问:
Real world applications of Pythagoras' Theorem
I have a school assignment, and it requires me to list a few of the real world applications of Pythagoras Theorem. However, most of the ones I found are rather generic, and not special at all.

What are some of the real world applications of Pythagoras' Theorem?

答: There was an older person nearby who used to use it in his job making window frames.  He needed them square or rectangular with 90 degree angles.  So he'd measure the two sides and a diagonal and if Pythagoras worked he'd got the 90 degree right.</p>


问:
Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[x]$-modules where $X$ acts as $A$ in $M$ and $B$ in $N$
Let $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$

Show that $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$ where $M,N=\mathbb R^2$ are $\mathbb R[X]$-modules and $X$ acts as $A$ in $M$ and $B$ in $N$.

I've completed the first part of this question which asked to calculate $(1-3X+X^2)\cdot\begin{pmatrix}1\\2\end{pmatrix}$ in both $M$ and $N$ for which I got $\begin{pmatrix}-1\\2\end{pmatrix}$ in both $M$ and $N$ (which is probably important in showing the part above that I can't do.)

However I'm struggling to use the fact that both answers were the same (if that's even relevant) to arrive at $\text {Hom}_{\mathbb R[x]}(M,N)=\{0\}$. I've tried fiddling around with the properties of homomorphisms of modules like $\varphi(rm)=r\varphi(m)$ for $r\in\mathbb R[X]$ and $m\in M$. I also don't really understand how we can define homomorphisms like these when $M$ and $N$ are essentially modules over different rings $\mathbb R[A]$ and $\mathbb R[B]$. I know the ring they're over is $\mathbb R[X]$ but I hope you get what I mean.

I couldn't find much about modules over polynomial rings involving a matrix online so I'm hoping someone here can help me. Thanks

EDIT: Corrected wrong entry in $B$

答:
One way to do this is as follows. First, the one dimensional modules for your ring are given by $\mathbb R_\lambda$ where $p(X)$ acts by $p(\lambda)$. These are pairwise non-isomorphic, and this is immediate. Because they are one dimensional, in fact this means there are no morphisms between them at all.

Your first module is $\mathbb R_2\oplus \mathbb R_3$ and the second one is $\mathbb R_1\oplus \mathbb R_0$. Moreover, $\hom(\mathbb R_2\oplus \mathbb R_3,\mathbb R_1\oplus \mathbb R_0)$ is just the sum of the four possible $\hom(\mathbb R_a,\mathbb R_b)$, which are all zero.

ContentLicense: CC BY-SA 4.0


问:
Looking for logistic function that passes through the origin and is asymptotic to y = k
I need to find a function with the following properties:

For all non-negative x,

F(0) = 0

lim x=> inf, F(x)=> k

F'(0) = 1

F"(x) < 0

I believe that the solution will take the form of a logistics function, but its been decades since I finished my math studies.

Body:
I think $F(x)=\arctan x$ will do fine.

$F(0) = \arctan 0 = 0$

$\displaystyle \lim_{x\to +\infty} F(x)=\frac{\pi}{2}$

$F'(0) = \left.\frac{1}{1+x^2}\right|_{x=0} = 1$

$F''(x)=\left(\frac{1}{1+x^2}\right)'= -\frac{2x}{(1+x^2)^2}< 0, \quad x>0 $


问:
is there any way to represent irrational numbers with a finite amount of integers?
I know that rational numbers can be represented with two integers $\frac{a}{b}$.
But is there any way to represent irrational numbers with an finite amount of integers?

My best guess is $\frac{a}{b} ^ \frac{c}{d}$.
it can represent any root of any number, but I don't know if it can represent things like $\sqrt{2}^\sqrt{2}$.

答:
Suppose you have an encoding scheme allowing you to express a subset $C\subseteq \mathbb{R}$ of real numbers using, for each, a finite number of integers. Then you have a bijection between $C$ and
$$
\cup_{k\in \mathbb{N}} \mathbb{N}^k
$$
and since this is countable, then so is $C$. But the irrational numbers are uncountable.


答: There are uncountably many irrational numbers but there are only countably many finite sets (or lists) of integers. So it is impossible to represent every irrational number using only finitely many integers.</p>

Title: is there any way to represent irrational numbers with a finite amount of integers?

Body:
I know that rational numbers can be represented with two integers $\frac{a}{b}$.
But is there any way to represent irrational numbers with an finite amount of integers?

My best guess is $\frac{a}{b} ^ \frac{c}{d}$.
it can represent any root of any number, but I don't know if it can represent things like $\sqrt{2}^\sqrt{2}$.


问:
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$
I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because

Although my answer is correct, I was apparently supposed to use some theory to answer this question.

I started by putting everything on common denominator and factoring out $n$:

$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$

then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$

$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$

$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$

$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$

$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$

$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$

$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$

to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,

and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.

Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?

答:
Fermat's Little Theorem

In particular, this polynomial admits a lot of simplification.
$$
\begin{align}
3n^5+5n^3+7n
&\equiv2n^3+n&\pmod3\tag{1a}\\
&\equiv2n+n&\pmod3\tag{1b}\\
&\equiv0&\pmod3\tag{1c}
\end{align}
$$
Explanation:

$\text{(1a)}$: reduce coefficients $\bmod3$

$\text{(1b)}$: apply Fermat's Little Theorem

$\text{(1c)}$: $3n\equiv0\pmod3$

$$
\begin{align}
3n^5+5n^3+7n
&\equiv3n^5+2n&\pmod5\tag{2a}\\
&\equiv3n+2n&\pmod5\tag{2b}\\
&\equiv0&\pmod5\tag{2c}
\end{align}
$$
Explanation:

$\text{(2a)}$: reduce coefficients $\bmod5$

$\text{(2b)}$: apply Fermat's Little Theorem

$\text{(2c)}$: $5n\equiv0\pmod5$

Thus, $3n^5+5n^3+7n\equiv0\pmod{15}$.

Binomial Polynomials

In general, any polynomial that takes only integer values can be written as an integer combination of binomial polynomials:
$$
\frac{3n^5+5n^3+7n}{15}=24\binom{n}{5}+48\binom{n}{4}+32\binom{n}{3}+8\binom{n}{2}+\binom{n}{1}\tag3
$$

The Computatation of the Coefficients

For a degree $d$ polynomial $P$,
$$
P(n)=\sum_{k=0}^dc_k\binom{n}{k}\tag4
$$
We can compute $c_n$ inductively, using
$$
P(n)-\overbrace{\sum_{k=0}^{n-1}c_k\binom{n}{k}}^{\substack{\text{using the $c_k$}\\\text{computed}\\\text{previously}}}=c_n\overset{\substack{1\\[3pt]\downarrow\\[3pt]\,}}{\binom{n}{n}}+\overbrace{\sum_{k=n+1}^dc_k\binom{n}{k}}^{\substack{\text{the binomial}\\\text{coefficients}\\\text{are $0$}}}\tag5
$$

$c_0=P(0):$
$$
c_0=\frac{3\cdot0^5+5\cdot0^3+7\cdot0}{15}=0
$$
$c_1=P(1)-c_0\binom{1}{0}:$
$$
c_1=\frac{3\cdot1^5+5\cdot1^3+7\cdot1}{15}-0\binom{1}{0}=1
$$
$c_2=P(2)-c_0\binom{2}{0}-c_1\binom{2}{1}:$
$$
c_2=\frac{3\cdot2^5+5\cdot2^3+7\cdot2}{15}-0\binom{2}{0}-1\binom{2}{1}=8
$$
$c_3=P(3)-c_0\binom{3}{0}-c_1\binom{3}{1}-c_2\binom{3}{2}:$
$$
c_3=\frac{3\cdot3^5+5\cdot3^3+7\cdot3}{15}-0\binom{3}{0}-1\binom{3}{1}-8\binom{3}{2}=32
$$
$c_4=P(4)-c_0\binom{4}{0}-c_1\binom{4}{1}-c_2\binom{4}{2}-c_3\binom{4}{3}:$
$$
c_4=\frac{3\cdot4^5+5\cdot4^3+7\cdot4}{15}-0\binom{4}{0}-1\binom{4}{1}-8\binom{4}{2}-32\binom{4}{3}=48
$$
$c_5=P(5)-c_0\binom{5}{0}-c_1\binom{5}{1}-c_2\binom{5}{2}-c_3\binom{5}{3}-c_4\binom{5}{4}:$
$$
c_5=\frac{3\cdot5^5+5\cdot5^3+7\cdot5}{15}-0\binom{5}{0}-1\binom{5}{1}-8\binom{5}{2}-32\binom{5}{3}-48\binom{5}{4}=24
$$
Thus, the converted polynomial is
$$
\frac{3n^5+5n^3+7n}{15}=0\binom{n}{0}+1\binom{n}{1}+8\binom{n}{2}+32\binom{n}{3}+48\binom{n}{4}+24\binom{n}{5}
$$


问:
If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has..?
I have a question that is as follows:

If $x$ and $\alpha$ are real, then the inequation $\log_2x+\log_x2+2\cos\alpha \leq0$ has:

(a) no  solution

(b) has exactly two solutions

(c) is satisfied by any real $\alpha$ and any real $x \in (0,1)$

(d) is satisfied by any real $\alpha$ and any real $x \in (1,\infty )$

The answer is given as (c)

Here's what I have tried, and I cannot proceed further ..... :

The given inequation is equivalent to:

$\log_2x+\frac{1}{\log_2x}+2\cos\alpha \leq0$

Now, if I let $x \in (1,\infty)$ then clearly $\log_2x >0$ and, by AM-GM inequality, I get that

$\log_2x + \frac{1}{\log_2x} \geq 2$

I also know that $2\cos \alpha \in [-2,2]$
This is what I've managed to analyse and cannot proceed further, plz help.

答:
$\log_2 x < 0$ if $x < 1$ so AM.GM wouldn't apply.  But $\log_x 2 =\frac 1{\log_2 x} < 0$ so if $x < 1$ then we'd have $\log_x 2 + \log_2 x = -(|\log_x 2| + \frac 1{|\log_2x|}) \le -2$ and as $2\cos \theta \le 2$ we will have $\log_x 2 + \log_2 x + 2\cos \theta$ will have solutions for all $x \in (0,1)$.  And $\cos \theta$ may be anything so any $\theta$ will do.

So $c$ is true and $a$ and $b$ are false.

We're done if we assume exactly one of the options is true but to see why $d$ is false:

And if $x > 1$ then well, we have $\log_x 2 + \log_2 x \ge 2 \ge 2\cos \theta$ which can only be solved if $\log_x 2 + \log_2 x = 2 = 2\cos \theta$.

That is only true if $\log_x 2 = 1=\cos \theta$ or $x = 2$ and $\theta = 2k \pi$.

So solutions exist but not for every $x \in (1, \infty)$ and not for every $\theta$.

So $d$ is false.

We can go further.  There's no solutions for $x \le 0$ as $\log_2 x$ and $\log_x 2$ would not be define.

Every $x\in (0,1); \theta \in \mathbb R$ is a solutions.

If $x = 1$ there is not solution as $\log_1 2$ is not defined or meaningful.

The solutions in $x > 1$ are $x=2; \theta = 2k\pi$.

So solutions are $(x, \theta) \in (0,1)\times \mathbb R \cup \{(1,2k\pi)|k\in \mathbb N\}$


问:
The Expected value for this continuous random variable
$$
F(y) = \begin{cases}
    0 & y < 1 \\
    1-y^{-3} & y \geq 1
\end{cases}
$$
$$
f(x) = \begin{cases}
    0 & x < 1 \\
    3x^{-4} & x \geq 1
\end{cases}
$$

(Original at https://i.stack.imgur.com/RYhnW.png)

I am trying to find the expected value of X, I know there is some rule for CRV and I believe that the expected value is either the exponent for the CDF which is $-3$ or it is either the exponent for the PDF which is $-4$, some explanation as to why this is would be very helpful, thank you in advance!

答:
Instead of trying to use a specific "rule", it is here simpler and more natural to use the definition and compute the expected value directly.

That is
$$
\mathbb{E}[X] = \int_{\mathbb{R}} xf(x)dx
= 3\int_{1}^\infty \frac{dx}{x^3}
= 3\left[-\frac{1}{2x^2}\right]^\infty_1
= \frac{3}{2}
$$


问:
A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$
Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.</p>

答:
The binomial expansion tells you the following $$(1+\frac{1}{n})^n= 1^n + n*1^{n-1}*\frac{1}{n}+\binom{n}{2}*1^{n-2}*(\frac{1}{n})^2 + \cdots +(\frac{1}{n})^n $$ $$= 1+ 1+ \frac{1}{2!}(1-{1\over n}) + \cdots + \frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1 - \frac{n-1}{n})$$ $$\lt 1+1+\frac{1}{2!}+\cdots+\frac{1}{n!} $$

I recommend you to do the calculation to see how it does this, it really helps in the future.

ContentLicense: CC BY-SA 3.0


问:
Reduced is equivalent to non-singular for commutative rings
Intro I've been (not successfully) trying to educate myself a bit about singular/nonsingular modules and I came by a proposition on Wikipedia that I find really interesting.

For commutative rings, being nonsingular is equivalent to being a reduced ring.

(https://en.wikipedia.org/wiki/Singular_submodule#Examples)

Question Can anyone give me some hints on how to prove this?

