5\}\}$

This maintains the action of $S_5$ on a set of size $10$ and extends it to $S_6$.

Title: Extending the action $S_5$ on $2$-subsets of $\{1,\cdots,5\}$ to an action of $S_6$.

Body:
The symmetric group $S_5$ acts on the set $\binom{5}{2}$ of ten $2$-subsets of $[5]=\{1,\cdots,5\}$. In The Finite Simple Groups (Wilson), problem 2.21 asks the reader to extend the group action $S_5\to S_{10}$ to an action $S_6\to S_{10}$.

There's probably a "hands-on" way to do this by writing down explicit cycle notations and relations and such, but is there a conceptual reason for this? Is this explained by some exceptional object, maybe like ${\rm Out}\,S_6$ somehow?

(For comparison, problem 2.24 asks to show things about a group generated by a couple of given permutations in $S_8$, without mentioning it's just ${\rm PSL}_2(\Bbb F_7)$ acting on the projective line $\Bbb F_7\Bbb P^1$, so hiding the story behind a problem seems like something the text would do.)


问:
Turn Disjunctive Normal Form into only AND and NOT statements
I'm working on implementing logic gates and want to implement an OR gate using only AND and NOT gates.

I came up with this canonical form for the OR function. $$Or(x, y) = (\lnot{x} \land y) \lor (x \land \lnot y) \lor (x \land y)$$ $$= y \lor (x \land \lnot y)$$

Then, working with DeMorgan's laws, I tried (following from above):

$$= \lnot(\lnot y) \lor \lnot(\lnot(x \land \lnot y))$$
$$= \lnot(\lnot y \land \lnot(x \land \lnot y))$$

But I can't figure out how to simplify it any further. Am I missing something or is this it?

答:
I would just start with the double negation and then apply De Morgan's law:

$$x\lor y = \lnot(\lnot( x\lor y)) = \lnot (\lnot x \land \lnot y)$$


问:
A subset of $\mathbb{R}$ that is dense in $\mathbb{R}$.
A subset $S$ of $\mathbb{R}$ is said to be dense in $\mathbb{R}$ if $\forall$ $\epsilon > 0$ and $x \in \mathbb{R}$, $\exists$ $s \in S$ such that $| x - s| < \epsilon$.

Using this definition, I would like to try to prove that the set $S = \{ a + b \sqrt{3} : a ,b \in \mathbb{Z} \}$ is dense in $\mathbb{R}$.

Any help would be appreciated.

I am aware that the set $S$ is closed under addition and multiplication. I'm not sure if that helps.

答: HINT: It follows from the fact that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$ whenever $r$ is irrational. (Here $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$.) This fact can be proved using the pigeonhole principle; I gave a proof here.


问:
A problem on related rates
Calculusman is demonstrating his strength by squeezing a rod of solid iron. Throughout the process, the radius is decreasing at the rate of 1 cm/s, but the rod always remains a cylinder, with constant volume. Find the rate at which the height is increasing in cm/s, at the moment the rod has a radius of 4 cm and a height of 10 cm.

I know that $\frac{dV}{dt} = 0 \frac{\text{cm}^3}{\text{s}}$, and $\frac{dr}{dt} = -1 \frac{\text{cm}}{s}$, but I don't know how I can use these definitions to my advantage. Help would be much appreciated!

答: Note that $$V = \pi r^2 h.$$ We now take the derivative of both sides to get $$V' = 2\pi r r' h + \pi r^2 h'.$$ As you noted $V' = 0$, $r' = -1$, and we are given that $r = 4$ and $h = 10$. Thus, putting our work together, we have that $$0 = 2\pi \cdot 4 \cdot (-1) \cdot 10 + \pi 16 h' \Rightarrow h' = 5 ~ \frac{\text{cm}}{\text{sec}}.$$

Title: A problem on related rates


问:
Vector space closed under scalar multiplication, what is the domain of c?
I was introduced to vector spaces recently, and when considering that they must be closed under multiplication.

$$c\mathbf{U} \in \boldsymbol{H} \text{ given } \mathbf{U} \in \boldsymbol{H}$$

What is the domain of $c$? Is it any element of any vector in $\boldsymbol{H}$?

We have primarily been talking about vectors in $\mathbb{R}^n$ so it has been handwaved as "any real number". But if it was an element, does that imply that you could define a vector space over $\mathbb{Q}^n$, $\mathbb{Z}^n$, or $\mathbb{N}^n$?

How does that relate to polynomial spaces or the vector space of all real-valued functions?

答:
A vector space is a triplet $(V,F, \psi)$ where $V$ is a commutative (Abelian) group and $F$ is a field and $\psi$ is a  special kind of function from $F\times V$ to $V.$

Before going further, a few standard conventions:

$+$ is used for the group-operation of $V$ and also for the addition-operation of $F.$

$ 0$ is used for the group-identity of $V$ and also for the additive-identity of $F.$

$1$ is used for the multiplicative-identity of $F.$

$fv$ is used for $\psi(f,v).$

The special properties of $\psi$ are:

If $f_1,f_2\in F$ and $v\in V$ then $(f_1+f_2)v=(f_1v)+(f_2v)$ and $f_1(f_2v)=(f_1f_2)v.$

If $f\in F$ and $v_1,v_2\in V$ then $f(v_1+v_2)=(fv_1)+(fv_2).$

If $f\in F$ and $v\in V$ then $[\,fv=0$ iff ($f=0$ or $v=0)\,].$

If $v\in V$ then $1v=v.$

This is called a vector-space over $F.$ But it is very common to refer to $V$ as the vector-space.

In some contexts $vf$ is defined to be $fv$ when $f\in F$ and $v\in V.$


问:
find $A_0$, $x_0$ such that the Gauss integration formula $\int_{-1}^{1} f(x)|x| \,dx\, \approx$ $A_0 f(x_0)$ will be with maximum accuracy degree
Hint: the formula will calculate the true value integrals of the form $\int_{-1}^{1} x^n |x| \,dx$ for all $n=0,1,...k$
such that $k$ is the accuracy degree.

How can I approach this?

An easy guess : $k=0$,$A_0 =1$  and $x_0 = 0$.

However, I don't know if the maximum value for $k$ is 0.

答:
Each $n$ yield an equation
$$
\int_{-1}^1 x^n |x| dx = A_0 x_0^n.
$$
On the other hand,
$$
\int_{-1}^1 x^n |x| dx = \begin{cases}
\frac{2}{n+2}, &n \text{ is even}\\
0, &n \text{ is odd}
\end{cases}
$$

So we have the following system of nonlinear equations for $A_0$ and $x_0$:
$$
1 = A_0, \quad n = 0\\
0 = A_0 x_0, \quad n = 1\\
\vdots\\
\int_{-1}^1 x^k |x| dx = A_0 x_0^k, \quad n = k.
$$
The first two equations immediately require $A_0 = 1, x_0 = 0$. If we truncate system at $n = 1$ then $A_0 = 1, x_0 = 0$ will be the solution. So $k$ is at least 1.

Let's check that $k$ cannot be greater than 1. If $k > 1$ then for $n = 2 \leq k$ the integration should be exact:
$$
\frac{2}{4} = A_0 x_0^2.
$$
But it is not true, so $k$ cannot be greater than $1$.


问:
Find the Laplace transform:$f(t) = \int_0^t {{e^{it}}\frac{{dt}}{{\sqrt {2\pi t} }}}$
Find the Laplace transform of the function:
$$\begin{array}{l}
 f(t) = \int_0^t {{e^{it}}\frac{{dt}}{{\sqrt {2\pi t} }}}  \\
 {\rm{Us}}e:\Gamma \left( {\frac{1}{2}} \right) = \sqrt \pi   \\
 \end{array}$$</p>

Body: $\mathcal{L}_{t\to s}\left\{\int_0^te^{it}\dfrac{dt}{\sqrt{2\pi t}}\right\}=\dfrac{\mathcal{L}_{t\to s}\left\{\dfrac{e^{it}}{\sqrt{2\pi t}}\right\}}{s}=\dfrac{\mathcal{L}_{t\to s-i}\left\{\dfrac{1}{\sqrt{2\pi t}}\right\}}{s}=\dfrac{1}{s\sqrt{2(s-i)}}$


问:
Gram-Schmidt Process: How does it fail?
I am confused about the Gram-Schmidt process for a linearly dependent set. I can't seem to grasp why it would cause the linearly dependent vectors to equal zero. Can someone please explain?</p>

答:
Try applying it to the vectors $\pmatrix{1\\0\\0}, \pmatrix{1\\1\\0}, \pmatrix{2\\1\\ 0}$. It should give you the "aha" you need.

