问:
Why $O(|x|^{m+1})$ in Taylor expansion of $f \in C^m$ is incorrect?
If this has been asked and answered before, I will delete the question.

Consider expanding $f(x+a)\in C^m$ for $m\in \mathbf{N}$ at $a=0$. Then writing $O(|x|^{x+1})$ in

$$f(x)=\sum_{k=0}^m \frac{f^{(k)}(0)}{k!} x^k + O(|x|^{m+1})$$

is not correct. Why is that?

答: Because it might be $o(\vert x\vert^m)$ but not $O(\vert x\vert^{m+1})$. Take as an example $f(x)=\vert x\vert^\frac{3}{2}$ on the reals. It's continuously differentiable with $f'(x)=\frac32\operatorname{sgn}(x)\vert x\vert^\frac12$ ($\operatorname{sgn}$ being the signum function, returning $1$ for positive numbers, $-1$ for negative ones, and $0$ for $0$) and it's already in expanded form for $C^1$ functions, since $f(0)=f'(0)=0$, so $f(x)=f(0)+f'(0)x+\vert x\vert^\frac32$. But clearly the remainder term $\vert x\vert^\frac32$ is not $O(\vert x\vert^2)$, just $o(\vert x\vert)$.


问:
How many possible ways are there to print 125 pages on 5 printers?
I think I understand this problem but I want to make sure I do get it thoroughly. In my mind, this problem is asking you to assign a single page to a single printer, but it doesn't really matter the order in which they are printed. I.E., we don't care if page 1 is sent to printer 1 first, or if printer one gets page 2 first and then page 1 is sent.

In my mind, then, we would simply take a page and say, how many choices do I have to print it? That would be 5 for the 1st page. Now page 2 also has 5 choices, and so on and so on, so that means the solution would be 5 x 5 x ... x 5 = 5 ^ 125.

Is this the correct answer or am I implicitly slipping in order some how and need to remove the number of permutations from this number? Thanks for your help in understanding this problem!

答:
I think there are 2 ways to think about this, since the problem itself is kind of vague.

Number 1: If each piece of paper counts as different, then it is true that they each have 5 choices, so the answer would be 5^125 as you said.

Number 2: Another possibility is that it is asking how many ways can you distribute 125 pages into 5 printers, which you would then use stars and bars with 125 stars and 4 bars to get 129 C 4.

ContentLicense: CC BY-SA 4.0


问:
How much of a proof can be left for the reader?
I'm an undergrad trying to publish a result concerning Conway's Base-13 function. Essentially, it's a direct proof that it's possible to formulate a closed-form expression for the function over specific subsets of $\mathbb{R}$.

Regardless, the proof requires defining a dozen or so functions, which, using finite arithmetic, serve as a variety of digit-manipulation operations.

I'm struggling to grasp how detailed the proof needs to be when it comes to proving that each of these functions do what I say they do, based on their definitions. Having dozens of them, which do relatively mundane things, I don't want the proof to be pages upon pages of lemma after lemma. That being said, I don't know how much can be left as an "exercise for the reader" either.

As a simple example, here's one of the first functions I define:

Definition: Let $D$, "the digit function", be defined as $D_b^n(x)=\big\lfloor\frac{x}{b^n}\big\rfloor-b\big\lfloor\frac{x}{b^{n+1}}\big\rfloor\ \forall x\in\mathbb{N}_0$, for any base $b\in\mathbb{Z}^{>1}$, and any digit-index $n\in\mathbb{N_0}$.

Aside from not knowing if this is good formatting/notation for definiting such a function, I'm uncertain how detailed of a proof I need to give to prove that this "digit function" does what I intend it do to. Which is, return the $n^{th}$ digit from the left of the radix point ($n=0$ being the units' place) for any base-$b$ expansion of a given integer $x$.

As I said, this is just a simple example. There are other functions further in the paper that do more complicated tasks such as replacing a specific digit with a decimal point, etc. I'm confident all these functions do what I've intended them to do, and any reader could prove such if they really wanted to.

Hence, for the sake of brevity, how much of a proof can be left for the reader? Is there a rule of thumb?

For more information on the situation, this is a follow-up to a previous question: Ambiguity of publishing as an undergrad.

答:
It depends on the reader. If the reader is "Conway", then "all of it" can be left to the reader.

If the reader is your fellow students, then more likely the answer is "some of it, but less than you think." And if you can communicate the main idea of the function, and why it should do what you want, then you may reasonably be able to omit the proof.

When you say "any reader could prove such if they really wanted to," think back to yourself in 8th grade, say. If you were like me, you could understand the ideas (generalizing from an example or two), but couldn't write a proof to save your life. So making a claim like this is an excuse for not thinking carefully, and I advise you to avoid it.

It often helps to start an article with your assumptions about the reader ("I assume the reader is familiar with the usual definitions from elementary differential geometry of surfaces in 3-space, including the first and second fundamental form; I'll use the notation from [ref1]."); the helps set expectations, and helps remind you of what you promised.


问:
Example of such family of functions in $L^{p}(\mathbb{R})$?
I'm preparing for the functional analysis test and I have a doubt on $L^{p}$ spaces. Kolmogorov criterion characterizes precompactness in $L^{p}(\mathbb{R}^d )$ for any $p \neq +\infty$, and says the following:

$H\subseteq L^{p}(\mathbb{R}^{d})$ is precompact if and only if:

$H$ is bounded

$\lim\limits_{R\rightarrow 0}\hspace{2mm}\sup\limits_{f\in H}\int\limits_{|x|\geq R}|f(x)|^{p}dx=0$

$\lim\limits_{|a|\rightarrow0}\hspace{2mm}\sup\limits_{f\in H}\|T_{a}f-f\|_{p}=0$

Our professors told us that usually this criterion fails on the third point, but so far I did a lot of exercises and it always worked (up to mistakes). Since it can't always be like that, as any arbitrary collection of functions can't be precompact, I'm asking:

Can you give me an example where such criterion fails? Is there a general idea one should check to see "at first glance" when this criterion indeed fails?

答:
One simple example of a collection which satisfies $1$ and $2$ but not $3$ is $B_\Omega$, the unit ball in $L^p(\Omega)$ for some bounded set $\Omega$, thought of as a subset of $L^p(\mathbb{R}^d)$. These are the functions in $L^p(\mathbb{R}^d)$ with norm at most $1$ and that are $0$ outside of $\Omega$. It is clear that $B_\Omega$ satisfies $1$ and $2$ but it is not compact as the unit ball in an infinite dimensional Banach space. Therefore $3$ cannot hold, which you may also verify directly by considering functions supported on very thin rectangles.

In general, given any collection of vectors it is usually relatively obvious if 1 and 2 hold and it is easy to see that a collection without one of these properties cannot be precompact. So, as I think about it, the real insight of the criterion is that among the collections which do not fail to be precompact in some "obvious" way, the precompact ones are exactly the ones satisfying 3.


问:
Use direct proof to prove the following identity.
Prove that $3n^2+7\leq n^3, \forall n\ge 4.$

My attempt

I could prove it by principle of mathematical induction.
How do I prove using direct proof method?

$3n^2+7=3n^2+4+2+1\leq3n^2+n+2n+1\leq n^3+3n^2+n+2n+1=(n+1)^3 (\because 4\leq n, 2\leq 2n) $

I could prove $3n^2+7\leq (n+1)^3, \forall n\ge 4.$
Please help me.

Body:
When $n\geq4$, we have $n^2\geq16>7$. So
$$3n^2+7<3n^2+n^2=4n^2\leq n\cdot n^2=n^3.$$


问:
Solution verification of a conditional probability question

Let $X$ and $Y$ be random losses with joint density function
$$f(x, y) = 2x \;\; 0<x, y<1$$
An insurance policy is written to cover the loss $X+Y$. The policy has a deductible of $1$. Calculate the expected payment under the policy.

My attempt:

Let $Z$ denote the payment under this policy.
$$Z=\begin{cases}
0 \; \;\;\; \text{if}\; 0<x+y<1, 0<x, y<1\\
X+Y-1 \;  \text{if}\; 1\le x+y\le 2, 0<x, y<1
\end{cases}$$

$E[Z] = 0+ E[X+Y-1\mid 1-x<y<1] \cdot P[1-x<y<1]$.

Note that
$E[X+Y-1\mid 1-x<y<1] = \int_0^1\int_{1-x}^1 (x+y-1)\cdot 2x \; \; \text{dy dx} = \frac{1}{4}$ and $P[1-x<y<1] = \int_0^1\int_{1-x}^1 2x \; \; \text{dy dx} = \frac{2}{3}$.

This is not the correct solution. The correct solution only computed $E[X+Y-1\mid 1-x<y<1]$ using the same formula as the one above but did not consider $P[1-x<y<1]$. Why is that approach correct (and my approach incorrect)?

答:
Your calculation of the conditional expectation is actually the complete calculation because in fact, $$\operatorname{E}[X+Y-1 \mid 1-X < Y < 1] = \frac{\int_{x=0}^1 \int_{y=1-x}^1 (x+y-1) 2x \, dy \, dx}{\Pr[1 - X < Y < 1]}.$$  This is because in order for the conditional expectation to be conditioned, it must be divided by the probability of observing the event $1 - X < Y < 1$.

Another way to see this is to write $$Z = \max(0, X+Y-1)$$ hence $$\begin{align}
\operatorname{E}[Z] &= \int_{x=0}^1 \int_{y=0}^1 \max(0, x+y-1) 2x \, dy \, dx \\ &= \int_{x=0}^1 \int_{y=0}^{1-x} 0 \, dy \, dx + \int_{x=0}^1 \int_{y=1-x}^1 (x+y-1) 2x \, dy \, dx. \end{align}$$


问:
How to get the value of $2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$ given two constrains?
The problem is as follows:

Given:

$\sqrt{\cos x}\cdot \sin y > 0$

and,

$\tan x\cdot \sqrt{\cot y} < 0$

Find:

$B=2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{-1}\\
2.&\textrm{5}\\
3.&\textrm{1}\\
4.&\textrm{-5}\\
\end{array}$

I'm confused on exactly how to use the given clues to solve the problem?

To me the source of confusion is how to use the absolute value in the question?

From the first given expression I'm getting this, assuming squaring both sides of the equation will not modify its order:

$\left(\sqrt{\cos x}\cdot \sin y\right)^2 > 0^2$

$\cos x \cdot \sin^2 y > 0$

$\left(\tan x\cdot \sqrt{\cot y}\right )^2 < 0^2$

$\tan^2 x\cdot \cot y < 0^2$

But the thing is this where I'm stuck, where to go from here?. There isn't known a relationship between those angles. If so then I believe trigonometric identities could be used to simplify the expression. Therefore, should these expressions be divided or what?.}

Can someone help me here on what should be done? and more importantly why?. It would help me the most is an answer which would explain how does absolute value is used here?.

答:
$a<0$ doesn't necessarily mean $a^2<0$.

For example, $-2<0$ but $(-2)^2 = 4 >0$.

Now,

$\begin{align}B &= 2|\csc x|\sin x + 3 |\cos  y|\sec y = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y}\\
\\& = 2\text{ sgn}(\sin x) + 3\text{ sgn}(\cos y)\end{align}$

Then, $\sqrt{\cos x} \sin y > 0 \Rightarrow \sin y > 0$ and $\cos x > 0$ as $\cos$ is under the square root.

Similarly, $\cot y>0$ and $\tan x<0$.

So we have,

$\sin y >0, \cot y = \frac{\cos y}{\sin y} > 0 \Rightarrow \boxed{\cos y >0}$

$\cos x>0, \tan x = \frac{\sin x}{\cos x}<0 \Rightarrow \boxed{\sin x<0}$

Thus,
$B = 2(-1) + 3(1) = 1$

Edit (based on comment)

As $\sin x<0 \Rightarrow |\sin x| = -\sin x$ and $\cos y >0 \Rightarrow |\cos y| = \cos y$.

So,

$\begin{align}B = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y} = 2 \frac{\sin x}{-\sin x}+3\frac{\cos y}{\cos y} = -2 + 3 = 1\end{align}$

Clarification for abs value:

Consider some numerical values. Let $x=1$, so $|x|=1=x$.

Again let $x=−1$. Now what is it's absolute value? It's again 1, right? $1=−(−1)$ or $|x|=−x$


问:
Probability of picking a specific element in a list
Suppose that we have a list with $n$ elements (where each element is a number between $1$ and $10$) and that each element is picked independently from the other elements in the list in a uniform random manner.

What would be the probability that the number $10$ is in the list? Would it be $\frac{1}{10} \cdot n$ since we have $n$ elements? I find it quite difficult to relate the probability to the list since we are dealing with $n$ elements so the length of the list is not fixed. A brief explanation would be appreciated.

答:
While it's true that each number has a 1/10 chance of being 10, you can't say that n numbers have a n/10 chance of having a 10, because that would mean that with 10 numbers, you are guaranteed a 10, which is clearly absurd.

The easiest way to solve this is to calculate the probability that 10 is NOT in the list.