My attempts I didn't really get far. I understand that the domain is non-singular. Because it has no non-zero zero divisors.

If I denote $Z$ the set of zero-divisors then clearly annihilators are subsets of $Z$. I even read a stronger statement in some article saying that that Annihilators in reduced rings are intersections of minimal primes (minimal primes are contained in $Z$ for reduced rings)

Also, it seems like a good direction to investigate when are ideals contained in $Z$ essential (which is a question I stumped upon also in different settings in past, so seems like an interesting problem on its own)

答:
Let $R$ be a commutative ring.

Lemma $\operatorname{Nil}(R) \subseteq \mathcal{Z}(R)$

Proof. Let $r \in R$ be nilpotent. Let $x \in R \setminus \{0\}$ be arbitrary. Let $n \geq 0$ be maximal such that $r^n x \neq 0$ (such an $n$ exists because $r$ is nilpotent and $r^0 x = x \neq 0$). Then $r^{n+1} x = 0$, so $r^n x \in \operatorname{ann}(r) \cap Rx$ tells us that $\operatorname{ann}(r) \cap Rx \neq 0$. We conclude that $\operatorname{ann}(r)$ is essential, so $r \in \mathcal{Z}(R)$. $\square$

This gives one direction of the proof – if $R$ is nonsingular, then $R$ is reduced. Here's the other direction (by contrapositive):

Suppose $R$ is not nonsingular, and let $z \in \mathcal{Z}(R) \setminus \{0\}$. Then $\operatorname{ann}(z) \cap Rz \neq 0$, so there is some $r \in R$ such that $rz \neq 0$ and $rz^2 = 0$. But then $(rz)^2 = r(rz^2) = 0$, so $rz$ is a nonzero nilpotent element, and thus $R$ is not reduced. $\square$


答:
I am trying to find the expected value of X, I know there is some rule for CRV and I believe that the expected value is either the exponent for the CDF which is −3 or it is either the exponent for the PDF which is −4, some explanation as to why this is would be very helpful, thank you in advance!

An expected value is determined by the parameters of a distribution, but not necessarily equal to them.   A well-known family of distributions will have a well known formula to use, but this isn't one of those.

So when in doubt, we go back to the basics and work it out.

$$\mathsf E(X)~{ = \int_\Bbb R x\,f_{\small X}(x)\,\mathrm d x\\ =\int_1^\infty 3x^{-3}\,\mathrm d x} $$

Title: The Expected value for this continuous random variable

Body:
$$
F(y) = \begin{cases}
    0 & y < 1 \\
    1-y^{-3} & y \geq 1
\end{cases}
$$
$$
f(x) = \begin{cases}
    0 & x < 1 \\
    3x^{-4} & x \geq 1
\end{cases}
$$

(Original at https://i.stack.imgur.com/RYhnW.png)

I am trying to find the expected value of X, I know there is some rule for CRV and I believe that the expected value is either the exponent for the CDF which is $-3$ or it is either the exponent for the PDF which is $-4$, some explanation as to why this is would be very helpful, thank you in advance!


问:
What gives the natural logarithm its name? What's "natural" about it?
I know that the natural logarithm is defined as $\ln(x)$ or $\log_{e}(x)$, where $e$ is the Euler number.

But what is so "natural" about it? Is there an explanation on why that name was chosen?

答: Base $e$ for exponentials and logarithms is distinguished by the fact that it is the unique base $c$ such that the derivative of $f(x)=c^x$ at $x=0$ is equal to one.


问:
Does pushout of schemes along formal neighborhoods exist in the category of schemes?
I have a question about gluing specific types of schemes which doesn't fit into any well-known gluing situation. Assume $C$ is an algebraic curve and $p$ a point on it. The formal completion of $C$ at that point is $k[[t]]$. Now imagine $\text{Spec}(k[[t]])\times C \times C$, $C\times \text{Spec}(k[[t]]) \times C$ and $C\times C \times \text{Spec}(k[[t]])$. We want to glue these in the obvious way i.e. glue the first and the second one along $\text{Spec}(k[[t]])\times \text{Spec}(k[[t]])\times C$, glue the first and the third one along $\text{Spec}(k[[t]])\times C \times \text{Spec}(k[[t]])$ and in the similar way glue the second and third one. I wonder whether the outcome is an scheme?

Intuitively the formal completion should behave like an open neighborhood but obviously it is not an open immersion.

答:
I will give a counterexample in a slightly simpler situation (two copies of $C$ instead of three) but the same argument can be made for your case.

Write $\operatorname{Sch}$ (resp. $\operatorname{Set}$) for the category of schemes (resp. sets). Pick $C=\operatorname{Spec} \mathbb C[t]$, call $p$ the origin in $C$ and put $C_p=\operatorname{Spec} \mathbb C[[t]]$. Suppose that there is a colimit $X$ in the category of schemes for the diagram $D$ formed by the two maps $C_p \times C_p \to C_p \times C$ and $C_p \times C_p \to C \times C_p$. We will derive a contradiction. There is a natural map $X \to C\times C$ by definition of the colimit. I claim that the functor of points of $X$ is the (Zariski) sheafification of the presheaf taking a scheme $T$ to the subset of $(C\times C)(T)$ consisting of morphisms $(f,g)$ such that either $f$ or $g$ factors through $C_p$. To prove this claim, call $X'$ this sheafification. Then $X'$ is the colimit of our diagram $D$ in the category of Zariski sheaves $\operatorname{Sch}^{op} \to \operatorname{Set}$. In particular, there is a natural map $X' \to X$ in this category (since $X$ is the colimit in a smaller category). Conversely, a morphism of schemes $T \to X$ gives a morphism $T \to C\times C$, which factors through $X'$ since $C_p \times C$ and $C \times C_p$ jointly surject onto $X$. So, we have $X=X'$, which proves the claim.

Now, pick an affine neighbourhood $U=\operatorname{Spec} A$ of $(p,p)$ in $X$. We can choose an affine neighbourhood $V=\operatorname{spec} B$ of $p$ in $C$ which is small enough so that $C_p \times V \to X$ and $V \times C_p \to X$ both factor through $U$. Identifying $C\times C$ with $\mathbb A^2_k=\operatorname{Spec}\mathbb C[t_1,t_2]$, we find that $A$ is a sub-$\mathbb C[t_1,t_2]$-algebra of $\mathbb C[[t_1,t_2]]$ which factors through both $\mathbb C[[t_1]]\otimes_\mathbb C B$ and $B \otimes_\mathbb C \mathbb C[[t_2]]$. Since $B$ is obtained from $\mathbb C[t]$ by inverting finitely many polynomials, $U=\operatorname{Spec} A$ is an open subscheme of $C \times C=\mathbb A^2_k$. This is incompatible with $U$ being contained in $X$, so we have our contradiction (for example because such an $U$ will contain many closed points which do not lie on one of the coordinate axes).


问:
orthogonal complement to range of a bounded operator closed iff bounded
Let $H,K$ be Hilbert spaces.

Suppose $T\in B(H,K)$.

Consider the restriction map of $T$, $T': \mathcal{N}^{\perp}(T) \rightarrow \mathcal{R}(T) $ .

Show that  $\mathcal{R}(T)$ is closed if and only if $T'$ is bounded below.

Atempt:

Suppose $\mathcal{R}(T)$ is closed, now I want to use the bounded inverse theorem. But I'm not sure how to show that  $\mathcal{N}^{\perp}(T)$ and $\mathcal{R}(T)$ are Banach. I feel like that's where the closedness of $\mathcal{R}(T)$ is used.

Now suppose $T'$ is bounded below. I know that $T'$ is a bijection. So the inverse exists. I also know that from here $T'^{-1}$ is also bounded below. So by continuity, $\mathcal{R}(T)= T'(\mathcal{N}^{\perp}(T)) = (T'^{-1})^{-1}(\mathcal{N}^{\perp}(T)) $ is closed.

Could someone please help me with the first part? And please let me know if you see a problem with second part.

答:
A closed subspace of   Banach space is always a Banach space and the orthogonal complement of any set is always closed. So the only thing you have to check is that $T'$ is a bijection.

If $T'x=0$ for some $x \in \mathcal N(T)^{\perp}$ then $x \in \mathcal N(T) \cap \mathcal N(T)^{\perp} =\{0\}$ so $x=0$. So $T'$ is injective.

Let $y \in \mathcal R{(T)}$. Then $y=Tx$ for some $x$ and we can write $x=x_1+x_2$ where $x_1 \in \mathcal N(T)$ and $x_2 \in \mathcal N(T)^{\perp}$. This gives $y=T'x_2$ so $T'$ is surjective.


问:
Calculate all possible combinations of 1, 2 & 3 which equal 67
My son (a third grader) was given this problem to solve. Is states that:

In a basketball game, one of the teams scored 67 points. If baskets can be worth 1 point for a free throw, 2 points for a field shot or 3 points for a 3 pointer, how did the team reach 67 points? How can you know that you have found all possibilities.

We can easily start out by doing 1 x 67 times, and then adding a 2 into the mix until we had mostly 2 and then starting with 3 and rounding that pattern back to introducing 1. We got up to 77 possible combinations when applying this simple pattern based solution. But that doesn't account for all the other combinations where any of the 3 numbers could be used which still equal 67.

It seems stars and bars may be a good way to deal with this problem, but I'm unclear on how to apply it. Would anyone be able to explain this? I'm not looking for the answer, but keen to understand how to calculate how many combinations there could be for any 3 sets of numbers to equal any given number.

Much appreciated!

答:
To summarize part of the discussion in the comments:

Personally, I think it is more tractable if the order does matter.  In that case we can work recursively.  Letting $a_n$ denote the answer with a total of $n$ we remark that, for $n>3$, we must have $$a_n=a_{n-1}+a_{n-2}+a_{n-3}$$
since the last score must be one of $1,2$ or $3$.  This resembles one standard interpretation of the Fibonacci numbers, where $F_n$ is defined as the number of ways to get a sum of $n$ using an ordered sequence of $1's$ and $2's$.  That is to say, the same problem only without the $3$ point shot.

That recursion is easily implemented in Excel, or whatever (you'll have to work out $a_1, a_2, a_3$ but that is not hard).  Of course, the terms grow very rapidly.

If order does not matter:

Then we are counting triples of non-negative integers $(a,b,c)$ with $3a+2b+c=67$.  We remark that $a\in \{0,\cdots,  22\}$.  For a fixed $a$ we are now trying to write $67-3a$ as $2b+c$.  How many ways are there for that?  Well, since $67-3a-c$ is even we can work out the parity of $c$ from that of $a$.  Indeed, they must have opposite parities.  It follows that the number of possible $c's$ is given by $$1+\Big \lfloor \frac{67-3a}2\Big \rfloor$$

Note:  I suggest that out by hand for several values of $a$ to convince yourself that it is correct.

That gives us the number of good triples for a fixed value $a$.  We get the answer by summing $$\sum_{a=0}^{22} \left(1+\Big \lfloor \frac{67-3a}2\Big \rfloor\right)=408$$

Should say, that is all a bit error prone.  To check, I did the computation using generating functions, here which confirms the result. (you can read off the coefficient of $x^{67}$ is $408$ If you are not used to Wolfram Alpha, you need to click "more terms" often enough to get to $x^{67}$).  Obviously that's not a useful method given the context, but I thought it was worth checking the result.


问:
Must there be some (a,b): p \divides q^a-r^b, where p,q, and r are distinct primes?
Let $p, q, r$ be distinct primes and $a,b$ be natural numbers greater than $1$. Is it necessarily the case that for all $p,q,r$ there exists some $a,b: p \mid{(q^a - r^b)}$?

答:
Yes. By Fermat's little theorem, $q^{p-1} \equiv 1 \pmod p$ and $r^{p-1} \equiv 1 \pmod p$, so $p \mid (q^{p-1} - r^{p-1})$.

If $p=2$, then $q$ and $r$ must be odd, so $q^2$ and $r^2$ are still odd, so $p \mid (q^2 - r^2)$. (This is done so that $a, b > 1$ as required.)


问:
Decreasing function
Given two increasing function $f(x)$ and $g(x)$ with all $x \ge 0$. Moreover, $f(x) > g(x)
$ for all $x \ge 0$ and $$\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - g(x)} \right) = 0.$$
Is $d(x) = f(x)-g(x)$ a decreasing function ?</p>

答:
Let $g(x)=e^x$, and let $f(x)= e^x+\frac{1}{6}(\sin^2 x+0.01) e^{-x}$.

Note that the derivative of $\frac{1}{6}(\sin^2 x+0.01) e^{-x}$ takes on positive values for arbitrarily large $x$, and negative value for arbitrarily large $x$.

And $f(x)$ is increasing when $x\ge 0$. This is because the absolute value of the derivative of $\frac{1}{6}(\sin^2 x+0.01) e^{-x}$ is "small" in comparison with the derivative of $e^x$.

The above is very close to the idea of coffeemath, with small differences of detail.

Title: Decreasing function


答:
An expontential function is one where it's rate of growth is proportional to its current value,  that is if it's current value at time $t$ is $g(t)$ then it's rate of growth is $c\cdot g(t)$ where $c$ is a constant value (and the same constant value for any time $t$).

Such functions are of the form $b^t$ and their rates of growth will be $c_b$, some constant based on the value of $b$.

