问:
Non-isomorphic curves with the same genus
I need to show that the curves $y^7=x^2(x-1)$ and $xy^3+zx^3+yz^3=0$ are not isomorphic even if the are both of genus 3. I have tried to show that they are a quartic and a hyperelliptic curve in order to show that they are in two different class of isomorphism.
I have normalized the first one with this sequence of blow-ups and transformations:

$x=x_1y_1, y=y_1     \rightarrow x_1^3y_1-y_1^5-x_1^2=0$

$x_1=x_2y_2, y_1=y_2 \rightarrow x_2^3y_2^2-y_2^3-x_2^2=0$

$x_2=x_3y_3,y_3=y_2  \rightarrow x_3^3y_3^3-y_3-x_3^2=0$

$x_3=\frac{X_3}{Z_3},y_3=\frac{Y_3}{Z_3} \rightarrow X_3^3Y_3^3-Y_3Z_3^5-X_3^2Z_3^4=0$

$y_3'=\frac{Y_3}{X_3}, z'_3=\frac{Z_3}{X_3} \rightarrow (y'_3)^3-(y'_3)(z'_3)^5-(z'_3)^4=0$

$y_3'=y_4z_4, z'_3=z_4\rightarrow y_4^3-y_4z_4^3-z_4=0$

The last one is nonsingular but is almost exactly the other, so what is wrong?
Thanks!

答:
The curve
$$
xy^3 + yz^3 + zx^3 = 0
$$
is the famous Klein quartic, see https://en.wikipedia.org/wiki/Klein_quartic. Its automorphism group has order 168, the maximal among curves of genus 3. In particular, it has an automorphism of order 7.

On the other hand, Theorem 2(a) of [Homma, Masaaki. Automorphisms of prime order of curves. Manuscripta Math. 33 (1980/81), no. 1, 99--109] shows that a curve $C$ of genus $g \ge 2$ with an automorphism $\sigma$ of prime order $2g + 1$ is birational to one of the curves
$$
y^{2g+1} = x^a(x-1)^b,
$$
where $a,b \ge 1$ and $a + b \le g + 1$. Moreover, the proof of Proposition 3.1 in the paper shows that $a + b$ is the degree of a covering $f \colon C \to \mathbb{P}^1$ which is equivariant for $\sigma$.

All this applies easily to the Klein quartic. One can choose the morphism $f$ to be the projection
$$
(x,y,z) \mapsto (x:y),
$$
then its degree is 3, hence $a + b = 3$, hence (after a possible change of coordinates) the Klein quartic is birational to the curve
$$
y^7 = x^2(x-1).
$$


问:
Showing $\frac{1}{f(z)}$ is holomorphic
Show that if $f(z)$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and $f(z)\rightarrow\infty$ as $z\rightarrow 0$ then f has a pole at $0$

My attempt:

f is holomorphic on $\mathbb{C}\setminus\{0\}$ so f is holomorphic on some ball radius $r$ about $0$ denoted $B(0,r)$

we know that $f(z)\rightarrow\infty$ as $z\rightarrow 0$ so $f$ is unbounded near $0$ which implies that $\frac{1}{f(z)}\rightarrow 0$ as $z\rightarrow 0$

My question:

Do we know that $\frac{1}{f(z)}$ extends to a holomorphic function on $B(0,r)$ thus showing $0$ is a pole. Or are there better ways to think about this question!

Any help would be amazing

答: So, $0$ is an isolated singularity, then it must be a removable singularity, a pole, or an essential singularity. But if it was a removable singularity, then the limit $\lim_{z\to0}f(z)$ would be a complex number. And, if it was an essential singularity, then, by the Casorati-Weierstrass theorem, the limit $\lim_{z\to0}f(z)$ would not exist. So, it must be a pole.

ContentLicense: CC BY-SA 4.0

Title: Showing $\frac{1}{f(z)}$ is holomorphic


问:
How to see whether a Boolean function with different quantifiers is true given a quantifier as an example?
P(x,y) is a Boolean function, and∀x∃yP(x,y) is true and the domain is nonempty, then witch ones must also be true?

```<code>1.∀x∀yP(x,y)
2.∃x∀yP(x,y)
3.∃x∃yP(x,y)
</code>```

Would anyone give me a hand on how to solve this problem? I had thought that I can simply make a Boolean function as an example but later realized that it would not cover all the cases. Thank you for all the help

答: Check that $p:\{0,1\}^2\rightarrow \{0,1\}$ defined by $p(1,0)=p(0,0)=1$ and $p(1,1)=p(0,1)=0$ satisfies the hypothesis but not 1, 2. As the domain is non-empty, there exists at least one pair $(x,y)$ in the domain, so it is straightforward  to see that $\forall x\,\exists y\,p(x,y)\Rightarrow  \exists x\,\exists y\,p(x,y)$ for all boolean functions, so only 3 must also be true.


问:
What is the difference between local coordinates and standard coodinates?
I am reading "An introduction to Manifolds" by Loring Tu. There it is written that we denote the standard coordinates in $R^n$ by $r_1,r_2,....r_n$. If $(U,\phi:U\to R^n)$ is a chart of a manifold, we let $x^i=r^i\circ\phi$ to be the ith component of $\phi$  and write $\phi=(x^1,...x^n)$ and $(U,\phi)=(U,x^1,...x^n)$ . My question is why they've used $r_1,r_2,....r_n$ and $(x^1,...x^n)$ separately? And what does the ith component of $\phi$ mean? And what is the difference between local coordinates and standard coordinates? I am also attaching the screenshot where this is written.

答:
Let's go through these questions one-by-one. Let's say that we're working in $\mathbb{R}^3,$ and we have a point $p=(p_1, p_2, p_3).$ Then, $r^i(p)=p_i.$ That is, $r^i$ just extracts the $i$th component of your point in $\mathbb{R}^3.$ Your chart takes a point on the manifold and returns a point in $\mathbb{R}^n,$ and what $x^i$ does is simply take the $i$th coordinate of this point. That is, if $p\in U\subset M,$ then $x^i(p)=r^i (\phi(p))$. Observe that $\varphi(p)\in\mathbb{R}^n,$ and then $r^i$ simply takes the $i$th component (since $\varphi(p)$ lives in $\mathbb{R}^n$, it is a vector with $n$ components). So, these functions give coordinates in a manner analogous to what we already understand in Euclidean space.

Now, let me explicitly answer your final question. The standard coordinates on $\mathbb{R}^n$ are the coordinate functions in $\mathbb{R}^n$. Local coordinates on $U$ are the coordinate functions defined by your chart in $\mathbb{R}^n$. That is, the components of your chart are translated into Euclidean coordinates.


问:
In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$

Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.

Is the triangle $KLM$ also an equilateral side?
In what relation are the area of triangle $KLM$ the area of triangle$ ABC$?

My attempt:

I can see that $KLM$ is an equilateral triangle.  But, why the fraction $\frac{1}{7}$ is the area ratio of the triangle $KLM$ to the triangle $ABC$?

答: Let $O$ be the center of triangle $ABC$ and observe a rotation about $O$ for $120^{\circ}$. Then \begin{eqnarray}A &\mapsto& B\\B &\mapsto& C\\C &\mapsto& A \end{eqnarray}and thus \begin{eqnarray}A_1 &\mapsto& B_1\\B_1 &\mapsto& C_1\\C_1 &\mapsto& A_1 \end{eqnarray} and so intersection of segements $AA_1$ and $BB_1$ goes to intersection of segements $BB_1$ and $CC_1$ i.e. $M\mapsto K$ and similary $K\mapsto L$ and $L\mapsto M$ which meanst that $KLM$ is also equilateral.


问:
How much algebraic geometry is there in complex geometry (for example, Demailly)?
I wonder how much of modern algebraic geometry (schemes, etc) is there in complex (algebro-analytic) geometry. What I mean is complex algebraic (analytic) geometry in the sense of Demailly, Lasarsfeld etc.

I wonder if scheme theory (as in Grothendieck's work) is actually useful there.

答:
Oka's work in complex geometry used "ideals of indefinite extent" (or similar wording) which were operationally sheaves of ideals.

And as other comments and answers have indicated, at the very least, the short-to-long exact sequences of sheaf theory give a very clean explanation of Chern classes and other basic things. R. Gunning's yellow Princeton book on Riemann surfaces gives a very persuasive introduction to this (though using a "covering" version of sheaf cohomology rather than "derived functor of global section functor", but this is relatively a secondary detail).

ContentLicense: CC BY-SA 3.0


问:
Looking for the name of a Rising/Falling Curve
I'm looking for a particular curve algorithm that is similar to to a bell curve/distribution, but instead of approaching zero at its ends, it stops at its length/limit. You specify the length of the curve of the curve and its maximum peak, and the plot will approach its peak at the midpoint of length(the middle) and then it curves downward to its end. As a math noob, I may not be making any sense. Here's an image of the curve I'm looking for:

答:
Thanks to J.M.'s help, I was able to create the appropriate curve using sine wave and length of $\pi$(since $\pi$ is the point at which the curve approaches zero in a positive phase). I graphed this example using javascript and flot:

```<code>var d1 = [];
for (var x = 0; x < Math.PI; x += 0.1)
    d1.push([x, Math.sin(x)]);
</code>```