Even simpler, try $\pmatrix{1\\1}, \pmatrix{2\\2}$.

Suppose you've been through $i-1$ steps of GS, and your vectors $v_1, \ldots, v_{i-1}$ have been converted into vectors
$$
u_1, \ldots, u_{i-1}
$$
which are all unit vectors (so $u_j \cdot u_j = 1$) and pairwise orthogonal, so $u_j \cdot u_k = 0$ for $j \ne k$, and whose span is the same as that of $v_1, \ldots, v_{i-1}$.

What happens, in GS, to the vector $v_i$, if it's in the span of $v_1, \ldots, v_{i-1}$ (i.e., if the vectors $v_1, \ldots, v_i$ are dependent, while $v_1, \ldots, v_{i-1}$ are independent?

Well, being in the span of $v_1, \ldots, v_{i-1}$ means that you're in the span of $u_1, \ldots, u_{i-1}$. So
$$
v_i = c_1 u_1 + \ldots + c_{i-1}u_{i-1}.
$$
What happens to $v_i$ when you apply the first step of GS? You convert
$v_i$ to $$v_i - (v_1\cdot u_1)u_1 + \ldots + v_1 \cdot u_{i-1}i_{i-1}),$$
right? Well, what is $v_1 \cdot u_1$?

\begin{align}
v_1 \cdot u_1
& = (c_1 u_1 + \ldots + c_{i-1}u_{i-1}) \cdot u_1 & \text{Why?} \\
& = c_1 u_1 \cdot u_1  + \ldots + c_{i-1}u_{i-1} \cdot u_1 \text{why?} \\& = 1c_1   + \ldots + 0c_{i-1} \text{why?} \\
& = c_1
\end{align}

So What do you get after that first step of GS?

Title: Gram-Schmidt Process: How does it fail?


问:
The definition of Arithmetic Function
Consider following functions (written on Page 120 of the book "Summing It Up" by Avner Ash and Robert Gross, 2016):
$$
a(n)= \begin{cases}
      0 & \text{if $n$ is not prime and} \\
      1 &  \text{if $n$ is prime.}
\end{cases}
$$
... It turns out that $a(n)$ is not a very easy function to work with. Numbers like to be multiplied, so our next guess is to use
$$
b(n) = \begin{cases}
      0 & \text{ if n is not a power of a prime and} \\
      1 &  \text{if n is a power of a prime.}
      \end{cases}
$$
This is better but still not so great. The problem is that $b(n)$ does not depend on which prime $n$ is a power of, if indeed $n$ is a prime power. Finally, we try
$$
\Lambda(n)=\begin{cases}0,&\text{if $n$ is not a power of a prime and}\\ \log p,&\text{if $n = p^m$ for some prime $p$.}\end{cases}
$$

QUESTION:

What is the meaning of

$a(n)$ is not a very easy function to work with. Numbers like to be multiplied (how this affect $a(n)$)

The problem is that $b(n)$ does not depend on which prime $n$ is a power of, if indeed $n$ is a prime power (what  $\Lambda(n)$ does different besides the output).

答:
There's a saying in mathematics: "A mathematical object is what it does." The authors of your textbook are trying to find a function that does something. What that "something" is is hard to say without reading the textbook and seeing what the function $\Lambda$ does in the proof or example that the textbook outlines. But to answer your questions directly:

$a(n)$ is a fine function to work with. Its values on the positive integers are $0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,\dots$; finding the value of $a(n)$ for an arbitrary $n$ is a question of open research: see Primality Test - Wikipedia for more information. However, we usually don't just define functions for the fun of it. We define functions for a purpose, be that educational or to help in a proof. The quote "numbers like to be multiplied" probably means that the proof or example in question has to do with multiplying numbers. The function $a$ is not very useful in that regard; in general we don't have $a(nm)=a(n)a(m)$ or $a(nm)=a(n)+a(m)$, or any other identities like that. In fact, unless $n$ or $m$ is equal to $1$, we have that $a(nm)=0$, so $a$ isn't very useful when it comes to multiplying numbers. $b$ doesn't have any such property either, but it does have a somewhat weaker property, that $b(n^k)=(b(n))^k$ for any natural numbers $n,k$. This makes it a bit more useful when it comes to multiplication, but it's hard to know why that function was chosen without knowing more about the application. In general, when we talk about multiplying numbers, we like to choose a multiplicative function; the page I linked shows many common examples of such functions.

You ask what $\Lambda$ does differently besides the output. That's a bit of a complicated question because what is the difference between any two functions with the same domain, except for the outputs? In this case, in the proof or example that your textbook is going over, it's important that if a number is a prime power, we know which prime we are raising to a power. I can't tell you why that is important unless you tell me what your textbook is doing with $\Lambda$. Usually, in number theory, logarithms are usually useful for converting products into sums, using the property that $\log(mn)=\log m+\log n$. This benefit is a little diluted when we set $\Lambda(n)=0$ whenever $n$ is not a prime power, but it is still beneficial. Maybe you're testing whether two numbers $m,n$ are powers of the same prime, which is only true when $\Lambda(mn)\neq 0$. Apparently, the value of the prime is important here, which is why we use $\Lambda$ instead of $b$.

I hope this answers your question. In general, numbers don't like to be multiplied unless we care about multiplying them. And we don't care about which prime $n$ is a power of, unless there's a reason we should. $\Lambda$ is what it does, and clearly we care about these things for some yet-unknown purpose. I'm sorry I can't give you more information, but if you want to leave a comment about this I can update the answer :)

Title: The definition of Arithmetic Function


问:
Find the vector $h_0$ in $\mathcal{H}$ such that $L(h) = \langle h, h_0 \rangle$
The following problem was taken from Conway's A course in Functional analysis Section 1.3 problem 2

Let $\mathcal{H} = l^2(\mathbb{N})$. If $N\geq1$ and $L:\mathcal{H}\rightarrow \mathbb{F}$ is defined by $L(\{\alpha_n\}) = \alpha_N$, find the vector $h_0$ in $\mathcal{H}$ such that $L(h) = \langle h, h_0 \rangle$ for every $h$ in $\mathcal{H}$.

My solution is below and I wanted to make sure it was correct:

The inner product on $l^2(\mathbb{N})$ is defined to be $\langle x, y\rangle = \sum_{n\geq1} x(n)\overline{y(n)}$. So if we let $h_0$ be the vector with all zeroes except for $h_0(N) = 1$. Then $\langle \{\alpha_n\}, h_0\rangle = \alpha_N$.

答: Yes, exactly. As disappointing as it may sound, I don't really have further comments about it, other than the observation that your $L$ is just a "projection", so the vector equivalent to it should be a vector in the "standard" basis of the space.


问:
How to show a subset of $C[a,b]$ is open?
I would like to show that $C[0,1] \setminus X = \{f(x) \in C[0,1] : f(x) < x\}$, where our metric is $d_{\infty}(f,g) = \text{sup}\{|f(x)-g(x)| : x \in [0,1]\}$, is open which would imply $X$ is closed but am I unsure as to how to go about showing this.

Any tips would be much appreciated.