The probability that any element is not 10 is 9/10. Therefore, the probability that all n numbers are not 10 is $(\frac{9}{10}$)^n, so by complementary counting, the probability of at least one 10 is $1$ - $(\frac{9}{10}$)^n.


问:
Prove that each multiple of 4 plus 3 is either prime or divisible by any $n\equiv3(\mod 4)$
I've been looking at a couple of questions in revision for an upcoming exam this year, and I've half solved a question, except for this bit.

I need to prove that for every multiple of 4, plus 3 (e.g: 3 x 4 + 3 = 15) is either a prime number or divisible by a number n for $n\equiv3(\mod 4)$

Thanks!

P.S. It's a bit vague, I don't know how to explain it, sorry!

答:
Suppose $n\equiv3\pmod4$ isn't prime. Consider its divisors. If any are $3\pmod4$, we are done. Otherwise, since $n$ is odd, all its divisors must be $1\pmod4$.

But if you multiply a bunch of numbers that are $1\pmod4$, the result will still be $1\pmod4$. This contradicts $n\equiv3\pmod4$.


Title: Use direct proof to prove the following identity.

Body:
Prove that $3n^2+7\leq n^3, \forall n\ge 4.$

My attempt

I could prove it by principle of mathematical induction.
How do I prove using direct proof method?

$3n^2+7=3n^2+4+2+1\leq3n^2+n+2n+1\leq n^3+3n^2+n+2n+1=(n+1)^3 (\because 4\leq n, 2\leq 2n) $

I could prove $3n^2+7\leq (n+1)^3, \forall n\ge 4.$
Please help me.

Body: $$n^3-3n^2-7=n^2(n-3)-7 \ge 4^2(4-3)-7=9>0.$$


问:
Clarification of my proof regarding closed nowhere dense sets
I have to prove that "A closed set $F$ of $\mathbb{R}^n$ is nowhere dense iff every point in $\mathbb{R}^n$ is a cluster point of its complement" ($\mathbb{R}^n$ being seen as a metric space with the usual metric).

Here it is my proof, in which I never use the hypothesis of $F$ being closed:

$\Rightarrow$ ] Let $x$ be a point of the space, and let's suppose there exists a number $r > 0$ such that the deleted neighbourhood $B_{r}(x)-\{x\}$ and the complement of $F$ are disjoint, i.e. $B_{r}(x)-\{x\} \subseteq F$. Since the set $U := B_{r}(x)-\{x\} $ is open (it is obtained from removing a finite number of points from an open set) and since in the euclidean space we can find points different from the center in every open ball, we can consider a point $z \in U $ and a number $s >0$ such that $\emptyset \not= B_{s}(z) \subseteq U \subseteq F$,
contradicting the fact F cannot contain any non-void open set.

$\Leftarrow$ ] Now let's suppose that every point in $\mathbb{R}^n$ is a cluster point of its complement, and suppose there exists $U \not= \emptyset $ open in $\mathbb{R}^n$ contained in $F$. Let then $z \in U$ and $r>0$ be such that $B_{r}(z) \subseteq U$. Then, $(B_{r}(z)-\{z\}) \cap F ^c \subseteq B_{r}(z) \cap F^c = \emptyset $,
thus $(B_{r}(z)-\{z\}) \cap F ^c = \emptyset $, contradicting $z$ being a cluster point of the complement of $F$.

I see nothing wrong on my proof, then I don't know if there's really a mistake or if the hypothesis of $F$ being closed is unnecessary.

答:
The problem is in the second half. You want to assume that $F$ is not nowhere dense in order to get a contradiction, so you suppose that $F$ contains a non-empty open set $U$. The problem is that when you say that a set $E$ is nowhere dense in a space $X$, you mean that the closure of $E$ has empty interior, not just that $E$ itself has empty interior. Thus, to say that your set $F$ is not nowhere dense in $\Bbb R^n$ you must assume that $\operatorname{cl}F$ contains a non-empty open set.

For an easy counterexample when $F$ is not closed just let $n=1$ and $F=\Bbb Q$.


问:
Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem
Let $ f(x) = \int_1^x \frac{\sqrt{t+1}}{t} \text{ d}t$

Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem

I can't choose $ g(x) $ and $ h(x) $ to use Squeeze Theorem.

Can someone give me a hint. Thanks

答: Hint: The integrand is larger than $\dfrac{\sqrt t}{t}.$


答:
Since for all $x\geqslant1$,

$$2(\sqrt x-1)=\int_1^x\frac{\sqrt t}{t}\,dt\leqslant \int_1^x\frac{\sqrt {t+1}}{t}\,dt = f(x),$$
and $2(\sqrt x - 1)\to \infty$, you get that $f(x)\to\infty$.

You don't really need an $h(x)$ in this case, since you are going to $\infty$. If you insist however you can take $$h(x) =2\sqrt2(\sqrt x-1) =\int_1^x\frac{\sqrt{2t}}{t}\,dt \geqslant f(x),$$
which obviously goes to $\infty$ also.

Body:
Let $ f(x) = \int_1^x \frac{\sqrt{t+1}}{t} \text{ d}t$

Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem

I can't choose $ g(x) $ and $ h(x) $ to use Squeeze Theorem.

Can someone give me a hint. Thanks


答:
You don't really need to find two functions, just one function that is smaller than your integrand.

First, notice that,
$$\frac{\sqrt{1+t}}{t}=\sqrt{\frac{1+t}{t^2}}=\sqrt{\frac{1}{t^2}+\frac{1}{t}}$$

Since $t>1$
$$\frac{1}{t^2}+\frac{1}{t}>\frac{2}{t^2}\implies\sqrt{\frac{1}{t^2}+\frac{1}{t}}>\frac{\sqrt{2}}{t}$$
$$$$
Now you might be able to do the rest.

Title: Prove that $ \lim_{x \to \infty}f(x) = \infty $ using Squeeze Theorem


问:
Do balls of finite radius have finite volume?
Let $M$ be a Riemannian manifold. For a point $p\in M$ and $r\ge0$, let $B_r(p)$ denote the closed ball of radius $r$ around $p$, i.e. the ball consisting of all points in $M$ that have distance less than or equal to $r$ from $p$ in the metric induced by the Riemannian metric. Do such balls always have finite volume? If not, what is a counter-example?

In case such a ball is compact, it necessarily has finite volume, because sufficiently small coordinate charts have finite volume and the ball could then be covered with finitely many of these. If $M$ is complete, the Hopf-Rinow theorem implies that these balls are compact, so a counter-example would have to be incomplete, but I don't know anything further.

答:
Consider the open unit ball in $\mathbb{R}^2$ with polar coordinates $(r,\theta)$ and a metric of the form $g=dr^2+f^2(r)d\theta^2$, where $f$ is a smooth positive function chosen so that $g$ is well behaved at the origin. The set $B_r(0)$ is the same as in the Euclidean metric, since radial geodesics are unchanged, but the volume differs; a simple computations shows
$$
\operatorname{vol}(B_r(0))=2\pi\int_0^{\min(1,r)}f(\rho)d\rho
$$
Thus $\operatorname{vol}(B_1(0))$ can be made to diverge by choosing a suitable $f$ which diverges as $r\to 1^-$.

I haven't computed it, but I suspect the curvature of such a metric also diverges as $r\to 1^-$, and that this specific behavior can be controlled by a lower bound on the curvature.

Edit:

Without any kind of completeness assumptions, one can construct even more pathological counterexamples. Here's one which is flat and simply connected.

Let $S$ be the universal cover of the punctured Euclidean plane the with projection $\pi:S\to\mathbb{R}^2\setminus\{0\}$ (equipped with the pullback metric), and let $\rho:S\to\mathbb{R}$ be the "radial" function defined by $\rho(p)=\|\pi(p)\|$. The set $\{p\in S:\rho(p)<r\}$ is an open submanifold of infinite volume, since it is an infinite-sheeted cover of a punctured open ball. Additionally, one can show by constructing paths which loop arbitrarily tightly around the origin that $d(p,q)\le\rho(p)+\rho(q)$. Thus, any ball $B_r(p)$ in $S$ with $r>\rho(p)$ has infinite volume.

With this kind of local pathological behavior in mind, I don't think the size of geodesic balls can be controlled by bounding the curvature alone.


答: How about considering $g(t) = \frac{1}{t}$ is easy to prove that $g(t) \leq \frac{\sqrt{1+t}}{t}$ for all $t \geq 1$. Now consider the function $G(x) = \int_1^x g(t)dt = \log(x)-\log(1)$. By monotonie of the integral it follows that $G(x) \leq f(x)$ And since $\log(x) \to \infty$ as $x \to \infty$ the we are done. I know is not the squeeze Theorem in its pure form but perhaps this helps.


答:
Can't get more direct than this: $n\ge 4 \Rightarrow n=a+4,\ a\ge 0$.

Substituting, we see $3n^2+7=3a^2+24a+53$ and $n^3=a^3+12a^2+48a+64$.

Subtracting, $n^3-(3n^2+7)=a^3+9a^2+24a+11>0,\ \forall a\ge 0$, QED


答:
Calculus based alternative approach:

$f(n) = n^3 - 3n^2 - 7 \implies f'(n) = 3n^2 - 6n = 3n(n-2).$

Therefore,

$f(4) > 0$.

$\forall n \geq 4, f'(n) > 0 \implies$ $f(n)$ is strictly increasing on $(4,+\infty).$


问:
How do i solve this integration question using the washer and shell method?
What is the volume of a solid enclosed by $y = (x-1)^2$ and $y = 4$ revolved around $x = - 3$?

I tried the washer method and the shell method and got different answers each time and I'm really confused please help!

My set up for the washer method was:

$$ \pi \int_0^4 {(4 + \sqrt{y} + 1)^2 - 4^2 \ \ dy} $$

My set up for the shell method was:

$$ 2 \pi \int_{-1}^3 {(x + 3)(4 - \sqrt{x} + 1)^2} \ \ dx $$

答:
Washer Method

The larger radius comes from the right side of the parabola $y = (x - 1)^2$, while the smaller radius comes from the left side of that parabola. Rewriting that parabola in terms of $x$, we have:

$$ y = (x - 1)^2 \Rightarrow x = 1 \pm \sqrt{y} . $$

Then, $ R(y) = (1 + \sqrt{y}) - (-3) $ and $ r(y) = (1 - \sqrt{y}) - (-3) $, where in both functions, the $ - (-3) $ comes from rotating about $x = -3$. These radii are also a bit easier to see graphically.

Then we have
$$ R(y) = 4 + \sqrt{y} , $$
$$ r(y) = 4 - \sqrt{y} . $$

Then, our formula for the volume is

$$ V_w = \pi \int_0^4 {R(y)^2 - r(y)^2 \ dy} $$
$$ = \pi \int_0^4 {(4 + \sqrt{y})^2 - (4 - \sqrt{y})^2 \ dy} $$
$$ = \frac{256\pi}{3} . $$

Shell Method

You've set up the radius of your cylindrical shells correctly:

$$ r(x) = x + 3 . $$

But, the height is just the difference in the $y$-coordinates of the two curves bounding your region:

$$ h(x) = 4 - (x - 1)^2 . $$

So, our volume is then

$$ V_s = 2 \pi \int_{-1}^3 {r(x)h(x) \ dx} $$
$$ = 2 \pi \int_{-1}^3 {(x + 3)(4 - (x - 1)^2) \ dx} $$
$$ = \frac{256\pi}{3} . $$


问:
Prior rounding scheme for efficient finite decimal multiplication
This concerns fixed precision finite decimal multiplication. It is about prior rounding which may be computationally beneficial as the result is only needed to be accurate at a specific Order of Magnitude (OM) given a particular rounding mode (RM).

Let, a and b be finite decimals, and an OM and RM be specified. How much prior rounding of a and b can be done to get a result equal to that obtained by multiplying a and b (a * b) exactly and then rounding using the specified OM and RM?

Is there a formula that works for all cases?

Example 1.

a=12.34567; b=67.89123; OM=-1; RM=Half_UP (HU)

With no prior rounding: a*b = 838.1627214741 ~ 838.2

(i) RM=HU for all rounding:

Prior round a(OM=-1), b(OM=-1): 12.3 * 67.9 = 835.17 ~ 835.2

Prior round a(OM=-2), b(OM=-2): 12.35 * 67.89 = 838.4415 ~ 838.4

Prior round a(OM=-2), b(OM=-3): 12.35 * 67.891 = 838.45385 ~ 838.4

Prior round a(OM=-3), b(OM=-2): 12.346 * 67.89 = 838.16994 ~ 838.2

Prior round a(OM=-3), b(OM=-3): 12.346 * 67.891 = 838.182286 ~ 838.2

(ii) RM=Half_Down (HD) to prior round a XOR b and RM = HU for the final round.

This is the same as (i) in this example (as neither a or b has 5 as the last digit at a significant place).

