问:
Can the expected value of a random variable be viewed as a particular linear transformation?
I teach a low-level probability and statistics class, and today, part of our discussion was of a couple of well-known properties that the expectation obeys. Namely, for random variables $X$ and $Y$ and a constant $a$, $E(aX) = aE(X)$ and $E(X + Y) = E(X) + E(Y)$.

A student who is also enrolled in a Linear Algebra course then recognized these properties as the same properties one asks for to get a linear transformation.

Thus, I'd like to know if one can view the expected value as a linear transformation from a vector space of random variables to $\mathbb{R}$. Is this an already-studied notion that's out there ? Perhaps one can link me to a source that might contain a discussion on this, as well.

Thank you !

答:
Yes.  Call the underlying probability space $\Omega$ and for simplicity let's assume that all random variables are real-valued.  Then as your student pointed out, the map $X \mapsto \mathbb{E}[X]$ is a linear transformation from the vector space $V = \{ \text{(measurable) functions } \Omega \to \mathbb{R} \}$ to the vector space $\mathbb{R}$.  The space $V$ has a very natural vector space structure.

(I put "measurable" in parentheses because whether or not you want to talk about that depends on the level of the probability course you're teaching)


问:
Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?
How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$</p>

答:
$$\int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \int_0^{\frac \pi 2} \frac{2}{3 - \cos 2x} dx = \int_0^\pi \frac{d\theta}{3 - \cos \theta } = \frac 12\int_0^{2\pi}\frac{d\theta}{3 - \cos \theta}$$

To evaluate $\displaystyle \int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta$ let $\displaystyle z = e^{i\theta} \implies \cos \theta = \frac{z^2 + 1}{2z}, d\theta = \frac{1}{iz}dz$

$$\int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta = \frac 2{i}\oint_{|z| = 1} \frac{1}{-z^2 + 6z - 1}dz $$

The poles are $3 - 2 \sqrt 2$ and $3  + 2 \sqrt 2$, and since $3 + 2 \sqrt 2  > 1$,

$$\frac 2{i}\oint_{|z| = 1} \frac{1 }{-z^2 + 6z - 1}dz = 4\pi \text{Res}\left[ \frac{1}{-z^2 + 6z - 1} , 3 - 2 \sqrt 2\right] = \frac{\pi}{\sqrt 2} $$

So $\displaystyle \int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \frac{\pi }{2 \sqrt 2}$

ContentLicense: CC BY-SA 3.0


问:
the Zassenhaus /Baker–Campbell–Hausdorff formula for cosine.
This question concerns the expansion of non-commutative algebra $[X,Y] \neq 0$ for two operators $X,Y$. One can think of $X$ and $Y$ as some matrices.

If $[X,Y] = 0$, we have

$$e^{t(X+Y)}= e^{tX}~  e^{tY}$$

If $[X,Y] \neq 0$, We know the Zassenhaus formula or the Baker–Campbell–Hausdorff formula: $$e^{t(X+Y)}= e^{tX}~  e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~
e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~
e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots$$

This expresses $e^{t(X+Y)}$ in terms of $e^{tX}$ and $e^{tY}$, and their further commutators $[X,Y]$.

Question: Do we have a similar form for $\cos(A+B)$ when $[A,B]=C \neq 0$? (We may take $[C,A]=[C,B]=0$ for the simplest case to extract the first order term.)

If $[A,B]=0$, we have
  $$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B).$$

If $[A,B]=C \neq 0$, do we have some similar expression like the Zassenhaus formula or the Baker–Campbell–Hausdorff formula:
  $$\cos(A+B)=\cos(A)\cos(B) \dots-\sin(A)\sin(B) \dots+ \dots$$

Can we express $\cos(A+B)$ in terms of $\cos(A)$,$\cos(B)$,$\sin(A)$,$\sin(B)$ and some function of $C$?

答:
Let us write the Zassenhaus formula in the form
$$
e^{X+Y}=e^Xe^Ye^{C_2(X,Y)}e^{C_3(X,Y)}\ldots,
$$
where $C_n(X,Y)$ satisfies $C_n(\alpha X,\alpha Y)=\alpha^nC_n(X,Y)$. The first couple of these are
$$
C_2(X,Y)=-\frac12[X,Y],\qquad C_3(X,Y)=\frac13[Y,[X,Y]]+\frac16[X,[X,Y]].
$$
Applying this to the definition of $\cos(A+B)$, we get
$$
2\cos(A+B)=e^{i(A+B)}+e^{-i(A+B)}\\=e^{iA}e^{iB}e^{-C_2}e^{-iC_3}e^{C_4}e^{iC_5}e^{-C_6}\ldots+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3}e^{C_4}e^{-iC_5}e^{-C_6}\ldots,
$$
where we have used the abbreviation $C_n=C_n(X,Y)$.
To get a sense of what is going on, let us assume $C_n=0$ for all $n>3$.
Thus we have
$$
e^{iA}e^{iB}e^{-C_2}e^{-iC_3}+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3}
\\=
(\cos A+i\sin A)(\cos B+i\sin B)e^{-C_2}(\cos C_3-i\sin C_3)\\
+(\cos A-i\sin A)(\cos B-i\sin B)e^{-C_2}(\cos C_3+i\sin C_3)\\
=2(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3+2(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3,
$$
yielding
$$
\cos(A+B)=(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3\\+(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3.
$$
It is now easy to see that the general formula is given by
$$
\cos(A+B)=\big[\cos(x_0+x_1-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}\big]\Big|_{x_0=A,x_1=B,x_2=C_2,x_3=C_3,\ldots},
$$
where it is understood that we expand $\cos(a+b-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}$ as if all the variables were simply numbers, with the terms ordered in such a way that the terms with $x_n$ come after all $x_1,\ldots,x_{n-1}$, and then substitute the matrices for the variables.

ContentLicense: CC BY-SA 4.0


问:
Inversion of harmonic points
Show that if four harmonic $A$, $B$, $C$ and $D$ points are inverted with respect to a circle whose center is another point on the same line as the first ones, we obtain another four harmonic points.

I tried to make some simple math, but it does not come out. I know $(A, B;\; C, D) = −1,$ but how can I get to $(A^\prime, B^\prime;\; C^\prime, D^\prime) = −1$.

答:
Conceptually this works because if the points are harmonic then $C$ and $D$ are inverses wrt the circle with diameter $AB$, and inversion in another circle $\Gamma$ preserves the relationship.  I.e. $C'$ and $D'$ are inverses wrt the circle with diameter $A'B'.$

But let's approach this from basic principles.

Let $PQ$ be the directed distance between $P$ and $Q$

$(A, B; C, D)=\dfrac{AC\cdot DB}{CB\cdot AD}$

let the circle $\Gamma$ have center $X$ and radius $r$

if $P'$ is the inverse of $P$   then $PX\cdot P'X=r^2$

then $AC = AX+XC = \dfrac{r^2}{A'X}+\dfrac{r^2}{XC'}
=r^2\dfrac{A'X+XC'}{A'X\cdot XC'}=-r^2\dfrac{A'C'}{A'X\cdot C'X}$.

Similarly for the other terms $DB,CB,AD.$

plug this into $(A, B; C, D)$ and simplify/cancel to get $(A', B'; C', D')$

Title: Inversion of harmonic points


问:
How to prove that the softmax function is not invariant under scalar multiplication.
I want to prove that the softmax function is not invariant under scalar multiplication.

How to continue from there to prove that S(x)i is not equal to S(xc)i ?

答:
I'll give you a direction you should follow.

Take instead the problem to solve to be that it is invariant to scalar multiplication.

A way to disprove it (i.e. that it is actually not scalar multiplication invariant) is to look at full example using a scalar of your choice (not 1 as it's not really a multiplication). (proof by contradiction to the opposite problem)


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答:
I recomend MAXIMA. Of home page of this CAS we have the apresentation

"Maxima is a system for the manipulation of symbolic and numerical
  expressions, including differentiation, integration, Taylor series,
  Laplace transforms, ordinary differential equations, systems of linear
  equations, polynomials, and sets, lists, vectors, matrices, and
  tensors. Maxima yields high precision numeric results by using exact
  fractions, arbitrary precision integers, and variable precision
  floating point numbers. Maxima can plot functions and data in two and
  three dimensions."

Title: Free online mathematical software


问:
Is always true for the formula of marginal density function from joint distribution?
This is the formula to help you get the marginal pdf from a joint distribution:

$$f(x)= \int_{}^{} f(x,y) \, dy\\$$
$$f(y)= \int_{}^{} f(x,y) \, dx\\$$

where the first integral is over all points in the range of (X,Y) for which X=x and the second inegral is over all points in the range of (X,Y) for which Y=y.

I have no problem to get the marginal in sample case, like f(x,y)=1/x, for 0<x<10, 0<y<5.

However, I don't think that those formula will work on the case below:
$$f(x,y)= \frac{8}{3}xy $$
$$for\;0\le x\le1,x\le y\le 2x$$

Y is dependent on X for Y's region.

Can anyone give me a key word or hint? And then I can search more for this.

答:
You can still find the marginal PDF - it's just going to be a piecewise function. Firstly - finding ${f(x) = \int_{x}^{2x}f(x,y)dy}$ is nice and straight forward. For ${f(y)}$ - we need to rewrite the bounds. If you do so, you'll notice it splits in two:
$$
0\leq y\leq 1,\frac{y}{2}\leq x\leq y
$$
$$
1 < y \leq 2,\frac{y}{2}\leq x < 1
$$
So
$$
f(y) = \left\{
\begin{array}{cc}
\int_{\frac{y}{2}}^{y}f(x,y)dx&\text{ if ${0\leq y\leq 1}$}\\
\int_{\frac{y}{2}}^{1}f(x,y)dx&\text{ if ${1 < y \leq 2}$}
\end{array}
\right.
$$
Notice now that if you integrate ${\int_{0}^{2}f(y)dy}$, we should get $1$ - and you see that
$$
\int_{0}^{2}f(y)dy = \int_{0}^{1}\int_{\frac{y}{2}}^{y}f(x,y)dxdy + \int_{1}^{2}\int_{\frac{y}{2}}^{1}f(x,y)dxdy = 1
$$


问:
Proof that 7 is irreducible in $\mathbb Z[\sqrt{15}]$.
I want to show that 7 is irreducible in $\mathbb Z[\sqrt{15}]$.  I define the field norm $N(a+b\sqrt{15}) = a^2-15b^2$.  If 7 can be factored into non-units, $7=\alpha\beta$, then $N(7)=49=N(\alpha)N(\beta)$.  If $N(\alpha)=1,49$ then $\alpha$ or $\beta$ is a unit so I only want to show that $N(\alpha)\ne 7$.  So suppose for contradiction that $N(\alpha)=7$.  If $\alpha=a+b\sqrt{15}$ then $N(\alpha)=a^2-15b^2=7$.

From here I'm not sure how to solve it.  This now seems to require some kind of number theory, yes?  Having almost no background in number theory, this makes me wonder if there is a better way to approach the problem or some general method for showing that something like this has no solutions.

But if I try to think of some solution in number theory, then I guess computing mod 15 is natural.

$$ a^2 \equiv_{15} 7 $$

I can then try to think of the square of every number from 0 to 14, but this seems like a bad solution method.

答: There is nothing wrong with your solution method, you can work out all of the squares modulo $15$ and conclude that no solution can occur.  Alternatively, you could have reduced modulo $5$ (less computation required), the squares modulo $5$ are just $\{\pm 1\}$, so as $7\equiv 2\mod 5$ it is not square mod $5$.  In particular, no solution to $a^2-16b^2=7$ exists for $a,b$ integers.


问:
How to prove the existence of non-abelian group of order $125$
The following question was part of my algebra assignment.

Let $G$ be a non abelian group. Can its order be $125 ?$

$p$ groups have non-trivial center. So, if $|Z(G)| =125$ and the group is abelian, if $|Z(G)| =25$ then $G/Z(G)$ is cyclic  and the group is abelian.But if $|Z(G)|= 5$ then G/Z(G) can be $\mathbb{Z}_5 \times \mathbb{Z}_5$  but how to prove that group is non-abelian in this case?

Any other approach of proving it non-abelian is also welcome.

答:
The Heisenberg group  modulo $5$ is a non-abelian group of order $125$.

It is the set of all $3 \times 3$ matrices with entries in $\mathbb F_5$ of the form
$$
\begin{pmatrix}
 1 & a & c\\
 0 & 1 & b\\
 0 & 0 & 1\\
\end{pmatrix}
$$


答:
Put $z=e^{i x}$; then the integral is equal to

$$i \oint_{|z|=1} dz \frac{z}{z^4-6 z^2+1} $$

There are four poles at

$$z = \pm \sqrt{2} \pm 1$$

only two of which are within the unit circle ($z = \pm (\sqrt{2}-1)$).  The residues from each of these poles are equal and are each

$$\frac{i}{4 (\sqrt{2}-1)^2-12} = \frac{-i}{8 \sqrt{2}}$$

The sum of the residues is double that residue.  The integral is then $i 2 \pi$ times that sum; I get

$$\int_0^{\pi/2} \frac{dx}{1+\sin^2{x}} = \frac{\pi}{2 \sqrt{2}}$$

Title: Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?

Body: How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$</p>


问:
Prove by induction $n!<4n^2+3$
How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?

Here´s my try:

The equation is valid for every $n\leq4$.

n=5: n!=5!=120>103

n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$

Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.

Any ideas how i could continue or whats wrong?

