问:
Lie algebra of $\mathrm{SL}(2, \mathbb{C})$ using method of differential geometry
I am asking a new question because people did not understand what  I meant and closed the other question.
I know that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ are traceless matrix and I can proof it. What I am interested is proof it using the methods of differential geometry.

Let $\gamma(t)$  be a curve on a manifold $M$ and $f$ a real function. We define the vector $V$ as the operator
$V\cdot f=\frac{d}{dt}f(\gamma )$

Now let $U$ be the set
$$
U:=\left\{\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \in \mathrm{SL}(2, \mathbb{C}) \mid a \neq 0\right\}
$$
and define the map
$$
\begin{array}{l}
x: \quad U \rightarrow x(U) \subseteq \mathbb{C}^{*} \times \mathbb{C} \times \mathbb{C} \\
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \mapsto(a, b, c)
\end{array}
$$

Te Lie algebra $\mathcal{L}(G),$ of a Lie group is defined by
$$
\mathcal{L}(G):=\left\{X \in \Gamma(T G) \mid \forall g, h \in G:\left(\ell_{g}\right)_{*}\left(\left.X\right|_{h}\right)=X_{g h}\right\}
$$
which can be  proved to be isomorphic to the Lie algebra $T_{e} G$ with Lie bracket
$$
[A, B]_{T_{e} G}:=j^{-1}\left([j(A), j(B)]_{\mathcal{L}(G)}\right)
$$
induced by the Lie bracket on $\mathcal{L}(G)$ via the isomorphism $j$
$$
\left.j(A)\right|_{g}:=\left(\ell_{g}\right)_{*}(A)
$$
Any $A \in T_{\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)} \mathrm{SL}(2, \mathbb{C})$ can be written as
$$
A=\alpha\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\beta\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)}+\gamma\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)},
$$
then we can show that $$j\left(\left(\frac{\partial}{\partial x^{1}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{1}}-x^{2} \frac{\partial}{\partial x^{2}}+x^{3} \frac{\partial}{\partial x^{3}}$$
$$j\left(\left(\frac{\partial}{\partial x^{2}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{1} \frac{\partial}{\partial x^{2}}$$

$$j\left(\left(\frac{\partial}{\partial x^{3}}\right)_{\left(\begin{array}{l}
1 & 0 \\
0 & 1
\end{array}\right)}\right)=x^{2} \frac{\partial}{\partial x^{1}}+\frac{1+x^2x^3}{x^1}\frac{\partial}{\partial x^{3}}$$
and so
$$
 j(A)=(\alpha x^1+\gamma x^2)\left(\frac{\partial}{\partial x^{1}}\right)+(-\alpha x^2+\beta x^1)\left(\frac{\partial}{\partial x^{2}}\right)+\left(\alpha x^3+\gamma\frac{1+x^2x^3}{x^1} \right) \left(\frac{\partial}{\partial x^{3}}\right) \tag 1
$$
On the other hand we can show that by other method  that the lie algebra of $\mathrm{SL}(2, \mathbb{C})$ consist of traceless matrices.

My question is how can we identify $(1)$ with a traceless matrix?

答:
The proof is still really the same as the standard one. But here it is using the definition of a tangent vector as a derivation.
$\newcommand{\SL}{\mathrm{SL}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\gl}{\mathrm{gl}}
\renewcommand{\sl}{\mathrm{sl}}$

Let $\sl(2,\C)$ denote the space of trace-free $2$-by-$2$ complex matrices. Consider the smooth map (which really consists of 4 separate scalar functions):
\begin{align*}
  f: \SL(2,\C) &\rightarrow \gl(2,\C)\\
  A &\mapsto \begin{bmatrix} a & b \\ c & d\end{bmatrix}
\end{align*}
Given any tangent vector $V \in T_I\SL(2,\C)$, there exists a curve $\gamma(t)$ such that $\gamma(0) = I$ and $\gamma'(0) = V$. Then
\begin{align*}
  Vf(I) &= \left.\frac{d}{dt}f(\gamma(t))\right|_{t=0}\\
        &= \begin{bmatrix} a'(0) & b'(0) \\ c'(0) & d'(0) \end{bmatrix}
\end{align*}
Differentiating $\det \gamma(t) = ac-bd = 1$ with respect to $t$ at $t = 0$, we get
$$
a'(0) + d'(0) = 0.
$$
This defines a linear map $T_I\SL(2,\C) \rightarrow \sl(2,\C)$.

Conversely, given any $A' \in \sl(2,\C)$, let $\gamma(t) = (I + tA')/(\det (I+tA'))$. Let $V = \gamma'(0)$. Then, a straightforward calculation shows that
\begin{align*}
  Vf(I) &= \left.\frac{d}{dt}f(\gamma(t))\right|_{t=0}\\
        &= A'.
\end{align*}
This shows that linear map is surjective.


问:
image of positive real number under field isomorphism
Claim:
If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a field isomorphism, then $f(x) >0$ for any $x>0$.

I know $f(x)$ can be determined by $f(1)$ the generator of $\mathbb{R}$ but not sure how to work in inequalities.

答: $\Bbb{R}$ admits no non-trivial field automorphisms, so your $f$ has to be the identity function. To see this: any automorphism of $\Bbb{R}$ must fix the rational numbers $\Bbb{Q}$. Also, the order on $\Bbb{R}$ is determined by its algebraic structure, because $x > y$ iff $\exists z(x - y = z^2)$. So any automorphism of $\Bbb{R}$ must fix $\Bbb{Q}$ and be order-preserving. As any element of $\Bbb{R}$ is the limit of a bounded ascending sequence of rationals and the limits of such sequences are unique, an automorphism has no choice but to fix every element of $\Bbb{R}$.

ContentLicense: CC BY-SA 4.0


问:
Existence of a "densest" sequence with natural density $0$

Determine whether there exists an increasing, infinite sequence of natural numbers $S$ with natural density $0$ such that, for almost all $n$ and any increasing, infinite sequence $P\neq S$ also with natural density $0$, $$\sum_{i=1}^{n}(\mathbb{I}_i(S)-\mathbb{I}_i(P))>0$$ (Note: Here, $\mathbb{I}_i$ represents the indicator function, which equals $1$ if $i$ is in sequence $S$ or $P$ and $0$ otherwise)

I came up with this problem while idly daydreaming, not from any textbook or math course. I'm not very experienced with this type of problem (as a high school senior, it's kind of out of my historical purview), but I think I have a solution, written below. If I've made any mistakes in notation, rigorousness, or otherwise, feel free to let me know.

答:
Suppose such a sequence $S$ exists. Then there is some natural number $k\notin S$ because the natural density of $S$ is zero. Inserting $k$ into the sequence yields a new sequence that violates the condition for all $n$. So no such sequence exists.