答:
The set $Y=\{f\in C[0,1]:\textrm{ }f(x)<x\textrm{ } \forall x\in [0,1]\}$ it's open because, take $f$ on this set, then $f(x)<x$ $\forall x\in [0,1]$ take $\epsilon=\min_{[0,1]}x-f(x)$, then suppose that $d_{\infty}(f,g)<\frac{\epsilon}{2}$ then
$$
|f(x)-g(x)|<\frac{\epsilon}{2}, \forall x\in [0,1]
$$

therefore $g(x)-x=g(x)-f(x)+f(x)-x<\frac{\epsilon}{2}-\epsilon<0$ then $g(x)<x$ $\forall g\in B(f,\frac{\epsilon}{2})$, thus $B(f,\frac{\epsilon}{2})\subset Y$


答: Take some $f\in C[0,1]\setminus X$. Then $f(x)<x$ on $[0,1]$. Since the identity function $i:x\mapsto x$ is continuous we can consider $\epsilon=d_\infty(f,i)$. Now consider $B(f,\frac{\epsilon}{2})=\{g\in C[0,1]:\sup\{|f(x)-g(x)|:x\in[0,1]\}<\frac{\epsilon}{2}\}$ and show that every $g$ in $B(f,\frac{\epsilon}{2})$ is also in $C[0,1]\setminus X$.

Title: How to show a subset of $C[a,b]$ is open?

Body:
I would like to show that $C[0,1] \setminus X = \{f(x) \in C[0,1] : f(x) < x\}$, where our metric is $d_{\infty}(f,g) = \text{sup}\{|f(x)-g(x)| : x \in [0,1]\}$, is open which would imply $X$ is closed but am I unsure as to how to go about showing this.

Any tips would be much appreciated.


答:
$$ \pi r^2 h = V= \text{constant} $$

Differentiate chain rule or product rule and simplify

$$\dfrac{r^2}{h} =\dfrac{-2rr'}{h'}$$

$$h'=\dfrac{-2r'h}{r}=\dfrac{-2\cdot -1 \cdot 10}{4} = 5 \text{cm/s}$$

Title: A problem on related rates

Body:
Calculusman is demonstrating his strength by squeezing a rod of solid iron. Throughout the process, the radius is decreasing at the rate of 1 cm/s, but the rod always remains a cylinder, with constant volume. Find the rate at which the height is increasing in cm/s, at the moment the rod has a radius of 4 cm and a height of 10 cm.

I know that $\frac{dV}{dt} = 0 \frac{\text{cm}^3}{\text{s}}$, and $\frac{dr}{dt} = -1 \frac{\text{cm}}{s}$, but I don't know how I can use these definitions to my advantage. Help would be much appreciated!


问:
Prove R is antisymmetric if $R^2$ is antisymmetric
Let $R^2$ be a binary relation on some set X. I need to prove or disprove the following claim:

"If $R^2$ is antisymmetric, then R is antisymmetric."

I know the opposite claim (if R is ANS then $R^2$ is ANS) is false, but I have struggled to find a counterexample for the former claim. Therefore I am leaning towards it being true, and wanted to prove it using the "set" definiton of antisymmetry, which says:

"R is antisymmetric  ⇐⇒  R ∩ R^(-1) ⊆ ∆(X)."    (∆(X) = {(x, x) ∈ X × X; x ∈ X} being a diagonal relation and R^(-1) being the inverse relation to R)

The claim would then look like this: "If $R^2$ ∩ ($R^2$)^(-1) ⊆ ∆(X), then R ∩ R^(-1) ⊆ ∆(X)."

Alternatively I could also use the more common definition of antisymmetry:

"R is antisymmetric ⇐⇒ (∀x, y ∈ X : xRy ∧ yRx ⇒ x = y)."

I have gotten stuck and can't seem to get to any conclusion. Thanks in advance.

答: You can't prove it, since it is false. On $\{1,-1\}$, consider the binary relation $R$ defined by $a\mathrel Rb$ if $ab=-1$. Then $a\mathrel{R^2}b\iff a=b$. So, $R^2$ is antisymmetric. But $R$ isn't (you have $1\mathrel R-1$ and $-1\mathrel R1$, but you don't have $1\mathrel R1$).


问:
Proving a statement
The question is

Prove that if $A\subset B$ and $B\subset C$, then $A\subset C$.

Now the logical equivalence should be

If $p\rightarrow q$ $ \land q \rightarrow r$, then $p\rightarrow r$.

Now I know that the statement "If $A\subset B$ and $B\subset C$, then $A\subset C$." is true.

However, the truth tables for $p\rightarrow q$ $ \land q \rightarrow r$ and $p\rightarrow r$ are not the same.

Am I doing something wrong here?

答:
It doesn't matter if $(p\to q)\land (q\to r)$ is false.  You are are only interested in the case where  $(p\to q)\land (q\to r)$ are true.  If $(p\to q)\land (q\to r)$ is false, but $p\to r$ is true, it is not a problem because that still means $(p\to q)\land (q\to r)\implies (p\to r)$ is true (even if $p\to r$ if false).

Compare it to $A \not \subset B; B\not\subset C$ and $A \subset C$.  (Can happen: $A=\{1,2\}, B=\{dog, cat\}$, $C=\{1,2,3\}$.)

Or $A\not \subset B; B\subset C$ and $A\subset C$.  (Can happen: $A=\{1,2\}; B= \{2,dog\}; C=\{1,2,dog\}$.

Or $A\subset B;B\not \subset C$ and $A \subset C$. (Can happen: $A=\{1,2\};B=\{1,2,dog\}; C=\{1,2,3\}$.)

The lines where $p\to q\land q\to r$ and $p \to r$ differ are in red

$\begin{array}\ p &q&r&p\to q&q\to r&(p\to q)\land (q\to r) &p\to r&(p\to q)\land (q\to r) \implies p\to r\\T&T&T&T&T&T&T&T\\ T&T&F&T&F&F&F&T\\T&F&T&F&T&\color{red}F&\color{red}T&T\\ T&F&F&F&T&F&F&T\\F&T&T&T&T&T&T&T\\ F&T&F&T&F&\color{red}F&\color{red}T&T\\F&F&T&T&T&T&T&T\\ F&F&F&T&T&T&T&T\end{array}$

This will correspond to the cases where $A\not \subset B; B\not\subset C$ but $A\subset C$ or where $A\not \subset B; B\subset C$ but

Title: Proving a statement


问:
Prove that, if $A\subset B$, then $A\cup B = B$.
Q1: If $A\subset B$ then $A\cup B=B$.

My attempt:

If $A\subset B$, it means $\forall x \in A $ we have $x\in B$

$A\cup B$ means that either $x\in A$ or $x\in B$ or both.

We already know that $\forall x \in A $ we have $x\in B$, thus $A\cup B$ will just be the set $B$.

Q2: If $(A\cup B)\subset (B\cup C) $ then $A\subset B$.

If $x\in A$ or $x\in B$ or both then $x\in B$ or $x\in C$ or both.

There may be some $x\in A$ such that $x\notin C$ but I don't understand how to write it formally.

Can someone please verify both of my attempts?

答:
The first question has already been answered here. The second claim is not true in general.

Take $A = \{1\}$, $B = \{2\}$ and $C = \{1,3\}$. Thus, $A\cup B = \{1,2\} \subset \{1,2,3\} = B\cup C$ but $A\not\subset B$.

Hopefully this helps.


问:
Prove Laplace's equation is rotation invariant
Prove that Laplace's equation $\Delta u = 0$ is rotation invariant; that is, if $O$ is an orthogonal $n\times n$ matrix and we define

$$       v(x) := u(Ox) \quad (x \in \mathbb{R}^n)$$
then $\Delta v = 0$.

I wanted to see if what I had below was correct and complete. Any feedback on rigor is greatly appreciated.

In cartesian coordinates, we have $\Delta u = \sum_{i = 1}^n \frac{\partial^2 u}{\partial x_i^2}$. Letting the $i$th row and $j$th column of the matrix $O$ be $a_{ij}$ we have

\begin{equation}
 \frac{\partial v}{\partial x_j} = \sum_{i =1}^n\frac{\partial u}{\partial x_i}a_{ij}
\end{equation}

Thus,
\begin{equation}
 \frac{\partial^2 v}{\partial x_j^2} = \sum_{i =1}^n\sum_{k = 1}^n\frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj}
\end{equation}

Therefore:
\begin{align}
\Delta v &= \sum_{j = 1}^n \frac{\partial^2 v}{\partial x_j^2}\\
    &= \sum_{j = 1}^n \left(\sum_{i =1}^n\sum_{k = 1}^n\frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj}\right)\\
    &= \sum_{i =1}^n\sum_{k = 1}^n\left(\sum_{j = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj}\right)\\
    &= \sum_{i =1}^n\sum_{k = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}\left(\sum_{j = 1}^na_{ij}a_{kj}\right)\\
    &=\sum_{i =1}^n\sum_{k = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}\delta_{i, k}\\
    &= \sum_{i =1}^n\frac{\partial^2 u}{\partial x_i^2} = 0
\end{align}

答:
A coordinate-free proof uses that $\triangle u = {\rm tr}(D\nabla u)$, where $\nabla u:\Bbb R^n \to \Bbb R^n$ is the gradient function of $u$ and the linear map $D(\nabla u)(x):\Bbb R^n \to \Bbb R^n$ is the Hessian of $u$ at $x$.