答:
Your work is mostly fine, but you can perfect your redacting.

What you want to show is that the set $A=\{n\in\mathbb N\mid n!<4n^2+3\}=\{0,1,2,3,4\}$

For that you need to verify the inequality for the a few numbers which you certainly did, but it needs to appear on your solution.

$\begin{cases}
n=0&0!=1<4\times 0^2+3=3&\text{ok}\\
n=1&1!=1<4\times 1^2+3=7&\text{ok}\\
n=2&2!=2<4\times 2^2+3=19&\text{ok}\\
n=3&3!=6<4\times 3^2+3=39&\text{ok}\\
n=4&4!=24<4\times 4^2+3=67&\text{ok}\\
n=5&5!=120\not<4\times 5^2+3=103&\text{not verified}\\
\end{cases}$

You are correctly suspecting that the inequality is not verified either for $n\ge 5$, so you should explicit the induction proposition :

$$P(n) : n! \ge 4n^2+3\qquad\forall n\ge 5$$

Base case $P(5)$ is verified.

So let assume $P(n)$ we get $(n+1)!=(n+1)n!\ge (n+1)(4n^2+3)$

And this is where you got stuck.

You need to make appear $4(n+1)^2+3$, instead I propose to subtract both quantities and show the difference is positive (this is easier to handle):

$\begin{align}(n+1)!-\bigg((4(n+1)^2+3\bigg)
&\ge (n+1)(4n^2+3)-\bigg(4(n+1)^2+3\bigg)
\\&\ge 4n^3+3n+4n^2+3-(4n^2+8n+7)
\\&\ge 4n^3-5n-4
\\&\ge 4n^3-5n-4n\qquad\text{ for } n\ge 1\implies 4\le 4n
\\&\ge n(4n^2-9)
\\&\ge 0 \qquad\qquad\qquad\qquad\text{ for } n^2\ge \frac 94\iff n\ge \frac 32
\end{align}$

Since $n\ge 5$ both conditions ($n\ge 1$ and $n\ge \frac 32$) are verified therefore you have proved your induction step.

Title: Prove by induction $n!<4n^2+3$


问:
Writing Out The Number of Zeros From 1 - 1,000,000
If you had to write out all the numbers between 1 and 1,000,000. How many zeros would you have to write in total.

I was given that the answer is 488,895. Can any one show the clearest way possible of getting this figure. The couple ways I have tried have given me an answer that is about half of that figure.

答:
How may zeroes appear from 1 to 1000000?

10% of 10,00,000 = 1,00,000

No of 1 digit numbers ×1 = 9×1 = 9

No of 2 digit numbers ×2 = 90×2 =180

No of 3 digit numbers ×3 = 900×3 = 2,700

No of 4 digit numbers ×4 = 9,000×4 = 36,000

No of 5 digit numbers ×5 = 90,000×5 = 4,50,000

No of 6 digit numbers from 1 to 1,00,000×6 = 1×6 = 6

Sum of (9+180+2,700+36,000+4,50,000+6) = 4,88,895 zeroes

Sum of each digit from 1 to  (10% of 10,00,000) = 4,88,895 zeroes.


问:
Coin Streak Probabilities in A Known Number of Coin Tosses
Please bear with me because I'll ask you with a very basic understanding of math.

During these hard times, I try to acquire and hone new skills so that I can be a little more competitive in finding a job after this whole situation dies down. I chose to learn Python because recently I'm interested in data analysis-related jobs. One day I found a question regarding probabilities and the goal was to write a code that returns the probabilities of a 6-streak (doesn't matter whether it's Head or Tail, HHHHHH or TTTTTT) turn up in a hundred coin tosses and the experiment is repeated up to one thousand times.

I write the code and got a 54.54 % probability that a 6-streak turn up in a hundred coin tosses.

My friends' answers were different and us having little to no background in math and coding only exacerbates the situation. I tried to find an analytical answer by reading similar problem like this, this, and this. After reading from those links and expecting enlightenment, however, it only made me more confused because I initially thought that the analytical answer would be just a mix of simple equations (e.g., plugging constant to the equation, multiplication, or division).

My questions to you all are:

Why can't I just use permutation and combination that I learned in high school to solve the above problem? Also, is 54.54% the right answer to my problem?

What I learned from previous links is many answers refer to Markov's rule or chain and I'm not familiar at all with it, why is that rule used to solve the probability problem? Is this Markov rule a single equation that solves all of the probability of coin tosses or are there any other similar approach?

Says that there are other similar questions that ask what is the probability of 6-streak appear, let's say: (a) 3 times, (b) exactly 3 times, and (c) no more than 3 times. What differs in the approach to each of the problems?

I want to learn more about this similar problem and all data-science related topics (e.g. statistics, big O, etc.). Where do I start if I have no experience nor knowledge at all related to the topics? Is there any curriculum that you guys can recommend?

I'd be really happy if the explanation is detail but the approach is like an explain-it-like-i'm-five type kind of approach. Lastly, sorry for the long rambling and any little input is greatly appreciated. Thank you!

答:
This is not the best forum for details about programming in a specific language.
If you did a simulation in Python, you should not expect simulation results
to match exact results from combinatorics.

In R, the procedure rle (for Run Length Encoding) shows the number and length
of runs in a sequence x of results. For example, here is one way to find the
length of the longest run of either 0s (say Tails) and 1a (say Heads)
among (a particular) 20 tosses of a fair coin. In this example the answer is six.

```<code>set.seed(2020)  <span class="math-container">$ for reprocucibility
x = sample(c(0,1), 20, rep=T); x
[1] 1 0 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 1 1 0
rle(x)
Run Length Encoding
 lengths: int [1:8] 1 1 1 6 4 2 4 1
 values : num [1:8] 1 0 1 0 1 0 1 0
rle(x)$</span>lengths
[1] 1 1 1 6 4 2 4 1
max(rle(x)$lengths)
[1] 6
</code>```

If I repeat this procdure many times, I can get a good idea of the
distribution of the length ML of longest run among 100 tosses, and thus of its mean.
The answer is $7.975 \pm 0.011.$ With $m = 10^5$ iterations one can expect
about two place accuracy. (With $m = 10^6$ I got $6.978\pm 0.004.)$

```<code>set.seed(1120)
ML = replicate(10^5, max( rle( sample(c(0,1), 100, rep=T))$lengthis ))
summary(ML)
     Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    3.000   6.000   7.000   6.976   8.000  22.000
mean(ML)
[1] 6.97582          # E(RL)
2*sd(ML)/sqrt(10^5)
[1] 0.01128483       # 95% margin of simulation error
hist(ML, prob=T, br=(2:22)+.5, col="skyblue2", main="Sim Dist'n of ML")
</code>```

I suppose analogous procedures are available in Plython. You seem to have good links for analytic solutions.

Body:
This is not the best forum for details about programming in a specific language.
If you did a simulation in Python, you should not expect simulation results
to match exact results from combinatorics.

In R, the procedure rle (for Run Length Encoding) shows the number and length
of runs in a sequence x of results. For example, here is one way to find the
length of the longest run of either 0s (say Tails) and 1a (say Heads)
among (a particular) 20 tosses of a fair coin. In this example the answer is six.

set.seed(2020)  $ for reprocucibility
x = sample(c(0,1), 20, rep=T); x
[1] 1 0 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 1 1 0
rle(x)
Run Length Encoding
 lengths: int [1:8] 1 1 1 6 4 2 4 1
 values : num [1:8] 1 0 1 0 1 0 1 0
rle(x)$lengths
[1] 1 1 1 6 4 2 4 1
max(rle(x)$lengths)
[1] 6

If I repeat this procdure many times, I can get a good idea of the
distribution of the length ML of longest run among 100 tosses, and thus of its mean.
The answer is $7.975 \pm 0.011.$ With $m = 10^5$ iterations one can expect
about two place accuracy. (With $m = 10^6$ I got $6.978\pm 0.004.)$

set.seed(1120)
ML = replicate(10^5, max( rle( sample(c(0,1), 100, rep=T))$lengthis ))
summary(ML)
     Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    3.000   6.000   7.000   6.976   8.000  22.000
mean(ML)
[1] 6.97582          # E(RL)
2*sd(ML)/sqrt(10^5)
[1] 0.01128483       # 95% margin of simulation error
hist(ML, prob=T, br=(2:22)+.5, col="skyblue2", main="Sim Dist'n of ML")

I suppose analogous procedures are available in Plython. You seem to have good links for analytic solutions.


答:
Assuming you want to prove that for each $n\geq 5$ one has $n! > 4n^2+3$ just note that it easily holds for $n=5$. Now assume that it holds up to $n \geq n-1 \geq \ldots 5.$ Then
$$(n+1)! = (n+1)n! > (n+1)(4n^2+3).$$
It is only left to prove that $(n+1)(4n^2+3) > 4(n+1)^2+3$ for each $n\geq 5.$ In fact,
$$(n+1)(4n^2+3)-4(n+1)^2-3 = 4n^3-5n-4.$$
So let $p(n) = 4n^3-5n-4.$ Note that $p(5) > 0.$ Moreover, $p'(n) = 12n-5 > 0,~\forall n\geq 5.$

Title: Prove by induction $n!<4n^2+3$

Body:
How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?

Here´s my try:

The equation is valid for every $n\leq4$.

n=5: n!=5!=120>103

n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$

Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.

Any ideas how i could continue or whats wrong?


问:
How to prove undecidability other than Rice Theorem?
Im studying Rice Theorem and I would like to verify its consistency. If I am able to prove de undecidability in other ways, the Rice Theorem would prove to be useful after all.

Im trying to find out two different ways to prove that the following Decision problem is undecidable:

Is it true that the Turing Machine $M$ accepts exactly a finite and odd number of words?

So far:

One way is using the Rice Theorem. Let $X$ be the property of being finite and have an odd number of words. Rice Theorem tells us that for every property of semi-decidable languages, the Test of Property is undecidable only if $X$ is non trivial. Now I need to verify if $X$ is non-trivial correct? I got stuck on that part.

Talking about other way to prove, I was thinking about Reductio ad absurdum. I have an idea but I don't know how to build this reduction.

If I have an entry $(M,w)$ of HALTING that produces $r(M,w) = M'$ of ODD.

If $(M,w)$ is an affirmative instance of HALTING then $M'$ accepts only one word and nothing more.
If $(M,w)$ is an negative instance of HALTING then $M'$ accepts no words at all.

Is that clear? Sorry, english is not my mother language, so translating the way we refer to problems and theorems can be tricky.

Thanks for everything.

答:
It doesn't really make sense to ask if you can avoid using Rice's theorem by writing down a reduction. This is because Rice's theorem is proved by writing down a reduction - or rather, a general recipe for constructing a reduction - already. Basically, what Rice's theorem does is save you the effort of defining essentially the same reduction over and over again. It's not avoiding reductions, but avoiding redundant effort.

For completeness, let's recall Rice's theorem and its proof. Rather than work with $HALT$, I'll work with the many-one-equivalent set of indices of machines which halt on input $0$ specifically, and I'll write "$M_e$" for the machine with index $e$.

Rice's theorem: Suppose $\mathcal{P}$ is any nontrivial property of c.e. languages. Then there is a many-one reduction of the halting problem to the set of indices of machines with property $\mathcal{P}$.

Proof: Suppose $\mathcal{P}$ is a nontrivial property of Turing machines. Let $M$ be some Turing machine which never halts on any input. Suppose WLOG that $M$ has property $\mathcal{P}$ (otherwise work with the complement of $\mathcal{P}$ instead). By nontriviality, fix some $N$ which does have property $\mathcal{P}$. We get a reduction of $HALT$ to the index set corresponding to $\mathcal{P}$ by - given a number $e$ - constructing a machine $A$ which behaves as follows:

On input $x$, $A$ first runs $M_e$ on input $e$; if it halts, $A$ then runs $N$ on input $x$ and does whatever $N$ does.

This $A$ behaves identically to $N$ if $e\in HALT$, and never halts on any input (= behaves identically to $M$) otherwise. That is, $A$ has property $\mathcal{P}$ iff $e\in HALT$. So we get a reduction of $HALT$ to our index set as desired. $\quad\Box$

The key point is that Rice's theorem applies to every nontrivial property of c.e. languages, so there's really no nuance in applying it. This is what makes it useful: when showing that an index set is undecidable, we don't have to think through the details of constructing a specific reduction.


问:
Forster exercise 17.3 on Riemann-Roch theorem
Let $X$ Riemann Surface of genus $g$, and $D$ is a divisor on $X$. I try to prove $l(D)=dim H^0 (X,O_D) ≤ 1+degD$ when $−1≤degD≤g−1$ Using the Riemann-Roch theorem and Serre duality. The notations here : $H^0(X,O_D)$ is obviously the $0$-th cohomology group of $X$ with respect to the sheaf $O_D$ where $O_D(U)=\{f∈\mathcal{M}(U)|ordx(f)≥−D(x)∀x∈U\}$ and $\mathcal{M}$ is the sheaf of meromorphic functions.

The statement of Riemann-Roch in O.Forster Book is : for a compact Riemann surface $X$ of genus $g$ and a divisor $D$ on it, $$dimH^0(X,O_D)−dimH^1(X,O^D)=1−g+degD.$$

I tried to use the Serre duality which asserts that $dimH^1(X,O_D)=dimH^0(X,Ω_{−D})$ where $Ω_{−D}$ is the sheaf of meromorphic $1$-forms that are multiples of $−D$. Or, use the the exact sequenece
$$0\to H^0(X,D−p)→H^0(X,D)→\mathbb{C}→0.$$ I find that, if $D=P-N$, where $P,N$ two effective divisors, then $l(D)≤ 1+deg(P)$. But i cannot find the finale result.