This means you can also adjust the length of the wave/curve by adding a multiplier to x:

```<code>var d2 = [];
var wavelength = 2
for (var x = 0; x < Math.PI; x += 0.1){
    var y =  Math.sin(x * wavelength)
        d2.push([x, y]);
}
</code>```

Here's an image of the result:

Body:
Thanks to J.M.'s help, I was able to create the appropriate curve using sine wave and length of $\pi$(since $\pi$ is the point at which the curve approaches zero in a positive phase). I graphed this example using javascript and flot:

var d1 = [];
for (var x = 0; x < Math.PI; x += 0.1)
    d1.push([x, Math.sin(x)]);

This means you can also adjust the length of the wave/curve by adding a multiplier to x:

var d2 = [];
var wavelength = 2
for (var x = 0; x < Math.PI; x += 0.1){
    var y =  Math.sin(x * wavelength)
        d2.push([x, y]);
}

Here's an image of the result:


问:
Looking for a heavy-tailed distribution
I want to find a probability distribution (given by its CDF $F$) such that
$$\liminf_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = 0 \quad \text{for all } \lambda > 0$$
and
$$\limsup_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = \infty \quad \text{for all } \lambda > 0.$$

The only thing that I could guess almost immediately is that this distribituion has to be heavy-tailed. Otherwise I have no clue.

答:
It suffices to find a non-negative integrable function $f$ such that for all $c>1$, there are sequences $(x_n'),(x_n'')$ tending to infinity such that you have $c^{x_n'}\int_{x_n'}^\infty f(x)\,\mathrm{d}x$ grows to infinity and  $c^{x_n''}\int_{x_n''}^\infty f(x)\,\mathrm{d}x$ decays to zero.

The trick to deal with all $c$ at the same time is to make $\int_{x_n''}^\infty f(x)\,\mathrm{d}x$ decay superexponentially, while making $\int_{x_n'}^\infty f(x)\,\mathrm{d}x$ decay only subexponentially, for instance by ensuring that $\int_{x_n'}^\infty f(x)\,\mathrm{d}x\geq 1/(2x'_n)$, while $\int_{x_n''}^\infty f(x)\,\mathrm{d}x\leq 2^{-2^{x''_n}}$.

I'll leave the details to you, but with a clever choice of $(x_n')$ and $(x_n'')$, you can ensure that for $A=\bigcup_n (x_n',x_n'')$ (the union of intervals), this is true for the function $f(x)=1/x^2\cdot \mathbf 1_A(x)$ (where $\mathbf 1_A$ is the indicator function of $A$).

Then you only need to rescale $f$ to ensure that it is a pdf. $F$ can be recovered by integrating the resulting function, if needed.

Title: Looking for a heavy-tailed distribution


问:
A sum of Euler products
Maybe someone have come across a particular sum related to Euler's product in the form:

$$S(x) = \frac{1}{5} (1 - \frac{2}{3}) + \frac{1}{7} (1 - \frac{2}{3})(1 - \frac{2}{5}) + \ ... \ = \sum_{p > 3}^{p < x} \ \frac{1}{p} \; \prod_{p_l > 1}^{p_l < p} \; (1 - \frac{2}{p_l})$$

where the sum and the products are over primes and we are interested in asymptotics at $x \to \infty$.

It is possible to approach it with estimating the products first (as $\sim \frac{1}{log^2 p_l}$) then summation over $p$ but it looks messy to keep track of the terms. Wonder if there can be alternative strategies?

答:
$\sum_p p^{-1} \prod_{2<q<p} (1-2/q)$ converges and $$\sum_{p> x }p^{-1} \prod_{2<q
 x }p^{-1} (C+o(1))(\log p)^{-2}$$ $$ =\sum_{n> x } (\log n)^{-1}n^{-1} (C+o(1))(\log n)^{-2}= \frac12(C+o(1)) (\log n)^{-2}$$
where $C=\lim_{x\to \infty} (\log x)^2 \prod_{2<q<x} (1-2/q)$ and $e^{2\gamma} C$ is the twin prime constant. We need the Mertens theorems.

Title: A sum of Euler products

Body:
$\sum_p p^{-1} \prod_{2<q<p} (1-2/q)$ converges and $$\sum_{p> x }p^{-1} \prod_{2<q<p} (1-2/q)=\sum_{p> x }p^{-1} (C+o(1))(\log p)^{-2}$$ $$ =\sum_{n> x } (\log n)^{-1}n^{-1} (C+o(1))(\log n)^{-2}= \frac12(C+o(1)) (\log n)^{-2}$$
where $C=\lim_{x\to \infty} (\log x)^2 \prod_{2<q<x} (1-2/q)$ and $e^{2\gamma} C$ is the twin prime constant. We need the Mertens theorems.


问:
How to analyze the following complex number?
$f(T) = \int_{0}^{1}e^{-iT\lambda(s)}ds$, where $\lambda(s)$ increases monotonously. I want to show that  $|f(T)|$ has an upper bound that decreases with $T$ for $T \gg 1$.

For example, if $\lambda(s) = s$ then $|f(T)| \le 2/T$.

This question comes from numerical analysis of my research. It's the expression of error analysis in adiabatic evolution. More specifically, the digital error has expression : $\sum_{k=1}^{L}e^{-iT\sum_{j<k}\lambda_{j}}/L$, where $\lambda_{j}$ equals $\lambda_{j} = \lambda(j/L)$. From numerical result I am sure it has an upper which bound decays with $T$. However, I don't know how to prove that. I wonder if there is related formulas in complex analysis and fourier series.

答:
Remark: This answer was for the first version of the question which stated that $|f(T)|^2$ was monotonic.

The result is not true. If $\lambda(s) = s$, then
\begin{equation}
f(T) = \frac{1}{i T}(1 - e^{-i T})
\end{equation}
hence $f(2k\pi) = 0$ for $k\in {\mathbb{N}^*}$ but $f$ is not identically $0$, hence $|f(T)|^2$ is not monotonic.


问:
Logarithm question involving different base
Calculate the values of $z$ for which $\log_3 z = 4\log_z3$.</p>

Body:
$\frac{\log z}{\log 3}=4\frac{\log 3}{\log z}$

$(\log z)^2=(\log 3)(4\log 3)$

$\log z=\sqrt{(\log3)(4\log 3)}$

$ z=10 \exp\left(\sqrt{4(\log 3)^2}\right)$


问:
Does this Make Sense ? Riemannian Manifold Directional Derivative.
Take a Riemannian Manifold $(M,g)$, where the tangent space at $p\in M$ is denoted $T_p M$. Let $f:M\to\mathbb{R}$.

Does it make sense to ask for the directional derivative of $f$ at $p$ in direction $v\notin T_p M$?

I'm guessing no?

答:
It doesn't, just like it does not make sense to ask for the value of $f$ at $x\notin M$.

Both might make sense if you actually have some extension of $f$ to a bigger set (or manifold) in mind, but with the data you describe, it's nonsense.


问:
Expected length of orbit for random permutation
Pick a permutation $\sigma$ of $\{1,2,\dots,n\}$ uniformly at random. Let $m$ be the smallest positive integer such that $\sigma^m(1)=1$. What is $\mathbb E[m]$?

I know that the expected number of cycles of length $k$ for a random permutation is $\frac{1}{k}$. This seems relevant (we want the expected length of the cycle containing $1$), but I'm not sure how.

I manually computed the answer for some small values:

$n=2$ the answer is $\frac{3}{2}$.

$n=3$ the answer is $2$.

$n=4$ the answer is $\frac{5}{2}$.

So from a (way too small) sample size, it looks like $\frac{n+1}{2}$?

答:
Choose the elements in the cycle of $1$ say $m-1$ order them and then the rest of the elements are other permutation so there are $(m-1)!(n-m)!$ of them. Then
$$\mathbb{E}[m]=\sum _{m=1}^nm\cdot P(\sigma \text{ has $1$ in a cycle of length m})=\sum _{m=1}^nm\cdot \frac{\binom{n-1}{m-1}\cdot (n-m)!(m-1)!}{n!}=\sum _{m=1}^n\frac{m}{n}.$$
Your conjecture is obtained by the Hockey-Stick identity.


问:
Complex analysis: calculate curve integral
I'm trying to calculate the following integral:

$$ \frac{1}{2\pi i}\int_{\gamma} \frac{\cos(z)}{z^7-2z^6}dz $$

Where $\gamma$ is the path that traverses the circle centered at $1/4$ with radius $1/2$ counter-clockwise.

It seems obvious that I should use Cauchy's integral formula here, but I cannot figure out how.

I can rewrite the integral as:

$$ \frac{1}{2\pi i}\int_{\gamma} \frac{z\cos(z)}{(z-2)}\frac{1}{(z-0)^7}dz $$

From this I find that the function $\frac{1}{(z-0)^7}$ has a pole at $z=0$, which is in $\gamma$, but $\frac{z\cos(z)}{(z-2)}$ has a zero at $z=0$. Finally, $ \frac{z\cos(z)}{(z-2)}$ has a pole at 2.

But from here I'm unsure about how I should proceed. It seems so simple, but I can't seem to find a way forward.

Any tips are appreciated.

答: If $f(z)=\frac{\cos z}{z-2}$, then\begin{align}\frac1{2\pi i}\int_\gamma\frac{\cos z}{z^7-2z^6}\,\mathrm dz&=\frac1{2\pi i}\int_\gamma\frac{f(z)}{z^6}\,\mathrm dz\\&=\frac{f^{(5)}(0)}{5!}\end{align}Or you can apply the residue theorem.


答: The probability that $\sigma(1)=1$ (so $m=1$) is $\frac{1}{n}$. Conditional on $\sigma(1)\neq1$, we know that $m=2$ with probability $\frac{1}{n-1}$, so the unconditional probability is $\frac{n-1}{n}\cdot\frac{1}{n-1}=\frac{1}{n}$. Continuing like this, we see that $\mathbb P(m=k)=\frac{1}{n}$ for any $1\leq k\leq n$. So $$\mathbb E[m]=\frac{1}{n}(1+2+\dots+n)=\frac{n+1}{2}.$$

Title: Expected length of orbit for random permutation

Body:
Pick a permutation $\sigma$ of $\{1,2,\dots,n\}$ uniformly at random. Let $m$ be the smallest positive integer such that $\sigma^m(1)=1$. What is $\mathbb E[m]$?

I know that the expected number of cycles of length $k$ for a random permutation is $\frac{1}{k}$. This seems relevant (we want the expected length of the cycle containing $1$), but I'm not sure how.

I manually computed the answer for some small values:

$n=2$ the answer is $\frac{3}{2}$.

$n=3$ the answer is $2$.

$n=4$ the answer is $\frac{5}{2}$.

So from a (way too small) sample size, it looks like $\frac{n+1}{2}$?


问:
Bounded Self-adjoint operator with finite spectrum
if $S$ be a self-adjoint bounded operator and $\sigma(S)=\left\{\lambda_{1}, \dots, \lambda_{n}\right\} .$ show that there exist projections operator $P_{1}, \dots, P_{n} \in B(H)$ that $\sum_{j=1}^{n} P_{j}=I$ and $P_i P_j=0$ if $j \neq k$ and $S=\sum_{j=1}^{n} \lambda_{j} P_{j}$.
I can see how we can define projection operator for each term in spectrum but I can't show that they are complete in the sense $\sum_{j=1}^{n} P_{j}=I$. actually its strange for me that in general self-adjoint bounded operator with finite spectrum have a complete set of projection operator.

答:
When you want to prove something about self-adjoint operators on a Hilbert space, your first instinct should be to think about the continuous functional calculus.

Recall that the functional calculus is an isometric $*$-morphism
$$\phi: C(\sigma(S)) \to B(H): f \mapsto f(S)$$
that sends the inclusion $z: \sigma(S) \hookrightarrow \mathbb{C}$ to $S$ and $1$ to $\text{id}_H$.

Since the spectrum is finite, it is discrete and the characteristic functions $f_i = \chi_{\{\lambda_i\}}$ are continuous on the spectrum. Let $P_i:= f_i(S) \in B(H)$. Now, note that on $\sigma(S)$, the following identities hold:
$$z = \sum_i \lambda_i f_i$$
$$f_i f_j = 0, \quad i \neq j$$
$$f_i^2 = f_i = \overline{f_i}$$

Hence, since the functional calculus is a unital $*$-morphisms that maps $z$ to $S$, we find that these identities are also satisfied for the images, and this is what you want.

Remark: The same is true (with the same proof) when we replace $B(H)$ with a unital $C^*$-algebra.


问:
Compute value of $\pi$ up to 8 digits
I am quite lost on how approximate the value of $\pi$ up to 8 digits with a confidence of 99% using Monte Carlo. I think this requires a large number of trials but how can I know how many trials?

I know that a 99% confidence interval is 3 standard deviations away from the mean in a normal distribution. From the central limit theorem the standard deviation of the sample mean (or standard error) is proportional to the standard deviation of the population $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}$

So I have something that relates the size of the sample (i.e. number of trials) with the standard deviation, but then I don't know how to proceed from here. How does the "8 digit precision" comes into play?

UPDATE

Ok I think I am close to understand it. From CLT we have $\displaystyle \sigma_{M} = \frac{\sigma}{\sqrt{N}}$ so in this case $\sigma = \sqrt{p(1-p)}$ therfore $\displaystyle \sigma_{M} = \frac{\sqrt{p(1-p)}}{\sqrt{N}}$

Then from the Bernoulli distribution, $\displaystyle \mu = p = \frac{\pi}{4}$ therefore

$$\sigma_{M}=\frac{\sqrt{\pi(4-\pi)}}{\sqrt{N}}$$ but what would be the value of $\sigma_{M}$? and then I have $\pi$ in the formula but is the thing I am trying to approximate so how does this work? and still missing the role of the 8 digit precision.

答:
The statistical analysis given in the other answer seems fine.

But the discussion of integrals is unnecessary, I think. The standard Monte-Carlo technique for estimating $\pi$ is to generate $N$ random points in a square centered at the origin -- say the square $[-1,1] \times [-1,1]$. After generating each point $(x,y)$, you can check to see if $x^2 + y^2 < 1$, which tells you whether that point lies in the unit circle. You keep track of the fraction $k$ of random points that lie within the unit circle. As $N$ gets larger, this fraction $k$ gives increasingly accurate estimates of the ratio $\text{(area of circle)} / \text{(area of square)}$. In other words $k \approx \pi/4$, when $N$ is large, so $\pi \approx 4k$.

The statistical analysis given in the other answer tells you how large $N$ must be in order to achieve the desired accuracy with a desired probability.

There is a fairly good explanation on this web page.

And another one on this page.

Title: Compute value of $\pi$ up to 8 digits


问:
Let $G=N\rtimes H$ and the conjugacy action of $H$ on $N$ divide $N$ in two orbits. Show $N$ is abelian (and more).
Given a finite group $G$, $N \trianglelefteq G$, and $H \leq G$, such that $G$ is the inner semidirect product of $N$ and $H$, and knowing that the conjugacy action of $H$ on $N$ divides $N$ in two orbits, prove that:

There exists a prime $p$ such that any nontrivial element of $N$ has order $p$

$N$ is abelian

It doesn't exist $M \trianglelefteq G$ nontrivial and properly contained in $N$.

My solution for $(1)$:
I know that $N$ is divided in two orbits, and since there's one orbit that contains only $1_G$, all the elements $n\in N:n \neq 1_G$ are in the same orbit, meaning that they all have the same order, since the action is the conjungancy one. It follows that the order of $N$ must be $p^\alpha$, $p$ prime. I've been stuck for some time on $(2)$ and $(3)$ though, so any hint would be really appreciated. Thanks.

答:
Thanks to the hints given in the comment section, I've been able to solve the problem, and I'm posting here the solution for any future reference.

For $(3)$, if there was $M \leq N\,,\,M\neq N$ normal in $G$,  it would mean that $h^{-1}mh \in M$ for every $h \in H$ and $m \in M$. But since $N$ is divided in only two orbits, it exists $x \in H$ such that $x^{-1}mx=n\,,\,n\in N - M$.

For $(2)$, I know that $|N|=p^\alpha$, and that there exists a normal subgroup of $X \leq N$ that has order $p^{\alpha-1}$, meaning that $N/X$ is abelian and $N' \leq X$. But $N'$ is characteristic in $N$, so for $(3)$ it must be trivial $\Rightarrow N$ is abelian.


问:
Proove that $ST$ has adjoint operator $(ST)^{*}$ and $(ST)^{*}=T^{*}S^{*}$
I have to prove this

Let be $V$ a inner product space over $\mathbb{C}$, with $T$ and $S$
lineal operators in $V$ with adjoints operators $T^{*}$ and $S^{*}$
respectively. Prove that $ST$ has adjoint operator $(ST)^{*}$ and
$(ST)^{*}=T^{*}S^{*}$

I've done it by this way:

Let be $\alpha, \beta \in V$. First we need to observe that:

\begin{align}
\left \langle ST \alpha,\beta \right \rangle&=\left \langle T\alpha,S^{*}\beta \right \rangle\\&=\left \langle \alpha,T^{*}S^{*}\beta \right \rangle
\end{align}
So, $ST$ has an adjoint lineal operator $T^{*}S^{*}$.

Now, to prove that $(ST)^{*}=T^{*}S^{*}$, I did this:

\begin{align}
\left \langle \alpha,T^{*}S^{*}\beta \right \rangle&=\left \langle T\alpha,S^{*}\beta \right \rangle\\&=\left \langle ST\alpha,\beta \right \rangle\\&=\left \langle \alpha,(ST)^{*}\beta \right \rangle \ \ \ \ \ \ \text{I have doubt in this last step, is it correct?}
\end{align}

\begin{align}
\therefore (ST)^{*}=T^{*}S^{*}
\end{align}

Is it correct my proof? I would really appreciate your help!

答:
So what you have shown is that $\langle \alpha, T^*S^*\beta\rangle = \langle \alpha, (ST)^*\beta\rangle$ for all $\alpha, \beta\in V$. We can rewrite this as saying that for all $\alpha, \beta\in V$ we have that
$$\langle \alpha, [T^*S^*-(ST)^*]\beta\rangle =0 $$
Fix $\beta\in V$ and choose $\alpha = [T^*S^*-(ST)^*]\beta$ then we can rewrite the inner product as
$$||[T^*S^*-(ST)^*]\beta||^2 =  \langle [T^*S^*-(ST)^*]\beta, [T^*S^*-(ST)^*]\beta\rangle = 0$$
Now a norm is nonnegative and is only equal to $0$ for the $0$ vector. Thus $||[T^*S^*-(ST)^*]\beta||^2=0$ implies that $[T^*S^*-(ST)^*]\beta=0$ which means that $$ T^*S^*\beta=(ST)^*\beta$$
Well $\beta\in V$ was arbitrary so the two operators are equal on the entire vector space, which means that $(ST)^*=T^*S^*$.


问:
simple summation example and simplyfing it
How this summation is near to $ \frac{n}{2}$.

$ \frac {1+2+3+4+...+(n-1)+(n-1)}{n} \sim \frac{n}{2} $

Body:
\begin{align*}\frac{1+2+\cdots+(n-1)+(n-1)}{n}&=\frac{[1+2+\cdots+(n-1)+n]-1}{n} \\
&=\frac{[n(n+1)/2]-1}{n} \\
&=\frac{n+1}{2}-\frac{1}{n}\sim\frac{n}{2}.\end{align*}


问:
Rank $ (a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0} $
If we work in a model of set theory where $2^{\aleph_0} = \aleph_5$, then Rank in ascending order the following cardinals:

$$
(a) {\aleph_4}^{\aleph_4}, (b) {\aleph_0}^{\aleph_4}, (c) {\aleph_4}^{\aleph_0}, (d) {\aleph_4}, (e) {\aleph_5}, (f) {\aleph_5}^{\aleph_0}
$$

If possible, specify whether there is a strict equality or inequality

My attempt:

the only one I could figure out was: (f) $ {\aleph_5}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} $.

What will be $\aleph_4$ here?

答:
$\aleph_4$ is simiply the fourth smallest infinite (well-ordered, if you work in ZF) cardinal.

From what you wrote, $\aleph_5=\aleph_5^{\aleph_0}$, so you have one equality.

By definition, $\aleph_4<\aleph_5$.

Clearly, $\aleph_4^{\aleph_0}\geq 2^{\aleph_0}=\aleph_5$.

On the other hand, $\aleph_5=\aleph_5^{\aleph_0}\geq \aleph_4^{\aleph_0}$.

For arbitrary cardinals $2\leq\lambda<\kappa$, $\kappa^\kappa=\lambda^\kappa$.

This is enough to find the order and specify all but one inequality, which is undecidable e.g. by Easton's theorem.


问:
Abstract vs Natural model - which one do I choose?
I have the problem described below:

Describe the model such that ∀x∀y (A(x,y) → x ≠ y) is false.
Does anyone have any hints on how to solve this problem?
I know that it is possible to solve this type of problem with an abstract model and a natural model,
which one do I chose?

答:
Here's a worked example that spells everything out. Please think about each step with some skepticism.

Try negating the sentence in first order logic and then take a look at it and see if any strategies for producing a model jump out.

$$ \forall x \forall y \mathop. A(x, y) \to x \ne y $$

rewrite in negative normal form

$$ \forall x \forall y \mathop. \lnot A(x, y) \lor x \ne y $$

negates to

$$ \exists c \exists d \mathop. A(c, d) \land c = d $$

We can skolemize (trivially), producing an equisatisfiable theory.

$$ A(c, d) \land c = d $$

So

$$ A(c,d) \;\;\text{and}\;\; c = d $$

So

$$ A(c, c) $$

So, if we pick $\{c\}$ as our domain of discourse and pick $\{(c, c)\}$ as our interpretation of $A$, we now have a model in which $\forall x \forall y \mathop. A(x, y) \to x \ne y $.

In order to check, note that our domain has a single element $c$.

$$ A(c, c) \to c \ne c $$
$$ \top \to \bot $$
$$ \bot $$

Therefore the original sentence is indeed false as desired.


问:
Graph with degree sequence (3,3,3,3,3,3)
Does there exist a graph with degree sequence (3,3,3,3,3,3)? I am pretty such graph does not exist since I have tried to draw one without success, but is there a way to prove it?</p>

答:
The two non-isomorphic graphs with these degree sequences can be generated using geng which comes with nauty.  The command is

