问:
How to solve Quadratic Equations with an Unknown C and other variables.
For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?

And also : If $x^2 +px + q$ is a factor of $x^3 + ax^2 + b$,

prove that:

$b = q(a-p)$

答:
The value of r could be anything, but I'm assuming that you mean r should be an integer and the quadratic should be factorable.

In that case, you can use the factoring to reverse engineer the quadratic. Let's say

$(3x+u)(x+w) = 3x^2 -11x +r$

where u and w are integers. Because of how multiplication of binomials works, the sum of u and w equals -11. In addition, they also are the factors of r. That being said, you can set up a system of equations to solve for r.

$u + 3w = -11$

$uw = r$

From there, some guessing and checking is necessary, but if you know your times tables, it shouldn't be too bad. Questions like these can definitely be tedious if they do not provide a point that exists on the curve.

As for the second part of your question, you can multiply the quadratic expression out with a generic binomial to organize your work. You should be able to see that for coefficient of x to the first power to equal zero in the cubic function, some measures must be taken. Let w be an integer.

$(x+w)(x^2 + px + q) = x^3 + ax^2 + b$

$x^3 + px^2 + wx^2 + qx + wpx + qw = x^3 + ax^2 + b$

For the coefficient of x to be zero in the cubic, the sum of q and wp must equal zero. However, w isn't in the cubic. We can find it, though. Let's simplify the equation first.

$px^2 + wx^2 + (q + wp)x + qw = ax^2 +b$

From here, we can already see that $p + w = a$ and $qw = b$. Therefore, we can substitute as follows.

$w = b/q$

$a = (p + b/q)$

Therefore, after canceling out the constants:

$(p + b/q)x^2 + (q + bp/q)x = ax^2

Now solve for b.

$a = p + b/q$

$b/q = a - p$

$b = q(a - p)$

Sorry for the long response and if it doesn't explain the material the way you wanted. I don't come on the stack as often as I used to, so I'm a bit out of practice.


答:
I'm not sure what you did, but it is wrong as $(1-\sqrt{3}\cdot i)^6=\boldsymbol{64}$ (Therefore the $\text{Arg}$ is $0$).

To solve this, let $z=(1-\sqrt{3}i)^6$. We'll turn $1-\sqrt{3}\cdot i$ into its polar form.

So, $|1-\sqrt{3}\cdot i|=\sqrt{1^2+(\sqrt{3})^2}=2$ like you've said. Also, $\text{Arg}(1-\sqrt{3}\cdot i)=-\dfrac{\pi}{3}$

Finally, $z=(2\text{cis}(-\dfrac{\pi}{3}))^6=2^6\text{cis}(-2\pi)=2^6\text{cis}(\boldsymbol{0})=2^6=64$

ContentLicense: CC BY-SA 4.0

Title: Arg of $(1-i\sqrt{3})^6$. Did I do it right?

Body:
Did I do it right?

$(1-i\sqrt{3})^6$

module = $\sqrt{1+3}=2$

$(1-i\sqrt{3})^6=2^6\cdot(\frac{1}{2}-\frac{\sqrt{3}}{2}i)^6$ $\implies$ Arg = $\frac{\pi}{3}$


问:
Prove that $x ^ {1 - x} (1 - x) ^ {x} \le \frac {1}{2}$
How do I prove $x ^ {1 - x}(1 - x) ^ {x} \le \frac{1}{2}$, for every $x \in (0, 1)$.

For $x = \frac {1}{2}$ the LHS is equal to one half. I tried studying what happens when $x \lt \frac {1}{2}$ and its correspondent, but to no result.

Body:
Let
$$ f(x)=\ln[x ^ {1 - x}(1 - x) ^ {x}]=(1-x)\ln x+x\ln(1-x) $$
and then
$$ f'(x)=-\ln x+\frac{1-x}{x}+\ln(1-x)-\frac{x}{1-x}, f''(x)=-\frac{1-x+x^2}{x^2(1-x)^2} .$$
Clearly $x=\frac12$ is the only point in $(0,1)$ such that $f'(x)=0$ since $f''(x)<0$ in $(0,1)$. Thus $f(x)$ obtains the max at $x=\frac12$, namely
$$ f(x)\le \ln(\frac12). $$
So
$$ x ^ {1 - x}(1 - x) ^ {x}\le \frac12. $$


问:
Trouble finding the level curves of $f(x,y)=\frac{2x}{x^2+y^2}$
I am trying to find the level curves of $f(x,y) = \frac{2x}{x^2+y^2}$ for $f(x,y) = {-1,0,1}$.

I can not find examples in Stewart Calculus for rational equations.

Any help would be greatly appreciated.

答:
In polar coordinates,

$$2r\cos\theta=lr^2$$ or

$$r=\frac2l\cos\theta.$$

This is a family of homothetic circles through the origin.

The case $l=0$ is degenerate as the axis $x=0$.


问:
Determine all chances of victory in a car race
Imagine a car racing championship (like Formula 1) made up of 13 drivers. The order of arrival score is defined as:

```
1st place: 23 points
2nd place: 20 points
3rd place: 18 points
4th place: 16 points
5th place: 15 points
6th place: 14 points
7th place: 13 points
8th place: 12 points
9th place: 11 points
10th place: 10 points
11th place: 9 points
12th place: 8 points
13th place: 7 points
```

The points are added for each pilot according to their finishing position. At the end of 4 races, whoever has the most points wins the championship. After the first two races, the partial result is:

```
| DRIVER   | RACE 01   | RACE 02   | RACE 03  | RACE 04  | POINTS    |
|----------|-----------|-----------|----------|----------|-----------|
|     A    |     20    |     23    |          |          |     43    |
|     B    |     18    |     18    |          |          |     36    |
|     C    |     15    |     20    |          |          |     35    |
|     D    |     23    |     11    |          |          |     34    |
|     E    |     14    |     15    |          |          |     29    |
|     F    |     13    |     13    |          |          |     26    |
|     G    |     11    |     15    |          |          |     26    |
|     H    |     16    |     9     |          |          |     25    |
|     I    |     12    |     12    |          |          |     24    |
|     J    |     8     |     16    |          |          |     24    |
|     K    |     7     |     10    |          |          |     17    |
|     L    |     9     |     8     |          |          |     17    |
|     M    |     10    |     0     |          |          |     10    |
```

Considering that no driver misses the last two stages, and the chances of victory or any other result are equal to all drivers, what are the possible combinations for each driver to be champion?

In other words, what are the minimum finishing positions that driver A needs to finish to be champion? And pilot B? And so on.

An example is the case in Formula 1 at the end of the 2008 championship, when driver Lewis Hamilton was in the first place and Felipe Massa in second in the standings. Observe the text taken from a news item of the time:

With five points ahead of Felipe Massa in the championship (84 to 79), Lewis Hamilton needs to make six more than the Brazilian in the
Chinese Grand Prix. In other words, in case of victory, the Englishman
needs to hope that the Ferrari driver will reach below fifth
position.

I'm trying to do some possibilities "by my hand" but this process can be enormous. I'm trying making this on R but I couldn't move forward.

Thanks!

答:
Consider the following cases:

If the driver is currently the best, then they needs to maintain their lead over the second best, and so we can determine how many points they can give up (and then distribute that across different races as desired.

If the driver is not currently the best, then they will need to win the best, who we may assume is first in a race that the driver doesn't win, or second in a race that the driver wins. This tells us the minimum number of first place finishes that are necessary.


问:
Are the names and symbols for common mathematical operators the same in every language?
Do all human languages that have arithmetic use +, -, ×, ÷, and ^?

How about sin, cos, tan, asin, acos, and atan?

答:
In Dutch, for the trigonometric and cyclometric functions, you will see

$\text{tg}$ instead of $\tan$, although $\tan$ is quite common

$\text{Bgtg}$, $\text{Bgtan}$, $\text{bgtan}$ instead of $\arctan$ or $\text{atan}$ because "arc" is "boog" in Dutch. Same for $\arcsin$ and the like.

Some other differences

$\text{ln}$ for the natural logarithm. This is probably not unusual but some English texts will use $\log$ for the natural logarithm.

One million is written $1.000.000$, a half in decimals is $0,5$. So the use of dots and commas is inverted w.r.t. English. This is true in most continental European languages by the way.

ContentLicense: CC BY-SA 3.0


问:
Supremum of nth terms of a sequence
Let us consider $x$ in $A = [0,1]$. Since $x$ is a limit point in $A$, we have many sequences converging to $x$.

Now let us construct a new sequence $(y_n)$ such that $y_n$ is the supremum of nth terms of all sequences which converges to $x$ and which are in $A$ (i.e. $0 \leq x_n \leq 1$).

This construction looks decent enough and seems obvious that $y_n$ converges to $x$. But I am confused that whether $y_n = 1$ or not. This is because I can have a sequence whose first term is 1 and that can converge to $x$, so we have $y_1 = 1$. Similarly there can be one more sequence whose 2nd term is 1, so we have $y_2 = 1$ and so on. So I will have $y_n = 1$ which will not converge to $x$.

What is wrong in the above reasoning?

Does $(y_n)$ converge to $x$ or $1$?

答:
For each $n\in\Bbb Z^+$ define a sequence $\sigma_n=\left\langle z_k^{(n)}:k\in\Bbb Z^+\right\rangle$ by setting

$$z_k^{(n)}=\begin{cases}
1,&\text{if }k=n\\
x,&\text{otherwise;}
\end{cases}$$

clearly $\sigma_n$ converges to $x$, and $z_n^{(n)}=1$, so $y_n=1$. Your sequence $\langle y_n:n\in\Bbb Z^+\rangle$ is therefore the constant sequence at $1$ and converges to $1$.

Had you considered only the strictly increasing sequences in $A$ that converge to $x$, you’d have found that $y_n=x$ for each $n\in\Bbb Z^+$, and hence $\langle y_n:n\in\Bbb Z^+\rangle$ converges to $x$.

Title: Supremum of nth terms of a sequence


问:
Having trouble finding the fault in my logic for this coding question(it's heavily math related)
So I have been doing some coding questions to practice my coding and I ran into this problem: http://www.usaco.org/index.php?page=viewproblem2&cpid=989

The general gist of the problem is that there is a person named Bessie is racing a length of $K\ge1$ meters. She starts at $0$ meters per second at the $0$ second mark, and then every second after that, she can either increase her speed by $1$ meter per second, stay at the same speed, or decrease her speed by $1$ meters per second. Her speed can not drop below zero(so if Bessie is at $0$ meters per second, she can not decrease her speed.).

Bessie wants to finish the race in an integral amount of seconds, either finishing at the $K$ meter mark or past the $K$ meter mark. But Bessie also doesn't want to finish the race too quick. At the instant when she finishes $K$ meters, she wants to be traveling at a speed of no more than $X\ge1$ meters per second. Bessie wants to know the minimum speed required to finish $K$ meters, given $K$ and $X$.

The logic I use to solve this problem only works for the first 4 test cases, and I'm sure it's not because of an coding error. So my logic is as follows:

Before we do anything, we first have to test whether or not a speed of $X$ meters per second can be reached, as the following solution assumes that $X$ meters per second is reachable.

To do so, we first note that the quickest way to get to $X$ meters per second is to increase the speed by $1$ each second for $X$ seconds. We then note that if after increasing $X-1$ times, if the distance covered is $<K$ meters, then it is guaranteed that $X$ meters per second is reachable. But if after increasing $X-1$ times the distance covered is $\ge K$, then we know that $X$ meters per second is unobtainable.

To calculate the distance covered after $X-1$ increases, we can calculate the following sum:$$1+2+\cdots+(X-2)+(X-1)$$which can be represented as$$\frac{X(X-1)}2$$. We want to test whether or not$$\frac{X(X-1)}2\ge K$$. If this inequality is false, then go to the solution under the gray line. If this inequality is true, then we know $X$ meters per second is unobtainable, and therefore we need to calculate how many increases are required to surpass $K$ meters. We will call this amount $n$. To find the value of $n$ that will cause the distance to go over $K$ meters, we first need to find the formula of the distance covered after $n$ increases. That can be represented with the sum $$1+2+3+\cdots+n=\frac{n(n+1)}2$$. So then we set this sum to be $<K$, then use this inequality to maximize $n$:$$\frac{n(n+1)}2<K\\\frac{n^2+n}2<K\\n^2+n<2K\\(n+1/2)^2-1/4<2K\\n+1/2<\sqrt{2K+1/4}\\n<\frac{\sqrt{8K+1}-1}2$$So the value of $n$ would be:$$n=\left\lceil\frac{\sqrt{8K+1}-1}2\right\rceil$$(without the ceiling function we would be calculating the largest amount of increases that doesn't surpass $K$ meters, instead of actually passing $K$ meters)

First we want to find the maximum speed in which Bessie can go at. Let's say that this maximum speed is $m$ and the target speed(the speed we want to have at $K$ meters) be $X$ meters per second(as stated in the problem). We can find the maximum speed by allowing Bessie to increase its speed every second until it reaches $m$, then immediately start decreasing her speed until she hits $X$ meters per second. We then know that the total distance traveled after this increase and decrease is(which I will denote as $d$): $$d=\underbrace{1+2+3+\cdots+m}_{\text{increasing speed}}+\underbrace{(m-1)+(m-2)+\cdots+(X+1)+X}_{\text{decreasing speed}}$$. We can then find the formula for this sum to be: $$d=m^2-\frac{X(X-1)}2$$. This sum has to be $\le K$(or else we can't decrease enough in time), so we have the following inequality:$$m^2-\frac{X(X-1)}2\le K$$. $K$ and $X$ are already given as inputs in the problem, so we just have to isolate $m$. We get that: $$m\le \sqrt{K+\frac{X(X-1)}2}$$(positive square root). To get the highest $m$, we just need to take the floor of the RHS so:$$m=\left\lfloor\sqrt{K+\frac{X(X-1)}2}\right\rfloor$$. Then if $d$ is $<K$, we need to find out the remaining distance that we need to cover. That is easy to calculate: $K-d$. From this we can calculate how many seconds we need to stay at $m$ meters per second(if we stay at a speed $<m$, we can always stay at a higher speed to potentially reach $K$ meters quicker. Not too sure about this logic though). Each second we stay at $m$ meters per second adds an extra $m$ meters to our distance. So we need to divide $K-d$ by $m$ to see how many times we need to add $m$ to $d$ to reach $K$(I will denote this as $s$). So we get that we need to stay$$s=\left\lceil\frac{K-d}m\right\rceil$$seconds at $m$ meters per second to pass $K$ meters. Then we need to calculate the amount of seconds that passed for traveling $d$ meters using the method stated. To calculate this, we need to count how many terms we added together in the sum. So we need to find the length of this list:$$1,2,3\dots,m,(m-1),(m-2),\cdots,(X-1),X$$This can be calculated with the following formula: $$2m-X$$So finally we calculate $$2m-X+s$$ to obtain the final answer.