The most "natural" such exponent would be the one where $c_b = 1$ and then rate of of $b^t$ would be the exact value of $b^t$ itself.

And that natural base and that natural exponent (the logarithmic inverse function that derives from it) all do exist and happens to be the base $e$.

===

Oh, guess an even more basic answer is there is, on the surface of things, no "natural" way of defining $b^x$ if $x$ is irrational. You certainly cant multiply $b$ times itself some irrational number of times.

But if we work backwards and define a function $LN(x) = \int_1^x \frac 1t dt$ so that $LN'(x) = \frac 1x$ (thus solving the question: If $(x^k)' = kx^{k-1}$ then the anti-deriviative of $x^k$ would be $\frac {x^{k+1}}{k+1}$ so the anti derivative of $x^{-1}$ should be .... $\frac {x^0}{0}???????$ ....?

Then $LN(x)$ is invertable and the inverse function we can call $EXP(x)$ so that $LN(EXP(x))= x$ and $EXP(LN(x)) = x$.

The we can define $b^x$ as $EXP(LN(b)\cdot x)$

That is..... $b^x= $ the value $M$ must be so that $\int_1^M \frac 1t dt= x\int_1^b \frac 1t dt$ would be

and (perhaps surprisingly the first time we ever see this) that satisfies all the properties we expect $b^x$ to have.  Namely:  $b^{x+y} = b^xb^y$ and if $x \in \mathbb N$ then $b^x= b$ times itself $x$ times.

If we define $e$ as $EXP(1)$ then $e^x = EXP(x)$ and $\log_e y = LN(y)$ and $e^x$ and $LN(y)$ are the most "natural" of all exponents and logs as they were the one all definitions are derived from.

Title: What gives the natural logarithm its name? What's "natural" about it?

Body:
I know that the natural logarithm is defined as $\ln(x)$ or $\log_{e}(x)$, where $e$ is the Euler number.

But what is so "natural" about it? Is there an explanation on why that name was chosen?


问:
Card game Probability: Having a second war after a first war
I was wondering about the probability of getting two wars.  The way the game works is, both players get $26$ cards, and flip one card over, whoever has the largest card value, takes both cards and puts them underneath the deck.

Now the tricky part comes when the two cards flipped over have an equal value, this is called a "war".  At a war, the players each flip another card on the top of their deck over, and whoever has the highest value card gets all the cards on the table.  My question is, what is the chance that these two cards also have equal value?  So, the probability of having two cards out of $52$, be equal, then the next two cards be equal.

I keep arriving at this number: $\frac{3}{49} + \frac{1}{49} = \frac{4}{49} = 0.0816$.

Thanks for the help!

答:
I think you can ignore everything except for the two flips that result in war. Given that there was a war on the first flip, the deck has lost two cards of equal rank.

Then $50$ cards remain. There are $12$ complete ranks of $4$ cards each. Each of those ranks has $\binom{4}{2}=6$ possible pairs. The $13$th rank is the one that was drawn for the first war, and there is only $1$ remaining pair of that rank.

Overall there are $(12\cdot6)+1=73$ pairs left in the deck of $50$ cards.

So the probability of a second war is $\frac{73}{\binom{50}{2}}=\frac{73}{1225}\approx5.96\%$


问:
Holomorphic vs differentiable (in the real sense).
Why a holomorphic function is infinitely differentiable just because of satisfying the Cauchy Riemann equations, but on the other side, a two variable real function that is twice differentiable is not infinitely differentiable?

I'm asking this for two reasons:

1) $\mathbb{R}^2$ is somehow an analog to the complex plane (they are isomorphic).

2) Holomorphic means that is differentiable in any direction of the complex plane, so why if a function is differentiable in any direction of the real plane then it is not infinitely differentiable?

答:
The usual presentation of holomorphic functions (and complex analysis in general) is misleading; the analogue of a holomorphic function to a function on the real line is one whose derivative is zero!

Recall one nice property of holomorphic functions: if a point $p$ lies inside a region $D$, which has a boundary $C$, then you can find the value of a holomorphic function $f(p)$ by only a contour integral about the path $C$.

The analogue of this property to the real line is as follows: let there be a region (an interval) $D$, whose boundary is two points $C_+$ and $C_-$.  Then the value of a function at a point $p \in D$ is characterized entirely by the value of the function at $C_+$ and $C_-$.  The only way this is possible is if $f$ is constant on $D$ and has identical values at $C_+$ and $C_-$.

(Note: it's possible that $f$ has a delta function source in the region instead, but this is just the analogue of a meromorphic function.)

This generalization--functions whose derivatives are zero--is what is ultimately more useful when you start looking at vector calculus and higher-dimensional real spaces.  Scalar fields that obey $\nabla \varphi = 0$ obey the exact same property of being fully characterized by their boundary values.

Similarly, the correct analogue on $\mathbb R^2$ is for a vector field $F$ to have vanishing divergence and curl: $\nabla \cdot F = \nabla \times F = 0$.  Such vector fields also obey the same basic property of holomorphic functions.


问:
Which $N$ are multiples of $2020$ with all distinct digits where swapping any two of them makes $N$ not a multiple of $2020$?

The question

I have attempted problem 3 from the Senior O-level Tournament of Towns paper, Fall 2020. The question is as follows:

A positive integer number N is divisible by 2020. All its digits are
different and if any two of them are swapped, the resulting number is
not divisible by 2020. How many digits can such a number N have?

Sergey Tokarev

My attempt

Let $n$ be the number of digits of $N$. I initially noticed that $4 \leq n \leq 10$, since for all digits of $N$ to be distinct we can have at most $10$ digits. This gave me the impression that this problem could be solved by investigating each of these cases.

Let us first write $N$ as $2020k = 2000k + 20k$. Let $2k$ have digits $d_1d_2\cdots d_m$, then $N = (d_1d_2\cdots d_m)000 + (d_1d_2\cdots d_m)0$.

Case n = 4: Then $2k = d_1$ and $N = d_1000 + d_10 = d_10d_10$, so $N$ must have a repeating digit.

Case n = 5: Then $2k = d_1d_2$ and $N =d_1d_1000 + d_1d_20 = d_1d_2d_1d_20$, so $N$ must have a repeating digit.

Case n = 6: Then $2k = d_1d_2d_3$ and $N = d_1d_2d_3000 + d_1d_2d_30 = d_1d_2(d_3 + d_1)d_2d_30$, so $N$ must have a repeating digit.

This argument no longer works for $n \geq 7$. But we do know that

$d_m \in \{0, 2, 4, 6, 8\}$ since $2k = d_1d_2\cdots d_m$, and

$N$ must end in $0$.

I used this to find numbers for $n = 7, 8, 9$ and $10$ with all distinct digits that are also divisible by $2020$ because they satisfy the observations listed above.

\begin{array}{|c|c|c|c|}
\hline
 n & 7 & 8 & 9 & 10\ \\ \hline
 N & 4981320 & 64975320 & 869753420 & 2537618940 \\ \hline
\end{array}

Claim: If $N$ is a multiple of $2020$ and has all distinct digits, then swapping any two digits results in a number $N'$ where $2020 \not \mid N'$.

proof. Let $N = a_na_{n-1}\cdots a_j\cdots a_i\cdots a_1$ and $N' = a_na_{n-1}\cdots a_i\cdots a_j\cdots a_1$ for any $i < j$. If $2020 \mid (N' - N)$ then $2020 \mid N'$. Suppose, WLOG, that $a_i > a_j$ so that $N' > N$, then
$$N' - N = 10^{j - 1}(a_i - a_j) - 10^{i - 1}(a_i - a_j) = (a_i - a_j)(10^{j - 1} - 10^{i - 1})$$

So $N' - N$ must be a multiple of $10$. Therefore, for the claim to hold, $N'$ must not be a multiple of $4$ or $101$, since $2020 = 2^2 \times 5 \times 101$. We know that $1 \leq a_i - a_j \leq 9$, thus we can safely conclude that $N' - N$ is never a multiple of $101$. Therefore, $2020 \not \mid (N' - N)$.

Therefore, $N$ can have $7, 8, 9$ or $10$ digits and we are done.

I think my solution is correct, but I don't like that I had to manually find $N$'s for $7 \leq n \leq 10$. Is there a different way to argue for those cases or a more systematic way to find $N$'s for those cases? Or if you have a different way to solve the problem, I would also like to see it.

答:
The error in your logic is that $N' - N = (a_i - a_j) (10^{i-1} ) ( 10^{j-i } - 1 ) $ could be a multiple of 101 when $ j-i = 4, 8$.

This yields a multiple of 2020 if $(a_i - a_j) (10^{i-1} )$ is a multiple of 20.

If the number has 7 or more digits, then swapping the 1st and 5th digit yields a difference of $(a_n - a_{n-4}) \times 9999 \times 10^{n-5} $, which is a multiple of 2020.

Hence there are no solutions with 7 or more digits.

Your arguments for $n = 4, 5$ are correct.

However, it fails for $n=6$ in the event that $ d_3 + d_1 \geq 10$, as the carry over makes the second digit $d_2 + 1$, so it need not be a repeated digit with the other $d_2$. (It could still repeat with another digit though).

As an explicit example, $ 2020 \times 377 = 761540$ has distinct digits.   I found it through trial and error.

If remains to find an example for $n=6$. If no such example can be found, then the answer to the question is "no digits" (which seems unlikely as an answer).

Suppose $N = a_6a_5a_4a_3a_2a_1$. Then, to ensure that the swap doesn't yield a difference that is a multiple of 2020,

In order to be a multiple of 101, we just need to concern ourselves with terms that are 4 apart.

We then need to check that $(a_6 - a_2)10$ and $(a_5 - a_1)$ are not multiples of 20.

These are satisfied by $761540$, so it is a valid example.

Hence the number must have 6 digits.

To find the counterexample without excessive trial and error, since (using OP's notation) $\overline{d_1d_2d_3} \times 1010 = \overline{d_1d_2(d_3+d_1)d_2d_30}$, then the conditions that we need are (I might have missed some):

$0, d_1, d_2, d_3 $ are distinct digits

$d_3$ is a multiple of 2

$d_1 - d_3$ is not a multiple of 2.

$d_2$ is not a multiple of 20.

$d_1 + d_3 \geq 11$   (cannot be 10 which leaves the digit of 0)

$d_2 + 1 \neq d_1, d_3, 10$

$d_1 + d_3 - 10 \neq d_2 $

These are satisfied by $ 754 = 377 \times 2$, which is why the example works.
Smaller values include $348$, which yields $351480$.


答:
If ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}$ then $(\log{\frac{a}{b}})*{\frac{c}{d}}=\frac{log{2}}{2}$, which implies
$\frac{a}{b}=2, \frac{c}{d}=\frac{1}{2}$.

But if ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}^\sqrt{2}$ then $\frac{c}{d}=\sqrt{2}$, and we would have found a rational number equal to $\sqrt{2}$. This is impossible, consequently there is no solution for ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}^\sqrt{2}$ among the rational numbers and, more generally, there can be no way of writing every real number using only a finite number of rational numbers and operations.


问:
Looking for an other method to compare $100!$ and $50^{100}$
There is a question asked me to compare $100! \ and \ 50^{100}$. I think more than 2 hours to solve it.
I show my work below, but I looking for other Ideas to prove the inequality.  Thanks in advance for any hints or new Ideas.

my work :
$$50 \times 50 >  1 \times 99 \times 2\\
50 \times 50 > \ 2 \times 98\\
50 \times 50 > \ 3 \times 97\\
50 \times 50 > \ 4 \times 96\\ \vdots \\
50 \times 50 > \ 48 \times 52=(50+2)(50-2)=50^2-4\\
50 \times 50 > \ 49 \times 51=(50+1)(50-1)=50^2-1\\
50 \times 50 \geq 50 \times 50 $$ then multiply them
$$50^{49}\times 50^{49}\times 50^2 >1.2.3...50....99.(2.50)\\50^{100}>100!$$

Remark: I use numerical approximation for $50^{100}\sim 7.8886\times 10^{169} $ and $100!=9.33236\times 10^{157}$ But is not interesting( Ithink).

Body:
Definitely quicker than 2 hours, plus one more method to have in arsenal (similar to this, by the way). At some point I will use
$$\log(1+x)\leq x, \forall x \geq 0$$
$$\log{2}<\frac{3}{4} \iff 16 < e^3$$
(in fact $2.6^3>16 \iff 13^3 > 5^3\cdot 4^2=5\cdot400$)
and $$\log{101}<\log{128}=7\log{2}<\frac{21}{4}$$

Let's look at $\sum\limits_{k=1}^{100}\log{k}$. Function $f(x)=\log{x}$ is monotone ascending, thus:
$$\sum\limits_{k=1}^{100}\log{k}=
\sum\limits_{k=1}^{100}\log{k} \cdot (k+1 - k)\leq \int\limits_{1}^{101}\log{x} dx =
x (\log{x}-1) \Big|_{1}^{101}=\\
101\log{101}-101+1=
101\log{101}-100=\\
100\log{100}+100\log{\left(\frac{101}{100}\right)}+\log{101}-100<\\
100\log{100}+100\log{\left(1+\frac{1}{100}\right)}+\frac{21}{4}-100\leq\\
100\log{100}+1+\frac{21}{4}-100=\\
100\log{50}+100\log{2}+\frac{25}{4}-100<\\
100\log{50}+100\cdot \frac{3}{4}+\frac{25}{4}-100<100\log{50}$$


问:
Why is this the trick to calculate $m\, 9$s $\times n\, 9$s?
Here is what observed for the results of $m\, 9$s $\times n\, 9$s ($m \leq n$)

$9\times9=81, 9\times99=891, 9\times999=8991$

$99\times99=9801, 99\times 999=98901, 99\times9999=989901$

$999\times999=998001,999\times 9999=9989001, 999\times99999=99899001$

Summarize the following trick  to get the result fast:

For example to get $999\times99999 =99899001$

Write down $n\,$$9$s $(n=5)$ first : $\to 99999$

Replace the $m\,$th $(m=3) 9\,$s with $8: \to 99899$

Attach $ (m-1)\, 0$s $(m-1=2): \to 99899 00$

Attach last  $1: \to  99899 00 1$

My question is how to prove this is true with algebra?