Could I please get a hint?

答:
If $D$ is non-negative ok, if not just change $D$ by $D'=div(f)+D$ where $f\in O_D$, then $l(D)=l(D')\leq 1+deg(D')=1+deg(D)$.

rmk$_1$: If $D$ and $D'$ are equivalent divisors on Riemann Surfaces then $O_D(X)\simeq O_{D'}(X)$ given by multiplication of a meromorphic function on $X$.

rmk$_2$: $div(f)$ is the divisor associated a meromorphic function, then $deg(div(f))=0$ (The degree of a holomorphic map is constant in the set of regular values).


问:
Extend an integral inequality from $C[0,1]$ to the whole $L^{2}([0,1])$ [related to operator norm].
Consider the following integral defined for $f\in L^{2}[0,1]$ and $x\in [0,1]$, that $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$

This shows that for $f\in C[0,1]$ and for any $x\in [0,1]$, we have $$|(T^{n}f)(x)|\leq \dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}|f(r)|dr\leq \dfrac{\|f\|_{\infty}}{n!}.$$ By definition of operator norm, this implies that the operator norm of $T^{n}$ $$\|T^{n}\|\leq \dfrac{1}{n!},$$ if the operator is on $C[0,1]$.

However, I want to prove the same bound of this operator norm on the whole $L^{2}([0,1])$.  Is there anyway for me to extend this proof to all of $f\in L^{2}([0,1])$? and to conclude $\|T^{n}\|\leq\frac{1}{n!}$

My attempt was to use density of continuous function of compact support in $L^{2}$ to find $g$ such that $f=g+\epsilon$ where $\epsilon>0$ is arbitrarily fixed. Then we bound the integral in the same way, and you will have a summation, the second term in the summation will go to $0$ when $\epsilon\searrow 0$. Therefore, we again have $$|(T^{n}f)(x)|\dfrac{\|g\|_{\infty}}{n!}.$$  But this cannot say anything to the operator norm $\|T\|$.

What should I do? Thank you!

Edit:

The overall purpose is to derive that $\|T\|\leq\frac{1}{n!}$. So if there is any other way to find this, it will also be really good.

Edit2: Proof

Okay, I confused myself in the first place. The operator norm has nothing to do with the sup-norm since $T:L^{2}\longrightarrow L^{2}$, so what we should do is to use Cauchy-Schwarz and $\|f\|_{L^{2}}$. See below my own answer to my post for details.

Also, Ruy gave a general theorem about how to compute the operator norm for Hilbert-Schmidt operator, and my computation will show you the idea about the proof. Basically, the absolute value of operator will be bounded by $\|f\|_{L^{2}}$ multiplication of the $L^{2}-$integral of the kernel, so we should expect such a theorem.

Thank you so much for all the users who helped me!!

答:
Okay, I guess in my way we will not be able to finally get $\frac{1}{n!}$, but a little bit worse bound $\frac{1}{\sqrt{n}(n-1)!}$, but it does not affect the general decaying I need in the end, so it is okay.

The following is my proof:

The $n-$fold iterate of $T$ is given by the formula $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$ This implies that, for $f\in L^{2}([0,1])$, by Cauchy-Schwarz, we have the following:
\begin{align*}
\Big|(T^{n}f)(x)\Big|^{2}\leq \Bigg(\dfrac{1}{(n-1)!}\int_{0}^{x}|x-r|^{n-1}|f(r)|dr\Bigg)^{2}&=\Bigg(\dfrac{1}{(n-1)!}\Bigg)^{2}\Bigg(\int_{0}^{x}|x-r|^{n-1}|f(r)|dr\Bigg)^{2}\\
&\leq\Bigg(\dfrac{1}{(n-1)!}\Bigg)^{2}\|f\|_{L^{2}([0,1])}^{2}\int_{0}^{x}|x-r|^{2n-2}dr,
\end{align*}
and  $$\int_{0}^{x}|x-r|^{2n-2}dr=\int_{0}^{x}(x-r)^{2n-2}dr=-\dfrac{1}{2n-1}\Big[(x-r)^{2n-1}\Big]_{0}^{x}=\dfrac{x^{2n-1}}{2n-1}.$$ As $x\in [0,1]$ and $n\geq 2$, we further have $$\int_{0}^{x}|x-r|^{2n-2}dr=\dfrac{x^{2n-1}}{2n-1}\leq \dfrac{1}{2n-1}\leq \dfrac{1}{n}.$$

Hence, $$|(T^{n}f)(x)|\leq \dfrac{\|f\|_{L^{2}([0,1])}}{\sqrt{n}(n-1)!},$$ and by definition of the operator norm of our operator $T:L^{2}([0,1])\longrightarrow L^{2}([0,1])$, this implies that $$\|T^{n}\|\leq \dfrac{1}{\sqrt{n}(n-1)!}.$$


问:
solving an equation from trigonometry
I am trying to solve the following trigonometric equation: $$\frac{\cot\theta+\csc\theta}{\tan\theta+\sec\theta}=\cot(\pi/4+\theta/2)\cot \theta/2$$ but unfortunately, after giving a lot of effort I am unable to solve the problem, I think I am missing some obvious steps so the problem is getting messed up but I wasn't able to point out my fault, I also tried to solve it by starting from RHS but I also failed in there.

I will be very much grateful if someone helps me.

答:
$$\frac{\cot \theta+\csc \theta}{\tan \theta+ \sec \theta}= \frac{1+\cos \theta}{1+\sin \theta}~\frac{\cos \theta}{\sin \theta}$$ $$=\frac{2\cos^2(\theta/2)}{[\sin(\theta/2)+\cos(\theta/2)]^2}\frac{\cos^2(\theta/2)-\sin^2(\theta/2)}{2\sin(\theta/2) \cos(\theta/2)}=\cot(\theta/2) \frac{\cos (\theta/2)-\sin(\theta/2)}{\cos (\theta/2)+\sin(\theta/2)}$$
$$=\cot(\theta/2) \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}=\cot(\theta/2) \tan(\pi/4-\theta/2)=\cot(\theta/2) \cot(\pi/4+\theta/2).$$
Use $\tan(\pi/2-z)=\cot z$.

Title: solving an equation from trigonometry


问:
if $n>5$ is prime,prove $(n-1)|(n-2)!$
Could you give a hint how to prove this?

if $n>5$ is prime, prove $(n-1)|(n-2)!$

答:
If $n$ is prime, then $n-1$ is composite unless $n=2 \text{ or } 3$. So the following line of argument begins by assuming $n>3$.

If $n>3$, then $n-1$ is composite, and it can be factored into two smaller integers, $(n-1)=ab$, each of which is smaller than $n-2$. This is true because an integer multiple of any positive integer $>1$ cannot equal the next larger integer (i.e. $k(n-2)\ne n-1$), so neither of $a,b$ can be as large as $(n-2)$.

If $n-1$ can be factored such that $a\ne b$, since $a,b<n-2$, they will each appear as separate terms in the product $(n-2)!$, and we are done. $a\mid (n-2)!$ and $b\mid (n-2)!$, so $ab\mid (n-2)!$, meaning $(n-1)\mid (n-2)!$. For odd primes, $n-1$ is even and can be always factored as $2$ times another number different from $2$, except when $n=5$. So far, the argument has been made assuming $n>3$, but now we must consider $5$ as a special case.

For $n=5$, where $n-1=4=2\times 2$ and $n-2=3$, the terms $1,2,3$ of $(n-2)!$ contain a factor of $2$ only once. In fact, $4\not \mid 3!=6$, so the proof fails for the case $n=5$, and we must require $n>5$ for it to be valid.

I hope that this not only shows that the proposition is true, but also explains why it must be conditioned on $n>5$

Title: if $n>5$ is prime,prove $(n-1)|(n-2)!$


问:
Question about conguence module m: If $n> 0$ and $n$ is not a multiple of $3$, prove that $a = 3^{2n} + 3^n + 1$ is divisible by $13.$
I would like to solve this problem using the idea of congruence module m, but I have no idea how to start. Could someone help me?

If  $n> 0$  and $n$ is not a multiple of $3$, prove that $a = 3^{2n} + 3^n + 1$ is divisible by $13$.

Body:
$3^3=27=2\times13+1\equiv1\bmod13$, so $3^{3n}\equiv 1 \bmod 13$,

so $3^{3n}-1=(3^n-1)(3^{2n}+3^n+1)\equiv 0\bmod13$,

which means $3^n-1\equiv0\bmod13$ or $3^{2n}+3^n+1\equiv0\bmod13$.

Can you take it from here?


问:
Give an example of a function $f$ with an essential discontinuity everywhere and yet $(f)^2$ is continuous everywhere

Note: $(f)^2$ means $(f \circ f)$. That is to say, $f$ composed with itself

To begin this problem, I consider $f(x)=\frac{1}{x}$

For $x \neq 0, f(x)$ is defined

But at $x=0,$ $f(x)=\frac{1}{x}$ has an essential discontinuity since $\lim_{x \to 0} \frac{1}{x}$ does not exist.

Notice however that $(f)^2=\frac{1}{\frac{1}{x}}=x$ and $\lim_{x \to 0} x=0$

This example works when $f$ is disontinuous at a point. How do I construct a function that is discontinuous for all values whose composition with itself is continuous for all values?

答: How's about $f(x)= x(-1)^{ \mathbb{1}_{x \in \mathbb{Q}}}$


答:
Pick your favorite dense subset $A\subseteq \mathbb{R}$ containing $0$ and $1$, such that its complement, $A^C$, is dense as well. Then $1_A$ is discontinuous everywhere and $1_A\circ 1_A$ is just constantly $1$.

($1_A$ is the function which is $1$ on $A$ and $0$ on $A^C$).

Title: Give an example of a function $f$ with an essential discontinuity everywhere and yet $(f)^2$ is continuous everywhere

Body:
Note: $(f)^2$ means $(f \circ f)$. That is to say, $f$ composed with itself

To begin this problem, I consider $f(x)=\frac{1}{x}$

For $x \neq 0, f(x)$ is defined

But at $x=0,$ $f(x)=\frac{1}{x}$ has an essential discontinuity since $\lim_{x \to 0} \frac{1}{x}$ does not exist.

Notice however that $(f)^2=\frac{1}{\frac{1}{x}}=x$ and $\lim_{x \to 0} x=0$

This example works when $f$ is disontinuous at a point. How do I construct a function that is discontinuous for all values whose composition with itself is continuous for all values?


答:
The simplest way to think in such a situation is to make a table of the data, working modulo $13$, i.e. in the ring $R=\Bbb Z/13$. The table is:
$$
\begin{array}{|r|c|l|}
\hline
n & 3^n\in R & 3^{2n}+3^n+1\in R\\\hline
0 & 1 & 1+1+1 = 3\\
1 & 3 & 9+3+1 = 13=0\\
2 & 9 & 3+9+1 = 13=0\\\hline
3 & 1 & 1+1+1 = 3\\
4 & 3 & 9+3+1 = 13=0\\
5 & 9 & 3+9+1 = 13=0\\\hline
\vdots & \vdots & \vdots
\end{array}
$$
and it repeats itself. The main point is to see that the powers of $3\in R$ repeat with period three, and then check these three cases.

Title: Question about conguence module m: If $n> 0$ and $n$ is not a multiple of $3$, prove that $a = 3^{2n} + 3^n + 1$ is divisible by $13.$

Body:
I would like to solve this problem using the idea of congruence module m, but I have no idea how to start. Could someone help me?

If  $n> 0$  and $n$ is not a multiple of $3$, prove that $a = 3^{2n} + 3^n + 1$ is divisible by $13$.


问:
Combinatorics Expected Value without Replacement
Suppose we have a deck of cards with 4 kings, 4 queens, and 4 jacks. Two cards are drawn uniformly at random without replacement. Let Q be the random variable representing the number of queens obtained. Find the expected value, variance, and standard deviation of Q.

To find the expected value, I would consider all possible cases.

2 queens are drawn: $\frac{4}{12}*\frac{3}{11} = \frac{12}{132}$

1 queen and 1 other: $\frac{4}{12}*\frac{8}{11} = \frac{32}{132}$

1 other and 1 queen: $\frac{8}{12}*\frac{4}{11} = \frac{32}{132}$

0 queens: $\frac{8}{12}*\frac{7}{11} = \frac{56}{132}$

Thus the expected Q is $2* \frac{12}{132} + 1*\frac{32}{132} + 1*\frac{32}{132} + 0*\frac{56}{132} = \frac{2}{3}.$

I am quite confused as to how I could continue this to find the variance and standard deviation of the distribution of the expected values. Any insights will be appreciated. Thank you.