(iii) RM=Down (D) to prior round a XOR b and RM=Up (U) to prior round the other, and RM=HU for the final round.

a(RM=D), b(RM=U)

Prior round a(OM=-2), b(OM=-2): 12.34 * 67.90 = 837.886 ~ 837.9

Prior round a(OM=-2), b(OM=-3): 12.34 * 67.892 = 837.78728 ~ 837.8

Prior round a(OM=-3), b(OM=-2): 12.345 * 67.90 = 838.2255 ~ 838.2

Prior round a(OM=-3), b(OM=-3): 12.345 * 67.892 = 838.12674 ~ 838.1

Prior round a(OM=-4), b(OM=-3): 12.3456 * 67.892 = 838.1674752 ~ 838.2

Prior round a(OM=-3), b(OM=-4): 12.345 * 67.8913 = 838.1180985 ~ 838.1

Prior round a(OM=-4), b(OM=-4): 12.3456 * 67.8913 = 838.15883328 ~ 838.2

Prior round a(OM=-5), b(OM=-4): 12.34567 * 67.8913 = 838.163585671 ~ 838.2

Prior round a(OM=-4), b(OM=-5): 12.3456 * 67.89123 = 838.157969088 ~ 838.2

a(RM=U), b(RM=D)

Prior round a(OM=-2), b(OM=-2): 12.35 * 67.89 = 838.4415 ~ 838.4

Prior round a(OM=-2), b(OM=-3): 12.35 * 67.891 = 838.45385 ~ 838.5

Prior round a(OM=-3), b(OM=-2): 12.346 * 67.89 = 838.16994 ~ 838.2

Prior round a(OM=-3), b(OM=-3): 12.346 * 67.891 = 838.182286 ~ 838.2

Prior round a(OM=-4), b(OM=-3): 12.3457 * 67.891 = 838.1619187 ~ 838.2

Prior round a(OM=-3), b(OM=-4): 12.346 * 67.8912 = 838.1847552 ~ 838.2

Prior round a(OM=-4), b(OM=-4): 12.3457 * 67.8912 = 838.16438784 ~ 838.2

Prior round a(OM=-5), b(OM=-4): 12.34567 * 67.8912 = 838.162351104 ~ 838.2

Prior round a(OM=-4), b(OM=-5): 12.3457 * 67.89123 = 838.164758211 ~ 838.2

Example 2.

a=99.00044=b; OM=-1; RM=HU.

With no prior rounding: a * b = 9801.0871201936 ~ 9801.1

(i) RM=HU for all rounding:

Prior round a(OM=-3), b(OM=-3): 99 * 99 = 9801 ~ 9801.0

Prior round a(OM=-4), b(OM=-3): 99.0004 * 99 = 9801.0396 ~ 9801.0

Prior round a(OM=-3), b(OM=-4): 99 * 99.0004 = 9801.0396 ~ 9801.0

Prior round a(OM=-4), b(OM=-4): 99.0004 * 99.0004 = 9801.07920016 ~ 9801.1
...

Example 3.

a=999999.0000999999; b=999.000999; OM=-1; RM=HU.

With no prior rounding: a * b = 999000000.0989009999999001 ~ 999000000.1

(i) RM=HU for all rounding:

Prior round a(OM=-4), b(OM=-4): 999999.0001 * 999.0010 = 999000001.0989001 ~ 999000001.1

Prior round a(OM=-4): 999999.0001 * 999.000999 = 999000000.0989010999 ~ 999000000.1
...

Considerations:

A good strategy might be to prior round one number up and the other down.

The aim of prior rounding a and b is to reduce the amount of computation. This may be futile for decimals with small numbers of digits or for when there is only a small amount of rounding, but there could be significant computational gains for decimals with very large numbers of digits and substantial amounts of rounding.

I have now considered prior rounding for the case of multiplication of integers and developed some Java code to start to answer this question. I realised that prior rounding only the largest integer is a good approach and I think that the most prior rounding that can be done of the larger number is to round to the magnitude that is the integer part of the square root of the magnitude of the larger number. So, if: the two numbers being multiplied are a=(100 consecutive 9s) and b=(50 consecutive 9s); OM=64. Then the most prior rounding of x that can be done is to effectively divide x by 100000000, multiply the result of this by y and then multiply by 100000000. This has made me realise that any computational benefits of prior rounding for multiplying integers of this magnitude are likely to be negligible, the numbers involved must be far larger to gain any benefit from the overhead of the prior rounding. Translating this to decimal numbers, the benefit of prior rounding is going to be dependent on the magnitude of the integer component of the larger number, the number of decimal places it has, and the order of magnitude at which a precise albeit rounded result is wanted.

Special cases with rounding typically involve the digits 4, 5 and 9.

Feedback:

Thanks in advance for any feedback (on how to improve the question, or what examples might help).

I do not plan to detail the examples further as I don't want to edit the question too much and I think now that prior rounding just one of the numbers is probably the best strategy.

I intend to post an answer and provide a link to the Java implementation in due course. I am hoping to post an answer or an update in February.

答:
The problem can be divided into cases:

Case 1: a and b are integers

Case 2: a is an integer AND b is a fraction

Case 3: a AND b are both fractions

Case 4: a is a fraction AND b is a decimal

Case 5: a AND b are both decimals

Where:

a fraction is a number x such that -1 < x < 1 and is expressed as a decimal.

a decimal has both an integer and fractional part

Case 5 is the general case. It can be split into a two part problem: Separate a into the integer and fractional parts. Multiply the integer part by b (an example of Case 2). Multiply the fractional part by the b (an example of Case 4). Add the two parts together and do a final round. (When calculating the two components a bit more precision is needed - calculate the parts to 3 extra digits of precision).

I developed some Java code and a test suite. The method names start "multiplyPriorRound" and can be found in the following classes:

https://github.com/agdturner/agdt-java-math/blob/master/src/main/java/uk/ac/leeds/ccg/math/Math_BigDecimal.java

https://github.com/agdturner/agdt-java-math/blob/master/src/main/java/uk/ac/leeds/ccg/math/Math_BigInteger.java

A suite of test cases are in the associated JUnit test classes with method names beginning suites "testMultiplyPriorRound":

https://github.com/agdturner/agdt-java-math/blob/master/src/test/java/uk/ac/leeds/ccg/math/Math_BigDecimalTest.java

https://github.com/agdturner/agdt-java-math/blob/master/src/test/java/uk/ac/leeds/ccg/math/Math_BigIntegerTest.java

I drew some pictures to help me figure out what to do. The key idea is to make the strings of digits smaller mostly by disregarding the smallest parts of any fractions. For each number it was important to figure out the OM of the digit furthest to the left and furthest to the right. The number of digits to the left of the OM wanted for the result precision is important for identifying if and where to round the other number.

OM =      *  9  8  7  6  5  4  2  1  0 -1 -2 -3 -4 -5 -6 -7 -8 -9  *

Case 1:
a =     *  *  *  *  *  *  *  *  *  *.
b =              *  *  *  *  *  *  *.

Case 2:
a =     *  *  *  *  *  *  *  *  *  *.
b =                                 . *  *  *  *  *  *  *  *  *  *

Case 3:
a =                                 . *  *  *  *  *  *  *  *  *  *
b =                                 . *  *  *  *  *  *  *  *  *  *

Case 4:
a =                                 . *  *  *  *  *  *  *  *  *  *
b =              *  *  *  *  *  *  *. *  *  *  *  *  *  *  *

Case 5:
a =     *  *  *  *  *  *  *  *  *  *. *  *  *  *  *  *  *  *
b =              *  *  *  *  *  *  *. *  *  *  *  *  *  *  *  *  *

Sorry this is not a complete self contained answer. I am not sure if anyone is interested in this, so I am going to leave this here for now and see what happens. I may detail things further in due course. I hope to get around to doing some timing experiments to compare the prior rounding multiplication with arbitrary precision multiplication. I may post about that in another forum appropriate for that. (The numbers may have to have a lot of digits before there is a significant advantage.) There are similar challenges for efficient prior rounding in division and in raising a decimal to the power of another decimal...


问:
The affine tangent cone and the number of tangents
Someone said "If you can draw the Affine Tangent Cone in the origin at the
$$y^2 = x^3 + x^2$$
Then you can know that when there is singularity, You will intuitively understand why it is possible to figure out the number of tangents by considering from that point to the affine tangent cone." So, If you know, could you explain this sentence's meaning by drawing picture?

答: Assume the variables are real. The function $x^2+x^3$ is  nonegative for $x\geq -1$, so the curve lies to the right of the vertical line $x=-1$ touching it in the single point $y=0$. As you move to the right it touches the line $x=c$ in two points unless $c=0$. The points have $y$-coordinates  $\pm \sqrt{c^3+c^2}$. From the implicit function theorem the only points that could have multiple tangents seems to be $(-1,0)$, and $(0,0)$ .  I think you can get your answer from there.


问:
number of all inconstant maps f from A to A
Let $ A=\{1, 2, 3,..., n\}$. Find the number of all nonconstant maps $f: A \rightarrow A$  for which $f(k) \le f(k + 1)$ and $f(k) = f(f(k + 1))$ for $k = 1, \dots, n-1$..</p>

答:
Start with a simple case to see what is going on.  If $n=2$, so that $A = \{1,2\}$, then there are only two non-constant maps:

$f(1) = 2, f(2) = 1$

$g(1) = 1, g(2) = 2$

but the first can't work because it breaks the rule $f(k) \leq f(k+1)$.  As for $g$, we see that $g(1) = 1$, but $g(g(2)) = g(2) = 2$, so it breaks the second rule.

So there are no maps that satisfy the needed conditions for $n=2$.  For $n=3$, there are more maps, but if you not sure what is happening it's good to work these things out to get an idea.

Hint:  draw two columns for A, like this

$1 \qquad 1$

$2 \qquad 2$

$3 \qquad 3$

and indicate the map by drawing arrows from the left column to the right.  Can you describe a rule for the arrows to match the first condition ($f(k) \leq f(k+1))?  What about the second condition?

ContentLicense: CC BY-SA 3.0


问:
Proving alternative of Integration by Parts from Stegun's Book and its Definite form
I only knew 1 kind of integration by parts given by this:
$$\int u\, \Bbb dv = uv - \int v \,\Bbb du\tag{1}$$
And i can understand where it came from.

And the definite integral form is:
$$\int_a^b u\, \Bbb dv = uv\,\Bigg|_a^b - \int_a^b v \,\Bbb du\tag{2}$$

From Stegun's Book, i found another interesting form of integration by parts given by:
$$\int uv \Bbb dx = \left(\int u\Bbb dx\right)v - \int\left(\int u \Bbb dx\right)\frac{\Bbb dv}{\Bbb dx} \Bbb dx\tag{3}$$.

Maybe i can derive it using the product rule of differentation:

$$\begin{align}
D_x\left[(u(x)v(x))\right] &= u'(x)v(x) + u(x)v'(x) \\
\int D_x\left[(u(x)v(x))\right] \Bbb dx&=
\int u'(x)v(x) + u(x)v'(x)\Bbb dx\\
uv &= \int v \Bbb du + \int u \Bbb dv
\end{align}$$

But what now?

My question is, what's the "definite" integral form of Eq. $(3)$? And how to prove it? (I confuse with $\frac{\Bbb dv}{\Bbb dx}$) and which parts are evaluated from $a$ to $b$ for the definite integral form?

Thanks in advance!

答: This is the same rule, it just uses different terms.  In the second formula, $u$ is now $v$, and $v$ has become $\int u\,dx$, or put otherwise, $dv$ has become $u\,dx$.


问:
Optimally played game

Consider a game played on the tree above, where the "cost" at a leaf is paid by P1 to P2.  Thus P1 wants the number to be as small as possible where P2 wants it to be large.  What is the cost paid by P1 to P2 if both players play optimally?

When a question asks "optimally", does it mean that the players play trying to get the maximum amount after each move, and that the opposite player is also playing optimally? What I mean is:

in this graph, player 1 has to move left, since 0 is his goal. Now, player 2 can move left next which gives him a chance to get to 4, however, it could also get him to 0. if he moves right, he guarantees at least a 1.

So does it mean that to play optimally, player 2 would go right in his first move, even tough he has "a chance" of getting  four if he goes the other way?