答:
Here's a setup to clarify $m=4,n=6$:
\begin{align}
9999 \times 999999 &= 999999 \times (10000-1)\\
&=9999990000 - 999999 \\
&= 9999990000 - 1000000 + 1 \\
&= 999\color{#0B3}{8}99000\color{#0B3}{1}\\
\end{align}

So looking at this semi-lexicographically, we shifted the $n$ $9$s by $m$ trailing zeroes, then switched the last zero to a $1$ and the $m^{th}$ $9$ to an $8$. Which is exactly what you found experimentally.


问:
Clarification of what is meant by loop in $S^{1}$
I am a little confused about what is meant by a loop in $S^{1}$. For example, let $f_{1}(t)= e^{2\pi it}$ defined on the interval $A= [0, 1/2]$ and $f_2 (t)=e^{2\pi i(1-t)}$  on the interval $B= [1/2, 1]$ then by the gluing lemma the path $\alpha (t)$ is $f_1(t)$ on $A$ and $f_2(t)$ on $B$ is continuous and $\alpha(0)=\alpha (1)$. But we don't have to stop there we can define a family of paths in $S^1$ such that don't traverse the whole circle $S^1$. How is this handled in the calculation of the fundamental group.

答: Loops do not have to be surjective (and, typically, they are not). In the context of the fundamental groups $\pi_1(X,x)$, one considers based loops, which are maps $c: [0,1]\to X$ such that $c(0)=c(1)=x$, where $x$ is the base-point.


问:
Showing a measurable function is uniformly continuous
Let $E \subset \mathbb{R}$ be a measurable set with $m(E) < \infty$. Define the functoin $f: \mathbb{R} \to \mathbb{R}$ by
$f(x) = m(E \cap (-\infty,x])$.

Prove that $f$ is uniformly continuous on $\mathbb{R}$.

Here is my attempt at a proof, but I am not confident in it.
We want to show $\forall x,y$ and $\epsilon > 0$ $\exists \delta > 0$ such that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$

Let $\delta = \epsilon$

Then,
$|x - y| < \delta$

$|m(E \cap (-\infty,x]) - m(E \cap (-\infty,y]|$

$= |m(E \cap ((-\infty,x] - (\infty,y]))|$

$\leq |m((-\infty,x] - (-\infty,y])|$

$= |m(-\infty + \infty,x - y])|$

Now, here is the step I'm quite unsure of (in fact I think it is wrong). Since this is simply our function evaluated at two different points, I think it's possible that the two infinities are the same level of infinity. So in that case, wouldn't $\infty - \infty = 0$?

$=|m(0,x - y)|$

$= |x - y - 0| = |x - y| < \delta = \epsilon$

I have two issues with my proof, but I can't seem to think of a different way to proceed. One I already mentioned (the infinity issue), and the other is that I don't see how my proof is any different than one for normal continuity. I know somewhere in here there needs to be an emphasis on why the function is uniformly continuous, but I don't see how I could include that. Can anyone shine some light on this? Thanks!

答:
Let $E \subset \mathbb{R}$ be a measurable set with $m(E) < \infty$. Define the functoin $f: \mathbb{R} \to \mathbb{R}$ by
$f(x) = m(E \cap (-\infty,x])$. Then $f$ is uniformly continuous on $\mathbb{R}$.

Proof:
We need to show that, for all $\epsilon > 0$ there is  $\delta > 0$ such that, for all $x,y \in \Bbb R$,  if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. (Note that the choice of $\delta$ must not depend on $x$ or $y$).

Given any $\epsilon > 0$, let $\delta = \epsilon$.  Then, for any  $x,y \in \Bbb R$, we can assume without loss of generality that $x\leqslant y$, and so $m(E \cap (-\infty,x]) \leqslant m(E \cap (-\infty,y])$. So,  we have that, if  $|x - y| < \delta$, then
\begin{align*}
|f(x) - f(y)| &= | m(E \cap (-\infty,x]) - m(E \cap (-\infty,y])|= \\
&= m(E \cap (-\infty,y]) - m(E \cap (-\infty,x]) = \\
&= m((E \cap (-\infty,y]) \setminus (E \cap (-\infty,x])) =\\
&= m (E \cap (( -\infty,y] \setminus (-\infty,x])) = \\
&= m(E \cap (x,y]) \leqslant \\
& \leqslant m((x,y]) = y-x = |x-y| <\delta = \epsilon
\end{align*}
(see Remark 1 for details).

So, $f$ is uniformly continuous on $\mathbb{R}$. $\square$

Remark 1:

We used in the proof above that if $A \subseteq B$ and $m(A)< \infty$, then $m(B) - m(A) = m(B \setminus A)$, where $B \setminus A$ is the set difference of $B$ and $A$, that is, $B \setminus A = \{p: p\in B \textrm{ and } p\notin A\} $.

We also used, assuming $x \leqslant y$, that $ ( -\infty,y] \setminus (-\infty,x] = (x, y]$. Let us prove it.

We have
\begin{align*}
( -\infty,y] \setminus (-\infty,x] & = \{ p \in \Bbb R : p \leqslant y \} \setminus \{ p \in \Bbb R : p \leqslant x \} = \\
& =  \{ p \in \Bbb R : p \leqslant y \textrm { and not }( p \leqslant x )\} = \\
& =  \{ p \in \Bbb R : p \leqslant y \textrm { and } p > x \} = \\
& = (x, y]
\end{align*}

Remark 2: We actually proved more. We proved that, for all $x, y \in \Bbb R$, $|f(x) - f(y)| \leqslant |x-y|$. So, $f$ is a Lipschitz continuous function.


问:
Lebesgue measure of an open unbounded set
In his book Frank Jones defines the Lebesgue measure of an open set $U\subset\mathbb{R}^n$ by

$$\lambda (U):=\sup\{\lambda(P):P\subset U, P \text{ special polygon}\}$$

where a special polygon $P$ is a union of compact rectangles with disjoint nonempty interiors, with lebesgue measure $\lambda(P)$ defined as the sum of the areas of the compact rectangles.

Jones write: In case $U$ is not a bounded set, then we interpret the definition to produce $\lambda (U)=\infty$.

Question: Is it possible to actually prove from the definition that $\lambda (U)=\infty$ if $U$ is unbounded?

答:
You have misquoted Frank Jones, which in PP.29-30 wrote (emphasis mine):

If $G \subset \Bbb R^n$ is an open set and if $G \ne \emptyset$, we define
$$\lambda(G) = \sup\{\lambda(P) \mid P \subset G, P\text{ a special polygon}\}.$$
[...] consider the least upper bound of the set of numbers $\lambda(P)$ for $P \subset G$. In case this set is bounded, then $0 < \lambda(G) < \infty$. In case it is not a bounded set of numbers, then we interpret the definition to produce $\lambda(G) = \infty$. In particular, $\lambda(\Bbb R^n) = \infty$.

You thought "it" refers to $G$, in which case it would make no sense to say "bounded set of numbers", because elements of $G$ are not numbers.

In actuality, "it" refers to the set of numbers $\lambda(P)$ for $P \subset G$, the set which you are taking supremum of. This is the set of numbers that Frank Jones is referring to, and in the case that this set is not bounded, then surely the supremum is $\infty$.


答:
The author writes:

If $G \subset \mathbb R^n$ is an open set and if $G \neq \emptyset$, we define
$$\lambda(G) = \sup \{\lambda(P) \mid P \subset G, P \text{ a special polygon}\}$$
Since $G$ is open and nonempty, it has interior points, and thus there are polygons contained in $G$. It is therefore possible to consider the least upper bound of the set of numbers $\lambda(P)$ for $P \subset G$. In case this set is bounded, then $0 < \lambda(G) < \infty$. In case it is not a bounded set of numbers, then we interpret the definition to produce $\lambda(G) = \infty$.

When he says "in case it is not a bounded set..." he's not referring to $G$, he's referring to the set $\{\lambda(P) \mid P \subset G, P \text{ a special polygon}\}$.

Title: Lebesgue measure of an open unbounded set

Body:
In his book Frank Jones defines the Lebesgue measure of an open set $U\subset\mathbb{R}^n$ by

$$\lambda (U):=\sup\{\lambda(P):P\subset U, P \text{ special polygon}\}$$

where a special polygon $P$ is a union of compact rectangles with disjoint nonempty interiors, with lebesgue measure $\lambda(P)$ defined as the sum of the areas of the compact rectangles.

Jones write: In case $U$ is not a bounded set, then we interpret the definition to produce $\lambda (U)=\infty$.

Question: Is it possible to actually prove from the definition that $\lambda (U)=\infty$ if $U$ is unbounded?


答:
This answer is in response to the OP's request, in comments below DreiCleaner's answer, to explain what is wrong in their (mis)calculation ${3\over49}+{1\over49}$.

The OP is quite right that if the third card flipped is matches the rank of the cards in the first war, then there is only a $1/49$ chance the fourth card will match the third. But this (the third card matching the first to) only happens with probability $2/50$ -- that is, the third card must be one of the $2$ cards of the same rank from the first war, out of $50$ cards available. Similarly, if the third card is different from the rank of the first war, which happens with probability $48/50$, then the fourth card will match it with probability $3/49$. So a correct calculation of the probability of a second war is

$${48\over50}\cdot{3\over49}+{2\over50}\cdot{1\over49}={24\cdot3+1\over25\cdot49}={73\over1225}$$

as in DreiCleaner's answer.

Title: Card game Probability: Having a second war after a first war

Body:
I was wondering about the probability of getting two wars.  The way the game works is, both players get $26$ cards, and flip one card over, whoever has the largest card value, takes both cards and puts them underneath the deck.

Now the tricky part comes when the two cards flipped over have an equal value, this is called a "war".  At a war, the players each flip another card on the top of their deck over, and whoever has the highest value card gets all the cards on the table.  My question is, what is the chance that these two cards also have equal value?  So, the probability of having two cards out of $52$, be equal, then the next two cards be equal.

I keep arriving at this number: $\frac{3}{49} + \frac{1}{49} = \frac{4}{49} = 0.0816$.

Thanks for the help!


问:
Continuous Mapping Theorem (CMT) for a sequence of random vectors
I need help proving the Continuous Mapping Theorem (CMT) for random vectors. I'm currently reading Econometric Analysis for Cross Section and Panel Data by Jeffrey M. Wooldridge (Chapter 3, pp. 40 - 41, 2nd edition). Unfortunately, he leaves it to the reader to prove most asymptotic results. Additionally, almost every other econometrics textbook I read simply states the result.

Definition 1: A sequence of random variables $x_n$ converges in distribution to a continuous random variable $x$ if and only if $\forall s \in \mathbb{R} \ \forall \epsilon >0   \ \exists N \ s.t. \  \forall n>N \; |Prob(x_n \leq s) - Prob(x \leq s)|<\epsilon$. We write $x_n \to^d x.$ [Note: A continuous random variable is one for which the cumulative distribution function is continuous.]

Definition 2: A sequence of K $\times$ 1 random vectors $\mathbf{x}_n$ converges in distribution to the continuous random $K \times 1$ vector $\mathbf{x}$ if and only if $\forall \mathbf{c} \in \mathbb{R}^{K}$ such that $\mathbf{c}^T\mathbf{c} = 1$, $\mathbf{c}^T\mathbf{x}_n \to^d \mathbf{c}^T\mathbf{x}$, and we write $\mathbf{x}_n \to^d \mathbf{x}.$

Theorem 1: Let $\mathbf{x}_n$ be a sequence of $K \times 1$ random vectors such that $\mathbf{x}_n \to^d \mathbf{x}$. If $\mathbf{g}:\mathbb{R}^k\to\mathbb{R}^{\ell}$ is a continuous function, then $\mathbf{g}(\mathbf{x}_n)$ $\to^d$ $\mathbf{g}(\mathbf{x}).$

Definition 3: A sequence of random variables $x_n$ is bounded in probability if and only if $\forall \epsilon>0 \ \exists b_{\epsilon}>0 \ \exists N \ s.t. \forall n>N \ Prob(|x_n|>b_{\epsilon})$. A vector $\mathbf{x}_n$ is bounded in probability if and only if the random variables which constitute the vector of random variables are themselves bounded in probability.

Theorem 2: If $\mathbf{x}_n \to^d \mathbf{x}$, where $\mathbf{x}$ is a $K \times 1$ vector, then $\mathbf{x}_n = O_p(1)$.