```<code>geng 6 9:9 -d3 -D3
</code>```

where we have 6 vertices, 9:9 means between 9 and 9 edges, -d3 means minimum degree 3 and -D3 means maximum degree 3.  These can be viewed using showg, but I prefer to write my own script to draw them:

Body:
The two non-isomorphic graphs with these degree sequences can be generated using geng which comes with nauty.  The command is

geng 6 9:9 -d3 -D3

where we have 6 vertices, 9:9 means between 9 and 9 edges, -d3 means minimum degree 3 and -D3 means maximum degree 3.  These can be viewed using showg, but I prefer to write my own script to draw them:


问:
$\mathcal{C}^{\infty}$ which are not analytical
During a course of functional analysis, we gave the following definition:

$$\mathcal{C}^{\infty}_{c}(\Omega)=\mathcal{C}^{\infty}(\Omega)\cap\mathcal{C}_{c}(\Omega)\qquad\Omega\subseteq\mathbb{R}^{d}\hspace{1.5mm}\text{open}$$

That is, the set of smooth function with compact support. We pointed out that such functions need not to be analytical. My question is, how do we know such functions are not also analytical? Given $f\in\mathcal{C}^{\infty}_{c}(\Omega)$, what is $f$ missing in order to be analytical?

答:
Analytic means that there is a power series that converges to that function (locally at every point).

The only power series that is identically zero in a region is the zero series.

That is, nonzero analytic functions cannot have compact support (ignoring arbitrary domain restrictions).


问:
Uniformly regular measure "Babiker"
A regular Borel (Radon) probability measures $\mu$ on compact Hausdorff space $X$ is called uniformly regular if:

There is a countable family $\mathcal{A}$ of compact $G_\delta$-subsets of $X$ such that for every open set $U\subseteq X$ and every $\epsilon >0$, there is $A\in\mathcal{A}$ such that $A\subseteq U$ and $\mu(U\setminus A)<\epsilon$,

or equivalently,

There is a sequence $\{U_n\}$ of open subsets of $X$ such that $\mu(K)=\displaystyle\lim_{n\to\infty}\mu(U_n)$ for every compact set $K\subseteq X$ and $K\subseteq U$.

I have seen in many papers, it is said:

"Clearly" Every measure $\mu$ on compact metric space is uniformly regular?

So I do not know how to prove this fact? anything idea is appreciated!

答:
In a metric space, closed sets are $G_\delta$ and there is $(U_k,k\in\Bbb N)$ a countable basis of open sets. For each $k$, consider $(K_{k,j},j\in\Bbb N)$ a sequence of compact subsets of $U_k$ such that $\mu(U_k\setminus K_{k,j})<2^{-k}j^{-1}$.

It is possible since in this context an open set is a countable union of closed sets.

Let $U$ open and $\varepsilon>0$. Then $U=\bigcup_{k\in I}U_k$ for some $I\subset\Bbb N$. There is $N$ such that $\mu(U\setminus\bigcup_{k\in I\cap [N]}U_k)<\varepsilon$. Now for $k\in [N]$, take $j$ such that $j^{-1}<\varepsilon$. Then $K:=\bigcup_{k=1}^NK_{k,j}$ gives what we want.

Title: Uniformly regular measure "Babiker"


问:
How to find Jordan normal form and basis?
I have a matrix:

$$A = \begin{pmatrix}-1&5&4\\ 4&-6&-6\\ -8&16&14 \end{pmatrix}$$

simplifying:

$$A = \begin{pmatrix}-1&5&4\\ 2&-3&-3\\ -4&8&7 \end{pmatrix}$$

What steps should I reproduce to find Jordan form and Jordan basis, I am really really confuswd without good step by step example!

What I really understand that I have to find eigenvalues, so I did:

$$\det|A-I\lambda| = \begin{vmatrix}-1 - \lambda&5&4\\ 2&-3-\lambda&-3\\ -4&8&7-\lambda \end{vmatrix}  $$

so I found cubic equation which is:

$$-\lambda^3+3\lambda^2-5\lambda+3 = 0$$
$$-(\lambda-1)^3-2(\lambda-1)$$

So as I can judge Jordan normal form will have three eigenvalues that equal to $\lambda = 1$ on its main diagonal, below main diagonal all the values should be "$0$". But I bet it is not all. I do not know how to proceed (If I am right at all), and do not know how to find Jordan's basis after :(

答:
I didn't find a satisfying answer writting about both algorithm to find basis and Jordan normal form and why this algorithm works, so I decide to write one by myself.

Suppose $f$ is a linear operator, whose characteristic polynomial factors into degree-one polynomial (repetition allowed), and its matrix representation under some basis is $A$. The algorithm to find Jordan normal form is as following:

Calculate the characteristic polynomial and all eigenvalues of $f$, denoted as $\lambda_1,...,\lambda_n$.

Calculate geometric multiplicity of each $\lambda_i$. The number of Jordan blocks of eigenvalue $\lambda$ is the geometric multiplicity of $\lambda$. The number of Jordan blocks of eigenvalue $\lambda$ of size $k$ is equal to $$2\cdot \dim(\ker(\lambda\cdot Id-f)^k)-\dim(\ker(\lambda\cdot Id-f)^{k-1})-\dim(\ker(\lambda\cdot Id-f)^{k+1}).$$
By these two information above, we can already determine the Jordan normal form.

Actually, there are more information given by algebraic multiplicity and minimal polynomial:
the Jordan block of the largest size of eigenvalue $\lambda$ is equal to the power of $(x-\lambda)$ in minimal polynomial of $\lambda\cdot Id-f$. And the sum of sizes of all Jordan blocks of eigenvalue $\lambda$ is equal to the algebraic multiplicity of $\lambda$.

To calculate the basis corresponding to the Jordan normal form, we need more work: from the calculation above, we can know that there are $m$ Jordan blocks of eigenvalue $\lambda$ and their sizes are $k_1,...,k_m$. Find a basis for $\ker(\lambda\cdot Id-f)=V_{\lambda}$, denoted as $b_1,...,b_m$. For each $b_i$, we recursively calculate $b_i^{j}$:

Let $b_i^{0}=b_i$,

Suppose we have calculated $b_i^{j}$, let $b_i^{j+1}$ be a vector s.t. $(\lambda\cdot Id-f)\cdot b_i^{j+1}=b_i^j$.

Such process will terminate at some $l$. Suppose we have got $b_i^{0},...,b_i^{l}$. We will see that they are linear independent and they are a basis corresponding to a Jordan block $J_l(\lambda)$. Perform this algorithm on each eigenvalue $\lambda$ and each eigenvector $b_i$. Put them together, we'll get a basis corresponding to the Jordan normal form.

Now, we wonder why this works. We'll firstly look at a special case -- nilpotent operator (i.e., an operator $f$ s.t. there exists $n\in\mathbb{N}^{>0}$ s.t. $f^n=0$).

$\textbf{Definition 1.}$ Suppose $V$ is a vector space and $f:V\to V$ is an linear operator. A $\textbf{cyclic subspace}$ associated to $f$ and $v$ is the subspace of $V$ spanned by $\{v,f(v),f^{2}(v),...\}$.