The problem is this only works for the first 4 test cases, and this strategy only works for certain values of $K$ and $X$ past test case 4, and is really close to the actual answer for other values(yes, I downloaded the test data), so I'm actually not too sure where I went wrong here. If you guys want the code I can also put it here, but this is more of a math problem, so I decided not to put the code for now.

答:
The error is in assuming that the form must be

We can find the maximum speed by allowing Bessie to increase its speed every second until it reaches $m$, then immediately start decreasing her speed until she hits $X$ meters per second.

Do you see why?

 For large $X$, if the previous second gets us really close to the finish line, then it might be possible that we have a smaller minimum speed so that we can slow down earlier and finish the race faster. Yes, the total distance covered will be smaller (but as long as it's longer than the race, we're fine).

In fact, assuming that the race must be of your form (increase or decrease every second, end with X), then there are some distances which do not work out (as you pointed out).

For example, with $X = 5$, $K = 40$, you'd calculate a max speed of $7 < \sqrt{ 40 + \frac{5\times 4}{2} }$. We verify this with $1+2+3+4+5+6+7+6+5 = 39 < 40 < 54 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 $.

However, she then can't run a increasing/constant/decreasing race with a max speed of 7, an end speed of 5, a distance of 40 since

$1 + 2 + 3 + 4+5+6+7+7+6 = 41 $.


答:
The argument of $1- i\sqrt{3}$ is $\arctan(-\sqrt{3})= -\frac{\pi}{3}$  So the argment of $(1-i\sqrt{3})^6$ is  $6\times-\frac{\pi}{3}= -2\pi$ which is equivalent to 0.

(Edited to correct negative sign, thanks to Ripi2.)


答:
Knowing that a number is a perfect cube, you can use a bit of simple logic to find the value of it quickly.

You begin by noting the number of digits in the resulting number - if $N$ is the number of digits in the perfect cube, then you know that the cube root will have $\left\lfloor\frac{N-1}3\right\rfloor+1$ digits.

Now, look at mod $10$. In this case, the sequence from 0 to 9 produces the sequence:
$$
0, 1, 8, 7, 4, 5, 6, 3, 2, 9
$$
As such, knowing that 2<->8 and 3<->7, and the others remain where they are, you can quickly identify the last digit of the number you're looking for. For instance, with the example you provided, $157464$ ends in 4 and has $N=6$ digits, so we're looking for a $\left\lfloor\frac{6-1}{3}\right\rfloor+1 = 2$ digit number, and it ends in 4.

In the case where you know that it's a 2 digit number, it's fairly easy to then use a bit of bounding to find the right value. Notice that
$$
50^3 = 125000
$$
and
$$
60^3 = 216000
$$
As such, you know that it has to be between $50$ and $60$. With the knowledge that the last digit is $4$, you quickly get $54$ as the final answer.

Where more digits are required, a more detailed method has to be used, of course. But the logic works the same in any case.

Title: How to guess cube root of a number?

Body:
I watched on tv that a guy could say the value of $\sqrt [ 3 ]{ x } $ for a given $x$,  in less than ten seconds. Moreover, he could do for $\sqrt [ 13 ]{ x } $, $\sqrt [ 103 ]{ x } $, $\sqrt [ 1003 ]{ x } $. I know that he was looking to the the last 3-4 digits of $x$. For example if the last digit of x is $1$, than the last digit of $\sqrt [ 3 ]{ x } $ must be $1$. However I could not model for the last 3-4 digits. So, the question is:

If you know that $0<\sqrt [ 3 ]{ x }<99 $ , what is the value of $\sqrt [ 3 ]{ x }$ for a given $x$

For example for $x=157464$, you can say $\sqrt [ 3 ]{ x }=54 $ by looking for three or two digits of $x$

Additionally build the same model for $\sqrt [ 13 ]{ x } $,$\sqrt [ 103 ]{ x }$,$\sqrt [ 1003 ]{ x }$

Note: $\sqrt [ 3 ]{ x } $ and $x$ are integers

Note2: $x$ was given as 3684849759365252680119420227475775049492674443146905344822745288298143046106331441531798764782169420286244212297417702761555740760756048067228412045874270238715327556678211597193757451059582259503545059959722118963270933846793818460752099206747525659014794031653506061992511860134802242876788750520888820110080505679940477828522656005828478195470651831585339407491541412547532043564019267830576862266475657742655079354385864558909932489824922657536960135959364334680734490825496314934021284389706462034993094441427484468431831131329495454329359968306231964741732502732936584806687832283240524973095558123557893421615630299392659144456999160880961090629674198135946098548815074663065403716229881899998734677466563511307098758301725180878228542151982532894727010449270694673832788117758121432660157632144863623257572958865327742110214978113074412669861204102215454568565209907518122414604893305540601040443389773145449494505681564919573412773764876132208183779019683251734342466854174739320538873007486268980279129645406862537226568433208719346636825106745675952144307470309490075574156853916253370585261188823302350410947564043990554122093606148767744

He could say that $\sqrt [ 1003 ]{ x }=14$ very quickly.


问:
Cumulative distribution function of $\sqrt{X^2+4Y^2}$ without $\chi$ distribution
I want to find the cumulative distribution function of the random variable $Z=\sqrt{X^2+4Y^2}$, where $X\sim N(0,4)$, $Y\sim N(0,1)$ are independent. I know it is equal to $\chi_{[0,\infty)}(t)\left(1-\exp\left(-\frac{t^2}{8}\right)\right)$, but I don't know how to show it.

My research has led me to the Chi-square distribution, however, this has never come up in class, and also the CDF for this distribution looks complicated enough on its own, without even taking the square root.

Noting that $$P(\sqrt{X^2+4Y^2}\le t) = 1-P(\sqrt{X^2+4Y^2} > t)$$
it would be sufficient to show
$$P(\sqrt{X^2+4Y^2} > t) = \exp\left(-\frac{t^2}{8}\right)$$
but I don't know how that could be done. How could I show this relation?

答:
Assuming that $X$ and $Y$ are independent random variables (a property that the OP has not admitted to as yet),
\begin{align}
F_{\sqrt{X^2+4Y^2}}(t) &= P(\sqrt{X^2+4Y^2}\leq t)\\
&= P\left(\sqrt{(X/2)^2+Y^2}\leq \frac t2\right) & \scriptstyle{\text{independent }}\frac X2, Y \sim N(0,1)\\
&= P\left(2 \text{ independent } N(0,1) \text{ RVs lie in disk of radius } \frac t2\right)\\
&= \iint_{\text{disc of radius }\frac t2} \frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right)\, \mathrm dx \,\mathrm dy\\
&=\int_0^{\frac t2}\int_0^{2\pi}\frac{1}{2\pi}\exp\left(-\frac{r^2}{2}\right)\, r\mathrm d\theta \,\mathrm dr\\
&= \int_0^{\frac t2} \exp\left(-\frac{r^2}{2}\right) r\,\mathrm dr\\
&= \left.-\exp\left(-\frac{r^2}{2}\right) \right\vert_0^{\frac t2}\\
&= 1 - \exp\left(-\frac{t^2}{8}\right)
\end{align}
without mentioning $\chi^2$ random variables etc at all.


问:
Finding the times when particles with positions $6\cos(2\pi t)\hat{i}+6\sin(2\pi t)\hat{j}$ and $-6\sin(2\pi t)\hat{i}-6\cos(2\pi t)\hat{j}$ meet

The 2 particles describe circular movements, with these equations of position:

$$\begin{align}
\vec{r_1}&=\phantom{-}6\cos(2\pi t)\hat{i}+6\sin(2\pi t)\hat{j} \\
\vec{r_2}&=-6\sin(2\pi t)\hat{i}-6\cos(2\pi t)\hat{j}
\end{align}$$
Find the 2 positions where both meet, and the first 2 times where they do it.

So this isn't a doubt about the problem, but rather about if they do really meet, i'm discussing this with one friend of mine, what we did was: Equated both vectors and after simplifying a little bit we got to $$\cos(2\pi t)=-\sin(2\pi t)$$ which answers are $$t=\frac38$$ and $$t=\frac78$$ is this correct or has the teacher messed up and there is not any meeting?

答:
from the unit circle definition, $$cos(\eta)=-sin(\eta)$$ whenever $$\eta = \frac {3\pi}4+n\pi$$ for $n\in \mathcal N$(natural numbers, in case you didn't know). Therefore, solving the equation $$2\pi t=\frac {3\pi}4 +n\pi$$ we get$$t=\frac 38 +\frac n2$$ for $n =1,2,3,...$.
So, yes, you are right, but not entirely, there are infinitely many solutions for $t$ corresponding to every natural number. But, if you want to know if they really meet, then all you must do is simply solve the equation, if the equation does not give you a contradiction (i.e., 1=2, etc.) then you must plug the value you get back into the original equation and see if you get the right values. For $-\sin$ and $\cos$ particularly, they do intersect infinitely often.


问:
Integer-valued metrics from a topological view point
I am working a bit in applied mathematics, but not in theoretical mathematics. I dabble a bit in general topology, but if possible would like to ask you to consider my level of know-how and add some intuition into your answer. I can handle some formulas, but since I am not a professional mathematician I am not fluent in all the lingo... so basic terminology is appreciated.

In data analytics / data science, we have premetrics, like Dynamic time warping, which can be calculated between two real-valued sequences, even if the sequences do not have the same length (so they are not vectors). As such, we can say that the space created by the premetric is a topological space.

When we are dealt with nominal-valued sequences (aka strings), like human words, we can have only equality as an operation at our disposal for comparing elements of the sequences. To calculate a metric distance between such, we have e.g. the Levenshtein distance. This metric only outputs integer-values as distances. This would create a metric space, I guess, however not in the normal sense that we have infinitesimal close neighbors for each sequence/point.

Usually a metric space would induce a topological space. Usually topological spaces concretize the notion of what "the closest next point" is. Here, the "closest next point" is not infinitesimally close. Can we still talking about topological spaces in this context?

I have not read anywhere that the codomain of the metric of a metric space needs to be a field (which integers are not), but are we still talking about a metric space for the Levenshtein distance?

How should I think about the fact that the Levenshtein distance have integers as output?

答:
This question is predicated on a false assumption, namely that

in the normal sense [of "metric space"] we have infinitesimal close neighbors for each sequence/point.

Most basically, there are no such things as "infinitesimally close points" in a metric space: the distance between any two points is always a (nonnegative) real number, and there are no infinitesimal real numbers.

Moreover, the claim "topological spaces concretize the notion of what "the closest next point" is" is wrong for the same reason: we don't think about "closest next points" almost ever.

Even ignoring that, it sounds like you're thinking of metric spaces without isolated points (or maybe complete metric spaces, or complete and connected metric spaces, or something similar). These are just a particular type of metric space; there's no requirement that all metric spaces "look like that."

Indeed, a metric space is merely any set $X$ equipped with a function $\delta: X\times X\rightarrow\mathbb{R}_{\ge 0}$ satisfying three basic properties:

$\delta(x,y)=0\leftrightarrow x=y$.

$\delta(x,y)=\delta(y,x)$.

$\delta(x,y)+\delta(y,z)\ge\delta(x,z)$.

The metric function $\delta$ could be (nonnegative-)integer valued; that's totally fine.

That said, there is a sense in which integer-valued metrics are "less topology-flavored" than one might expect.

Suppose $(X,\delta)$ is a metric space with $ran(\delta)\subseteq\mathbb{Z}$. Then the topology on $X$ induced by $\delta$ is discrete: every set is open. In particular, any two integer-valued metrics on the same set yield the same topology on that set.

This means that if I want to compare integer-valued metrics - or more generally, discrete metric spaces, which are spaces where for each point $x$ there is some  $u>0$ such that every point other than $x$ is at distance at least $u$ from $x$ - I have to use "finer-grained" notions than the usual ones coming from topology. But that doesn't mean that integer-valued metrics aren't allowed, it just means that I might have to be careful about what questions I ask about them if I want to get meaningful answers.


问:
Is the real coordinate space in polar coordinates the same as the real coordinate system for the plane?
I understand one could transition from one to the other but I would like to know if they are the same two dimensional space. Why I ask?

In a topology book I am reading  the polar coordinates can be represented as one dimensional verticle lines from 0 to pi and map from one plane to another plane. Whereas the polar coordinate plane has a different visualization of concentric rings expanding outward.

Maybe a subtle point but I would like to ask.

答:
Yes, cartesian and polar coordinates account for the same space, just different notation. Every point on a cartesian plane can be represented with a polar coordinate and vice versa. However, due to the way angles work, there are infinitely many ways to represent one distinct point with polar coordinates, if you allow the range of the angle to be all real numbers.