The chain rule says that $$\langle \nabla(u\circ O)(x) ,v\rangle = D(u\circ O)(x)(v) = Du(Ox)\circ DO(x)(v) = Du(Ox)(Ov) = \langle \nabla u(Ox),Ov\rangle = \langle O^\top \nabla u(Ox),v\rangle,$$ so $\nabla(u\circ O)(x) = O^\top \nabla u(Ox)$. Apply $D$ to get $$D\nabla (u\circ O)(x)(v) = O^\top D\nabla u(Ox)O(v).$$ Thus $$\triangle(u\circ O)(x) = {\rm tr}(O^\top D\nabla u(Ox) O) = {\rm tr}(OO^\top D\nabla u(Ox)) = {\rm tr}(D\nabla u(Ox)) = (\triangle u)(Ox),$$by cyclic invariance of trace and $OO^\top = {\rm Id}$. So $\triangle(u\circ O) = (\triangle u)\circ O$. The conclusion is that $u$ is harmonic if and only if $u\circ O$ is harmonic for one (and hence every) orthogonal map $O$.


问:
Let $X$ be a non-negative random variable. Show $(EX)(E\frac{1}{X})\geq 1$
Let $X$ be a non-negative random variable. Show $(EX)(E\frac{1}{X})\geq 1$. What does this say about the correlation between $X$ and $\frac{1}{X}$.

Further show that if $f,g$ are positive borel measurable functions  such that $f(x)g(x)\geq1$ then $Ef(X)Eg(X)\geq 1$

I suppose I could start with proving the second part of the question and then prove the first part as a consequence.

I know I'm going to ultimately use Holder's inequality for this one, but I'm struggling to clear the $p$ and $q$ when I integrate $$1=\int_\Omega 1 dP\leq \int_\Omega f(X)g(X) dP\leq \left(\int_\Omega f(X)^pdP\right)^{1/p}\left(\int_\Omega g(X)^qdP\right)^{1/q}$$

Is the product of those integrals at the end less than or equal to $$\left(\int_\Omega f(X)dP\right)\left(\int_\Omega g(X)dP\right)?$$

If so, how could I prove this?

Thanks in advance for any suggestions.

答:
Note that $f(X)g(X)\geqslant 1\iff \sqrt{f(X)g(X)}\geqslant 1$, therefore the Cauchy-Schwarz inequality give us

$$
1= \left(\int_{\Omega }1\mathop{}\!d P\right)^2\leqslant \left(\int_{\Omega }\sqrt{f(X)g(X)}\mathop{}\!d P\right)^2\leqslant \left(\int_{\Omega }f(X)\mathop{}\!d P\right)\left(\int_{\Omega }g(X)\mathop{}\!d P\right)
$$


问:
Calculate center of mass using integration
How to calculate center of gravity having figure described by those equations with integration? This can and should be solved by numerical analysis.

$y < -x + 1\\ y > x^2 - 1\\ y < \sin(x)$

Firstly I found bounding points for integration:

$-x + 1 = \sin(x) \\ x_0 = -0.63673265$

$sin(x) = -x + 1 \\ x_1 = 0.51097343$

$-x + 1 = x^2 - 1 \\ x_2 = 1$

Then I've calculated integral of $g(x)$ from $x_0$ to $x_1$:

$\int_{x_0}^{x_2} |x^2 - 1| dx = 1.2173486603176453$

Then integral of $h(x)$:

$\int_{x_0}^{x_1} |\sin(x)| dx = 0.32334348721908174$

Then integral of $f(x)$:

$\int_{x_1}^{x_2} |-x + 1| dx = 0.11957349302898969$

So the area between is equal to
$$\int_{x_0}^{x_1} |\sin(x)| dx + \int_{x_1}^{x_2} |-x + 1| dx - \int_{x_0}^{x_2} |x^2 - 1| dx = -0.7744316800695739$$

Which is something wrong I guess.

答:
HINT:

You have neatly found three roots as cutting points/limits $(x_0,x_1,x_2) $.

Next area evaluation of definite integrals by integrating the difference of the functions between two curves one at a time in between the limits you found.

$$ A= \int_{x_0}^{x_1} (h-g) dx +\int_{x_1}^{x_2} (f-g) dx $$

For x-moment

$$ \int_{x_0}^{x_1} x (h-g) dx +\int_{x_1}^{x_2} x (f-g) dx $$

Divide by A to get CG x coordinate.

For y-moment

Take average $y$ between $ (h/g  \quad f/g) $ curves:

$$ \int_{x_0}^{x_1} y (h-g) dx +\int_{x_1}^{x_2} y(f-g) dx $$

Divide by A to get CG y coordinate.


答:
Quick way to show that $S$ is dense in $\Bbb{R}$:

$S = \{ a + b \sqrt{3} : a ,b \in \mathbb{Z} \}$ contains the element $\zeta := 2 - \sqrt{3}$

Since $S$ is closed under addition and multiplication, $\zeta^n$ is in $S$ for every $n \geq 2$

For any $x \in \Bbb{R}$, $x$ is at most $\zeta^n/2$ away from some element of $\zeta^n \Bbb{Z} \subset S$

Since $0 < \zeta < 1$, $\zeta^n \to 0$ as $n \to \infty$.

Title: A subset of $\mathbb{R}$ that is dense in $\mathbb{R}$.

Body:
A subset $S$ of $\mathbb{R}$ is said to be dense in $\mathbb{R}$ if $\forall$ $\epsilon > 0$ and $x \in \mathbb{R}$, $\exists$ $s \in S$ such that $| x - s| < \epsilon$.

Using this definition, I would like to try to prove that the set $S = \{ a + b \sqrt{3} : a ,b \in \mathbb{Z} \}$ is dense in $\mathbb{R}$.

Any help would be appreciated.

I am aware that the set $S$ is closed under addition and multiplication. I'm not sure if that helps.


问:
Finding the density function of this point uniformly picked at random
Consider a point picked uniformly at random from the area inside the shape with the coordinates (-2,0), (0,2), (2,0), (0,-2).

I need to find "the density function of the x coordinate".

The solution is for [-2,2]: $f(x)dx=P(X \in dx) = \frac{2*(2-|x|)dx}{4*(\frac{1}{2}*2*2)}=\frac{1}{4}(2-|x|)dx$.

So $f(x)=\frac{1}{4}(2-|x|)$ on $[-2,2]$ and otherwise $f(x)=0$.

It is $\frac{1}{4}(2+x)$ for $[-2,0]$ and $\frac{1}{4}(2-x)$ for $[0,2]$.

I found a formula for $U=\frac{X-a}{b-a}$ for $(a,b)$ for uniform distribution, but I don't think this is it, because then the solution to the question would be $\frac{X+2}{4}$ without $X$ being an absolute value. Plus it doesn't work with other similar questions.

Please help, it's a really weird exercise question on the textbook that doesn't give us any hint on how to solve it. I don't know how they got the $\frac{2*(2-|x|)dx}{4*(\frac{1}{2}*2*2)}$.

Also I have no idea how the solution's $f(x)$ means $f(x)=0$ when $x$ is not in $[-2,2]$

答:
The formula $\frac{x-a}{b-a}$ is for the density of a uniform distribution on $[a,b]$. But the $x$-coordinate of your randomly chosen point from the shape is not uniform on $[-2, 2]$. The $x$-coordinate is more likely to be close to zero than it is to be close to $-2$ or $2$.