答:
So far so good. Let $X$ be the random variable for the number of queens. Then the corresponding probabilities are

\begin{align*}
    &
P(X=2)=\frac{12}{132}=\frac1{11} \\  &P(X=1)=\frac{32}{132}+\frac{32}{132}=\frac{64}{132}=\frac{16}{33} \\ &P(X=0)=\frac{56}{132}=\frac{14}{33} \\
\end{align*}

Therefore the pdf is

$$f_X(x)=\begin{cases}\frac{14}{33}, \ \textrm{if} \ x=0 \\ \frac{16}{33}, \ \textrm{if} \ x=1 \\ \frac{1}{11}, \ \textrm{if} \ x=2 \\ 0, \ \textrm{elsewhere}  \end{cases}$$

The variance then is

$$Var(X)=\mathbb E(X^2)-\left[\mathbb E (X) \right]^2 =\sum_{x=0}^2 x^2\cdot f_X(x)-\left[\sum_{x=0}^2 x\cdot f_X(x) \right]^2 $$

Hints: $\mathbb E(X^2)=\frac{28}{33}, \ \left[\mathbb E (X) \right]^2=\left(\frac23\right)^2=\frac{4}{9}$


问:
If sequence of derivatives converges uniformly is function sequence increasing?
Let $g_n : \mathbb{R} → \mathbb{R}$ be a sequence of functions where $g_n'(x)$ converges to $1+x^2$ uniformly does $g_n$ necessarily converge to an increasing function?

I claimed the answer was no, since we do not know about the convergence of $g_n$ at any point in $\mathbb{R}$ what would be the correct answer?

答: Let $g_n(x) = a_n+ x+ \frac{1}{3}x^3 $. You can choose $a_n$ to be any sequence of numbers.


问:
Symmetry within the Lebesgue exponents of Young's convolution inequality
Young's convolution inequality states that if $f\in L^p(X,\mu), g\in L^q$, and
$$
1/p+1/q=1/r+1,
$$
then $\|f\|_p\|g\|_q\geq\|f\ast g\|_r.$

Now, if we write $\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q},$ then it is easy  to deduce that
$$
\frac{1-\theta}{p}+\frac{\theta}{q}=1.
$$
This relation somehow tells us that $1/r$ is the "reflection" of the number $1$ in the midpoint of $1/p$ and $1/q$, because we are swapping $1-\theta$ and $\theta$.

Is there an intuitive, possibly geometric explanation of this observation? Why the inequality works this way?

答:
I'm not sure if this is sufficiently intuitive or geometric, but it might be helpful for thinking about the play of exponents in Young's inequality.

Now the exponent condition can equivalently be written as $1/p' + 1/q' = 1/r'$, where the primes indicate dual exponents. Here we can draw in the Hausdorff-Young inequality, which gives that the Fourier transform maps $L^s$ into $L^{s'}$ for $1 \leq s \leq 2$.

So at least if $1 \leq p, q \leq 2$, then $f \in L^p$ and $g \in L^q$ implies that $\hat{f} \in L^{p'}$ and $\hat{g} \in L^{q'}$; and of course the Fourier transform converts convolution to multiplication, so $(f \ast g)^{\wedge} = \hat{f} \hat{g}$. By Hölder's inequality, we can then see that $(f \ast g)^{\wedge} \in L^{r'}$, which is at least consistent with $f \ast g \in L^r$.


答:
The simplest way is to note that there are 600,000 of each of 1-9 from 000,000 to 999,999, a total of 5,400,000.

Now, the number of digits of all numbers from 1-999,999 is:

$1 \times 9 + 2 \times 90 + \cdots +6 \times 900000 = 9\times 654321 = 5,888,889$.

Therefore the number of zeros is 6 + 5,888,889-5,400,000 = 488,895.

Title: Writing Out The Number of Zeros From 1 - 1,000,000

Body:
If you had to write out all the numbers between 1 and 1,000,000. How many zeros would you have to write in total.

I was given that the answer is 488,895. Can any one show the clearest way possible of getting this figure. The couple ways I have tried have given me an answer that is about half of that figure.


问:
Question concerning graphs
Is it true, that in an undirected single graph, if there are two adjacent cut vertices, the edge connecting those two cut vertices is a cut edge?
I think that this is indeed the case, but I'm struggling a bit with proving this fact. Could someone give me some tips or a general method for approaching graph related proofs?</p>

答:
This is not true. A simple way to make a vertex $v$ int a cut vertex is to add a leaf adjacent to $v$ and nothing else. Using this construction, you should have no trouble finding a counterexample; the smallest I can think of has $5$ vertices and $5$ edges.

Some general tips! Try looking at small cases or small examples and seeing if a statement holds. If it does hold, try to think about what makes it true, and if that has to happen in all cases; this could lead you to a proof, or to a counterexample.

Keep in mind equivalent ways to state a property. For this question, it's useful to remember that an edge is a cut edge if and only if it is not part of any cycle.

Title: Question concerning graphs


问:
All 2-groups have cyclic commutator subgroups?
I have seen that groups of order 8 and order 16 have commutator subgroup cyclic. I want to know if that can be extended to all 2-groups or if there is a reason why this happens.</p>

答:
No: if $G$ and $H$ are groups, then $[G\times H,G\times H]= [G,G]\times[H,H]$. So all you need to do is take the direct products of two $2$-groups with nontrivial commutator subgroup to immediately get one with noncyclic commutator subgroup. In particular, $D_8\times D_8$ (where $D_8$ is the dihedral group of order $8$) is an example.

What you noticed is a consequence of one might call the "law of small groups": there aren't enough groups of small order to avoid coincidences.
Any nonabelian group of order $p^3$ has cyclic commutator subgroup (equal to the center) of order $p$. Something similar happens with groups of order $p^4$. There just isn't "enough room" for a noncyclic commutator subgroup. Once you get to $p^6$ you can use the easy trick I start with above.

But you also get examples with order $p^5$.  Take the group of all $4\times 4$ upper triangular unipotent matrices and entries in $\mathbb{F}_p$, of the form
$$ \left(\begin{array}{cccc}
1 & 0 & * & *\\
0 & 1 & * & *\\
0 & 0 & 1 & *\\
0 & 0 & 0 & 1
\end{array}\right).$$
Here the commutator subgroup is isomorphic to $C_p\times C_p$. The group is generated by $3$ elements, two of which commute with each other but not with the third generator.


问:
Relations; aRb if ab=k^2
Hi I am supposed to prove that R is an equivalence relation on N (natural number)

We define the relation R on N as aRb if $ab=k^2$ for some k that belongs to N.

I know that I should prove that the relation is reflexive, symmetric and transitive.

The relation is reflexive if aRa for some a that belongs to N.
a^2 is a natural number and therefore a^2 = k^2 and aRa.

The relation is symmetric if aRb then bRa for all a and b that belongs to N.

The relation is symmetric because if aRb then $ab=k^2$ for some k that belongs to Z. This is equivalent to $ba= k^2$ and therefore bRa.

I'm not sure about Transitive tho. I must prove that if aRb and bRc then aRc. How can I do that?

答: Hint. Try to prove transitivity when $a$ and $b$ are both powers of $2$. A few examples should show you how. Then think about the fundamental theorem of arithmetic.

Title: Relations; aRb if ab=k^2


答: Since you know that aRb and bRc, you can conclude that there are natural numbers $m$ and $n$ such that $ab=m^2$ and $bc=n^2$.  Your job is to prove that $ac$ is also the square of some natural number involving $m$ and $n$. (Hint: multiply those two equations together.)

Title: Relations; aRb if ab=k^2

Body:
Hi I am supposed to prove that R is an equivalence relation on N (natural number)

We define the relation R on N as aRb if $ab=k^2$ for some k that belongs to N.

I know that I should prove that the relation is reflexive, symmetric and transitive.

The relation is reflexive if aRa for some a that belongs to N.
a^2 is a natural number and therefore a^2 = k^2 and aRa.

The relation is symmetric if aRb then bRa for all a and b that belongs to N.

The relation is symmetric because if aRb then $ab=k^2$ for some k that belongs to Z. This is equivalent to $ba= k^2$ and therefore bRa.

I'm not sure about Transitive tho. I must prove that if aRb and bRc then aRc. How can I do that?


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
Ya, right, so strange, so how about to use Chinese count? Maybe a little convenient. :D

```
  # |   English |           Chinese
----------------------------------------------
  0 |      zero |            零
  1 |       one |            一
  2 |       two |            二
  3 |     three |            三
  4 |      four |            四
  5 |      five |            五
  6 |       six |            六
  7 |     seven |            七
  8 |     eight |            八
  9 |      nine |            九
 10 |       ten |            十
 11 |    eleven |            十|一
 12 |    twelve |            十|二
 13 | thir|teen |            十|三
 14 | four|teen |            十|四
 15 |  fif|teen |            十|五
 16 |  six|teen |            十|六
 17 |seven|teen |            十|七
                             十|八
                             十|九
 20 |    twenty |            二|十
 21 |twenty-one |            二|十|一
 22 |twenty-two |            二|十|二

100,000,000 | 1 hundred million  |  一|亿 （1 followed by eight zeros）
OK, let me show you the 999,999,999.   九亿九千九百九十九万九千九百九十九
```

Just imagine that what if we use this to do math. ;-P

Title: Why do the French count so strangely?

Body:
Ya, right, so strange, so how about to use Chinese count? Maybe a little convenient. :D

  # |   English |           Chinese
----------------------------------------------
  0 |      zero |            零
  1 |       one |            一
  2 |       two |            二
  3 |     three |            三
  4 |      four |            四
  5 |      five |            五
  6 |       six |            六
  7 |     seven |            七
  8 |     eight |            八
  9 |      nine |            九
 10 |       ten |            十
 11 |    eleven |            十|一
 12 |    twelve |            十|二
 13 | thir|teen |            十|三
 14 | four|teen |            十|四
 15 |  fif|teen |            十|五
 16 |  six|teen |            十|六
 17 |seven|teen |            十|七
                             十|八
                             十|九
 20 |    twenty |            二|十
 21 |twenty-one |            二|十|一
 22 |twenty-two |            二|十|二

100,000,000 | 1 hundred million  |  一|亿 （1 followed by eight zeros）
OK, let me show you the 999,999,999.   九亿九千九百九十九万九千九百九十九

Just imagine that what if we use this to do math. ;-P


问:
Largest expected number of birthday's in a single day
I have 1200 friends in Facebook. What is the largest number of them I can (probabilistically) expect to have their birthday on the same day? In general, what would it be if I had N friends?</p>

答:
Here is a partial (asymptotic) answer.

Let's make some simplifying assumptions:

There are 365 days in every year

Every birthday is independent and uniformly distributed throughout the year

Let $X_1, \dots, X_N$ be the birthdays of your $N$ friends, so according to the above assumptions the $X_i$ are iid distributed uniformly on the set $\{1,2,\dots, 365\}$.  Define $D_j := \frac{1}{N} \#\{ n : X_n = j \}$ to be the proportion of birthdays that fall on day $j$.  The goal is to calculate $\mathbb{E}(\max_{1 \leq j \leq 365} D_j)$.

Proposition: $\mathbb{E}(\max_j D_j) \to \frac{1}{365}$ as $N \to \infty$.

Proof: First note that $\max_j D_j \geq \frac{1}{365}$ always holds, essentially by something like the pigeonhole principle.  Next apply a union bound to conclude
$$
\mathbb{P}\left(\max_j D_j > \frac{1}{365} + \epsilon\right) \leq \mathbb{P}\left(D_1 > \frac{1}{365} + \epsilon\right) + \dots + \mathbb{P}\left(D_{365} > \frac{1}{365} + \epsilon\right).
$$
Notice now that we can also express $D_j$ as the average $\frac{1}{N} \left( 1_{X_1 =j} + \dots 1_{X_N = j} \right)$.  By assumption 2, these indicators are iid with mean $\frac{1}{365}$.  Therefore by the weak LLN, $\mathbb{P}\left(D_{j} > \frac{1}{365} + \epsilon\right) \to 0$ as $N \to \infty$ for any $\epsilon > 0$ and for each $j$.

Thus we have shown that $\mathbb{P}\left(\max_j D_{j} > \frac{1}{365} + \epsilon\right) \to 0$ as $N \to \infty$ for any fixed $\epsilon > 0$.  We already know that $1 \geq\max_j D_j \geq \frac{1}{365}$ deterministically, so this implies that $\mathbb{E}(\max_j D_j) \to \frac{1}{365}$ for the following reason: we know that $\max_j D_j$ is a random variable which takes values in $[\frac{1}{365}, \frac{1}{365} + \epsilon]$ with probability $1-o(1)$ and values in $(\frac{1}{365} + \epsilon, 1]$ with probability $o(1)$.  Therefore
$$
\frac{1}{365} \leq \mathbb{E} (\max_j D_j) \leq (\frac{1}{365} + \epsilon)(1+o(1)) + 1 \cdot o(1)
$$
Taking $N \to \infty$ and then $\epsilon \to 0$ gives the result. $\square$

Note: It's possible to get a more quantitative estimate for finite $N$ by applying a large deviation inequality like Hoeffding's inequality instead of the weak LLN.


问:
Another round table problem involving 12 chairs and 5 persons
I spent a couple of hours thinking about the following excercise:

Given a round table with 12 chairs, how many possibilites exist to place 5 persons at this chair. There should be at least one chair between each two persons. The chairs are not numbered and the possibilities only differing in the rotation of the table are equal.

I think there are the following three possible patterns of free chairs (0) and persons (1)

1 0 1 0 1 0 1 0 1 0 0 0

0 0 1 0 0 1 0 1 0 1 0 1

0 0 1 0 1 0 0 1 0 1 0 1

every other combination of persons and chairs could be shifted to one of these three.

So there is one last thing to do, the possibilities of the 5 persons sitting on 5 chairs, this leads to 5!.

So my answer is:

$
3 * 5! = 360
$

There is a bonus task, I should give a generall solution für n chairs and k persons. It would be great to get some advice on this bonus task, because I got stuck thinking about this.