For example: the point (1,1) on a cartesian plane is the polar point $(\sqrt2, (\pi+n2\pi)/4)$ where n is any integer.


问:
A question on von Neumann algebra
Let $M$ be a von Neumann algebra and let $\pi: M \rightarrow B(H)$ be a representation. Suppose that $N$ is a von Neumann subalgebra of $M$ such that $E$ is a faithful conditional expectation from $M$ to $N$. Is it true that if $\pi$ is faithful on $N$ implies that $\pi$ is faithful on $M$?

答:
No.

Let $M=M_2(\mathbb C)\oplus\mathbb C$, and $N=\mathbb C\,I_3$.

Let $\pi:M\to  M_2(\mathbb C)$ given by $\pi(A\oplus b)=b\,I_2$. Take $E(A,b)=\tfrac23\,\operatorname{Tr}(A)+\tfrac13\,\operatorname{Tr}(b)$.

Then $E$ is a faithful conditional expectation onto $N$, $\pi$ is faithful on $N$, and $M_2(\mathbb C)\oplus0\subset\ker\pi$.

Title: A question on von Neumann algebra


问:
Why the radius of the ball needs to be rational? Second countability
In proving that $\mathbb{R}$ is second countable with the usual metric, one says that each member of the family of open balls $\mathcal{B}$ is centered in each rational $x\in V$ for $V\subseteq \mathbb{R}$ with rational radius.
Examples are given here and here. To fix the center as a rational is needed to see that $V = \cup_x B_r(x)$ is countable.

Why the radius needs to be rational? For example in $(0,\pi)$, there can be balls with radius proportional to a fraction of $\pi$, hence irrational radius.

A topological space $X$ is said to be second countable if there exists some countable family $\mathbb{B}$ of open sets such that each open set $V\subseteq X$ can be expressed as the union of some subfamily of $\mathbb{B}$

答: Sure, you could use radii that are rational multiples of $\pi$ as well. It doesn't matter. You want to prove that a countable basis exists, so the most natural countable basis to find is the one with rational radii.


答:
Your computations indeed appear to be correct. Your results for $T_1,T_2$ can be confirmed using the Kronecker product, since we have
$$
T_2 = I_2 \otimes T, \quad T_1 = T \otimes I_2.
$$

Title: Operators $T_i$ in matrix form.

Body:
Let $V$ be a $K$-vector space and let $T$ be a $K$-linear map $T: V \otimes V \to V \otimes V$. Denote by $T_i$ the linear operator on $V^{\otimes n} \to V^{\otimes n}$ which acts on the $i$th and $(i+1)$th factors and acts trivially on other factors. I would like to understand the matrix form of $T_i$. For example, let $V$ be $2$ dimensional and $e_1, e_2$ be its standard basis. Suppose that with respect to $e_1\otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, $T$ is
\begin{align*}
\left[ \begin {array}{cccc} \mu_{{1,1}}&\mu_{{1,2}}&\mu_{{1,3}}&\mu_{
{1,4}}\\  \mu_{{2,1}}&\mu_{{2,2}}&\mu_{{2,3}}&\mu_{{2
,4}}\\  \mu_{{3,1}}&\mu_{{3,2}}&\mu_{{3,3}}&\mu_{{3,4
}}\\  \mu_{{4,1}}&\mu_{{4,2}}&\mu_{{4,3}}&\mu_{{4,4}}
\end {array} \right] .
\end{align*}
Let $n=3$. I obtained that, with respect to $$e_1 \otimes e_1 \otimes e_1, e_1 \otimes e_1 \otimes e_2, e_1 \otimes e_2 \otimes e_1, e_1 \otimes e_2 \otimes e_2, e_2 \otimes e_1 \otimes e_1, e_2 \otimes e_1 \otimes e_2, e_2 \otimes e_2 \otimes e_1, e_2 \otimes e_2 \otimes e_2,$$
\begin{align*}
T_1 =  \left[ \begin {array}{cccccccc} \mu_{{1,1}}&0&\mu_{{1,2}}&0&\mu_{{1,3
}}&0&\mu_{{1,4}}&0\\ 0&\mu_{{1,1}}&0&\mu_{{1,2}}&0&
\mu_{{1,3}}&0&\mu_{{1,4}}\\ \mu_{{2,1}}&0&\mu_{{2,2}
}&0&\mu_{{2,3}}&0&\mu_{{2,4}}&0\\ 0&\mu_{{2,1}}&0&
\mu_{{2,2}}&0&\mu_{{2,3}}&0&\mu_{{2,4}}\\ \mu_{{3,1}
}&0&\mu_{{3,2}}&0&\mu_{{3,3}}&0&\mu_{{3,4}}&0\\ 0&
\mu_{{3,1}}&0&\mu_{{3,2}}&0&\mu_{{3,3}}&0&\mu_{{3,4}}
\\ \mu_{{4,1}}&0&\mu_{{4,2}}&0&\mu_{{4,3}}&0&\mu_{{4
,4}}&0\\ 0&\mu_{{4,1}}&0&\mu_{{4,2}}&0&\mu_{{4,3}}&0
&\mu_{{4,4}}\end {array} \right],
\end{align*}
\begin{align*}
T_2 = \left[ \begin {array}{cccccccc} \mu_{{1,1}}&\mu_{{1,2}}&\mu_{{1,3}}&
\mu_{{1,4}}&0&0&0&0\\ \mu_{{2,1}}&\mu_{{2,2}}&\mu_{{
2,3}}&\mu_{{2,4}}&0&0&0&0\\ \mu_{{3,1}}&\mu_{{3,2}}&
\mu_{{3,3}}&\mu_{{3,4}}&0&0&0&0\\ \mu_{{4,1}}&\mu_{{
4,2}}&\mu_{{4,3}}&\mu_{{4,4}}&0&0&0&0\\ 0&0&0&0&\mu_
{{1,1}}&\mu_{{1,2}}&\mu_{{1,3}}&\mu_{{1,4}}\\ 0&0&0&0
&\mu_{{2,1}}&\mu_{{2,2}}&\mu_{{2,3}}&\mu_{{2,4}}\\ 0
&0&0&0&\mu_{{3,1}}&\mu_{{3,2}}&\mu_{{3,3}}&\mu_{{3,4}}
\\ 0&0&0&0&\mu_{{4,1}}&\mu_{{4,2}}&\mu_{{4,3}}&\mu_{
{4,4}}\end {array} \right].
\end{align*}
Are these computations correct? Thank you very much.


答:
EDIT: per the OP's comment to ndhanson3's answer, it seems the sticking point is:

Why does the set $A$ of balls with rational center and arbitrary (positive) radius not demonstrate the second countability of $\mathbb{R}$ (with the usual topology)?

Well, we have to check two things. First, we have to show that $A$ is in fact a base for the topology in question. This is a bit tedious to do, but it isn't hard, so let's just assume we've done it.

The issue however is the countability requirement: to show that $\mathbb{R}$ is second-countable we need a countable base, not just an arbitrary base. And when we say "countable base" we really do mean "countable base:" the base $A$ above may "feel countable" (in some sense there are countably many "types of" element of $A$), but since it's not literally a countable set it doesn't help us.

(Why isn't $A$ countable? Well, think about the set $\{B_r(42): r\in\mathbb{R}_{>0}\}$. This is a subset of $A$, and it's "as big as" the set of positive reals; now, is the set of positive reals countable or uncountable?)

By contrast, since "countable times countable equals countable" if we look at a set of balls with rational centers and rational radii, we do indeed wind up with a countable set - since the set of positive rational numbers (unlike the set of positive real numbers) is countable.

OK, why specifically the rationals?

This is indeed unnecessary - it's just a convenient choice.

All we actually need to show second-countability is:

A countable dense set $P$ of points in our space.

A countable set $D$ of positive real numbers which is "downwards-dense" in the sense that for each $\epsilon>0$ there is some $\delta\in D$ with $\epsilon>\delta$.

As soon as we have such $P$ and $D$, we can show that our space is second-countable: the set of balls with radii in $D$ and with center in $P$ is a countable base of our space. (Moreover, note that $D$ doesn't depend on our space at all: we can fix a $D$ to use in every situation, so that a metric space is second-countable if it has a countable dense set. This is not true for general topological spaces.)

At this point it's just a question of what's most natural for you. For spaces "based on" $\mathbb{R}$, the usual choice of $P$ is something "based on" $\mathbb{Q}$ (e.g. $\mathbb{Q}$ itself for $\mathbb{R}$, $\mathbb{Q}^2$ for $\mathbb{R}^2$, etc.). Meanwhile, the usual choices for $D$ that I've seen are $\mathbb{Q}_{>0}$, $\{{1\over n}: n\in\mathbb{N}\}$, and $\{2^{-n}: n\in\mathbb{N}\}$ (the former is by far the most common in my experience but every so often one of the latter two is a bit more convenient). But we could also use something like $\{q\pi: q\in\mathbb{Q}_{>0}\}$: it just has to be countable and downwards-dense.

Title: Why the radius of the ball needs to be rational? Second countability

Body:
In proving that $\mathbb{R}$ is second countable with the usual metric, one says that each member of the family of open balls $\mathcal{B}$ is centered in each rational $x\in V$ for $V\subseteq \mathbb{R}$ with rational radius.
Examples are given here and here. To fix the center as a rational is needed to see that $V = \cup_x B_r(x)$ is countable.

Why the radius needs to be rational? For example in $(0,\pi)$, there can be balls with radius proportional to a fraction of $\pi$, hence irrational radius.

A topological space $X$ is said to be second countable if there exists some countable family $\mathbb{B}$ of open sets such that each open set $V\subseteq X$ can be expressed as the union of some subfamily of $\mathbb{B}$


问:
Can $e^{\frac{1}{\ln(x)}}$ be simplified?
I would like to find the inverse function of $$f(x) = \ln(x) + \frac{1}{\ln(x)}$$
but I got stuck when trying to remove the exponents:
$$e^x = y \cdot e^{\frac{1}{\ln(x)}}$$

答:
Let $u=\ln x$, to simplify the algebra. Then we have
$$y=u+\frac{1}{u}$$

Multiply this by $u$ to get a quadratic equation in $u$.

Solve for $u$ in terms of $y$.

Now put $x=e^u$.


Title: Can $e^{\frac{1}{\ln(x)}}$ be simplified?

Body:
I would like to find the inverse function of $$f(x) = \ln(x) + \frac{1}{\ln(x)}$$
but I got stuck when trying to remove the exponents:
$$e^x = y \cdot e^{\frac{1}{\ln(x)}}$$

Body:
$$y=\ln(x)+\frac1{\ln(x)}$$

$$\ln^2(x)-y\ln(x)+1=0$$
$$\ln(x)=\frac{y\pm\sqrt{y^2-4}}{2}$$

$$x=\sqrt{e^{y\pm\sqrt{y^2-4}}}.$$


问:
Homomorphism between $\mathbb{Z}_m$ and $\mathbb{Z}_n$ if $m\mid n$
How can I show $\phi:\mathbb{Z}_m \to \mathbb{Z}_n$ defined by

$$
\phi(k) = k\mod{n}
$$

satisfies $\phi(k +_m l) = \phi(k)+_n \phi(l)$ when $n\mid m$.

I have tried using remainder terms but the solution gets messy as I have mod n's and m's and cant get rid of them.

答:
You know that $nq=m$ for some integer $q$.

Then you have that $m\in (n)$, in other words $m\equiv_{n}0$.

So if you have $a\equiv_{m}b$, this implies that $a-b\equiv_{m} 0$, and $a-b\equiv_{n} 0$, and therefore $a\equiv_{n}b$.

If you apply this same argument to $k+l\equiv_{m} k'+l'$, you will have that $k+l\equiv_{n} k'+l'$, and you will get your prove.


问:
trouble understanding some parts of integration,
what does $dx$ mean in: $$\int f(x)dx$$ and is it true that, if $\frac{df(x)}{dx} = g(x)$, then: $$\int g(x)dx = \int df(x)$$

答:
take the XY coordinate system. Integration means the calculation of area basically. So when we use $A=l*b$ in mensuration usually here our breadth is the infinitesimal change in the value of $x$ that is $dx$ into the value of length which is on the $y$ axis and it’s the function $f(x)$.

The entire essence is calculation of area of the continuous bodies. Integration means adding up those infinitely small areas that you calculated to form the entire surface.


问:
getting started self study
I am 14 and currently in year 10 (uk). I am very interested in maths and intend to pursue it. I currently can do C1, C2, C3 and parts of C4 edexcel exam board a level mathematics. However, when i learn something new i have trouble truly remembering it and this is the same for the a level maths i have done. I have used a variety of resources up to now including khan academy, mit open courseware for calculus and quite a few more including the edexcel textbooks which i find awkward to use. I would appreciate a suggested order to learn each topic from beginning a levels to end of bachelors degree and also i would appreciate good resources or books to use. The reason i wish for resources to a level basics is because i would like to relearn it so ensure it is safe in my mind.

Thanks in advance.

答:
I can recommend you the book I liked very much when I was your age:

"What is mathematics?"

by R. Courant and H. Robbins.

If I may give you one advice: do not concentrate on remembering math. Concentrate on understanding it. Some people claim that very good memory is rather a handicap in mathematics (a little joke, but I find something true in this statement:)

Title: getting started self study


问:
Prove that the solutions to an ODE are non-periodic oscillating.
I have the ODE $$\ddot{x}+ \dot{x}+k^2x=0$$
where the dot means derivative with respect to $t$ ($x$ is a function of $t$) and $k$ is a constant such that $k>\frac{1}{2}$

prove that the solutions to the ODE are non-periodic and oscillating.