The density $f(x) \, dx$ can be approximated by finding $P(X \in dx)$, which is a shorthand for $P(x \le X < x + dx)$. Can you draw a picture of this sub-region inside your shape? It should look like a thin vertical strip with width $dx$ and height $2(2-|x|)$. The area of this region is approximately $(2-|x|) \, dx$. The area of the entire shape is $8$. Since $X$ is uniform on the shape, the probability is the area of the strip divided by the area of the shape. So, $P(x \le X < x + dx) = \frac{2(2 - |x|) \, dx}{8}$.


答:
Here is a solution using Jensen's inequality and the fact that $x\mapsto \frac{1}{x}$ is convex on $(0,\infty)$.

Jensen's inequality implies:

$$\frac{1}{\mathbb{E}[X]} \leq \mathbb{E}[\frac{1}{X}],$$
which immediately solves the first part. For the second part we can again use Jensen's inequality to state, that
$$\frac{1}{\mathbb{E}[g(X)]} \leq \mathbb{E}[\frac{1}{g(X)}]\leq \mathbb{E}[f(X)]$$
and conclude, that $1 \leq \mathbb{E}[g(X)] \mathbb{E}[f(X)]$.

Remark: Technically you would need to handle the cases $\mathbb{E}[X] = \infty$ and $\mathbb{E}[g(X)] = \infty$ separately, but the statements are quite easy to verify in these cases.

Body:
Let $X$ be a non-negative random variable. Show $(EX)(E\frac{1}{X})\geq 1$. What does this say about the correlation between $X$ and $\frac{1}{X}$.

Further show that if $f,g$ are positive borel measurable functions  such that $f(x)g(x)\geq1$ then $Ef(X)Eg(X)\geq 1$

I suppose I could start with proving the second part of the question and then prove the first part as a consequence.

I know I'm going to ultimately use Holder's inequality for this one, but I'm struggling to clear the $p$ and $q$ when I integrate $$1=\int_\Omega 1 dP\leq \int_\Omega f(X)g(X) dP\leq \left(\int_\Omega f(X)^pdP\right)^{1/p}\left(\int_\Omega g(X)^qdP\right)^{1/q}$$

Is the product of those integrals at the end less than or equal to $$\left(\int_\Omega f(X)dP\right)\left(\int_\Omega g(X)dP\right)?$$

If so, how could I prove this?

Thanks in advance for any suggestions.


问:
Derivative with multiplication and division
So I have the following homework. I don't want the answer, only point me in the right direction please. Thanks.

I'm stuck in the product rule. Do I apply the product rule twice or just one time after applying the quotient rule?

$$
\frac{(2x^7-x^2)(x-1)}{(x+1)}
$$

I get this.
$$
\frac{(x+1)\frac{d}{dx}(2x^7-x^2)(x-1)-(2x^7-x^2)(x-1)\frac{d}{dx}(x+1)}{(x+1)^2}
$$

By writing this down here I think I answered my own question. I should apply the product rule to the first $$(2x^7-x^2)(x-1)$$ but not to the second one, since the second one (after the '-' sign) doesn't have the $$\frac{d}{dx}$$ am I right?

Sorry for any confusion.

答:
Why not just expand the product in the numerator so that you have $$\frac{2x^{8} - x^{3} - 2x^{7} + x^{2}}{x+1}$$

Then you just apply the quotient rule, no problems.

More to the point of your question, though, I only see one application of the product rule. Apply the quotient rule, and in taking the derivative of the expression in the numerator, you use the product rule. As you said in your post, I don't see any product in the second expression to apply the product rule to, so I think you've got it.


问:
Summation Translation
I translated a sum of terms being multiplied into two summations being multiplied, but I'm not sure if the summation's I've made matches the sum I originally had.

Original Equation: $\frac{26!}{8!(26-8)!} + \frac{26!}{7!(26-7)!}(\frac{26!}{1!(26-1)!}) + \frac{26!}{6!(26-6)!}(\frac{26!}{2!(26-2)!}) + \frac{26!}{5!(26-5)!}(\frac{26!}{3!(26-3)!}) = 264,517,825$

My Summations: $\sum_{n=8}^5 \frac{26!}{n!(26-n)!} * \sum_{i=0}^3 \frac{26!}{i!(26-i)!}$

My calculator enters overflow mode when I try to check with that.

答:
The two are not equal. You started with something of the form

$$a_1b_1+a_2b_2+a_3b_3+a_4b_4\tag{1}$$

and changed it to

$$(a_1+a_2+a_3+a_4)(b_1+b_2+b_3+b_4)\,.\tag{2}$$

If you multiply out $(2)$, you get all $4\cdot4=16$ terms of the form $a_ib_j$ with $1\le i,j\le 4$, not just the ones with $i=j$ that appear in $(1)$. You can see very easily that the two aren’t the same by letting all $8$ of the variables be $1$: then $(1)$ evaluates to $4$, but $(2)$ evaluates to $4\cdot 4=16$.

Title: Summation Translation


答:
You need only one sum and not a product of two sums.

$$ \sum _{n=5} ^{8} ( \frac{26!}{n! (26-n)!} . \frac{26!}{(8-n)!(26-8+n)!})  $$

Title: Summation Translation

Body:
I translated a sum of terms being multiplied into two summations being multiplied, but I'm not sure if the summation's I've made matches the sum I originally had.

Original Equation: $\frac{26!}{8!(26-8)!} + \frac{26!}{7!(26-7)!}(\frac{26!}{1!(26-1)!}) + \frac{26!}{6!(26-6)!}(\frac{26!}{2!(26-2)!}) + \frac{26!}{5!(26-5)!}(\frac{26!}{3!(26-3)!}) = 264,517,825$

My Summations: $\sum_{n=8}^5 \frac{26!}{n!(26-n)!} * \sum_{i=0}^3 \frac{26!}{i!(26-i)!}$

My calculator enters overflow mode when I try to check with that.


问:
How to prove $a^{1729}\:=_{1729}\:\:a$? I know it's a Carmichael number, but how can we show it directly without knowing it's a Carmichael number?
I already proved it if $1729$ divides $a$, or if it doesn't divide $a$ but it's Greatest Common Divisor is not equal to $1$ (we get that the left side is equal to $0$ and the right side is equal to $0$ thus in these cases $a^{1729}\:=_{1729}\:\:a$ is a true statement).

But I can't find a proof for $1729$ does not divide $a$ and it's Greatest Common Divisor is equal to $1$.

Body:
$1729 = 7*13*19$ and $a^6 \equiv 1 \pmod 7$ so $a^{7*13*19}\equiv a^{1*1*1}\equiv a\pmod 7$.

And $a^{12}\equiv 1 \pmod {13}$ so $a^{7*13*19}\equiv a^{7*1*7}\equiv a^{49}\equiv a^{48+1} \equiv a \pmod {13}$

And $a^{18}\equiv 1 \pmod {19}$ so $a^{18}\equiv 1 \pmod {19}$ so $a^{7*13*19}\equiv a^{7*13*1}\equiv a^{91}\equiv a^{90 + 1}\equiv a \pmod{19}$.

So by Chinese Remainder Theorem $a^{1729}\equiv a \pmod{7*13*19}$.


答:
The original equation is $$\binom{26}{8}+\binom{26}{7}\cdot\binom{26}{1}+\binom{26}{6}\cdot\binom{26}{2}+\binom{26}{5}\cdot\binom{26}{3}$$

Expanding out your sum gives: $$\bigg[\binom{26}{1}+\binom{26}{2}+\binom{26}{3}\bigg]\bigg[\binom{26}{5}+\binom{26}{6}+\binom{26}{7}+\binom{26}{8}\bigg]$$ these quite clearly don't match, your summation is significantly larger.

