问:
$\bar{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(x)}$?
Here the over line means algebraic closure. That is to say, if $E/K$ is a field extension, then $\bar{K}$ denotes the algebraic closure of $K$ in $E$. It is easy to imagine $\bar{\mathbb{Q}}$: all (complex) roots of rational polynomials. However, what is $\overline{\mathbb{Q}(x)}$ by definition exactly? It is confusing if I still use the definition "all roots of something"...

Thank you in advance for your contribution and help!

答:
The algebraic closure of a field $K$ is the smallest algebraically closed field $\bar{K}$ containing $K$; that is, $\bar{K}$ is algebraically closed, contains $K$, and if $E$ is any algebraically closed field such that $K\hookrightarrow E$, then also $\bar{K}\hookrightarrow E$. This definition can be applied with $K=\mathbb{Q}$ or with $K=\mathbb{Q}(x)$, the field of rational functions in 1 variable over $\mathbb{Q}$. You can think of $\overline{\mathbb{Q}(x)}$ as solutions in $y$ to equations of the form
$$f_n(x)y^n+\ldots+f_1(x)y+f_0(x) = 0,$$
where $f_i(x)$ are rational functions in $x$ over $\mathbb{Q}$. This means that the solution $y$ will also be a function in $x$, but not necessarily a rational one.

There is a theorem that states that an algebraically closed field is determined up to isomorphism by its transcendence degree over its prime subfield. The prime subfield of $K$ is the smallest subfield $k\hookrightarrow K$. The transcendence degree of $K/k$ is the cardinality of the largest subset of $K$ which is algebraically independent over $k$. In both your cases, the prime subfield is $\mathbb{Q}$. $\bar{\mathbb{Q}}$ is by definition algebraic over $\mathbb{Q}$, so it's transcendence degree is 0. $\overline{\mathbb{Q}(x)}$ is algebraic over $\mathbb{Q}(x)$, which clearly has transcendence degree 1 over $\mathbb{Q}$. Thus the transcendence degree of $\overline{\mathbb{Q}(x)}/\mathbb{Q}$ is still 1. Therefore, $\bar{\mathbb{Q}}$ and $\overline{\mathbb{Q}(x)}$ cannot be isomorphic.


问:
Particular case of Green’s theorem
Suppose we have $u(r)=\sum_{\lambda=1}^{\infty} a_{\lambda} u_{\lambda}(r) \quad 0 \leq r \leq a$ in this article Introduction to R-matrix theory in atomic
physics

they say that
$$\int_{0}^{a}\left[u_{\lambda} \frac{d^{2} u}{d r^{2}}-u \frac{d^{2} u_{\lambda}}{d r^{2}}\right] dr=\left[u_{\lambda}\frac{du}{d r}-u \frac{du_{\lambda}}{d r}\right]_{r=a}$$

using Green’s theorem.
What is this  Green’s theorem they are talking?

答:
One can only speculate, but perhaps they mean the 1D case of Green's first identity $\int_U (f \Delta g+ \nabla  f \cdot \nabla g) d V=\int_{\partial U} f \nabla g \cdot n dS$. In 1D this is (in the "indefinite" form) $\int  (f g''+f'g') dx=fg'$, so that

$$\int fg''-gf''dx=\left(\int fg''+f'g' dx\right)-\left(\int gf'' +f'g'dx\right)=fg'-gf'$$

This is of course an overkill in 1D, but maybe they were specializing from a higher dimensional case.

ContentLicense: CC BY-SA 4.0

Title: Particular case of Green’s theorem


问:
A question of Number Theory+Binomial Theorem
Let the sum of all divisors of the form $2^{p} \cdot 3^{q}$ (p,q are positive integers) of the number $19^{88} - 1$ be $\lambda$ . Find unit digit of $\lambda$

I noticed one fact that $2^{p} \cdot 3^{q}$ means any number which is not of form $6k\pm 1$.

Tried to expand it $(19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$ then found that $19^{11}+1$ has one factor as $20$ but this approach didn't work.

Then I tried to expand it using Binomial Theorem but it turned out be very lengthy process.

答:
We just need to find the largest factor of $19^{88}-1$ of the form $2^p3^q$. The rest shall follow from considering its factors.

Using the factorization $19^{88}-1 = (19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$, we first check the value of $p$. Taking mod $4$:

$$19^{88}+1 \equiv (-1)^{88}+1 \equiv 2 \equiv 19^{44}-1 \pmod 4$$

$$19^{11}+1 \equiv (-1)^{11}+1 \equiv 0 \pmod 4$$

$$19^{11}-1 \equiv (-1)^{11}-1 \equiv 2 \pmod 4$$

Now we check mod $8$ for $19^{11}+1$: $19^{11}+1 \equiv 361^5 \times 19 + 1\equiv 1^5 \times 3 + 1 \equiv 4 \pmod 8$.

This shows that the greatest power of $2$ that divides $19^{88}$ is $2 \times 2 \times 2 \times 4 = 2^5$.

Now for $q$, notice that:

$$19^{k}+1 \equiv 1^k+1\equiv 2 \pmod 3$$

the factor of $3^q$ comes from the factor $19^{11}-1$. Now we use the factorization:

$$19^{11}-1 = (19-1)(19^{10}+19^9+19^8+\dots +19 + 1)$$

The latter factor $\equiv 11 \pmod 3$. Hence the greatest power of $3$ that divides $19^{11}-1$ is $9$ (from $19-1=18$).

Hence the largest factor of $19^{88}-1$ of the form $2^p3^q$ is precisely $2^53^2$.


问:
Suggest an Antique Math Book worth reading?
I'm not a math wizard, but I recently started reading through a few math books to prepare myself for some upcoming classes and I'm starting to really get into it. Then I noticed a few antique math books at a used bookstore and bought them thinking that, if nothing else, they would look cool on my bookshelf. But as it turns out, I enjoy both reading and collecting them. I find myself constantly browsing used book stores, thrift stores, antique stores ect. looking for the next book to add to my library.

So do you know of an antique book that you found interesting, helpful, or historically relevant?

(Just some insight- some of the books I have that I like are: The Laws of Thought by George Boole; Mathematical Methods of Statistics by Harald Cramer; and Introduction to Mathematical Analysis by F.L. Griffin. I've also enjoyed reading online about probablity, logic, and math history. But any area of mathematics is fine as I'm still discovering which areas interest me.)

答:
I Second Euclid's Elements (Green Lion Edition is nice; Byrne's edition mentioned above is beautiful.

There's also a Dover edition of The Works of Archimedes.

Green Lion also has a nice publication of Apollonius' Conics that is more accessible than Dover's edition of Archimedes.

Although I have far to go in both books, I also second Hardy's Course in Pure Mathematics, as well as Inequalities.

I love Geometry and the Imagination by Hilbert which is mentioned above. It's not too hard to find the German Anshauliche Geometrie as well as the English edition.

And, lastly, Casey's A Sequel to the Elements of Euclid (late 19th century). U. Michigan Libraries has a nice reprint.

Some books on my wishlist would be (as above) Cauchy's Cours d'Analyse, original works of Möbius, Omar Khayyam's early mathematical explorations, Pappus, Ptolemy ...

I have looked for some of the books on my wishlist, but have sofar located only prohibitively expensive copies.

ContentLicense: CC BY-SA 3.0


问:
Prove that the set S is an orthogonal set
Let $V=\left\{\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in \mathbb{R}\right\}$ be the vector space of $2 \times 2$ real matrices endowed with the inner product
$$
\langle A, B\rangle=\operatorname{tr}\left(A^{T} B\right)
$$
for any $2 \times 2$ real matrices $A$ and $B .$ Let $S=\left\{\left[\begin{array}{cc}1 & 1 \\ -1 & -1\end{array}\right],\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right],\left[\begin{array}{cc}1 & -1 \\ 1 & -1\end{array}\right]\right\} .$

Prove that S is an orthogonal set and check if S is an orthonormal set.

Find an orthogonal basis for V containing Set S

I planned on using Gram-Schmidt process but I am stuck with the first step as I checked if the Set S is a basis by checking if the matrices are linearly independent.

$A=\left[\begin{array}{cccc}1 & 1 & 1 & 0 \\ 1 & -1 & -1 & 0 \\ -1 & -1 & -1 & 0 \\ -1 & 1 & -1 & 0\end{array}\right]$

$\operatorname{rref}(A)=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$

Which shows that the matrices are linearly dependent and can't be the basis. If it so, How can I make it such that the Set S can be a basis should I remove a one of the matrices or add? Any help would be appreciated. Thank you.

答:
There is no need for them to be a basis (though they are not, in fact, linearly dependent - I have no idea where that matrix $A$ came from, but it's not relevant - in particular, we have only three vectors). We just want to show that they are orthogonal. That is: we wish to show that, for any $A, B \in S$, we have $\langle A,B\rangle = 0$.

So, let's check that. We have

\begin{align*}\left\langle\begin{bmatrix}1 & 1 \\ -1 & -1\end{bmatrix},\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\right\rangle &= \mathbf{tr}\begin{bmatrix}1&1\\-1 & -1\end{bmatrix}^T\begin{bmatrix}1&-1\\-1&1\end{bmatrix} \\&=\mathbf{tr}\begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\\&=\mathbf{tr}\begin{bmatrix}2&-2\\2&-2\end{bmatrix}\\&= 0,\end{align*}
and similarly for the other pairs.

To check if it's orthonormal, you then need only check that each element is normal. For example, we have

\begin{align*}\left\langle\begin{bmatrix}1&1\\-1&-1\end{bmatrix},\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\right\rangle &= \mathbf{tr}\begin{bmatrix}1 & -1\\1 & -1\end{bmatrix}\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\\&=\mathbf{tr}\begin{bmatrix}2 &2 \\2 &2\end{bmatrix}\\&= 4\\&\neq 1\end{align*}

To extend to an orthogonal basis, you can first extend to any spanning set, say by appending $$\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}$$ to the end, then sift that down to a basis (which will consist of $S$ plus exactly one of those four matrices, call whichever matrix it is $X$, and finally orthogonalise by replacing $X$ by $$X - \mathbf{proj}\left(X,\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\right) - \mathbf{proj}\left(X,\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\right) - \mathbf{proj}\left(X,\begin{bmatrix}1&-1\\1&-1\end{bmatrix}\right)$$

where $$\mathbf{proj}(A,B) = \dfrac{\langle A,B\rangle}{\langle B,B\rangle}B$$


问:
Is there a closed-form expression for $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$?
I would like to ask if for $|x|<1$, we can express the product $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$ as a function $f(x)$. I tried to use Weierstrass factorization theorem, but without much success.

I would really appreciate a reference or a solution.

答: As already noticed in comments, the expression can be obtained from the infinite products for $\Gamma$ (either Euler's one, or Weierstrass's one): $$\Gamma(1+z)=\prod_{n=1}^\infty\frac{(1+1/n)^z}{1+z/n}=e^{-\gamma z}\prod_{n=1}^\infty\frac{e^{z/n}}{1+z/n},$$ and the "algebraic" $1-x/n^3=(1-x^{1/3}/n)(1-e^{2\pi i/3}x^{1/3}/n)(1-e^{-2\pi i/3}x^{1/3}/n)$, giving $$\prod_{n=1}^\infty\left(1-\frac{x}{n^3}\right)=\Big(\Gamma(1-x^{1/3})\Gamma(1-e^{2\pi i/3}x^{1/3})\Gamma(1-e^{-2\pi i/3}x^{1/3})\Big)^{-1}.$$ This easily applies to more general "rational infinite products", as outlined here.


问:
Show that $\prod_p1+\frac{1}{p^{4s}}+\frac{1}{p^{5s}}$ converges
I want to show that
$$\prod_p1+\frac{1}{p^{4s}}+\frac{1}{p^{5s}}$$
for the Re(s)$\geq\frac{1}{3}$. ( $p$ prime)
Should I compare it to the Riemann zeta function, although that only converges for Re(s)>1?

答:
Absolute convergence of the infinite product.

Note that $\sum n^a$ converges for $a<-1$ by the integral test.

Let $p_n$ be the $n$th prime.  Let $s>0$ and write $\sigma = \mathrm{Re }(s)$. As $n \to \infty$,
\begin{align}
p_n &\sim n\log n
\\
|p_n^{-4s}| &= p_n^{-4\sigma} \sim n^{-4\sigma}(\log n)^{-4\sigma}
\\
|p_n^{-5s}| &\sim n^{-5\sigma}(\log n)^{-5\sigma}
\\
|p^{-4s}+p^{-5s}| &\sim n^{-4\sigma}(\log n)^{-4\sigma}
\end{align}
This means that series
$$
\sum_n \left(p_n^{-4s}+p_n^{-5s}\right)
$$
converges absolutely if $-4\sigma<-1$, that is $\sigma>1/4$.  Thus the infinite product
$$
\prod_p (1+p_n^{-4s}+p_n^{-5s})
$$
converges absolutely if $\text{Re}(s)>1/4$.


问:
Checking if the mean obtained is within the interval $\mu_{ \bar X} ± 2\sigma_{ \bar X }$
I have the following question:

A soft-drink machine is regulated so that the amount of drink
dispensed averages $240$ milliliters with a standard deviation of $15$
milliliters. Periodically, the machine is checked by taking a sample
of $40$ drinks and computing the average content. If the mean of the
$40$ drinks is a value within the interval $\mu_{ \bar X}  ± 2\sigma_{
\bar X }$, the machine is thought to be operating satisfactorily;
otherwise, adjustments are made. The com- pany official found the mean
of $40$ drinks to be $\bar x = 236$ milliliters and concluded that the
machine needed no adjustment. Was this a reasonable decision?