What I tried is the following.

I solved the characteristic polynomial equation $$\lambda^2+\lambda+k^2=0$$
which gave me $$ \lambda_{1,2} = \frac{-1 \pm 2 \sqrt{k^2-\frac{1}{4}}}{2} $$
Now since $k>\frac{1}{2}$
the solutions are real so the ODE has no periodic solutions, but I can't understand how to prove that the solutions oscillate, what does it even mean for the solutions to oscillate?

答:
Given the equation

$\ddot x + \dot x + k^2x = 0, \tag 1$

the characteristic equation is indeed

$\lambda^2 + \lambda + k^2 = 0, \tag 2$

but the "quadratic formula" used to find $\lambda_{1,2}$ is misstated; it should read

$\lambda_{1,2} = \dfrac{-1 \pm \sqrt{1 - 4k^2}}{2}; \tag 3$

from this we see that when $k > 1/2$ the values of $\lambda_{1,2}$ are complex conjugate numbers with real part $-1/2$, hence the solutions exponentially decay to $0$; however the presence of a non-vanishing imaginary component in the $\lambda_{1,2}$ forces the solutions to venture
back and forth across $0$ whilst they decay; thus the solutions are non-periodic and oscillating.

We can in fact write down explicit solutions to (1); they serve to reify and illustrate the words of the preceding paragraph.  Writing (3) in the form

$\lambda_{1, 2} = -\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1}, \tag 4$

we may express two relatively simple solution as

$x_\pm(t) = \exp \left (\lambda_{1,2} t\right ) = \exp \left (\left (-\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1} \right)  t\right )$ =
$\exp \left (-\dfrac{1}{2}  t\right )\exp \left (\pm i \dfrac{1}{2}\sqrt{4k^2 - 1} t\right )$
$= \exp \left (-\dfrac{1}{2}  t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \pm i \sin \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right )\right). \tag 5$

Since (1) is a real differential equation, the real and imaginary parts of $x_\pm(t)$ are individually solutions, thus for example

$x(t) =  \exp \left (-\dfrac{1}{2}  t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \right) \tag 6$

is a solution which exhibits both exponential decay and oscillation.


问:
What is the probability for 3 different groups of people having different birthdays?
There are $15$ students. Assume that there are $365$ days in a year.

What is the probability that $3, 5$ and $7$ of them have the same birthday on any three days of a year, with these three groups of students have a different birthday?

$$
P\left(3\right)=\frac{365}{365}\times\frac{1}{365}\times\frac{1}{365}=\frac{1}{133225}
$$
$$
P\left(5\right)=\frac{364}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}=5.6187\times{10}^{-11}
$$
$$
P\left(7\right)=\frac{363}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}=4.206\times{10}^{-16}
$$

should  I change the denominator for $P(5)$ to $364$  and $P(7)$ to $363$ as after each group the available days for birthday decreases?

Can someone please explain it to me?

答:
Tortar's comment makes the very fair point that perhaps the OP wants to compute the following probability.

There are three distinct pertinent days in the 365 day year, day-1, day-2, and day-3.

Of the 15 people, exactly 3 of the people have day-1 as their birthday.

Of the 15 people, exactly 5 of the people have day-2 as their birthday.

Of the 15 people, exactly 7 of the people have day-3 as their birthday.

In my opinion, the easiest way to attack this problem is to express the probability as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$

with

$$D = (365)^{(15)}.$$

$N$, which represents the # of distinct ways that the scenario can occur, will be computed as the product of factors.

The # of ways that the 3 days can be selected is

$$N_1 = 365 \times 364 \times 363.$$

The # of ways that the 15 people can be partitioned into groups of
3, 5, and 7 is

$$N_2 = \binom{15}{3} \times \binom{12}{5}.$$

Final answer is

$$\frac{N_1 \times N_2}{D}.$$


问:
Two definitions of order convergence
I came accross two different definitions of order convergence, one (Def 1) given by my professor (I can't find any same ones elsewhere), and one (Def 2) which can be found in any materials available on the Internet, such as here.

$\{p_k\}$ is an $\alpha- $order convergent sequence, where $p_k \rightarrow p$, iff

Def 1
\begin{equation}
\exists \lambda>0,\  \exists k^* \ge 0,\ \forall k\ge k^*,\
||p_{k+1}-p|| \le \lambda ||p_k-p||^{\alpha}
\end{equation}

Def 2
\begin{equation}
\exists \lambda>0,\  \exists \alpha > 0, \lim_{k\to+\infty} \frac{||p_{k+1}-p||}{||p_k-p||^{\alpha}} = \lambda
\end{equation}

How are the two equivalent to each other?

答:
The second definition isn’t quite right: you’re defining $\alpha$-order convergence, so $\alpha$ should be a free variable. It should say simply that there is a $\lambda>0$ such that $\lim\limits_{k\to\infty}\frac{\|p_{k+1}-p\|}{\|p_k-p\|^\alpha}=\lambda$.

And it is definitely stronger than the first definition. Let

$$p_k=\left(\frac12\right)^{2^k}$$

for $k\in\Bbb N$; clearly $\langle p_k:k\in\Bbb N\rangle$ converges to $0$, and

$$\left|\left(\frac12\right)^{2^{k+1}}\right|\le\left|\left(\frac12\right)^{2^k}\right|$$

for all $k\in\Bbb N$, so by the first definition the sequence is $1$-convergent with $\lambda=1$. It is also $2$-convergent:

$$\left|\left(\frac12\right)^{2^{k+1}}\right|=\left|\left(\frac12\right)^{2^k}\right|^2\,.$$

However,

$$\frac{\left|\left(\frac12\right)^{2^{k+1}}\right|}{\left|\left(\frac12\right)^{2^k}\right|}=\left(\frac12\right)^{2^{k+1}-2^k}=\left(\frac12\right)^{2^k}$$

converges to $0$ as $k$ increases without bound, so the sequence is not $1$-convergent by the second definition. It is $2$-convergent by the second definition:

$$\frac{\left|\left(\frac12\right)^{2^{k+1}}\right|}{\left|\left(\frac12\right)^{2^k}\right|^2}=1$$

for all $k\in\Bbb N$.

It’s not hard to check that if a sequence is $\alpha$-convergent by the second definition, it is $\beta$-convergent by the first definition whenever $0<\beta\le\alpha$.

Title: Two definitions of order convergence


问:
Kelvin transform and the fractional Laplacian
For $0< \alpha< 2$ the fractional Laplacian of order $\alpha$ is defined up to multiplicative constants by
\begin{equation*}
 (-\Delta)^{\alpha/2}u(x) = \int_{\mathbb R^n}\frac{u(x) - u(y)}{|x - y|^{n + \alpha}}\; dy,
\end{equation*}
where the integral is understood in the principal value sense and, for my purposes, $u:\mathbb R^n \to \mathbb R$ is as smooth and rapidly decaying as you want. For such $u$, define the Kelvin transform of $u$ to be
s\begin{equation*}
 v(x) = \left(\frac 1{|x|}\right)^{n + \alpha}u(\bar x),
\end{equation*}
where $\bar x = \frac{x}{|x|^2}$.

My question is about the relationship between the fractional Laplacian $(-\Delta)^{\alpha/2}$ and the Kelvin transform. Is there a direct proof for the equality
\begin{equation*}
 \left((-\Delta)^{\alpha/2}v\right)(x) = \left(\frac{1}{|x|}\right)^{n + \alpha}\left((-\Delta)^{\alpha/2}u\right)(\bar x)?
\end{equation*}
where by ``direct'' I mean only using the above definition of $(-\Delta)^{\alpha/2}$ and possibly a slick change of variable? \

Here is an attempt that is not quite giving me what I want but will hopefully provide a starting point for suggestions. Below I use the change of variable $y = \bar z = \frac z{|z|^2}$, $dy = \frac{1}{|z|^{2n}}dz$ together with the equality
\begin{equation*}
 |\bar x - \bar z| = \frac{|x - z|}{|x||z|}.
\end{equation*}
\begin{eqnarray*}
 \left((-\Delta)^{\alpha/2}u\right)(\bar x)
 & = &
 \int_{\mathbb R^n}\frac{u(\bar x) - u(y)}{|\bar x - y|^{n + \alpha}}dy
 \\
 & = &
 \int_{\mathbb R^n}\frac{u(\bar x) - u(\bar z)}{|\bar x - \bar z|^{n + \alpha}}\frac 1{|z|^{2n}}dz
 \\
 & = &
 \int_{\mathbb R^n}\frac{u(\bar x) - u(\bar z)}{|x - z|^{n + \alpha}}\frac{(|x||z|)^{n +\alpha}}{|z|^{2n}}dz
 \\
 & = &
 |x|^{n + \alpha}
 \int_{\mathbb R^n} \frac{\left(\frac{1}{|z|}\right)^{n - \alpha}u(\bar x) - \left(\frac 1{|z|}\right)^{n - \alpha} u(\bar z)}
 {|x - z|^{n+ \alpha}}dz
 \\
 & = &
 |x|^{n + \alpha}
 \int_{\mathbb R^n} \frac{\left(\frac{|x|}{|z|}\right)^{n - \alpha}v(x) - v(z)}{|x - z|^{n+ \alpha}}dz.
\end{eqnarray*}

Edit:

As it turns out  the above computation is just one step away from the desired equality. The function $\Phi(x) = |x|^{n - \alpha}$ is the fundamental solution for $(-\Delta)^{\alpha/2}$ in the sense that (still up to a multiplicative constant)
\begin{equation*}
 (-\Delta)^{\alpha/2}\Phi = \delta_0.
\end{equation*}
Therefore, continuing the above computation for $x\neq 0$ gives
\begin{eqnarray*}
 \left((-\Delta)^{\alpha/2}u\right)(\bar x)
 & = &
 |x|^{n + \alpha}\int_{\mathbb R^n} \frac{\left(\frac{|x|}{|z|}\right)^{n - \alpha}v(x) - v(z)}{|x - z|^{n+ \alpha}}dz
 \\
 & = &
 -v(x)|x|^{2n}\int_{\mathbb R^n} \frac{|x|^{\alpha - n} - |z|^{\alpha -n}}{|x - z|^{n +\alpha}}dz
 \\
 &&
 +
 |x|^{n + \alpha}\int_{\mathbb R^n}\frac{v(x) - v(z)}{|x - z|^{n + \alpha}}dz
 \\
 & = &
 -v(x)|x|^{2n}\left((-\Delta)^{\alpha/2}\Phi\right)(x) + |x|^{n + \alpha}\left((-\Delta)^{\alpha/2}v\right)(x)
 \\
 & = &
 |x|^{n + \alpha}\left((-\Delta)^{\alpha/2}v\right)(x),
\end{eqnarray*}
which is the desired equality. However, the only proof that I know for the fact that $(-\Delta)^{\alpha/2}\Phi = \delta_0$ relies on the Fourier transform of $\Phi$. Thus, this entire computation is still not as ``direct'' as I would like, but I suppose it will do for now.

答:
SUMMARY.

I am going to prove that
$$\tag 1
(-\Delta)^{\frac\alpha 2}_x u = \lvert \bar x \rvert^{-\alpha-d}(-\Delta)_{\bar x}^\frac\alpha 2\bar u,$$
where $\bar x = \frac x {\lvert x \rvert^2}$ and $\bar u(\bar x)=\lvert \bar x \rvert^{\alpha -d}u(x).$

INTRODUCTION.

I have noticed quite a few times that the fractional Laplacian with positive exponent $(-\Delta)^{\alpha/2}$ is much messier than its inverse
$$
(-\Delta)^{-\alpha/2} v(x)=C\int_{\mathbb R^d} \frac{ v(y) }{ \lvert x- y \rvert^{d-\alpha}}\,dy.$$
So in this answer I am going to work with the latter, which is also known as Riesz potential; see the paper of Kwaśnicki 10 definitions of fractional Laplacian, Theorem 1.1 (i). There, you will also find the exact value of the constant $C>0$, which depends on $d$ and on $\alpha$. After the derivation of the correct transformation formula for $(-\Delta)^{-\alpha /2}$, taking inverses will yield (1).

The Riesz potential is a proper convolution, with a locally integrable kernel. On the other hand,  $(-\Delta)^{\alpha/2}$ is given by a crazy singular integral. This is the reason I give to myself to justify why $(-\Delta)^{-\alpha/2}$ is better behaved than $(-\Delta)^{\alpha/2}$.

PROOF.

We apply the change of variable
$$
\bar x=\frac x {\lvert x\rvert^2}, \quad \bar y=\frac y {\lvert y\rvert^2},\quad v(x)=\bar v (\bar x) W(\bar x), $$
where the weight $W$ is to be determined. Using the Jacobian formula $dy=\lvert \bar y \rvert^{-2d}d\bar y$ and the "chordal distance formula" (terminology mutuated from the stereographic projection)
$$
\lvert \bar x - \bar y \rvert = \frac{ \lvert x - y \rvert}{\lvert x\rvert \lvert y\rvert}, $$
we obtain
$$
(-\Delta)^{-\alpha/2}v(x)= C\int_{\mathbb R^d} \frac{W(\bar y)\bar v (\bar y) \lvert \bar x\rvert^{d-\alpha}}{\lvert \bar y\rvert^{d+\alpha}  \lvert \bar x- \bar y\rvert^{d-\alpha}}\, d\bar y, $$
so letting
$$W(\bar y)=\lvert \bar y\rvert^{d+\alpha}$$
yields
$$\tag 2
(-\Delta)^{-\alpha/2}_xv=\lvert \bar x \rvert^{d-\alpha} (-\Delta)^{-\alpha/2}_{\bar x}(\lvert \bar x\rvert^{d+\alpha}v) .$$

Now we show that (2) implies (1). We let $u=(-\Delta)^{-\alpha/2}v$, so that $v=(-\Delta)^{\alpha/2}u$ and so (2) reads
$$
u=\lvert \bar x\rvert^{d-\alpha}(-\Delta)^{-\alpha/2}_{\bar x}\left( \lvert \bar x\rvert^{d+\alpha} (-\Delta)^{\alpha/2}_x u\right), $$
from which we conclude
$$
(-\Delta)^{\alpha/2}_{\bar x} \left( \lvert \bar x \rvert^{\alpha - d} u\right) = \lvert \bar x \rvert^{d+\alpha} (-\Delta)^{\alpha/2}_x u, $$
which is (1). The proof is complete.