To solve this, note that $\binom{26}{0}=1$, and so the original equation can be seen as: $$\binom{26}{8-0}\cdot\binom{26}{0}+\binom{26}{8-1}\cdot\binom{26}{1}+\binom{26}{8-2}\cdot\binom{26}{2}+\binom{26}{8-3}\cdot\binom{26}{3}$$

And we can write this in sum notation as: $$\sum_{n=0}^3 \binom{26}{8-n}\cdot\binom{26}{n}=\sum_{n=0}^3\frac{26!^2}{n!\cdot (8-n)!\cdot (26-n)!\cdot (18+n)!}$$

Title: Summation Translation


问:
How do we prove the continuity of the inner product?
Let $\lim x_{k}=a$, $\lim y_{k}=b$ in $\mathbb{R^n}$ then it must be proved that $\lim \langle x_{k}, y_{k}\rangle = \langle a, b\rangle$, where $\langle a, b \rangle$ is the inner product in $\mathbb{R^n}$

The idea I have is that $ \langle x_{k}, y_{k} \rangle$ $=$ $\sum_{i=1}^{n} x_{i}^k y_{i}^k$ then $\lim \langle x_{k},y_{k}\rangle = \lim \sum_{i=1}^{n} x_{i}^k y_{i}^k$ $=$ $\lim x_{1}^ky_{1}^k+...+\lim x_{n}^ky_{n}^k$

That would be fine and being like that how I continue?

答:
Proposition

Given a sequence $x_{k} = (x_{1}^{k},x_{2}^{k},\ldots,x_{n}^{k})\in\mathbb{R}^{n}$, it converges to $a = (a_{1},a_{2},\ldots,a_{n})$ iff $x_{j}^{k}$ converges to $a_{j}$.

Proof

Let us prove the implication $(\Rightarrow)$ first.

Suppose that $x_{k}\in\mathbb{R}^{n}$ converges to $a$. Let $\varepsilon > 0$. Then there exists a natural number $N_{\varepsilon}\in\mathbb{N}$ s.t.
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow \|x_{k} - a\| \leq\varepsilon
\end{align*}

But we do also know that $|x_{j}^{k} - a_{j}| \leq \|x_{k} - a\|$. Hence we conclude that $x_{j}^{k}\to a_{j}$, and we are done.

Let us now prove the converse implication $(\Leftarrow)$.

We shall assume that $x_{j}^{k}$ converges to $a_{j}$. Let $\varepsilon > 0$. Then there exists $N_{\varepsilon_{j}}\in\mathbb{N}$ s.t.
\begin{align*}
k\geq N_{\varepsilon_{j}} \Rightarrow |x_{j}^{k} - a_{j}| \leq \varepsilon/n
\end{align*}
Consequently, if we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}},\ldots N_{\varepsilon_{n}}\}$, we conclude that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow \|x_{k} - a\| \leq |x_{1}^{k} - a_{1}| + |x^{k}_{2} - a_{2}| + \ldots + |x^{k}_{n} - a_{n}| \leq n\times\frac{\varepsilon}{n} = \varepsilon
\end{align*}
and we are done.

Proposition

Let $a_{k}\in\mathbb{R}$ and $b_{k}\in\mathbb{R}$ be sequences s.t $a_{k}\to a$ and $b_{k}\to b$. Then $a_{k} + b_{k}$ converges to $a + b$.

Proof

Let $\varepsilon > 0$. Then there are natural numbers $N_{\varepsilon_{1}}$ and $N_{\varepsilon_{2}}$ such that
\begin{align*}
\begin{cases}
k\geq N_{\varepsilon_{1}} \Rightarrow |a_{k} - a| \leq \varepsilon/2\\\\
k\geq N_{\varepsilon_{2}} \Rightarrow |b_{k} - b| \leq \varepsilon/2
\end{cases}
\end{align*}

Thus if we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}}\}$, one has that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow |a_{k} + b_{k} - a - b| \leq |a_{k} - a| + |b_{k} - b| \leq \varepsilon/2 + \varepsilon/2 = \varepsilon
\end{align*}
whence we conclude that $a_{k} + b_{k} \to a + b$.

Proposition

Let $a_{k}\in\mathbb{R}$ and $b_{k}\in\mathbb{R}$ be sequences such that $a_{k}\to a$ and $b_{k}\to b$. Then $a_{k}b_{k}$ converges to $ab$.

Proof

Let us consider the expression:
\begin{align*}
|a_{k}b_{k} - ab| = |a_{k}b_{k} - ab_{k} + ab_{k} - ab| & \leq |b_{k}||a_{k} - a| + |a||b_{k} - b|\\\\
& \leq |B||a_{k} - a| + |a||b_{k} - b|
\end{align*}

where $|b_{k}|\leq B$ (since $b_{k}$ converges). Since $a_{k}\to a$ and $b_{k}\to b$, let $\varepsilon > 0$. Then there are natural numbers $N_{\varepsilon_{1}}$ and $N_{\varepsilon_{2}}$ such that
\begin{align*}
\begin{cases}
k\geq N_{\varepsilon_{1}} \Rightarrow |a_{k} - a| \leq \displaystyle\frac{\varepsilon}{2(|B| + 1)}\\\\
k\geq N_{\varepsilon_{2}} \Rightarrow |b_{k} - b| \leq \displaystyle\frac{\varepsilon}{2(|a| + 1)}
\end{cases}
\end{align*}

If we take $N_{\varepsilon} = \max\{N_{\varepsilon_{1}},N_{\varepsilon_{2}}\}$, one has that
\begin{align*}
k\geq N_{\varepsilon} \Rightarrow |a_{k}b_{k} - ab| \leq \left(\frac{|B|}{2(|B| + 1)} + \frac{|a|}{2(|a| + 1)}\right)\varepsilon \leq \varepsilon
\end{align*}
whence we conclude that $a_{k}b_{k}\to ab$.

Solution

Based on the above-mentioned results, if $x_{k}\to a$ and $y_{k}\to b$, then $\langle x_{k},y_{k}\rangle$ converges to $\langle a,b\rangle$.

Hopefully this helps.


问:
Continuity of "dotted" function
$f(x)=1$; $f:\{1,2\} \rightarrow \{1\}$ apparently this function is continous. But we can't neither argue with a $\delta-Neighbourhood$ nor with two sided $lim_{+-}$. What would be the right approach here? Thanks!

答:
Two approaches based on point set topology come to mind, which feel more natural than an approach based on analysis

If the topology you give $X = \{1, 2\}$ is the one induced from $\mathbb{R}$, then you end up with the discrete topology, so that the open sets are $\{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$. Using the "open-set" definition of continuity, can you show that all functions from a discrete space to any topological space are continuous? Open-set continuity: a function $f: X\to Y$ is continuous iff for any open set $V$ of $Y$, the set $\{x | f(x) \in V\}$ is open in $X$.

The function you've chosen is constant. An equivalent definition of continuity is the closed-set definition, where a function $f: X\to Y$ is continuous iff for any closed set $C$ of $Y$, the set $\{x | f(x) \in C\}$ is closed in $X$. Using this definition, can you show that all constant between two topological spaces are continuous?

Title: Continuity of "dotted" function


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

I am completely stuck!

答: Hint: a skew-symmetric matrix commutes with its transpose, and so it is diagonalizable. Your block matrix on your right hand side is also skew-symmetric, and so it is also diagonalizable.


问:
How can I find the value of the coefficients of an inverse exponential graph?
If I have a set of discrete data points and I want to find the line of best fit of the form $$y=Ae^{-Bx}$$

then I can do the following:
$$\ln(y)=-Bx+\ln(A)$$

Therefore, if I plot $(x,\ln(y)$), the gradient is the $-B$ coefficient of my line of best fit and my $y$-intercept is $\ln(A)$ of my line of best fit.

Note: The above method, finds the values of the coefficient of my line of best fit using all data points, and not just choosing two and solving a set of equations.

Now, if I have a set of data points and I want to fit a function of the following form:
$$y=A(e^{-Bx}-1)+C$$

how can I find the value of the coefficients? The method above would not work for this scenario. Instead how could I identify the value of the coefficients $A, B$ and $C$?

Note: ${A,B,C \in \mathbb R }$ and $x$ and $y$ can be either positive or negative.

答:
I presume you are already aware that by doing a linear regression on $\log y$ you are minimizing
the relative error of $y$ , i.e. $\Delta y/y$ and not just $\Delta y$.

That said, in linear regression the line is always passing through the barycenter of the cloud of points $\left( {\overline x ,\,\overline y } \right)$.