So not only does such a sequence not exist, but in fact for any sequence $S$ of density $0$ and any natural number $N$, there exists a sequence $P$ of density $0$ such that for almost all $n$ you have
$$\sum_{i=1}^{n}(\mathbb{I}_i(S)-\mathbb{I}_i(P))<N.$$


问:
Base Case Strong Induction Terrence Tao question
Terrence Tao formulates strong induction as follows: "(Strong principle of induction). Let $m_o$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number . Suppose that for each $m \geq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0 \leq m' \lt m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true since in this case, the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $m \geq m_0$. "

Since we are assuming that $P(m')$ is true for all $m_0 \leq m' \lt m$. Where in this statement does it suggest to prove the base case first? To me all its suggesting is to simply suppose $P(m')$ is true for all $m_0 \leq m' \lt m$ without proving a base case?

答:
This is often a useful approach for using induction to prove a statement's contrapositive.  Assume toward a contradiction that $P(m)$ is false for some $m$.  Then there must be a least $m$ for which it is false.  Tao's implication, though, demonstrates that if $P(m')$ is true for all smaller $m'$, then it must also be true for $P(m)$, which means that $m$ was not in fact the smallest possible counterexample, establishing a contradiction.

Thus, the set of counterexamples to $P(m)$ has no least element and therefore must be empty.


问:
Why is this true by Markov's Inequality?
In the beginning of page 7 of this paper I'm reading, it says that the mass that $P_n$ assigns to each $B_j$ is at most $Le^{-2^{2j-3}}$ by Markov's inequality, but I don't understand how Markov's inequality implies that.

I usually think of Markov's inequality as $P(X>a) <= EX/a$ but how does that tell us anything about the most mass assigned to $B_j$? Is there something I'm missing here?

答:
So, Markov's inequality is a very general inequality, and this is an application of it. (You might hear this one as Chernoff, though this is a special case of that too in that it is applied to empirical distributions.) While reading this paper, do note that $P_n$ is a probability distribution that puts a mass of $1/n$ on each $x_i$, so we can talk about probabilities and expectations under it. In particular, in their notation,

$$L = E_{P_n} e^{\frac{|X|^2}{2 d \sigma^2}} $$

where we have $X$ a random sample chosen by $P_n$. Now:

\begin{align*}
P_{P_n}(X \in B_j) &\leq P_{P_n}(|X|^2 \geq d(2^{2(j-1)} \sigma^2)) \\
&=P_{P_n}(e^{\frac{|X|^2}{2 \sigma^2 d}} \geq e^{2^{2j - 3}}) \\
&\leq Le^{-2^{2j - 3}}
\end{align*}

where the last inequality is Markov's inequality indeed.


问:
Why is $f(x) = x\phi(x)$ injective, i.e. $x\phi(x) = y\phi(y)\Rightarrow x = y$
I noticed that $f(x) = x\phi(x)$ seems to be one-to-one, where $\phi(x)$ is Euler's Phi function.