Edit:

The persons are distinguishable.

答:
Your method works, but here is an approach that will generalize better. (Everything I do here can be done for $k$ people sitting in $n$ chairs.)

Imagine each person sitting at the table rudely puts their feet on the chair to their left. This should always be possible if there's a gap between any two people. Conversely, if it is possible, then the gap is ensured.

So we can imagine that each person sitting at the table occupies two chairs in this way, and we are trying to arrange $5$ people and $2$ empty chairs around the circle.

To deal with rotations, the easiest thing to do is to fix the position of person 1. Then we are trying to arrange the remaining $4$ people and $2$ empty chairs in a sequence that can no longer be rotated, and this is easy: it's $\frac{6!}{2!} = 360$.

(There are $6$ people and empty chairs to rearrange in $6!$ ways, but we divide by the $2!$ ways to rearrange the empty chairs, which doesn't change the outcome.)


问:
Show that $-\left(\min_{w\in C}(w^\top s+\frac12\|w\|_2^2)\right)$ is convex or show that it is concave
$$f(s)=-\left(\min_{w\in C}(w^\top s+\frac12\|w\|_2^2)\right)$$ where $C$ is a compact and convex set in $\mathbb{R}^n$

I believe that the function is concave, but I need to show that it is convex (which could be false). Please give me any result. I am thinking of using this link which answers that the minimum of a parametric convex function is convex to prove that it is concave.

答:
Given any family of convex functions, the pointwise supremum is convex. See the answer here: Prove the supremum of the set of affine functions is convex

(It assumes that the domain is compact, but the proof does not use this, and in any case one can always assume the domain is compact by restricting to a line segment.)

Multiplying by $-1$, we obtain that the pointwise infimum of concave functions is concave.

Affine functions are concave, hence so is the infimum in the definition of $f(s)$, so that $f(s)$ is convex. This argument does not require that $C$ is convex.

Note: it is not automatic that the minimum in your question exists (this can fail for certain $C$ if the term $\frac12 \lVert w \rVert^2$ is not there). But you can write it as
$$f(s) = \frac12 \lVert s \rVert^2 - \inf_{w \in C} \frac12 \lVert s + w \rVert^2 $$

and this infimum is attained because $C$ is closed. Moreover, we see that
$$f(s) = \frac12 \lVert s \rVert^2 - \frac12 d(s, -C)^2 \,.$$


问:
Deciding whether a theory $T$ is $\aleph_0$-categorical?
I'm trying to solve the following exercise from my lecture notes:

For $n ≥ 1$, let $L_n$ be the language with $n$ unary predicate symbols
$P_1, . . . , P_n$. Show there exists a theory $T_n$ asserting that each $P_i$
is infinite, the $P_i$ are disjoint ($¬(∃x)(P_1(x)\wedge P_2(x))$, etc.) and there are
infinitely many elements not in any $P_i$, $i ≤ n$.

Show $T_n$ is $ℵ_0$-categorical. Conclude that $T_n$ is complete.

How many models of cardinality $ℵ_1$ (up to isomorphism) does $T_2$ have?

Let $T = \bigcup_n T_n$. Show that $T$ is a complete theory.

Is $T$ $\aleph_0$-categorical? If not how many countable models does $T$ have?

I've done everything apart from number 5.

It's easy to show that each $T_n$ is $ℵ_0$-categorical as in part 2, because if you have a countably infinite model, and each of the $P_i$ are infinite then they must also be countably infinite and you can take bijections between them. Moreover, if we take any countably infinite model to have domain $A$, then since $A\backslash(P_1\cup ...\cup P_n)$ is infinite this will also be countably infinite, and so you can take a bijection between these parts of any two countably infinite model. It's easy to check that the function defined on the domain of a whole model as the union of these bijections (which is well defined since the $P_i$ are disjoint) will be an isomorphism between any two countably infinite models, hence $T_n$ is $ℵ_0$-categorical.

I'm not sure if I can do the same in part 5 - each of the $P_i$ will still be countably infinite and so I could do the same thing for them, taking bijections between their interpretations under any two models. But I'm not sure if in this case the subset of the domain of a model not in any $P_i$ will still have to be infinite - there are infinitely many not in any finite selection of the $P_i$ because $T$ contains all the $T_n$, but I'm not convinced that means there will be infinitely many not in any of the $P_i$.

I'd really appreciate any advice you could offer - if the the subset of the domain of a model not in any $P_i$ will still have to be infinite, an explanation of why would be really helpful. If this might not be infinite, then I can see that presumably in this case there would be $\aleph_0$ countable models (up to isomorphism) as you could take a bijection between this subset of any possible size and use it to get an isomorphism as above, but I'm not sure how to show it might not be infinite - I can't think of an obvious counterexample.

答:
Your intuition is absolutely correct; we can make the set $M=\bigcup_{0<i\in\omega}\{i\}\times\mathbb{Z}$ into an $L$-structure by realizing each $P_i$ as $\{i\}\times\mathbb{Z}$. This is a model of $T$; each $P_i$ is infinite, the $P_i$ are pairwise disjoint, and for any $n\in\omega$ there are infinitely many elements not in $\bigcup_{0<i\leqslant n}P_i=\bigcup_{0<i\leqslant n}\{i\}\times\mathbb{Z}$.

On the other hand, we can realize $\{\star\}\cup M$ as an $L$-superstructure of $M$ by asserting $\star\notin P_i$ for any $0<i\in\omega$, and this will again be a model of $T$ for the same reasons as $M$. However, these two structures cannot be isomorphic, as one has an element not in any $P_i$ and the other does not. In fact there is no $L$-embedding at all from $\{\star\}\cup M$ to $M$, as there is nowhere to map $\star$ to.

In this way, there are indeed $\aleph_0$ countable models of $T$ up to isomorphism, with one unique model for any possible countable size of the complement of $\bigcup_{0<i\in\omega}P_i$.


问:
Categorical characterisation of fields and groups
Given a monoidal category $\mathcal{V}$, is there a categorical characterisation of an object in $\mathcal{V}$ such that it is a group if $\mathcal{V} = Set$ and a field if $\mathcal{V} = Ab$? Or does anybody know of useful references? This would be much appreciated!

Ofcourse I do not expect it by simple means of extra data as groups cannot be descriped operadically, but rather by universal properties or something similar.

EDIT: As noted in the comments: I am looking for something different than group objects, as they do not encompass fields. So I am looking for some categorical framework in which both can fit.

答:
As far as I know there is no such characterization. Groups are, of course, the group objects in $\text{Set}$; to my mind the cleanest way to characterize fields is that they are the simple objects in the category of commutative rings (the ones with no nontrivial quotient objects, where by "quotient object" I mean "effective epimorphism.")

Groups and fields are just not actually that analogous. Fields are required to be commutative and inverses are only defined for nonzero elements, so they aren't actually "algebraic." And there's no way to talk about group objects in a monoidal category that isn't cartesian; the closest you get is talking about Hopf algebras, which are very different from fields.


问:
Compute the Maclaurin expansion of $\frac{1 - x \cosh(a)}{1 - 2 \cosh(a) x + x^2}$
The exercise

Compute the Maclaurin expansion of $\frac{1 - x \cosh(a)}{1 - 2 \cosh(a) x + x^2}$ at $0$ for any order $n \in \mathbb{N}$.

My try

I remarked $1 - 2 \cosh(a) x + x^2 = (x-e^a)(x-e^{-a})$, not sure it helps.

One also has, setting $u(x) := x^2 - 2 x \cosh(a)$, $\frac{1}{1-u(x)} = 1 + u(x) + u(x)^2 + \dots + u(x)^n + o(u(x))$ as $u \underset{0}{\rightarrow} 0$. We also have $o(u(x)) = o(x)$ when $x \to 0$.

I could compute the expansion for $n = 1, 2, 3...$ but having a clean formula for any $n$ seems hard to me.

I am almost sure there is a trick that would ease a lot the resolution of the exercise.

Thanks for the help.

答: Your fraction is$$\frac{1-x(e^a+e^{-a})/2}{(x-e^a)(x-e^{-a})}=\frac12\sum_\pm\frac{1}{1-e^{\pm a}x}=\sum_{n\ge0}x^n\cosh na.$$


问:
If $\frac{3\pi}{4} < x < \pi$ then $\sqrt{\csc^2 x + 2 \cot x}$ is =?
Q: If $\frac{3\pi}{4} < x < \pi$ then $\sqrt{\csc^2 x + 2 \cot x}$ is equal to what?

A: $\sqrt{ \csc^2 x + 2 \cot x}$
= $\sqrt{1 + \cot^{2} x + 2 \cot x}$
= $\sqrt{(1 + \cot x)^{2}}$
= $\pm (1 + \cot x)$

How do I move from here? The given answer is $- 1 - cotx$. From the inequality all I can conclude is $-1-cotx<0$. Thank you for any help.

答:
Hint

$$1+\cot x=\dfrac{\sin x+\cos x}{\sin x}$$

Clearly $\sin x>0$ here

Now $f(x)=\sin x+\cos x=\sqrt2\sin\left(x+\dfrac\pi4\right)$$

and $\dfrac{3\pi}4+\dfrac\pi4<x+\dfrac\pi4<\pi+\dfrac\pi4$

So, $f(x)<0$


问:
$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$
My original question is

$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}):\Bbb Q]=p_1p_2p_3$ ? where $p_1,p_2,p_3,q_1,q_2,q_3$ are all distinct prime numbers.

But I would rather change the question to $[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$ i.e. $x^{p_2}-q_2$ is irreducible over $\Bbb Q(\sqrt[p_1]{q_1})$. Any hints or ideas?

Edit: I don't know whether the above statement is true or not. But I guess it's true but I have no idea how to prove it.

答:
First, I don't think this question deserves the downvotes, as you make a reasonable attempt to replace your problem with a simpler one to understand it, and I understand what you're asking.

Here is a hint to help you think about your simpler problem, and I believe it will also get you to an answer to your original question.  Consider the field lattice diagram consisting of $ F = \mathbb{Q} $, the two extensions $ E_1 = \mathbb{Q}(\sqrt[p_1]{q_1}) $ and $ E_2 = \mathbb{Q}(\sqrt[p_2]{q_2} $), and the full field $ K = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}) $.  Then (assuming they are non-trivial) these extensions $ E_1/F $ and $ E_2/F $ have index $ p_1 $ and $ p_2 $.  Now here is the punchline:
$$
[K : F] = [K : E_1][E_1 : F] = [K:E_2][E_2:F]
$$
i.e. $ [K:F] $ is divisible by both $ p_1 $ and $ p_2 $.  This means $ [K : F] $ is divisible by $ p_1 p_2 $ (!).  I believe you can take it from here to conclude that $ K = \mathbb{Q}(\sqrt[p_1]{q_1})(\sqrt[p_2]{q_2}) $ is a degree-$p_2$ extension of $ E_1 = \mathbb{Q}(\sqrt[p_1]{q_1}) $.


问:
Conjecture on cycle length and primes
That thanks for Peter Košinár's answer,I change the conjecture a lot.

For positive odd $a$, let $b = A179382((a+1)/2)$,let $b_1 = znorder(Mod(10,a))$,If $b = (a-1)/(2^c)$ and $b_1 = (a-1)/c_1$ for some $c,c_1 > 0$,then $a$ is prime.

Are there grounds to consider this plausible?

答:
Consider $a=92673 = 3^2 \times 7 \times 1471$. I believe that $b=724$ in this case, which gives us $c=7$. Yet, $a$ is (obviously) not a prime.

Three other counterexamples have been identified subsequently, the largest of them also avoids the $a,b\not\equiv 0\pmod 3$ wriggle:

$c = 5$ and $a = 143713 = 137\times 1049 = 2^5\times 3^2\times 499 + 1$

$c = 5$ and $a = 4110529 = 809\times 5081 = 2^6\times 3\times 79\times 271 + 1$

$c = 6$ and $a = 28688897 = 1009\times 28433 = 2^9\times 137\times 409 + 1$

Title: Conjecture on cycle length and primes


问:
$J:=M+\langle a\rangle $ how can I show that this ideal is generated by $\langle m_1,..., m_n,a \rangle$ and that $M \subsetneq J$?
This is what I want to prove:

Every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ implies that every ideal in $R$ is finitely generated.

I found this answer here Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.

Proof. We take an ideal $I$ of $R$ and we define the set
$$F:=\{J: J\text{ finitely generated ideal of } R  \text{ and } J\subseteq   I\}.$$
We can see that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). So, from hypothesis, there exists a maximal element $M\in F$.

We want to show that $I=M$.

Obviously, $M\subseteq I$, because $M\in F$.
We suppose now that $M\subsetneq I$. Then, there exists an element $a\in I\backslash M\iff a\in I$, with $a\notin M$.

We consider the ideal
$$J:=M+\langle a\rangle.$$
Then, if $M=\langle m_1,...,m_n \rangle$, we have $J=M+\langle a\rangle=\langle m_1,..., m_n,a \rangle \implies J\in F$

Question: Why $M\neq M+\langle a \rangle $?

If  $M= M+\langle a \rangle$ $\underline{ \text{and}}$ $R$ has $1_R$, then
$a=1_R\cdot a \in M+\langle a \rangle =M $, contradiction.

But what happens if $R$ hasn't unity?