答:
Other posts are great +1 for them, but it still took 30 mins to get the whole and missing parts to automate it to a computer program, so adding a comm wiki here:

Picture representation of the play

```<code>P1(min)                                   1*
P2(max)                  1*                                 3
P1(min)            0            1*                   2             3
P2(max)         4     0     1*     3             3      2      4      3
A[0]           4 2   0 0   1 0    3 2           2 3    1 2    4 1    3 2

Above, we have to do bottom-up.

</code>```

One question remains though:

Since we need to do bottom-up, who plays 1st while playing bottom-up?

Here, since the P1 plays 1st from (top-down), we need find who will play first if bottom-up.

So calculate Math.ceil(Math.log2(N))

If EVEN, OTHER player (P2) plays first when bottom-up.

If ODD, SAME player (P1) plays first when bottom-up.

N = number of costs/leafs = 16 here

Math.ceil(Math.log2(16)) = 4

4 is EVEN

so P2 plays first from bottom-up

If costs/leafs were only 9

then also it is 4

P2 plays first

If costs/leafs were only 8

then it is 3

P1 (same player this time) plays first

Please update this post if anything is anything.

Title: Optimally played game

Body:
Other posts are great +1 for them, but it still took 30 mins to get the whole and missing parts to automate it to a computer program, so adding a comm wiki here:

Picture representation of the play

P1(min)                                   1*
P2(max)                  1*                                 3
P1(min)            0            1*                   2             3
P2(max)         4     0     1*     3             3      2      4      3
A[0]           4 2   0 0   1 0    3 2           2 3    1 2    4 1    3 2

Above, we have to do bottom-up.

One question remains though:

Since we need to do bottom-up, who plays 1st while playing bottom-up?

Here, since the P1 plays 1st from (top-down), we need find who will play first if bottom-up.

So calculate Math.ceil(Math.log2(N))

If EVEN, OTHER player (P2) plays first when bottom-up.

If ODD, SAME player (P1) plays first when bottom-up.

N = number of costs/leafs = 16 here

Math.ceil(Math.log2(16)) = 4

4 is EVEN

so P2 plays first from bottom-up

If costs/leafs were only 9

then also it is 4

P2 plays first

If costs/leafs were only 8

then it is 3

P1 (same player this time) plays first

Please update this post if anything is anything.


问:
Proving combinatoric identity using the inclusion-exclusion principle
I need to prove the identity ${\displaystyle \sum_{k=0}^{n}(-1)^{k}{2n-k \choose k}2^{2n-2k}=2n+1}
 $ using the inclusion-exclusion principle. It was hinted to think about the number of ways to color the numbers ${1,..,2n}$  in the colors red and blue such that if $i$ is colored red so is $i-1$. Well, it is clearly why the number of ways to do it is $2n+1$, but I don't manage to show why it also equal to the left side of the equation.

Anyone can give me a hint about the sets I should define in order to use the inclusion-exclusion principle? Thanks!

答: There are $2^{2n}$ colourings in total, $2^{2n-2}{2n-1\choose 1}$ ways to pick a pair of adjacent numbers and colour the first blue and the second red and the rest any colour, $2^{2n-4}{2n-2\choose 2}$ ways to pick two pairs of numbers and have both with the first color blue and the second colour red and the rest any colour, and so on. </p>


问:
Determining which function grows the most over time
I have five logarithmic functions, 1 for each country I'm investigating. x = an integer value for years, y = share of 1-year-olds vaccinated against rotavirus.

So if x = 2, y would tell you how many 1-year-olds have been vaccinated against rotavirus (%) in the country's second year of progress.

Time Interval of x (Years)
a
b
y = a + bln(x)

Morocco
1-9
32.4
33.8
y = 32.4 + 33.8ln(x)

Sudan
2-9
22.3
34.7
y = 22.3 + 34.7ln(x)

Rwanda
3-9
41.8
27.9
y = 41.8 + 27.9ln(x)

Ghana
3-9
36.6
28.3
y = 36.6 + 28.3ln(x)

Botswana
3-9
75.4
6.18
y = 75.4 + 6.18ln(x)

This is for a school project, and I'm interested in determining which function grows the most over the course of its time interval -> so which country makes the most progress in vaccinations for the years of progress recorded.

I'm totally new to calculus but I want to include a calculus concept to determine this. I was looking into derivatives for calculating rates of change but I'm not sure if that's the simplest and most effective approach for what I want to do - would appreciate a suggested calculus approach potentially other than this that isn't too difficult to learn as a beginner!

答:
Given a function of the form $f(x) = a + b \ln(x) $, we can break it up into a couple of smaller functions.

First, the derivative (the function that describes the instantaneous rate of change) of the sum of two functions is the sum of their derivatives.

So, we can break up the function into the sum of two smaller functions. Let $f_1(x) = a$ and let $f_2(x) = b \ln(x) $.

Then, we have

$$ f(x) = f_1(x) + f_2(x) $$
$$ \Rightarrow f'(x) = f_1'(x) + f_2'(x) .$$

Here, the prime notation denotes the derivative of the function, so $f'(x)$ is the derivative of $f(x)$.

Now, $f_1(x) = a$ is a constant function - graphically, it's a horizontal line through $y = a$.

The rate of change of that function between any two points is always, since $y_2 - y_1 = 0$ for any two points on the graph. For that reason, the instantaneous rate of change is also 0 everywhere.

That gives us

$$ f'(x) = f_1'(x) + f_2'(x) = 0 + f_2'(x) = f_2'(x) . $$

Now, we need the derivative of $f_2(x)$. The derivative of the natural log function is $\frac{1}{x}$. This is in fact the way some mathematicians choose to define the natural log (and, by extension, the number $e$).

We can also use the fact that scaling a function by a multiple also scales its derivative in the same way. Given a constant $c$ and a function $g$, we have

$$ \frac{d}{dx}cg(x) = c \frac{d}{dx}g(x) . $$

Here, $ \frac{d}{dx} g(x) $ is another way to denote the derivative of $g(x)$ - it's equivalent to $g'(x)$.

This gives us:

$$ f_2'(x) = \frac{d}{dx} b \ln(x) $$
$$ = b \frac{d}{dx} \ln(x) $$
$$ = \frac{b}{x} . $$

So, putting this back into our equation above, we have

$$ f'(x) = \frac{b}{x} . $$

Notice, this is independent of $a$, since all $a$ does in our function is shift the curve up or down - it doesn't affect the actual rate of change of the function. That's why the derivative of the constant term was 0.

Now, applying this to the problem at hand, we want to know the instantaneous rate of change in year 9 (Matthew's advice for how to incorporate calculus). So, all we have to do is plug the value $ x = 9 $ into our formula for $f'(x)$.

For example, in the case of Morocco, the instantaneous rate of change in year 9 is:

$$ f'(9) = \frac{33.8}{9} . $$
$$ \approx 3.76 . $$

You can interpret that to mean that in year 9, the rate at which $y$ is increasing is approximately 3.76 percentage points per year. So, by year 10, you could expect that the proportion of one-year olds vaccinated in Morocco would increase by 3.76 percentage points.


问:
Are these 2 graphs isomorphic (with illustrations and counter examples that appear not to preserve adjacencies)?
The following labeled slide was presented in a first year discrete mathematics course. However,  the adjacencies don't appear to be preserved (see the second illustration below).

If the vertices in the above graph on the left are arbitrarily labeled A to F, and we select the arbitrary vertex A for the vertex labeled A in the graph on the right, there appears to be no way to label the adjacent vertices such that it will not create an edge that connects two vertices (the other two A-adjacent vertices) that are not adjacent in the original graph.

Of the 3 nodes that that are connected to vertex A, whichever is chosen to be the endpoint at the bottom right of the supposedly isomorphic graph, will lead to a situation where the other two nodes adjacent to A, will also be adjacent to each other, despite this not being the case in the original graph.

If these are really isomorphic, how can the second graph given in the slide preserve all the adjacencies of the original graph? Obviously, the degrees of the vertices and the sum of degrees of the graphs are the same, and the second graph is a bijection of the first, but what about the adjacencies?

答: You are right that the two given graphs are not isomorphic, so there is a mistake. An easy way to tell them apart is that the graph on the left is bipartite: edges only go from $\{A,C,E\}$ to $\{B,D,F\}$. The graph on the right contains a $3$-cycle, so it is not bipartite.


问:
Can we treat $\operatorname{span}()$ as an operator by taking the span of both sides of a relation?
Can we treat $\operatorname{span}$ as an operator by taking the span of both sides of a relation?

For example, does $A\subset B$ imply that $\operatorname{span}(A) \subset \operatorname{span}(B)$?

答:
The span() will map a set to a set. To be technical you will need some analogue of addition and scalar multiplication on the power sets of the space and then show that the span() is a linear map with respect to those analogues. Then it will be an operator in the more generalized sense, but still not in the strict sense.

Differing fields have more rigorous or more loose requirements to being an operator.

Also, the example given fails unless you allow equality in the subset operator. The counterexample is to take A to be a single non-zero vector v and B to be v together with 2v.


问:
Induction Not Starting from $\boldsymbol{0}$
I am reading the proof for the proposition below:

For all $\boldsymbol{a},\boldsymbol{m},\boldsymbol{n} \in \mathbb{N}$, if $\boldsymbol{a} > \boldsymbol{0}$ and $\boldsymbol{m} < \boldsymbol{n}$ then $\boldsymbol{a} \cdot \boldsymbol{m} < \boldsymbol{a} \cdot \boldsymbol{n}$.

It is assumed that the following theorem can be used in the proof:

For all $\boldsymbol{a},\boldsymbol{m},\boldsymbol{n} \in \mathbb{N}$, if $\boldsymbol{m} < \boldsymbol{n}$ then $\boldsymbol{a} + \boldsymbol{n} < \boldsymbol{a} + \boldsymbol{n}$.

The proof was constructed using mathematical induction, and the induction was conducted on $\boldsymbol{n}$. The author argued that the smallest $\boldsymbol{n}$ for $\boldsymbol{m} < \boldsymbol{n}$ is $\boldsymbol{m}^{+}$, so that the starting point for the induction should be $\boldsymbol{m}^{+}$ instead of $\boldsymbol{0}$. I am a little confused by this statement, although intuitively it makes sense. Below is my confusion.

To prove the theorem using mathematical induction, of course we should construct a set
\begin{equation}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{a} > \boldsymbol{0} \textrm{ and } \boldsymbol{m} < \boldsymbol{n} \textrm{ implies } \boldsymbol{a} \cdot \boldsymbol{m} < \boldsymbol{a} \cdot \boldsymbol{n}\right\},
\end{equation}
and try to prove $A = \mathbb{N}$ using the induction theorem ($\boldsymbol{0} \in A$ and for all $\boldsymbol{n} \in A$, $\boldsymbol{n}^{+} \in A$). In this case, $\boldsymbol{0} \in A$ is trivial, as $\boldsymbol{m} < \boldsymbol{0}$ has to be false. Then why did the author use $\boldsymbol{m}^{+}$ as the starting point for induction? Is it necessary to verify the first $\boldsymbol{n}$ where $\boldsymbol{m} < \boldsymbol{n}$ is true? Is it OK not to use $\boldsymbol{0}$ as the starting point for induction?

答: Given any integer $a,$ observe that the statement $P(n)$ holds for all integers $n \geq a$ if and only if the statement $P(m + a)$ holds for all integers $m \geq 0.$ Consequently, the Principle of Mathematical Induction can be invoked to show that (1.) if $P(n)$ holds for the integer $n = a$ and (2.) $P(n + 1)$ holds whenever $P(n)$ holds for some integer $n \geq a,$ then $P(n)$ holds for all integers $n \geq a.$ Ultimately, this says precisely that the base case for a proof by mathematical induction need not be the case that $n = 0.$


问:
Show $\mathbb{Q}(\sqrt{p}\: |\: p\text{ prime})$ is a normal extension of $\mathbb{Q}$
I tried to show that $\mathbb{Q}(\sqrt{p}\: |\: p\text{ prime})$ is a normal extension of
$\mathbb{Q}$.

For finite extension we have that normal Extensions are splitting fields and vice versa,
but how can I argue in the case of an infinite extension ?

Is there an argument that the infinite union of normal extensions is normal (If yes, how can I proof that?) or is there a simple argument for that example?

答:
Any $\Bbb{Q}$-algebra homomorphism from $\mathbb{Q}(\sqrt{p}\: |\: p\text{ prime})$ to its algebraic closure sends $\sqrt{p}$ to $\pm\sqrt{p}$, whence it sends $\mathbb{Q}(\sqrt{p}\: |\: p\text{ prime})$ to itself which is normal over $\Bbb{Q}$.