$\textbf{Claim 1.}$ Suppose $f$ is a nilpotent operator and $v\in V$. If $k$ is the least positive number s.t. $f^k=0$, then $\{v, f(v),...,f^{k-1}(v)\}$ is a basis of $Z(f,v)$.

$\textbf{Claim 2.}$ If $f$ is a nilpotent operator, then there exists $v_1,...,v_n\in V$ s.t. $$V=Z(f,v_1)\oplus Z(f,v_2)\oplus\dots\oplus Z(f,v_n).$$
Note that if we take $\{v_1,f(v_1),...,f^{k_1-1}(v_1),...,v_n,f(v_n),...,f^{k_n-1}(v_n)\}$ as  basis of $V$, then the matrix representation of $f$ under this basis is
\begin{pmatrix} J_{k_1}(0) & &\\
                         & \ddots & \\
                         &   & J_{k_n}(0).\\
   \end{pmatrix}

Now, look at an arbitrary linear operator $f:V\to V$. We have the following lemma.

$\textbf{Lemma 1.}$ (Fitting's Lemma) If $V$ is a finite-dimensional vector space and $f:V\to V$ is a linear operator. Then, $V_N=\cup_{k=1}^{\infty} \ker(f^k)$ and $V_I=\cap_{k=1}^{\infty} \ker(f^k)$ are two $f$-invariant subspaces s.t. $f\upharpoonright_{V_N}$ is nilpotent and $f\upharpoonright_{V_I}$ is invertible. $V=V_N\oplus V_I$. Moreover, if $U$ and $W$ are two $f$-invariant subspaces of $V$ s.t. $V=U\oplus W$, $f\upharpoonright_U$ is nilpotent and $f\upharpoonright_W$ is invertible, then $U=V_N$ and $W=V_I$.

Once we have Lemma 1, we can use induction to prove the existence of Jordan normal form. However, we need to be careful that it is possible for $V_N$ to be zeros space. Therefore, for an arbitrary $f$, we use Fitting's lemma for $\lambda\cdot Id-f$, where $\lambda$ is an eigenvalue of $f$. Then, the kernel of $\lambda\cdot Id-f$ is nontrivial. Thus, we can apply Fitting's lemma and induction hypothesis on $V_I$ of it. This is also why we need to require the characteristic polynomial of $f$ to factor into product of linear forms.

(I'll write about this point in a bit more detailed way when I have time.)


问:
(Non)necessity of the axiom of choice in proving associativity of the tensor product
I am trying to prove the associativity of the tensor product $\otimes$ of vector spaces, along the line with the sketch discussed on a previously-asked question, but I got stuck on the last point.

Question in brief

Do we need the axiom of choice to prove associativity of the tensor product of vector spaces which are not necessarily finitely generated (along this line, in particular)? If not, why not?

What I have managed to do

Hereafter let $X,Y,Z$ be a vector space (on a field $K$). I have managed to show that a map $$(y,z)\mapsto (x\otimes y)\otimes z$$ is bilinear for any fixed $x\in X$, and (after some argument) I realized that this eventually induces a unique bilinear map $$\psi_1:(x\in X,t\in Y\otimes Z)\mapsto A_x(t),$$
where $A_x: Y\otimes Z\rightarrow (X\otimes Y)\otimes Z$ is a linear map which satisfies $$A_x(y\otimes z)=(x\otimes y)\otimes z.$$ Then, it follows from the universal property of the tensor product that there uniquely exists a linear map $F_1:X\otimes(Y\otimes Z)\rightarrow (X\otimes Y)\otimes Z$ such that $$\psi_1(x,t)=A_x(t)=F_1(x\otimes t).$$ Similarly one can induce a bilinear map $$\psi_2:(s\in X\otimes Y,z\in Z)\mapsto B_z(s)$$
where $$B_z(x\otimes y)=x\otimes(y\otimes z),$$
and show that there uniquely exists a linear map $F_2:(X\otimes Y)\otimes Z\rightarrow X\otimes(Y\otimes Z)$ such that $$\psi_2(s,z)=B_z(s)=F_2(s\otimes z).$$
Now the remaining task is to show that $F_1$ is an isomorphism from $X\otimes(Y\otimes Z)$ to $(X\otimes Y)\otimes Z$ (and $F_2$ from $(X\otimes Y)\otimes Z$ to $X\otimes(Y\otimes Z)$). I can easily observe that $$F_2 \circ F_1(x\otimes(y\otimes z))=F_2((x\otimes y)\otimes z)=x\otimes(y\otimes z),$$
$$F_1\circ F_2((x\otimes y)\otimes z)=F_1(x\otimes(y\otimes z))=(x\otimes y)\otimes z.$$

My concern

The above observation, however, does not immediately suggest that $F_1\circ F_2=\mathrm{id}$. The problem is that whether we can extend this observation to the cases where the tensor $t\in Y\otimes Z$ cannot be expressed in terms of the tensor product of two vectors.

I am aware that the thing would be way easier if the vector spaces $X,Y,Z$ were finitely generated: In this case, bases would be guaranteed to exist and every tensor could be expressed in terms of linear combination of some "basis tensor" (tensor product of basis vectors in $Y$ and $Z$), as $Y\otimes Z$ would be also finitely generated (though proof would be necessary for completeness). Then bilinearity of the tensor product and linearity of $F_1,F_2$ would enable me to finish the job.

Unfortunately, however, existence of basis in general vector spaces is equivalent to the axiom of choice (if my knowledge is correct), which may be rejected in some systems of mathematics. Currently I have not come out with a way to work around this issue. I looked up some literature, but explicit mention to the axiom of choice was not found. So it is quite likely that I am missing something very elementary and eventually we do not need the axiom to prove the associativity, but what is the missing point?

答:
You do not need bases for this proof (e.g. associativity continues to hold for tensor products of modules which are not necessarily free). Every element of a tensor product can be expressed as a finite sum of pure tensors (this does not require bases, just the explicit construction of the tensor product) so you can use (bi)linearity.

Alternatively, if you are comfortable with the universal property of the tensor product, here is to my mind a much cleaner proof: $X \otimes (Y \otimes Z)$ and $(X \otimes Y) \otimes Z$ are naturally isomorphic because they represent the same functor, namely the functor of trilinear maps out of $X \times Y \times Z$. (So secretly I am reducing to associativity of the cartesian product.) This proof generalizes with no modification to modules.


问:
When are $1/n!$ repeating decimal with single digit repetend
Find all $n$ where $1/n!$ is repeating decimal with single digit repetend(for example $0.4111111...$ but not $0.412121212...$) but cannot be expressed as a terminating decimal (for example $0.9999999$ doesn't count since it can be expressed as terminating decimal $1.0$)

From my experiment, it seems like once $n$ has prime factor 7, it won't work.

I am not sure how to express "repeating decimal with single digit repetend". It seems a sufficient condition is $10/n! - 1/n! = p/q$ but it's not a necessary condition.

答:
If you know that $\frac{1}{9} = 0.1111\cdots$, then a number of the form $0.a_1a_2 \cdots a_k bbb\cdots$ is equal to $$\frac{[a_1 a_2 \cdots a_k]}{10^k} + \frac{b}{9 \cdot 10^k} = \frac{9[a_1 a_2 \cdots a_k] + b}{9 \cdot 10^k},$$
where $[a_1 a_2 \cdots a_k]$ denotes the $k$-digit number with digits $a_1,\ldots, a_k$.

If this equals $\frac{1}{n!}$ for some $n$, then $9 \cdot 10^k$ must be divisible by $n!$. This supports your observation that things don't work if $n \ge 7$, since $9 \cdot 10^k$ is not divisible by $7$. It thus suffices to check $n=1,\ldots, 6$ manually.


问:
Study of the number of the orbits of $S_4$ acting in $X$ with $|X|=36$
The objective is to check some statements and these are questions that I was unable to answer:

Consider the number of the orbits of $S_4$ acting in $X$ with $|X|=36$

i) Can we have exactly $15$ orbits?

ii) The number of orbits must divide $24$?

iii) Can we have only one orbit?

I tried to apply the Burnside's Lemma, since it involves the number of orbits in the action. But I really don't know how I can get the results.

For the first question I see that $\sum |X^g|=15\cdot24$, but how do I know this is impossible ?

For the second question I don't see why it should divide $24$, but I'm not sure I can say that the answer is no.

For the third one, if I have only one orbit I suppose that $\forall x,y \in X\,\,\ \exists g \in S_4$ such that $g\cdot x = y$, but how can I use that (if true) ?

I'm a begginer in Group Theory and I imagine these are easy questions, but I can't see how to answer then. I'm sorry in advance for any English or Math errors.

Thank you.

答:
Sketch solution: $|S_4|=4!=4\cdot3\cdot2=24$ and we know that it acts on $X$, whose number of elements is $36$.

The order of a generic orbit of an element $x\in X$ is such that $$|\mathcal O(x)|=|S_4:Stab(x)|\implies|\mathcal O(x)|\big|24.$$
This means that $|\mathcal O(x)|\in\{1,2,3,4,6,8,12,24\}$ and the action induces a partition on the set $X$.

A possible partition, in order to have $15$ orbits, could be $(2,2,2,2,2,2,2,2,2,2,2,2,2,2,6)$.

The action defines an equivalence relation in $X$ and the $\textbf{order}$ of each orbit (which are equivalence classes respect to the relation $a\sim b$ if $\exists\sigma\in S_4:b=\sigma*a$) divides the order of the group $S_4$.

The number of the orbits is one if the action is transitive, so, as you wrote, if $\forall a,b\in X$ $\exists\sigma\in S_4$such that $b=\sigma*a$ and in this case there is only one equivalence class (the partition is $(36)$).

Since the order of an orbit is a divisor of $|S_4|$, we can't have only one orbit.


问:
Where can I find the history of mathematical symbols in the order they were introduced?
Lately I've wondered how I could learn about the history of mathematical symbols in the order they were introduced in the chain of human being. For example the symbol for Union in set theory (∪) is taught as if it were universal, but the date of it's first inscription is obscure.

I would like to read something which shows the timeline of historical unfolding of mathematical symbols (their use and application) to frame it within it's historical background so to provide a more rapid introduction to everything, including computers in the present day.

PS: I am vision impaired and struggle with symbols because text-to-speech programs do not vocalize them. However, I have no problem hearing sentences. However, the time is out of joint, sentences surrounding symbols alone (historical knowledge) is tricky to find. I am looking for a great list or a great book.

Thank you

答:
Check out Florian Cajori, "A History of Mathematical Notations."

https://www.amazon.com/History-Mathematical-Notations-Dover-Mathematics/dp/0486677664

This question might also get good answers on the History of Science and Math stackexchange.

https://hsm.stackexchange.com/


问:
Analytic solution to $\alpha , \beta \in \mathbb{R}$ such that $\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)$?
Are there any $\alpha , \beta \in \mathbb{R}$ such that $$\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)?$$

The trivial solutions are $(\alpha, \beta)=(0, \dfrac{\pi}{2})$. But are there more? I dont see an obvious way of tackling this problem but I keep coming back to it because it seems interesting to me (look at the graph of the solutions in desmos: https://www.desmos.com/calculator/ms8ad8cxqt).

I tried simplifying matters by trying out the case where $\alpha = \beta$ where we get that $$\cos \alpha \cdot \sin \alpha = \cos \bigl(\sin (\alpha ^{2})\bigr)$$ and using the fact that $2\sin \alpha \cos \alpha = \sin 2\alpha$ we really want to find $\alpha \in \mathbb{R}$ such that $$\dfrac{1}{2}\sin 2\alpha = \cos \bigl(\sin (\alpha ^{2}) \bigr),$$ though I have to admit that this does not make the problem easier (it seems to me at least).

Has anyone looked at this problem before and have a solution or maybe some hints or suggestions on how to go further in solving the problem?

答:
There are infinitely many solutions but it is hard to write them down.
Take $\alpha=2\pi t$,  $\beta=\frac{\pi}{2}t$ and look at the sign of
$$f(t)=\cos(2\pi t)\cdot \sin(\frac{\pi}{2}t) - \cos(\sin({\pi^{2}}t^{2}))$$
when $t$ runs over integers.
The second term is always between $(-1,1)$ since $\sin()$ is zero only at multiplicities of $\pi$ and $\pi$ is irrational. At the same time if $t = 4k+1$ the first term is $+1$ and if $t = 4k+3$ it is $-1$.
In words, $f(t)$ changes sign infinitely many times.


答:
A positive rational number has terminating decimal expansion iff it can be expressed as $x=A/10^k$ for some $A,k \in \mathbb{N}$.

A rational number has a length-1 periodic decimal expansion iff it's not in the form above, but  $10x-x=9x=A/10^k$. Then, if $x=1/m$  we require that $m = 2^a 5^b 3^c$ with $a \ge 0$, $b\ge 0$ and  $c \ge 1$.

That is, it must be a multiple of $3$, and it can be a multiple of $2$ and $5$, with no other prime factor.