Here is my correction to this answer:

Assume that every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ relative to the partial order of set inclusion. Let $\mathcal{A}$ be any ideal of $R$.Define
$$F:=\{I: I\text{ finitely generated ideal of } R  \text{ and } I\subseteq \mathcal{A}\}.$$

Now, It is clear from the definition of $F$ that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). Now, by our assumption, there exists a maximal element $M\in F$ relative to the partial order by set inclusion.

We want to show that $\mathcal{A} = M$. If $M\neq \mathcal{A}$, then there is some $\langle a \rangle \in A\setminus M$, and then $M\cup \langle a \rangle$ is a finite subset of $A$ which is a proper superset of $M$. But $M$ was maximal, so no such $\langle a \rangle$ can exist, so $\mathcal{A}=M$.

Still I have a wrong step in my correction, which is considering this $M\cup \langle a \rangle$ as an ideal, I know that the union of 2 ideals is not necessarily an ideal unless one of them is contained into the other, I feel like I should consider the ideal
$J:=M+\langle a\rangle $ instead, but then how can I show that this ideal is generated by $\langle m_1,..., m_n,a \rangle$? How can I show that it contains the ideal $M$ and not equal to it i.e. $M \subsetneq J$? that  could anyone help me correct that step in my proof please?

答:
From what I can discern from your comments, the following question remains unresolved: given a commutative ring $R$ (possibly without unity $1_R$), an ideal $M \subseteq R,$ and an element $a \in R - M,$ why is it true that $M \subsetneq M + aR?$ On the contrary, let us assume that $M = M + aR.$ Consequently, for every element $ar$ in $aR,$ there exist elements $m, n \in M$ such that $m = n + ar$ so that $ar = m - n.$ By hypothesis that $M$ is an ideal, it follows that $ar = m - n \in M$ so that $aR \subseteq M.$ Obviously, this is a contradiction if $R$ is unital: we would have that $a = a \cdot 1_R \in M.$

Unfortunately, there is no problem here if the ring is non-unital. For instance, it is trivial to see that $4 \mathbb Z = 4 \mathbb Z + 4 \mathbb Z = 4 \mathbb Z + 2(2 \mathbb Z)$ for $R = 2 \mathbb Z,$ $M = 4 \mathbb Z,$ and $a = 2.$ Explicitly, we have that $4i = 4 \cdot 0 + 2(2i)$ for all integers $i$ so that $4 \mathbb Z \subseteq 4 \mathbb Z + 2(2 \mathbb Z)$ and $4i + 2(2j) = 4(i + j)$ for all integers $i$ and $j$ so that $4 \mathbb Z + 2(2 \mathbb Z) \subseteq 4 \mathbb Z.$ Of course, $2 \mathbb Z$ is a non-unital ring; $4 \mathbb Z$ is an ideal in $2 \mathbb Z$ because $4i - 4j = 4(i - j) \in 4 \mathbb Z$ and $(2i)(4j) = 4(2ij) \in 4 \mathbb Z$ for all integers $i$ and $j$; and $2$ is not in $4 \mathbb Z$ by definition.

Consequently, the assumption that $R$ is a unital ring is crucial to the proof that $M \subsetneq M + aR.$ But it is standard to assume that all rings are unital; a non-unital ring is more commonly referred to as a rng (no "$i$" means no unity).


问:
Is an unlabeled graph $G$ characterized by the degrees of endpoints of edges in $G$?
An unlabeled graph is just an isomorphism class of graphs. I am trying to make a way to store an unlabeled graph in a computer without necessarily charactering it with a labeled graph.

For instance, take the labeled graph $G=(V,E)$, where $V=\{A,B,C,D\}$ and $E=\{\{A,B\},\{B,C\},\{B,D\},\{C,D\}\}$. Suppose I want to store this in a computer but without assigning labels to the vertex set, so I am merely storing the structure of the graph.

I know the degree sequence of $G$ is $D=(3,2,2,1)$. From here, I can create an multiset of 2-sets of $D$ that corresponds the edges in $G$: $E'=\{\{1,3\},\{2,3\},\{2,3\},\{2,2\}\}$. Note every 2-set in $E'$ is an edge in $G$, but written in terms of the degrees of the adjacent vertices, rather than the indices of the adjacent vertices. In other words $\{u,v\} \in E \iff \{\deg u, \deg v\} \in E'$.

It is clear any graph isomorphic to $G$ will have this same $E'$.
Is it true that every unlabeled graph will have a unique $(D,E')$?

Is this a way to uniquely characterize every unlabeled graph? If not, am I missing something?

答:
This approach will not work: in particular, any two $k$-regular $n$-vertex graphs will have the same multiset with $\{k,k\}$ occurring $\frac{kn}{2}$ times. The complete bipartite graph $K_{3,3}$ and the skeleton of a triangular prism are both $3$-regular graphs on $6$ vertices; these are just one of many pairs of graphs you will not be able to distinguish.

In general, if you were hoping for an easy way to do this, you're out of luck, because it would correspond to an easy way to check if two graphs are isomorphic. (We don't currently know of a polynomial-time algorithm to do this.)

If you are looking for a way to do this with software, a graph isomorphism tool such as bliss or nauty can generate a "canonical labeling" for a graph, which will be the same for two isomorphic graphs.


问:
Extend an integral inequality from $C[0,1]$ to the whole $L^{2}([0,1])$ [related to operator norm].
Consider the following integral defined for $f\in L^{2}[0,1]$ and $x\in [0,1]$, that $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$

This shows that for $f\in C[0,1]$ and for any $x\in [0,1]$, we have $$|(T^{n}f)(x)|\leq \dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}|f(r)|dr\leq \dfrac{\|f\|_{\infty}}{n!}.$$ By definition of operator norm, this implies that the operator norm of $T^{n}$ $$\|T^{n}\|\leq \dfrac{1}{n!},$$ if the operator is on $C[0,1]$.

However, I want to prove the same bound of this operator norm on the whole $L^{2}([0,1])$.  Is there anyway for me to extend this proof to all of $f\in L^{2}([0,1])$? and to conclude $\|T^{n}\|\leq\frac{1}{n!}$

My attempt was to use density of continuous function of compact support in $L^{2}$ to find $g$ such that $f=g+\epsilon$ where $\epsilon>0$ is arbitrarily fixed. Then we bound the integral in the same way, and you will have a summation, the second term in the summation will go to $0$ when $\epsilon\searrow 0$. Therefore, we again have $$|(T^{n}f)(x)|\dfrac{\|g\|_{\infty}}{n!}.$$  But this cannot say anything to the operator norm $\|T\|$.

What should I do? Thank you!

Edit:

The overall purpose is to derive that $\|T\|\leq\frac{1}{n!}$. So if there is any other way to find this, it will also be really good.

Edit2: Proof

Okay, I confused myself in the first place. The operator norm has nothing to do with the sup-norm since $T:L^{2}\longrightarrow L^{2}$, so what we should do is to use Cauchy-Schwarz and $\|f\|_{L^{2}}$. See below my own answer to my post for details.

Also, Ruy gave a general theorem about how to compute the operator norm for Hilbert-Schmidt operator, and my computation will show you the idea about the proof. Basically, the absolute value of operator will be bounded by $\|f\|_{L^{2}}$ multiplication of the $L^{2}-$integral of the kernel, so we should expect such a theorem.

Thank you so much for all the users who helped me!!

答:
If you are satisfied with a weaker bound $\frac1{\sqrt{n}(n-1)!}$ then we can use Cauchy-Schwarz on the $\|T^nf\|_\infty$ estimate:

\begin{align}
\|T^nf\|_2 &\le \|T^nf\|_\infty \\
&\le \frac1{(n-1)!} \sup_{x\in[0,1]}\int_0^1 |x-r|^{n-1}|f(r)|\,dr\\
&\le \frac1{(n-1)!} \sqrt{\sup_{x\in[0,1]}\int_0^1 |x-r|^{2n-1}\,dr} \sqrt{\int_0^1 |f(r)|^2\,dr}\\
&= \frac1{(n-1)!} \sqrt{\frac1{2n-1}}\|f\|_2\\
&\le \frac1{\sqrt{n}(n-1)!}\|f\|_2.
\end{align}


问:
Linear arrangement of 4 eagles, 3 horses and 5 dragons in which no two dragons get placed side by side
I want to check my answer because I am not sure the answer key is correct.

Consider there are 12 creatures: 4 eagles, 3 horses and 5 dragons. They need to be placed in a row in which no two dragons get placed side by side. How many arrangements are there? Note that each creature must be considered as a distinct object.

My attempt:

The eagles and horses must be placed first as depicted as numbers below.

__ 1 __ 2 __ 3 __ 4 __ 5 __ 6 __ 7 __

A line represents a seat for a dragon.

There are $7!$ ways to place eagles and horses and $\frac{8!}{(8-5)!}=\frac{8!}{3!}$ ways to place dragons. So there are
$$
\tfrac{4}{3}(7!)^2
$$
ways to arrange those creatures in which no 2 dragons get placed side by side.

Is it my combinatorial argumentation correct?

答: Yes it is perfect, and couldn't have done better.The answer key is wrong</p>


答:
I think your "score" is the degree sequence.

I doubt that the degree sequence characterizes an arbitrary graph.

This is from https://faculty.math.illinois.edu/~kumbhat2/DSFpairs.pdf

The degree sequenced(G) of a graph G is the list (d1,d2,...,dn) of
degrees of the vertices o G, written in non increasing order. We say a
graph class C has a degree sequence characterization if it is possible to
determine whether or not a graph G is in C based solely on the degree
sequence of G. Degree sequence characterizations do not exist for most
graph classes, but are extremely useful when they do exist.

Title: Is an unlabeled graph $G$ characterized by the degrees of endpoints of edges in $G$?

Body:
An unlabeled graph is just an isomorphism class of graphs. I am trying to make a way to store an unlabeled graph in a computer without necessarily charactering it with a labeled graph.

For instance, take the labeled graph $G=(V,E)$, where $V=\{A,B,C,D\}$ and $E=\{\{A,B\},\{B,C\},\{B,D\},\{C,D\}\}$. Suppose I want to store this in a computer but without assigning labels to the vertex set, so I am merely storing the structure of the graph.

I know the degree sequence of $G$ is $D=(3,2,2,1)$. From here, I can create an multiset of 2-sets of $D$ that corresponds the edges in $G$: $E'=\{\{1,3\},\{2,3\},\{2,3\},\{2,2\}\}$. Note every 2-set in $E'$ is an edge in $G$, but written in terms of the degrees of the adjacent vertices, rather than the indices of the adjacent vertices. In other words $\{u,v\} \in E \iff \{\deg u, \deg v\} \in E'$.

It is clear any graph isomorphic to $G$ will have this same $E'$.
Is it true that every unlabeled graph will have a unique $(D,E')$?

Is this a way to uniquely characterize every unlabeled graph? If not, am I missing something?


问:
Parametric equations for given line
How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.

答:
A line will be parameterized by $y\colon x\mapsto ax +b$. For instance, to solve 1), you want
$$
\begin{align*}
1 &= a\cdot3 +b \\
4 &= a\cdot(-5) +b
\end{align*}
$$
since $(3,1)$ and $(-5,4)$ must be of the form $(x,y(x))$ for the desired line. Solve this system to get the coefficients $a$ and $b$.</p>

Title: Parametric equations for given line


问:
$I-M$ is an non-invertible matrix
Why if $M = (m_{ij})$, with dimension $n \times n$, is a stochastic matrix, then the matrix $I-M$ must be non-invertible? I think if we echelon $I-M$, then eventually we will get a zero line (i.e. $\det(I-M)=0$), but I am not sure how I can prove it.

答:
Each column of your matrix sums to one so we have
$$(I-M)^T\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix} = (I-M^T)\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix} - M^T\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix} = \begin{bmatrix} 1 - \sum_{j=1}^n m_{j1} \\ 1 - \sum_{j=1}^n m_{j2} \\ \vdots \\ 1 - \sum_{j=1}^n m_{jn}\end{bmatrix} = 0$$
so the kernel of $(I-M)^T$ is nontrivial, which implies that $(I-M)^T$ is singular. Now of course $I-M$ is singular as well.

Title: $I-M$ is an non-invertible matrix


问:
Prove $(2−1)^2−4$ is not a perfect cube for any integer
Last week, my teacher asked us to prove this:
Prove that $(2−1)^2−4$ is not a perfect cube for any integer
I tried this approach:$y=2k$ and $y=2k+1$
but I couldn't prove it this way.

答:
$$(2 y - 1)^2 - 4=(2y-1+2)(2y-1-2)=(2y+1)(2y-3)$$
$(2y-3)$ and $(2y+1)$ are coprime. For euclidean algorithm

$\gcd (2y+1,2y-3)=\gcd(2y+1,4)=1$

because $(2y+1)$ is odd and is coprime with $4$

So the two numbers have no common factors. Thus or both are cubes or their product cannot be a cube.

As the difference is $4$ we see that $2y-3$ and $2y+1$ cannot be both cubes because beside the case $1^3-0^3=1$ the minimum difference between two cubes is $2^3-1^3=7$

Therefore $(2 y - 1)^2 - 4$ cannot be a cube.


问:
matrix multiplication can be thought of as taking partial derivatives?
I'm struggling with a sentence in the following text: ross.

It says that the action $y=Ax$ is

like performing many inner products, one for each row of $A$ -

which I agree with. But then it says:

Generally, the matrix elements can be thought of as the partial derivatives

$A_{ij} = \frac{dy_i}{dx_j}$

Why is that true?