问:
Example of an infinite sequence of irrational numbers converging to a rational number?
Are there any nice examples of infinite sequences of irrational numbers converging to rational numbers?

One idea I had was the sequence: $ 0.1001000010000001\cdots,0.1101000110000001\cdots,\cdots,0.1111000110000001\cdots,$ etc.

Where the first term in the sequence has ones in the place $i^2$ positions to the right of the decimal point. $(i=1,2,3,\dots)$ For the second term, we keep all the ones from the first term and add one in place of the first zero after the decimal point. We then add ones $i^3$ places after the decimal point. (The logical sum 1+1=1, i.e a number $i^2=j^3$ spaces after the decimal place has the value 1).

It is clear that this will converge to $1/9$, and I don't think the decimal expansion repeats at all.

答: If you want it to converge to a specific rational number $x$, then the sequence $\{x(\sqrt{2})^{1/n}\}$. All terms of the sequence are irrational but as $n\rightarrow\infty$, the sequence converges to $x$.


问:
Explanation regarding eigenvectors and eigenvalues
In PCA algorithm, $C$ the covariance matrix is defined as $\frac1{N-1}XX^T$, where $X$ is a matrix of size $d\times n$. Also, the rank of $X$ is $\min(d-1,n)$. (Lets assume)

Now, since $C$ is symmetric and PSD, all its eigenvalues are real and non negative.

We know that number of eigenvalues of a $d \times d$ matrix is $d$.

Now my question(s) are,

(i) how do we know that there are exactly, $r=\min(d,n-1)$ linearly independent eigenvectors corresponding to $r$ non-zero eigenvalues?

(ii) and that the rest $0$ eigenvalues correspond to $(d-r)$, linearly independent eigenvectors.

(iii) and also that even together, all these eigenvectors are linearly independent. ((iv) are they all orthogonal).

I am saying this because in PCA algorithm, $$CV=V\Lambda$$ where $V$ is $d\times d$ eigenvector (at columns) matrix, which is orthogonal, and $\Lambda$ is a $d\times d$ eigenvalue matrix.

Note that to prove the claims, I cannot use SVD, since it is introduced later, and is derived using the facts I wrote above (in some way).

答:
Every symmetric matrix is orthogonally diagonalizable.

The easiest way to prove this is using the SVD, but if you don't want to use the SVD, you can start by proving the easier fact that if a matrix has two distinct eigenvalues, then the corresponding eigenvectors are orthogonal.


问:
How to find a positive semidefinite matrix satisfying the following condition?
Consider such a problem:
$$\mathbf A\mathbf x=\mathbf b$$
where the vectors $\mathbf x$ and $\mathbf b$ are known.

I want to find a positive semidefinite matrix $\mathbf A$ satisfying the aforementioned equation.

What conditions do $\mathbf x$ and $\mathbf b$ need to satisfy?

Thanks a lot.

答:
Let us give a separate treatment for the case $n=2$.

We are going to show that there exists an infinite number of spd (symmetric positive definite) matrices $A$ such that:

$$\underbrace{\begin{pmatrix}a&b\\b&c\end{pmatrix}}_A \underbrace{\binom{x_1}{x_2}}_x=\underbrace{\binom{y_1}{y_2}}_y \tag{1}$$

(we have changed $b$ into $y$ in order to avoid ambiguities with the lower left entry of $A$)

under the necessary condition found by @Fenris:

$$\text{Dot product:} \ \ x.b=x_1y_1+x_2y_2 > 0 \tag{2}$$

(please note that we take $>0$ instead of $\ge 0$).

WLOG, one can assume that the two vectors have unit norms

$$x_1^2+x_2^2=y_1^2+y_2^2=1 \tag{3}$$

(indeed, once we have found a solution $A$, we just have to multiply it by $\dfrac{\|b\|}{\|x\|}$.)

It is now easy to check that for any $d$, we have:

$$\underbrace{\begin{pmatrix}(dx_2^2+(y_1x_1-y_2x_2))&(x_1y_2+x_2y_1 - d x_1x_2)\\(x_1y_2+x_2y_1 - d x_1x_2)&(dx_1^2-(y_1x_1-y_2x_2))\end{pmatrix}}_A\binom{x_1}{x_2}=\binom{y_1}{y_2}\tag{4}$$

with $$\operatorname{trace}(A)=d \ \ \text{and} \ \ \det(A)=d(x_1y_1+x_2y_2)-\underbrace{(y_1^2+y_2^2)}_1.\tag{5}$$

If we take

$$d > \dfrac{1}{x_1y_1+x_2y_2},\tag{6}$$

The RHS in (6) being positive (see (2)), we have $\det(A)>0$ and $\operatorname{trace}(A)>0$.

Therefore, if (6) is fulfilled, $A$ given by (4) is symmetric positive definite.

Remarks:

How has $A$ in (2) been found ? By writing condition (0) under the form:

$$\begin{pmatrix}x_1&x_2&0\\0&x_1&x_2\\1&0&1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}y_1\\y_2\\d\end{pmatrix} \ \iff
\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}x_1&x_2&0\\0&x_1&x_2\\1&0&1\end{pmatrix}^{-1}\begin{pmatrix}y_1\\y_2\\d\end{pmatrix}
$$
giving the entries $a,b,c$ of $A$ in (2), $d$ bringing its very useful degree of freedom.

For readers acquainted with pole-polar relationship, here is a nice graphical interpretation.

(1) can be interpreted as the fact that point $(x_1,x_2)$ is the pole of line $xy_1+yy_2=k$ with respect to the conic curve (an ellipse) with equation

$$ax^2+2bxy+cy^2=k_1\tag{7}$$

In fact, due to normalization (3), it is the particular case of tangent-point of tangency relationship... under the condition that the conic defined by (7) passes through point $(x_1;x_2)$, which means that

$$k_1=ax_1^2+2bx_1x_2+cx_2^2$$

The figure below represents different ellipses with equation (7), according to the value of $d$.

Fig. 1: The case where $x=(x_1,x_2)=(4,3)$ and $y=(y_1;y_2)=(2,1)$ (giving, by normalization, $x'=(0.8;0.6)$ materialized by a little circle); the (tangent) line has equation $xy_1+yy_2=k$, $k$ being taken such that this line passes through $x'$. The different ellipses correspond to different values of $d$.


问:
Unsure about double integral bounds of integration in polar coordinates
I'm trying to convert the bounds of integration to polar coordinates but I'm stumped on one of the bounds.

$$\int_{x=0}^{6}\int_{y=\frac{1}{\sqrt{3}}x}^{\sqrt{8x-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$

The only thing that left me stumped was converting $y=\frac{1}{\sqrt{3}}x$ to polar.

Right now I have $\int_{\theta=0}^{\frac{\pi}{6}}\int_{?}^{8\cos{\theta}}r^2\,dr\,d\theta$

Where do I go from here?  It's nowhere in my notes and I'm having a tough time finding anything online about it. Thank you!

答:
Let's do a quick plot:

Now it should be obvious that $\theta$ goes from $\frac\pi6$ to $\frac\pi2$ and $r$ goes from $0$ to $8\cos\theta$


问:
Bounded set absolute value of the difference
Let $\emptyset\neq A\subseteq\mathbb R$.
Prove that $$
\sup(\left\{ \,\left|x-y\right|\;:\;x,y\in A\,\right\})=\sup A-\inf A$$
and prove the following as the first step:
$$\sup(\left\{ \,\left|x-y\right|\;:\;x,y\in A\,\right\})=\sup(\left\{ \,x-y\;:\;x,y\in A\,\right\}).
$$
I've tried proving the assumption using the strict definition of the supremum, but all I could do is prove the assumption without the needed "first step". any idea on how to do that?

答:
Hint

$$\sup\limits_{\{(x,y)\in A^2\}}\vert x-y\vert= \sup\limits_{\{(x,y)\in A^2 \mid x \ge y\}}(x-y)= \sup\limits_{\{(x,y)\in A^2\}}(x-y)$$

for the first step.

And

$$x-y \le \sup A -\inf A$$ for any $(x,y) \in A^2$ for the conclusion.


答:
For $r,$ your lower bound is simply zero. Please sketch and see.

You have a circle with radius $4$ at center $(4,0)$. You are integrating over the region above the line $y = \frac {x} {\sqrt3}$ and below the circle. In polar form the radius will go from zero to $8 \cos \theta$. The bound on $\theta$ ensures you are integrating over the circular segment.

EDIT: I just noticed bound of your $\theta$. That has to be from $\pi/6 \,$ to $\pi/2$.

Title: Unsure about double integral bounds of integration in polar coordinates

Body:
I'm trying to convert the bounds of integration to polar coordinates but I'm stumped on one of the bounds.

$$\int_{x=0}^{6}\int_{y=\frac{1}{\sqrt{3}}x}^{\sqrt{8x-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$

The only thing that left me stumped was converting $y=\frac{1}{\sqrt{3}}x$ to polar.

Right now I have $\int_{\theta=0}^{\frac{\pi}{6}}\int_{?}^{8\cos{\theta}}r^2\,dr\,d\theta$

Where do I go from here?  It's nowhere in my notes and I'm having a tough time finding anything online about it. Thank you!


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
It seems when looking at
\begin{align*}
\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0\tag{1}
\end{align*}
in full generality, which means in this context considering any four points in $\mathbb{R}^3$ there is rather some (not so deep) combinatorial meaning than any deep geometrical meaning. Nevertheless there are beautiful geometrical visualisations when considering special cases instead.

Essentially one-dimensional: Note the identity (1) which is stated in $\mathbb{R}^3$ is essentially a relationship of four points in $\mathbb{R}$. When taking for instance the projection mapping $\pi_1(X)=X_1$ where $X=(X_1,X_2,X_3)$, the identity (1) boils down to

\begin{align*}
&\color{blue}{(C_1-B_1)(D_1-A_1)}+\color{red}{(A_1-C_1)(D_1-B_1)}+(B_1-A_1)(D_1-C_1)\\
&\qquad=\ \ \color{blue}{C_1D_1-B_1D_1-A_1C_1+A_1B_1}\\
&\qquad\quad\color{red}{-C_1D_1\qquad\qquad\qquad\quad\ -A_1B_1+A_1D_1+B_1C_1}\tag{2}\\
&\qquad\qquad\qquad\ +B_1D_1+A_1C_1\qquad\qquad-A_1D_1-B_1C_1\\
&\qquad=0
\end{align*}
We observe in (2) the terms in each dimension cancel away independently from terms of other dimensions. From this we might conclude that any essence of geometrical information should be already contained within each coordinate dimension separately.

Moreover the order of Points $A_1,B_1,C_1$ and $D_1$ is also irrelevant. We may assume $A_1\leq B_1\leq C_1\leq D_1$ or any other of the $4!=24$ permutations, since any change in the order may change the sign from $X_1Y_1$ but then $-X_1Y_1$ does also change the sign.

Beautiful special cases: As @darigrinberg indicated in the comment section we have for instance Ptolemy's theorem
which claims the identity (1) in the one-dimensional case using lengths $|\overline{BC}|,   |\overline{AD}|$, etc, for the special case that the $4$ points $A,B,C,D\in\mathbb{R}^2$ are cyclic quadrilateral.  A nice proof of this theorem is based upon circle inversion and triangle similarities as shown in the figure below.

The four points $A,B,C,D$ are elements of the blue circle with point $D$ being the center of the red circle. By circle inversion the points $A,B,C$ are mapped to a line giving $A',B',C'$. We so derive $|\overline{DA}||\overline{DA'}|=|\overline{DB}||\overline{DB'}|=|\overline{DC}||\overline{DC'}|$ and find this way similar triangles as the shaded pair and two more pairs from which the theorem follows. This proof is nicely shown here by Adam Hrankowsi which will soon be available in When Am I Ever Gonna Use This Stuff?.

Note the Pythagorean theorem and the Cosine rule follow from special cases of Ptolemy's theorem.

Combinatorial aspects: We look again at the vectorised identity (1) and consider the products
\begin{align*}
&\overrightarrow{\color{blue}{B}C} \cdot \overrightarrow{\color{blue}{A}D}\tag{3}\\
&\overrightarrow{C\color{blue}{A}} \cdot \overrightarrow{\color{blue}{B}D}\tag{4}\\
&\overrightarrow{\color{blue}{A}\color{blue}{B}} \cdot \overrightarrow{CD}\tag{5}\\
\end{align*}

The points $A$ and $B$ for example occur in three variations: $A$ and $B$ both at the left side of the vector in (3) which can be encoded as $(L,L)$. $A$ at the right side and $B$ at the left side in (4), encoded as $(R,L)$ and $A$ and $B$ both at the same side, encoded as $(0,0)$. This relationship holds for each of the six possible selections of two points.