The reason I am posting this is that I solved it in a different way than the solution provided and I want someone to verify my solution, I did the following:

I applied the central limit theorem since the central limit theorem as I think the central limit theorem tells how many standard deviations away from the mean, I got a value of $-1.68$ which is between $-2$ and $2$ so the decision is correct.

In the solution given they find the sample standard deviation, multiply it by $2$ and add/subtract it from the population mean to see if $236$ falls in this range, the decision was correct too but I am not very sure about my reasoning of the central limit theorem so I want someone to correct me if I am wrong and the right decision answer was just by luck.

答:
The two methods are arithmetically equivalent.

You found $-2 \le \dfrac{236-240}{\frac{15}{\sqrt{40}}} \le 2$ where the middl term is your $-1.6865 $

while the book found $-2\times \frac{15}{\sqrt{40}} \le 236-240 \le 2\times \frac{15}{\sqrt{40}}$, where the middle term is $-4$ and the outside tems $\pm4.7434$

so both concluded the difference was not significant


问:
Find equation of this plane
Find the equation of a plane that goes through the intersection of the planes $x_1+5x_2+x_3-5=0$ and $x_1-3x_3-2x_3=0$ and contains the point $(1,1,0)$ .

I calculated the cross product of the given plane to obtain the intersection and it gave me $-7x_1+3x_2-8x_3$, but I'm stuck there.

Title: Find equation of this plane

Body:
Given
$$\alpha:x_1+5x_2+x_3-5=0,\;\beta:x_1-3x_2-2x_3=0$$
Any linear combination of the two planes passes through the intersection line
$$\lambda(x_1+5x_2+x_3-5)+\mu(x_1-3x_2-2x_3)=0\tag{1}$$
plug the coordinates of the point $(1,1,0)$
$$\lambda(1+5+0-5)+\mu(1-3-0)=0\to \lambda=2\mu$$
substitute in $(1)$
$$2\mu(x_1+5x_2+x_3-5)+\mu(x_1-3x_2-2x_3)=0$$
$$2(x_1+5x_2+x_3-5)+(x_1-3x_2-2x_3)=0\to 3x_1+7x_2-10=0$$


问:
If the function preserves angle, then it is holomorphic
I'm reading some problem in Stein Shakarchi's complex analysis.

The question says,

Conversely, prove the following: suppose $f:\Omega\to \mathbb{C}$ is a
complex-valued function, that is real-differentiable at
$z_0\in\Omega$, and $J_f (z_0)\neq 0$. If $f$ preserves angles at
$z_0$, then $f$ is holomorphic at $z_0$ with $f'(z_0)\neq 0$.

I have trouble understanding the question. What is $J_f(z_0)$ and the meaning of real-differentiability? And why do they matter? Can you please explain some context of this question? Also, the question doesn't specify that $\Omega$ is a subset of complex plane, or even metric space. Do I assume that it is a subset of complex plane?

答:
What is $J_f(z_0)$ and the meaning of real-differentiability?

Any complex function $\Bbb C\to\Bbb C$ can be viewed as a function $\Bbb R^2\to\Bbb R^2$. You can ask whether this function is differentiable, in the conventional multivariable calculus sense. That's "real differentiability".

Being real differentiable at $z_0$ means that there is a linear map $A:\Bbb R^2\to\Bbb R^2$ such that
$$
f(z_0+z)\approx f(z_0)+Az
$$
for a particular formalized meaning of $\approx$. The matrix representation of this linear map is $J_f(z_0)$. It's called the Jacobian matrix, and it happens to consist of the four partial derivatives of $f$ at $z_0$ (in short, $\operatorname{Re}f$ differentiated with respect to $x$ and $y$, and $\operatorname{Im} f$ differentiated with respect to $x$ and $y$, where $z=x+iy$).

Being real-differentiable is a much less strict requirement than being holomorphic / analytic / complex differentiable. For instance, the complex conjugation map is real differentiable (its Jacobian matrix is $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$ everywhere). However, it is not holomorphic, as it doesn't preserve angles (it reverses them).

Essentially, a function is complex differentiable iff the linear map $A$ corresponds to multiplication by a complex number (that's what a linear map $\Bbb C \to \Bbb C$ is, so it meshes nicely with the conventional definition of "differentiable"), and this complex number is "the derivative". There are many, many linear maps $\Bbb R^2 \to \Bbb R^2$ that do not do this. You are asked to show that a non-zero linear map which preserves angles actually does correspond to multiplication by a (non-zero) complex number. That's a different way to view your problem.

And yes, I think it is safe to assume $\Omega \subseteq \Bbb C$. They really should've specified that, but unfortunately they didn't. Maybe they said so a paragraph or two earlier? This excerpt is clearly written as a follow-up to something else (with the "Conversely" at the start).


问:
How to factor $x^5-1$ for binary codes
Currently, I have that I can do:

$$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1) $$

Then I thought I could do:

$$ x^5-1 = (x-1)(x(x+1)(x^2+1)+1) $$

However I want to use the factorization to find all cyclic codes in $[5, k]$. For that, I need to find all irreducible factors of $ x^5-1 $. I am not seeing if I can actually reduce $x^4+x^3+x^2+x+1$ any further

答:
Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already
$$
x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ .
$$
To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. The irreducible factors in degrees one and two are $x$, $(x+1)$, and $(x^2+x+1)$. The factors in degree one are easily excluded. (Since $f(0)=f(1)=1$.) And the rest by division with rest of $f$ by $(x^2+x+1)$ is $x+1$, so the factor in degree two is also excluded.