答:
Consider a linear function $y:\mathbb{R}^n\longrightarrow \mathbb{R}^n$.
$$y_i = \sum_{l=1}^n A_{il}x_l$$then
$$\frac{\partial y_i}{\partial x_j} = \frac{\partial }{\partial x_j}\left(\sum_{l=1}^n A_{il}x_l\right) = \sum_{l=1}^n A_{il}\frac{\partial x_l}{\partial x_j} = \sum_{l=1}^n A_{il}\delta_{lj} = A_{ij}$$


问:
Proof: If convergence of Dirichlet series in one point, then uniform convergent in a sector
I'm currently reading the proof the theorem: if a Dirichlet serie converges at some point, $s_0$, then the serie is uniformly convergent in a sector around that point. (Montgomery and Vaughan: Multiplicative Number Theory I, Thr 1.1).

But there are small things I don't understand.

So, first we define $R(u)=\sum_{n>u}a_n n^{-s_0}$ to be the remainter term of a Dirichlet serie. We note that $a_n=(R(n-1)-R(n))n^{s_0}$. Then the proof say: "so by partial summation:

\begin{align}
\sum^{N}_{n=M+1}a_nn^{-s}&=\sum^{N}_{n=M+1}(R(n-1)-R(n))n^{s_0-s}\\
&=R(M)M^{s_0s}-R(N)N^{s_0-s}-\sum^{N}_{n=M+1}R(n-1)((n-1)^{s_0-s}-n^{s_0-s})"
\end{align}

Maybe someone can explain why the last equal-sign holds.

The next question is how I can justify this bound:

$$\int^{\infty}_{M} u^{\sigma_0 - \sigma -1} \,du \leq \frac{1}{\sigma - \sigma_0}$$

when $|R(u)|\leq\epsilon$ for all $u\geq M$ and if $\sigma>\sigma_0$?

答:
It is possible to skip that summation by parts. Similar to what the book did, let's define

$$
R(u)=\sum_{n>u}{a_n\over n^{s_0}}
$$

where $s_0$ is chosen such that the series

$$
\alpha(s)=\sum_{n=1}^\infty{a_n\over n^s}
$$

converges. Now, let's pick two positive integers $M<N$, then by Riemann-Stieltjes integration (a detailed account is available in section 7.8 of Tom M. Apostol's Mathematical Analysis), we have

$$
\begin{aligned}
\sum_{n=M+1}^N{a_n\over n^s}
&=\sum_{M<n\le N}{a_n\over n^{s_0}}{1\over n^{s-s_0}}=\int_M^Nx^{s_0-s}\mathrm d[\alpha(s_0)-R(x)] \\
&=[\alpha(s_0)-R(x)]x^{s_0-s}|_M^N-\int_M^N[\alpha(s_0)-R(x)]\mathrm d(x^{s_0-s}) \\
&=R(M)M^{s_0-s}-R(N)N^{s_0-s}+\color{red}{\alpha(s_0)(N^{s_0-s}-M^{s_0-s})} \\
&-\color{blue}{\int_M^N\alpha(s_0)\mathrm d(x^{s_0-s})}+\int_M^NR(x)\mathrm d(x^{s_0-s})
\end{aligned}
$$

It can be verified that the red and the blue components cancelled each other out, leaving us

$$
\sum_{n=M+1}^N{a_n\over n^s}=R(M)M^{s_0-s}-R(N)N^{s_0-s}+\int_M^NR(x)\mathrm d(x^{s_0-s})
$$

For the second question, we have

$$
\int_M^\infty x^{\sigma_0-\sigma-1}\mathrm dx=\left.{x^{\sigma_0-\sigma}\over\sigma_0-\sigma}\right|_M^\infty={M^{\sigma_0-\sigma}\over\sigma-\sigma_0}
$$

When $\sigma>\sigma_0$ there is $\sigma_0-\sigma<0$, which implies $M^{\sigma_0-\sigma}\le1$, leaving us the following bound:

$$
\int_M^\infty x^{\sigma_0-\sigma-1}\mathrm dx\le{1\over\sigma-\sigma_0}
$$


问:
Inequalities for Inner products.
Are there special versions of Inequalities for inner products?

Thank you very much.

答:
You start with the Cauchy-Schwarz inequality:
$$
-2 \langle \alpha(x-y),\beta(v-w)\rangle \le 2 |\langle \alpha(x-y),\beta(v-w)\rangle| \\
\le 2 \Vert \alpha(x-y) \Vert \cdot \Vert \beta(v-w) \Vert \\
= 2 \bigl( \frac{1}{\sqrt \delta }\Vert \alpha(x-y) \Vert \bigr) \cdot
\bigl( \sqrt \delta \Vert \beta(v-w) \Vert \bigr)
$$
and then apply Young's inequality with $a = \frac{1}{\sqrt \delta }\Vert \alpha(x-y)$, $b= \sqrt \delta \Vert \beta(v-w)$ and $p=q=2$.

Title: Inequalities for Inner products.


问:
divergence in polar coordinates
For a vector field $X$, the divergence in coordinates is given by $\nabla\cdot X=\sum_n\frac{X^i}{\partial x^i}$. In polar coordinates, the metric is $\begin{bmatrix}1 & 0\\ 0 & r^2\end{bmatrix}$, and so $\frac{1}{\sqrt{g(\frac{\partial}{\partial r},\frac{\partial}{\partial r})}}\frac{\partial}{\partial r}=\frac{\partial}{\partial r}$ and $\frac{1}{\sqrt{g(\frac{\partial}{\partial\theta},\frac{\partial}{\partial\theta})}}\frac{\partial}{\partial\theta}=\frac{1}{r}\frac{\partial}{\partial\theta}$ are unit vectors. Then for $X=X_{r}\frac{\partial}{\partial r}+X_{\theta}\frac{\partial}{r\partial\theta}$,  $\nabla\cdot X=\frac{\partial X_r}{\partial r}+\frac{\partial}{\partial\theta}\frac{X_{\theta}}{r}=\frac{\partial X_r}{\partial r}+\frac{1}{r}\frac{\partial X_{\theta}}{\partial\theta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?

答:
If you look at equation (114) here https://mathworld.wolfram.com/CylindricalCoordinates.html

you can see that you said is essentially right but you also need to add on the "connection" terms with the Christoffel symbols. Then you simplify and you get the usual formula.

Title: divergence in polar coordinates


问:
Combinations without repeats with n = k
I'd like to generate all possible unique combinations of a set of characters.
For example, with the set [a,b], we'd get [[a,a],[a,b],[b,b]].
I've been looking to find even a formula giving me the number of possibilities, without success... Is it possible ?
Of course I could bruteforce the answer, but that's not efficiency, nor even possible once we start ramping up the numbers.

Thanks for your time.

答: I think you are looking for the formula for the number of combinations (order matters) with repetition of elements, where the number of objects chosen is the same as number of objects given. This is given by $\frac{(k+n-1)!}{(k)!(n-1)!}$. When they are equal you can just reduce this to $\frac{(2n-1)!}{n!(n-1)!}$, which for $n=2$ (like with the example you posted) is $\frac{3!}{2!1!} = 3$.

Title: Combinations without repeats with n = k


答:
The mechanical way to find the parametric equation for any lines going through two points $A$ and $B$ is to write it in vector form, for some constant $\mu\in\mathbb{R}$:

$$\vec{l}=\overrightarrow{OA}+\mu\overrightarrow{AB}$$

For instance, in your case, for the first line, we have: $A(3,1)$ and $B(-5,4)$, therefore the equation of our line is:

$$\vec{l_{1}}=\begin{pmatrix}3 \\ 1\end{pmatrix}+\mu\begin{pmatrix}-5-3 \\ 4-1\end{pmatrix}=\begin{pmatrix}3 \\ 1\end{pmatrix}+\mu\begin{pmatrix}-8 \\ 3\end{pmatrix}$$

If you want the non-vector parametric equations you simply take the $x$ and $y$ components separately, i.e.:

$$\begin{align}x&=3-8\mu \\ y&= 1+3\mu\end{align}$$

Title: Parametric equations for given line

Body:
How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.


问:
After how much time the same configuration will repeat?

A ball rotates at a rate $r$ rotations  per second and simultaneously revolves around a stationary point $O$ at a rate $R$ revolutions per second $(R<r)$. The rotation and revolution are in the same sense. A certain point on the ball is in the line of the centre of the ball and point $O$ at a certain time. This configuration repeats after a time

$(1)\ $ $\frac{1}{r-R}$

$(2)\ $ $\frac{1}{R}-\frac{1}{r}$

$(3)\ $ $\frac{1}{r+R}$

$(4)\ $ $\frac{1}{R}+\frac{1}{r}$

How do I solve this aptitude question? Thanks for your time.

Source $:$ CSIR NET JUNE $2019.$

答:
As rotation and revolution are both in the same direction (say, counterclockwise), the point will come in the line connecting centers after one complete rotation. Say that happens after part revolution of angle $\theta$ against one full revolution ($2\pi$). In that time the rotation has completed by $(2\pi + \theta)$ (more than one rotation of ($2\pi$).

One rotation takes $\frac{1}{r}$ second and one revolution takes $\frac{1}{R}$ second. Equating the time taken -

$\displaystyle \frac{2\pi + \theta}{2\pi r} = \frac{\theta}{2\pi R}$

Now please find $\theta$ and then plug in either Right Hand side or in Left Hand side to find the time.


问:
Proof for $ax + by \equiv 0 \pmod c$ implies $x\equiv 0 \pmod c$ and $y\equiv 0 \pmod c$ when $a,b$ are co-prime
I believe this is true, but I'm not sure if I'm making a mistake in my rather informal proof below.

Proof sketch:
First, $ax+by \equiv 0\pmod c \Leftrightarrow \exists k \in \mathbb{Z}, ax+by = kc$.

Next, since $a,b$ are co-prime, we know that the only solutions to $ax+by = kc$ must be multiples of the Bezout coefficients $(s,t)$ such that $as + bt = 1$.
This means that $x = s(kc)p$ and $y=t(kc)q$ for some $p,q\in \mathbb{Z}$, which implies $x\equiv 0 \pmod c$ and $y\equiv 0\pmod c$.(This is the part I'm unsure about.)

Am I making any mistakes or do you see an easier way to prove this?

答:
The claim is not true.

E.g. $ 2 \times 1 + 3 \times 1 \equiv 0 \pmod{5}$.

Do you see where you made a reasoning error?


答:
Let's take an example:
$$\mathbf{y} = A\mathbf{x},$$
where
$$A=\begin{bmatrix}1&2\\3&4\end{bmatrix}.$$

So, for instance, if $\mathbf{x}=(1,1)$, we have
$$\mathbf{y} = \begin{bmatrix}1&2\\3&4\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$$

The partial derivative $\frac{\partial y_i}{\partial x_j}$ describes what happens to $y_i$ if you change $x_j$, leaving the rest constant. Let's see what happens if you increase $x_1$ by $1$, that is
$$\mathbf{y} = \begin{bmatrix}1 & 2\\ 3 & 4 \end{bmatrix} \begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}4\\10\end{bmatrix}.$$
As you can see, the $y_1$ coordinate increases by $1$, and the $y_2$ coordinate increases by $3$. This directly corresponds to the elements $A_{11}$ and $A_{21}$ in the matrix, and that is why we can consider them as partial derivatives. You can check for yourself that if you increase $x_2$ by $1$, then $y_1$ will increase by $2$ (corresponding to $A_{12}$), and $y_2$ will increase by $4$ (corresponding to $A_{22}$).

Title: matrix multiplication can be thought of as taking partial derivatives?

Body:
I'm struggling with a sentence in the following text: ross.

It says that the action $y=Ax$ is

like performing many inner products, one for each row of $A$ -

which I agree with. But then it says:

Generally, the matrix elements can be thought of as the partial derivatives

$A_{ij} = \frac{dy_i}{dx_j}$

Why is that true?


问:
Induction for modular arithmetic
I've been trying this question for a while now: Let $a_0$ and $a_1$ be integers. For an integer $n \geq2$ define $a_n=a_{n-1}+a_{n-2}$. Prove that if $a_2\equiv6a_0\pmod {19}$ and $a_3\equiv6a_1 \pmod{19}$, then $a_{n+2}\equiv6a_n \pmod {19}$ for any non-negative integer $n$. I did the following:

We use strong induction to prove this.

Base cases: $n=0$ and $n=1$, which are trivial to prove

Inductive step: For any non-negative integer $k$, $P(k) \implies P(k+1)$.

What should I do now? I've wrote the simple parts of the proof but can't seem to get the inductive step right. Any help is appreciated.

答:
Take $n\ge 2$.

For strong induction, assume $ a_k\equiv 6 a_{k-2}$ for all $k$ between $2 $ and $n+1$.

Now $a_{n+2}=a_{n+1}+a_{n}$, and by strong induction $a_{n+1}\equiv6a_{n-1}$ and $a_{n}\equiv6a_{n-2}$.

Therefore, $a_{n+2}\equiv6a_{n-1}+6a_{n-2}=6a_n.\square
$

Title: Induction for modular arithmetic


问:
how to calculate $d\Omega(f)$ here
the question was to find $d \Omega(f)$ with :

$$ \Omega : (E,[.]) \to (F,||.||) \\f \to -f'' +f^3$$

$
[f] = |f'(0)| + ||f''|| $
; $ ||f|| = Sup_{[0,1]}|f(x)| $

the answer is given to me like this :

$ d\Omega(f)(h)= 3f^2h - h'' $
 , without any more details

i want to know what is the role of h (i thought it's df at first) here , and how they calculated this answer

note :

it is true that $$lim_{[h]\to 0} {||\Omega(f+h) - \Omega(f) - d\Omega(f)(h)||\over [h]}=0$$

but i wanna know how they got that result and the role of (h) in the formula ? i struggled alot with this definition to understand it !