I need rigorous proofs for Theorems 1 and 2. This problem has been frustrating me for a couple days now, so any help would go a long way.

Thanks.

CS

答:
For Theorem 1:

Let $x_n$ be defined on the probability space $(\Omega, \mu)$.
Your Definition 2 for convergence in probability of a sequence of random vectors says that for any half space $H$ of $\mathbb{R}^k$, i.e. $H = \phi^{-1}(r)$ for some linear functional $\phi: \mathbb{R}^k \rightarrow \mathbb{R}$ and $r \in \mathbb{R}$, $\mu(x_n^{-1}(H)) \rightarrow \mu(x^{-1}(H))$. ($\phi$ is inner product with $c$ in your definition.)

Now if $g: \mathbb{R}^k \rightarrow \mathbb{R}^l$ is linear, then Theorem 1 is immediate: For any half space $H \subset \mathbb{R}^l$, $g^{-1}(H)$ is again a half space of $\mathbb{R}^k$. So $\mu(x_n^{-1}(g^{-1}(H))) \rightarrow \mu(x^{-1}(g^{-1}(H)))$.

The case $g$ is just measurable takes a little doing. Definition 2 implies the following: for any closed convex $C \subset \mathbb{R}^k$, $\mu(x_n^{-1}(C)) \rightarrow \mu(x^{-1}(C))$. This can be shown by writing $C$ as the countable intersection of polygons and use continuity-from-above of the pushforward measures. Now take any half space $H \subset \mathbb{R}^l$. Consider the measurable set $g^{-1}(H)$. The pushforward measure $\mu$ induced by $x$ is regular. So it can be approximated from below by some compact $K \subset g^{-1}(H)$. In turn, $K$ can be covered by finite rectangles $C_1,\cdots, C_m$. Since  $\mu(x_n^{-1}(C_i)) \rightarrow \mu(x^{-1}(C_m))$ for $i = 1,\cdots,m$, Theorem 1 holds.

For Theorem 2:

Let $B_b$ denote the closed cube centered at the origin of radius $b$ in $\mathbb{R}^k$.

Your Definition 3 says that, for all $\epsilon > 0$, there exists $b> 0$ and $N \in \mathbb{N}$ such that $\mu(x_n^{-1}(B_b)) > 1 - \epsilon$ for all $n \geq N$.

$B_b$ is convex and closed. So $\mu(x_n^{-1}(B_b)) \rightarrow \mu(x^{-1}(B_b))$ for any such cube. By the regularity of the pushforward measure again, $\mu(x^{-1}(B_b)) \rightarrow 1$ as $b \rightarrow \infty$. So Theorem 2 holds.

I am an econ grad student myself. Hope this helps.


问:
Conditional probability red balls
One box contains four black balls and five red balls.
Two balls are pulled out without putting them back.

What is the probability that the second ball is red if the first ball was black.

I'm not sure if my solution is correct:

All possible events $ \Omega:=\{(\text{red},\text{red}),(\text{red},\text{black}),(\text{black},\text{red}),(\text{black},\text{black})\} $

Event A := "Second ball is red." $ :=\{(\text{red},\text{red}),(\text{black},\text{red})\} $

Event B := "First ball is black." $ :=\{(\text{black},\text{red}),(\text{black},\text{black})\} $

Probabilitytree:

Then I have $ A\cap B=\{(\text{black},\text{red})\} $ and

$ P_B(A)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{4}{9}\cdot \frac{5}{8}}{\frac{4}{9}\cdot \frac{5}{8}+\frac{4}{9}\cdot \frac{3}{8}}=\frac{5}{8} $

答:
The first ball was black. So there remain

five $\color{red}{\text{red}}$ balls, and

three black balls.

The probability to extract a red ball is now $\displaystyle\frac{\bf 5}{\bf 5 +\bf 3}=\frac 58$.

In your picture, it is the $5/8$ marked after the first branching to black, going into the red ball.

Title: Conditional probability red balls


问:
Absolute Value Inequality Implication
From my knowledge of the properties of absolute value inequalities, I'm failing to see why:

$$|f(x)-A|<1  \rightarrow |f(x)|<|A| + 1$$

The only approach I know is to split up the two possible cases:

$$ (1) :  f(x) - A> -1$$

and

$$(2) : f(x) - A < 1$$

So $f(x) < A+1$, but how does this imply $|f(x)|<|A| + 1$ ?

答:
The triangle inequality says $|a+b|\leq |a|+|b|$. From this, setting $a=(x-y)$ and $b=y$, we get
$$|x| \leq |x-y|+|y|$$
from which we deduce
$$|x|-|y|\leq |x-y|.$$
This inequality gives you the result you want:
$$|f(x)|-|A| \leq |f(x)-A| \lt 1$$
so from $|f(x)|-|A| \lt 1$
you immediatly get
$$|f(x)|\lt |A|+1.$$
In fact, the following holds: try to verify it:
$$\Bigl| |x|-|y|\Bigr| \leq |x\pm y| \leq |x|+|y|.$$

Title: Absolute Value Inequality Implication


答:
More generally: This is a consequence of the so-called reverse triangle inequality, itself a consequence of the triangle inequality:
$$
||x|-|y|| \leq |x-y|, \qquad x,y\in\mathbb{R} \tag{$\dagger$}
$$
Apply it to $f(x)$ and $A$ to get
$$
||f(x)|-|A|| \leq |f(x)-A|
$$
which implies $|f(x)|\leq |f(A)|+|f(x)-A|$.

The reverse triangle inequality, very useful, is quite immediate given the triangle inequality: from
$$
|x| = |x-y+y|  \leq |x-y|+|y|
$$
and
$$
|y| = |y-x+x|  \leq |x-y|+|x|
$$
we get $||x|-|y||= \max( |x|-|y|, |y|-|x| ) \leq |x-y|$.

Title: Absolute Value Inequality Implication

Body:
From my knowledge of the properties of absolute value inequalities, I'm failing to see why:

$$|f(x)-A|<1  \rightarrow |f(x)|<|A| + 1$$

The only approach I know is to split up the two possible cases:

$$ (1) :  f(x) - A> -1$$

and

$$(2) : f(x) - A < 1$$

So $f(x) < A+1$, but how does this imply $|f(x)|<|A| + 1$ ?


问:
Inequality using Bernoulli's inequality?
Let $0<b<a<1$. Does the following equality hold?

$b^{a-b} \geq \frac{b}{a}$

I feel that this can be proved by Bernoulli's inequality, but cannot figure out exactly how. Please help!

答:
Let $a - b = \epsilon \in (0,1)$. The inequality becomes $b^{\epsilon} \geq \frac{b}{b + \epsilon}$. By inverting both sides this becomes equivalent to
$\left( \frac{1}{b} \right)^{\epsilon} \leq 1 + \frac{\epsilon}{b}.$

Now clearly $\frac{1}{b} > 1$ so Bernoulli's inequality with exponent in $(0,1)$ gives
$$\left( 1 + \left(\frac{1}{b}-1\right) \right)^{\epsilon}\leq 1 + \epsilon \left(\frac{1}{b}-1\right) = 1 + \frac{\epsilon}{b} - \epsilon < 1 + \frac{\epsilon}{b}, $$

which is what we wanted.


问:
How to color a board such that each square has exactly two neighbors of the opposite color
Let $m≥2$, $n≥2$ integers. We want to color the squares of an $m × n$ board with black and white so that each square has exactly two neighbors of the other color. Determine all the values of $m$ and $n$ for which it is possible to do such a coloring.
Clarification: Neighboring squares are those that have a common side.

I found a way to color the square when $m$ and $n$ are both even. I think that if $m$ or $n$ is odd, then it is impossible to color de board. But i can't prove it. Could someone help me, please?

答:
Here's a convenient trick: Flip the color of every other square.  I.e. if a square has coordinates $(x,y)\in \mathbb{Z}^2$, flip it's color if $x+y$ is even.  Then this modified board will be such that every square has two neighbors of the same color.  This means that each square is on a loop of identical color.

Now look at one side of the rectangle, say the bottom.  Each square on the bottom has at least one neighbor of the same color also on the bottom (since it has only one neighbor on the board that's not on the bottom).  I claim the contiguous blocks of same-colored squares on the bottom can only have even length: If a contiguous block of squares $1, 2, \dots, n$ has the same color, say black, then squares $2, 3, \dots, n-1$ must have white neighbors above them (see picture below; grey squares are undetermined by this argument).  Number these white squares $2,3, \dots, n-1$ just like their neighbors below.  Then white squares $3,4,\dots,n-2$ must have black neighbors above them.  And so on inductively until we are left with just 2 squares at positions $n/2, n/2+1$.  This shows you can only have even blocks on the boundary of the board, and hence the boundary itself must have even length, being a sum of possibly several such blocks.  Hence the board must have even dimensions $m$ and $n$.


问:
To prove that the intersection of the given sets is not empty
Below is a question that I came across recently.
Suppose $G = \text{conv}\{k_{1}, . . . ,k_{n}\}$ be a convex set in $R^{d}$. Then we have to prove that the intersection of $F_{i}=\{p :(-1/d)k_{i}+ p\in G\}$ is not empty.

$\textbf{What I have tried}$: Given $  G =\text{conv}\{k_{1}, . . . ,k_{n}\}$ hence it can be written as convex combinations of all $k_{1},...,k_{n}$:
$G=\{\sum_{j=1}^{n}\lambda_{j}k_{j};\lambda_{j}\geq 0;\sum_{j=1}^{n}\lambda_{j}=1\}$.

Now $F_{i}=\{p :(-1/d)k_{i}+ p\in G\}=\{\sum_{j=1}^{n}\lambda_{j}k_{j}+(1/d)k_{i}\}$ (I have skipped writing the convexity conditions inside the bracket but its there).

Now by $\sum_{j=1}^{n}\lambda_{j}k_{j}+(1/d)k_{i}$, I believe the author is talking about Minkowski sum. So $F_{i}$ is basically a translation of the convex hull $G$ by $k_{i}/d$.

Now I have to show that $\cap_{i=1}^{n}F_{i}\neq\phi\Rightarrow\cap_{i=1}^{n}\Big\{\sum_{j=1}^{n}\Big(\lambda_{j}k_{j}+(1/d)k_{i}\Big)\Big\}\neq\phi$. This is where I got stuck. Could anyone please help me on how to proceed from here.

Thanks in advance.

$\textbf{P.S.}$: $\textbf{I have edited out the question because earlier there was a mistake in the question.}$

答:
Let's suppose $G$ has nonempty interior, so $n \geq d + 1$.
For each set $S \subset \{1, \dotsc, n\}$ with $\lvert S \rvert = d+1$, consider
the point $p_S = \sum_{i \in S} \frac{k_i}{d}$.
For any given $j \in S$,
$$(-1/d)k_j + p_S = \sum_{i \in S \setminus \{j\}} \frac{1}{d} k_i \in G.$$

Therefore $p_S \in \bigcap_{j \in S} F_j$.

So we have a collection of $n$ convex sets $F_i$, and the intersection of any $d+1$ of them is nonempty. By Helly's Theorem, the intersection of the whole collection is nonempty.


问:
Understanding subsitution by parts of $\int_0^{1/4}x\sin^{-1}(4x)dx$
I am doing an Integration by Parts exercise and can't wrap my head around one of the answers:

Consider the definite integral

$$\int_0^{1/4}x\sin^{-1}(4x)dx$$

We select $u = \sin^{-1}(4x)$.

We select $dv = xdx$.

Consider the known formula $$\int udv = uv - \int vdu$$

Then, after integrating by parts, we obtain the integral $$\int_0^{1/4}vdu = \int_0^{1/4}f(x)dx$$ on the right hand side where
$$f(x) = \frac{x^2}{2} \cdot \frac{1}{(\sqrt{1-(4x)^2})} \cdot 4$$

I get it all up to this point, but then the exercise says this:

The most appropriate subsitution to simplify this integral is $x = g(t)$ where $$g(t) = \frac{\sin(t)}{4}$$

It's not clear to me what do they mean by "simplify this integral". What's there to simplify and why?

And of course, why $\frac{\sin(t)}{4}$? I can almost see what's going on because that integral looks a lot like the differentiation of $\arcsin(x)$, but still failing to make the connection.

What is it that I'm missing? What was the train of thought behind that substitution?

Body:
Integrate by parts twice as follows

\begin{align}
\int_0^{1/4}x\sin^{-1}(4x)dx
&=\frac12 \int_0^{1/4}\sin^{-1}(4x)d(x^2)\\
&=\frac12x^2\sin^{-1}(4x)\bigg|_0^{1/4}-2\int_0^{1/4} \frac{x^2}{\sqrt{1-16x^2}}dx\\
&=\frac\pi{64}+\frac1{16}\int_0^{1/4} \frac{x}{\sqrt{1-16x^2}}d(1-16x^2) \\
&=\frac\pi{64}-\frac1{16}\int_0^{1/4} \frac{1}{\sqrt{1-16x^2}}dx\\
&=\frac\pi{128}
\end{align}


问:
If $A \cong I$ in a monoidal category, then any arrow $X \otimes A \to Y \otimes A$ decompose into $(X \to Y) \otimes 1$
The question is originated from the answer of this post. Let me make it precise first. Suppose $(\mathcal{A}, \otimes, I)$ is a monoidal category and $A\in \mathcal{A}$ is an object isomorphic to $I$. Then for any morphism $g:X \otimes A \to Y \otimes A$ in $\mathcal{A}$ there is a unique morphism $f:X\to Y$ such that $g = f \otimes 1_A$.