A computer algebra system is useful in such cases, for instance sage delivers the factorization:

```<code>sage: R.<x> = PolynomialRing(GF(2))
sage: factor(x^5+1)
(x + 1) * (x^4 + x^3 + x^2 + x + 1)

sage: # note that x^2 + 1 is reducible...
sage: factor(x^2+1)
(x + 1)^2
</code>```

Title: How to factor $x^5-1$ for binary codes

Body:
Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already
$$
x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ .
$$
To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. The irreducible factors in degrees one and two are $x$, $(x+1)$, and $(x^2+x+1)$. The factors in degree one are easily excluded. (Since $f(0)=f(1)=1$.) And the rest by division with rest of $f$ by $(x^2+x+1)$ is $x+1$, so the factor in degree two is also excluded.

A computer algebra system is useful in such cases, for instance sage delivers the factorization:

sage: R.<x> = PolynomialRing(GF(2))
sage: factor(x^5+1)
(x + 1) * (x^4 + x^3 + x^2 + x + 1)

sage: # note that x^2 + 1 is reducible...
sage: factor(x^2+1)
(x + 1)^2


问:
Example of Lebesgue measure zero
Let $A=[a,b] \times \{c\}$, where $b>a.$ Then $\forall \epsilon >0$ $$ A\subset I_k = [a,b]\times\biggr[c-\frac{\epsilon}{2(b-a)},c+\frac{\epsilon}{2(b-a)}\biggr].$$

Now $\mu(I_k)=\epsilon.$ A has measure zero if $\forall\epsilon>0$, there exists compact intervals $I_k$ such that $A\subset \cup_{k=1}^{\infty} I_k$  and $\sum_{k=1}^{\infty}\mu(I_k)=\mu(I_k)<\epsilon.$ Can I now use the arbitrariness of the first $\epsilon'$ to claim that $\mu(I_k)=\epsilon'<\epsilon\enspace ?$, and thus giving $A$ zero measure. Supposedly this is correct, but both of these $\epsilon$ are arbitrary and this confuses me.

答:
You wish to prove that $A$ has measure $0$, which is the same as proving that for any $\varepsilon>0$, $\mu(A)\leq\varepsilon$. Take any $\varepsilon$ and take $$I_\varepsilon=[a,b]\times\left[c-\frac{\varepsilon}{2(b-a)},c+\frac{\varepsilon}{2(b-a)}\right]$$
Clearly $A\subset I_\varepsilon,\forall\varepsilon>0$ so $\mu(A)\leq\mu(I_\varepsilon)=\varepsilon$, and we just proved that $\mu(A)=0$.

So you don't need to pick two arbitrary $\varepsilon$ just one.

Title: Example of Lebesgue measure zero


答:
Comment:

The bound of this product can be found usig Weierstrassn inequality:

If $a_1, a_2, a_3, \ldots,a_n$ are real positive integers less than unity, and:

$S_n=(a_1+a_2+a_3+ \cdots+a_n)<1$

then:

$1-S_n<(1-a_1)(1-a_2)(1-a_3)  \cdots (1-a_n)<\frac 1 {1+S_n}$

Where we can let:

$a_n=\frac x {n^3}$

Title: Is there a closed-form expression for $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$?

Body:
I would like to ask if for $|x|<1$, we can express the product $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$ as a function $f(x)$. I tried to use Weierstrass factorization theorem, but without much success.

I would really appreciate a reference or a solution.


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答:
This is connected to some of the other answers, but I thought it might be worth mentioning to show the utility of approximation formulas.  The first two derivatives of $ \ f(x) = x^{1/2} \ $ at $ \ a = 1000 \ $ are  $ \ f'(1000) = \frac{1}{2} \cdot 1000^{-1/2} \ $ and $ \ f''(1000) = -\frac{1}{4} \cdot 1000^{-3/2} \ $ (we won't need these to be evaluated).  The Taylor series for  $ \ x^{1/2} \ $ about $ \ a = 1000 \ $ is then

$$f(x) \ = \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (x - 1000) \ - \ \frac{1}{4\cdot 1000^{3/2}} (x-1000)^2 \ + \ \ldots   $$

Hence,

$$999^{1/2} \ + \ 1001^{1/2} \ \approx$$

$$[ \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (-1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (-1)^2 \ + \ \ldots   $$

$$ + \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (1)^2 \ + \ \ldots \ ] $$

$$= \ 2  \cdot  1000^{1/2} \  - \ \frac{2}{4\cdot 1000^{3/2}}  \ - \ \ldots \ , $$

with all the terms in odd powers of $ \ (x-1000) \ $ cancelling and all of the higher-order terms in even powers also being negative.  Because the two square-roots are being added, it is not sufficient to only carry this calculation to linear terms ("first-order") ; this is hinted at by the fact that OP found by calculator how close the two values are.

ADDED: The linear terms are sufficient, though, to show that $ \ \sqrt{1001} \cdot \sqrt{999} < 1000 \ . $


问:
Calculate with residue classes in $\mathbb{Z}/{7\mathbb{Z}}$
I want to calculate the residue class in $\mathbb{Z}/7\mathbb{Z}$, when $a = \{ b ~|~ b \equiv_7 a\}$.

Is my following calculation right?

$\bar{2} \cdot \bar{3} + \bar{4} \cdot \bar{3} = \bar{6} + \bar{12} = \bar{18} - \bar{14} = \bar{4}$

This feels like a simple calculation but I have trouble understanding this part " when $a = \{ b ~|~ b \equiv_7a\}$.  "

Ty in advance.

答:
I guess this definition wants to introduce the bar notation, i.e.
$$\bar a\,:=\,\{b\in\Bbb Z:b\equiv_7 a\}\,.$$
And the important thing is that the operations $+,-,\cdot$ are well defined on these elements, meaning that $a\equiv_7a',\ b\equiv_7b'\implies a\cdot b\equiv_7a'\cdot b'$, and similarly for the other operations.

So it doesn't matter which representatives you execute the operation on, you will always end up in the same residue class.

Your calculations seem correct.


答:
The answer to the question in the title is no because $\mathbb{Q}(x)$ cannot be embedded into $\overline{\mathbb{Q}}$:

If $\alpha \in \overline{\mathbb{Q}}$, then $\mathbb Q(\alpha)$ has finite dimension over $\mathbb Q$. But $\mathbb{Q}(x)$ has infinite dimension over $\mathbb Q$.

Title: $\bar{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(x)}$?

Body:
Here the over line means algebraic closure. That is to say, if $E/K$ is a field extension, then $\bar{K}$ denotes the algebraic closure of $K$ in $E$. It is easy to imagine $\bar{\mathbb{Q}}$: all (complex) roots of rational polynomials. However, what is $\overline{\mathbb{Q}(x)}$ by definition exactly? It is confusing if I still use the definition "all roots of something"...

Thank you in advance for your contribution and help!


问:
Finding the value of this infinite product
The following question is from my analysis assignment and I was unable to solve it.

Find the value of $\prod_{n=1}^{\infty}(1+ 1/n^2)$ .

I tried to do some algebraic manipulation but I was not sucessful in that.Also, If the terms doesn't appear to me to cancel in a fashion.

I am at loss of ideas so I thought I should post here.

答:
$$\prod_{k=1}^{\infty} \left(1+\frac{z^2}{k^2}\right)=\frac{\sinh \pi z}{\pi z}.$$ Put $z=1$ to get $$\prod_{k=1}^{\infty} \left(1+\frac{1}
{k^2}\right)=\frac{\sinh \pi}{\pi}$$

Edit: An Elementry Proof:

$$F_n(x)= x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)......(x-n\pi)(x+n\pi)$$ $\sin x$, is finite and has infinitely many zero. $F_n(x)$ has $2n+1$ zeros but as $n\to \infty$ $\lim_{n\to \infty} F_n(x)\to \infty$, so $F_n$ cannot represent $\sin x$. But $$ G_n(x)=x(1-\frac{x}{\pi})(1+\frac{x}{\pi}) (1-\frac{x}{2\pi})(1+\frac{x}{2\pi}).....(1-\frac{x}{n\pi})(1+\frac{x}{n\pi}).~~~~~(1)$$ So $G_n$ can represent $\sin x$ when $G_n$ converges as $n\to \infty$, it also gives $\frac{\sin x}{x}=1$ when $x \to 0$. Thus Eq. (1) gives the correct representation of $$\frac{\sin x}{x}=\prod_{n=1}^{\infty} \left(1-\frac{x^2}{n^2\pi^2}\right) \implies \frac{\sinh \pi z}{ \pi z}=\prod_{n=1}^{\infty} \left(1+\frac{z^2}{n^2}\right) $$
Here we took $x=iz.$


问:
Let $g$ maps from a unit disk $D$ to itself with the modulo of $g(z) \leq 1/2$. Prove that $|g(z)-g(0) \leq |z|$ for all $z \in D$.
I'm still confused with this question.  Let $g$ maps from a unit disk $D$ to itself with the modulo of $g(z) \leq 1/2$.  Prove that $|g(z)-g(0) \leq |z|$ for all $z \in D$.

What I did is this

As $g(z)$ is analytic in $D$, $g(z)$ is continuous and differentiable in $D$.  Then
$\left|\frac{g(z) - g(0)}{z-0} \right| \leq \epsilon$.  Choose $\epsilon = 1/2$.  Then we have

$|g(z) - g(0)| \leq |z-0|\epsilon$

I don't know how to continue with this. I just had a feeling it might be of $\epsilon = 1/2$ but I am confused with the proof now.  Can you help me? I would much appreciate that.  Thanks.

答: Let $h(z)=g(z)-g(0)$. Then $|h(z)| \leq 1$ so   $h$ maps the unit disk to its closure and vanishes at $0$. By Schwarz Lemma we get $|h(z)| \leq |z|$.


问:
What is the derivative of $x \times y$ with respect to $x$
In physics class, we saw that for an unit vector $e$, and a random vector $p$, $$\frac{\partial}{\partial \vec x}\hat e \cdot(\vec x \times \vec p) = \hat e \times \vec p.$$ It seems to be taken for granted and I don't know how to reach the solution. Is there a general formula for this? And how can we derive it?

答:
What you seem to mean is that if we fix two 3D vectors $\vec{v}$ and $\vec{p}$ and define the function of a 3D vector $\vec{x}$ by $f(
\vec{x})= \vec{v}\cdot (\vec{x}\times \vec{p})$ then we have for the gradient of $f$ the formula $\vec{\nabla} f(\vec{x})=\vec{p}\times \vec{v}$ (note the ordering!).
This is indeed true, the high-brow reason being that since $f$ is linear, the (row vector) derivative of $f$ is $f$, meaning

$$\vec{\nabla}  f\cdot \vec{\delta}=Df(\vec{\delta})=f(\vec{\delta})=
\vec{v}\cdot (\vec{\delta}\times \vec{p})=\vec{\delta}\cdot (\vec{p}\times\vec{v})$$

for all vectors $\vec{\delta}$ and so indeed $\vec{\nabla}  f(\vec{x})=\vec{p}\times \vec{v}$, a constant vector field.

You can also deduce this from the (corrected) equality for $\hat{e}$ that you have by writing $\vec{v}=|\vec{v}|\hat{v}$ and observing that compared with formulas for $\hat{v}$, both the left and the right hand side get multiplied by the same constant $|\vec{v}|$ and so remain equal.


问:
Why is boundary information so significant? -- Stokes's theorem
Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?

I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?

答:
When a function has zero derivative within a volume, its values are determined by sources outside the volume, which are measured only by values of the field on the boundary.

Got that?  Here's the mathematical explanation.

There are many functions $f$ that obey the basic differential equation $\nabla F = 0$.  We could be talking about a scalar field, or if we extend the idea of the "gradient" to act on vector fields, then this condition is that the field is both divergenceless and curlless.  Let's not fret on how the gradient can be extended in this fashion--that's a lot of tensor stuff that we don't really need to worry about.  Let's just keep in mind that this is the condition being imposed.

We often say that equations of the type $\nabla F = j$ represent a field $F$ that permeates space with a source $j$.  Let's understand first how that source $j$ determines the field.

To do that, we first need the Green's function for $\nabla$.  We'll call it $G$, and it obeys $\nabla G = \delta$, the Dirac delta function.

This is where the generalized Stokes theorem comes in.  It tells us that

$$\oint_{\partial M} G(r-r') \, dS' \, F(r') =- \int_M \delta(r-r') \, dV' \, F(r') + (-1)^{n-1} \int_M G(r-r') \, dV' \, j(r')$$

Or, equivalently,

$$F(r) = (-1)^{n-1} \oint_{\partial M} G(r-r') \, \hat n |dS'| \, F(r') + \int_M G(r-r')  j(r') \, |dV'|$$

It tells that, at any point $r$, the function $F(r)$ is determined by the values on a boundary plus the values of its derivative throughout the volume.

Moreover, only when that derivative is zero is the function determined entirely by its values on the boundary.

In physics, this is common in electromagnetism problems: there may be no currents in a given volume, but there is some source outside the volume instead.  These produce values of the field on the boundary, which in turn affect values within the volume.

Edit: In short, if you know a field's derivative everywhere, you should be able to reconstruct the field itself.  Stokes' theorem allows us to divide that information into two categories: sources outside a given volume, and sources inside.  Information from sources outside is entirely captured by the surface integral; information from sources inside has to be computed through the volume integral.

Functions entirely determined by their surface values are those that, in particular, have no sources within the volume.


问:
Min value of $a\sec \theta +b\cos \theta$
Min value of$$a\sec \theta +b\cos \theta$$ for $$\theta \in (-90^\circ,90^\circ) $$

I have trouble in approaching this problem .Could only a method/hint be given ? I would to try it myself.

答:
You are looking for the minimum value of function
$$f(\theta)=a\sec (\theta) +b\cos (\theta) \qquad \text{with}\qquad -\frac \pi 2 \leq \theta \leq \frac \pi 2 $$ Its derivative is

$$f'(\theta)=a \tan (\theta ) \sec (\theta )-b \sin (\theta )=\sin(\theta )\big[a \sec ^2(\theta )-b \big]$$

So, beside the trivial $\theta$, for other roots we need $a\,b >0$ and in such a case $$\theta_*=\sec ^{-1}\left(\sqrt{\frac{b}{a}}\right)$$ (remember that the function is even). So, now we have
$$f(0)=a+b \qquad \text{with} \qquad f''(0)=a-b$$
$$f(\theta_*)=2\sqrt{ab} \qquad \text{with} \qquad f''(\theta_*)=2(b-a) \sqrt{\frac{b}{a}} $$
So, by the second derivative test, if $a > b$ the minimum value is $(a+b)$ at $\theta=0$ and if $a < b$ the minimum value is $2\sqrt{ab}$ at $\theta=\theta_*$.


问:
Why the moment-generating function does not always exist
Why the moment-generating function $M(t)=E\left(e^{tx}
\right)$
does not always exist while the characteristic function $\varphi(\epsilon)=E\left(e^{i\epsilon x}\right)$
exists for any random variable X?

答:
Counter example is Cauchy distribution.

moment generating funciton does not exist , but characteristic function exists.

Since $E(e^{tx})$ is not absolutely convergent


问:
Spectrum of operator $T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$ defined by $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$
What is the point, continuous and residual spectrum of the following operator?

$T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$, $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$, where $\text{dom}(T) = \left\{(x_{n})_{n\in \mathbb{N}} \in \ell^{2} |
 (nx_{n})_{n\in \mathbb {N}} \in \ell^{2} \right\}$

My thoughts so far: Since $T$ is self-adjoint it must be that $\sigma_{r}(T) = \emptyset$. Now considering the point spectrum, I tried to find the eigenvalues of $T$: $Tx = \lambda x \iff (n-\lambda) x_{n} = 0$. But then I would have $x_{n}=1 $ for $n = \lambda$ and $x_{n}=0$ otherwise, so the corresponding eigenvectors are not in $\text{dom}(T)$, right? So $\sigma_{p}(T) = \emptyset$? Can anyone provide hints for $\sigma_r(T)$ or $\sigma_c(T)$, please?

答:
You are correct that $T$ is self-adjoint and thus $\sigma_r(T) = \emptyset$ (and also $\sigma(T) \subset \mathbb R$).
Your reasoning for $\sigma_p(T) = \emptyset$ is only partially correct (as the other answer indicates) because it only works if $\lambda \in \mathbb N$ (otherwise $n = \lambda$ doesn't work!).
But the sequence
$$
x_n = \begin{cases}
1, & \text{for } n = \lambda, \\
0, & \text{else.}
\end{cases}
$$
is in $\text{dom}(T)$, as
$$
n x_n = \begin{cases}
\lambda, & \text{for } n = \lambda, \\
0, & \text{else.}
\end{cases}
$$
and thus
$$
\| (n x_n)_{n \in \mathbb N} \|_{\ell^2}
= \sqrt{\sum_{n = 0}^{\infty} (n x_n)^2}
= \lambda
< \infty.
$$
This shows that $\mathbb N \subset \sigma_p(T)$ and equality if you argue that for $\lambda \not\in \mathbb N$ there is no solution to $(n - \lambda) x_n = 0$ for all $n \in \mathbb N$ and the solution above is the only one for $\lambda \in \mathbb N$.

We have
$$
\sigma_c(T)
:= \{ \lambda \in \mathbb R: T - \lambda \text{ injective}, \text{ ran}(T - \lambda) \subsetneq \ell^2 \text{ dense} \}.
$$
Inorder for $T - \lambda$ to be injective, $(T - \lambda) x = ((n - \lambda) x_n)_{n \in \mathbb N} = 0$ has to imply $x = 0$.
We have seen above that this is true for all $\lambda \in \mathbb R \setminus \mathbb N$.


问:
How to find the norm of a vector if the definition of the inner product is not given?
I'm not really sure with this question because whenever these kinds of problems appears, it always comes with an integrals definition of the inner product but this time we are only given this:

$$ x(t) = x_{1}t^{2} + x_{1}t + x_{0} $$
$$ y(t) = y_{1}t^{2} + y_{1}t + y_{0} $$

in P2[t], defined
$$ <x(t),y(t)> = x_{2}y_{2} + x_{1}y_{1} + 3x_{0}y_{0} $$

Then the question is that if
$ x(t) = t^{2}-2, y(t) = t+1,s(t)= 3t^{2} -t$

compute for $||x(t)||$. What do I do with the the definition of $<x(t),y(t)>$ to solve this probem? Or i dont think for this problem? I'm thinking of writing x(t) in coordinate vectors by using a standard basis for P2[t] then from that solve the length using this:
enter image description here

Edit:  I just realized the image I attached above is the dot product.

答:
By the definition of the norm of an inner-product space,

$\|p(t)\| = \sqrt{\langle p(t),p(t)\rangle}$.


问:
Compute the volume of the solid bounded by $x=0, x=\frac{\pi}{2}, z=0,z=y, y=\cos(x).$

Compute the volume of the solid bounded by $x=0, x=\frac{\pi}{2}, z=0,z=y, y=\cos(x).$

I'm not sure if I'm mistaken here, but isn't this just $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{y}\int_{0}^{\cos(x)} \ dy \ dz \ dx = \int_{0}^{\frac{\pi}{2}}\int_{0}^{y} \cos(x) \ dz \ dx = \int_{0}^{\frac{\pi}{2}}y\cos(x) \ dx = y$$

this doesn't seem right to me. Is there a problem regarding the limits of $y$?

The picture of the $xy$-plane looks like a $\cos(x)$ bounded between $0$ and $\frac{\pi}{2}$

答:
Since $z$ depends on $y$ and $y$ depends on $x$, you should integrate first on $z$, so the integral should be
$$\int_0^{\pi/2}\int_{0}^{\cos(x)}\int_{0}^ydz\,dy\,dx=\int_0^{\pi/2}\int_{0}^{\cos(x)}y\,dy\,dx=\int_0^{\pi/2}\frac{\cos(x)^2}{2}dx=\pi/8.$$


问:
For which x does $\sum\limits_{i=0}^{n}x^i=O(n)$ hold?
I'm stuck with this exercise: I have to find for which $x$ the estimate $\displaystyle\sum\limits_{i=0}^{n}x^i=O(n)$ holds.

It seems intuitive to me that this must be the case for all $x \in (0,1)$ but proving this seems to be beyond my abilities.

I tried some different approaches like the usual $\displaystyle \lim\limits_{x\to\infty}\frac{f(x)}{g(x)} = \text{some finite value}$ with $f(x)$ the formula for the partial sums. I tried the same thing with l'Hôpital's rule. I also tried to argue that the highest exponent of the sum must be $x^n$ and therefore I can just say that this holds for all $0 < x < 1$, but that doesn't seem very convincing to me.

I am out of ideas how to solve this problem and everything I try feels wrong to me, I hope someone in this community can help me.

答:
Hint: If $0\leq x\leq 1 $ then $x^i\leq 1$ so that $$\left|\sum_{i=0}^n x^i\right|\leq n+1.$$

For $x>1$ notice this is a geometric series and that $$\sum_{i=0}^n x^i= \frac{x^{n+1}-1}{x-1}.$$  Then we are then comparing $x^{n}$ to $n$.


问:
Exercise 2.4 Fulton's Algebraic Curves
I am looking at exercise 2.4 in William Fulton's "Algebraic Curves". It asks to prove that if $X\subset \mathbb{A}^n$ is nonempty affine variety, then the following are equivalent

$X$ is a point

$\Gamma(X)=k$

$dim_k\Gamma(X)<\infty$

I have a problem with 3 implying 1. Suppose $X$ is the union of two points, say $1,2\in\mathbb{A}(\mathbb{C})$. Then $I(\{1,2\})=((x-1)(x-2))$ so

$\Gamma(X)=\mathbb{C}[x]/I(X)=\mathbb{C}[x]/((x-1)(x-2))$, and I am pretty sure that this has finite dimension as a $\mathbb{C}$ vector space. But yet our original variety is not a point. Does the question mean to have $X$ is a finite union of points instead? Though that you make number 2 not quite correct.

Thanks for the help

答: By corollary 4 to Hilbert's Nullstellensatz, $V(I(V))=V$ is a finite set, say $\{P_1,\dots,P_r\}$. Hence $V=\{P_1\}\cup \dots  \cup \{P_r\}$. Since $V$ is an affine variety, $r=1$, and thus $V$ is a point.</p>

Title: Exercise 2.4 Fulton's Algebraic Curves


问:
Uniformaly integrability of submartingale and stopped submartingale
I learnt a proposition on my probability course:

If $\{X_n\}$ is a uniformly integrable $(F_n)$ submartingale, and $T$ is a $(F_n)$ stopping time, then $X_{\min\{n,T\}}$ is uniformly integrable $(F_n)$ submartingale and $E(|X_T|)<\infty$.

I propose a proof:

Since $\{X_n\}$ is a uniform integrable $(F_n)$ submartingale and $T$ is a $(F_n)$ stopping time, we have $X_{\min\{n,T\}}$ is a $(F_n)$ submartingale (which is integrable).

Therefore, $E(|X_T|)\leq \liminf_{n\rightarrow\infty}E(|X_{\min\{n,T\}}|)<\infty$ by Fatou's lemma.

$\sup_n\int_{|X_{\min\{n,T\}}|>M}|X_{\min\{n,T\}}|dP\leq\int_{\{|X_T|>M}|X_T|dP+\sup_n\int_{|X_{n}|>M}|X_n|dP<\infty$ by Dominated convergence theorem and uniformly integrability of $X_n$.

(By DCT, $\int_{\{|X_T|>M}|X_T|dP\rightarrow 0$ as $M\rightarrow\infty$.)

My questions are:

Is this proof correct?

It seems that to show $X_T$ is integrable, we do not need $X_n$ is uniformly integrable.

答: To answer 2), I believe you need to use the fact that $X_n$ is uniformly bounded to say that $\lim \inf E(|X_{\min\{n,T\}}|) < \infty$. I do this below following a template from my course. The rest looks good to me! You actually helped me for c), so thanks!


问:
How do I evaluate this gamma function integral?
Is the following statement true? If so, can someone help me see why this is the case?

$\int_0^T x^k \exp(-x/x_0) dx = \Gamma(k + 1)x_0^k$

Body:
$$I=\int_0^T x^k\,e^{-\frac{x}{x_0}} dx=x_0^{(k+1)}\int_0^{\frac T{x_0}} t^k\,e^{-t}\,dt$$
$$I=x_0^{(k+1)} \Big[\Gamma (k+1)-\Gamma
   \left(k+1,\frac{T}{x_0}\right)\Big]$$


问:
$ 3^2 + 4^2 = 5^2 $. Are there more of these equations?
I found this equation and it actually makes sense once you figure it out:
$$ 3^2 + 4^2 = 5^2 $$
$$ 9 + 16 = 25 $$
$$ 25 = 25 $$

Are there any more of these kinds of equations?

Here's a list I made of all I can find:

$ 3^2 + 4^2 = 5^2 $

$ 12^2 + 5^2 = 13^2 $

答: There is an infinite amount of such numbers, for example for any integer $n$ we have that $(3n)^2+(4n)^2=(5n)^2$, but there is also an infinite amount of elements $(x,y,z)\in\mathbb Z^3$ such that $\text{gcd}(x,y,z)=1$ and $x^2+y^2=z^2$. You can find more on Wikipedia.


答:
These numbers are called Pythagorean triples, as they satisfy Pythagoras Theorem. More generally, they are solutions of the Diophantine equation $a^2 + b^2 = c^2$. Here are a few more:

$8^2 + 15^2 = 17^2$

$24^2 + 7^2 = 25^2$

$40^2 + 9^2 = 41^2$

A larger list can be found on Wikipedia. There are a lot of formulae to generate Pythagorean triples. According to Euclid's formula, the numbers
$$a = m^2 - n^2, b = 2nm, c = m^2 + n^2$$
form a pythagorean triple $a^2 + b^2 = c^2$ for all combinations of $m$ and $n$.

Note that multiples of $a, b, c$ i.e. $ka, kb, kc$ also satisfy the equation. When $a, b, c$ have a highest common factor of 1, the triple is called a primitive pythagorean triple.

You can find out more about pythagorean triples in the mentioned links.

Title: $ 3^2 + 4^2 = 5^2 $. Are there more of these equations?

Body:
I found this equation and it actually makes sense once you figure it out:
$$ 3^2 + 4^2 = 5^2 $$
$$ 9 + 16 = 25 $$
$$ 25 = 25 $$

Are there any more of these kinds of equations?

Here's a list I made of all I can find:

$ 3^2 + 4^2 = 5^2 $

$ 12^2 + 5^2 = 13^2 $


问:
Discrete mathematics show anisymmetry
We let $(A,\leq)$ be an ordered set. We define a relation $\preceq$ on AxA by:
$$(a_1,a_2)\preceq(b_1,b_2) <=>[(a_1\neq b_1) ∧(a_1\leq b_1)]∨ [(a_1=b_1)∧(a_2\leq b_2)]$$
I think we have that $(a_1\neq b_1) ∧(a_1\leq b_1)->a_1<b_1$. So we can reduce it to:
$$(a_1,a_2)\preceq(b_1,b_2) <=>[a_1<b_1]∨ [(a_1=b_1)∧(a_2\leq b_2)]$$
Then I have to show that $\preceq$ is an order-relation. I have shown that we have reflextivity and transitivity. But I got some problem to show that we got antisymmetry.

If I check for the second condition it works while:
$$((a_1=b_1)\wedge(a_2\leq b_2))\wedge((b_1=a_1)\wedge(b_2\leq a_2)) =>(a_1,a_2)=(b_1,b_2)$$
But if I check for the first condition I got:
$$(a_1<b_1)\wedge(b_1<a_1) => emty?$$
Can someone help me proving the antisymmetry?

答:
For antisymmetrie, consider the premise

$[(a_1=b_1)\wedge (a_2≤b_2)]\wedge [(b_1=a_1)\wedge (b_2≤a_2)]$.

Since $a_2\leq b_2 \wedge b_2\leq a_2$ is equivalent to $a_2=b_2$, it follows that $(a_1,a_2) = (b_1,b_2)$.

Title: Discrete mathematics show anisymmetry


答: The identity $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ shows you how Pythagorean triples arise.


问:
Putting 10 distinguishable people into 2 groups so that no group is empty
In how many ways can we put 10 distinguishable people into 2 groups so that no group is empty? Order does not matter in a group and groups are not named.

So I've done some research and people seem to say that this is a Stirling problem of the second kind.
This is the same as the number of ways to distribute n distinct balls into k identical boxes so that no box is empty = $S(n, k)$

In particular $S(n=10,k=2) = 511$

Is this correct? Also is there a more logical approach to this problem?

I think there that are $2^{10}$ ways to place $10$ distinguishable balls into $2$ boxes. However, since the boxes are indistinguishable, we are overcounting by $2!$. Therefore, $2^{10}/2 = 2^9$

Using inclusion-exclusion, we subtract the case where $1$ box has all the balls and the other has $0$. There is only $1$ way this can happen since the teams are indistinguishable and we don't care about order.

Therefore, $512-1=511$

答:
That is correct. Stirling Number of the second kind or the Principle of Inclusion Exclusion  is the right approach for such problems. But just consider this in your specific case and simpler cases like this can be solved without using Stirling Number of the second kind.

Every person has two choices of groups and so across $10$ people, there are $2^{10}$ choices. But given the groups are not distinguishable, the choices are $\frac{1}{2} \times 2^{10} = 512$.

Now there is only one way all of them form just one group and hence,

Number of ways to form two non-empty groups $ = 512 - 1 = 511$.


问:
Can you suggest me a good book for self-study of analytic geometry 1?
I'm stuyding mathematics alone, but I plan to enter in the university in the near future, I went to the university website and it suggests Analytic Geometry 1 as part of their curricula, the topics I'll need to cover are in this PDF File, sorry if I didn't translate it but I did so because I believe you won't have much trouble since there are a lot of cognates.

I'm aware that there may be other questions asking the same thing, but I'm afraid on the topics being different because I'm from Brazil (Maybe there's something more, something less, don't know). This book should also be adequate for self-study due to my current conditions. Thanks for your help.

答: Calculus and Analytic Geometry by George B. Thomas is a classic introduction. There have been a lot of editions, which do vary some in their content, so be careful if you buy a copy to buy one that has Analytic Geometry in the title. </p>


答:
This is not an answer, just a comment to supply more data.

I used Pari/GP to detect the polynomial-structure in your C-table, and find the following table containing additional info for the column $n=1$:

```<code>  m \ n|    1     2     3      4      5      6      7      8  |
    -  +    -     -     -      -      -      -      -      -  +
    1  |    2    34    98    194    322    482    674    898  |   // ????? see note
    2  |   34   194   482    898   1442   2114   2914   3842  |
    3  |   98   482  1154   2114   3362   4898   6722   8834  |
    4  |  194   898  2114   3842   6082   8834  12098  15874  |
    5  |  322  1442  3362   6082   9602  13922  19042  24962  |
    6  |  482  2114  4898   8834  13922  20162  27554  36098  |
    7  |  674  2914  6722  12098  19042  27554  37634  49282  |
    8  |  898  3842  8834  15874  24962  36098  49282  64514  |
    -  +  ...   ...   ...    ...    ...    ...    ...    ...  +
          ???
</code>```

This table is produced by the small Pari/GP function

Cpol(m,n='x)= (4*m*n)^2-16*m*n+2   \\ reproduces OP's table for m,n>=2

or, a bit more symmetric,

Cpol(m,n='x)= 4*(2*m*n-1)^2 - 2   \\ reproduces OP's table for m,n>=2

$\qquad \qquad$ (leaving n in the function-call indeterminate gives the $m$'th polynomial in $x$)

Surprisingly, the first row, (for $m=1$) does not agree with your first row; and the same might occur, if you would supply a first column with $n=1$ of your data (however looking numerically for solutions when $n=1$ agrees with the first column here).

I'm checking further to see what's going on with this.

update1

I have looked at the solutions for $n=1$ and $m \gt 1$ .

First, with a chosen $m$, I take, from sequential search, the first 6 solutions for $(x,y,z)$. We find, that assuming we have 2 interlaced subsequences, that subsequences $x_a$ and $x_b$ have a simple recursion rule; for $x_a$ and $x_b$ we have
$$x_{a,k+2}=C_{m,1}\cdot x_{a,k+1} - 1 \cdot x_{a,k} \\
  x_{b,k+2}=C_{m,1}\cdot x_{b,k+1} - 1 \cdot x_{b,k} \\\tag {1.x}$$

The empirical data agree with the $C_{m,n}$-values from my proposal.

For the interlaced $y$- and $z$- sequences we have to include a constant value $D_{y,m}$ and $D_{z,m}$ in the recursion formula:
$$ y_{a,k+2}=C_{m,1}\cdot y_{a,k+1} - 1 \cdot y_{a,k} + D_{y,m}\\
  y_{b,k+2}=C_{m,1}\cdot y_{b,k+1} - 1 \cdot y_{b,k} + D_{y,m}\\ \tag {1.y}$$
and
$$ z_{a,k+2}=C_{m,1}\cdot z_{a,k+1} - 1 \cdot z_{a,k} + D_{z,m}\\
  z_{b,k+2}=C_{m,1}\cdot z_{b,k+1} - 1 \cdot z_{b,k} + D_{z,m}\\ \tag {1.z}$$

The complete recursion can be done in a matrix-formula, where the recursion-depth can be formalized by an initial matrix $I$  and the multiplicatormatrix $M$ such that
$$ S_{h+1} = I \cdot M^h $$
gives the $h$'th iteration.