Then, if $m=n!$,  we must have $n \in \{ 3, 4, 5, 6 \}$

Title: When are $1/n!$ repeating decimal with single digit repetend

Body:
Find all $n$ where $1/n!$ is repeating decimal with single digit repetend(for example $0.4111111...$ but not $0.412121212...$) but cannot be expressed as a terminating decimal (for example $0.9999999$ doesn't count since it can be expressed as terminating decimal $1.0$)

From my experiment, it seems like once $n$ has prime factor 7, it won't work.

I am not sure how to express "repeating decimal with single digit repetend". It seems a sufficient condition is $10/n! - 1/n! = p/q$ but it's not a necessary condition.


问:
When do we consider a boolean expression "fully simplified"? (Eg, is $B'+ BA'C$ the simplest form of $AB'C + A'BC+ A'B'C+A'B'C'+AB'C'$?)
Edit:

I have simplified a boolean expression (see below) and there still some terms who share the same element (it is very hard to simplify more). So,

When do we consider a boolean expression to be "fully simplified" (can't be simplified anymore)?

Here is an example:

$$AB'C + A'BC+ A'B'C+A'B'C'+AB'C'$$
where ($'$) means "not". For this example, I think the expression can not be simplified more than this: $$B'+ BA'C$$

So, is there any way to make sure the expression can't be simplified any more?

Thanks in advance

答:
In this case there is a further simplification you can make, viz to $B + A'C$.

In general, the problem of finding the simplest of the expressions that are equivalent to a given boolean expression is a hard one. 3-SAT is NP-complete, and that is only a subset of the general problem. In 3-SAT, the expression is the "and" of some terms, each of which is the "or" (or $+$ in your notation) of three variables, any of which might be negated. Something like this: $$(A+B+C)(A'+C'+D)(C+D'+E)$$ Then the task is to find whether or not the given expression is "satisfiable" or not. "The expression is not satisfiable" is equivalent to "The expression is equivalent to False". If the expression is not equivalent to False, 3-SAT does not ask for the simplest equivalent expression. So the general problem of finding the simplest equivalent expression is at least as hard as 3-SAT.


问:
Counting Sequences (recurrence)
Compute the number of sequences of real numbers $a_1, a_2, a_3, \dots, a_{16}$ satisfying the condition that for every positive integer $n$,
$$
a_1^n + a_2^{2n} + \dots + a_{16}^{16n} = \left \{ \begin{array}{ll} 10^{n+1} + 10^n + 1 & \text{for even } n \\ 10^n - 1 & \text{for odd } n \end{array} \right. .
$$
Source: NIMO Summer 2012

A solution of the problem proceeds by the recurrence:

It's a recurrence, so it's $a_{n+1}=100a_n \pm 99$. Adding consecutives, gives $a_n=-a_{n-1} +100(a_{n-2}+1)$.
Characteristic equation: $x^3+x^2-100(x+1)=0$.

In particular, I wish to know how this recurrence is established or is there any other way to approach this problem.

Thankyou!

答:
(Fill in the gaps as needed. If you're stuck, state what you've tried and why it doesn't seem to work.)

First, an alternative solution/approach to the stated portion.

Let $b_i = a_i ^i$.

Let the sum of the $n$th powers be denoted as $ s_n = \sum b_i ^n$.

Claim: More generally, for any set of $ \{ s_i \}_{i=1}^{16}$, there is a unique solution (up to permutation) in complex numbers to $ \sum_{i=1}^{16} b_i ^n  = s_n$ for $n = 1$ to 16.

Proof: Use Newton's identities to determine the coefficients of the degree 16 polynomial who roots are $b_i$.

Thus, $ \{b_i\}$ are the roots, including multiplicity, of the polynomial, so are uniquely determined.

Claim: By observation, the solution is $\{ b_i \} = \{ 10, 10, 10, 10, 10, 10, -10, -10, -10, -10, -10, -1, 0, 0, 0, 0 \}$.

Proof: Well, you can verify it after the fact.

To guess it, using wishful thinking for now, given the simplicity of the setup, we hope that they are of the form $\pm 10, \pm 1 \pm 0$,
We're lucky that this works out, and that previous claim tells us the solution is unique.

Notes

In fact, all that we needed was the first 16 equations. We didn't need them for all $n$. (This should be known at the start, as it follows from Newton's identities.)

The (more) challenging part is to count the number of solutions for $ \{ a_i ^i\}$. It is mostly just case checking, as Evan demonstrated in the link.

Explanation of what they were thinking of

Note: There are lots of errors in what they wrote, so it's not surprising that you can't make sense of it. For now, ignore it.

Claim: Observe that $ s_{n+2} = 100 s_n + (-1)^{n+1} 99$ .

Proof: Splitting by $n$ odd/even, it is natural to consider $s_{n+2} - 100 s_n$, and we observe that it is equal to $(-1)^{n+1} 99$.

Claim: $s_{n+3} - 100 s_{n+1}  = (-1)^{n+2} 99 = - (s_{n+2} - 100 s_n)$.

Proof: Direct.

Corollary: Considering the recurrence in $s_n$, the characteristic equation is $ 0 = x^3 + x^2 - 100x - 100 = (x+1)(x-10)(x+10)$.

Corollary: $s_n = 6\times 10^n + 5 \times (-10^n) + 1\times (-1)^n$, by checking $ n = 1, 2, 3$.

Corollary: $\{ b_i \} = \{ 10, 10, 10, 10, 10, 10, -10, -10, -10, -10, -10, -1, 0, 0, 0, 0 \}$.

Errors in their writeup, which should be clear from reading the above:

The identity is for $s_n$, not for $a_n$.

The identity should have been $ s_{n+2} = 100 s_n \pm 99$, not $ s_{n+1}$.

That is not the characteristic equation for their stated recurrence. It is the characteristic equation for my stated recurrence.

Title: Counting Sequences (recurrence)


问:
inverse of preservation of intervals theorem
is inverse of preservation of intervals theorem true?
preservation of intervals theorem is true so I have this question if map of every interval by f still be an interval can f be continuous?</p>

答: No. Consider $f(x) = \begin{cases}\sin \frac 1x,&x \ne 0\\0,&x = 0\end{cases}$.


问:
Uniform distribution on the n-sphere.
I have the next RV:

$$\underline{W}=\frac{\underline{X}}{\frac{||\underline{X}||}{\sqrt{n}}}$$
where $$X_i \tilde \  N(0,1)$$
It's a random vector, and I want to show that it has a uniform distribution on the n-sphere with radius $$\sqrt{n}$$.

I understand that it has this radius, just calculate it. But I don't understand from calculating the CDF how to I arrive at uniform distribution.

Thanks in advance, MP.

答:
You can prove that the $n$-dimensional Gaussian is invariant under transformation by $T$ for any orthogonal matrix $T$. (This is well known)

For orthogonal $T$ we have $\|TX\|=\|X\|$ hence $T\underline W = \sqrt n\frac{TX}{\|TX\|}$. Therefore the distribution of $W$ is also invariant under transformation by $T$

So the distribution of $W$ is invariant under any isometry of the sphere and
the uniform distribution is the only distribution on the sphere that satisfies this condition.

Title: Uniform distribution on the n-sphere.


问:
Alignment of multiple bodies on circular paths
Below is a problem that I came across in which I don't know the answer. It's as follows:

$\\$

Problem:

Suppose there are $n-2$ planets currently at rest (stationary) - all positioned on the positive side of the x-axis (line) of the euclidean plane. They will soon orbit on a circular motion around the sun. Each planet $k$ will travel at the same speed and have an orbit of length $k$ where $k=3,4,...,n$ and $n>4$. All orbits will lie on the euclidean plane and will be centred at the origin (where the sun is positioned). Suppose now that all planets leave their initial position of rest at the same time. From now on, two or more planets are considered to be aligned if they lie on the positive side of the x-axis (line) at the same time in their respective orbits. Now consider all the planets that won't be aligned with planet $n$ during its first orbit. How many of them will be aligned with at least one planet that will be aligned with planet $n$ (during the first orbit of planet $n$)?

$\\$

Additional Information:

In the previous version of this problem, I realised I had made the mistake of letting planet $n$ orbit the sun more than once which made it quite obvious that every other planet would be aligned with planet $n$ after a certain number of orbits - hence you have the second comment below and the answer (from 1st January) that were driven by this mistake. So I've now corrected the mistake by letting planet $n$ orbit the sun once only - which was my original intent.

答:
So, when planet $n$ meets the positive side of the $x$-axis the second time then it happens at the first orbit. Then planet $k<n$ is aligned with planet $n$ during its first orbit iff $k|n$. Then a planet $\ell$ is aligned with such planet $k\ne \ell$ (during the first orbit of planet $n$)  iff the least common multiple $[k,\ell]$ of $k$ and $\ell$ is at most $n$.

Now consider all the planets that won't be aligned with planet $n$ during its first orbit. How many of them will be aligned with at least one planet that will be aligned with planet $n$ (during the first orbit of planet $n$)?

The above follows that a planet $\ell$ belongs to the required set $S_n$ iff $\ell \not|n$ and ( $(\ell,n)\ge 3$ or $\ell q\le n$), where $q\ge 3$ is the smallest prime divisor of $n$ or $4=q|n$. Let $s_n=|S_n|$. When $n$ is prime then $s_n=0$. A quest to find $s_n$ is an integer challenge, but the respective sequence is not in OEIS.


问:
Evaluate the volume of the solid bounded by $z=8-x^2 - y^2, z = x^2 + y^2, x = 1, y=\sqrt{3} x, y=0$

I am currently stuck on how to find the radius as shown in my xy-projection it is cut by the plane x=1. However I tried ignoring that part as I cannot find a solution for it and first tried to find an answer without that given plane. I arrived at this answer.$ \int_0^ \frac{\pi}{3}  \int_0^2  \int_0^ \frac{\pi}{3} rdzdrd\theta $

How will I be able to determine the radius if it's projection is shaped like so? Any insight would be appreciated.

答:
The region is in first octant, bound by planes $y = 0, x = 1$ and $y = \sqrt3 \, x$ and is bound below and above by paraboloids $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$ respectively.

So the integral is pretty straightforward if we set up in the order $dz$ first. In cartesian coordinates,

$\displaystyle \int_0^{1} \int_0^{x \sqrt3} \int_{x^2+y^2}^{8 - x^2 - y^2} dz \, dy \, dx$

In cylindrical coordinates,

$\displaystyle \int_0^{\pi/3} \int_0^{1/cos \theta} \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta$

As you may have already observed, paraboloids intersect each other at $z = 4$, the plane $y = \sqrt3 \, x$ cuts the paraboloids between $0 \leq z\ \leq 8$ but the plane $x = 1$ cuts the paraboloids only between $z = 1$ and $z = 7$. So $x, y$ is bound by paraboloid and the plane $y = \sqrt3 \, x$ for $0 \leq z \leq 1, 7 \leq z \leq 8$. Between $1 \leq z \leq 7$, it has both parts where $x, y$ are bound by the plane(s) or is bound by the paraboloids. So trying to integrate over $dz$ last will make the integral complicated and you will have to split it (or use min function). But if you are interested in seeing it, let me know.


问:
When can nets be replaced by sequences?
For now, my understanding is very unclear. So please help me.

My first question is : Can a sequential space(e.g. a pseudometrizable space) be understood to be a space where any topological property(e.g. open, compact, and so on) can be described in terms of sequences instead of nets? If my question is unclear, then, in other words, when can sequences replace nets?

I see an example in Wikipedia: In sequential spaces, a function is continuous if and only if it is sequentially continuous. So, in this case, we only need to consider sequences, not nets, to determine whether a function is continuous. I am reading so many things containing sequential spaces, uniform spaces, uniform properties to obtain a clear understanding, which is not that successful until now. All wiki or stack exchange posts seem to be perversely dodging my tantalizing question.

My second question is: An bounded linear operator $T$ acting between two Banach spaces is called compact if for any bounded sequence $\{x_n\}$, the sequence $\{Tx_n\}$ contains a convergent subsequence. In deriving this definition among many other alternative definitions, one use the fact that a metric space is totally bounded if any infinite sequence contains a Cauchy subsequence. In this case, "a sequence" is used instead of "a net", which means that it suffices to consider "sequences" instead of "nets" to determine whether a space is totally bounded. But, total boundedness is not a topological property.(Here, I came to know the concept of "uniform property" and "uniform space". But, these concepts seem to rather interrupt resolving my question.)

Another example I found in Wikipedia is : A metric space is complete if and only if it is sequentially complete. Thus, in this case, one only needs to consider Cauchy sequences, instead of Cauchy nets. Here, completeness is not a topological property. But, we can replace nets with sequences. Is it also due to the fact that any metric space is a sequential space?

I am guessing that the reason for a sequence to be used instead of a net is that the space concerned is a (pseudo-)metric space. I'd like to know the super-clear and simple reason or logic behind this, without being further covered by advanced concepts, which tends to make the issue more complicated. Thanks a lot.