In particular, I'm writing some numerical python code and the line I have looks something like

```<code>sorted([n*phi(n) for n in range(1,1000)])
</code>```

and there are no duplicates in the list.

First, is it one-to-one?

Second, if it is, is there a simple proof sketch?

答:
Let $x, y$ be integers $ > 1 $ such that $ x \varphi(x) = y \varphi(y) $. We'll try to show $ x = y $.

Let's write the prime factorisations of $ x, y $ as $ x = p_1 ^{\alpha_1} \ldots p_k ^{\alpha_k} $ $ ( p_1 < \ldots < p_k ; \text{ each } \alpha_i > 0 ) $, and $ y = q_1 ^{\beta_1} \ldots q_l ^{\beta_l} $ $ ( q_1 < \ldots < q_l ; \text{ each } \beta_j > 0 ) $. Now $ x\varphi(x) = y\varphi(y) $ becomes $ ( p_1 ^{\alpha_1} \varphi(p_1 ^{\alpha_1}) ) \ldots ( p_k ^{\alpha_k} \varphi(p_k ^{\alpha_k}) ) $ $ = ( q_1 ^{\beta_1} \varphi(q_1 ^{\beta_1}) ) \ldots ( q_l ^{\beta_l} \varphi(q_l ^{\beta_l}) ) $.

Now notice the exponent of $ p_k $ in the prime factorisation of LHS is $ \alpha_k + ( \alpha_k - 1 ) $ $ = 2 \alpha_k - 1 $ [ Because for $ i < k $ neither $ p_i ^{\alpha_i} $ nor $ \varphi(p_i ^{\alpha_i}) ( = p_i ^{\alpha_i - 1} (p_i - 1)) $ carries a prime factor $ p_k $, we need only consider $ p_k ^{\alpha_k} \varphi(p_k ^{\alpha_k}) $, which gives exponent $ 2 \alpha_k - 1 $ ]. Similarly the exponent of $ q_l $ in the prime factorisation of RHS is $ 2 \beta_l - 1 $.

Looking at LHS = RHS and using uniqueness of prime factorisation, we get $ \{ p_k = q_l ; \, 2 \alpha_k - 1 = 2 \beta_l - 1 \} $, i.e. $ \{ p_k = q_l ; \, \alpha_k = \beta_l \} $. So we can cancel $ p_k ^{\alpha_k} \varphi(p_k ^{\alpha_k}) $ in LHS with $ q_l ^{\beta_l} \varphi(q_l ^{\beta_l}) $ in RHS, and now continuing similarly onto the remaining factors gives $ \{ l=k; p_k = q_l, \alpha_k = \beta_l ; $ $ \ldots ; p_1 = q_1, \alpha_1 = \beta_1 \} $, i.e. that $ x = y $, as needed.


问:
Linking matrices to linear equations
I'd love advice on connecting two understandings of matrices as linear transformations.
If we have a system like
2x - y = 1 and
x + y = 5 ,
then we can understand it as

a scalar x times the vector (2, 1) plus a scalar y times the vector (-1, 1) results in the vector (1, 5). You can see that visually. You're trying to find how many copies of (2, 1)T plus how many copies of (-1, 1) will help you to reach (1, 5). But how do I match that with the matrix as linear transformation interpretation, that sending (1, 0) --> (2, -1) and (0, 1) --> (1, 1) results in sending (x, y) to (1, 5)? Obviously, it works out, but they feel like two very different things. In one case, x and y are scalar multipliers of the two component vectors, and in the other case (x, y) is a vector itself that gets transformed to (1, 5). How can I visually connect these two?

答:
The trouble I was having was showing visually. I found a start elsewhere and connected the dots, so I'm sharing this in case it's useful. I think my sticky point was that I kept thinking of the coefficients as the multiplicative factors instead of the variables as the scalars. For example, say that we have the matrix that represents the equations 5x+1y=21 and 2x+3y = 26. Another way we can think of this is

Take x copies of a length of 5

Take y copies of a length of 1

Together, those sum to 21

Take x copies of a length of 2

Take y copies of a length of 3

Those sum to 26.
This matches up with writing the equation as
x * (5, 2) + y * (1, 3) = (21, 26) and you can draw this on graph paper.
Then, in terms of linear transformations, you can think of that coefficient matrix as sending (1, 0) to (5, 2) and (0, 1) to (1, 3). So then the linear system, written with matrix multiplication times the vector (x, y) can be interpreted as looking for some original vector (x, y) that gets transformed to (21, 26) when (1, 0)--> (5, 2) and (0, 1) --> (1, 3). But why, right? Because in the first case, there was no vector (x, y). Instead, x and y were seen as scalars acting on vectors that represented the coefficient of the respective terms. But, we can write the vector (x, y) in terms of the basis vectors  (1, 0) and (0, 1). In other words, (x, y) = x * (1, 0) + y * (0, 1). And you can draw this on graph paper, as some number of (1, 0)s and some number of (0, 1s). But we already know where those basis vectors go to, so substitute those in. As a result, the transformed vector would be x * (5, 2) + y * (1, 3). But we already did that right? We took x copies of (5, 2) and y copies of (1, 3) when we were just thinking about linear equations, trying to guess how many copies of (5, 2) and (1, 3) would get us to (21, 26). As an earlier poster said, the fact that this is a linear transformation, that you can either transform and then add or else add and then transform, makes this possible.

Title: Linking matrices to linear equations


问:
$\sum \frac{n^2+1}{n+2}$ using root test
If I'm only using root test, then is $\sum \frac{n^2+1}{n+2}$ useful to root test? I make it $(\frac{n^2+1}{n+2})^{\frac{1}{n}}$ simply, and it's turning to 1. Then there is impossible to see if the test works. ¿Yes? or no.

答: You are right about the value of the limit in this particular case, the root test is inconclusive. However, note that $\frac{n^2 + 1}{n + 2} \geq 1$ for all $n \geq 2,$ so that your series does not converge. Another way to see this is to note that $\lim_{n \to \infty} \frac{n^2 + 1}{n + 2} = \infty.$ I hope this helps. :)


问:
How do I model the Expected Value in a trial and error until find some value?
The problem I was thinking is: Supose you have m pieces of cake and  n of them are spoiled, I start picking one piece, if it's normal I take it out and continue doing this until I find one spoiled.

What is the expected value of cakes I catch before the spoiled? I thought it could be something geometric but the probabilities change. How can I attack this problem?

答:
You can also compute it directly

$$E(X)=\sum_{i=1}^{n-m}i\frac{n-(i-1)}{m}\prod_{j=1}^{i-1}(1-\frac{n-j}{m})$$


问:
What is the meaning of $S^1$ or $S^2$?
So I'm just getting into topology and I see spaces such as $S^1$ or $S^2$  used a lot.

What is the meaning of $S^1$ or $S^2$? Furthermore, what is  meant by the cartesian product $S^1 \times S^1$ or $S^2 \times S^2$?

I've looked everywhere and it's a difficult question to Google.

All I know is that it has to do with circles. Like a two-dimensional torus is composed of two circles and a three dimensional torus is composed of three circles.

I know this website is for higher-order, computational/proof-based questions but I don't know where else to go to find this information

答:
In general, $\mathbb{S}^n$ denotes the $n$-dimensional unit sphere, i.e, the set of points in $\mathbb{R}^{n+1}$ with magnitude $1$:

$$\mathbb{S}^n = \{v \in \mathbb{R}^{n+1} \ \vert \ \|v\| = 1 \}$$

And so we have:

$$\mathbb{S}^n \times \mathbb{S}^m = \{(v, w) \in \mathbb{R}^{n+m+2} \ \vert \ v \in \mathbb{S}^n, w \in \mathbb{S}^m \}$$

And the topology of these sets is the standard one, i.e the subspace topology.

Title: What is the meaning of $S^1$ or $S^2$?


问:
Uniform convergence of $(f_n)_{n \in \Bbb{N}}$ on $D_1 \cup D_2$
I want to show using the Cauchy criterion that if $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on both $D_1$ and $D_2$, then it must be uniformly convergent on $D_1 \cup D_2$