Solutions:

Let $m=2$.
The first few empirically found solutions for $(x,y,z)$ are
empirically the following sequences, obviously by 2 recursive intertwined  subsequences:

```<code>x=  1  2  23  64   781  2174  26531   73852
y=  1  3  33  91  1105  3075  37521  104443
z=  1  2  17  46   553  1538  18761   52222
</code>```

The matrix $M$ is
$$ M = \small{ \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&C_{m,1}\end{bmatrix}
= \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&34\end{bmatrix} }$$

The initmatrix $I$ is composed of the first two empirical values $[x_a,y_a,z_a]$ and $[x_b,y_b,z_b]$, and additionally prepended by the -at the moment- unknown constants $D$ taken as zero:
$$ I_{\text{init}}=\small  \begin{bmatrix}
0&x_{a,1}&x_{a,2} \\ D_y&y_{a,1}&y_{a,2}\\D_z&z_{a,1}&z_{a,2} \\
\hline
0&x_{b,1}&x_{b,2} \\ D_y&y_{b,1}&y_{b,2}\\D_z&z_{b,1}&z_{b,2}
\end{bmatrix} =
\begin{bmatrix}
0&1&23\\0&1&33\\0&1&17 \\ \hline
0&2&64\\ 0&3&91\\0&2&46
\end{bmatrix}
$$
If we compute $S_2 = I \cdot M$ we get a wrong result for $y_{a,3},z_{a,3},y_{b,3},z_{a,3}$ - but this differences are then simply the new constant values for $D$ and this gives then
$$ \small I= \begin{bmatrix}
0&1&23\\-16&1&33\\-24&1&17 \\ \hline
0&2&64\\ -16&3&91\\-24&2&46
 \end{bmatrix}$$

With this we can compute the set of solutions using $S_{h+1}=I \cdot M^h$ for $h=0..7$ (or arbitrarily more); I show this having the two subsets as submatrices:
$$\small \begin{bmatrix}
 o: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ \hline
 x_a: & 1 & 23 & 781 & 26531 & 901273 & 30616751 & 1040068261 & 35331704123 & \cdots \\
 y_a: & 1 & 33 & 1105 & 37521 & 1274593 & 43298625 & 1470878641 & 49966575153 & \cdots \\
 z_a: & 1 & 17 & 553 & 18761 & 637297 & 21649313 & 735439321 & 24983287577 & \cdots \\
 x_b: & 2 & 64 & 2174 & 73852 & 2508794 & 85225144 & 2895146102 & 98349742324 & \cdots \\
 y_b: & 3 & 91 & 3075 & 104443 & 3547971 & 120526555 & 4094354883 & 139087539451 & \cdots \\
 z_b: & 2 & 46 & 1538 & 52222 & 1773986 & 60263278 & 2047177442 & 69543769726 & \cdots
 \end{bmatrix}$$

Note that we could even insert negative values for $h$ to continue the matrix of solutions to the left side.

Let $m=3$.
The first few empirically found solutions for $(x,y,z)$ are
empirically the following sequences, obviously again by 2 recursive intertwined  subsequences:

```<code>x=    1  3  69  287  6761  28123
y=    1  4  85  352  8281  34444
z=    1  2  29  118  2761  11482
</code>```

The matrix $M$ is
$$ M = \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&98\end{bmatrix} $$

The initmatrix $I$ is made as before and this gives then
$$  I =  \small \begin{bmatrix}
 0 & 1 & 69 \\
 -48 & 1 & 85 \\
 -80 & 1 & 29 \\ \hline
 0 & 3 & 287 \\
 -48 & 4 & 352 \\
 -80 & 2 & 118
 \end{bmatrix}$$

With this we can compute the set of solutions using $S_{h+1}=I \cdot M^h$ for $h=0..7$ (or arbitrarily more); again I show this having the two subsets as submatrices:
$$\small \begin{bmatrix}
 o: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ \hline
 x_a: & 1 & 69 & 6761 & 662509 & 64919121 & 6361411349 & 623353393081 & 61082271110589 \\
 y_a: & 1 & 85 & 8281 & 811405 & 79509361 & 7791105925 & 763448871241 & 74810198275645 \\
 z_a: & 1 & 29 & 2761 & 270469 & 26503121 & 2597035309 & 254482957081 & 24936732758549 \\ \hline
 x_b: & 3 & 287 & 28123 & 2755767 & 270037043 & 26460874447 & 2592895658763 & 254077313684327 \\
 y_b: & 4 & 352 & 34444 & 3375112 & 330726484 & 32407820272 & 3175635660124 & 311179886871832 \\
  z_b: &2 & 118 & 11482 & 1125038 & 110242162 & 10802606758 & 1058545220042 & 103726628957278
 \end{bmatrix} $$

Title: An infinite family of generalized Pell equations

Body:
Let $m,n,x,y,z$ be positive integers such that $y=m(z-1)+1$ and $yz-1=n(y^2-x^2)$.

If you fix $m$ and $n$ then this is a generalized Pell equation in $x$ and $z$ (i.e., a quadratic Diophantine equation of the sort that can be solved by this website).

I have 2 specific conjectures, which I have checked computationally for $m,n\leq100$, $z\leq10^7$.

Conjecture 1: If $n\geq2$ then $x,y,z$ are all odd.

Conjecture 2: If $(m,n)\neq(1,1)$ then $z$ is indivisible by $4$.

A proof or counterexample of either of these conjectures would be much appreciated.

The table below gives the first four $(x,y,z)$ solutions for fixed $(m,n)$:

$$\begin{array}{c|c|c|c}&n=1&n=2&n=3\\\hline m=1&
\begin{array}{c}(1,1,1)\\(1,2,2)\\(1,3,3)\\(1,4,4)\end{array}&
\begin{array}{c}(1,1,1)\\(5,7,7)\\(29,41,41)\\(169,239,239)\end{array}&
\begin{array}{c}(1,1,1)\\(9,11,11)\\(89,109,109)\\(881,1079,1079)\end{array}\\\hline m=2&
\begin{array}{c}(1,1,1)\\(2,3,2)\\(23,33,17)\\(64,91,46)\end{array}&
\begin{array}{c}(1,1,1)\\(167,193,97)\\(32397,37409,18705)\\(6284851,7257121,3628561)\end{array}&
\begin{array}{c}(1,1,1)\\(439,481,241)\\(211597,231793,115897)\\(101989315,111723697,55861849)\end{array}\\\hline m=3&
\begin{array}{c}(1,1,1)\\(3,4,2)\\(69,85,29)\\(287,352,118)\end{array}&
\begin{array}{c}(1,1,1)\\(417,457,153)\\(200993,220177,73393)\\(96878209,106124761,35374921)\end{array}&
\begin{array}{c}(1,1,1)\\(1053,1117,373)\\(1215161,1288873,429625)\\(1402294741,1487358181,495786061)\end{array}
\end{array}$$

In the comments, Will Jagy observed that if $n\geq2$ then the $x$ values satisfy a linear recurrence $x_{n+2}=Cx_{n+1}-x_n$ (and you can do something similar for $y$ and $z$). The table below gives the values of $C$:

$$\begin{array}{c|c|c|c|c|c|c|c|c|c}&n=2&n=3&n=4&n=5&n=6&n=7&n=8&n=9&n=10\\\hline m=1&6&10&14&18&22&26&30&34&38\\\hline m=2&194&482&898&1442&2114&2914&3842&4898&6082\\\hline m=3&482&1154&2114&3362&4898&6722&8834&11234&13922\\\hline m=4&898&2114&3842&6082&8834&12098&15874&20162&24962\\\hline m=5&1442&3362&6082&9602&13922&19042&24962&31682&39202\\\hline m=6&2114&4898&8834&13922&20162&27554&36098&45794&\\\hline m=7&2914&6722&12098&19042&27554&37634&49282&&\\\hline m=8&3842&8834&15874&24962&36098&49282&&&\\\hline m=9&4898&11234&20162&31682&45794&&&&\\\hline m=10&6082&13922&24962&39202&&&&&\end{array}$$

Notice the symmetry!

Gottfried Helms observed the formula $C=4(2mn-1)^2-2$ for $m,n\geq2$.

When $m=1$, it looks like $C=4n-2$ for $n\geq2$.

This answer proves the $m=1$ case of conjecture 2, using methods from Diophantine approximation.

Body:
This is not an answer, just a comment to supply more data.

I used Pari/GP to detect the polynomial-structure in your C-table, and find the following table containing additional info for the column $n=1$:

  m \ n|    1     2     3      4      5      6      7      8  |
    -  +    -     -     -      -      -      -      -      -  +
    1  |    2    34    98    194    322    482    674    898  |   // ????? see note
    2  |   34   194   482    898   1442   2114   2914   3842  |
    3  |   98   482  1154   2114   3362   4898   6722   8834  |
    4  |  194   898  2114   3842   6082   8834  12098  15874  |
    5  |  322  1442  3362   6082   9602  13922  19042  24962  |
    6  |  482  2114  4898   8834  13922  20162  27554  36098  |
    7  |  674  2914  6722  12098  19042  27554  37634  49282  |
    8  |  898  3842  8834  15874  24962  36098  49282  64514  |
    -  +  ...   ...   ...    ...    ...    ...    ...    ...  +
          ???

This table is produced by the small Pari/GP function

Cpol(m,n='x)= (4*m*n)^2-16*m*n+2   \\ reproduces OP's table for m,n>=2

or, a bit more symmetric,

Cpol(m,n='x)= 4*(2*m*n-1)^2 - 2   \\ reproduces OP's table for m,n>=2

$\qquad \qquad$ (leaving n in the function-call indeterminate gives the $m$'th polynomial in $x$)

Surprisingly, the first row, (for $m=1$) does not agree with your first row; and the same might occur, if you would supply a first column with $n=1$ of your data (however looking numerically for solutions when $n=1$ agrees with the first column here).

I'm checking further to see what's going on with this.

update1

I have looked at the solutions for $n=1$ and $m \gt 1$ .

First, with a chosen $m$, I take, from sequential search, the first 6 solutions for $(x,y,z)$. We find, that assuming we have 2 interlaced subsequences, that subsequences $x_a$ and $x_b$ have a simple recursion rule; for $x_a$ and $x_b$ we have
$$x_{a,k+2}=C_{m,1}\cdot x_{a,k+1} - 1 \cdot x_{a,k} \\
  x_{b,k+2}=C_{m,1}\cdot x_{b,k+1} - 1 \cdot x_{b,k} \\\tag {1.x}$$

The empirical data agree with the $C_{m,n}$-values from my proposal.

For the interlaced $y$- and $z$- sequences we have to include a constant value $D_{y,m}$ and $D_{z,m}$ in the recursion formula:
$$ y_{a,k+2}=C_{m,1}\cdot y_{a,k+1} - 1 \cdot y_{a,k} + D_{y,m}\\
  y_{b,k+2}=C_{m,1}\cdot y_{b,k+1} - 1 \cdot y_{b,k} + D_{y,m}\\ \tag {1.y}$$
and
$$ z_{a,k+2}=C_{m,1}\cdot z_{a,k+1} - 1 \cdot z_{a,k} + D_{z,m}\\
  z_{b,k+2}=C_{m,1}\cdot z_{b,k+1} - 1 \cdot z_{b,k} + D_{z,m}\\ \tag {1.z}$$

The complete recursion can be done in a matrix-formula, where the recursion-depth can be formalized by an initial matrix $I$  and the multiplicatormatrix $M$ such that
$$ S_{h+1} = I \cdot M^h $$
gives the $h$'th iteration.

Solutions:

Let $m=2$.
The first few empirically found solutions for $(x,y,z)$ are
empirically the following sequences, obviously by 2 recursive intertwined  subsequences:

x=  1  2  23  64   781  2174  26531   73852
y=  1  3  33  91  1105  3075  37521  104443
z=  1  2  17  46   553  1538  18761   52222

The matrix $M$ is
$$ M = \small{ \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&C_{m,1}\end{bmatrix}
= \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&34\end{bmatrix} }$$

The initmatrix $I$ is composed of the first two empirical values $[x_a,y_a,z_a]$ and $[x_b,y_b,z_b]$, and additionally prepended by the -at the moment- unknown constants $D$ taken as zero:
$$ I_{\text{init}}=\small  \begin{bmatrix}
0&x_{a,1}&x_{a,2} \\ D_y&y_{a,1}&y_{a,2}\\D_z&z_{a,1}&z_{a,2} \\
\hline
0&x_{b,1}&x_{b,2} \\ D_y&y_{b,1}&y_{b,2}\\D_z&z_{b,1}&z_{b,2}
\end{bmatrix} =
\begin{bmatrix}
0&1&23\\0&1&33\\0&1&17 \\ \hline
0&2&64\\ 0&3&91\\0&2&46
\end{bmatrix}
$$
If we compute $S_2 = I \cdot M$ we get a wrong result for $y_{a,3},z_{a,3},y_{b,3},z_{a,3}$ - but this differences are then simply the new constant values for $D$ and this gives then
$$ \small I= \begin{bmatrix}
0&1&23\\-16&1&33\\-24&1&17 \\ \hline
0&2&64\\ -16&3&91\\-24&2&46
 \end{bmatrix}$$

With this we can compute the set of solutions using $S_{h+1}=I \cdot M^h$ for $h=0..7$ (or arbitrarily more); I show this having the two subsets as submatrices:
$$\small \begin{bmatrix}
 o: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ \hline
 x_a: & 1 & 23 & 781 & 26531 & 901273 & 30616751 & 1040068261 & 35331704123 & \cdots \\
 y_a: & 1 & 33 & 1105 & 37521 & 1274593 & 43298625 & 1470878641 & 49966575153 & \cdots \\
 z_a: & 1 & 17 & 553 & 18761 & 637297 & 21649313 & 735439321 & 24983287577 & \cdots \\
 x_b: & 2 & 64 & 2174 & 73852 & 2508794 & 85225144 & 2895146102 & 98349742324 & \cdots \\
 y_b: & 3 & 91 & 3075 & 104443 & 3547971 & 120526555 & 4094354883 & 139087539451 & \cdots \\
 z_b: & 2 & 46 & 1538 & 52222 & 1773986 & 60263278 & 2047177442 & 69543769726 & \cdots
 \end{bmatrix}$$

Note that we could even insert negative values for $h$ to continue the matrix of solutions to the left side.