Reducing this relationship to one coordinate by the projection map we observe: Whenever two points $X_1, Y_1$ are encoded as $(L,L)$ or  $(R,R)$ the product $X_1Y_1$ has different sign than the product resulting from a constellation $(L,R)$ or $(R,L)$. Checking each pair of points in (1) for this relationship we obtain
\begin{align*}
\begin{array}{cccccc}
(A_1,B_1)&(A_1,C_1)&(A_1,D_1)&(B_1,C_1)&(B_1,D_1)&(C_1,D_1)\\
\hline
(L,L)&(L,R)&(0,0)&(0,0)&(L,R)&(R,R)\tag{6}\\
(R,L)&(0,0)&(R,R)&(L,L)&(0,0)&(L,R)\\
(0,0)&(L,L)&(L,R)&(R,L)&(R,R)&(0,0)
\end{array}
\end{align*}
showing that each pair of points results in two products which cancel away, while $(0,0)$ means it does not produce a product at all.

Conclusion: In order to provide a geometric visualisation of any four points $A,B,C,D$ in $\mathbb{R}^3$ it should appropriately encode the information stated in table (6).


问:
Determine all prime numbers p such that p^2+2 is also a prime number
Determine all prime numbers p such that p^2+2 is also a prime number.

I have only found 3, and I think that's the only number but I have trouble writing proof for this. But I know that if p is a prime number other than 3 then p^2+2 will be divisible with 3.

答:
Primes larger than $3$ cannot be divisible by either $2$ or $3$. Numbers with this property, which include all primes other than $2$ or $3$, have the form $6k\pm 1$.

$p^2+2=(6k\pm 1)^2+2=6k^2\pm 12k+3=3(2k^2\pm 4k+1)$, showing that $3\mid (p^2+2)$ when $p\ne 2,3$.

Simple arithmetic checking shows that only $p=3$ has the property that $p^2+2=11$ yields a prime.


问:
Fitting group of a direct product
Given the group $G=N\oplus M$, show $F(G)=F(N)\oplus F(M)$, where $F(G)$ denotes the Fitting group of $G$ (the product of all nilpotent normal subgroups).

I would say the inclusion $\supseteq$ is trivial, as $F(N)\oplus F(M)$ is itself a nilpotent normal subgroup of $G$, by characterization of the Fitting group it must be contained in $F(G)$, but the other inclusion is what worries me.

答:
If $A\triangleleft G$ is a nilpotent normal subgroup of $G$, then
$$\pi_N(A) = \{n\in N\mid \exists m\in M\text{ such that } (n,m)\in A\}$$
is a nilpotent normal subgroup of $N$ (being the image of a nilpotent normal subgroup under a surjective homomorphism), and likewise $\pi_M(A)$ is a nilpotent normal subgroup of $M$. Clearly, $A\leq \pi_N(A)\oplus \pi_M(A)\leq F(N)\oplus F(M)$. Thus, every nilpotent normal subgroup of $G$ is contained in $F(N)\oplus F(M)$.

Title: Fitting group of a direct product


问:
Why <HF,⊆> is simply isomorphic to the lattice of finite subsets of a countable set, such as the collection of finite subsets of N under inclusion.
<
<HF,⊆> is simply isomorphic to
the lattice of finite subsets of a countable set, such as the collection of
finite subsets of N under inclusion.
why?Can we have <HF,⊆> is simply isomorphic to finite subsets of N under devide relation <= to <N,|>

答:
Suppose $X$ is a countable set, and let $[X]^{<\omega}$ be the set of finite subsets of $X$. Consider a bijection $f: X\rightarrow HF$. This bijection induces a map $$\hat{f}: [X]^{<\omega}\rightarrow [HF]^{<\omega}$$ given by $$A\mapsto \{f(a): a\in A\}.$$ Moreover, $\hat{f}$ "doesn't change subsethood" in the sense that $$A\subseteq B\iff \hat{f}(A)\subseteq \hat{f}(B)$$ for all $A,B\in[X]^{<\omega}$. Put another way, $\hat{f}$ is an isomorphism between the structures $([X]^{<\omega},\subseteq)$ and $([HF]^{<\omega},\subseteq)$.

Note that $\hat{f}$ is just a restriction to finite subsets of the usual function $\mathcal{P}(X)\rightarrow\mathcal{P}(HF)$ induced by $f$.

But here's the cute bit: it turns out that $HF=[HF]^{<\omega}$. This may look weird at first and it's certainly not true that $X=[X]^{<\omega}$ in general - it's a special feature of $HF$. So we have in fact $$([X]^{<\omega},\subseteq)\cong([HF]^{<\omega},\subseteq)=(HF,\subseteq).$$ (That "$=$" instead of "$\cong$" isn't a typo, they are literally the same object.)


问:
Linear algebra unitary operator problem
I am not undertanding what I have to do in letter (c) of this problem.. I just wrote $\alpha_1=\begin{bmatrix} \cos\phi & -\sin\phi\\ \sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\cos\phi \\ \sin\phi\end{bmatrix}$

and $\alpha_2=\begin{bmatrix} \cos\phi & -\sin\phi\\
\sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-\sin\phi \\ \cos\phi\end{bmatrix}$

答:
Let $e_1 , e_2$ be standart basis vectors for $R^2$

$\epsilon_1=n_1e_1+n_2e_2$ ,$\epsilon_2=m_1e_1+m_2e_2$

$U_\phi(e_1)=\begin{pmatrix} \cos\phi \\ \sin\phi\end{pmatrix}$ , $U_\phi(e_2)=\begin{pmatrix} -\sin\phi \\ \cos\phi \end{pmatrix}$

Let $v=c_1\alpha_1+c_2\alpha_2$ then

$U_\theta(v)=U_\theta c_1\alpha_1+U_\theta c_2\alpha_2$

$=c_1U_\theta\alpha_1+c_2U_\theta\alpha_2=c_1U_\theta U_\phi\epsilon_1+c_2U_\theta U_\phi\epsilon_2$$=U_\theta U_\phi(c_1\epsilon_1+c_2\epsilon_2)$

$=U_\theta U_\phi(c_1n_1e_1+c_1n_2e_2+c_2m_1e_1+c_2m_2e_2)=(c_1n_1+c_2m_1)U_\theta U_\phi e_1+(c_1n_2+c_2m_2)U_\theta U_\phi e_2$

$=(c_1n_1+c_2m_1)\begin{pmatrix} \cos\phi+\theta \\ \sin\phi+\theta\end{pmatrix}+(c_1n_2+c_2m_2)\begin{pmatrix} -\sin\phi+\theta \\ \cos\phi+\theta \end{pmatrix}$

$\begin{pmatrix}  (c_1n_1+c_2m_1)\\ (c_1n_2+c_2m_2) \end{pmatrix}=\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} \begin{pmatrix} c_1 \\ c_2\end{pmatrix}$

so we can say that

$(c_1n_1+c_2m_1)\begin{pmatrix} \cos\phi+\theta \\ \sin\phi+\theta\end{pmatrix}+(c_1n_2+c_2m_2)\begin{pmatrix} -\sin\phi+\theta \\ \cos\phi+\theta \end{pmatrix}=\begin{pmatrix} \cos\phi+\theta & -\sin\theta+\phi \\ \sin \phi+\theta & \cos \phi + \theta  \end{pmatrix}\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} \begin{pmatrix} c_1 \\ c_2\end{pmatrix}$

so our matrix is $\begin{pmatrix} \cos\phi+\theta & -\sin\theta+\phi \\ \sin \phi+\theta & \cos \phi + \theta  \end{pmatrix}\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} $

Title: Linear algebra unitary operator problem


问:
Interesting question regarding congruences of prime power sums

For prime $p$ and $a \in Z$, let $l$ be any prime dividing $a^{p-1}+a^{p-2}+\cdots+a+1$. Show that the order of $a$ modulo $l$ is either $1$ or $p$, and therefore that either $l=p$ or $l \equiv 1 \pmod p$

Does anyone have any pointers with this question? I'm afraid I really don't have any kind of ideas whatsoever...

答: Notice that the sum is $\frac{a^p-1}{a-1}$. Thus $\ell|(a^p-1)$, so $a^p\equiv 1\pmod{\ell}$. By definition of order, the order $o(a)$ must divide $p$, it follows that $o(a)\in \{1,p\}$.


问:
Lacking some steps in a proof regarding set of units of $\mathbb{Z}[\sqrt{3}].$
I am trying to prove that:

$\mathbb{Z}[\sqrt{3}]^* = \{ x \in \mathbb{Z}[\sqrt{3}]: |N(x)| = 1 \}.$

Where $N$ is the norm function defined as follows:

Define the norm function $N:\mathbb{Q}[\sqrt{3}] \rightarrow \mathbb{Q}$ by $N(a_{1} + a_{2}\sqrt{3}) = \operatorname{det} \phi (a_{1} + a_{2}\sqrt{3}) = a_{1}^2 - 3 a_{2}^2.$

My trial is below:

Let $x \in \mathbb{Z}[\sqrt{3}]^*,$ then $x = a + b\sqrt{3}$ and there exists $y = c + d\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*$ such that $$xy = (a + b\sqrt{3})(c + d\sqrt{3})= (ac + 3bd) + \sqrt{3}(bc + ad) = 1,$$
Hence, $ac + 3bd = 1$ and $bc + ad = 0$

But then I do not know how to complete, could anyone help me please in continuing the proof?

答:
Assuming that $\mathbb{Z}[\sqrt{3}]^*$ is the group of units of the ring $\mathbb{Z}[\sqrt{3}]$, here is a roadmap:

Let $U=\{ x \in \mathbb{Z}[\sqrt{3}]: |N(x)| = 1 \}$

Prove that $\mathbb{Z}[\sqrt{3}]^* \subseteq U$ using that $N(xy)=N(x)N(y)$

Prove that $U \subseteq \mathbb{Z}[\sqrt{3}]^*$ using that $N(x)=x\bar x$


问:
How to calculate Assouad Dimension of sets
I am trying to work with sets $F$ as subsets of $\mathbb{R}^n$ and their Assouad dimension. I am a bit stuck on how to apply the definition as it asks that

We find the infimum of possible $\alpha$

We show for all $x \in F$ there exists a constant $C>$ such that for all $0<r<R$ the following holds

$$N_r(B(x,R)\cap F) \leq C\left(\frac{R}{r}\right)^\alpha$$

This is the definition I am using, which is from Fraser's Assouad Dimension and Fractal Geometry.

My issue is more related to HOW I can show these things. Of course I can let $x \in F$ and $0<r<R$, but then how do I show a constant that I choose will satisfy the inequality, moreover, that the $\alpha$ I choose is the infimum of possibilities.

The only example that Fraser provides does not do a direct proof but uses some sort of contradiction argument, which I am still not sure how?

Any advice or examples would be appreciated. Thank you!

答:
I think your (1) and (2) are in the wrong order.

For a fixed $\alpha$, the claim (2) might be true or false. The Assouad dimension is the infimum of all $\alpha$ such that (2) is true.

This definition allows you to bound the Assouad dimension in the following ways (which may be useful when computing the exact value of the dimension is difficult).

If are able to show (2) holds for some fixed $\alpha$, then the Assouad dimension is $\le \alpha$.

If you are able to show (2) is false for some fixed $\alpha$, then the Assouad dimension is $\ge \alpha$.


问:
Proving that if $\mathrm{ker}\, RTR^{-1}= \mathrm{ker}\, RSR^{-1}$ then $\mathrm{ker} T = \mathrm{ker}S$.
So I'm currently studying linear algebra and came across this interesting question about proving that if there are two vector spaces $V$ and $W$ with the following transformations: $S, T: V\to V$ are lineartransformations and $R:V\to W$ is an isomorphism, and if  $\mathrm{ker}\, RTR^{-1}= \mathrm{ker}\, RSR^{-1}$ then $\mathrm{ker} T = \mathrm{ker}S$.

I'm not exactly sure how to prove this. One way that came to mind is using the fact that $R$ is isomorphic such that $RR^{-1}=I$ and multiplying each side of $RTR^{-1}$ to get $T$ and just following through by stating that $\therefore$ $T=S$.

I don't think it's a valid proof so I was wondering if there is another way to prove this.

答:
The kernel of $RTR^{-1}$ is the subspace of all vectors $x$ such that $RTR^{-1}x=0$. Similarly for $RSR^{-1}$. We are given that these two kernels are equal. We then want to show that if $y$ is in the kernel of $T$ (i.e., $Ty=0$) then $y$ is in the kernel of $S$ (i.e., $Sy=0$). We also want to show if $y$ is in the kernel of $S$ then $y$ is in the kernel of $T$, but this will be the same argument. Together these two will give us that the kernels of $S$ and $T$ are equal.

So how do we show that if $Ty=0$ then $Sy=0$? By multiplying by $R$ and $R^{-1}$ in a `smart' way. Let $z=Ry$, then $y=R^{-1}z$. So we have that $Ty=TR^{-1}z = 0$. Multiply by $R$ on the left of both sides to get $RTR^{-1}z=R0=0$. Now we know that z is in the kernel of $RTR^{-1}$ so it must be in the kernel of $RSR^{-1}$, which means $RSR^{-1}z=0.$ Can you go from here to show that $Sy=0$?


问:
Number $3 + i$ is an image of a number $\sqrt{2} + 2\sqrt{2}*i$ turned around a point 0. Am I doing it right?
Number $3 + i$ is an image of a number $\sqrt{2} + 2\sqrt{2}*i$ turned around a point 0 by which angle?

So I check that $2\sqrt{2}*x=1 ->x= \frac{\sqrt{2}}{4}$

Then: $\sqrt{2}*\frac{\sqrt{2}}{4}=\frac{\sqrt{1}}{2}$

$\frac{\sqrt{1}}{2}=/=3$

So it is not an image of that number turned around a point 0 by any angle. Am I right?

答:
Rotation in the complex plane is multiplication by a number on the unit circle, i.e. multiplication by a number of the form $a+bi$ such that $a^2+b^2=1$.