Equivalently it is the splitting field of an infinite set of polynomials $x^2-p\ | \ p$ prime.


问:
If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$
I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?

答:
UPDATE: the proof seems incorrect as b=4, a=2, fail when n=4 (check the comments below).

First, there is a counter example when n=1, where a=6 and b=13 (as $ (6+1) | (13+1) $ ). However I shall continue for all n>1.

If $(a^n+n) | (b^n+n)$ then there is a natural number k $\geq 1$ such that $b^n+n=k(a^n+n)$. Clearly $a=b \mbox{ iff } k=1$. We shall proceed to prove that k is always 1 for all n>1.

Continuing from $b^n+n=k(a^n+n)$,

$b^n+n \equiv 0 \quad (\mbox{mod } k)$

If we allow n to vary for the same a and b, then the only value for k that will make that equation hold for all n>1 is k=1, which implies that a=b.

ContentLicense: CC BY-SA 2.5

LastEditorDisplayName: anonymous


问:
Finite groups with exactly one maximal subgroup
I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated.

Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.

Now, does this mean exactly one conjugacy class of maximal subgroups? If it really means exactly one maximal subgroup, then what can we say about a finite group that possesses exactly one class of (at least two) maximal subgroups that are all conjugate to each other?

答:
If $G$ has only one conjugacy class of maximal subgroup, $M$, and is not itself a $p$-group, then for every prime $p$, $M$ contains a Sylow $p$-subgroup $P$ of $G$.  But then by Lagrange, $|M|$ has order divisible by every prime power dividing $|G|$, so $|M|$ is a multiple of $|G|$.  Since $|M| \leq |G|$, one must have $|M| = |G|$ and $M = G$, a contradiction.  If $G$ is a $p$-group and one conjugacy class of maximal subgroup, then every maximal subgroup is normal, and so $G$ must have exactly one maximal subgroup, and so is cyclic as before.

In particular, every finite group has at least two classes of maximal subgroups.  The non-abelian group of order 6 has exactly two classes.

It is not too uncommon for a finite group to have only one class of maximal subgroup that is not normal, and one that is normal (so two classes total): every finite primitive solvable group is like this.  Depending on your definition, it might just be by definition.  But there are a few definitions that make it sound exciting.


问:
finding the volume of a cube, why doesn't my solution work
Question: the distance from a corner to the centre of a cube is 6. what is the volume of the cube?

Answer: 332.55

I drew a figure of a cube and two lines from a corner to their opposite corner and labeled the middle of the cube e. the triangle aec is right angled because the 2 lines make 4 equal angles when they intersect at e. knowing this and that the distance from a corner to e is 6, I calculated ac to be 8.49 or √72 using the Pythagorean theorem. Since I know ac is 8.49, I calculated x to be 6. 6³ is 216, so the volume of the cube is 216. Please tell me where I went wrong, Thank You.

答:
The angle $\angle AEC$ is NOT $90^{\circ}$, which is why your approach doesn't quite work.

Another approach -- the distance between two opposite corners is the diameter of the sphere, i.e. $12$. This implies the side of the cube is $12/\sqrt{3}$ (can you see why?), which gives the correct answer.


问:
How to find this conditional probability
I am struggling with the question below and would appreciate any help.

Let $Z$ be a random variable with a continuous distribution having density
$$
f_{Z}(z)=\left\{\begin{array}{ll}
2 z & \text { if } 0 \leq z \leq 1 \\
0 & \text { otherwise }
\end{array}\right.
$$
Let $X$ and $Y$ be two random variables, independent given $Z=z$ satisfying
$$
\begin{array}{l}
\mathbf{P}(X=1 \mid Z=z)=z=1-\mathbf{P}(X=0 \mid Z=z)
\end{array}
$$
and
$$
\begin{array}{l}
\mathbf{P}(Y=1 \mid Z=z)=z=1-\mathbf{P}(Y=0 \mid Z=z)
\end{array}
$$
Find the conditional probability
$$
\mathbf{P}(Z \leq 1 / 2 \mid X=1 \text { and } Y=1).
$$

I tried to solve it below:
\begin{align*}
&\mathbf{P}(Z \leq 1 / 2 \mid X=1 \text { and } Y=1)\\&=\frac{\mathbf{P}(X=1,Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)}{\mathbf{P}(X=1,Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)+\mathbf{P}(X=1,Y=1|Z>1/2)\mathbf{P}(Z>1/2)}\\
&=\frac{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)}{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)+\mathbf{P}(X=1|Z>1/2)\mathbf{P}(Y=1|Z>1/2)\mathbf{P}(Z>1/2)}\\
&=\frac{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)1/4}{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)1/4+\mathbf{P}(X=1|Z>1/2)\mathbf{P}(Y=1|Z>1/2)3/4}\\
\end{align*}
I know $X$ and $Y$ are Bernoulli random variables given $Z$ but I don't know how to proceed from here. I think the right answer is 0.125 based on my simulation.

答:
I think I know how to solve it now,
$$
\begin{array}{l}
P(Z \leq 1 / 2 \mid X=1 \text { and } Y=1) &=\dfrac{P(X=1, Y=1\mid Z \leq 1 / 2 )P(Z \leq 1 / 2 )}{P(X=1,Y=1)}\\
&=\dfrac{\int_0^{1/2}z\cdot z\cdot 2z dz}{\int_0^{1}z\cdot z\cdot 2z dz}\\
&=1/16.
\end{array}
$$


问:
Is infinity a branch point of $\sin(\sqrt{z})$?
Does the function $f(z) = \sin(\sqrt{z})$ have a branch point at infinity? I'm confused because infinity is an essential singularity of $\sin(z)$, so I'm not sure how to do the usual $z\to w=1/z$ substitution and analyse the resulting function $f(w)$ at zero... Any help would be appreciated.

答:
$\frac{\sin(z^{1/2})}{z^{1/2}}$ is entire,

$\frac{\sin(z^{-1/2})}{z^{-1/2}}$ has an isolated (essential) singularity at $0$

thus, $\sin(z^{-1/2})$ has a branch point at $0$ and $\sin(z^{1/2})$ has a branch point at $\infty$.


问:
Every Maximal ideal contains either a or b
Let R be a commutative ring and if ab belongs to Jacobson radical of R then every maximal ideal of R contains either a or b. How to prove this?
Suppose not then let M be a maximal ideal that doesn't contain either a or b.
But ab belongs M
Then i couldn't arrive at a contradiction</p>

答:
Hint: note that

$$
J(R) = \bigcap_{\mathfrak m  \text{ maximal}} \mathfrak m
$$

and that maxmimal ideals are prime.

 Pick an element $ab \in J(R)$. Then $ab$ lies in each maximal $\mathfrak m$, which in particular is prime. Hence by primality $ab \in \mathfrak m$ implies either $a \in \mathfrak m$ or $b \in \mathfrak m$.


答:
The 2 diagonal lines [through the center of the cube] make 4 angles but the lines are not on a plane.

Let $x$ be the length of an edge of the cube. Then the length of a diagonal of a side is $x\sqrt{2}$. Furthermore, there is a right triangle where the legs are 1. the diagonal of a side [such as say from the lower-right to upper-left corner of the side facing us in your picture], 2. an edge--such as the lower edge of the right side of the cube, and where the hypo is a diagonal across the entire cube.

This gives us $x^2 + 2x^2 = (6+6)^2$ which gives us $x =\sqrt{48}$; and $x^3=48\sqrt{48}$ is the answer.

Title: finding the volume of a cube, why doesn't my solution work

Body:
Question: the distance from a corner to the centre of a cube is 6. what is the volume of the cube?

Answer: 332.55

I drew a figure of a cube and two lines from a corner to their opposite corner and labeled the middle of the cube e. the triangle aec is right angled because the 2 lines make 4 equal angles when they intersect at e. knowing this and that the distance from a corner to e is 6, I calculated ac to be 8.49 or √72 using the Pythagorean theorem. Since I know ac is 8.49, I calculated x to be 6. 6³ is 216, so the volume of the cube is 216. Please tell me where I went wrong, Thank You.


问:
Does there exist a smooth stochastic process with nonzero quadratic variation?
Pick a sequence $b_i\in C^\infty_c(\Bbb R)$ and define $S_t=\sum_{i=1}^N X_ib_i(t)$ where $N$ is a discrete r.v. and the $X_i$'s are i.i.d. standard normals. It is clear that $S_t$ is a.s. smooth regardless of the distribution of $N$. However, when I try to compute its quadratic variation I run into the formula
$$\Bbb E[(S_t-S_r)^2]=\sum_n p_n\sum_{i=1}^n(b_i(t)-b_i(r))^2=(t-s)^2\sum_n p_n\sum_{i=1}^nb_i'(c_i)^2$$
Which looks like I would have to pick the $b_i$'s really carefully in order to make the above behave nicely. It appears this is a problem of balancing the upper bounds on the derivatives of realizations of a process. If these upper bounds have a supremum the variation vanishes, but if too many of them blow up, so does the variation. Is there a smooth stochastic process with nonzero, finite variation?

答:
Assuming "smooth" means at least "continuous and finite variation," the quadratic variation will be $0$.  That is because if $0 = t_0 < t_1 < \cdots < t_n = T$ is a partition of $[0,T]$ then $$\sum_{i=1}^n (S_{t_i}-S_{t_{i-1}})^2 \le \sup_{i \le n}|S_{t_i}-S_{t_{i-1}}| \cdot \sum_{i=1}^n |S_{t_i}-S_{t_{i-1}}| \le \sup_{i \le n}|S_{t_i}-S_{t_{i-1}}| \cdot V_T$$

where $V_T$ is variation of $S$ over $[0,T]$.  Taking the limit as the mesh goes to $0$ then gives $0$ by the continuity of $S$.


问:
Relationship between meager subsets of $\mathbb{R}$ and $\omega^\omega$.
I'm reading this book about structure of the real line. One of the things that's proven there is that $\mathfrak{b}\leq non(\mathcal{M}).$ And the proof goes as follows:

For every $f\in\omega^\omega$ let us define $K_f$ as follows:
$$K_f=\lbrace g\in\omega^\omega:\ g\preceq f\rbrace.$$
We observe that
$$K_f=\bigcup_n \bigcap_{k>n} \lbrace g\in\omega^\omega:\ g(k)\leq f(k)\rbrace.$$
So for every $f$ a set $K_f$ is a union of nowhere-dense sets a thus - first category subset of space $\omega^\omega$.

Suppose that $\kappa<\mathfrak{b}$ and that $\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq\omega^\omega$. Then there exists a function $f\in\omega^\omega$ such that
$$(\forall \alpha<\kappa)(f_\alpha \preceq f).$$
But then we have
$$\lbrace f_\alpha:\ \alpha<\kappa\rbrace\subseteq K_f,$$
so $non(\mathcal{M})\geq\mathfrak{b}$.

My main question is why are we proving this for $\omega^\omega$ and not for $\mathbb{R}$? Are meager sets in $\omega^\omega$ somehow connected to meager sets of $\mathbb{R}$?

答:
Often $\omega^\omega$ is easier to work with than $\mathbb{R}$, and despite their obvious topological differences for many purposes in logic they're basically equivalent:

They're (low-level) Borel isomorphic, so descriptive set theoretic questions above the first few levels of the Borel hierarchy can't distinguish between them.

There's a homeomorphism between $\omega^\omega$ and $\mathbb{R}\setminus\mathbb{Q}$; since $\mathbb{Q}$ is "small" in $\mathbb{R}$ (= countable, and so a fortiori meager and null), this means that many of their cardinal invariants are equal.

Moreover there's also a continuous bijection from $\omega^\omega$ to $\mathbb{R}$ - see e.g. here (whose inverse of course is badly discontinuous). However, in my experience that's actually less useful than the bulletpoint above.

In particular, for the question in the OP the second bulletpoint above guarantees that we can answer the question for $\mathbb{R}$ by answering it for $\omega^\omega$.


问:
Let $F$ be a field. Prove that for every integer $n \geq 2$, there exist $p,q \in F$ such that $x^2 + x + 1$ divides $x^n + px + q$.
How to start the proof? I only know when $n = 2$ the remainder is $(p-1)x+q-1$ but I'm not sure.
Thanks for help

答:
You are asked to prove existence of $p,q$. So let's make this as simple as possible.

$$x^n +x+1 = p(x)[x^2+x+1] + ax+b$$ for some $a,b \in F$ and some $p(x) \in F[x]$.  Do you see why this is. You don't need to know precisely what $a$ and $b$ are, just that $x^n +x+1$ can be written in this form for some $a,b \in F$.

If $a$ and $b$ are both 0 then we are done.

Otherwise, note the following: $$x^n +x+1 = p(x)[x^2+x+1] +(ax+b).$$ So subtracting $ax+b$ from both sides gives $$x^n+n+1 -(ax+b) = p(x)[x^2+x+1].$$ Thus $x^2+x+1$ divides $x^n +x +1 -(ax+b)$. So take $p =1-a$ and $q=1-b$.


答:
In general $-p x - q$ should be the remainder for the division of $x^n$ by $x^2+x+1$.  But we can make this explicit:

If $n \equiv 0 \mod 3$ then $x^n -1$ works

If $n\equiv 1 \mod 3$ then $x^n - x$

If $n \equiv 2 \mod 3$ then $x^n + x + 1$

Title: Let $F$ be a field. Prove that for every integer $n \geq 2$, there exist $p,q \in F$ such that $x^2 + x + 1$ divides $x^n + px + q$.

Body:
How to start the proof? I only know when $n = 2$ the remainder is $(p-1)x+q-1$ but I'm not sure.
Thanks for help


答:
In general, if $s,t$ are polynomials, then $t = qs+r$ with $r = 0$ or $\deg r < \deg s$. In particular there always exists $r$ with $r = 0$ or $\deg r < \deg s$ such that $t-r$ is divisible by $s$.