Let $m=3$.
The first few empirically found solutions for $(x,y,z)$ are
empirically the following sequences, obviously again by 2 recursive intertwined  subsequences:

x=    1  3  69  287  6761  28123
y=    1  4  85  352  8281  34444
z=    1  2  29  118  2761  11482

The matrix $M$ is
$$ M = \begin{bmatrix} 1&0&1 \\ 0&0&-1\\0&1&98\end{bmatrix} $$

The initmatrix $I$ is made as before and this gives then
$$  I =  \small \begin{bmatrix}
 0 & 1 & 69 \\
 -48 & 1 & 85 \\
 -80 & 1 & 29 \\ \hline
 0 & 3 & 287 \\
 -48 & 4 & 352 \\
 -80 & 2 & 118
 \end{bmatrix}$$

With this we can compute the set of solutions using $S_{h+1}=I \cdot M^h$ for $h=0..7$ (or arbitrarily more); again I show this having the two subsets as submatrices:
$$\small \begin{bmatrix}
 o: & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ \hline
 x_a: & 1 & 69 & 6761 & 662509 & 64919121 & 6361411349 & 623353393081 & 61082271110589 \\
 y_a: & 1 & 85 & 8281 & 811405 & 79509361 & 7791105925 & 763448871241 & 74810198275645 \\
 z_a: & 1 & 29 & 2761 & 270469 & 26503121 & 2597035309 & 254482957081 & 24936732758549 \\ \hline
 x_b: & 3 & 287 & 28123 & 2755767 & 270037043 & 26460874447 & 2592895658763 & 254077313684327 \\
 y_b: & 4 & 352 & 34444 & 3375112 & 330726484 & 32407820272 & 3175635660124 & 311179886871832 \\
  z_b: &2 & 118 & 11482 & 1125038 & 110242162 & 10802606758 & 1058545220042 & 103726628957278
 \end{bmatrix} $$


问:
Sphere volume. Calculation formula from school textbook.
I had a question from the course of school stereometry: why is the volume of a ball calculated exactly according to the formula $V=\frac{4}{3} \pi R^3$, why is $4/3$ used?

I did not find the necessary information in the textbook and on the Internet.
I will be glad to have a good explanation.

答:
Below, see a half ball, a cone, and a cyclinder (all sideways and therefore looking like a half disk, a triangle, a rectangle).
The all have radius $r$ and height $r$.
Look at the plane at heihgt $h$ above the ground (dotted line).
The intersection with the cylinder is of course a circle of radius $r$ and area $\pi r^2$;
the intersection with th econe is a circle of radius $h$ and area $\pi h^2$;
the intersection with the half ball is a circle of radius $\sqrt{r^2-h^2}$ (use Pythagoras to find the radius) and area $\pi(r^2-h^2)$.
Hence - independently of $h$ - the coross section area of the half ball plus the cross section area of the cone equals the cross section area of the cylinder:
$$\pi(r^2-h^2)+\pi h^2=\pi r^2. $$

Convince yourself that this means that the same equation holds for the volumes:
$$ V_{\text{half sphere}} + V_{\text{cone}} = V_{\text{cylinder}}.$$
You certainly know that the volume of the cylinder is $\pi r^3$ and that the volume of the cone is $\frac 13\pi r^3$ (where the factor $\frac13$ comes in exactly the same way as it does for pyramids).
Therefore, the volume of the half sphere is $\frac23\pi r^3$ and the volume of the full sphere is
$$ \frac43\pi r^3.$$


问:
Inverse Laplace transform is not giving the same result with 2 different theorems
I'm trying to transform this image function back to its original :

$$ F(s) = \frac{5}{s^2 -4s-32} (I)$$

I tried first to use the convolution theorem :
$\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$

$=> 5 . \mathcal{F}^{-1}(\frac{5}{s^2 -4s-32}) = 5\mathcal{F}^{-1}(\frac{1}{(s-8)(s+4)}) = 5 e^{4t} (*)
$(Inverse Laplace transform)

With the partial fraction decomposition I got a different result :

$\frac{A}{s-8} + \frac{B}{s-4}   => A = 5/12 $   and  $ B=-5/12 $ which will lead to a different result
$ F(s) = \frac{5}{12(s-8)} -\frac{5}{12(s+4)}  $
=> $\mathcal{F}^{-1}$(II)$ $= $ \frac{5}{12}(e^{8t} - e^{-4t} ) (**)$(Inverse Laplace transform)

What am I doing wrong here ? Why are $(*)$ and $(**)$ not having the same result ?

答: You mistakenly write that f*g is just (f times g), when instead you meant to use $f*g = \int_0^t f(z)g(t-z) dz$.


问:
What's the point of removing $1$ from the list of primes and having an empty product in the fundamental theorem of arithmetic?
I was reading about how we stopped considering the number $1$ as prime since this would mean the fundamental theorem of arithmetic wouldn't apply any more (since we can repeat the number $1$ over and over again and then the factorization wouldn't be unique) however, this got me wondering about the unique factorization of $1$ itself and I found out that we call it the "empty product" but what is the point of this? If the point from removing $1$ from the list of the prime numbers was to not make it an exception while stating this theorem and some others but this added another definition to make up for it so it feels like there was no point in doing it in the first place. Is there any other reason why we removed $1$ from the list of primes and made this?

答:
There is actually a more fundamental reason why we exclude $1$. Primes are usually defined (in any integral domain $R$) as $p \in R$ such that whenever $p | ab$ then $p | a$ or $p | b$.

However, this definition always excludes units. Since units divide everything, every unit would be trivially a prime. This is because we generally want to aim for unique prime factorization. That is, for any $n \in R$ we want a unique list of distinct primes $p_1,..., p_k$ and positive integers $\alpha_1,..., \alpha_k$ such that

$$ n = \prod_{i=1}^k p_i ^{\alpha_i} $$

If $1$ was a prime, then we would have $1 = 1^2 = 1^3 = ...$ so we lose uniqueness. Unique prime factorization can be useful for many arguments and if we count primes as units we would always have infinitely many factorizations.


问:
Computing $\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^2}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx$

Compute the integral $$\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^²}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx$$

Transforming to spherical coordinates I have that $$\int_{0}^{2}\int_{0}^{2 \pi}\int_{0}^{\sqrt{4-r^2\sin^2\varphi}} r^2\cos^2 \varphi\ \cdot r^3 \ d\varphi \ d\theta \ dr$$

but this is very hard to compute, I feel like I'm missing something here? The additional $r$ to form the cubic comes from the Jacobian.

答:
What you are integrating over is 1/8th of a ball of radius 2. It will be easier to integrate over the entire ball and divide by 8 (since this function is "even").

$8\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^²}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx = $

$\int_{0}^{\pi / 2}\int_{0}^{2\pi}\int_{0}^{2}r^2 sin(\varphi)(r^2\cdot cos^2(\varphi))\cdot r^2  \ dr \ d\theta \ d\varphi = $

$\int_{0}^{\pi / 2}\int_{0}^{2\pi}\int_{0}^{2}r^6 sin(\varphi)\cdot cos^2(\varphi)  \ dr \ d\theta \ d\varphi = $

$\int_{0}^{\pi / 2}\int_{0}^{2\pi}\int_{0}^{2}r^6 sin(\varphi)\cdot (1-sin^2(\varphi))  \ dr \ d\theta \ d\varphi $.

It should be easy to end the calculation from here because it reduced to calcullating one dimentional integral over $\varphi$.


问:
Contradiction between OEIS and factordb.com
OEIS A014545 says 1+13494## is a prime number but factordb.com says it is composite, where n## is the product of first n primes on factordb. Which is correct?

edit:
Sorry if my question is not appropriate here. I found some small prime numbers of the form 1+n##, and looked them up in OEIS to find this number. Then I also looked it up in factordb and found that it says not a prime. I wondered if there was a mistake in the OEIS or factordb, so I asked this question. I want to know whether factordb or OEIS is correct.