$\textbf{Attempt}$
Let $k:A \to I$ be an isomorphism. Given any such $g$, there is an arrow
$$X \xrightarrow{\rho^{-1}} X \otimes I \xrightarrow{1 \otimes k^{-1}} X \otimes A \xrightarrow{g} Y \otimes A \xrightarrow{1 \otimes k} Y \otimes I \xrightarrow{\rho} Y$$
(and I think that's the only way to construct a morphism $X \to Y$). But I don't see how this map tensoring with the identity gives you $g$. I believe it is just some sort of combining different diagrams together + the interchange law, as you normally do proofs in monoidal category. I have no clue on the uniqueness part of the morphism either. Any help is appreciate.

答: A briefer, though perhaps less satisfying, argument: since $A$ is isomorphic to $I$ and by definition $(-)\otimes I$ is naturally isomorphic to the identity functor, also $(-)\otimes A$ is naturally isomorphic to the identity functor, say via $\alpha$. Then we have a natural isomorphisms $\mathrm{Hom}(X,Y)\to \mathrm{Hom}(X,A\otimes Y)$ and $\mathrm{Hom}(X,A\otimes Y)\to \mathrm{Hom}(A\otimes X,A\otimes Y)$ given by post- and pre-composition with $\alpha$.


问:
Polynomial such that $e^{2i\pi P(n)} \rightarrow 1$
here is a  problem i’ve been having quite a lot of trouble with .

Let $P$ be a polynomial such that the sequence $e^{2i\pi P(n)}$ converges to 1 $(i^2=-1).$

Show that $\forall n ,P(n)$ is an integer.

Taking the imaginary part we are left with some $P$ polynomial such that

$$\sin(2\pi P(n))\rightarrow0$$
But a half angle factorization also gives  that: $$\sin(\pi P(n))\rightarrow 0$$
And we wish to force P to , i guess even if it’s not equivalent, to have only integers coefficients.

May you help me please.

答:
Write $f(x)= \exp(2\pi ix)$. If $P(n)=a_1n+a_0$ is linear, you could use  $$f(a_1)=f(P(n+1)-P(n))= \frac{f(P(n+1))}{f(P(n))} \to 1 \; \text{as} \; n \to \infty$$
to deduce that $a_1$ is an integer, whence so is $a_0$.  This approach generalizes, but one needs to use the representation of a polynomial in terms of binomials.

Write a degree $d$ polynomial as
$P(x)=\sum_{k=o}^d a_k {x\choose k}$ (Every polynomial has a unique representation as such a combination of binomial coefficients.)

Claim:    The assumption $f(P(n)) \to 1$ as $ n \to \infty$ implies that all $a_j$ in the representation $P(x)=\sum_{k=o}^d a_k {x\choose k}$ are integers.
(This then  implies that $P(m)$ is an integer for all integer $m$.)

Proof By induction on $d$. We just need the induction step. Given $P$ that satisfies the assumption,
$$Q(x):=P(x+1)-P(x)=\sum_{k=1}^d a_k {x \choose k-1} $$
also satisfies $f(Q(n)) \to 1$ as $n \to \infty$, so by the induction hypothesis,
$a_1, \ldots , a_d$ are integers. It then   follows that $f(a_0)=1$ so $a_0$ is also an integer.


问:
What gives the natural logarithm its name? What's "natural" about it?
I know that the natural logarithm is defined as $\ln(x)$ or $\log_{e}(x)$, where $e$ is the Euler number.

But what is so "natural" about it? Is there an explanation on why that name was chosen?

答:
Another way of looking at it is that, to find the instantaneous rate of change of a function, f(x) we take the limit of (f(x+h)- f(x))/h as h goes to 0.  In particular, for $f(x)=a^x$, that is $\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)$.  Notice that we have $a^x$, with no h, times $\left(\frac{a^h- 1}{h}\right)$ which has no x.

The limit of that is $a^x\left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)$.

Whatever a or x is that limit is a number so we can say:

The derivative of $a^x$ is a number, that depends on a, times $a^x$.

Now what is that number?  Look at some examples.  If a= 1 then a^h= 1 for all h so $\frac{1- 1}{h}= 0$.

If a= 2 and suppose h= 0.001, then $a^h= 2^{0.001}= 1.000693387$ (hallelujah for calculators!) so $2^h- 1= 0.000693387$ and $\frac{2^h- 1}{0.001}= 0.6934. The limit as h goes to 0 is close to 0.69.

If a= 3, h= 0.001, then $a^h= 3^{0.001}= 1.001099$, $3^{0.001}-1= 0.001099$ and $\frac{3^{0.001}- 1}{0.001}= 1.099$. Again the limit as h goes to 0 is close to 1.099.

Notice that for x= 2, that value is less than 1 and for x= 3, that limit is greater than 1.  Don't you suspect that there is an a, between 2 and 3, such that it is exactly 1?

For one more check, try a= 2.7.  $a^h= 2.7^{0.001}= 1.00099374521$, $2.7^{0.001}- 1= 0.00099374521$, and $\frac{2.7^{0.001}- 1}{0.001}= 0.99374521$.

Now that's awfully close to 1 and still just a little less than 1.  We can conjecture that there is a number, slightly larger than 2.7, which I am going to call, for no particular reason, $e$, for which $\lim_{h\to 0}\frac{e^h- 1}{h}= 1$.   So the derivative of $e^x$ is just $e^x$ again!  The function $f(x)= e^x$ is its own derivative!

Of course, any time we have a function, f(x), would like to have it "inverse" function, $f^{-1}(x)$, so that we can solve equations- to solve f(x)= a, take $f^{-1}(f(x))= x= f^{-1}(a)$.

$ln(x)$, the "natural logarithm" is defined as that inverse function to $e^x$.  It is "natural" because it is the inverse to the "natural" exponential $e^x$.

One caveat, e, a positive number, to any power, is positive so while its domain is "all real numbers" its range is "all positive numbers.  That means that the domain of ln is positive numbers, ln(-1) is undefined (until you move to the complex numbers).


问:
Why are Fuchsian groups interesting?
I am recently reading the book "Fuchsian groups" by Katok and now on Chapter $2$. I am curious about why Fuchsian groups are interesting. I look it up online and find answers here. Those are great answers. However, I am at an undergrad level now and I don't get a very clear picture. Hyperbolic geometry is interesting since there are some great properties, like hyperbolic Gauss-Bonnet. Would someone like to provide an example, e.g. rubric cube when studying group theory?

Appreciate any resources and helps!

Addendum: Fuchsian groups are discrete subgroups of PSL($2$, $\mathbb{R}$) or automorphism of the upper-half plane.

答: Fuchsian groups appear in several areas of mathematics, including complex analysis, number theory, topology. Number theorists really like Fuchsian groups like $PSL(2, {\mathbb Z})$ (and its relatives) and many deep number-theoretic questions can phrased in terms of such groups. For instance, Shimura-Taniyama Conjecture (proven by Andrew Wiles), which, in turn, implies Fermat's Last Theorem, can be formulated in terms of certain subgroups of  $PSL(2, {\mathbb Z})$ called congruence subgroups. I can provide more details on this in case you are interested.

Title: Why are Fuchsian groups interesting?


问:
Given a basis, is there a unique inner product such that that basis is orthogonal?
I know there is a unique way to define an inner product such that the basis are orthonormal. Given a basis $(e_1, e_2, e_3,\dots,e_n),$ we can use the Kronecker delta function $(e_j,e_k)=0$ if $j$ is not equal to $k$, and $1$ if $j$ and $k$ are the same. The inner product defined this way is unique. Now if instead of orthonormal, we only required an inner product such that the basis is orthogonal, is the inner product unable to be unique because this requirement isn't strict enough?

答: You can just scale a given inner product by any positive real number.</p>


问:
What's the intuition behind augmenting paths in residual graphs? (Ford-Fulkerson)
I've tried looking everywhere online about the intuition behind augmenting paths in residual graphs and all I have really gotten is how to systematically solve the problem of finding max flow without any real understanding why it works.

My understanding is that the augmenting path theorem (Ford-Fulkerson, 1956) says A flow f is a max
flow if and only if there are no augmenting paths.

But why is this true? How can I intuitively see this?

I also do understand that residual graphs have edges that "send flow backwards" as a way to "undo" a possibly bad path. But in the examples I saw, a path is "undone" and then never used again, we just end up having less capacity to flow through. But I'm most likely just not entirely understanding the back edges of residual graphs.

答:
Say we have an augmenting path
$$
    s \to a \to b \gets c \to d \to t
$$
with a backward edge $b \gets c$. The logic of using it to improve the flow is that:

First, we can straightforwardly transport more flow from $s$ to $a$ to $b$.

We were previously transporting some flow from $c$ to $b$, but now that we've gotten that flow to $b$ in another way, we don't need to transport as much flow from $c$ to $b$. This frees up some flow from $c$ to send elsewhere.

Then, we can send the freed-up flow from $c$ to $d$ to $t$.

The logic in point 2 is generally the logic behind backward edges.

This explains why augmenting paths help us increase the value of the flow. By itself it doesn't prove the theorem which says that augmenting paths are essentially the only way to increase the value of the flow; if no augmenting path exists, then the flow cannot be increased.

The logic here is that if no augmenting path exists, then in the residual graph, there is no path from the source $s$ to the sink $t$. We can divide the vertices into two sets: $S$, the set of vertices reachable from $s$ in the residual graph, and $T$, the vertices not reachable from $s$. There are no edges from any $v\in S$ to $w \in T$ in the residual graph, or else we'd be able to reach $w$ from $s$, too.

This means that every arc from $S$ to $T$ in the network must be at full capacity, and every arc from $T$ to $S$ must have zero flow. This means that the net flow leaving $S$ to go to $T$ is the largest it could possibly be, and in particular the flow going from $s$ to $t$ cannot be increased.


答:
I think the suggestion is implying that it is "simpler" to deal with an integral that has $\sin(t)$ in it, as opposed to $\sin^{-1}(4x)$.

If you substitute $x=\frac{\sin(t)}{4}$, then you will be able to use

$$\sin^{-1}(4x) = \sin^{-1}(\sin t) = t.$$

After the substitution, you will still have to use integration by parts, and it might help to use a trig identity ($(\sin t)(\cos t) = \frac12 \sin 2t$), but arguably this is simpler than the two integration by parts solution that was posted earlier.

Title: Understanding subsitution by parts of $\int_0^{1/4}x\sin^{-1}(4x)dx$

Body:
I am doing an Integration by Parts exercise and can't wrap my head around one of the answers:

Consider the definite integral

$$\int_0^{1/4}x\sin^{-1}(4x)dx$$

We select $u = \sin^{-1}(4x)$.

We select $dv = xdx$.

Consider the known formula $$\int udv = uv - \int vdu$$

Then, after integrating by parts, we obtain the integral $$\int_0^{1/4}vdu = \int_0^{1/4}f(x)dx$$ on the right hand side where
$$f(x) = \frac{x^2}{2} \cdot \frac{1}{(\sqrt{1-(4x)^2})} \cdot 4$$

I get it all up to this point, but then the exercise says this:

The most appropriate subsitution to simplify this integral is $x = g(t)$ where $$g(t) = \frac{\sin(t)}{4}$$

It's not clear to me what do they mean by "simplify this integral". What's there to simplify and why?

And of course, why $\frac{\sin(t)}{4}$? I can almost see what's going on because that integral looks a lot like the differentiation of $\arcsin(x)$, but still failing to make the connection.

What is it that I'm missing? What was the train of thought behind that substitution?


问:
Calculate Area between $\sin(x)$ and $\sin(2x)$
Let $A$ be the area of region $R$ bounded by $y = \sin(x)$ and $y = \sin(2x)$ over $[0,\pi]$. What is the area of $A$?

Here's some work that I did:

$$\int\limits_{0}^{\frac{\pi}{3}}\sin(2x) - \sin(x)\mathrm{d}x + \int\limits_{\frac{\pi}{3}}^{\pi}\sin(x) - \sin(2x)\mathrm{d}x$$

where $\frac{\pi}{3}$ is where the graphs intersect. I ultimately got $A = \pi$.

Can someone check my answer? Thank you

Body:
$$A_{1} = \int_{0}^{\frac{\pi}{3}} (sin(2x)-sin(x)) dx$$

$$A_{1} = \int_{0}^{\frac{\pi}{3}} sin(2x) dx - \int_{0}^{\frac{\pi}{3}} sin(x) dx$$

$$A_{1} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$$

$$A_{2} = \int_{\frac{\pi}{3}}^{\pi} (sin(x)-sin(2x)) dx$$

$$A_{2} = \int_{\frac{\pi}{3}}^{\pi} sin(x) dx - \int_{\frac{\pi}{3}}^{\pi} sin(2x) dx$$

$$A_{2} = \frac{3}{2} - \left(-\frac{3}{4}\right) = \frac{9}{4}$$

$$A = A_{1} + A_{2} = \frac{5}{2}$$


问:
definition of discriminant and traces of number field.
Let $K=\Bbb Q [x]$ be a number field, $A$ be the ring of integers of $K$.
Let $(x_1,\cdots,x_n)\in A^n$. In usual, what does it mean $D(x_1,\cdots,x_n)$? Either $\det(Tr_{\Bbb K/ \Bbb Q} (x_ix_j))$ or $\det(Tr_{A/ \Bbb Z} (x_ix_j))$? Or does it always same value? I searched some definitions, but it is not explicitly stated.</p>

答:
I'm not entirely certain what $\operatorname{Tr}_{A/\mathbb{Z}}$ is, but the notation $D(x_1,\dots,x_n)$ or $\Delta(x_1,\dots,x_n)$ usually means the discriminant of $K$ with respect to the basis $x_1,\dots,x_n$, so I would say it's most likely the former.