答:
There are not any easy shortcuts or simple general rules here.  Whether you can replace sequences by nets in some situation depends on the details of the situation and how exactly you are using the sequences or nets.  In particular, properties related to compactness tend to be very finnicky about when they can be captured using sequences.  For instance, a metric space is compact iff it is sequentially compact, but this is a quite nontrivial theorem (much harder than, for instance, the theorem that sequences detect continuity of maps on a metric space).  And this theorem does not hold for arbitrary sequential (or even first-countable) spaces: for instance, the ordinal space $\omega_1$ is first-countable and sequentially compact, but is not compact.

As a general rule of thumb, you can always replace nets with sequences when thinking about (pseudo)metric spaces.  I don't know any counterexamples to this, but I also don't know any precise theorem that would cover every possible situation in which you might want to replace nets with sequences.  And once you leave metric spaces to more general spaces like first-countable or sequential spaces, things get much more complicated.

Title: When can nets be replaced by sequences?


问:
Given a $4\times 4$ symmetric matrix, is there an efficient way to find its eigenvalues and diagonalize it?
I have a $4\times 4$ matrix $$A=\left(\begin{array}{cccc}8 & 11 & 4 & 3\\11 & 12 & 4 & 7\\4 & 4 & 7 & 12\\3 & 7 & 12 & 17\end{array}\right).$$ I want to do the things I describe below.

Find the eigenvalues.

Find a  unitary matrix $P$ (if there is any) so that the matrix $(P^{-1})AP$ is diagonal.

Find (if there are any) an identity matrix $Q$ and an upper triangular matrix $R$ so that $A=QR$.

Comments (item by  item)

I want to know if there is a better way than calculating $\det(A-\lambda I)$.

Well for this I think I have the answer as the matrix A is symmetric that means that it has 4 distinct eigenvectors that are orthogonal with each other also P a matrix composed by using the eigenvectors as columns gives us that $(P^{-1})AP$ = with the diagonal form of A. And P is unitary as if we take the inner product of all the eigenvectors with each other we get 0 since they are orthogonal with each other. Is there a flaw to the way i am thinking?

I tried to solve this using the Gram–Schmidt process I found the first column of Q but then the numbers get too big and gets hard to compute. I have been thinking maybe symmetric matrices have some better way for QR decomposition

Body:
This is what I get in a session with Pari-GP:

? M=[8,11,4,3;11,12,4,7;4,4,7,12;3,7,12,17]
%1 =
[8 11 4 3]

[11 12 4 7]

[4 4 7 12]

[3 7 12 17]

? charpoly(M)
%2 = x^4 - 44*x^3 + 340*x^2 + 1096*x - 569
? qfsign(M)
%3 = [3, 1]
? qfjacobi(M)
%4 = [[-2.7711067095643205005042713725054278775,
       0.45792421148127108739250530588654614018,
       13.784936206564400480761522475780657126,
       32.528246291518648932350243590838224611]~,
      [0.56546566894067682530206059221727842481,
       0.48618155778071956234867882744958918213,
       0.54955525973510618817381976083023906288,
       0.37664981962144319750903998279270607039;
      -0.46810465013958985829800140380345548297,
      -0.47042263023491137214267988086021197332,
       0.55128143386128172202765139532340765196,
       0.50563758380920349835989610695765410760;
      -0.54210521905381975449864154038806958910,
       0.64436536825618266659441254842846620785,
      -0.31320991082067338855031289595942215904,
       0.43910676996647412411072995204550132449;
       0.40896032233185550464871681388665378512,
      -0.35654356540099667600450300414305218092,
      -0.54403800321396212602733917384888354204,
       0.64003967985456572726102891872973617133]]

The first four numbers after $\mathtt{qfjacobi(M)}$ are the eigenvalues, the rest is a matrix $P$ that diagonalizes $A$. I don't think it's possible to explicitly compute the roots of the characteristic polynomial, in this case.


问:
Proof that $\sum_{p=0}^\infty \frac{(Ct)^p (\sqrt{p})^p}{p!} \leq Ae^{Bt^2}$ for some constants $A, B > 0$
Fix $C > 0.$ I need to prove that there are constants $A, B> 0$ (depending on $C$) such that

$$\sum_{p=0}^\infty \frac{(Ct)^p (\sqrt{p})^p}{p!} \leq Ae^{Bt^2}$$

for every $t>0.$

Sorry for not showing any work, but I don't really know how to tackle this. Thanks.

答:
Let $f(t)$ denote the sum. Then for $t > 0$,

\begin{align*}
f'(t)
&= \sum_{p=1}^{\infty} \frac{C^p p^{p/2}}{(p-1)!} t^{p-1} \\
&= C + \sum_{p=0}^{\infty} \frac{C^{p+2} (p+2)^{(p+2)/2}}{(p+1)!} t^{p+1} \\
&= C + C^2 t \sum_{p=0}^{\infty} \frac{C^{p} p^{p/2}}{p!} \left( \frac{(p+2)^{(p+2)/2}}{p^{p/2}(p+1)} \right) t^{p}.
\end{align*}

Since

$$ \lim_{p \to \infty} \frac{(p+2)^{(p+2)/2}}{p^{p/2}(p+1)} = e $$

is finite, it follows that $M = \sup_{p \geq 0} \frac{(p+2)^{(p+2)/2}}{p^{p/2}(p+1)}$ is also finite. From this, we get

$$ f'(t) \leq C + C^2 M t f(t) $$

Writing $B = C^2M/2$ and using the above bound, it follows that

$$ (e^{-Bt^2}f(t))' = e^{-Bt^2}(f'(t) - 2Btf(t)) \leq Ce^{-Bt^2}, $$

and so,

$$ e^{-Bt^2}f(t) - f(0)
\leq \int_{0}^{t} C e^{-Bs^2} \, \mathrm{d}s
\leq \int_{0}^{\infty} C e^{-Bs^2} \, \mathrm{d}s. $$

Setting $A = f(0) + \int_{0}^{\infty} C e^{-Bs^2} \, \mathrm{d}s$, we therefore obtain

$$ f(t) \leq A e^{Bt^2}  $$

as desired.


问:
Intersection of subspaces: Does there exist for a subspace a non trivial subspace such that that their intersection has dimension 1?
More specifically: Let $V$ be an $n$ dimensional vector space over a finite field $F_{q}$. Let $S\subset V$ be a $k$-dimensional ($k> 1$) subspace. Does there exist an $m$-dimensional subspace $W\subset V$, with $m=\dim(W)>1$ such that $\dim(S\cap W)=1$?

答:
Hint. Let $T$ be the span of $S$ and $W$. Then
$$
\dim(T) = \dim(S) + \dim(W) - \dim(S \cap W)
$$

Whenever you can satisfy this equality there will be such a $W$.

The fact that the field is finite is irrelevant.


问:
Orthogonal transformations and metric tensor: prove that $U^{T} g U=g$
I'm attempting to prove that, if $U=(U^i_j)_{n \times n}$ is the matrix associated to an orthogonal transformation and $g=(g_{ij})_{n \times n}$ is the metric tensor, then

$$U^{T} g U=g \tag{1}$$

My attempt

Applying the condition that an orthogonal transformation must preserve the scalar product, $\big(U(\boldsymbol{x}), U(\boldsymbol{y})\big) = (\boldsymbol{x}, \boldsymbol{y})$, I have found that that

$$ \sum ^n _{i=1} \sum ^n _{j=1} \sum ^n _{k=1} \sum ^n _{l=1} g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l}= \sum ^n _{i=1}\sum ^n _{j=1} g_{i j} x^{i} y^{j} \tag{2}$$

Using the Einstein summation convention and changing in the RHS the indexes $i \rightarrow k, j \rightarrow l$

$$g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l} = g_{kl} x^{k} y^{l} \tag{3}$$

Since the terms $x^{k} y^{l}$ appears at both sides, if we could identify $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$, this would correspond to the matrix product $U^{T} g U=g$, and the proof would be complete.

But I'm stuck here, because I don't see how we can match parts of different sums... Could someone tell me how could it be justified from $(3)$ that $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$ (if it is actually possible to do this)?

答: @PaulSinclair is right. Suppose $x^k$ ($y^l$) is $0$ except for being $1$ when $k=m$ ($l=n$) in our coordinates, so $g_{ij}U^i_mU^j_n=g_{mn}$.


问:
Conformal Mapping with Neuman Conditions
I'm trying to solve the added question with Neumann conditions instead of Dirichlet. Also, I could not figure it out how to obtain $x^2$ at (4).

I would be pleased if anyone could help. Thank you.

答:
The curve $xy=3$ is transformed to $v=3$ in u-v plane. Eq.(4) is obtained by solving algebra equation $x^4-u x^2-9=0$.

Suppose you have a Neumann conditions $\psi_y=30$ and $y \psi_x+x \psi_y=10 e^x$, then you will have $\psi_v=30$, and $ \psi_v=10 e^x$ with $x$ in Eq.(4).


答:
it is impossible because of the assumption which is $W\subset V$

$W\subset V \Rightarrow W \cap V=W$

$W \cap V=W \Rightarrow dim(W \cap V)=dim(W)>1$ which means $dim(W) \neq1$

Title: Intersection of subspaces: Does there exist for a subspace a non trivial subspace such that that their intersection has dimension 1?

Body: More specifically: Let $V$ be an $n$ dimensional vector space over a finite field $F_{q}$. Let $S\subset V$ be a $k$-dimensional ($k> 1$) subspace. Does there exist an $m$-dimensional subspace $W\subset V$, with $m=\dim(W)>1$ such that $\dim(S\cap W)=1$?


问:
The negative square root of $-1$ as the value of $i$
I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i =  (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i =  -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i =  -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

答:
You are correct that $\sqrt{-1} = -i$, but you are not correct when you say that this fact is neglected--it's used all the time. E.g. the quadratic formula:

$$x^2+x+1=0$$

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

$$x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$$

The plus-or-minus is there because $\sqrt{-3}$ has two roots: $i\sqrt{3}$ and $-i\sqrt{3}$.


问:
What is the right change of variables which allows to obtain this expression?
Let
$$f(x) =\frac{1}{\omega_{N-1}} \frac{\log\left(\frac{d}{|x|}\right)}{(\log j)^{1/N}}\quad\mbox{ if }\quad d/j <|x|< d$$
where $\omega_{N-1}$ denotes the area of the unit sphere in $\mathbb{R}^N, j\geq 2$ and $d>0$.
Why the integral $\int f(x)^k dx$ can be written in this way
$$\int f(x)^k dx = \frac{\omega_{N-1}^{1-k/N} d^N}{(\log j)^{k/N}} \int_{1/j}^1 (-\log s)^k s^{N-1} ds?$$

Could anyone please please help me? I guess there is a change of variables somewhere, but I am not able to understand what is the right choice.

Thank you in advance.

答:
Switching to polar coordinates,
\begin{align*}\int_{d/j<|x|<d}\left(\frac{1}{\omega_{N-1}}\frac{\log\left(\frac{d}{|x|}\right)}{(\log j)^{1/N}}\right)^k\, dx&=\frac{\omega_{N-1}^{-k}}{(\log j)^{k/N}}\int_{d/j<|x|<d}\left(\log\left(\frac{d}{|x|}\right)\right)^k\, dx\\
&=\frac{\omega_{N-1}^{-k}}{(\log j)^{k/N}}\int\limits_{S^{n-1}}\int\limits_{d/j}^j\left(\log\left(\frac{d}{r}\right)\right)^kr^{N-1}\, drd\sigma
\end{align*} Let $s(r)=r/d.$ Observe that $s(d/j)=1/j$, $s(d)=1,$ and $d\ ds=dr.$ Using this change of variables,
\begin{align*}
\frac{\omega_{N-1}^{-k}}{(\log j)^{k/N}}\int\limits_{S^{N-1}}\int\limits_{d/j}^j\left(\log\left(\frac{d}{r}\right)\right)^kr^{N-1}\, drd\sigma&=\frac{\omega_{N-1}^{-k}}{(\log j)^{k/N}}\int\limits_{S^{N-1}}\int\limits_{1/j}^1\left(\log(s^{-1})\right)^k(sd)^{N-1}d\, dsd\sigma\\
&=\frac{\omega_{N-1}^{-k}d^N}{(\log j)^{k/N}}\int\limits_{S^{N-1}}\int\limits_{1/j}^1\left(-\log s\right)^ks^{N-1}\, dsd\sigma\\
&=\frac{\omega_{N-1}^{1-k}d^N}{(\log j)^{k/N}}\int\limits_{1/j}^1\left(-\log s\right)^ks^{N-1}\, ds
\end{align*}
The power of the area is slightly off from what you stated, but I see no way of getting a power of $N$ in there, unless you meant to define your function a little bit differently.