答:
Accoridng to PFGW, the number is $3$-PRP. This does not prove the primality but chances stand very good.

```<code>PFGW Version 4.0.1.64BIT.20191203.Win_Dev [GWNUM 29.8]

145823#+1 is 3-PRP! (120.3835s+0.0233s)

Done.
</code>```

This number can be proven prime with the p-1-method. Perhaps someone does this.

UPDATE : The number passed a miller rabin test with $5$ random bases.

Body:
Accoridng to PFGW, the number is $3$-PRP. This does not prove the primality but chances stand very good.

PFGW Version 4.0.1.64BIT.20191203.Win_Dev [GWNUM 29.8]

145823#+1 is 3-PRP! (120.3835s+0.0233s)

Done.

This number can be proven prime with the p-1-method. Perhaps someone does this.

UPDATE : The number passed a miller rabin test with $5$ random bases.


问:
How many elements will be there in a group so that it has $4$ subgroups of order $3$?
I m thinking that a group of order $3$ will be of the form $\langle a \rangle=\{1,a,a^2=a^{-1} \}$, $ a^3=e$ therefore $a^2=a^{-1}$.
Therefore $\langle a \rangle=\{1,a,a^{-1} \}$.

Then there will be $3$ cyclic groups different then "$a$" ie "$b$", "$c$", "$d$" with no element same as any element of "$a$". So each group will have $2$ elements of order $3$. in total $4\cdot 2$ elements. Is this reasoning correct that all subgroups should be distinct?

答: Note that $S_4$ and $A_4$ have exactly $4$ subgroups of order $3$, the Sylow $3$-subgroups. So your conclusion that the group should have $8$ elements is not valid.


问:
AB = I <--> BA = I, proof not valid?
Ok so I'm right at the end of the first chapter in my book which means I just know some matrix/vector operations, what it means for a matrix to be invertible and what the transpose of a matrix is. Add to that some of the basic rules regarding the inverse and transpose of a matrix, such as $(AB)^{-1} = B^{-1}A^{-1}$

One questions asks me to prove

$AB = I \iff BA = I$

I googled but all I could find was proofs that used concepts I've never heard of, like determinants. The reason I'm asking for help is because I think the book's answer is wrong. Here it is for $AB = I \implies BA = I$:

$B = BI$

  $B(AB) = (BA)B$

  Post-multiply this $B = (BA)B$ by $B^{-1}$:

  $BB^{-1} = (BA)BB^{-1}$

  $I = (BA)I = BA$

  We have $BA = I$

He proves $BA = I \implies AB = I$ in a similar way.  The thing I am turning against is that it is assumed that $B$ is invertible, that $B^{-1}$ exists. The question never said anything about $B$ being invertible. But if $B^{-1}$ actually exists then no proof is needed since then by the definition of an invertible matrix, $A$ is the inverse of $B$ and $AB = BA = I$. So am I wrong or is the proof in the book not really correct? Or is it a valid assumption that $B$ is invertible?

答:
We assume that $AB = I$ and we want to prove that $BA$ will be $I$ as well. I'll prove by contradiction.

Let's assume that $AB = I$, but $BA = C \neq I$(a $C$ which is not the identity matrix). Then left-multiplying both sides by $A$ it gives $A(BA) = A(C)\,\,\, != A(I)$. Also since matrix multiplication is associative we can say $(AB)A = AC \,\,\,  != A$. Now, according to our initial assumption, $AB = I$, so we can say $IA = AC \,\,\, != A$, but it is $A\,\,\,  != A$ which is a contradiction, so the $C$ must be $I$.

Title: AB = I <--> BA = I, proof not valid?


问:
Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$
Can somebody help me with this.

Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$

I have no idea how to start with this.  This is my first time encounter with the problem.  I'm sorry.  If you could, I will appreciate this.

What I did by the way is to expand the product.  However, is there a way to do this in a way that it does not like doing the expanding classical form?

答: Your product is $\prod_n\frac{(n+1)/(n+2)}{n/(n+1)}$, which telescopes to $2$.


问:
Given a diagonalizable triangular matrix $A$, its eigenvector matrix $X$ is triangular
How would I prove that given a diagonalizable triangular matrix $A$, its eigenvector matrix $X$ is triangular?
I've tried using $AX=X\Lambda$ (where $\Lambda$ is the diagonal eigenvalue matrix) and various properties of matrix multiplication, but I still have no clue where to continue from any of my starting points. Help will be much appreciated! Thanks!

EDIT: $(A-\lambda I)x=0$ has also been tried by me now, but to no avail.

答:
An intuitive understanding of this problem

There is a hacky proof below but I think this intuition has more value.

Let's take a left-triangular matrix. Fist thing to note about an $n$-dimensional left triangular matrix: the result of applying this matrix to a $n$ dimensional point $p \in \mathbb{F}^n$ is equivalent to the cumulative result of applying the upper-left submatrices to $p$. ($\mathbb{F}$ here can be complex or real).

To see this, try this $\mathbb{F}^3$ matrix:

$A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{pmatrix}$

I take a point $p \in \mathbb{F}^3$. I can directly $Ap$ to find the result of this matrix action on $p$. But note the result:

$\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} x \\ x + 2y \\ 3x + 2 + z\end{pmatrix}$.

Instead of applying the matrix in one shot, I could start with the first dimension of point $p$ and apply the upper left $1 \times 1$ matrix. This gives the result of the $1$-st dimension of the full matrix!

Now I can move to two dimensions and apply the upper left $2 \times 2$ matrix to $p_2$. The first dimension of the existing result won't change. The 2nd dimension will also be available now.

Continue doing this and note that when you have $k$ dimensions of your result, adding a dimension to the matrix and to the point, only changes the $k+1$-th dimension.

This is because of the zeros which limit the matrix's action on the "higher dimensions."

Once this point is super-clear, you can see that at each new additional dimension, we can only have an additional eigenvector. The existing eigenvectors will remain the same. Because each additional dimension $k$ gives you a new eigenvector in $k$ dimensions. So we will have eigenvectors in each dimension from $1$ all the way to $n$.

Postscript: A geoemtric understanding

A triangular matrix is actually a shear action. The 2D case is the easiest to understand:

$\begin{pmatrix} 1 & 0 \\ 2 & 1\end{pmatrix}$

You can use this tool or this tool to understand how this matrix acts.

The shearing in $n$ dimension has one invariant plane in $n-1$ dimensions i.e. the points on this plane don't rotate. They may translate but only within this plane.

You can see how each additional dimension to the input matrix is actually an additional shearing action in the new dimension. So each additional dimension results in a $1+$ dimensional invariant plane.

Inductive proof

I have an inductive proof that could work. It's not very neat and it can definitely be improved on.

$A$ is triangular, so we know that the diagonal values of $A$ are the eigenvalues of $A$.

Now, let's say that $A$ is left triangular/lower triangular. We use the characteristic equation $A - \lambda I$ now. Since we know $\lambda = a_{ii}$ for $i \in [1,n]$ ($a_{ii}$ are the diagonal elements):

$\begin{pmatrix} a_{11} & 0 & 0 & \ldots \\ a_{21} & a_{22} & 0 & \ldots
\\
\vdots \end{pmatrix} \mathbf{v} = a_{ii} \mathbf{v}$

$\begin{pmatrix} v_1 a_{11} \\ v_1a_{21} + v_2a_{22} \\ v_1a_{31} + v_2a_{32} + v_3a_{33} \\ \vdots \end{pmatrix} = \begin{pmatrix} a_{ii} v_1 \\ a_{ii}v_2 \\ a_{ii}v_3   \end{pmatrix}$

We only need this final system of equations to infer a few things:

$v_1 a_{11} = a_{ii} v_1$ and $v_1 \neq 0$ only if $a_{ii} = a_{11}$. So apart from the eigenvalue of $a_{11}$ this component of the eigenvector will be zero

A similar line of reasoning for the second row: $v_1a_{21} + v_2a_{22} = a_{ii}v_2$; say $a_{ii} \neq a_{11}$, then we already know $v_1 = 0$. But now if we want $v_2 \neq 0$, we also need $a_{ii} = a_{22}$.

For any $a_{ii}$ where $i > 2$, $v_2$ = 0 (note that for $a_{11}$, $v_2 = \frac{v_1a_{21}}{a_{11} - a_{22}}$)

extend this line of reasoning for the rest of the $v_i$ and we will find eigenvectors with

$v_1 = 0$

$v_1 = v_2 = 0$

$v_1 = v_2 = \ldots = v_{n-1} = 0$

We can do a similar analysis for upper triangular/right triangular $A$, leading to the same conclusion.

TODO/Future work

I think we can use properties of diagonalization to show that for a triangular $A$, there is a triangular $P$, such that $P^{-1}AP = D$. Triangularity is retained through multiplication and inversion. If $A, P$ are triangular, then $P^{-1}AP$ is triangular (both multiplication and inversion retains triangularity), but we have to prove that $P^{-1}AP$ is both upper and lower triangular i.e. diagonal.


问:
Proof using trigonometry that circle circumference is $2 \pi R$
Using trigonometry, I would like to prove that the circumference of a circle is $2\pi$ times its radius. Can someone help please?</p>

答:
HINT:

If we divide an $n$ sided equilateral polygon into $n$ triangles (which are generated by the lines to centre from the vertices), the angle at the centre $\frac{2\pi}n$

Let each side be $x$ and the distance from the centre to any vertex be $r$

The rest two angles are same $=\theta$(say)

So, $2\theta=\pi-\frac{2\pi}n\implies \theta =\frac\pi2-\frac\pi n$

Now, using Sine Law of triangle, $$\frac r{\sin \left(\frac\pi2-\frac\pi n\right) }=\frac x{\sin \frac{2\pi}n}$$
$$\implies x=\frac{\sin \frac{2\pi}n}{\cos \frac\pi n }=2r\sin\frac\pi n  $$

Now, the circumference $=n\cdot 2r\sin\frac\pi n $

Now, the polygon will become a circle if $n\to\infty$


问:
Proving $A^T B A$ is positive definite or semidefinite, when $B$ is positive definite
Let's assume we have a positive definite matrix called $B$. (Here I assume the definition of the positive definite matrix stating $B$ should be symmetric.)

How could we conclude that $A^T B A$ is positive definite or semidefinite (that is, all eigenvalues are nonnegative) for any matrix $A$?

I tried to use the theorem which states that "for any vector $v \neq 0$ and positive definite or semidefinite matrix $A$, always $v^T A v \ge 0$ " and put $A^T B A$ in place of $A$ in the theorem, but I wasn't successful in reaching a conclusion.

答:
Set $w=Av,$ then

$$v^T(A^TBA)v=(Av)^TB (Av)=w^TBw \ge 0.$$


问:
Prove an homomorphism on groups
How can I show that the map $\theta:S_4\rightarrow S_3$ defined by $θ(g) = g^∗$ is an homomorphism.
I've been trying this problem and I can only prove for all the cases which is exhausting, how can I show directly that it's an homomorhpism.

$g^*$ is the the permutation induced on the partitions. That is with a permutation we substitute each element on the partition by its' next element on the permutation.

For example,if
$$A = \{\{1, 2\}, \{3, 4\}\}; \\
B = \{\{1, 3\}, \{2, 4\}\}; \\
C = \{\{1, 4\}, \{2, 3\}\}.$$
and if g is the cyclic permutation (1, 2, 3, 4), then $g^∗ = (A, C)(B)$.

答:
An action of a group $G$ on a set $E$ gives rise to a group morphism: $G\to S(E)$, sending $g\in G$ to the permutation $\sigma_g: x\in E\mapsto g\cdot x\in E$.

You can check that this map is indeed a morphism just using the group action axioms (the fact that $\sigma_g$ is bijective comes from the fact that an inverse is $\sigma_{g^{-1}}$, again using the definition on a group action)

Your map is exactly the morphism you obtain with the action you describe in your post...

Title: Prove an homomorphism on groups


问:
Wave rider function
I'm looking for an example of a 'wave-like' function $f : \mathbb{R} \longrightarrow \mathbb{R}$ with the following properties:

deterministic

smooth (continously differentiable)

periodic but with varying period lengths $p_i$, $p_i \neq p_j$

mean period length $p=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i=-n}^n p_i$ is finite but not $0$ (periods don't get constantly larger or smaller)

I tried various combinations of sine functions but that seems more tricky than I thought.

A non-trivial example might be the Riemann zeta function (normalized to have unit average spacing), i.e., $f(x) = \text{Re}\zeta(\frac{1}{2}+\mathsf{i} \delta(x))$.

Background: The question is related to how random the roots of a deterministic function can be?

答: I would try a function like $\sin(x\arctan(x))$ since the periods get shorter as you aproach infinity but never aproach 0.

Title: Wave rider function


问:
An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$
Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1)+C,
$$
so
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx=\frac{\pi}{4}.
$$
Is there an alternative way to calculate this integral?

答: A trick: use $\int_0^1f(x)dx=\int_0^1f(1-x)dx$ to average two versions of your integral, reducing it to$$\tfrac12\left(\int_0^1\tfrac{xdx}{x^2+(1-x)^2}+\int_0^1\tfrac{(1-x)dx}{x^2+(1-x)^2}\right)=\tfrac12\int_0^1\tfrac{dx}{x^2+(1-x)^2}.$$


问:
If $f(x)$ is equal to $x$ or $1-x$ when $x$ is rational/irrational, then $f$ is continuous only at $x=1/2$
How can I resolve the task below:

Given $f(x)=
\begin{cases}
 x, &x\in \mathbb{Q}\text{  }\\
 1-x, &x\notin \mathbb{Q}\text{ (irrational)}
\end{cases}$, $0 \leq x \leq 1$.

Show $f(x)$ is continuous only in $x=\dfrac{1}{2}$.

答:
You have for any $x$ that
$$\limsup_{t\to x} f(x) = \max\{x, 1 - x\}$$
and
$$\liminf_{t\to x} f(x) = \min\{x, 1 - x\}.$$
The two right quantities match precisely at $x = 1/2$.</p>


问:
Set theory and Proofs
Proof that the statement is true
For all non-empty sets A and B, A×B=B×A⇒A=B.
I have tried to come up with the logical order but I cannot seem to find it</p>

答:
Fix $b_0 \in B.$

If $a \in A$, then $(a,b_0) \in A \times B= B \times A.$ This gives $a \in B.$ Hence we have shown that

$$ A \subseteq B.$$

A similar proof gives

$$ A \supseteq B.$$

Title: Set theory and Proofs


答:
Your integral is $ \, \displaystyle \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^²}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx$

To do this integral in spherical coordinates, let's first understand the region we are integrating the function over, which is part of a sphere of radius $2$ centered at the origin in the first octant.

From the upper bound of $z, \, z^2 = 4 - x^2 - y^2 \implies x^2 + y^2 + z^2 = 4$ is the sphere. We also notice that the lower bounds of $x, y, z$ are zero which means it is part of the sphere in the first octant. Now that we know our region, let's convert this into spherical coordinates,

$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$

Also $dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

So our triple integral of the given function over this region in spherical coordinates will be -

$\displaystyle \int_{0}^{\pi/2}\int_{0}^{\pi/2}\int_{0}^{2} \rho^6 \cos^2 \phi \, \sin \phi \,\ d\rho \, d\phi \, d\theta = \frac{64 \pi}{21}$

Title: Computing $\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^2}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx$

Body:
Compute the integral $$\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{\sqrt{4-x^2-y^²}}z^2(x^2+y^2+z^2) \ dz \ dy \ dx$$

Transforming to spherical coordinates I have that $$\int_{0}^{2}\int_{0}^{2 \pi}\int_{0}^{\sqrt{4-r^2\sin^2\varphi}} r^2\cos^2 \varphi\ \cdot r^3 \ d\varphi \ d\theta \ dr$$

but this is very hard to compute, I feel like I'm missing something here? The additional $r$ to form the cubic comes from the Jacobian.


问:
Is there a nice expression for the Fourier transform of a dot product of two functions?
Suppose we have two vectors $A(x)$ and $B(x)$ on $\mathbb{R}^n$ and we wish to take the Fourier transform of their dot product, that is, we wish to evaluate

\begin{equation}
\mathcal{F}\{A(x) \cdot B(x) \}.
\end{equation}