Here's what I have:

$(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_1$, therefore, $\forall \varepsilon \gt 0, \exists  n_0$ such that if $ n,m \ge n_0$ then $\vert f_n(x) - f_m(x) \vert \lt \frac{\varepsilon}2$.

Similarly, $(f_n)_{n \in \Bbb{N}}$ is uniformly convergent on $D_2$, therefore, $\forall \varepsilon \gt 0, \exists  k_0$ such that if $ p,q \ge k_0$ then $\vert f_p(x) - f_q(x) \vert \lt \frac{\varepsilon}2$.

Hence letting all $n, m, p, q \ge max\{n_0, k_0\}$:

$$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \le \vert f_n(x)- f_m(x)\vert + \vert f_p(x) - f_q(x)\vert \lt \frac{\varepsilon}2 + \frac{\varepsilon}2 = \varepsilon$$

Then it would follow that $$\vert \vert f_n(x)- f_m(x)\vert -\vert f_p(x) - f_q(x)\vert \vert \lt \varepsilon $$

Which I want to think implies that $(f_n)_{n \in \Bbb{n}}$ is convergent on $D_1 \cup D_2$

I don't know if what I did was correct, or if I made a mistake somewhere, so I would appreciate any feedback on my take and some hints on what direction my proof should go assuming it went wrong.

答:
Let me suggest one possible way to prove desired. We know that $f_n \to f$ uniformly on $A$, iif $\lim\limits_{n \to \infty}\sup\limits_{x\in A}|f_n(x)-f(x)|=0$.

So, we can consider 2 sequences $x_n=\sup\limits_{x\in A}|f_n(x)-f(x)|$ and $y_n=\sup\limits_{x\in B}|f_n(x)-f(x)|$. Knowing, that $x_n \to 0$ and that $y_n \to 0$, we need $z_n=\max(x_n,y_n)\to 0$. Now this is more easy to achieved, because for $\forall \varepsilon \gt 0$ we simply take $N=\max(N_1,N_2)$ from limit definition for $x_n$ and $y_n$


问:
Permutations of repetition for selection from a set of 11 colours with 5 sections to colour
I am attempting a generative art project of which I am creating all possible permutations of a template using 11 colours. There are 5 sections that can be coloured.

I have calculated the possible permutations to be 11^5 giving 161,051 different possibilities. I wish to grade them in rarity via their statistical probability of occurring,

i.e all 5 colours the same being the rarest, next being 4 the same, 3 etc etc.

So essentially I am looking for all ways to mix colours = [a, b, c, d, e, f, g, h, i, j, k] with pieces = [r, s, x, y, z]

Doing so has returned the following numbers:

11 of 5/5 the same colour

550 of 4/5 ...

11,000 of 3/5 ...

94,050 of 2/5 ...

55,440 of 1/5 ...

And I am coming up short with being able to calculate the odds of it returning a final result within each bracket (my maths is a little rusty) and initially thought that the values would grow larger within each bracket as the number of the same colours reduced, but as can be seen there is more in the second to last than the last. I'm thinking this may be due to combining two such as 3/5 also having the possibility to also contain a 2/5 or two 1/5 separate permutations within it?

Would anyone be able to help me calculate the correct probabilities of each tier, as currently I'm getting more in the uncommon category than the common one?

If so thank you in advance!

答:
Here's an explanation for why each of the counts is correct, case-by-case.

$5/5$: if all of the sections are the same color, there are $11$ ways to pick the color and therefore $\boxed{11}$ permutations.

$4/5$: there are $11$ ways to pick Color $A$, the one that occupies $4$ sections; $10$ ways to pick some other Color $B$; and $5$ ways to pick the lone section colored with Color $B$, giving $11\cdot10\cdot5 = \boxed{550}$ permutations.

$3/5$: there are two cases. Either there are three colors $A$, $B$, and $C$, such that $A$ colors $3$ sections and $B$ and $C$ color one each, or there are two colors $A$ and $B$, where $A$ colors $3$ sections and $B$ colors two sections.

In the first case, there are $11$ ways to pick Color $A$, $10$ ways to pick Color $B$, and $9$ ways to pick Color $C$. There are $5$ ways to choose the section with Color $B$ and $4$ ways to choose the section with Color $C$, and the other sections can be left for Color $A$. However, this double-counts by a factor of $2$, since we can reach a permutation twice by swapping our choices of $B$ and $C$ and swapping the sections we choose. Therefore, this case includes $11\cdot10\cdot9\cdot5\cdot4 \cdot (1/2) = 9{,}900$ permutations.

In the second case, there are $11\cdot10$ ways to pick Colors $A$ and $B$, and $\binom{5}{2} = 10$ ways to pick the $2$ sections to be colored by Color $B$. Therefore, this case includes $11\cdot10\cdot10=1{,}100$ permutations, yielding a total of $\boxed{11{,}000}$ permutations for $3/5$.

$1/5$: We have to pick a different color for each section, so there are $11\cdot10\cdot9\cdot8\cdot7 = \boxed{55{,}440}$ of these permutations.

$2/5$: this is just all the other cases, so there are $161{,}051-11-550-11{,}000-55{,}440=\boxed{94{,}050}$ of these.

In general, this counter-intuitiveness only arises because there are just not enough colors for duplicates to be uncommon. On the low end–if there are only $5$ colors, for example–it'd seem pretty unlikely to manage to hit a different color each time. If there were $4$ colors, it'd be outright impossible! Meanwhile, on the high end–say, if there are $1{,}000$ colors–hitting the same color twice would be pretty unlikely. So, the "crossover" point, where using the same color twice becomes less likely than using different colors for each section, is somewhere in the middle, and it just so happens that $11$ colors is not enough to make duplicating colors less likely.

In fact, we can find the number of colors such that using a different color each time would be more frequent than using the same color twice. The probability of using all different colors is $$ \frac{n(n-1)(n-2)(n-3)(n-4)}{n}. $$ (To see why, there are $n$ ways to pick the first section's color, $n-1$ ways to pick the second section's color, etc.) Numerically solving $$ \frac{n(n-1)(n-2)(n-3)(n-4)}{n} > \frac{1}{2}$$ yields $n > 16.01$, so you'd have to have at least $17$ colors for those "common" arrangements to be more likely than not.


答:
Given an infinite set $S$ with density $0$, to construct a set $P$ of density $0$ whose summatory function grows asymptotically faster:

For all $k\ge 0,a\ge 0$, $2^k-a\in P$ if $2^{k/2} (\sum_{n\in S,n\le 2^{k+1}} 1)^{1/2}\ge a$

Title: Existence of a "densest" sequence with natural density $0$

Body:
Determine whether there exists an increasing, infinite sequence of natural numbers $S$ with natural density $0$ such that, for almost all $n$ and any increasing, infinite sequence $P\neq S$ also with natural density $0$, $$\sum_{i=1}^{n}(\mathbb{I}_i(S)-\mathbb{I}_i(P))>0$$ (Note: Here, $\mathbb{I}_i$ represents the indicator function, which equals $1$ if $i$ is in sequence $S$ or $P$ and $0$ otherwise)

I came up with this problem while idly daydreaming, not from any textbook or math course. I'm not very experienced with this type of problem (as a high school senior, it's kind of out of my historical purview), but I think I have a solution, written below. If I've made any mistakes in notation, rigorousness, or otherwise, feel free to let me know.


问:
combinations question help