Here $t = x^n$ and $s = x^2+x+1$. The condition $\deg r < 2$ can be expressed by saying that $r = px+q$ for some $p,q \in F$. So, to sum up, there are always $p,q$ such that

$$
t-r = x^n-px-q
$$

is divisible by $s = x^2+x+1$ (you can replace $p$ and $q$ with $-p$ and $-q$ for this to fit the original question).

A concise way to say this is that $F[X]/(s)$ is always a $F$-vector space with basis $\{[1],[X],\ldots, [X]^{d-1}$} where $d = \deg s$ (here's a proof), hence for any $t \in F[X]$ there exist $a_0,\ldots, a_{d-1}$ such that

$$
a_0+a_1x+\ldots+a_{d-1}x^{d-1} \equiv t\pmod{s}.
$$


问:
Homeomorphisms between lens spaces and Poincaré conjecture
It is known that $L(p,q)\cong L(p,q')$ if and only if $q\equiv \pm (q')^{\pm 1}\mod p$ where

$$
L(p,q)\cong{\mathbb{S}^3}/\mathord{\sim}
$$
is the quotient space generated by the $\mathbb{Z}_p$-action
$$
\rho(z_1,z_2)=(\zeta z_1,\zeta^q z_2)
$$

where $\zeta$ is a pth root of unity and $p$ and $q$ are coprimes.

Right to left implication can be proven using Reidemeister torsion (https://www3.nd.edu/~lnicolae/Torsion.pdf page 104). However, left to right proof is supposed to be verified easily by the reader. Does someone know how to construct the homeomorphism between $L(p,q)$ and $L(p,q')$ assuming the congruence on $q$ and $q'$?

EXTRA QUESTION:
Does the Poincaré conjecture imply that every closed compact orientable 3-manifold with $\pi_1(X)\cong \mathbb{Z}_p$ is a lens space?

答:
For two roots of unity $\lambda,\mu$ define $\rho_{\lambda,\mu}: (z,w)\mapsto (\lambda z, \mu w)$. Then $\rho_{\lambda,\mu}$ and $\rho_{\lambda,\bar\mu}$ are conjugate via the orthogonal map
$$
(z,w)\mapsto (z, \bar{w}).
$$
This shows that $L(p,q)$ is homeomorphic to $L(p,-q)$.

Next, if $q$ is coprime to the orders of $\lambda,\mu$ then the subgroup generated by $\rho_{\lambda,\mu}$ is the same as the one generated by $(\rho_{\lambda,\mu})^q$.

Lastly, assuming that $p$ is coprime to $q$, $q'$,
$qq'\equiv 1$  mod $p$, and

$$
\lambda=\exp(i 2\pi/p), \mu= \exp(i 2\pi q/p),
\mu'= \exp(i 2\pi q'/p),
$$
then
$$
\lambda^{q'}= \exp(i 2\pi q'/p),
$$
$$
\mu^{q'}= \exp(i 2\pi /p).
$$
In other words, the subgroup generated by $\rho_{\lambda,\mu}$ is the same as generated by
$\rho_{\mu',\lambda}$. However, $\rho_{\mu',\lambda}$ is
conjugate in $U(2)$ to $\rho_{\lambda,\mu'}$ via
$$
(z,w)\mapsto (w,z).
$$
This proves that $L(p,q)$ is homeomorphic to $L(p,q')$.

Lastly, for your extra question, PC is not enough, you need SSFC. The later was proven by Perelman in the same paper(s) where he proves the Poincare Conjecture.


问:
Solving a triangle using the law of tangents
I was asked to solve a triangle with sides 10 and 13 and an included angle of 35 using the law of tangents.

The law of tangents is $\frac{a+b}{a-b}=\frac{\tan{\left(\frac{1}{2}\left(\alpha+\beta\right)\right)}}{\tan{\left(\frac{1}{2}\left(\alpha-\beta\right)\right)}}$.

$a,b,\text{ and }c$ are the sides of the triangle. $\alpha,\beta,\text{ and }\gamma$ are the angle.s

In this case, $a=13,\ b=10,\ \text{and }\gamma=35$.

$$\frac{23}{3}=\frac{\tan(\frac12(180-35))}{\tan(\frac12(\alpha-\beta))}$$

Solving for $\alpha-\beta$ gives me $\alpha-\beta=44.948417$

Can I solve for this using my knowledge of $\alpha+\beta$ and setting up a system of equations?

答:
After that you have $\alpha + \beta =145^{\circ}$ and $\alpha - \beta =$ (something numerical or an expression involving the arctangent of an angle). Just solve as linear simultaneous equations. That will give you $\alpha$ and $\beta$ in degrees. Finally, use another appropriate angle pairing to solve for the remaining side using the same method.

Of course, cosine rule from the start is the much more natural way here.


问:
Scale value needed to shrink one circle to match another?
I have a couple of circles that share the same center point. Both are already scaled down from a larger circle.

I know each circle's radius rather easily. I only have the larger circles "scale" value. Given each circle can have a random scale value that makes the circle larger and smaller (and gives it a new radius), I'm trying to figure out how much more I need to scale that outer circle down to match the smaller circle's radius.

On a programming level, I've tried scaling down the larger circle at very large iterations until match -- but that is really tedious. And I'd like to understand the math behind this all.

Does anyone know how I might go about solving this?

答:
I may have misinterpreted the question, but you could just get the radius of your "standard" circle with scale = 1 and then use the radius of B to figure out it's scale

$$r_1 = \frac{r_A}{s_A} = \frac{1.936}{0.498} = 3.887$$

$$s_B = \frac{r_B}{r_1} = \frac{0.637}{3.887} = 0.164$$


问:
A hermitian positive semi-definite matrix with all entries on the complex unit circle has rank one
Can someone think of an easy way to prove (or disprove) the above conjecture?

For $2\times 2$ matrices, the conjecture is trivially true. For $3\times 3$ matrices, the non-negativity of the determinant can be used to show that the conjecture holds. But I don't know the answer for higher dimensional matrices. Maybe some kind of induction argument is required to prove the conjecture.

答:
Let $A$ be $HPSD$ and $n\times n$

then $A$ has all ones on its diagonal (why?)

with eigenvalues in the usual ordering

$\lambda_1 \geq \lambda_2\geq ... \geq \lambda_n \geq 0$

$n^2$

$= \Big(1+1+...+1\Big)^2 $

$= \Big(\text{trace}\big(A\big)\Big)^2 $

$= \big(\lambda_1+ \lambda_2+...+\lambda_n\big)^2$

$\geq \lambda_1^2+ \lambda_2^2+...+\lambda_n^2 $

$= \text{trace}\big(A^2\big)  $

$=\Big \Vert A\Big \Vert_F^2$

$= n^2$

this inequality is met with equality

$\implies 0=\lambda_2=...=\lambda_n$

$\implies \lambda_1 = n$

and $A$ is diagonalizable, hence it is rank 1.


问:
Prove that ${\mathbb{Q}}/{\mathbb{Z}}$ is isomorphic to $\bigoplus_p \mathbb{Z}[p^{\infty}]$
Prove that ${\mathbb{Q}}/{\mathbb{Z}}$ is isomorphic to $\bigoplus_p \mathbb{Z}[p^{\infty}]$

I know we will use $A_p$'s for solution. But i dont know how to do the isomorphism between ${\mathbb{Q}}/{\mathbb{Z}}$ and $\bigoplus_p \mathbb{Z}[p^{\infty}]$.

答:
Hint: the exponential map

$$
\mathbf{exp} \colon t \in \Bbb R \mapsto e^{i2\pi t} \in  S^1
$$

is a surjective group homomorphism with kernel $\Bbb Z$, hence

$$
S^1 = \Bbb R/\Bbb Z.
$$

Via this correspondence, elements of $\Bbb Q / \Bbb Z$ are complex elements of absolute value $1$ of the form $e^{i2\pi p/q}$ with $p/q \in \Bbb Q$. Can you think of a way to embed $\Bbb Z[p^\infty]$ in $S^1$?

 Note that elements of the form $e^{i2\pi r}$ with rational $r$ are roots of unity, and conversely roots of unity also have this expression. Maybe you can 'classify' roots of unity in terms of primes?

Hint': Note that $\Bbb Z[p^\infty]$ is the $p$-torsion of $\Bbb Q/ \Bbb Z$, then use primary decomposition. This may be easier to do with the aforemention description (or not).

Hint'': show that if $\xi_n$ and $\xi_m$ are primitive roots of unity of coprime orders $n$ and $m$, then $\xi_n\xi_m$ is primitive of order $nm$. Deduce that any root of unity is a finite product of powers of primitive roots of unity of order $p^r$ for $p$ prime and $r \geq 0$. Use that to construct a map $\bigoplus_p \Bbb Z[p^\infty] \to S^1$ whose image is (the isomorphic copy of) $\Bbb Q / \Bbb Z$, prove that it is also injective.


问:
What's new in higher dimensions?
This is a very speculative/soft question; please keep this in mind when reading it. Here "higher" means "greater than 3".

What I am wondering about is what new geometrical phenomena are there in higher dimensions. When I say new I mean phenomena which are counterintuitive or not analogous to their lower dimensional counterparts. A good example could be hypersphere packing.

My main (and sad) impression is that almost all phenomena in higher dimensions could be thought intuitively by dimensional analogy. See for example, this link:

What this implies (for me) is the boring consequence that there is no new conceptual richness in higher dimensional geometry beyond the fact than the  numbers are larger (for example my field of study is string compactifications and though, at first sight, it could sound spectacular to use orientifolding which set a loci of fixed points which are O3 and O7 planes; the reasoning is pretty much the same as in lower dimensions...)

However the question of higher dimensional geometry is very related (for me) to the idea of beauty and complexity: these projections to 2-D of higher dimensional objects totally amazes me (for example this orthonormal projection of a 12-cube) and makes me think there must be interesting higher dimensional phenomena...

I would thank anyone who could give me examples of beautiful ideas implying “visualization” of higher dimensional geometry…

答:
"What this implies (for me) is the boring consequence that there is no new conceptual richness in higher dimensional geometry beyond the fact than the numbers are larger."

This turns out to be incorrect and should serve as a warning that intuition can lead us astray.

One of the best examples of a completely new property that emerges in higher dimensions involves optimal sphere packing. It transpires that dimensions 8 and 24 are special and different from other dimensions. There exists a lattice in dimensions 8 and 24 which specifies an optimal universal packing:

https://arxiv.org/pdf/1607.02111.pdf

It also turns out that dimension 4 has the unique property that orbits in 4-D space with 3 spatial dimensions are stable, while in higher-dimensional space they are not.

https://physics.stackexchange.com/questions/50142/gravity-in-other-than-3-spatial-dimensions-and-stable-orbits

"In four dimensions, stable orbits are possible. For example, the Earth moves around the sun in the stable orbit, the moon around the Earth in the stable orbit and these are sorts of planar orbits. In higher dimensions, achieving stable orbits is very difficult and generally stable planet orbits don’t exist. And so, it is interesting that somehow Einstein’s equation is telling us that four dimensions are actually the right number of dimensions for life to exists, given that life really depends on the planet moving around the sun and so forth. Four dimensions is really a Goldilocks number. It’s not so too constrained like three dimensions where effectively there is no freedom and no dynamics, but it’s also not too free so that you just have complete chaos, many different solutions, no stable orbits and so on."

http://serious-science.org/higher-dimensional-gravity-6741

Mathematically, certain dimensions associated with symmetry groups have unique properties not found in other numbers of dimensions. For example, the symmetry group E8 (quite famous among mathematicians) describes a space of 248 dimensions. The properties of the E8 Lie Group are unique.  More here:

http://serious-science.org/higher-dimensional-gravity-6741

And here:

https://en.wikipedia.org/wiki/E8_(mathematics)

The group of Clifford algebras has various special properties not found in groups of other size (i.e., dimension):

https://en.wikipedia.org/wiki/Clifford_algebra

Since a given dimension number is generally dual to a Galois group, groups of different size (i.e., dimension) exhibit different mathematical properties. It turns out the group S6 is the only one with a non-trivial outer automorphism, which makes it unique among groups.

https://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups

Topology and groups share a close connection, as do the solution spaces of differential equations and abstract algebras. Different numbers of dimensions have special properties, depending on the abstract algebra or differential equation or topological manifold concerned.  In other words, certain dimensions like 4 or 8 or 24 or 6 exhibit unique properties depending on the situation in the same way that a specific card like an ace can exhibit a special property in the context of the rules of poker.

Many of the answers above gives examples of the unique behavior of various numbers of dimensions, and there exist many more examples throughout mathematics and physics and chemistry and dynamical systems.

Title: What's new in higher dimensions?


问:
Need help with $\int_0^\infty x^{-\frac{3}{2}}\ \text{Li}_{\sqrt{2}}(-x)\ dx$
I need help with solving this integral:
$$\int_0^\infty x^{-\frac{3}{2}}\ \text{Li}_{\sqrt{2}}(-x)\ dx,$$
where $\text{Li}_{s}(z)$ is the polylogarithm.</p>

答:
$$\int_0^\infty x^{-\frac{3}{2}}\ \text{Li}_{\sqrt{2}}(-x)\ \mathrm dx=-2^{\sqrt{2}}\pi.$$

Proof:

Use formula (3) from here to get an integral representation of the polylogarithm:
$$\text{Li}_s(-x)=-\frac{1}{\Gamma(s)}\int_0^\infty\frac{k^{s-1}}{\frac{e^k}{x}-1}\mathrm dk.$$
Then, changing the order of integration,