The angle of that rotation can be computed using inverse trig functions, for example, $\tan^{-1}(b/a)$ if $a > 0$.

So the equation you should be trying to solve is
$$(3+i) \cdot (a+bi) = \sqrt{2} + 2 \sqrt{2} \, i
$$


问:
Triple Integral over a Superquadric
How do I solve the following:

$\iiint _V (x^2+y^2+z^2) dxdydz $, where $V=\{x^{2n}+y^{2n}+z^{2n} \leq 1\}$ is a superquadric?

答:
I found an online scan of the first or second edition of Whittaker and Watson, A Course of Modern Analysis. It includes Dirichlet's method, pages 191-193 of
See if this works, links to the individual page 191, page 192, page 193.

There was a way to magnify the image, here are the beginning set-up and the final calculation. For the current problem $f=1.$ To take the integral of $x^2$ (on the positive octant) I took
$$ f=1 \; , \; \; \alpha = \beta = \gamma = 2n \; , \; \; a=b=c=1 \; , \; \; p=3, q=1,r=1 $$
Then
$$ \frac{p}{\alpha}  +   \frac{q}{\beta}  + \frac{r}{\gamma}  =  \frac{5}{2n}   $$
so that the one-variable integral on the right is $  \frac{2n}{5} $

For the integral over the whole body of $x^2 + y^2 + z^2$  we get
$$  8 \cdot  \frac{\Gamma \left( 1 + \frac{3}{2n} \right)\Gamma \left( 1 + \frac{1}{2n} \right)^2}{\Gamma \left( 1 + \frac{5}{2n} \right)}  $$
Here I have used the recurrence relation
$$ z \; \Gamma(z) = \Gamma(1+z)  $$
which is 6.1.15 in Abramowitz and Stegun. Also found it in  the Wikipedia page.

This gives the correct figure for the sphere ($n=1$) which is $4 \pi / 5.$  It also gives the correct limit as $n$ goes to positive infinity, namely $8$

Title: Triple Integral over a Superquadric


问:
Lebesgue Integral of 1 over $\mathbb{R}$
We've been studying Lebesgue integrals in my analysis class and I'm unsure how to evaluate,
$$\int_{\mathbb{R}} 1 d\lambda(t). $$
I know it's a simple function, is $\lambda(\mathbb{R})=+\infty$? If so, would this integral just be infinity?

答: By definition the Lebesgue integral of a non-negative function $f$ is, as an extended real number, the least upper bound of the set $\left\{\sum\limits_{x\in\operatorname{range}\phi}x\lambda(\phi^{-1}(x))\,:\,\phi\text{ has finite range }\land0\le\phi\le f\right\}$, with the convention that $0\times (+\infty)=0$ and $\text{(not minus infinity)}+\infty=+\infty$. In your case $\phi(x):=1$ has finite range and $\sum\limits_{x\in\operatorname{range}\phi}x\lambda(\phi^{-1}(x))=1\cdot\lambda(\Bbb R)=+\infty$ and $\phi$ is smaller or equal to the integrand. Therefore $\int 1\,d\lambda\ge+\infty$.


问:
Weaker/Stronger Topologies and Compact/Hausdorff Spaces
In my topology lecture notes, I have written:

"By considering the identity map between different spaces with the same underlying set, it follows that for a compact, Hausdorff space:

$\bullet$ any weaker topology is compact, but not Hausdorff

$\bullet$ any stronger topology is Hausdorff, but not compact "

However, I'm struggling to see why this is. Can anyone shed some light on this?

答: Weaker implies compact and stronger implies Hausdorff is a general fact (an open cover in the weaker one is an open cover in the original too, pick there a finite subcover and you conclude. If you can separate two point in the original topology you can do it in the richer one because you can use the same open sets).Then suppose $ (X, \tau)$ is the topological space from which we start, if $\sigma$ is a weaker topology on $X$ and $(X,\sigma)$ is compact-Hausdorff, then: $Id: (X,\tau) \rightarrow (X,\sigma)$ is a continous bijection between compact-Hausdorff spaces, thus an homeomorphism. As it's the identity we get $\sigma=\tau$; similarly for the other case.</p>


答: Wrong.  Convert both numbers to polar form $(re^{i\theta})$.  For both numbers $r=\sqrt{10}$, so one is a rotation of the other.  For the first number $\theta  =arctan(1/3)$, while for the second $\theta =arctan(2)$.  Look up the angles to get the difference.

Title: Number $3 + i$ is an image of a number $\sqrt{2} + 2\sqrt{2}*i$ turned around a point 0. Am I doing it right?

Body:
Number $3 + i$ is an image of a number $\sqrt{2} + 2\sqrt{2}*i$ turned around a point 0 by which angle?

So I check that $2\sqrt{2}*x=1 ->x= \frac{\sqrt{2}}{4}$

Then: $\sqrt{2}*\frac{\sqrt{2}}{4}=\frac{\sqrt{1}}{2}$

$\frac{\sqrt{1}}{2}=/=3$

So it is not an image of that number turned around a point 0 by any angle. Am I right?


问:
Investigating convergence under the condition $|a_{n+1}-a_n| \le t^n$ for some number $0<t<1$
Problem: Let $(a_n)_n$ be a sequence with the property that, for some number $0 < t < 1$ and all $n \in N$, we have $|a_{n+1}-a_n| \le t^n$. Investigate whether the sequence converges.

I am not sure about the answer but I thought maybe $a_n=log(n)$ type of a sequence would satistfy this inequality and still diverge because its growth rate decreases as $n$ gets bigger. I can see that the condition will be $|log(n+1)-log(n)| = |log(\frac{n+1}{n})| \le t^n$ but I cannot find a compatible base for the log and I cannot show that this holds for all $n$.

Thanks for your effort and time.

答:
hint

Assume there exists $ t\in(0,1)$ such that

$$\Bigl(\ln(1+\frac 1n)\Bigr)^\frac 1n\le t$$

or,

$$\frac 1n\ln\Bigl(\ln(1+\frac 1n)\Bigr)<\ln(t)<0$$

but, using Taylor expansion
$$\ln(1+\frac 1n)=\frac 1n(1+\epsilon(n))$$
and
$$\ln(\frac 1n(1+\epsilon(n)))=$$
$$\ln(1+\epsilon(n))-\ln(n)$$
we prove easily that
$$\lim_{n\to\infty}\frac 1n\ln\Bigl(\ln(1+\frac 1n)\Bigr)=0$$
So, the condition $$|a_{n+1}-a_n|<t^n$$ is not satisfied when $ a_n=\ln(n) $.


答:
Let $\epsilon>0$ and let $N$ be the smallest natural number such that

$$\frac{t^N}{1-t}<\epsilon$$

For any $N\leq m<n$ we have

$$|a_n-a_m|=|(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+...+(a_{m+1}-a_m)|$$

$$\leq |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+...+|a_{m+1}-a_m|$$

$$\leq t^{n-1}+t^{n-2}+...+t^m=\frac{t^m-t^n}{1-t}<\frac{t^m}{1-t}\leq \frac{t^N}{1-t}<\epsilon$$

Thus, $a_n$ is Cauchy and therefore converges.

Title: Investigating convergence under the condition $|a_{n+1}-a_n| \le t^n$ for some number $0<t<1$

Body:
Problem: Let $(a_n)_n$ be a sequence with the property that, for some number $0 < t < 1$ and all $n \in N$, we have $|a_{n+1}-a_n| \le t^n$. Investigate whether the sequence converges.

I am not sure about the answer but I thought maybe $a_n=log(n)$ type of a sequence would satistfy this inequality and still diverge because its growth rate decreases as $n$ gets bigger. I can see that the condition will be $|log(n+1)-log(n)| = |log(\frac{n+1}{n})| \le t^n$ but I cannot find a compatible base for the log and I cannot show that this holds for all $n$.

Thanks for your effort and time.


问:
How do I describe this graph with an equation?
Spawn Graph

I need to describe this graph as an equation for my game in unity 3D, it describes the spawn of enemies in time. At first is kind of like a sigmoidal/power curve, but after some time it turns into a spline that goes up and down but overall increases slowly with time.

Can you help?

答:
Your quick sketch is not sufficient to understand what you want.

$$x-2\frac{\sin(5x)}x$$


问:
What is the difference between a clique and a simplex?
I have seen several descriptions of simplicial complexes and clique complexes as being a combination of simplexes and cliques.
I have heard descriptions of cliques being a subnetwork or subgraph wherein all nodes/vertices are 'connected' to each other.  I have also heard similar descriptions of simplices or definitions of the simplex.

Are these the synonyms, or is there a subtle difference between the two?

For example, is a simplex simply a subgraph wherein there are no unconnected vertices, but not strictly requiring all-to-all connections? Does the clique require all-to-all connections for the subgraph?

答:
They are very different types of object, but there is a relation between them.

In graph theory, an $n$-clique is a graph with $n$ vertices, and an edge joining each vertex to each other vertex.  Note that a graph, in the language of graph theory, consists only of a set of "vertices" and a set of "edges", each edge corresponding to a pair of distinct vertices.

In geometry or algebraic topology, an $n$-simplex is a subset of a Euclidean space which is the convex hull of $n+1$ affinely independent points (called the vertices of the $n$-simplex).

As a polytope, the $n$-simplex has edges which join each vertex to each other vertex.  Thus the vertices and edges of the $n$-simplex correspond to a graph which is an $n+1$-clique.

The use of the same words "vertex" and "edge" in both contexts is not really an accident, as polytopes provided some of the important motivating examples for graph theory.  But in the context of graph theory, the geometry is stripped away and all you have is an abstract set of "vertices" and "edges" that are pairs of vertices.


问:
A property of a prime divisor of a number consisting of 1s

For $n>0$ let $A(n) = \underbrace{111 \ldots 11}_{n}$. Prove that if $A(n)$ is divisible by a prime number $p>3$, then $\gcd(n, p-1) > 1$.

It is no huge discovery that if $n$ is even, then $2$ is a common divisor of $n$ and $p-1$, thus the implication holds. I don't know how to justify the general case though, so I would appreciate some hints.

答:
If $p(>3)$ divides $\underbrace{111 \ldots 11}_n, p$ divides $\underbrace{999\ldots 99}_n\implies p$ divides $(10^n-1)$

$\implies 10^n\equiv1\pmod p\implies ord_p{10}$ divides $n$

Again, using Fermat's Little Theorem, $10^{p-1}\equiv1\pmod p\implies ord_p{10}$ divides $p-1$

$\implies ord_p{10}$ divides $(n,p-1)$

If $(n,p-1)=1,ord_p{10}$ divides  $1\implies ord_p{10}=1\implies 10^1\equiv1\pmod p\implies p$ divides $9$ which is impossible as $p>3$


问:
Probability of $k$ fixed points for a random function from and to $\{1,..,n\}$
I would like to derive the probability mass distribution $p_k$ of the number $k$ of fixed points of a random function from $A:=\{1,..,n\}$ to the same set.

I proceed computing the number of derangements $D_n$, as per a random permutation, obtaining:
$$ D_n = n^n - \binom n1 n^{n-1} + \binom n2 n^{n-2} - .. (-1)^k\binom nk n^{n-k}=\sum_{i=0}^n (-1)^i\binom ni n^{n-i}.$$

Then I compute the number of random functions with $k$ fixed points $n_k$ as the product of $\binom nk$ and $D_{n-k}$, as if to say that each function is composed by a $k$-subset of $A$ times a deranged $n-k$-disposition of the remaining $n-k$ elements of $A$:
$$n_k=\binom nk \sum_{i=0}^{n-k} (-1)^i\binom {n-k}i (n-k)^{n-k-i},$$

so that
$$p_k = \frac{n_k}{n^n}=\binom nk \sum_{i=0}^{n-k} (-1)^i\binom {n-k}i \frac{(n-k)^{n-k-i}}{n^n}.$$

To me it makes sense, but a quick check on Python shows me that $\sum_{k=0}^n p_k\neq 1$.

Any thoughts?

答:
As JMoravitz pointed out in a comment, to find the number of functions with $\ k\ $ fixed points you need to multiply $\ {n\choose k}\ $ by the number of functions with no fixed points on $\ n-k\ $ elements—namely $\ (n-1)^{n-k}\ $—, not the number of derangements. Thus,
\begin{align}
p_k&=\frac{{n\choose k}(n-1)^{n-k}}{n^n}\\
&= {n\choose k} \left(\frac{1}{n}\right) ^k\left(1-\frac{1}{n}\right)^{n-k}\ .
\end{align}
That is, the distribution of the number of fixed points is $\ Binomial\big(n, \frac{1}{n}\big)\ $.  Another way of seeing this is that each of the numbers $\ 1,2,\dots,n\ $ is independently and equally likely to be a fixed point with probability $\ \frac{1}{n}\ $.


问:
Proving a linear map has a solution if and only if its solution is in the null space of the adjoint map
Let $U,V$ be finite dimensional vector spaces over a field $\mathbb{F}$ and $T\in L(U,V)$. I want to prove that for any $v\in V$, the equation $T(u)=v$ has a solution for some $u\in U$ if and only if for any $v'\in V$ such that $T'(v')=0$, we have $\langle v,v'\rangle=0$.

I am having trouble seeing how I could show either direction of this statement. I know that if we have such a $v'\in V'$, then this $v'\in N(T')$, but I cannot see how this will imply that there is such a $u\in U$ that solves $T(u)=v$. Moreover, the other direction seems even more absurd to me. I would really appreciate any help here as I am stuck on how to prove this.

Note: Prime notation means it is the dual space or adjoint space if the prime is on a linear map.