So I would suggest that you operate in this way
$$
\eqalign{
  & y = A\left( {e^{\, - Bx}  - 1} \right) + C = Ae^{\, - Bx}  + D  \cr
  & \quad  \Downarrow   \cr
  & \left( {y - \bar y} \right)
 = \left( {Ae^{\, - B\,\bar x} } \right)e^{\, - B\left( {x - \bar x} \right)}  + \left( {D - \bar y} \right)  \cr
  & \quad  \Downarrow   \cr
  & \left\{ \matrix{
  D - \bar y \hfill \cr
  \Delta y = A'e^{\, - B\Delta x}  \hfill \cr}  \right. \cr}
$$


问:
Monotone Convergence Theorem applied to the limit of a real-valued function

I have a given function $f(x,\omega)\geq 0$, where
$x\in\mathbb{R}$ is a parameter and $\omega\in\Omega$, some sampling
space. $f(x,\omega)$ is increasing in $x$ for any $\omega$. I am interested in showing that

$$\lim_{x\to\infty}\int_\Omega
f(x,\omega)d\omega=\int_\Omega \lim_{x\to\infty} f(x,\omega)d\omega.$$

What I have done thus far:

Pick any sequence $\{x_n\}_{n\in\mathbb{N}}$, such that $x_n\uparrow\infty$ when $n\to\infty$. Define $f_n(\omega)\equiv f(x_n,\omega)$ for every $n\in\mathbb{N}$, and $f(\omega)\equiv \lim_{n\to\infty}f_n(\omega)=\lim_{x\to\infty}f(x,\omega)$. Then, as $f_n(\omega)\geq 0$ and $f_n(\omega)\uparrow f(\omega)$, the Monotone Convergence Theorem implies that

$$\lim_{n\to\infty}\int_\Omega f_n(\omega)d\omega=\int_\Omega f(\omega)d\omega.$$

What I don't know is how to go back from having the MCT working for every sequence $\{x_n\}_{n\in\mathbb{N}}$, such that $x_n\uparrow\infty$ when $n\to\infty$, to working for $x\to\infty$, as the latter technically is an uncountable index. I found this post but I find the answers highly unprecise with the details. Any help is greatly appreciated.

答:
Let $I_x:=\int_{\Omega }f(x,\omega )\mathop{}\!d \omega $, then if for all sequences $\{x_n\}$ that diverges to infinity you have that $\lim_{n\to \infty }I_{x_n}=I$, for some $I\in \mathbb{R}$, then $\lim_{x\to \infty }I_x=I$.

Proof: if it would not be true then, by the logical negation of a convergent functional limit, there exists some $\epsilon >0$ such that for any $\delta >0$ exists some $x_\delta >\delta $ such that $|I_{x_\delta }-I|>\epsilon $. Now taking $\delta =n$ for each $n\in \mathbb N $ we have a sequence $\{x_n\}$ that diverges to infinity but $|I_{x_n}-I|>\epsilon $ for all $n$, what contradicts the assumption that $\lim_{n\to \infty }I_{x_n}=I$.$\Box$

By last its easy to see that if $x<y\implies I_x\leqslant I_y$ for all $x,y\in \mathbb{R}$ then $\lim_{n\to \infty }I_{x_n}=\sup_{x\in \mathbb{R}}I_x$ for all sequences $\{x_n\}$ diverging to infinity (you can prove it again by contradiction, assuming that $\lim_{n\to \infty }I_{x_n}<\sup_{x\in \mathbb{R}}I_x$ and using the definition of supremum).


问:
How many vertices of degree 3 or more can a tree have at most?
It is known that a tree $T=(V,E)$ has at least $\Delta$ leaves, where $\Delta$ is the maximum degree of $T$. But how many vertices of specific degree at least $k$ can a tree have at most? I'm particularly interested in $k=3$, but also in general. Is there an easy way to see or count this?

Let $s$ be the number of vertices of degree 3 or more in any tree. I think it's easy to see that $s < |V|$, but this is probably not tight.

答:
If we denote the number of vertices as $n$, from the hand shaking lemma and the fact that the number of edges in a tree is $n-1$, we have
$$\sum_{v \in V} d_n = 2 \vert E \vert = 2(n-1)$$
If $s_k$ vertices have degree at-least $k$, then we have
$$2(n-1) = \sum_{v \in V} d_n \geq s_k \times k +(n-s_k)$$
Hence, we get that
$$s_k(k-1) \leq n-2 \implies s_k \leq \dfrac{n-2}{k-1} \implies s_k \leq \left\lfloor\dfrac{n-2}{k-1} \right\rfloor$$
This bound is tight, i.e., you can construct a tree with exactly $\left\lfloor\dfrac{n-2}{k-1} \right\rfloor$ vertices of degree $k$.

Hence, in the case of $k=3$, we have $s_3 \leq \left\lfloor\dfrac{n}2 \right\rfloor - 1$.

I hope I understood your question correctly.


问:
Determine if estimator is unbiased
I need to find out if the following estimator for a regression with no intercept (ie. $Y_i = \beta X_i + \epsilon_i$) is unbiased.

$$\hat{\beta} = \frac{\sum x_i^2y_i}{\sum x_i^2}$$

I'm given $\bar{x},\bar{y} $ both = 0. I know I need to find if $\mathbb{E}[\hat{\beta}] = \beta $, but I keep getting stuck. Any help would be really appreciated!

Update: I have changed the summation signs to be more clear, sorry this is my first time using this site and I'm still working on getting the code correct.

答:
\begin{align}
\mathbb{E}\big(\hat{\beta}\big)&= \mathbb{E}\left(\frac{\sum_i X_i^2\big(\beta X_i+\epsilon_i\big)}{\sum_iX_i^2}\right)\\
&= \beta\,\mathbb{E}\left(\frac{\sum_i X_i^3}{\sum_iX_i^2}\right)+ \mathbb{E}\left(\sum_i\epsilon_i\left(\frac{X_i^2}{\sum_j X_j^2}\right)\right)\\
&=\beta\,\mathbb{E}\left(\frac{\sum_i X_i^3}{\sum_iX_i^2}\right)+ \sum_i\mathbb{E}\big(\epsilon_i\big) \mathbb{E}\left(\frac{X_i^2}{\sum_j X_j^2}\right)\\
&= \beta\,\mathbb{E}\left(\frac{\sum_i X_i^3}{\sum_iX_i^2}\right)
\end{align}
Thus $\ \frac{\sum_i x_i^2y_i}{\sum_ix_i^2}\ $ will be an unbiassed estimator of $\ \beta\ $ if and only if $\ \mathbb{E}\left(\frac{\sum_i X_i^3}{\sum_iX_i^2}\right)=1\ $, which will not usually be the case.  Note, however, exactly the same technique will show $\ \frac{\sum_i x_i^2y_i}{\sum_ix_i^3}\ $ to be an unbiassed estimator.  Are you sure you have the right exponent in the denominator of your putative estimator?

Title: Determine if estimator is unbiased


问:
Transform a complex function in (x,y) into one in z?
I was working on a problem that asked us to find the harmonic conjugate of $u=x^3-3xy^2$.

After determining that $u$ is indeed harmonic by using Laplace's equation $\nabla u^2=\frac{\partial ^2}{\partial x^2}\left(u\right)+\frac{\partial ^2}{\partial y^2}\left(u\right)=0$, I determined its harmonic conjugate to be $v(x,y)=3x^2y-y^3+C$ by using the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

Now, the complex function is $f(x,y)=u(x,y)+iv(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$.

I would like to transform this into $f(z)$ instead of $f(x,y)$. How would I go about doing this?

I have tried to substitute the equations $x=\frac{z+\overline{z}}{2}$ and $y=\frac{z-\overline{z}}{2i}$:

$f(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$, which becomes:

$f(z)=\left(\frac{z+\overline{z}}{2}\right)^3-3\left(\frac{z+\overline{z}}{2}\right)\left(\frac{z-\overline{z}}{2i}\right)^2+i\left[3\left(\frac{z+\overline{z}}{2}\right)^2\left(\frac{z-\overline{z}}{2i}\right)-\left(\frac{z-\overline{z}}{2i}\right)^3\right]+iC$.