Let $N > 10$ be an integer. We have $N$ one-dollar bills that we split between three people. No person can be given less than two one-dollar bills, and everyone gets a whole number of bills (no cents). In how many ways can we do this?

This is how I tried solving the problem. Before splitting, if we give each of the three people two (2) one-dollar bills, then we'll be starting with $N-6$ total bills. Since order doesn't matter, there could be $\displaystyle{{N-6}\choose3}$ ways to do this. Is this right or am I missing something? Any help would be appreciated.

答:
You're right about starting with figuring out how many ways to give out $N-6$ bills.
The method you need is called the "stars and bars" argument.  Imagine the three people are lined up.  There are two dividers separating people.  There are $N-6$ bills.  Together there are $N-4$ things.  How many ways can you arrange the two dividers among them?  ${N-4 \choose 2}$

Title: combinations question help


问:
$X$ is a r.v. given $f(x)=2x\chi_{[0,1]}(x)$ a density function calculate $Y=X^2$.
QUESTION: Let $X$ be a random variable with density function given by $f(x)=2x\chi_{[0,1]}(x)$, for all $x\in \mathbb{R}$ (where $\chi_{[0,1]}$ is the indicator function). Calculate the density function of $Y=X^2$.

MY ATTEMPTY: First of all, let's check that $f$ is a density function.

Clearly we have $f(x)\geq 0$ once the indicator function is give by $\chi_{[0,1]}(x)= 1$ if $x\in [0,1]$, and $\chi_{[0,1]}(x)= 0$ otherwise. Then $f(x)=2x$ if $x\in [0,1]$ and $f(x)= 0$ otherwise. Thus, indeed $f(x)\geq 0$, for all $x\in \mathbb{R}$.

$\int_{-\infty}^{\infty} f(x)dx=\int_{0}^{1} 2x dx= 1$

From 1 and 2 it follows that f is a density function.

Now, we are going to calculate $Y=X^2$.
We have that $f_Y(y)=F'_Y(y)$ if $F'_Y$ exists. Hence,

\begin{align*}
F_Y(y)&=P(Y\leq y)\\
 &=P(X^2\leq y)\\
 &=P(X\leq \sqrt{y})
\end{align*}
$\implies F_Y(y)=F_X(\sqrt{y})$ this means that the diferenciability of Y works where X works.

Now, defining, $h(y)=\sqrt{y}$ we have that $h'(y)=\frac{1}{2\sqrt{y}}, \; \forall y\in \mathbb{R}.$

By the chain rule follows

\begin{align*}
F'_Y(y)&=F'_X(y)\cdot \frac{1}{2\sqrt{y}}, \; \forall y \in \mathbb{R}\\
 &=\frac{1}{2\sqrt{y}} f_X(\sqrt{y})
\end{align*}

where
$$
    f_X(h(y))=\left\{\begin{array}{cc}
    1 &\;\mbox{if}\; y \in [0,1]\\
    0&\;\mbox{otherwise}
 \end{array}
   \right.
$$
$$\iff 0\leq \sqrt{y}\leq 1 \iff 0\leq y \leq 1$$

Therefore, $Y$ is a continuous function and its density is given by
$$
    f_Y(y) = f_X(h(y)). |h'(y)|=\left\{\begin{array}{cc}
    \frac{1}{2\sqrt{y}} &\;\mbox{if}\; y \in [0,1]\\
    0&\;\mbox{otherwise}
 \end{array}
    \right.
$$

MY DOUBT: Is my answer right? Would you please check it for me? My question is precisely because I don't understand where I have to use $f(x)=2x$ when I'm calculating $Y=X^2$.

答:
It's incorrect. Since $f_X(x) = 2x \cdot \chi_{[0,1]}(x) \ \forall x \in \mathbb{R}$ and $h(y) = \sqrt{y}$, it follows that $$ \begin{aligned} f_Y(y) = f_X( \sqrt{y} )  \cdot \frac{1}{2 \sqrt{y}} = 1 \ \forall y \in [0,1] \ \end{aligned}$$

and $f_Y(y) = 0$ otherwise, therefore $f_Y = \chi_{[0,1]}$.


问:
Check whether the series converges...

Check whether the series converge or diverge; $\sum_{n=1}^\infty(3^{\frac{1}{n^2}}-1)\csc^2(n).$

By the way, As $n\to \infty(3^{\frac{1}{n^2}}-1)\csc^2(n)=1.$
inconclusive.

I don't have an idea. How to check this series?

Any help will be appreciated. Thanks!

答:
The series is all positive terms, so no Alternating Series Test.
The $\csc^2 (n)$ factor essentially oscillates chaotically between $1$ and $\infty$
so there is no point beyond which the terms stay reliably below a given $\epsilon$. The factor of $(3^{\frac{1}{n^2}}-1)$ goes to zero rapidly, but I don't think there is any limit to how high the cosecant factor can spike.  So by the Divergence test, it diverges.

Title: Check whether the series converges...


问:
What exactly is going on here? (order of limits and substitutions give different results)
Consider this expression:

$$(-1)^{G+T+1} 2^{G-2T-2} G (G-1)(G-2) \frac{(G-T-2)!}{T!(G-2T)!}$$

Here, $G,T\in \mathbb{Z}$.

If we set $G = 2T$, it becomes

$$(-1)^{T+1}(2T-1) .$$

If we now set $T=0$, this evaluates to one. So far so good. Note that since $G=2T$, this obviously corresponds to $G=0$.

But if I try to set $G=0$ first, the original expression evaluates to zero. To make it worse, if I first set $T=0$, the original expression becomes

$$(-1)^{G+1} 2^{G-2}(G-2)$$

and it evaluates to one half when setting $G=0$.

I understand that taking limits in different order cannot always be done and I understand that, technically speaking, expressions of type $n!/(n-1)!$ aren't always strictly equal to $n$. Could someone shed some more light on what exactly is going on here? I'm a big boy and I don't like being confused by elementary stuff :)

答:
For the first expression (when you first plug in $G = 2T$), the problem is a domain issue with the simplification. Substituting $G=2T$, we have $$ (-1)^{G+T+1}2^{G-2T-2}G(G-1)(G-2)\frac{(G-T-2)!}{T!(G-2T)!} = (-1)^{3T+1}2^{-2}(2T)(2T-1)(2T-2)\frac{(T-2)!}{T!}. $$

However, as you've pointed out, this last fraction $\frac{(T-2)!}{T!}$ doesn't always simplify to $\frac{1}{T(T-1)}$. In fact, it doesn't even make sense when $T -2 < 0$, without extending the definition of $!$ with something like the gamma function.

Similarly, if you first set the value of $T$ to $0$, then it becomes $$ (-1)^{G+T+1}2^{G-2T-2}G(G-1)(G-2)\frac{(G-T-2)!}{T!(G-2T)!} = (-1)^{G+1}2^{G-2}G(G-1)(G-2)\frac{(G-2)!}{G!}, $$ and you run into the same problems.

Without further context on the purpose of this expression (if it's just for recreation, that's fine too!), there isn't a clear "best" way to resolve the domain issue when $G - T - 2 < 0$, $T < 0$, or $G - 2T < 0$. Here, the first is the one that is causing problems.


问:
Find the least $n$ such that $f(n)/g(n)=4/7$

Let $f(n)$ and $g(n)$ be functions satisfying $$f(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 1+f(n+1) \hspace{14pt} \text{otherwise}} $$
$$g(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 2+g(n+2) \hspace{14pt} \text{otherwise}}$$
for positive integers $n$. Find the least positive value of $n$ such that $\dfrac{f(n)}{g(n)}=\dfrac47$.

I found the values for $f$ and $g$ till $n=9$, I noticed that $f(n)=g(n)$ happened at $n=1,2,4,5,7,9$ and for some values it also happened that if $f(a)=b$, then $f(b)=a$. I don't know if that is useful in anyway. I can't see how to approach the problem.

Please give some hints.

答:
Computing the Functions

Define the "next square root" function
$$
s(n)=\left\lceil\sqrt{n}\right\rceil\tag1
$$
and the "next square root with the same parity" function
$$
p(n)=2\left\lceil\frac{\sqrt{n}-[n\text{ odd}]}2\right\rceil+[n\text{ odd}]\tag2
$$
where $[\cdots]$ are Iverson Brackets.

Then
$$
\begin{align}
f(n)&=s(n)+s(n)^2-n\tag{3a}\\
g(n)&=p(n)+p(n)^2-n\tag{3b}
\end{align}
$$
$\frac{f(n)}{g(n)}=\frac47$ when
$$
n\in\{258,488,790,1164,1610,2128,2718,3380,4114,4920,5798,6748,7770,8864\}
$$
That is all for $n\le10000$.

However, this approach relies on brute force computation to figure out when $\frac{f(n)}{g(n)}=\frac47$.

Mathematica Code