My question is is there a nice expression for this in terms of the Fourier transforms of $A(x)$ and $B(x)$?

So far I have determined that if we let $A(x) = (a_1, a_2, \dotsc , a_n) $ and similarly for $B(x)$, this can be written as the sum of the convolutions of individual elements

\begin{equation}
\mathcal{F}\{A(x) \cdot B(x) \} = \sum_{j=1}^n \mathcal{F}\{a_j\} \star\mathcal{F}\{b_j\},
\end{equation}

but ideally I would prefer to write this in terms of the FTs of the vectors themselves. I suspect there might not be a nicer expression, but I am very rusty with FTs and convolutions and my googling skills are not turning up anything useful.

(As an aside, the reason I am interested in this is I wish to find the FT of the poisson bracket of two functions which I am writing as

\begin{equation}
\{H, \rho\} = \nabla H \cdot \Omega \nabla \rho
\end{equation}

so if anyone knows any specific results about FTs of poisson brackets that might also be of help.)

Thanks.

答:
Supposing the dot-product is the Euclidean inner product, you could apply the convolution theorem to each component, i.e. $F(A\cdot B) = F(\sum_i A_i B_i) = \sum_i F(A_i)*F(B_i)$, where $*$ means convolution, $F$ means Fourier transform, and subscripts indicate components of the vectors.

This may not be easier to compute than the original form in some cases.


问:
How to understand the residue $\text{Res}(\frac{e^{iz}}{\sinh(z)},0) = \frac{e^{iz}}{\cosh(z)}\bigg|_{z=0}$
I am reading a solution of a complex integral using residue, in one step it syas

$$\text{Res}(\frac{e^{iz}}{\sinh(z)},0) = \frac{e^{iz}}{\cosh(z)}\bigg|_{z=0}$$

I cannot understand how to get the RHS from the formula for polynomial case.

Can anyone please let me know how to get the RHS?  Thanks!

答:
Suppose $z_0$ is a simple pole of $f(z)=\frac{P(z)}{Q(z)}$. We have

$$\text{Res}(z_0, f) = \frac{P(z)}{Q'(z)} \bigg|_{z = z_0}.$$

This is easily proven using that

$$\text{Res}(z_0, f) = \lim_{z \rightarrow z_0} (z-z_0) \frac{P(z)}{Q(z)} = \lim_{z \rightarrow z_0} \frac{P(z)}{\tfrac{Q(z)-Q(z_0)}{z-z_0}} =  \frac{P(z_0)}{Q'(z_0)}.$$

Observe that this onluy works when $z_0$ is a simple pole, otherwise the expression above would also have a singularity.


问:
Studying for qualifying exams. Complex analysis identity
I am currently studying for my qualifying exam and I am having real difficulty with the following question. I need to find the value of the following integral for the path below. $$\displaystyle\int\limits_\gamma \frac{2z+1}{e^{\pi z}-1}dz$$ I have attempted to use the residue theorem but I don't know how to break up the regions of the integral, or how to sum them. I am also having difficulty applying said theorem to this case. I have solved such things with unit circle paths but this one is tough. Any ideas on how to solve this?

**The path:

答:
You could use residue theorem to compute this.

Residue Theorem: $$ \int_\gamma \frac{2z+1}{e^{\pi z}-1} dz = 2\pi i \sum\limits_{k=1}^n Res(f,a_k)$$ where $a_k$ are all possible singular
points inside $\gamma$, and $Res(f,a_k)$ is the residue at $a_k$.

Here for this integrand $f=\frac{2z+1}{e^{\pi z}-1}$, there are only 2 singular points inside your $\gamma$, i.e. $a_1 = 0$, $a_2 = 2i$. We need to find the residue of them respectively.

For $a_1 = 0$, since $e^{\pi z}-1 = \sum_{n=1} \frac{(\pi z)^n}{n!}$, thus $Res(f,a_1) = \frac{2*0+1}{\pi} = \frac{1}{\pi}$.

For $a_2 = 2i$, since $e^{\pi z}-1 = e^{\pi (z-2i)} - 1 = \frac{(\pi z-2\pi i)^n}{n!}$, thus $Res(f,a_2) = \frac{2* 2i +1}{\pi}$.

So
$$ \int_\gamma \frac{2z+1}{e^{\pi z}-1} dz = 2\pi i (\frac{1}{\pi} + \frac{4i+1}{\pi}) = -8+4i$$


答: Try letting two arbitrary elements belong to each non-empty set. Next, proceed to work with the definition of the Cartesian product. Remember that the cartesian product outputs a set of 2-tuples and is defined as $\{(a,b) |  a \in A , b \in B\}$ for all $a \in A$ and for all $b \in B$. This set can shown to be equal by the given assumption that $ A \times B = B \times A $ under the definition of the Cartesian product. Finally, you should then be able to show that the two sets are equal.

Title: Set theory and Proofs

Body:
Proof that the statement is true
For all non-empty sets A and B, A×B=B×A⇒A=B.
I have tried to come up with the logical order but I cannot seem to find it</p>


问:
The relationship between plane curves and the derivative of the Wronskian
I have found a theorem but I did not understand the proof. I'm looking for a clarification of the proof or a different proof.

Let $f_1, f_2, f_3$ be the three components of a curve in $R^3$ parameterized by $t$. Let $W$ be the Wronskian matrix of $f$. whose $(i,j)$ entry is $f_j^{(i-1)}$.

Theorem: If $|W'| = 0$ then $f$ is a planar curve i.e. it satisfies $a_1 f_1 + a_2 f_2 + a_3 f_3 + a_0 = 0$ for some constant $a_i$s.

The proof goes something like this:

We can find $b_i$ which are ordinary functions of $t$ such that $W' b = 0$, that is, $\sum b_i f_i' = 0, \sum b_i f_i'' = 0, \sum b_i f_i''' = 0$.

Although the proof does not explain why such functions $b_i$ exist, I imagine I could watch the null space of $W'$ evolve smoothly with time, and select an evolving vector $b$ living inside that nullspace.

 By takng the derivative of the equations, we can arrive at $\sum b_i'f_i' = 0, \sum b_i f_i'' = 0$. Then the proof says something I don't understand: "The $b$'s are proportional respectively to the cofactors of the elements of the last row in the determinant of $W'$. The same is true of the $b'$'s"

Then we get $\frac{b_1'}{b_1} = \frac{b_2'}{b_2} =\frac{b_3'}{b_3} = \phi(t)$ and then $b_i = a_i e^\phi$ so $\sum a_i e^\phi f_i' = 0$ so $e^\phi\sum a_i f_i' = 0$ and then we remove the exponential and integrate to get the final result.

答:
$W'=W(f'), f=(f_1,f_2,f_3)$

If a set of analytic functions has a wronskian of $0$ then the functions are linearly dependent, since the $0^{th}$ to $n-1^{th}$ derivatives of each function can be considered as components of a vector, and checking if the determinant is $0$ is a standard test of linear dependence. If they are linearly dependent, there will be one coefficient for each function, $n$ in total, which is roughly why the derivatives up to $n-1$ contain enough information to determine linear dependence/independence.

$W'=0$ implies that there exist $a_1,a_2,a_3$ such that $a_1f_1'+a_2f_2'+a_3f_3'=0$. Integrating gives the required result.


问:
Calculating the order of poles of $\frac{\sin(3z)}{z^2}$
I have the function $f(z)=\frac{\sin(3z)}{z^2}$ and want to calculate the order of the poles.

If we look at $\frac{1}{f}=0$ we have that $\frac{z^2}{\sin(3z)}=0$ and hence I would assume the pole would be at $z_0=0$ and it has order 2. However, Wolfram says the order is 1?

The order is obviously equal to 1, as when I calculate the residue, I have the Laurent series as
$$f(z)=\frac{3}{z}-\frac{9}{2}z+\frac{81}{40}z^3-\dots$$
and the residue is 3. And this can also be calculated  $$\lim_{z\to 0}(z)\frac{\sin(3z)}{z^2}=\lim_{z\to 0}\frac{\sin(3z)}{z}=3$$. However this method only works for simple poles, so the order must be 1.

I'm just trying to figure how we can calculate the order of the poles of the function directly as 1 instead of 2?

答: Your Laurent series shows you the order of the pole correctly. Since it "starts" at the $z^{-1}$-term, i.e. does not have any terms of more negative powers of $z$, the order is $1$.


问:
What's the point of removing $1$ from the list of primes and having an empty product in the fundamental theorem of arithmetic?
I was reading about how we stopped considering the number $1$ as prime since this would mean the fundamental theorem of arithmetic wouldn't apply any more (since we can repeat the number $1$ over and over again and then the factorization wouldn't be unique) however, this got me wondering about the unique factorization of $1$ itself and I found out that we call it the "empty product" but what is the point of this? If the point from removing $1$ from the list of the prime numbers was to not make it an exception while stating this theorem and some others but this added another definition to make up for it so it feels like there was no point in doing it in the first place. Is there any other reason why we removed $1$ from the list of primes and made this?

答:
It is generally useful to define empty products (and sums, intersections, etc). If we have any associative and commutative binary operation on a set $S$ such that there is a neutral element with respect to this binary operation, we define the empty operation to return the neutral element. For instance:

$\sum_{x\in\emptyset}x=0$, the empty sum is $0$.

$\prod_{x\in\emptyset}x=1$, the empty product is $1$.

$\bigcup_{A\in\emptyset}A=\emptyset$, the empty union is the empty set.

$\bigotimes_{V\in\emptyset}V=\mathbb F$, the empty tensor product is the base field.

The last one may be a bit abstract, but I wanted to show that there are more than just the three "obvious" operations where we do this. The reason why we do this is that this way, if $A,B$ are disjoint subsets of $S$, then we  have the simple rule that
$$\sum_{x\in{A\cup B}}x=\sum_{x\in A}x+\sum_{x\in B}x.$$
The same goes for all other types of commutative, associative binary operations. This is important in different fields of mathematics. For instance, we need the empty tensor product to define a useful tensor algebra. So defining an empty operation is not something new, done specifically to salvage the unique prime factorization theorem (though that would be worthwile all by itself). It's just a natural definition to choose, and it's useful elsewhere as well.


答:
The order of a pole is the greatest positive integer $m$ for which $a_{-m}\neq 0$ in the Laurent series, so after you calculate the Laurent series you're done; simply observe $a_{-1}\neq 0$ yet $a_{k}=0$ for $k\leq -2$.

Here's a way to "see" the order of a pole at a glance. If $f$ and $g$ are analytic in a disk about $z_0$, and $f$ and $g$ have zeroes of order $n$ and $m$ at $z_0$ respectively, then $f/g$ has a removable singularity at $z_0$ when $n\geq m$, and a pole at $z_0$ of order $m-n$ otherwise. This is not too hard to show from the definitions. In this case, $\sin(3z)$ has an order $1$ zero at $z=0$ while $z^2$ has an order $2$ zero at $z=0$. Thus $\sin(3z)/z^2$ has a simple pole at $z=0$.

Title: Calculating the order of poles of $\frac{\sin(3z)}{z^2}$

Body:
I have the function $f(z)=\frac{\sin(3z)}{z^2}$ and want to calculate the order of the poles.

If we look at $\frac{1}{f}=0$ we have that $\frac{z^2}{\sin(3z)}=0$ and hence I would assume the pole would be at $z_0=0$ and it has order 2. However, Wolfram says the order is 1?

The order is obviously equal to 1, as when I calculate the residue, I have the Laurent series as
$$f(z)=\frac{3}{z}-\frac{9}{2}z+\frac{81}{40}z^3-\dots$$
and the residue is 3. And this can also be calculated  $$\lim_{z\to 0}(z)\frac{\sin(3z)}{z^2}=\lim_{z\to 0}\frac{\sin(3z)}{z}=3$$. However this method only works for simple poles, so the order must be 1.

I'm just trying to figure how we can calculate the order of the poles of the function directly as 1 instead of 2?


问:
Is "cofunctor" an accepted term for contravariant functors?
People are used to the prefix co- flipping arrows in a concept1, and I have seen people using cofunctor to mean a functor that flips arrows, i.e. that takes $A \to B$ to $FB \to FA$. I know this concept by the name contravariant functor, which I believe is standard, and contrasted with a covariant functor which doesn't flip arrows – the latter making the name cofunctor particularly unhelpful.

Nevertheless, Wiktionary and Wikipedia (last sentence in the contravariance paragraph) both back up this usage, though uncitedly, so it seems to have some traction. Is there any substance to this usage? Are there any authors who support it, or mention it?

If it is just incorrect, has anyone authoritative dismissed it in a citeable way, such that it could be banished from Wikipedia in future?

Edit: I've also heard contrafunctor used to mean a contravariant functor. Does this mean there are people who use cofunctor to mean covariant functor?

1 What do I mean by "concept" here? Well, an ncept with the arrows flipped, of course.

答: I might be wrong, but I don't think it is completely right to call a controvariant functor "cofunctor" (at least if we want to stick to the convention of using the particle co- to mean and evoke duality) because the sentence "$T$ is a functor" is clearly self-dual, as Mac-Lane explicitely pointed out in his Categories for the Working Mathematician. So, given a functor $T$ a "cofunctor", interpreted as the dual concept, would be just $T$ itself.


问:
Find the direction cosines of the a vector in the plane of $\vec b=<2, 1, 3>$ and $\vec c=<-1, 2, 1>$ and perpendicular to $\vec c$
Let the vector $\vec a =\langle x, y, z\rangle$

Since it is perpendicular to $\vec c$
$$-x+2y+z=0$$

And $$\begin{vmatrix} x &y&z \\ 2&1&3 \\ -1&2&1 \end{vmatrix}$$

$$\implies -5x-7y+5z=0$$ since a, b, c are coplanar

So $y=0$

Then $x=y$

So the answer should be $\left\langle\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2}\right\rangle$ , but that is not correct.

What is wrong with my solution?

The answer is $\left \langle\frac{15}{\sqrt {369}} , 0, \frac{12}{\sqrt {369}} \right \rangle $

答:
Note that the projection of $\vec b$ onto the direction perpendicular to $\vec c$ is

$$\vec b - (\vec b\cdot\vec c)\frac{\vec c}{|c|^2}
= <2, 1, 3>- \frac12<-1, 2, 1>=<\frac52,0,\frac52>
$$
Normalize to get $\vec a = <\frac1{\sqrt2},0,\frac1{\sqrt2}>$.


问:
Specific parametrization of a sphere
I would like to parametrize the region
\begin{equation}
M=\{(x,y,z):x^2+y^2+z^2 \leq 1 \text{ with } z \leq x+y \}
\end{equation}
and calculate the flux $\iint_S F\cdot n \ dS$ where $S=\partial M$. The problem is'nt the flux but this parametrization. With spherical coordinates i know how to write the sphere, but the inequality
\begin{equation}
z\leq x+y
\end{equation}
stays
\begin{equation}
\cos(\varphi)\leq\cos(\theta)\sin(\varphi)+\sin(\theta)\sin(\varphi)
\end{equation}
and im stucked at this part. My geometric intuition tells me that $\theta$ is dependent of $\varphi$, but  how write this? Probably in the flux part we are going to use divergence theorem, because the surface is an sphere with lid $z=x+y$, so we need a parametrization here in order to write $\iiint_M div(F) \ dV$, right?

答:
Having an insight: my vector field is $F=(-2xy,y^2,3z)$, so $div(F)=3$. Divergence theorem yields
\begin{equation}
\iint_S F \cdot n \ dS=\iiint_M div(F)\ dV=3\iiint_M dV.
\end{equation}
But the plane $z=x+y$ divides the sphere in two equal parts, hence
\begin{equation}
3\iiint_M dV=3\dfrac{Vol(B_1(0))}{2}=2\pi.
\end{equation}

The result is ok, but the main problem of parametrization remains. I remember seeing somewhere the question of whether or not there is a parameterization for a given surface, does anyone know anything about it?

Title: Specific parametrization of a sphere


答:
$$I=\int\frac{x}{x^2+(1-x)^2}\,dx=\int \frac{x}{2x^2-2x+1}\,dx=\int \frac{x}{2(x-a)(x-b)}\,dx$$
$$I=\frac 1{2(a-b)}\int \Big[\frac a{x-a}-\frac b{x-b} \Big]\,dx$$
$$I=\frac 1{2(a-b)}\Big[a \log(|x-a|)-b\log(|x-b|)\Big]$$ Now $a=\frac{1-i}2$, $b=\frac{1+i}2$,
$$I=\frac {1-i}4\left(\log
   \left(x-\frac{1+i}{2}\right)+i \log
   \left(x-\frac{1-i}{2}\right)\right)$$

Use the bounds, play with the complex numbers to get the answer.

Title: An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$

Body:
Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1)+C,
$$
so
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx=\frac{\pi}{4}.
$$
Is there an alternative way to calculate this integral?