After all, $x_1,\dots,x_n\in A\subset K$, so it still makes sense to talk about the trace of these as elements of $K$, and that is what the definition of the discriminant of $K$ with respect to a basis is (assuming that $x_1,\dots,x_n$ are indeed a basis for $K/\mathbb{Q}$!).


问:
Prove that $0\le\int_0^1\log(u){\rm d}x+\frac1{2\pi^2}\int_0^1\frac1{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x$
I want to prove a special case of a functional inequality stated in a book. To be specific, let $u(x)$ being a positive, differentiable function on $[0,1]$ with unit mass (i.e., $\int_0^1 u(x){\rm d}x = 1$), then we want to show that
$$0 \leq \int_0^1 \log(u){\rm d}x + \frac{1}{2\pi^2}\int_0^1 \frac{1}{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x.$$
In a book, the author resorts to the construction of certain solutions of one-dimensional heat equation and used the eigen-expansion (hence the arguments are very lengthy and involved). Since I am only interested in this special inequality stated above, may I know whether there exists a simpler and more elegant proof ?

答:
OK, after looking back at the various different proofs of the classical log Sobolev inequality, there is at least one way we can vanquish this. It turns out it is just a special case of the inequality stated (and proved) in section 2.1. of the lecture notes "Entropy Methods and Related Functional Inequalities" written by Daniel Matthes. So here is the hammer:

Assume that $\phi:\mathbb{R}_+ \to \mathbb{R}$ is convex such that $(\phi'')^{-\frac{1}{2}}$ is concave, let $\psi$ be such that $(\psi'(s))^2 = \phi''(s)$. Then we have that
$$ \int_0^1 \phi(u){\rm d}x - \phi\left(\int_0^1 u{\rm d}x\right) \leq \frac{1}{2\pi^2}\int_0^1 (\psi(u)_x)^2{\rm d}x \quad (*)$$
holding for all smooth, positive functions $u$ on $[0,1]$. Notice that if $\phi(s) = \frac 12 s^2$ and $\psi(s) = s$, $(*)$ becomes the usual Poincare inequality
$$\int_0^1 u^2{\rm d}x - \left(\int_0^1 u{\rm d}x\right)^2 \leq \frac{1}{\pi^2} \int_0^1 (\partial_x u)^2{\rm d}x$$
in Daniel's notes the choice is $\phi(s) = s\log(s)$ with $\psi(s) = 2s^{\frac 12}$, leading to the well-known log Sobolev inequality, here we just need to take $\phi(s) = -\log(s)$ with $\psi(s) = \log(s)$, and the game is over.
Remark: I definitely welcome any other different approaches!


问:
The equation $2x=a$ has no solution in an infinite additive cyclic group generated by $a$

Q: If $G$ is an infinite additive cyclic group with the generator element $a$, prove that the equation $2x=a$ has no solution in $G$, $x \in G $.

Answer:

Since $G$ is an infinite cyclic group then $|a|= \infty$ (where $|a|$ denotes the order of $a$)

Since $x \in G $ there for $x=ia$

$$2x=a  \Rightarrow 2ia=a$$

$$a(2i-1)=0$$

i'm stuck here so i don't know how to prove that this is unsolvable

答: $(2i-1)a=0$ implies the order of $a$ is finite. That's a contradiction.


问:
Morphisms from quotient ring

Let $f(x)$ be a polynomial in $\mathbb{Z}[x], \left<f(x)\right>$ be the ideal it generates. Let $R$ be a ring. Prove that giving a ring homomorphsim $\mathbb{Z}[x]/\left<f(x)\right> \rightarrow R$ is equivalent to give a root of $f(x)$ in $R.$ Also, prove that the function $\phi \mapsto \phi(\bar{x})$ from set of homomorphism to the set of roots of $f(x)$ in $R$ is bijective.

I have shown the first part of the question. But I am stuck on the bijectivity part. Any hint would be appreciated.

答:
Injectivity: since there is only one ring homomorphism $\mathbb Z \to \mathbb Z[x]/(f(x))$, namely the composition $\mathbb Z \hookrightarrow \mathbb Z[x] \twoheadrightarrow \mathbb Z[x]/(f(x))$, any ring homomorphism $\mathbb Z[x]/(f(x)) \to R$ is completely determined by its value in $\bar x$.

Surjectivity: Take $\alpha \in R$ such that $f(\alpha)=0_R$. Then prove that the evaluation map $$\operatorname{ev}_\alpha : \mathbb Z[x] \to R; \ p(x) \mapsto p(\alpha)$$ factors through $\mathbb Z[x] \twoheadrightarrow \mathbb Z[x]/(f(x))$.

Title: Morphisms from quotient ring


答:
You have shown that $\lvert a\rvert$ divides the absolute value of $2i-1$. But the order of $a$ is infinite.

Alternatively, $G\cong\Bbb Z$ is free of rank one and so has no torsion (but this jumps the gun a bit).

Title: The equation $2x=a$ has no solution in an infinite additive cyclic group generated by $a$

Body:
Q: If $G$ is an infinite additive cyclic group with the generator element $a$, prove that the equation $2x=a$ has no solution in $G$, $x \in G $.

Answer:

Since $G$ is an infinite cyclic group then $|a|= \infty$ (where $|a|$ denotes the order of $a$)

Since $x \in G $ there for $x=ia$

$$2x=a  \Rightarrow 2ia=a$$

$$a(2i-1)=0$$

i'm stuck here so i don't know how to prove that this is unsolvable


答: Using additive notation: if $a$ generates the group $G$ and $2x = a$ for some $x \in G$, then, because $a$ generates $G$, $x = ma$ for some $m \in \Bbb{Z} \setminus \{0\}$, giving us that $2ma = a$. So $G$ cannot be infinite, since we must have $(2m-1)a = 0$.


答: I agree with @JorgeRael:  $5/2$

Title: Calculate Area between $\sin(x)$ and $\sin(2x)$

Body:
Let $A$ be the area of region $R$ bounded by $y = \sin(x)$ and $y = \sin(2x)$ over $[0,\pi]$. What is the area of $A$?

Here's some work that I did:

$$\int\limits_{0}^{\frac{\pi}{3}}\sin(2x) - \sin(x)\mathrm{d}x + \int\limits_{\frac{\pi}{3}}^{\pi}\sin(x) - \sin(2x)\mathrm{d}x$$

where $\frac{\pi}{3}$ is where the graphs intersect. I ultimately got $A = \pi$.

Can someone check my answer? Thank you

Body: I agree with @JorgeRael:  $5/2$


问:
A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$
Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in  \mathbb{N}$.</p>

答:
In this answer, and in this answer for $x=1$, it is shown, using the binomial theorem, that for $x\ge0$,
$$
\begin{align}
\left(1+\frac xn\right)^n
&=\sum_{k=0}^n\binom{n}{k}\frac{x^k}{n^k}\\
&=\sum_{k=0}^n\left(\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dots\frac{n-k+1}{n}\right)\frac{x^k}{k!}\\
&\le\sum_{k=0}^n\frac{x^k}{k!}
\end{align}
$$
where the inequality is strict for $x\gt0$ and $n\gt1$.

Setting $x=1$ gives your result.


问:
Nontrivial periodic solutions of second order ODE
We consider the ODE
$$x''(t)+x^3(t)=0.$$
Clearly, $x\equiv 0$ is a solution. But also, numerically solving this ODE shows that nontrivial periodic solutions exist, too. I am interested in proving the statement that the above ODE has a nontrivial periodic (say, $2\pi$ periodic) solution. How does one go about proving such a statement?

答:
Let $x'(t)=p(t)$, then the quantity $H$ is a constant of motion

$$
H=\frac{p^2}{2}+\frac{x^4}{4}
$$

As we can verify

$$
\frac{dH}{dt}=pp'+x^3x'=x''x'+x^3x'=x'(x''+x^3)=0
$$

The term in $( ...)$ vanishes by virtue of the differential equation. The phase portrait is given by the closed curves $\frac{p^2}{2}+\frac{x^4}{4}=\text{const}$, so the motion is periodic.

By the way, your equation has an 'exact' solution. Since $H$ is constant, we can write

$$
p^2=2H-x^4/2 \\
x'= \pm\sqrt{2H-x^4/2}
$$

This can be integrated to find $t(x)$

$$
t(x)=\int \frac{dx}{\sqrt{2H-x^4/2}}
$$

Which is an elliptic integral. Finally, in principle, $t (x)$can be inverted to find $x(t)$.


问:
Irreducibility of $\operatorname{Spec}$ $\iff$ primality of nilradical
The following is an exercise (Problem 1.19) in Atiyah and Macdonald's Introduction to Commutative Algebra text :

A topological space $X$ is said to be irreducible if $X \neq \emptyset$ and if every pair of non-empty open sets in $X$ intersect, or equivalently if every non-empty open set is dense in $X$. Show that $\operatorname{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal.

I have a specific question about the solution to this problem found here.

Let $X_f$ be the set of prime ideals of $A$ that do not contain $f \in A$, and let $X_g$ be the set of prime ideals of $A$ that do not contain $g \in A$. Suppose $\operatorname{Spec}(A)$ is irreducible. Then the author says that $X_f \cap X_g = \emptyset \Rightarrow X_f$ is empty or $X_g$ is empty. Why does this implication hold ? Is this just a general fact about two prime ideals -- that if their intersection is empty, then one of them must be empty ?

I tried to show this implication by the contrapositive. Suppose that $X_f$ and $X_g$ are both nonempty. Then there is at least one prime ideal of $A$ that does not contain $f$, and there is at least one prime ideal of $A$ that does not contain $g$. However, I don't know how this implies that $X_f \cap X_g \neq \emptyset$ (that is, that there is at least one prime ideal that doesn't contain $f$ nor $g$).

Thanks !

答: A topological space being irreducible means that any two nonempty open sets intersect non-trivially.</p>


答:
The rationale behind that substitution is that $\sqrt{1-(4x)^2}$ becomes something nicer - namely, $|\cos x|$, which is just $\cos x$ on the interval of integration.

I think it would have been clearer if they made that substitution at the beginning, rather than at the end. That way, you could see that $\sin^{-1}(4x)$ becomes $t$, and the original integral would become $$\frac{1}{16}\int_0^{\pi/2}t\sin(t)\cos(t)dx$$ which is easier to integrate in my opinion.


问:
Conditional probability red balls
One box contains four black balls and five red balls.
Two balls are pulled out without putting them back.

What is the probability that the second ball is red if the first ball was black.

I'm not sure if my solution is correct:

All possible events $ \Omega:=\{(\text{red},\text{red}),(\text{red},\text{black}),(\text{black},\text{red}),(\text{black},\text{black})\} $

Event A := "Second ball is red." $ :=\{(\text{red},\text{red}),(\text{black},\text{red})\} $

Event B := "First ball is black." $ :=\{(\text{black},\text{red}),(\text{black},\text{black})\} $

Probabilitytree:

Then I have $ A\cap B=\{(\text{black},\text{red})\} $ and

$ P_B(A)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{4}{9}\cdot \frac{5}{8}}{\frac{4}{9}\cdot \frac{5}{8}+\frac{4}{9}\cdot \frac{3}{8}}=\frac{5}{8} $

答:
Your solution is correct but overly worked.

You can read the answer off the tree.  It is the probability on the second branch off the first root.  You calculated this as $5/8$ , being the probability for drawing one among the five red balls when selecting from the eight balls remaining after removing a red ball.

That was all you needed to do!

There was no need to fill out the rest of the tree to use the definition for conditional probability.

You used the value you already had for $\mathsf P(A\mid B)$ to calculate $\mathsf P(A\mid B)$.

$$\color{blue}{\mathsf P(A\mid B)}=\dfrac{\mathsf P(B)~\color{blue}{\mathsf P(A\mid B)}}{\mathsf P(B)~\mathsf P(A^\complement\mid B)+\mathsf P(B)~\color{blue}{\mathsf P(A\mid B)}}$$

Title: Conditional probability red balls


问:
Is $y'-1=e^{x^2}y$ separable or linear?
Consider differential equation $y'-1=e^{x^2}y$. I ask two questions, is it separable, and is it linear. First of all I think is separable, because there's $e^{x^2}$ and $y$, so it's separate by dividing. Second of all is it linear differential equation. For this, I think is not linear, because it's $e^{x^2}$ for coefficient..