答:
Assume first that $T(u)=v$, and $f\in V':\,T'(f)=0$, then
$$f(v)=\langle f,v\rangle=\langle f,\,T(u)\rangle=\langle T'(f),\,u\rangle=0 \, .$$
Note that this also implies $ f|_{R(T)}=0\implies T'(f)=0$ where $R(T)$ is the range of $T$.

For the other direction, if $v$ is not in the range of $T$, we need to find a $f\in V'$ such that $T'(f)=0$ but $f(v)\ne 0$.

Pick an arbitrary basis $b_1,\dots,b_r$ of  $R(T)$, and extend it by $c_1=v,\,c_2,\dots,c_{n-r}$. Then define $f\in V'$ so that
$$f(b_i)=0,\ f(c_1)=1,\ f(c_j)=0\ (j\ge 2)\,.$$


问:
Find the limit of the function without L'Hôpital's rule
I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!

答:
HINT

We can use that

$$\left(\frac{\cos(ax)}{\cos(bx)}\right) ^\tfrac{1}{x^2}=\left(\frac{1-(1-\cos(ax))}{1-(1-\cos(bx))}\right) ^\tfrac{1}{x^2}$$

and then use that

$$1-(1-\cos(ax))^\tfrac{1}{x^2}=\left(1-(1-\cos(ax))^\tfrac1{1-\cos(ax)}\right)^{a^2\tfrac{1-\cos(ax)}{(ax)^2}}$$

$$1-(1-\cos(bx))^\tfrac{1}{x^2}=\left(1-(1-\cos(bx))^\tfrac1{1-\cos(bx)}\right)^{b^2\tfrac{1-\cos(bx)}{(bx)^2}}$$

and refer to standard limits.


答: In terms of $t:=\lim_{y\to0}\frac{\ln\cos y}{y^2}$, write your limit as $L=\exp\ell$ with$$\ell:=\lim_{x\to0}\frac{\ln\tfrac{\cos ax}{\cos bx}}{x^2}=t(a^2-b^2).$$We now prove $t=-\tfrac12$ so $L=\exp\tfrac{b^2-a^2}{2}$. In terms of $z:=1-\cos y$,$$t=\underbrace{\lim_{z\to0}\frac{\ln(1-z)}{-z}}_{1}\underbrace{\lim_{y\to0}\frac{\cos y-1}{y^2}}_{-\tfrac12}.$$

Title: Find the limit of the function without L'Hôpital's rule

Body:
I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!


问:
Suggestions for books on combinatorics
I am very weak in combinatorics and I have not got my understanding clear to think a problem in combinatorics and come up with a logic/analysis.I want to develop both the interest and the understanding so that I can proceed further and solve questions.Are there books on combinatorics that are both interesting and has good problems(problems of all difficulties)</p>

答:
Here are three books that I like:

Introductory Combinatorics by Brualdi. It covers theory well, though not in an encylopaedic way, and has lots of examples.

A Path to Combinatorics for Undergraduates by Andreescu. This books pretends to be for undergraduates, but like most of Andreescu's works, it is for olympiad or contest preparation. There is nothing "undergraduate" about it and can be approached by a motivated school student.

Combinatorics: A Problem-Based Approach by Mladenović. There is a huge selection of problems in this book and it is fairly comprehensive.

Title: Suggestions for books on combinatorics


问:
Suppose $R$ be a ring and $I$ an ideal in R. prove that $rad(I)=R$ if and only if $R=I$.
Suppose $R$ be a commutative ring and $I$ an ideal in R. prove that $rad(I)=R$ if and only if $R=I$.

If $R=I$ then $\forall x \in R : x^1 =x \in I$ so $rad(I)=R.$

Can you help me to prove the other side? If it is incorrect please say a counterexample. Thanks.

$$rad(I)= \{ x\in R | \exists n ; x^n\in I \}.$$

Footnote : $R$ has unity.

答:
This is false in rings without $1$: take any ring $R$ with the zero multiplication. Then $\mathrm{rad}(I)=R$ for all ideals $I$, but any subgroup of the underlying abelian group is an ideal. So any abelian group with proper subgroups and the zero multiplication would be a counterexample.

Note: The material below is way too complicated; I somehow got the idea that $\mathrm{rad}(I)$ had been defined as the ideal generated by the set of elements $a$ such that $a^n\in I$ for some $n$; the argument below can be used to show that this is in fact equal to that set (or equivalently, to show that the set of such elements is an ideal). So, feel free to ignore it.

For commutative rings with $1$, let $x\in R$; then there exist $f_1,\ldots,f_n\in\mathrm{rad}(I)$ and elements $a_1,\ldots,a_n\in R$ such that $1=a_1f_1+\cdots+a_nf_n$. The general summand of $(a_1f_1+\cdots+a_nf_n)^N$ looks like
$$\frac{N!}{(i_1!)\cdots(i_n!)}(a_1^{i_1}f_1^{i_1})\cdots(a_n^{i_n}f_n^{i_n}),\qquad\text{where }i_1+\cdots+i_n=N.$$

We know there exists $r_i\gt 0$ such that $f_i^{r_i}\in I$. So pick a sufficiently large $N$ to ensure that each term of $(a_1f_1+\cdots+a_nf_n)^N$ is necessarily in $I$.

Grrr... somehow thought the radical was the ideal generated by the elements whose power lies in $I$, and that it hadn’t been shown it was equal to the elements... this is overly complicated.


问:
General term of two sequences
I'm trying to find the general term of $(1,0,-3,0,5,0,-7,\ldots)$ and $(0,2,0,-4,0,6,\ldots)$. I think I can manipulate the general term of the sequence $(1,0,-1,0,1,\ldots)$, but I can't figure it out, Any ideas are welcome. I'm trying to understand sequences on $\Bbb R^2$.

答:
As it was pointed out in the comments, it would be a good idea to use $\sin$ and $\cos$, since we know they can generate $0,1$ and $-1$. We'll use that

$\sin(\frac{\pi}{2})=1,\sin(\pi)=\sin(2\pi)=0,\sin(\frac{3\pi}{2})=-1,\sin(x+2\pi)=\sin(x)$, from what we get $\sin(n\frac{\pi}{2})=1,0,-1,0,1,\ldots$

and

$\cos(\frac{\pi}{2})=\cos(\frac{3\pi}{2})=0,\cos(\pi)=-1,\cos(2\pi)=1,\cos(x+2\pi)=\cos(x),\cos(x+\pi)=-\cos(x)$, from what we get $\cos(n\frac{\pi}{2}+\pi)=-\cos(n\frac{\pi}{2})=0,1,0,-1,0,1,\ldots$

Now, for the first one, if we want $(1,0,-1,0,1,\ldots)$ to be $(1,0,-3,0,5,\ldots)$ we need to multiply the terms of the sequence by its position, since then we get $(1·1,0·2,-1·3,0·4,1·5,\ldots)=(1,0,-3,0,5,\ldots)$

To do this we just need to multiply the $n$-th term by $n$. So $(1,0,-3,0,5,\ldots)$ is given by $n\sin(n\frac{\pi}{2})$

I'll leave to you the second one.

Title: General term of two sequences


问:
Why is this "iterated compactification" construction on $\Bbb{R}^4$ so different from $S^4$?
So the standard method of compactifying $\mathbb{R}^4$ is to append a point at infinity and then map $\mathbb{R}^4$ to $S^4$ using stereographic projection with the point at infinity representing the "north-pole" used in the stereographic construction.

The construction motivating this question is as follows:

Start by performing the standard compactification to $S^2$ on the copy of $\mathbb{R}^2$ comprising, say, the $xy$-plane. This transforms $\mathbb{R}^4$ so as to be a (trivial) plane bundle over $S^2\backslash \{\infty\}$ with a single point, $\infty$, which does not uphold a plane. Denote this sphere $S_0$ and its point at infinity by $\infty_0$. For each plane attached at some $p \in S_0\backslash\{\infty_0\}$, append a new point at infinity $\infty_p$ and perform the compactification process again. We now have uncountably many spheres $S_p$, one for each $p \in S_0 \backslash\{\infty_0\}$. Now "roll" each $S_p$ along $S_0$ so that in the end $\infty_p$ is identified with $\infty_0$; we can do this because there is a path $\gamma_0$ in $S_0$ from $p$ to $\infty_0$ and another path $\gamma_p$ in $S_p$ from $p$ to $\infty_p$, so "roll" $S_p$ along $S_0$ such that at time $t$ we have $\gamma_0(t)\sim \gamma_p(t)$. This realizes our space as something that seems a lot like an uncountable wedge of spheres.

̶I̶ ̶t̶h̶i̶n̶k̶ ̶t̶h̶e̶ ̶r̶e̶s̶u̶l̶t̶i̶n̶g̶ ̶s̶p̶a̶c̶e̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶c̶o̶m̶p̶a̶c̶t̶,̶ ̶b̶u̶t̶ ̶I̶ ̶c̶o̶u̶l̶d̶ ̶b̶e̶ ̶w̶r̶o̶n̶g̶.̶

(This was silly, it is not.)

These constructions can't be homotopic since $\pi_2(S^4)=0$ whereas the uncountable wedge of spheres would need to have uncountable $\pi_2$, but this clearly isn't a regular ol' wedge of spheres (I underestimated how weird the topology would be).

I am wondering if there is some intuition for why this space is so different from $S^4$ when their underlying constructions seem very similar.

答:
Call the uncountably infinite wedge of spheres $X$ and let $\Phi: \Bbb{R}^4 \rightarrow X$ be the map realizing the construction in the problem statement.

Per Paul Sinclair's comments, $X$ is second countable. I was not initially convinced of this because it seemed we couldn't have a base yielding a neighborhood $N_p$ of $\infty_p$ satisfying $N_p \cap S_q = \emptyset$ for all $p\neq q$, but such sets don't correspond to any open set in our plane bundle and so cannot be members of the quotient topology. Rather, the open neighborhoods of $\infty_p$ will be uncountable unions of "would-be" open disks on the spheres (i.e. disks that would be open if we thought about equipping each sphere with the standard topology and then looked at the topology of their uncountable wedge), so specifying a neighborhood of any $\infty_p$ is the same as specifying a closed set whose intersection with the affine plane corresponding to $S_p$ is bounded. It is clear that $X$ is Hausdorff.

Let $P_0$ denote the cartesian plane, $\kappa_0: P_0 \rightarrow S_0$ the compactification map. Moreover, given $\kappa_0(x_p,y_p):=p\in S_0$, let $P_{p}$ be the plane attached to $\kappa_0(p)$ and let $\kappa_p : P_p \rightarrow S_p$ be its compactification. To show $X$ is not compact, consider the net $(N,\leq_{pre})$ defined by $$N:=\{z_{\alpha}\in S_{p_{\alpha}} \; | \; z_{\alpha} =(\kappa_{p_{\alpha}}\circ \kappa_0)(x_{p_{\alpha}},y_{p_{\alpha}},0,0), p_{\alpha} \in P_0\}$$ and preorder $$z_{\alpha}\leq_{pre} z_{\beta}
 \Leftrightarrow x_{p_{\alpha}}^2+y_{p_{\alpha}}^2 \leq x_{p_{\beta}}^2+y_{p_{\beta}}^2$$

Such a net cannot accumulate at a point in $Im(\Phi)$ since this would imply there is a $q\in \Bbb{R}^4$ such that eventually (with respect to the preorder $\leq_{pre}$) all $(\kappa_{\alpha}\circ \kappa_0)^{-1}(z_{\alpha})$ are in $B(1,q)$ whereas I can select large enough $x$ and $y$ such that the distance between $(x,y,0,0)$ and $q$ exceeds $\epsilon$.

$N$ also cannot accumulate at the wedge point. Letting $A:=P_0\times [-1,1]^2\subset \Bbb{R}^4$, note that $\overline{\Phi(A)}^c$ is a neighborhood of wedge point and disjoint from $N$.

Hence, the net has no accumulation point and $X$ is not compact.

I am still curious if there is an interesting reason why these spaces end up being so different when the procedures for constructing them seem similar.


问:
How many Assumptions can I make in Proof by Induction?
Let's say I want to prove by mathematical induction that a statement involving $n$ is true for all $n\in\mathbb N$. A classic example is proving that
$$\sum_{r=1}^nr=\frac{1}{2}n(n+1)$$
The base step is trivial. This requires only $1$ assumption to prove by induction; namely that for a given $n=k$ the statement holds and then we go on to prove that this implies the statement is also true for $n=k+1$. But what if one assumption is not enough to prove the general case? Let's say to prove the general case I needed to assume the statement held for $2$ values: $n=k-1$ and $n=k$ and then I used this to prove that the result follows for $n=k+1$. Would that be ok?

More generally, for a proof by induction must I only make one assumption or am I permitted to use many? Please explain why, whatever the answer is.

Thank you for your help.

答: Of course you can! But then you must have two base cases. For example, if you show that ${n=k}$ and ${n=k+1}$ both being true would imply that ${n=k+2}$ is true, and you have two base cases (such as ${n=0}$ and ${n=1}$ for example) then that would imply the case for ${n=2}$ is true. Then it being true for ${n=1}$ and ${n=2}$ would imply it's true for ${n=3}$. Then for $4$, then for $5$... etc etc. It's the same principle


问:
Closed form for $1-2+3-4+\cdots(-1)^{n-1}n$
How do I find a closed formula for the following?

$$\sum_{i=1}^n(-1)^{i-1}i$$

If $n$ is odd number, I can express it as $\frac{n+1}{2}$; if $n$ is even, then the expression if $\frac{-n}{2}$, but how do you find a expression that can express it in one formula?

答:
We can use $(-1)^n$ to distinguish positive and negative terms:
$$
\frac{-n\cdot(-1)^n+\frac{1-(-1)^n}2}2=\frac{1 + (-1)^{n+1}(1+2n)}{4}
$$