问:
Solve the system of equations for $\sqrt{xy}$
$$x + y\sqrt{x} = \frac{95}{8}$$
$$y + x\sqrt{y} = \frac{93}{8}$$
$$x, y \in \mathbb{R}$$

I can't solve this system of equations I got asked in a group. I added and substracted them to find the following equations but I don't really know what to do with them.

$$\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{xy}\right) = \frac{1}{4}$$

$$\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right) - 2\sqrt{xy} = \frac{47}{2}$$

This is all I have. I checked the answer using Wolframalpha and it was $\frac{15}{4}$ for $\sqrt{xy}$.

I have been looking for this question's solution but didn't find any related question/solution on Stack Exchange or Web. Sorry if it is duplicate, thank you in advance.

答:
It is not very probable that being $x$ and $y$ not squared the two equations give a rational and besides they differ by $\dfrac14$ so we can put in a first try
$$(x,y)=(\frac{t^2}{4},\frac{s^2}{4})$$ so we get the system
$$t(2t+s^2)=95=5\cdot19\\s(2s+t^2)=93=3\cdot31$$ The solution $(t,s)=(5,3)$ comes out immediately.
Thus $$\sqrt{xy}=\frac{15}{4}$$


问:
Introduction to finite differences
I'm looking into solving the wave equation for underwater sound as described in the Naval Underwater Systems Center's technical report Numerical Solutions of Underwater Acoustic Wave Propagation Problems (PDF link via dtic.mil).

The basic idea is obvious, but I'm looking for an introduction and/or tutorial that describes the matrix formulations and how they simplify the calculation.

答: I always recommend Finite Difference Methods for Ordinary and Partial Differential Equations: Steady-State and Time-dependent Problems by Randall LeVeque. It is a very accessible text and gives overviews and enough examples of Finite Difference Methods for you to get the hang of the method as well as covering consistency, convergence and stability questions.

Title: Introduction to finite differences


问:
are two the following functors isomorphic
Consider the following two functors that go $C^{op}\times C\to Set$, where $X$ is an object in $C$

$(A,B)\mapsto Set(C(X,A)\times A,B)$

$(A,B)\mapsto Set(C(X,A),Set(A,B))$

Are these two functors isomorphic? (i.e. is there an natural transformation between them that is an isomorphism?)
I want to think that they are due to $Set(A\times B,C)\cong Set(A,Set(B,C))$, but I haven't grasped all the details.

答:
If we have a fixed functor $U:C\to Set$ (typically the 'underlying set' functor)  and a fixed object $X\in C$, then the cited natural isomorphism only implies
$$Set(C(X,A)\times UA,\, UB)\ \cong\ Set(C(X,A),\, Set(UA,UB))\,.$$
In case $U$ admits a left adjoint $F$, we get that $Set(UA,UB)$ is further naturally isomorphic to $C(FUA,\, B)$, and by the way, if the unit $1_C\to FU$ of this adjunction is invertible (which is rarely the case, at least in categories of algebraic structures), then it will also be naturally isomorphic to $C(A,B)$.


问:
"Every proper subset of the indiscrete space is nowhere dense"

Every proper subset of the indiscrete space is nowhere dense.

This is a claim from Sasho Kalajdzievski's An illustrated introduction to topology and homotopy (section 3.4, Example 3, p. 59). I think this is wrong, because $A = \{0\}$ is a proper subset of the indiscrete space $(X, \tau) = (\{0, 1\}, \{\varnothing, \{0,1\}\})$, but $\overline{A} = \{0,1\} \,\therefore\, \mathrm{int}\,(\overline{A}) = \{0,1\}$. Am I right?

答: The author should have written that every nontrivial proper subset is dense, or something along those lines.  This appears to be an error.</p>


问:
Change of intervals for numerical integration?
(Question is at the bottom)

Suppose we have an integral
$
\int_c^d f(t) dt
$. We then define a linear function $\lambda$ of $t$ such that if $t$ traverses $[c,d]$, $\lambda(t)$ traverses $[a,b]$:
$$
\lambda(t) = \frac{b-a}{d-c}t + \frac{ad-bc}{d-c}
$$
Now in the integral $\int_a^b f(x) dx$, we make the change of variable $x = \lambda(t)$. Then $dx = \lambda'(t) dt = (b-a)(d-c)^{-1} dt$ and so we have
$$
\int_a^b f(x) dx = \frac{b-a}{d-c} \int_c^d f(\lambda(t)) dt
$$

Here is the question: Why is $dx = \lambda'(t) dt = (b-a)(d-c)^{-1} dt$? Specifically, the part $dx = \lambda'(t) dt$. I thought that $dx$ was simply a notation saying "we integrate w.r.t $x$" :/

答: Think of $dt$ as a small (infinitesimal) change in $t$ then the small change in $x$ is $\frac{b-a}{d-c}dx$, which is what it means bt integrating with respect to $x$.


问:
When are $1/n!$ repeating decimal with single digit repetend
Find all $n$ where $1/n!$ is repeating decimal with single digit repetend(for example $0.4111111...$ but not $0.412121212...$) but cannot be expressed as a terminating decimal (for example $0.9999999$ doesn't count since it can be expressed as terminating decimal $1.0$)

From my experiment, it seems like once $n$ has prime factor 7, it won't work.

I am not sure how to express "repeating decimal with single digit repetend". It seems a sufficient condition is $10/n! - 1/n! = p/q$ but it's not a necessary condition.

答:
Well.... Let's think.

$\frac 1{n!}=0.a_1a_2.....a_m\overline{kkkkk}= \frac {a_1a_2....a_m}{10^m} + \frac k{9*10^m}=$

$\frac {9\cdot a_1a_2....a_m + k}{9\cdot 10^m}$

So $n! = \frac {9*10^m}{9\cdot a_1a_2...a_m + k}\in \mathbb Z$

This means the only prime factors of $n!$ are $2,3$ and $5$(!!!).  That means $n< 7$. And to not terminate that means $n! \ge 3$ and $n\ge 3$.

There are only a finite number to check. $\frac 1{3!} = \frac 16 = 0.1\overline 6$. And $\frac 1{4!}=\frac 1{24}=0.041\overline 6$ and so on.

If there are more than $1$ repeating decimal, say $w$ of them, we will get $n! = \frac {\underbrace{99...9}_w\cdot 10^m}{something + somethingelse}$ and those $n!$ will have only prime factof of $2,5,3$ and the prime factors of $\underbrace{111......1}_w$ but as all numbers not divisible by $2$ or $5$ will divide some $\underbrace{111...1}$ that will always be possible.  (Although particular $w$s will put severe restrictions).

(For example to have only $2$ repeating can not happen as $11|99$ but $7\not\mid 99$.  But $7\&11\&13|999999$ so $\frac 1{7!}$ through $\frac 1{16!}$ all have $6$ repeating digits.)


问:
Pure states and minimal projections
Probably it is trivial or already discussed, but I just wanted to pose a simple question about pure states.

${\bf Lemma \; 1 \colon}$ Let $\mathcal{A}$ be a $C^*$-algebra acting irreducibly on some Hilbert space $\mathcal{H}$ and let $\varphi(x)=(\xi | x \xi)$ be some vector state. If $\xi$ is cyclic, then $\varphi$ is pure.

${\bf Proof \colon}$ It suffices to notice that by $\overline{\mathcal{A} \xi} = \mathcal{H}$ we have that the identity representation is equivalent to the GNS representation $(\pi_\varphi, \mathcal{H}_\varphi, \xi_\varphi)$. But such a representation is irreducible if and only if $\varphi$ is pure and the assertion follows.

A related lemma is the following.

${\bf Lemma \; 2 \colon}$ Let $\mathcal{M}$ be a von Neumann algebra and let $\varphi$ be some faithful normal state in $\mathcal{M}_*^+$. If $\varphi$ is pure then its support $s(\varphi)$ is minimal.

${\bf Proof \colon}$ We can identify $\mathcal{M}$ with its GNS representation associated to $\varphi$, so that $\varphi(x) = (\xi | x \xi)$. By the identity $s(\varphi) = [\mathcal{M}' \xi]$ we have that $s(\varphi)$ is a rank one projection, hence it is minimal.

${\bf Question \colon}$ Is it true that a normal state is pure if and only if its support is minimal?

Clearly it could be useful that a state $\omega$ on a $C^*$-algebra $\mathcal{A}$ is pure if and only if it is the only state vanishing on its left kernel

$$ \mathcal{L}_\omega = \{ x \in \mathcal{A} \colon \omega(x^*x)=0 \} \,.$$

I think that the decomposition of a linear functional $\varphi \in \mathcal{M}^*$ in normal and singular part could be useful too.

Any hint? Thank you in advance.

答:
The result is true, but in a rather trivial fashion. Namely, if a von Neumann algebra has a pure normal state, then the state is supported on a type I central summand. This is simply because minimal projections can only exist in type I central summands.

Let $\varphi:M\to\mathbb C$ be a pure normal state. Because it is pure, its GNS representation $\pi_\varphi$ is irreducible. The normality of $\varphi$ implies that $\pi_\varphi$ is normal; so its image is sot-closed. Thus $\pi_\varphi(M)=B(H_\varphi)$ is type I. As $\pi_\varphi$ is faithful on $pMp$, we get that $pMp$ is isomorphic to $\pi_\varphi(M)$ and so type I. The state $\varphi$ is pure on $pMp$, so it has to be supported on a minimal subprojection of $p$; but $p$ was the support of $\varphi$. It follows that $pMp=\mathbb C p$, so $p$ is minimal.

The takeout is that pure states are largely irrelevant to von Neumann algebra, because a pure state on a von Neumann algebra with no type I summand cannot be normal.

Title: Pure states and minimal projections


答:
The square roots with the properties we know are used only for Positive Real numbers.

We say that $i$ is the square root of $-1$ but this is a convention. You cannot perform operations with the usual properties of radicals if you are dealing with complex numbers, rather than positive reals.

It is a fact that, $-1$ has two complex square roots. We just define one of them to be $i$.

You have to regard the expression $i=\sqrt {-1}$ just as a symbol, and not do operations.

Counterexample:
$$\dfrac{-1}{1}=\dfrac{1}{-1}\Rightarrow$$
$$\sqrt{\dfrac{-1}{1}}=\sqrt{\dfrac{1}{-1}}\Rightarrow $$
$$\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}\Rightarrow$$
$$\dfrac{i}{1}=\dfrac{1}{i}\Rightarrow$$
$$i^2=1\text { ,a contradiction}$$

Title: The negative square root of $-1$ as the value of $i$

Body:
I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i =  (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i =  -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i =  -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.


问:
Measurability of reminder term in Taylor expansion about a point in a Brownian motion
Suppose that $f: \mathbb R \to \mathbb R$ is twice continuously differentiable and that $\mathrm B$ is a one-dimensional Brownian motion. Using Lagrange's form of the reminder in the Taylor expansion of $f(\mathrm{B}_t(\omega))$ about $f(\mathrm{B}_s(\omega))$ we have that for each $\omega \in \Omega$,
$$
f\left(\mathrm{B}_t(\omega)\right) = f\left(\mathrm{B}_s(\omega)\right) + f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) + \frac{1}{2}f''\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2
$$
for some point
$$
\xi(\omega) \in \left [\mathrm{B}_s(\omega) , \mathrm{B}_t(\omega)\right].
$$

I would like to show that the map
$$
\omega \mapsto \xi(\omega)
$$
is measurable.

In the case of a first order Taylor expansion it is possible to prove measurability as follows [following an argument by saz https://math.stackexchange.com/questions/896394/mean-value-theorem-inside-the-expectation)]. For each $\omega$ we have from Taylor's theorem
$$
f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) = f'\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)
$$
for some point
$$
\xi(\omega) \in \left [\mathrm{B}_s(\omega) , \mathrm{B}_t(\omega)\right].
$$
Now since
$$
f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) = \int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r,
$$
we get that
$$
\xi(\omega) = \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r.
$$
Since
$$
s, t \mapsto \int_s^t f'(r) d r,
$$
subtraction and division are continuous maps, and composition of continuous and measurable maps are measurable, it follows that
$$
\omega \mapsto \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r
$$
is measurable.

Is it possible to do something similar with $\xi(\omega)$ above?

Most grateful for any help provided!

答:
As $$
f\left(\mathrm{B}_t(\omega)\right)= f\left(\mathrm{B}_s(\omega)\right) + f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) + \frac{1}{2}f''\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2
$$
it follows that

$$
f''\left(\xi(\omega)\right) = \frac{2 (f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) - f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) ) }{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2}
$$
where $\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2 \ne 0$ a.s.

Hence $f''\left(\xi(\omega)\right)$ is measurable as a quotient of measurble expressions.

Also we may get that $\xi(\omega) = F(B_t, B_s)$, where $F$ may be found from $$
\xi(\omega) = \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r.
$$
As $F$ is continious a.s. we get that $\xi$ is measurable.


问:
Show that every mapping $f: X \to Y$ is continuous where $(X,d)$ is a metric space with the discrete metric

Let $(X,d)$ be a metric space with the discrete metric and let $(Y,d')$ be an arbitrary metric space. Show that every mapping $f: X\to Y$ is continuous.