But this seems to simplify with a $\overline{z}$ in the numerator. It is my understanding that an analytic function will not have $\overline{z}$? I am not sure what I am doing wrong.

答:
Notice the expression

$$\displaystyle z^3 = (x + iy)^3 = x^3 + 3ix^2 y - 3xy^2 - iy^3.$$

From here one sees that

$$\displaystyle f(x,y) = (x^3 - 3xy^2) + i(3x^2 y - y^3) + C = z^3 + C.$$


问:
Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$
Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?

答:
Integrate by parts

\begin{align}
& \int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} dx\\\overset{IBP} = & ab\int_0^{\infty} \frac1x
\left(\frac1{1+ax}- \frac1{1+b x} \right) dx
 = ab \int_0^{\infty}
\left(-\frac a{1+ax}+ \frac b{1+b x} \right) dx \\
=& ab\ln\frac{1+bx}{1+ax}\bigg|_0^\infty = ab\ln\frac ba
\end{align}


问:
Hypergeometric distribution mean.
We have a box with 40 balls inside it, one for each child. 15 black balls and 25 white balls.
The reward for getting a black ball is noted Rb, and for white ball is noted Rw, where Rb > Rw.
I'm the 20th child in line to randomly pick a ball, and 2 children before me already got a black ball.
The last child in line gives me an offer that if I get a black ball and he doesn't, he'll pay me 2$ to switch balls with him.

What is my expected reward if I don't accept the deal?
What is my expected reward if I accept the deal?

I thought about using the mean formula of hipergeometric distribution but it does not take into account the 2 black balls chosen so its clearly not the way to go (I think).
What am I missing here? Can anyone help please?

答:
By the time you get to pick, there are 13 black and 8 white balls left. If you don't accept the deal, then your expected winnings are simply

$$\boxed{ \frac{13}{21} \; R_B + \frac{8}{21} \; R_W }$$

Suppose you accept the deal. Now there are three cases to consider:

Case (1): $\quad$you draw White, the deal has no effect and you earn $R_W$;

$ \qquad \qquad \; \; $ the probability is $\dfrac{8}{21} $

Case (2): $\quad$you and the other child both draw Black, the deal has no effect and you earn $R_B$;

$ \qquad \qquad \; \; $ the probability is $\dfrac{13}{21}\cdot \dfrac{12}{20} = \dfrac{13}{35}$

Case (3): $\quad$you draw Black, the other child draws White, the deal holds and you earn \$($2+R_W$);

$ \qquad \qquad \; \; $ the probability is $\dfrac{13}{21}\cdot \dfrac{8}{20} = \dfrac{26}{105}$

Thus your expected winnings if you do accept the deal are

$$\begin{align}
& \; \frac{8}{21} \; R_W + \frac{13}{35} \; R_B + \frac{26}{105} \; (2+R_W)\\
\\
= \quad & \boxed{ \frac{22}{35} \; R_W + \frac{13}{35} \; R_B + \frac{52}{105} }
\end{align}$$

Title: Hypergeometric distribution mean.


问:
Probability of getting the flu
Suppose that the probability of exposure to the flu during an epidemic is .6. Experience has
shown that a serum is 80% successful in preventing an inoculated person from acquiring the
flu, if exposed to it. A person not inoculated faces a probability of .90 of acquiring the flu if
exposed to it. Two persons, one inoculated and one not, perform a highly specialized task in a
business. Assume that they are not at the same location, are not in contact with the same people,
and cannot expose each other to the flu. What is the probability that at least one will get the flu?

The correct answer is 0.5952, I can't find my mistake.

A: vaccinated person catching the flu

B: unvaccinated person catching the flu

F: exposed to flu

thus, I must find P(A ∪ B)

P(F)=0.60

P(A | F)=0.20

P(B | F)=0.90

A, B ⊆ F

$P(A│F)=(P( A∩F))/(P(F))→P(A│F)=P( A)/(P(F))→ 0.20=P( A)/0.60→P(A)=0.12$

$P(B│F)=(P( B∩F))/(P(F))→P(B│F)=P( B)/(P(F))→ 0.90=P( B)/0.60→P(B)=0.54$

$P(A ∪ B)=P(A)+ P(B) -P(A∩B)=0.12+0.54-0=0.66$

Ans: 0.66

答:
The two events of $A$ and $B$ catching the flu are independent, meaning that $P(A\cap B)=P(A)P(B)$, not zero. (In the context of the question, you can't rule out the possibility that both of them catching it just because they don't come into contact)

Hence, $$P(A\cup B)=0.12+0.54-(0.12)(0.54)=0.5952$$ as required.

Title: Probability of getting the flu


问:
If $G$ is finite of odd order, then the product of all its elements, in any order, is an element of the commutator subgroup $G'$.
This is Exercise 3.36 of Roman's "Fundamentals of Group Theory: an Advanced Approach". According to Approach0, it is new to MSE.

The Details:

Neither semidirect products nor presentations are covered in the book so far.

Definition: The commutator subgroup or derived subgroup $G'$ of a group $G$ is the subgroup of $G$ generated by all commutators $$[a,b]:=aba^{-1}b^{-1}$$ (for $a,b\in G$). A group $G$ is perfect if $G=G'$.

The Question:

If $G$ is a finite group of odd order, then the product of all its elements, in any order, is an element of the commutator subgroup $G'$.

Thoughts:

If $G$ is abelian, then $$\prod_{g\in G}g=e\in G'$$ trivially.

If $G$ is perfect, there is nothing to prove.

By Lagrange's Theorem, $G$ has no involutions, so there are no elements $x\in G$ such that $x=x^{-1}$.

Putting the identity to one side, then, elements come in pairs: $y$ and $y^{-1}$. Let's separate $G$ into any $\frac{n-1}{2}$ elements $g$ in a set $P$; and their inverses $g^{-1}$, a set $N$. Note that $P\cap N=\varnothing$. Now one product of all elements of $G$ is of the form

$$e\left(\prod_{g\in P}gg^{-1}\right)=e\in G',$$

for any order of the elements of $P$.

So much for the theoretical approach . . .

Consider when $G=\Bbb Z_7\rtimes \Bbb Z_3$, given by the presentation

$$\langle a,b\mid  a^7, b^3, bab^{-1}=a^4\rangle,$$

which I believe to be the smallest, nonabelian group of odd order, according to group names.

Here $G'\cong \Bbb Z_7$, again, according to group names.

The elements of $G$ are of the form $a^ib^j$ for $i\in \{0,\dots, 6\}$ and $j\in\{0,1,2\}$. Again, taking the identity aside, there is, essentially, up to $20!$ different products of all elements of $G$; there might be subtleties that lower that upper bound, but still: it's a lot: that's what I'm getting at. Consider $[a^{\alpha_1}b^{\beta_1}, a^{\alpha_2}b^{\beta_2}].$ This is looks like it would be messy to compute in abstraction. There ought to be a slick way of doing this.

Please help :)

答: Hint: $G/G'$ is abelian, the image of the product of elements in any order by the quotient map $G\rightarrow G/G'$ is the neutral element of $G/G'$, a fact that you have noticed


问:
Estimating total number of players based a rank subset
I recently played a lottery game where there was a unique "first prize". A rank is randomly attributed to each player. You win if you get the first rank, otherwise you lose. So, my friends and I received a rank. I received 199, one received 133, another 45, etc.
The total number of player is never disclosed. How can I estimate the number of players who joined the game, and thus, the probability of winning? Note that a player cannot buy more than a ticket.

I initially had the feeling that it's about fitting a binomial cumulative density function, but the variable is clearly not iid as rank duplicates are impossible.

答:
Thanks to @lulu for pointing me in the right direction. So, to sum up, the number of player can be estimated with three methods. Let $N$ be the number of player, $m$ the highest rank (furthest from 1) and $k$ the number of samples we got:

Frequentist approach: $N=m+\frac{m}{k}-1$

Bayesian median: $N=m+\frac{m\: ln(2)}{k-1}$

Bayesian mean: $N=(m-1)\frac{k-1}{k-2}$