```<code>s[n_] := Ceiling[Sqrt[n]]
q[n_] := If[OddQ[n], 1, 0]
p[n_] := 2 Ceiling[(Sqrt[n] - q[n])/2] + q[n]
f[n_] := s[n] + s[n]^2 - n
g[n_] := p[n] + p[n]^2 - n
Flatten@Position[Table[f[n]/g[n], {n, 1, 10000}], 4/7]
</code>```

A More Understandable Approach

Note that
$$
(s(n)-1)^2\lt n\le s(n)^2\tag4
$$
When $s(n)-n$ is even, the next square root with the same parity is $p(n)=s(n)$ and so $f(n)=g(n)$.

Thus, to get $\frac{f(n)}{g(n)}=\frac47$, we need $s(n)-n$ to be odd. This means that the parity of $n$ and $s(n)$ are different; therefore, the next square root with the same parity is
$$
p(n)=s(n)+1\tag5
$$
Furthermore,
$$
\begin{align}
g(n)&=f(n)+2s(n)+2\tag{6a}\\
\frac34f(n)&=2s(n)+2\tag{6b}\\
3n&=3s(n)^2-5s(n)-8\tag{6c}\\[6pt]
&=(3s(n)-8)(s(n)+1)\tag{6d}
\end{align}
$$
Explanation:

$\text{(6a)}$: apply $(5)$ to $(3)$

$\text{(6b)}$: $g(n)=\frac74f(n)$

$\text{(6c)}$: apply $(3)$

$\text{(6d)}$: factor

$(4)$ and $\text{(6c)}$ give
$$
\overbrace{3s(n)^2-6s(n)+3}^{3(s(n)-1)^2}\lt\overbrace{3s(n)^2-5s(n)-8}^{3n}\tag7
$$
which says we need $s(n)\gt11$.

$\text{(6d)}$ guarantees that $n$ is even since $3s(n)-8$ has the same parity as $s(n)$ and $(s(n)+1)$ has the opposite parity. Since $s(n)-n$ needs to be odd, $s(n)$ needs to be odd.

$\text{(6d)}$ requires that $s(n)\equiv2\pmod3$.

Thus, if $s\gt11$, $s$ is odd, and $s\equiv2\pmod3$, then $n=\frac{(3s-8)(s+1)}3$ will give $\frac{f(n)}{g(n)}=\frac47$.

Therefore, let $s=6k+5$, and then $\frac{f(n)}{g(n)}=\frac47$ precisely when
$$
\bbox[5px,border:2px solid #C0A000]{n=2(k+1)(18k+7)\quad\text{and}\quad k\ge2}\tag8
$$
When $k=2$, we get $n=258$ as the smallest $n$ so that $\frac{f(n)}{g(n)}=\frac47$.

Body:
Computing the Functions

Define the "next square root" function
$$
s(n)=\left\lceil\sqrt{n}\right\rceil\tag1
$$
and the "next square root with the same parity" function
$$
p(n)=2\left\lceil\frac{\sqrt{n}-[n\text{ odd}]}2\right\rceil+[n\text{ odd}]\tag2
$$
where $[\cdots]$ are Iverson Brackets.

Then
$$
\begin{align}
f(n)&=s(n)+s(n)^2-n\tag{3a}\\
g(n)&=p(n)+p(n)^2-n\tag{3b}
\end{align}
$$
$\frac{f(n)}{g(n)}=\frac47$ when
$$
n\in\{258,488,790,1164,1610,2128,2718,3380,4114,4920,5798,6748,7770,8864\}
$$
That is all for $n\le10000$.

However, this approach relies on brute force computation to figure out when $\frac{f(n)}{g(n)}=\frac47$.

Mathematica Code

s[n_] := Ceiling[Sqrt[n]]
q[n_] := If[OddQ[n], 1, 0]
p[n_] := 2 Ceiling[(Sqrt[n] - q[n])/2] + q[n]
f[n_] := s[n] + s[n]^2 - n
g[n_] := p[n] + p[n]^2 - n
Flatten@Position[Table[f[n]/g[n], {n, 1, 10000}], 4/7]

A More Understandable Approach

Note that
$$
(s(n)-1)^2\lt n\le s(n)^2\tag4
$$
When $s(n)-n$ is even, the next square root with the same parity is $p(n)=s(n)$ and so $f(n)=g(n)$.

Thus, to get $\frac{f(n)}{g(n)}=\frac47$, we need $s(n)-n$ to be odd. This means that the parity of $n$ and $s(n)$ are different; therefore, the next square root with the same parity is
$$
p(n)=s(n)+1\tag5
$$
Furthermore,
$$
\begin{align}
g(n)&=f(n)+2s(n)+2\tag{6a}\\
\frac34f(n)&=2s(n)+2\tag{6b}\\
3n&=3s(n)^2-5s(n)-8\tag{6c}\\[6pt]
&=(3s(n)-8)(s(n)+1)\tag{6d}
\end{align}
$$
Explanation:

$\text{(6a)}$: apply $(5)$ to $(3)$

$\text{(6b)}$: $g(n)=\frac74f(n)$

$\text{(6c)}$: apply $(3)$

$\text{(6d)}$: factor

$(4)$ and $\text{(6c)}$ give
$$
\overbrace{3s(n)^2-6s(n)+3}^{3(s(n)-1)^2}\lt\overbrace{3s(n)^2-5s(n)-8}^{3n}\tag7
$$
which says we need $s(n)\gt11$.

$\text{(6d)}$ guarantees that $n$ is even since $3s(n)-8$ has the same parity as $s(n)$ and $(s(n)+1)$ has the opposite parity. Since $s(n)-n$ needs to be odd, $s(n)$ needs to be odd.

$\text{(6d)}$ requires that $s(n)\equiv2\pmod3$.

Thus, if $s\gt11$, $s$ is odd, and $s\equiv2\pmod3$, then $n=\frac{(3s-8)(s+1)}3$ will give $\frac{f(n)}{g(n)}=\frac47$.

Therefore, let $s=6k+5$, and then $\frac{f(n)}{g(n)}=\frac47$ precisely when
$$
\bbox[5px,border:2px solid #C0A000]{n=2(k+1)(18k+7)\quad\text{and}\quad k\ge2}\tag8
$$
When $k=2$, we get $n=258$ as the smallest $n$ so that $\frac{f(n)}{g(n)}=\frac47$.


问:
Smallest projective subspace containing a degree $d$ curve
Is it true that the smallest projective subspace containing a degree $d$ curve inside $\mathbb{P}^n$ has dimension at most $d$? If not, is there any bound on the dimension? Generalization to varieties?