答:
I am assuming that $E=C^2([0,1])$ and $F=C([0,1])$.  Also, if $T$ is a linear transoformation, we gonna use the symbol $\langle T,x\rangle =T(x)$. Remember that the derivative of a function $F$ between two Banach spaces $X$ and $Y$  is a function $F':X\to L(X,Y)$ between $X$ and the set of all linear bounded transformations between $X$ and $Y$, $L(X,Y)$.

Write $T_1(f)=f^3$ and $\langle T_2, f\rangle =-f''$. As you can see, $T_2$ is a linear transformation, hence the derivative of $T_2$ is constant and equal to $T_2$ for all $f$, which implies that $$\langle T'_2(f),h\rangle=\langle T_2,h\rangle =-h''$$

Let's calculate $T'_1$. Fix some $h\in E$ and note that for $t$ real

\begin{eqnarray}\tag{1}
 \frac{T_1(f+th)(x)-T_1(f)(x)}{t} &=& \frac{(f(x)+th(x))^3-f(x)^3}{t}      \\
  \end{eqnarray}

For eaxh fixed $x\in [0,1]$, we can apply the mean value theorem to the function $y^3$ to conclude that $$\tag{2}\frac{(f(x)+th(x))^3-f(x)^3}{t}=\frac{3y(t,x)^2th(x)}{t}$$

where $y(t,x)\in [f(x),f(x)+th(x)]$ (the last interval is not oriented). Note that for each fixed $x$, if we let $t\to 0$, we conclude that $y(t,x)\to f(x)$, so, from $(1)$ and $(2)$ we have that $$\lim_{t\to 0}\frac{(f(x)+th(x))^3-f(x)^3}{t}=3f(x)^2h(x)$$

This is de Gateux derivative of $T_1$ with respect to the direction $h$. Take your conclusions from here.

Title: how to calculate $d\Omega(f)$ here


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
For $k\ge 1$ and $0<x<2^k$, suppose$$(2^{2k}-x^2) |  (2^{2k}-1)$$for some $x>1$, and hence that $2^{2k}-x^2$ is composed only of the prime factors of $2^{2k}-1$ but lacks one or more of them (or perhaps contains all of the distinct factors but with one or more of them to a lesser power).

Thus let$$2^{2k}-1=(2^k-1)(2^k+1)=pqr\cdot stu$$and suppose, for some $x>1$, that$$2^{2k}-x^2=pq\cdot stu$$and hence divides $2^{2k}-1$.

Then since for $x\ge1$,$$2^{2k}-x^2=(2^k-x)(2^k+x)$$and hence the sum$$(2^k-x)+(2^k+x)=2^{k+1}$$then$$pqr+stu=pq+stu=2^{k+1}$$which is impossible.

Hence it is clear that removing one or more prime factors from either $2^k-1$ or $2^k+1$, while leaving the other addend intact, must make their sum less than $2^{k+1}$. Still less can their sum be $2^{k+1}$ if one or more prime factors are removed from both $2^k-1$ and $2^k+1$.

The remaining possibility is that removing one or more prime factors from both $2^k-1$ and $2^k+1$ and re-arranging the remaining prime factors, might yield a sum $=2^{k+1}$.

E.g. for $k=6$, $(2^k-1)(2^k+1)=63\cdot65=3^2\cdot7\cdot5\cdot13$, and$$3^2\cdot7+5\cdot13=2^{k+1}=2^7$$Removing one $3$-factor and re-arranging the four remaining distinct prime factors in the seven possible ways we get$$3\cdot7+5\cdot13=86$$$$3\cdot5+7\cdot13=106$$$$3\cdot13+5\cdot13=74$$$$3+7\cdot5\cdot13=458$$$$5+3\cdot7\cdot13=278$$$$7+3\cdot5\cdot13=202$$$$13+3\cdot5\cdot7=118$$Noteworthy here is that all sums are odd multiples of $2^1$.

For $k=10$, removing one of the $5$-factors from $2^{2k}-1=1023\cdot1025=3\cdot11\cdot31\cdot5^2\cdot41$, I find the fifteen possible sums of two addends containing the remaining five distinct primes are all odd multiples of $2^2$.

If it could be shown, then, that all such sums dividing $2^{2k}-1$ are odd multiples of some power of $2$, and hence not equal to $2^{k+1}$, it would follow that, for $k\ge1$ and $x>1$, $2^{2k}-x^2$ does not divide $2^{2k}-1$.

This of course is not a full answer, but it lays out a possible approach.


问:
Let E be the standard basis and S another basis for $\mathbb{R}^2$. Assume the matrix M is understood to be in terms of E. Find $[M]_S.$
The question: Let $E = \left\{\begin{pmatrix} 1\\ 0\end{pmatrix}, \begin{pmatrix} 0\\ 1\end{pmatrix} \right\}$ be the standard basis for $\mathbb{R}^2$. The set $S = \left\{\begin{pmatrix} 1\\ 2\end{pmatrix}, \begin{pmatrix} -3\\ 2\end{pmatrix} \right\}$ is another basis for $\mathbb{R}^2$, Assume the matrix $M =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$ is understood to be in terms of $E$. Find $[M]_S.$

I am really confused on what this is asking of me and how I should approach this. I think that it is saying that I have $[M]_E$, so then to find $M$. I have tried
$
\left[
\begin{array}{cc|cc}
3 & 5 & 1 & 0 \\
-3 & 2 & 0 & 1 \\
\end{array}
\right]
\rightarrow
\left[
\begin{array}{cc|cc}
1 & 0 & \frac{2}{21} & \frac{-5}{21}\\
0 & 1 & \frac{1}{7}& \frac{1}{7}\\
\end{array}
\right]
$, so that I am left with $M = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}$. I then used this new $M$ to solve for $[M]_S$ by doing $MS = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}
\begin{bmatrix}
1 & -3\\
2& 2\\
\end{bmatrix}=
\frac{1}{21}
\begin{bmatrix}
-8 & -16\\
9& -3\\
\end{bmatrix}
$.

But this seems extremely wrong. My professor didn't do a great job going over bases, so any advice would be appreciated.

答:
$$[M]_E =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$$
means that the linear map $M$ acts on the basis $E$ as

$$M\begin{bmatrix}1 \\ 0 \end{bmatrix} = 3\begin{bmatrix}1 \\ 0 \end{bmatrix}-3\begin{bmatrix}0 \\ 1 \end{bmatrix}, \quad M\begin{bmatrix}0 \\ 1 \end{bmatrix} = 5\begin{bmatrix}1 \\ 0 \end{bmatrix}+2\begin{bmatrix}0 \\ 1 \end{bmatrix}$$
or in general
$$M\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}3x+5y \\ -3x+2y \end{bmatrix}.$$

You have to find a matrix $$[M]_S = \begin{bmatrix} \alpha & \gamma \\ \beta & \delta\end{bmatrix}$$
such that
$$M\begin{bmatrix} 1 \\ 2\end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 2\end{bmatrix} + \beta \begin{bmatrix} -3 \\ 2\end{bmatrix}, \quad M\begin{bmatrix} -3 \\ 2\end{bmatrix} = \gamma \begin{bmatrix} 1 \\ 2\end{bmatrix} + \delta \begin{bmatrix} -3 \\ 2\end{bmatrix}.$$


问:
Gaussian integral with a complex exponent
In physics, one often encounters the integral:
$$I = \int_{-\infty}^\infty e^{-\frac{1}{2}zx^2}dx $$
for $z\in\mathbb{C}$. Inputting this into Mathematica, I get that $I=\frac{\sqrt{2\pi}}{\sqrt{z}}$ so long as $\text{Re}(z)>0$. I'm a little confused by this (isn't $\sqrt{z}$ at risk of being multivalued since there are two square roots?), so I would like to derive it.

I followed a derivation which claims to do this in polar coordinates. Squaring the above and switching to polar coordinates using $r^2=x_1^2+x_2^2$, one finds that:
$$ I^2= \int_{-\infty}^\infty e^{-\frac{1}{2}zx_1^2}dx_1 \int_{-\infty}^\infty e^{-\frac{1}{2}zx_2^2}dx_2 =\int_0^{2\pi}\int_0^\infty re^{-zr^2/2} dr d\phi$$
which, under a change of variables $u=zr^2/2$ yields:
$$I^2=\frac{2\pi}{z}$$
So far so good. It is the next step that confuses me, where the square root is taken on both sides to yield:
$$I=\frac{\sqrt{2\pi}}{\sqrt{z}}$$
Suppose $z\in \mathbb{R}$ - then I'm fine with this, since $I$ must be real and positive, so this can only hold for $z>0$ and it must be the positive root, so there's no ambiguity. But I don't understand (a) how to derive the above result for $z\in\mathbb{C}$, and (b) why the expression $\sqrt{z}$ for $z\in\mathbb{C}$ is well-defined at all. How do I resolve these concerns?

答:
The square root of a complex number is well-defined, as long as you make sure there is no ambiguity on its argument.

If $z = re^{i\theta}$ ($r > 0, \theta \in (-\pi, \pi]$, then $\sqrt{z} := \sqrt{r} e^{i\theta/2}$


答:
In this case, the square of of $z$ is defined as the analytic function $f(z)$ in the half-plane
$$
\Omega=\{x+iy: x>0\}
$$
which is simply connected, and misses $0$, and $f$ coincides with the square root of positive reals, i.e., $f(x)=\sqrt{x}$, for $x>0$.

Hence
$$
f(r\mathrm{e}^{i\theta})=r^{½}\mathrm{e}^{i\theta/2}
$$
when $r>0$ and $\theta\in(-\pi/2,\pi/2)$.

Title: Gaussian integral with a complex exponent

Body:
In physics, one often encounters the integral:
$$I = \int_{-\infty}^\infty e^{-\frac{1}{2}zx^2}dx $$
for $z\in\mathbb{C}$. Inputting this into Mathematica, I get that $I=\frac{\sqrt{2\pi}}{\sqrt{z}}$ so long as $\text{Re}(z)>0$. I'm a little confused by this (isn't $\sqrt{z}$ at risk of being multivalued since there are two square roots?), so I would like to derive it.

I followed a derivation which claims to do this in polar coordinates. Squaring the above and switching to polar coordinates using $r^2=x_1^2+x_2^2$, one finds that:
$$ I^2= \int_{-\infty}^\infty e^{-\frac{1}{2}zx_1^2}dx_1 \int_{-\infty}^\infty e^{-\frac{1}{2}zx_2^2}dx_2 =\int_0^{2\pi}\int_0^\infty re^{-zr^2/2} dr d\phi$$
which, under a change of variables $u=zr^2/2$ yields:
$$I^2=\frac{2\pi}{z}$$
So far so good. It is the next step that confuses me, where the square root is taken on both sides to yield:
$$I=\frac{\sqrt{2\pi}}{\sqrt{z}}$$
Suppose $z\in \mathbb{R}$ - then I'm fine with this, since $I$ must be real and positive, so this can only hold for $z>0$ and it must be the positive root, so there's no ambiguity. But I don't understand (a) how to derive the above result for $z\in\mathbb{C}$, and (b) why the expression $\sqrt{z}$ for $z\in\mathbb{C}$ is well-defined at all. How do I resolve these concerns?


问:
Forster exercise 17.3 on Riemann-Roch theorem
Let $X$ Riemann Surface of genus $g$, and $D$ is a divisor on $X$. I try to prove $l(D)=dim H^0 (X,O_D) ≤ 1+degD$ when $−1≤degD≤g−1$ Using the Riemann-Roch theorem and Serre duality. The notations here : $H^0(X,O_D)$ is obviously the $0$-th cohomology group of $X$ with respect to the sheaf $O_D$ where $O_D(U)=\{f∈\mathcal{M}(U)|ordx(f)≥−D(x)∀x∈U\}$ and $\mathcal{M}$ is the sheaf of meromorphic functions.

The statement of Riemann-Roch in O.Forster Book is : for a compact Riemann surface $X$ of genus $g$ and a divisor $D$ on it, $$dimH^0(X,O_D)−dimH^1(X,O^D)=1−g+degD.$$

I tried to use the Serre duality which asserts that $dimH^1(X,O_D)=dimH^0(X,Ω_{−D})$ where $Ω_{−D}$ is the sheaf of meromorphic $1$-forms that are multiples of $−D$. Or, use the the exact sequenece
$$0\to H^0(X,D−p)→H^0(X,D)→\mathbb{C}→0.$$ I find that, if $D=P-N$, where $P,N$ two effective divisors, then $l(D)≤ 1+deg(P)$. But i cannot find the finale result.

Could I please get a hint?

答:
If $D$ is not effective, $h^0(X,D) = 0$, so we can assume $D$ effective. By Riemann-Roch and Serre duality,

$$
h^0(X,D) = 1 + \deg(D) - g + h^0(X, \Omega_{-D})
$$

The inequality follows if $h^0(X, \Omega_{-D}) \le g$. But $h^0(X, \Omega) = g$ in the first place, and imposing zeroes (twisting down by $D$ effective) can only decrease $h^0$.