答:
Your equation can be written as
$$y'=1+\mathrm{e}^{x^2}y =:f(x,y). $$
Suppose $f(x,y)=M(x)N(y)$ is separable. $f$ then satisfies the PDE

$$f f_{xy} = f_x f_y (=M N M' N'). $$

However, we get

$$f f_{xy}-f_x f_y=2 xe^{x^2} \not\equiv 0.$$

This proves the equation is not separable.

It is linear, however.

Title: Is $y'-1=e^{x^2}y$ separable or linear?


问:
What is the meaning of this formula (y2-y1) * cos ((x1+x2) /2)?

Can anyone please explain the meaning of the cos((y1+y2)/2) in this formula please?

Note: the constant 6371 is the earth's radius

答:
Here's a picture of the earth with the top sliced off at latitude $\theta$:

The circle around the earth at latitude $\theta$ has radius $R \cos(\theta)$ where $R$ is the radius of the earth.  Just as its radius is multiplied by $\cos(\theta)$ compared to the equator, its circumference is multiplied by $\cos(\theta)$.  The difference between longitudes of points at a given latitude tells us what fraction of the circle at that latitude lies between the points.  So the distance between points
at longitudes $longitudeA$ and $longitudeB$ and the same latitude $\theta$ is $\cos(\theta)$ times what that distance would be if the points were at the equator.

It's more complicated if the points are at different latitudes as well as different longitudes, but as an approximation the "x" part of the distance can be taken to be what the distance would be if we took the latitude to be the average of the latitudes of the two points.


问:
I need a limit definition for the Hessian, does this work?

Let $f: \mathbb{R}^n \to \mathbb{R}$ be of class $C^2$. Let $x$ be a non-degenerate critical point of $f$. Prove that there is an open neighborhood of $x$ which contains no other critical points of $f$.

$\textbf{Proof:}$  Suppose the contrary. Then for every $n \in \mathbb{Z}_+$ there is an $x_n \in B\left(x,\frac{1}{n}\right)$, which is a critical point of $f$ not equal to $x$. So $x_n$ converge to $x$. Now let $v_n = \frac{x_n - x}{||x_n-x||}$. Let us just assume $v_n$ converges to some $v \in \mathbb{S}^{n-1}$ (since $v_n$ is a bounded sequence, it has a convergent subsequence). So by the definition of the Frechet derivative, and since $x_n \to x$, we have that
\begin{align*}
0 &= \lim_{n\to\infty}  \frac{||Df(x_n) - Df(x) - D^2f(x)(x_n -x)||}{||x_n - x||}\\
  &=\lim_{n\to\infty}  \frac{||- D^2f(x)(x_n -x)||}{||x_n - x||}\\
  &=\lim_{n\to\infty}  ||D^2f(x)v_n||\\
  &= D^2f(x)v,
\end{align*}
where we use the fact that $Df(x) = Df(x_n) = 0$ for all $n \in \mathbb{Z}_+$ ($x$ and $x_n$ are critical points). This implies that $D^2f(x)v = 0$, but since $x$ is a nondegenerate point, $D^2f(x)$ is invertible, and since $v \ne 0$, we have a contradiction. Thus, there is an open neighborhood of $x$ that contains no other critical points of $f$.

Does this approach work? I am not sure if I used a correct definition for the Hessian. Also, I did not use that the Hessian is continuous, which worries me. Regardless of whether this is correct or not, is there a direct approach, or even maybe a more elegant one?

答:
That argument looks fine to me. Another way would be to use Morse's lemma to say that after a coordinate change, for some $m$ one has $f(x+h) = f(x) + \sum_{i = 1}^m h_i^2 -
\sum_{i = m+1}^n h_i^2$ for small enough $|h|$.
Therefore if $h_i \neq 0$, ${\partial f \over \partial x_i}(x + h)$ is nonzero in the new coordinates.


问:
Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.

Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$

So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.

Something i’m not sure whether it’s true:

$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$

Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?

答:
My english is not that good, I will try to explay the main steps, I hope you can understand.

$3(2^m)=(n^2-4)=(n-2)(n+2)$

so $3|n-2$ or $3|n+2$.

If $3|n-2$, then $n=3x+2$ for some $x\geq0$.

So, going back to the first equation with $n=3x+2$, we have $2^m=x(3x+4)$. So

$x=2^k$, for some $k$ satisfying $0\leq k \leq m$.

This means that. we have $2^{k-m}=3.2^k+4$, and so,
$3.2^k=4(2^{k-m-2}+1)$.

Finally $3.2^{k-2}=2^{k-m-2}+1$. And you analyse the conditions to both sides be odd or even.

The same idea, you can work with the case 3|n+2.


问:
Homology of free loop space and Hochschild cohomology
I am looking for honest proof of the following isomorphism.
For a simply connected space $X$, let $LX$ be its free loop space.
Then
$$H_{\ast}(LX) \simeq HH^\ast(C^\ast(X),C_\ast(X))$$

I have looked Jones' approach on ''Cyclic homology and equivariant homology'' that
given a map
$$f_k : \Delta^k \times LX \to X^{k+1}$$
defined by
$$(x_1,\cdots,x_k)\times \gamma \mapsto (\gamma(0),\gamma(x_1),\cdots,\gamma(x_k))$$

its cochain map
$$f_k^\ast : C^\ast(X)^{\otimes k+1} \to C^{\ast -k}(LX)$$
fit together to define a chain map from the Hochschild complex to the free loop space
$$f^\ast : CH_\ast(C^\ast(X)) \to C^\ast(LX)$$
which is a chain homotopy equivalence, so that we get an isomorphism
$$HH_\ast(C^\ast(X),C^\ast(X)) \simeq H^\ast(LX)$$

Many people simply say that ''dualizing'' the statement we get the relation between homology of loop space and Hochschild cohomology, but what is the honest procedure to dualize?

On Cohen and Voronov's Notes on String topology, they just take dual of $f^\ast_k$ and they claim that
$$C_\ast(LX) \simeq Hom(C^{\ast}(LX),\mathbb{Z})$$
and
$$Hom(C^\ast(X)^{\otimes k+1},\mathbb{Z}) \simeq Hom(C^\ast(X)^{\otimes k},C_\ast(X))$$
which also means
$$C_\ast(X) \simeq Hom(C^{\ast}(X),\mathbb{Z})$$

However, there is no reason to be true that those groups are either isomorphic or quasi-isomorphic.

Please somebody help me.

答:
Let’s use the following grading:

On a dual complex, $$((C_*)^\vee)^i=Hom(C_*,\mathbb{Z})^i=Hom(C_i,\mathbb{Z}).$$

On the Hochschild chain complex of $C^*(X)$, $$CH_i(C^*(X),C^*(X))=\bigoplus_{i_1+\cdots+ i_k=i+k} C^{i_1}(X)\otimes\cdots\otimes C^{i_k}(X).$$
Therefore $CH_*$ is honestly $\mathbb{Z}$-graded, and the Hochschild differential increases degree: $\partial^*+\delta: CH_*\to CH_{*+1}$.

With the above grading, there is an isomorphism
$$H^i(LX)\cong HH_i(C^*(X),C^*(X))\ \ \ \forall i$$ in case $X$ is simply-connected.

The validity of dualizing the above isomorphism could come from the Universal Coefficient Theorem. For simplicity let’s work over $\mathbb{Z}$, and assume that homology groups are finitely generated in each degree.

For a degree-decreasing complex $\{\cdots \to C_*\to C_{*-1}\to \cdots\}$, form its dual complex $C^*=(C_*)^\vee$, then
$$F(H^i)\cong F(H_i),\ \ \ T(H^i)\cong T(H_{i-1}),$$
where $F,T$ denotes free part and torsion part, respectively. Similarly, if $\{\cdots \to C^*\to C^{*+1}\to \cdots\}$ is a degree-increasing complex, its dual complex $\bar{C}_*=(C^*)^\vee$ has
$$F(\bar{H}_i)\cong F(H^i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1}).$$
As a consequence, take $\bar{C}_*= C^{\vee\vee}_* $ to be the cocochain complex obtained by doubly dualizing $C_*$, we have
$$F(\bar{H}_i)\cong F(H_i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1})\cong T(H_i).$$
Therefore $C_*\to C_*^{\vee\vee}$ is a quasi-isomorphism.

By this MO question I think the finite generation assumption is satisfied in our situation.

Since $CH^*(C^*(X),C^*(X)^\vee)$ is exactly the dual complex of $CH_*(C^*(X),C^*(X))$, where the differential on the latter is degree-increasing,
we have
\begin{align*}
T(H_i(LX))&\cong T(H^{i+1}(LX))\\ &\cong T(HH_{i+1}(C^*(X),C^*(X)))\\ &\cong T(HH^{i}(C^*(X),C^*(X)^\vee)).
\end{align*}

The free parts are of course identified in the same degree, so
$$H_i(LX)\cong HH^{i}(C^*(X),C^*(X)^\vee).$$

Finally, the $C^*(X)$-bimodule quasi-isomorphism $C_*(X)\to C_*(X)^{\vee\vee}=C^*(X)^{\vee}$ induces an isomorphism
$$HH^{i}(C^*(X),C_*(X)))\cong HH^{i}(C^*(X),C^*(X)^{\vee}),$$ and we are done.


问:
Deconstructing the numerator in derivatives
I'm trying to deconstruct $\frac{d(\sqrt{x^2+y^2})}{dt}$ into possibly $\frac{dx}{dt}$ and $\frac{dy}{dt}$, but I have no idea where to start or what to search.

答:
The chain rule says $\frac{d f(g(t))}{dt}=f'(g(t))g'(t)$.  In this case, take $f(z)=\sqrt z$ and $g(t)=x(t)^2+y(t)^2$ so that
$$\frac{d\sqrt{x(t)^2+y(t)^2}}{dt}=\frac1{2\sqrt{x(t)^2+y(t)^2}}\cdot\frac d{dt}(x(t)^2+y(t)^2)$$
and the rest follows pretty quickly with another application of the chain rule.  I would hesitate to call $f$ a numerator, I'd say it's the function which your taking the derivative of, not sure if there's a slick term like "integrand".


问:
Smooth functions with linearly independent gradients extend to coordinate charts (reference request)
This feels obvious, but I can't find a reference for it. Suppose I have a smooth $n$-dimensional manifold $M$, and I have $k$ smooth functions $f_1,\ldots,f_k \in C^\infty(M)$ such at some point $x \in M$, the gradients $\nabla f_1,\ldots,\nabla f_k$ are linearly independent at $x$ (and therefore in a neighbourhood of $x$). Can I find $n-k$ functions $f_{k+1},\ldots,f_n \in C^\infty(M)$ and a set $U \subset M$ containing $x$ such that $(U,(f_1,\ldots,f_n))$ forms a smooth coordinate chart for $M$ (i.e. is contained in the maximal atlas for $M$)?

答: As @Deane points out in the comments, this is an application of the canonical submersion theorem. It appears in e.g. Guillemin and Haine's Differential Forms, Theorem 4.2.14.


问:
Riemann sum with negative coefficient and positive literal

$ f(x)=x^2-1$ with an interval of $[0,1]$

i have this equation and made each step to solve this problem, but at the time of applying the special rule when $\sum_{k=1}^{n}k^2= \frac{n(n+1)(2n+1)}{6}$ but i think i messed up a value in my equations because i got a value that doesn't match with the one in my guidebook, here are my steps, please try to explain where i got wrong please, thank you.

here is my work since LATEX is pretty hard rn for me at the moment.

答:
$$ \int_0^1 f(x)dx=\lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right) $$
And
$$ \begin{aligned} \frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)&=\frac{1}{n}\sum_{k=1}^n\left(\frac{k^2}{n^2}-1\right) \\
&=\frac{1}{n^3}\sum_{k=1}^n k^2-1 \\
&=\frac{n(n+1)(2n+1)}{6n^3}-1
\end{aligned}$$
Letting $n\rightarrow +\infty$ gives $$ \int_0^1 f(x)dx=\frac{2}{6}-1=-\frac{2}{3} $$


问:
Classification of fourth-order PDE
I'm dealing with the following PDE:

\begin{equation} k\frac{\partial^2 w}{\partial x^2} + k\frac{\partial^2 w}{\partial y^2} - \frac{\partial^4 w}{\partial x^2 y^2} = 0 \end{equation}

I have unsuccessfully flipped through the entire Handbook of Linear PDE by Polyanin, but I can't seem to find a "name" nor a solution. The closest thing I found is the Rayleigh bar vibration equation,

\begin{equation*} \frac{\partial^2 w}{\partial x^2} - \frac{\partial^2 w}{\partial y^2} - \frac{\partial^4 w}{\partial x^2 y^2} = 0 \end{equation*}

Although my equation resembles a Laplace operator rather than a d'Alembert one. So far I haven't been able to find a suitable change of coordinates to convert the first equation into a more familiar one, but I'm wondering if someone with more experience may point me in some direction (or instantly recognize the equation as some well-known PDE).

答:
Your equation is equivalent to the system
$$
 kw-w_{yy} = v, \ \ \ kv-v_{xx} = k^2 w
$$
in the sense that a smooth solution of either one gives a smooth solution of the other. But the system seems somewhat unusual in that it is sort of two linked ODEs, where $x$ is a parameter in the equation that has $y$ derivatives, and vice versa. I think your equation is hyperbolic due to the 4th derivative, in spite of having the Laplacian terms.

Title: Classification of fourth-order PDE