For $d=1$ this is obvious. I think for the case that the curve is an embedding of $\mathbb{P}^1$ this is also true: Suppose the embedding is given by $n$ degree $d$ homogeneous polynomials $f_0,\dots,f_n$. For each $0\leq i\leq d$, let $p_i=(c_{0,i},\dots,c_{n,i})$ where $c_{j,i}$ is the coefficient of $x^iy^{d-i}$ in $f_j$ (or we ignore $p_i$ if all $c_{j,i}$ are zero). Then the curve is contained in the projective subspace spanned by all $p_i$.

答:
I think your observation is correct for curves.

Given a curve $C$ in $\mathbb{P}^n$ satisfying $C$ is not contained in any projective subspace of $\mathbb{P}^n$, WLOG we may assume $C$ is irreducible. Let $\tilde C$ be the normalization of $C$, then we have a regular map $\phi: \tilde C\rightarrow \mathbb{P}^n$ which is an embedding outside a finite subset of $\tilde C$.

Since $C$ is not contained in any projective subspace, $\phi$ is the map induced by the linear series $L$ of all hyperplane divisors. It is clear that $\dim L=\dim \mathbb{P}^n=n$, so for any hyperplane divisor $D$, we have $h^0(\tilde C,D)\geq \dim L+1=n+1$. On the other hand, $h^0(\tilde C,D)\leq \deg(D)+1$ since $D$ is an effective divisor, so $\deg C=\deg D\geq n$.

Any generalizations to higher dimensional varieties?

Guess: for a $k$-dimensional variety $X$ embedded in $\mathbb{P}^n$, if $X$ is not contained in any projective subspace, then $\deg X\geq n-k+1$.

ContentLicense: CC BY-SA 3.0


问:
Gradient of 2 layer network
I am considering a simple $2$-layer network with $m$ training data $(x^i, y^i)$ data whose cost function is

$$\ell(w,\alpha, \beta) := \sum_{i=1}^{m} \left( y^i - \sigma(w^Tz^i) \right)^2$$

where

$$\sigma(x) := \frac{1}{1+e^{-x}}$$

is the sigmoid function and $z_1^i = \sigma(\alpha^T x^i)$, and $z_2^i = \sigma(\beta^T x^i)$. The gradient is

$$\nabla_w \ell(w, \alpha, \beta) = - \sum_{i=1}^m 2(y^i - \sigma(u^i))\sigma(u^i)(1-\sigma(u^i)) z^i$$

where $(u^i) = w^Tz^i$

I am trying to derive the gradient by taking derivative over the given cost function, but am not able to do so. After applying chain rule, i get stuck at the term $\sigma(w^Tz^i)$ Any one could step me through please?

答:
The cost function was

$$\ell(w,\alpha, \beta) := \sum_{i=1}^{m} \left( y^i - \sigma(w^Tz^i) \right)^2$$

Now take derivative
$$=2 \sum_{i=1}^{m} \left( y^i - \sigma(w^Tz^i) \right)D(\sigma(w^Tz^i))$$

We note that
$$D(\sigma(w^Tz^i))=\sigma(w^Tz^i)(1-\sigma(w^Tz^i))D(w^Tz^i)$$
$$=\sigma(w^Tz^i)(1-\sigma(w^Tz^i))z^i$$

And done...

(Note that we used the usual identity for sigmoid, the derivative is $f(x)(1-f(x))$.)

Title: Gradient of 2 layer network


答: You are right about the first part of your answer. Since each person automatically receives at least $2$ dollars, the problem is equivalent to finding the number of ways $N - 6$ can be split as a sum of three non-negative numbers. I will provide a different method which does not rely on knowing the stars and bars argument. Let $S_3(N - 6)$ be this number. In a similar manner, we may define $S_2(n)$ to be the number of ways in which $n$ may be split as the sum of two non-negative integers, where the order matters. Note that $S_2(n) = n + 1$ for any $n \geq 0.$ It follows that $S_3(N - 6) = \sum_{0 \leq k \leq N - 6} S_2(k) = \sum_{0 \leq k \leq N - 6} (k + 1) = \frac{(N - 5)(N - 4)}{2} = \binom{N - 4}{2}.$ This is because we imagine giving the first person $k$ bills. Then we will be forced to split $N - k$ bills between the remaining two. Generalising this method leads to an interesting discovery. Let us denote by $S_n(p)$ the number of ways in which we can write $p$ as the sum of $n$ non-negative integers. Also define by convention $S_0(0) := 1,$ $S_0(l) := 0$ if $l > 0$ and note that $S_1(l) = 1$ for all $l \geq 0.$ Then by the same reasoning as above, we get that for any $0 \leq k \leq n,$ the following equality holds: $$S_n(p) = \sum_{0 \leq j \leq p} S_{n - k}(p - j) S_k(j) = S_{n - k} \ast S_k (p),$$ where $\ast$ denotes the convolution product. It is somehow interesting to find the convolution product in a problem where it isn't obvious how it might sneak up on us a priori. I hope this helps. :)

Title: combinations question help

Body:
Let $N > 10$ be an integer. We have $N$ one-dollar bills that we split between three people. No person can be given less than two one-dollar bills, and everyone gets a whole number of bills (no cents). In how many ways can we do this?

This is how I tried solving the problem. Before splitting, if we give each of the three people two (2) one-dollar bills, then we'll be starting with $N-6$ total bills. Since order doesn't matter, there could be $\displaystyle{{N-6}\choose3}$ ways to do this. Is this right or am I missing something? Any help would be appreciated.


问:
Given any finite surjections $f:X\to Y$ and $g:X\to Z$ with $|Y|<|Z|$ does there always $\exists x\in g^{-1}[Z\setminus Y]:|f^{-1}[\{f(x)\}]|>1$?
Given any three finite sets $X,Y,Z$ satisfying $1\leq |Y|<|Z|\leq |X|$ as well as surjections $f:X\to Y$ and $g:X\to Z$ is it always true that $\exists x\in g^{-1}[Z\setminus Y]:|f^{-1}[\{f(x)\}]|>1$?

Or equivalently must there always be distinct $p,q\in X$ such that $f(p)=f(q)$ and $g(q)\in Z\setminus Y$?

答:
Here is a counterexample.

Let $X=Z=\{0,1,2\}$ and $Y=\{0,1\}$. Let $f(0)=f(1)=0$ and $f(2)=1$, and let $g$ be the identity function. The only $x\in X$ such that $g(x)\in Z\setminus Y$ is $2$, and it is the only element of $X$ whose image under $f$ is $1$.


问:
Existence of a particular function
Given positive integers $n,d,\ell\ge1$ such that $n\ge 1+(d-1)\ell$, I must find a a function
$$
a=a(n,d,\ell),
$$
such that
\begin{align*}
0< a(n,d,\ell)&<1,\\
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&\le a(n,d,\ell),
\end{align*}
for all $n,d,\ell\ge1,\ n\ge1+(d-1)\ell.$

I've been able to solve only the case $\ell=1$. In such a case, the function
$$
a(n,d,1)=\frac{n-d}{n-d+1}
$$
satisfy the requests.