$$\int_0^\infty x^{-p}\ \text{Li}_s(-x)\mathrm dx=-\frac{1}{\Gamma(s)}\int_0^\infty\int_0^\infty\frac{x^{-p}\,\ k^{s-1}}{\frac{e^k}{x}-1}\mathrm dx\ \mathrm dk=\\\frac{\pi}{\Gamma(s)\sin \pi p}\int_0^\infty e^{k(1-p)}\ k^{s-1}\mathrm dk=\frac{\pi}{(p-1)^s \sin \pi p}$$


问:
Matrix that represents a stretching along $X$-axis by a factor $2$ followed by a clockwise rotation of $60$ degrees.
I was given to find a matrix that represents a stretching along  $X$-axis  by a factor $2$ followed by a clockwise rotation of $60$ degrees.

Basically I have to translate and rotate a point.

So I can use the following matrix.

$\begin{bmatrix} 2\cos(2\pi - \frac{\pi}{3}) & 0\\
 0 & \cos(2\pi - \frac{\pi}{3}) \end{bmatrix}$

Have I gone wrong anywhere?

答:
Short answer is no - you need to go about this in two steps

The stretch - This can be defined the matrix

$$M_S = \begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$$

The rotation - This is a standard rotation matrix

$$M_R = \begin{bmatrix}\cos (- \frac{\pi}{3}) & -\sin(-\frac{\pi}{3}) \\ \sin(-\frac{\pi}{3}) & \cos(-\frac{\pi}{3})\end{bmatrix}$$

Now your transformation would be described as

$$T = M_RM_S$$


问:
Why are we treating du/dx as a fraction?
When I was learning implicit differentiation in my class I was told to not think of $dy/dx$ as a fraction. Now I am doing integration by $u$-substitution and we treat $du/dx$ as a fraction and solve for $dx$. Why is this?

答:
In the 19th century mathematicians became extremely squeamish about anything that is not logically rigorous, and they toss around the word "intuitive" promiscuously without noticing that that term admits a variety of different meanings to refer to non-rigorous things.

In many contexts, thinking of $dx$ as an infinitely small increment of $x$ and $dy$ as the corresponding infinitely small increment of $y$ is very useful.

Some instances of treating $dy/dx$ as if it were just an ordinary fraction can be justified by the chain rule, and among those are the things done in $u$ substitution.


问:
Vector Khintchine Inequality
Suppose that $X_1,\ldots,X_n$ are fixed vectors in $\mathbb{R}^{d}$ and $\epsilon_1,\ldots,\epsilon_n$ are Rademacher random variables.  Is it the case that there are constants $A_p,B_p$ so that

$$
A_{p}\sqrt{\sum_{i=1}^{n}||X_{i}||_r^{2}}\le \left(\mathbb{E}||\sum_{i=1}^{N}\epsilon_{i}X_{i}||_r^p\right)^{1/p}\le B_{p}\sqrt{\sum_{i=1}^{n}||X_{i}||_r^{2}}
$$

I want a version of the Khintchine inequality that works for $\mathbb{R}^{d}$ equipped with the $r$ norm instead of $\mathbb{C}$ with $|\cdot|$.  If $r$ has to be $2$, that's fine.

答: For simplicity, take $p=1$; otherwise use some Jensen-type argument to reduce to the case $p=1$. The sought-for quantity corresponds to the Rademacher complexity of the dual unit-ball $B^*_r := \{w \in \mathbb R^d \mid \|w\|_{r^*} \le 1\}$, where $r^* \in [1,\infty]$ is the harmonic conjugate of $r$, i.e $1/r + 1/r^* = 1$. Upper-bounds of the form you seek have been established in this book (see Lemmas 26.9 and 26.10), for the cases $r \in \{2,\infty\}$.

Title: Vector Khintchine Inequality


问:
General solution of differential equation of order 3
Please ,how to find that the general solution of  $u'''(t)=e(t) , t\in [0,1]$ is given by
$u(t)=c_0+c_1t+c_2 t^2 +\frac12 \int_0^t (t-s)^2 e(s) ds$

$e:(0,1)\rightarrow \mathbb{R}$, and $e\in L(0,1)$.

i think that it is general homogeneous solution + particular solution ,

the general homogeneous is $c_0+c_1 t+ c_2 t^2$ , but i dont know how to finde that the paricular solution is $\frac12 \int_0^t (t-s)^2 e(s)ds $

Please;

Thank you

答:
The particular solution

$$u(t)=\frac{1}{2}\int^{1}_{0}(t-s)^2e(s)ds $$

is the Green's function solution $G(t,s)$ for this problem.


问:
Where does the Normal Distribution of Error terms come from, and why?
In a Simple or Multiple Linear Regression Model, we are given the assumption that the error terms, $\epsilon$, are distributed like so:

$ε ∼ N(0, I_nσ^2).$

Can anybody explain where the mathematical logic comes from to derive this assumption, or what the intuition is behind it?

答:
It comes mainly from the requirements of the classical statistical theory. For example, in econometrics you usually assume only uncorrelated errors with constant variance and zero expectation. These assumptions are suffice to derive the best linear unbiased estimator (Gauss-Markov theorem). However, this is not enough for the likelihood-based statistical inference that relies on a specified probability distribution. The most straightforward distribution that satisfies the aforementioned conditions is multivariate normal. In such a case, the uncorrelated error terms become independent and thus the estimation and inference becomes very simple. Before the computer era, mathematical simplicity was crucial, since non-analytical derivations were very hard and even impossible, hence closed-form solutions were much desirable.

In practice, normality assumed merely as approximation, if assumed at all, and much of the inference relies on large-sample theory, i.e., the asymptotic distributions of the estimators. However, note that at least one assumption is really essential for the very analysis. The finite variance assumption is crucial, otherwise your estimators (either OLS or ML) will not converge to finite values. The zero expectation assumption should not bother you, since it is mere technical. If you have non-zero constant expectation you can add it to the intercept term, if it depends on the explanatory variable, you should change your model (conditional expectation of $Y$ given $X$). In both cases, you can obtain the zero expectation after a proper modification. The uncorrelated errors assumption can be dropped by using GLS or GLM.

To sum it up, normal distribution is straightforward model to work with that is usually plausible to assume where linear regression models are applicable. However, you should always bear in mind that as more assumptions you have, the more mistakes you prone to do. Thus, you have various tests of model assumptions and model-misspecifcation, that checks the validity of your assumptions and quantifies the cost of error.


问:
Proof for every permutation will have a cyclic notation to it or not?
Assuming you do not know about Cyclic Notation Let's first see what it is
 Suppose we have some Permutation of Length 4 which is $[4,1,2,3]$ So We can map it by saying Number at Position $1$ is 4 , Number at Position $2$ is $1$ , and so on

So It can be written that

$1\xrightarrow{}4$

$2\xrightarrow{}1$

$3\xrightarrow{}2$

$4\xrightarrow{}3$

So if we try to make a Graph out of the given connections we Get a Cycle out of it which is as follows

$1\xrightarrow{}4\xrightarrow{}3\xrightarrow{}2\xrightarrow{}1$

 can be translated in Cyclic Notation as

$(1,4,3,2) $

I was watching this video about Cycle Notations of a permutations. I am a beginner in Abstract Algebra , It has not been easy for me to visualize it that why is it always True that A Permutation will have a cyclic notation , Isn't It possible that We do not have a cycle being made ?

答:
Not every permutation is a cycle, but every permutation is by necessity a product of disjoint cycles. For example, consider the permutation $$\begin{bmatrix}1&2&3&4&5&6&7&8&9\\3&4&2&6&1&5&8&7&9\end{bmatrix}$$ To break into cycles, just pick a number a follow its path:

$1 \to 3 \to 2 \to 4 \to 6 \to 5 \to 1$. This gives the cycle $(132465)$. This includes everything from $1$ to $6$, so the next unaccounted for value is $7$:

$7 \to 8 \to 7$. This gives the cycle $(78)$. This accounts for $7$ and $8$, leaving $9$.

$9 \to 9$, which is the cycle $(9)$.

So in all, $[342615879] = (132465)(78)(9)$, a product of three disjoint cycles.

And it always has to work like this. When you start following the trail for $1$, there only a finite number of values for it to pass through. It has to repeat a value at some time. Further that value has to be $1$, because if the first repeat was some later value $n$, then the predecessor of $n$ on its first and second visits would be different, and that is not allowed in a permutation. So the trail from $1$ eventually closes in a cycle. Now pick any number not in that cycle, and follow it instead. Eventually it must lead back to the original number for the same reason as before. So you get another cycle. This continues until you run out of numbers to trace.

So every finite permutation has to be a product of disjoint cycles.


问:
How do i solve this integration question using the washer and shell method?
What is the volume of a solid enclosed by $y = (x-1)^2$ and $y = 4$ revolved around $x = - 3$?

I tried the washer method and the shell method and got different answers each time and I'm really confused please help!

My set up for the washer method was:

$$ \pi \int_0^4 {(4 + \sqrt{y} + 1)^2 - 4^2 \ \ dy} $$

My set up for the shell method was:

$$ 2 \pi \int_{-1}^3 {(x + 3)(4 - \sqrt{x} + 1)^2} \ \ dx $$

答:
Substitute $x\mapsto x-3$, then $x\mapsto r$. Then the region in question is
$$
y=(r-4)^2\le4\tag1
$$
The shell method:
$$
\int_2^6\overbrace{\left(4-(r-4)^2\right)}^{\text{width of the shell}}\overbrace{\ \ 2\pi r\,\mathrm{d}r\ \ \vphantom{4^2}}^{\substack{\text{length$\,\times$}\\\text{thickness}\\\text{of the shell}}}\tag2
$$

The washer method:

For a given $y$, the outer radius is $4+\sqrt{y}$ and the inner radius is $4-\sqrt{y}$.
$$
\int_0^4\left(\vphantom{\left(\sqrt{y}\right)^2}\right.\!\overbrace{\pi\left(4+\sqrt{y}\right)^2}^{\substack{\text{area of disk}}}-\overbrace{\pi\left(4-\sqrt{y}\right)^2}^{\substack{\text{area of hole}}}\left.\vphantom{\left(\sqrt{y}\right)^2}\!\right)\overbrace{\ \quad\mathrm{d}y\ \quad\vphantom{\left(\sqrt{y}\right)^2}}^{\substack{\text{thickness}\\\text{of the disk}}}\tag3
$$

Both of the integrals equal $\frac{256\pi}3$.


问:
Is a series (summation) of continuous functions automatically continuous?
I'm being asked to show that a given series (of rational functions) converges uniformly on a given disc, and then and asked to use this fact to show that integrating its limit function (i.e. a summation of rational functions) along a given contour results in a given series.

In the solution provided, the arugment is that each of the rational function is continuous, thus the uniform convergence implies that the limit function is continuous too.

I don't understand the need to invoke the uniform-convergence-implies-continuous-limit theorem. Isn't a series of rational functions already automatically continuous?

答:
Consider the following sequence:

$$f_1(x) = x \\
f_2(x) = x^2 - x \\
f_3(x) = x^3 - x^2 ...$$

Each function is a rational function (even a polynomial) and obviously continuous.
Now look at the series $\sum_n f_n(x)$ in the interval $[0, 1]$. This converges (pointwise) to $0$ in $[0, 1)$ and to $1$ at $x=1$, which is not continuous.

Historical note: The following is taken from the Wikipedia article List of incomplete proofs.

In his Cours d'analyse of 1821, Cauchy "proved" that if a sum of continuous functions converges pointwise, then its limit is also continuous. However, Abel observed three years later that this is not the case. For the conclusion to hold, "pointwise convergence" must be replaced with "uniform convergence". There are many counterexamples. For example, a Fourier series of sine and cosine functions, all continuous, may converge to a discontinuous function such as a step function.


问:
What is the meaning of Matlab's ODE solver naming conventions?
MatLab has a number of built-in ODE solvers and are named using the convention odepq, for exampleL ode45, ode23, etc.

I understand that $p$ here stands for the order of the numerical method. For example, when using Euler's method, this is an order 1 because you only need to substitute the value of the function once before going on the update equation. As opposed to 2 times for Heun's and midpoint method (hence order 2). Classical RK method is order 4.

What I don't understand is the $q$ part. What is it?

答: 5 is for the error estimate order - RK4's local truncation error is order of $O(h^5)$.


问:
Explain the difficulty of find primes with the same last digit
Refer to https://oeis.org/A340800 to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase.  Is there an explanation for this?  How can the size of primes have any influence on the last digit of the following primes?</p>

答:
Supposing the last digits of primes form a random sequence, from the set 1,3,7,9.

Let the prime we want end in a 1.  The next $n-1$ primes must end with something else, and the $n$th end with a 1.  This happens with probability $$\left(\frac34\right)^{n-1}\frac14$$