I got a bit confused about this, initially, I thought that it would come quite directly from the definition, but I'm not sure how to proceed since I'm not given $f$ to work with. Perhaps there is some other definition for continuity that I'm not familiar? I've used the one that states $\forall \varepsilon >0 \ \exists \delta >0$ s.t. $d(x,y) < \delta \Longrightarrow |f(x)-f(y) | < \varepsilon.$

答: Hint: What happens if you set $\delta=\frac12$?


问:
Understanding Direct Sums and Product (Vector Spaces)
On p.21 of A Course in Homological Algebra - Hilton and Stammbach. there is a proof for a (possibly infinite) collection of modules that there is an isomorphism $\eta$

$\eta: \text{Hom} (\bigoplus_{j \in J} A_j, B) \to \prod_{j \in J} \text{Hom} (A_i, B)$

For $\psi: \bigoplus A_j \to  B$ define  $\eta(\psi) = (\psi i_j)_{j \in J}$
where the $i_j$ are the injections  $i_j: A_j \to \bigoplus A_j$

I'm fine with seeing this as a bijection, and with its linearity.

The proof continues to describe the projections associated with the product, saying the projections $\pi_j : \prod_{j \in J} \text{Hom} (A_i, B) \to \text{Hom} (A_i, B)$
are given by $\pi_j \eta = \text{Hom} (i_j, B) $

I don't see this last part at all (it appears to equate a map with a collection of maps), nor do I have an idea what it should be. Assistance would be appreciated.

答:
Page 16 of the book:

Let $\beta : B_1 \to B_2$ be a homomorphism of $\Lambda$-modules. We can assign to a homomorphism $\varphi : A \to B_1$, the homomorphism $\beta \varphi : A \to B_2$, thus defining a map $$\beta_* = \operatorname{Hom}_\Lambda(A,\beta) : \operatorname{Hom}_\Lambda(A,B_1) \to \operatorname{Hom}_\Lambda(A,B_2).$$

$\dots$

On the other hand, if $\alpha : A_2 \to A_1$ is a $\Lambda$-module homomorphism, then we can assign to every homomorphism $\varphi : A_1 \to B$ the homomorphism $\varphi\alpha : A_2 \to B$, thus defining a map $$\alpha^* = \operatorname{Hom}_\Lambda(\alpha,B) : \operatorname{Hom}_\Lambda(A_1,B) \to \operatorname{Hom}_\Lambda(A_2,B).$$


问:
For there to be cycling, is it necessary that the solution is degenerate? (LP)
In Linear Programming, I'm trying to answer the question:

For cycling to occur, is it required that the solution is degenerate?

In the SIMPLEX method, we say that a solution $x$ is degenerate if $x_i = 0$ for some $i$ in the basis.

Cycling is when one of the previously visited basis in the SIMPLEX algorithm, appears again before reaching the optimal solution.

My answer is that it doesn't have to. Meaning that cycling can occur without the solution being degenerate, but I can't come up with an explanation or a proof. I guess I could say so because after all the computations performed after each pivotting, there's no reason why the solution has to be degenerate again.

答:
Yes, it has to be degenerate so that it “thinks” it can improve the solution by moving another variable into the basis. If that new solution is also degenerate and happens to take us to another degenerate solution that leads back to our original degenerate solution, then we have an infinite cycle.

See here as well: https://ocw.mit.edu/courses/sloan-school-of-management/15-053-optimization-methods-in-management-science-spring-2013/tutorials/MIT15_053S13_tut07.pdf


问:
Reference for Probability Theory and Central Limit Theorem
I was wondering what is a good source / reference to study probability theory with proofs of the CLT and Berry-Esseen Theorems? I am pretty well-versed in the foundations of measure theory but less so in probability (i.e. stuff like characteristic functions, moments, etc) so if there's a text that suits my specific pre-requisite well please let me know, thanks!</p>

答:
Most graduate level texts on probability will prove some flavor of CLT. There is one text I know of that also proves Berry-Esseen:

“Probability Theory” by Alexander Borokov

https://www.springer.com/cda/content/document/cda_downloaddocument/9781447152002-t1.pdf?SGWID=0-0-45-1403085-p175128623


问:
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$

I attempted to solve this question using binomial coefficients, saying:

$(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$

and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance.

I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question.

Could you please explain to me how to solve this question and how to solve similar questions in the future?

答:
This is a USAMTS problem. The official solution can be found here. You are actually on the right track.

Below is my solution. Notice that I used the same notation as in the original problem.

My proof:

We prove by induction that: If $m$ is odd then
$$x^m = a_m\sqrt n + b_m \sqrt{n+1}, na_m^2+1 = (n+1) b_m^2 $$
If $m$ is even then
$$x^m = c_m + d_m \sqrt{n(n+1)}, c_m^2 = n(n+1) d_m^2 +1 $$

When $m=1$ it's true.

If it's true for $m$, we prove it's true for $m+1$.

Case 1: If $m$ is odd, then
$$x^{m+1} = (a_m\sqrt n + b_m \sqrt{n+1}) (\sqrt n + \sqrt{n+1})$$
$$=na_m + (n+1) b_m + (a_m+b_m) \sqrt{n(n+1)}$$
So $c_{m+1} = na_m+(n+1)b_m, d_{m+1} = (a_m+b_m)$ and
$$c_{m+1}^2-n(n+1)d_{m+1}^2-1$$
$$= n^2 a_m^2 + 2n(n+1)a_mb_m + (n+1)^2b_m^2 - n(n+1)a_m^2 - 2n(n+1)a_mb_m - n(n+1)b_m^2 - 1$$
$$
= - n a_m^2+(n+1)b_m^2 - 1 = 0
$$
Therefore $x^{m+1}$ is groovy.

Case 2: If $m$ is even then
$$x^{m+1} = (c_m+d_m\sqrt{n(n+1)})(\sqrt n + \sqrt{n+1}) $$
$$=(c_m+(n+1)d_m)\sqrt n + (c_m+nd_m)\sqrt{n+1}$$
So $a_{m+1} = c_m+(n+1)d_m, b_{m+1} = c_m + nd_m$ and
$$na_{m+1}^2+1 - (n+1) b_{m+1}^2$$
$$=nc_m^2+2n(n+1)c_md_m+n(n+1)^2 d_m + 1 - (n+1)c_m^2 - 2n(n+1) c_md_m - (n+1) n^2 d_m^2$$
$$=-c_m^2 + (n+1) d_m^2 + 1=0$$
Again $x^{m+1}$ is groovy.


问:
Counting colored integer partitions with a strictness condition
For any $n \in \mathbb{N}$, let us first consider $P_2(n)$, the number of 2-colored integer partitions of $n$. Recall that these are enumerated by $(\lambda^{(1)},\lambda^{(2)})$, where $\lambda^{(1)}_1 \geq \cdots \geq \lambda^{(1)}_k \geq 1$ and  $\lambda^{(2)}_1 \geq \cdots \geq \lambda^{(2)}_\ell \geq 1$ are two ordinary partitions, and
$\sum_{j} \lambda_j^{(1)}+\sum_j \lambda_j^{(2)}=n$. The generating $q$-series for $P_2(n)$ is  clearly $\prod_{i \geq 1} \frac{1}{(1-q^i)^2}$.

Question: Is there a similar (closed) formula if we require that parts of different color are not allowed to be equal, that is $\lambda_i^{(1)} \neq \lambda_j^{(2)}$ for all i and j? Let us denote the number of such 2-colored partitions of $n$ by $P^s_2(n)$. For example $P^s_2(3)=8$, with explicit partitions: $$\{3,2+1,1+1+1,3',2'+1',1'+1'+1',2'+1,2+1'\},$$ where for simplicity I used $'$ to distinguish colors.

A few more values are (typo corrected):
$$\sum_{n \geq 0} P_2^s(n)q^n=1+2q+4q^2+8q^3+14q^4+24q^5+....$$

I would expect there is a closed formula for this $q$-series. Pointers to the literature are much appreciated.

答:
Notice that $\frac{1}{(1-q^i)^2}$ takes care of how many times is $i$ in the partition where the square takes care of in which of the two color partitions it goes. Now, you want is $\frac{2}{1-q^i}-1=1+2q^i+2q^{2i}+\cdots 2q^{n\cdot i}+\cdots$ because you have two options for the same kind of part(you either move all of them to one or the other except if the number is not a part(that is why we subtract 1)). So the $q-$series is $$\prod _{i\geq 1}\left (\frac{2}{1-q^i}-1\right )=\prod _{i\geq 1}\frac{1+q^i}{1-q^i}=(1+2q+2q^2+2q^3+\cdots)\cdots (1+2q^n+2q^{2n}+\cdots)\cdots$$
$$=\sum _{n\geq 0}q^n\sum _{a_1\cdot 1+a_2\cdot 2+\cdots +a_n\cdot n=n}2^{\text{# of }a_i\neq 0}.$$
The following (ugly) sage code

```<code>def tal(n):
    s = 0
    for k in range(0,n+1):
        P = Partitions(k)
        Q = Partitions(n-k)
        for x in P:
            for y in Q:
                ss = 1
                for i in x:
                    if i in y:
                        ss = 0
                        break
                if ss==1:
                    #print(x,y)
                    s+=1
    return s
for n in range(1,10):
    print(n,tal(n))
</code>```

gives 2 4 8 14 24 40 64 100 154 which agrees with what I have explained according to OEIS

Body:
Notice that $\frac{1}{(1-q^i)^2}$ takes care of how many times is $i$ in the partition where the square takes care of in which of the two color partitions it goes. Now, you want is $\frac{2}{1-q^i}-1=1+2q^i+2q^{2i}+\cdots 2q^{n\cdot i}+\cdots$ because you have two options for the same kind of part(you either move all of them to one or the other except if the number is not a part(that is why we subtract 1)). So the $q-$series is $$\prod _{i\geq 1}\left (\frac{2}{1-q^i}-1\right )=\prod _{i\geq 1}\frac{1+q^i}{1-q^i}=(1+2q+2q^2+2q^3+\cdots)\cdots (1+2q^n+2q^{2n}+\cdots)\cdots$$
$$=\sum _{n\geq 0}q^n\sum _{a_1\cdot 1+a_2\cdot 2+\cdots +a_n\cdot n=n}2^{\text{# of }a_i\neq 0}.$$
The following (ugly) sage code

def tal(n):
    s = 0
    for k in range(0,n+1):
        P = Partitions(k)
        Q = Partitions(n-k)
        for x in P:
            for y in Q:
                ss = 1
                for i in x:
                    if i in y:
                        ss = 0
                        break
                if ss==1:
                    #print(x,y)
                    s+=1
    return s
for n in range(1,10):
    print(n,tal(n))

gives 2 4 8 14 24 40 64 100 154 which agrees with what I have explained according to OEIS


问:
What is the term used for a "factorial" of prime numbers ? Is there such a definition?
A "factorial" of a given number $n$ is defined as a product of positive integers up to that number. What is the term used if those positive integers are prime numbers? Is there a term for that (like a product of primes or a factorial of primes)?

For example if I calculate the product of first five primes it would be $2 \cdot 3\cdot 5\cdot 7\cdot 11= 2310$. What do I call $2310$ as? Should I simply call it : "product of first $n$ primes from up to 11" or "factorial of primes up to $11$" ?

答: The term you are looking for is, fittingly enough, primorial. The product of the primes up to $p_n$ (the first $n$ primes) is typically denoted $p_n\#$.


问:
Proving the formula for $\prod_{k=1}^n \cos(kx)$
I'm trying to find a closed-form for the following product:

$$ \prod_{k=1}^n \cos(kx)$$

Here is my attempt at it:

$$ \prod_{k=1}^n \cos(2kx) = \prod_{k=1}^n \frac{ e^{i2kx} + e^{-i2kx} }{2} = \frac{(e^{2ix} - e^{-2ix}) \prod_{k=1}^n  e^{i2kx} + e^{-i2kx}  }{2^n (e^{i2x}- e^{-i2x})}= \frac{1}{2^n (e^{i2x}- e^{-i2x})} ( e^{4inx} - e^{-4inx} )= \frac{1}{2^n} \frac{\sin(4nx)}{\sin(2x)} $$

The above was simplified by using the algebraic identity $(a+b)(a-b)=a^2 -b^2$ repeatedly in the product. Now sub:$  x \to \frac{x}{2}$

$$ \prod_{k=1}^n \cos( kx)= \frac{1}{2^n} \frac{\sin(2nx)}{\sin x}$$

I'm not sure if I'm doing something wrong, but when I plug $ x= \frac{\pi}{n}$, I'm not getting the result as shown in this MSE post

Edit: Whoops! The algebraic manipulation doesn't hold after k=3 as correctly pointe od out by @leon. However, still am looking for a closed form.

答: Your result seems to be wrong for $n>1$, because the limit as $x\to 0$ of the left hand side is $1$, whereas the limit as $x\to 0$ of the r.h.s is $\frac{2n}{2^n}$