So it is harder to belong in a later slot of A340800, so the first one in each slot tends to be higher.


答:
Let $$\mathfrak P(x,n)=\Bbb P(\exists m\leq n \in \Bbb N : x+10m \text{ is prime} | x \text{ is prime})$$

The relative density of primes decreases as numbers get larger. Because of this, $\mathfrak P$ is very high for small $x$ and generally decreasing in $x$. At the same time, it is increasing in $n$ (Every time you add another potential number that might be prime, the probability increases).

One can see from this exactly what the OEIS results show, that if a prime is small, the next prime with the same last digit is very likely not much bigger, and for a large prime, you'll likely need a larger $n$ to satisfy, which implies the next prime will be further away.

Title: Explain the difficulty of find primes with the same last digit

Body: Refer to https://oeis.org/A340800 to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase.  Is there an explanation for this?  How can the size of primes have any influence on the last digit of the following primes?</p>


问:
Why is the fact that two sets can be homeomorphic but not isometric implies completeness is not a topological property?
Completeness is not preserved by homeomorphism, but what is the relationship with that and isometric mappings?</p>

答: By definition a topological peopery is one which is perserved under homeomorphisms. Completeness is not a topological property : $(-1,1)$ with the usual metric is homeomorphic to $\mathbb R$ with the usual metric under the map $x \to \tan (\frac {\pi} 2 x)$. While $(-1,1)$ is incomplete, the real line is complete. Under an isometric bijection completness is preseved.


问:
Solving $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$.

Find solutions to $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$.

So I have that $V_{xx}(x,t)=e^tf''(x)$ and $V_t(x,t)=e^tf(x)$

But from the equation I get $e^tf''(x)=e^tf(x)$?

So I should have $f''(x)=f(x)$?

I'm not really sure what to be doing from here, or if what I've done even makes sense.

But solving $y''-y=0$
I get $r^2-r=r(r-1)=0$
So I have $r=0,r=1$
So $f(x)=c_1e^{x}+c_2$ is a solution?

But this doesn't seem to work with me given equations. Since if $V(x,t)=c_1e^te^x+c_2e^t$

then $V_t=c_1e^te^x+c_2e^t\neq V_{xx}$

答:
You made a mistake when writing out the characteristic polynomial of $y''-y=0$.
The correct polynomial is $r^2 -1$. Otherwise what you have done is correct.


问:
existence of certain continuous function $\overline{f}$

Let $(X,d_X)$ and $(Y, d_Y)$ be complete metric spaces (so $X$ and $Y$ are nonempty). Let $A\subseteq X$ be nonempty. A function $f : A\to Y$ is uniformly continuous if given $\epsilon > 0,\exists \delta > 0$ so that $d_Y(f(x), f(y)) < \epsilon$ whenever $d_X(x,y) < \delta$ for $x,y \in A.$ Show that if $f : A \to Y$ is uniformly continuous then there is a unique continuous function $\overline{f} : \overline{A} \to Y$ so that $\overline{f}\vert_A = f$ (here $\overline{A} $ is the closure of $A$, or intersection of all closed sets containing $A$). Moreover, $\overline{f}$ is uniformly continuous.

I think I can show that $\overline{f}$, if it exists, must be unique. Suppose $g_1, g_2 : \overline{A}\to Y$ are such that $g_i\vert_A = f$ for $i=1,2.$ Then for all $x \in \overline{A}, \exists (x_n)\subset A, x_n \to x$ so $g_1(x_n)\to g_1(x)$ and $g_2(x_n)\to g_2(x).$ But $g_1(x_n) = g_2(x_n) = f(x_n)$ for all $n$ and so by the uniqueness of limits $g(x) = f(x).$

However, I'm not sure how to show the existence of $\overline{f}.$

I think I can show that if $\overline{f}$ exists, it is uniformly continuous by taking advantage of the continuity of $f,$ the definition of every element in $\overline{A}$ as the limit of a sequence in $A,$ and the uniform continuity of $\overline{f}.$

答:
Hint:

Define $$\overline{f}(x)=f(x) \text{ if } x\in A$$ and

$$\overline{f}(x) = \lim_{x_n\rightarrow x} f(x_n)$$ (with $(x_n)\in A$ is an arbitrary sequence converging to $x$) if $x \not \in A$.
You should be able to finish from this point on.


问:
$\sum_{n=1}^\infty \frac{\sin nx}{n+1}$ is not continuous at $0$ and $2\pi$?
$\sum_{n=1}^\infty \frac{\sin nx}{n+1}$ is not continuous at $0$ and $2\pi$? It is easy by Dirichlet test that the above sum is continuous on $(0,2\pi)$. What about the endpoint case $0$ and $2\pi$? Computation show it is not continuous, but what is the rigourous proof?

答:
The sum can be computed explicitly. Let $\color{blue}{0<x<2\pi}$. By Abel's theorem $$\sum_{n=1}^\infty\frac{\sin nx}{n+1}=\Im\lim_{r\to1^-}\sum_{n=0}^\infty\frac{r^{n+1}}{n+1}e^{inx}=\Im\lim_{r\to1^-}\big(-e^{-ix}\log(1-re^{ix})\big),$$ and since $\lim\limits_{r\to1^-}\log(1-re^{ix})=l(x)-is(x)$ with $$l(x)=\log\left|2\sin\frac x2\right|,\qquad s(x)=\frac{\pi-x}{2},$$ we get $$\sum_{n=1}^\infty\frac{\sin nx}{n+1}=s(x)\cos x+l(x)\sin x.$$ This holds for all $x$ if we assume the RHS at $x=0$ to be $0$ (that is, if we let $s(0)=0$ and extend $l(x)\sin x$ by continuity) and if we let $s(x+2\pi)=s(x)$ for all $x$.

Now the discontinuity is apparent: the sum tends to $\pm\pi/2$ as $x\to\pm 0$.


问:
Why is the Derangement Probability so Close to $\frac{1}{e}$?
A derangement is a permutation $\sigma$ of $\{1,2,3,\dots,n\}$ such that $\sigma(i) \neq i$ for every $i$.  A common application of inclusion/exclusion in undergraduate combinatorics and probability classes is to compute the number of derangements, and in the process show that the probability a random permutation is a derangement approaches $\frac{1}{e}$ for large $n$.

There's also a "standard" intuition for this probability, which goes roughly as follows:  Let $E_i$ be the event that $\sigma(i)=i$.

1) For a given $i$, the probability of $E_i$ is exactly $\frac{1}{n}$.

2) If $n$ is large, than these events should be "nearly" independent ($E_i$ occurring means that $\sigma(i) \neq j$, making it a tiny bit more likely that $\sigma(j)=j$, but this shouldn't have much of an effect for large $n$), so we'd expect the probability none of the $E_i$ occur to be roughly $\left(1-\frac{1}{n}\right)^n$.

3) For large $n$, $\left(1-\frac{1}{n} \right)^n \approx \frac{1}{e}$.

Now the last approximation alone already has an error proportional to $\frac{1}{n}$.  What's suprising then is that, after working through the inclusion/exclusion, you find that the probability is not just approximately $\frac{1}{e}$, but incredibly close -- the error is less than $\frac{1}{(n+1)!}$.

Is there some alternative intuitive explanation for the $\frac{1}{e}$ asymptotic probability that gives a sense of why the convergence is so fast?

答: In fact a much stronger $e$-related statement is true: let $X_i$ denote the number of $i$-cycles in a random permutation on $n$ elements. Then for fixed $k$, as $n \to \infty$ the random variables $X_1, X_2, ... X_k$ are asymptotically independently Poisson with rates $1, \frac{1}{2}, ... \frac{1}{k}$. This observation about derangements is a special case applied to $X_1$. See this blog post for details, which proves this fact as a corollary of the exponential formula. The convergence rate is presumably also controlled by the exponential formula but I haven't worked out the details. </p>


问:
I've noticed some relationships with cosine and square root.

Yesterday I've noticed some relationships with cosine and square root. Anything interesting about it?

I was trying to find the smallest width on an hexagon with radius $1.0$ and I noticed that I could get it both by $\sqrt 3$ and the or two times the cosine of $30^\circ$.

I wondered if there would be more matches and so I notice these even though might look silly...

$2 \times \cos 90^\circ = \sqrt 0$

$2 \times \cos 60^\circ = \sqrt 1$

$2 \times \cos 45^\circ = \sqrt 2$

$2 \times \cos 30^\circ = \sqrt 3$

$2 \times \cos 0^\circ = \sqrt 4$

Is there anything interesting to know about these?

Thanks

答:
The last figure was also teached to me mnemotecnically to memorize the value of trigonometric functions on certain angles, notice that you can also calculate the tangent by ignoring dividing 2 this way:

Take the 0 and the 4 of the fist column (0 radians), divide and take the root, you have: $tan (0)= \sqrt(\dfrac{0}{4})=0$

For the second column we have $\dfrac{\pi}{6}$, take the 1 and the 3, divide and take the root again: $tan (\dfrac{\pi}{6})=\sqrt(\dfrac{1}{3})=\dfrac{\sqrt(1)}{\sqrt(3)}=\dfrac{1}{\sqrt 3}$

Similarily, $tan(\dfrac{\pi}{4})=\sqrt(\dfrac{2}{2})=\sqrt(1)=1$

$tan(\dfrac{\pi}{3})=\sqrt(\dfrac{3}{1})=\sqrt 3$

Finally, take for the 4th column ($\dfrac{\pi}{2}$) 4 and $0$.  $tan(\dfrac{\pi}{2})=\sqrt(\dfrac{4}{0})!$ wich is in agreement of the fact that $tan$ is not defined on $\dfrac{\pi}{2}$ and thus cannot be.


问:
How to find the area of the image of the unit disc?

Let $f:\Bbb R^2 \to \Bbb R^2$ be defined by $$f(x,y)=(x+y,2x+ay)$$
Please examine the image of the unit square $[0,1]\times[0,1]$, when $a=1,2$ and find the area of the image of the unit disc $x^2+y^2\le 1$ when $a=3$.

For $a=1,2,3$, $f$ is a linear transformation.

When $a=1$, $f(x,y)=\begin{bmatrix}1&2\\1&1\\ \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix}$ and $\text{Im}f=\{x(1,1)+y(2,1):x,y\in[0,1]\times[0,1]\}$.

When $a=2$, $f(x,y)=\begin{bmatrix}1&2\\1&2\\ \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix}$ and $\text{Im}f=\{(x+2y)(1,1):x,y\in[0,1]\times[0,1]\}$.

But I have no idea how to find the area of the image of the unit disc $x^2+y^2\le 1$ when $a=3$. Appreciate any suggestion.

答: This is a linear transformation. The areas are multiplied by the Jacobian (the determinant of the transformation) which is $1$. So the area remains the same, equal to $\pi$.


问:
Multivariable Calculus - Maximum direction derivative

We have f(x,y,z) at point $m$ the directional derivative is maximum
at the direction of vector $u=(1,1,1)$ And is equal to $\sqrt{27}$

find for which a the directional derivative for f in point m in
direction of vector (1,2,a) is 5?

Since the D'D  at point $m$ is maximum at the direction of vector $u=(1,1,1)$ that means $\triangledown f(m)$ is equal to the direction of vector $u$ and the direction of $u$ is $\frac{u}{\left \| u \right \|}=\frac{1}{ \sqrt{3}}$ that is supposed to mean that gardient vector for $\triangledown f(m)=\alpha \vec{u}$ for some $\alpha\epsilon  \mathbb{Z}$

Am I correct so far? and if that's so i dont know how to continue from here

答:
If the directional derivative of $f$ is at a maximum in some direction then $\nabla f$ points in that direction.

What is $\nabla f$

$\nabla f = \mu (1,1,1)\\
\nabla f \cdot (\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})  = \sqrt {27}\\
\mu\sqrt 3 = \sqrt {27}\\
\mu = 3\\
\nabla f = (3,3,3)$

What is the directional derivative in the direction $(1,2,a)$?

$\nabla f\cdot \frac {(1,2,a)}{\|(1,2,a)\|} = \frac {9+3a}{\sqrt {5+a^2}}$

We have been told that this equals $5$

$\frac {9+3a}{\sqrt {5+a^2}} = 5\\
9+3a = 5\sqrt {5+a^2}$

Square both sides and solve the quadratic for $a.$

$81 + 54a + 9a^2 = 125 + 25a^2\\
16a^2 - 54a + 44 = 0\\
2(8a - 11)(a-2)=0\\
a = \frac {11}{8}, 2$


问:
How to determine a coplaner perpendicular vector?
How do I determine a perpendicular vector of a given vector on a given plane ? I can't think of any way put the theories of dot product in this problem.</p>

答: Take the cross product of the given vector and the normal vector of the plane. The resulting vector is perpendicular to both vectors, so it is perpendicular to the given vector and lies on the plane as required.</p>


答: Your vector satisfies two conditions: it is perpendicular to the normal vector of the given plane (since it is parallel to the given plane) and it is perpendicular to the given vector. Just cross-multiply those vectors to get your answer.</p>

Title: How to determine a coplaner perpendicular vector?

Body: How do I determine a perpendicular vector of a given vector on a given plane ? I can't think of any way put the theories of dot product in this problem.</p>


问:
How to check if two matrices have the same column space?
Consider the two matrices :
$$
\begin{bmatrix}
1 & 2 \\
2 & 4
\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}
1& 2\\
3 & 6
\end{bmatrix}
$$

How do I check if they have the same column space?

they do not seem to have the same echelon form,

答: The echelon form is obtained by performing operations on rows, but you're interested in columns. So you do something similar - you take the transpose, then perform the row operations to bring your matrices into echelon form, and check if you get the same thing. The two matrices will have the same column space if and only if their transpose will have the same reduced echelon form (up to possible rows of zeros, if the matrices are not of the same size).