Any help is appreciated.

答:
This might be a bit bulky, but I found a solution. TL;DR, for all $\ell>1$, you can have: $$a(n,d,\ell)= {n!\prod_{g=0}^{\operatorname{floor}(\frac{n-d}{\ell-1})}(1+\frac{n-d}{\ell-1}+g(\ell-2))\over1+(1+\frac{n\ell-d\ell}{\ell-1})!\prod_{h=0}^{d-1}(n-h-(d-1-h)(\ell-1))}$$
This works for all $\ell>1$, but if $\ell>2$, then you could "simplify" it to$$a(n,d,\ell)=\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$

The way I derived this answer was by considering your condition that$$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)\leq a(n,d,\ell)$$ I noticed it looked a lot like recursion because if you have a working set of positive integers $(n,d,\ell)$ that satisfy $n\geq1+(d-1)\ell$, and if $d\geq2$, then $(n-1,d-1,\ell)$ also satisfies that condition. The proof for that is trivial, but we needed this information to realize that$$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)\leq a(n,d,\ell)$$ $$\Rightarrow$$ $$\frac{n-1}{n-1-(d-2)(\ell-1)}a(n-2,d-2,\ell)\leq a(n-1,d-1,\ell)$$

So, initially I thought it would be an exponential, but that led me nowhere, so I looked harder at the condition $n\geq1+(d-1)\ell$ a little more. I focused at the graph of $1\leq d\leq\frac{n+\ell-1}{\ell}$ in cartesian coordinates to find patterns. I used $d$ as the $y$-axis, and $n$ as the $x$-axis, but that's besides the point. I chose this space because $n$ and $d$ would be changing by $-1$ while $\ell$ wouldn't change at all. Since you need the function to be within the bounds of $0$ and $1$, I knew that the line $d=\frac{n+\ell-1}{\ell}$ would have to be where $a(n,d,\ell)=1$ because $n$ and $d$ had to decrease by a factor of at least $\frac{n-(d-1)(\ell-1)}{n}$ every step down and to the left.

From here, I knew that the inverse was true, that the function would increase by at most $\frac{n}{n-(d-1)(\ell-1)}$ every step up and to the right, so I derived the numerator of our function$$\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}\Rightarrow{n!\over(n-d)!\prod_{h=0}^{d-1}(n-h-(d-1-h)(\ell-1))}$$
Remember that we need the line $d=\frac{n+\ell-1}{\ell}$ to be 1, and we move up and to the right by one step each. So we need to divide our numerator by where that line and $d=n+n_0-d_0$ meet where $n_0$ and $d_0$ are arbitrary inputs (it'll cancel out later). The two lines intersect at $(1+\frac{n_0\ell-d_0\ell}{\ell-1},1+\frac{n_0-d_0}{l-1})$.

All we have to do now is take our numerator and replace $n$ and $d$ with $1+\frac{n_0\ell-d_0\ell}{\ell-1}$ and $1+\frac{n_0-d_0}{\ell-1}$ respectively, resulting in the denominator being:$${\frac{(1+\frac{n_0\ell-d_0\ell}{\ell-1})!}{((1+\frac{n_0\ell-d_0\ell}{\ell-1})-(1+\frac{n_0-d_0}{\ell-1}))!}\over\prod_{h=0}^{(1+\frac{n_0-d_0}{l-1})-1}(1+\frac{n_0\ell-d_0\ell}{\ell-1}-h-((1+\frac{n_0-d_0}{l-1})-1-h)(\ell-1))}$$
After some simplification, we get$${(1+\frac{n_0\ell-d_0\ell}{\ell-1})!\over(n_0-d_0)!\prod_{g=0}^{\frac{n_0-d_0}{l-1}}(1+\frac{n_0-d_0}{\ell-1}+g(\ell-2))}$$
which is now our denominator.

We have a lot of junk right now, so lets put them together to see what we can cancel out.$${{n!\over(n-d)!\prod_{h=0}^{d-1}(n-h-(d-1-h)(\ell-1))}\over{(1+\frac{n_0\ell-d_0\ell}{\ell-1})!\over(n_0-d_0)!\prod_{g=0}^{\frac{n_0-d_0}{l-1}}(1+\frac{n_0-d_0}{\ell-1}+g(\ell-2))}}$$
It turns out that $n_0$ and $d_0$ can be replaced with just $n$ and $d$, and the proof is trivial. That lets us simplify a little bit so we get our (almost) final answer of:$$a(n,d,\ell)= {n!\prod_{g=0}^{\frac{n-d}{\ell-1}}(1+\frac{n-d}{\ell-1}+g(\ell-2))\over(1+\frac{n\ell-d\ell}{\ell-1})!\prod_{h=0}^{d-1}(n-h-(d-1-h)(\ell-1))}$$
The reason we're not done yet is because the product has a fractional upper bound. That allows it to reach greater than 1. To fix this, we add the floor function to the upper product bound so that it truly matches what we want to do. We'll also add a $+1$ denominator so that the range is exactly less than one if the numerator and denominator happen to be the same.$$a(n,d,\ell)= {n!\prod_{g=0}^{\operatorname{floor}(\frac{n-d}{\ell-1})}(1+\frac{n-d}{\ell-1}+g(\ell-2))\over1+(1+\frac{n\ell-d\ell}{\ell-1})!\prod_{h=0}^{d-1}(n-h-(d-1-h)(\ell-1))}$$ That is our final answer. We could make it continuous. Although I wasn't able to simplify it by hand to get rid of the product sign, Wolfram Alpha helped plenty. This gives the other final answer of$$a(n,d,\ell)=\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$ where $z!$ is shorthand for $z!=\Gamma(z+1)$ and $\Gamma(z)$ is the gamma function, which is continuous by definition.

Let's see how it tests under the conditions, shall we? Let's start with the easy bit where $$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)\leq a(n,d,\ell)$$
We'll evaluate the left side. Let's start with simplifying the function by substituting the inputs to get $$a(n-1,d-1,\ell)=\frac{(n-1)!(\frac{(n-1)-(d-1)\ell+(d-1)+1}{\ell-2})!(\ell-2)^{1+\frac{(n-1)-(d-1)\ell}{\ell-1}}}{(1+\frac{(n-1)\ell-(d-1)\ell}{\ell-1})!(\frac{(n-1)-(d-1)}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
We'll get rid of those pesky parentheses to get $$\frac{(n-1)!(\frac{n-1-d\ell+\ell+d-1+1}{\ell-2})!(\ell-2)^{1+\frac{n-1-d\ell+\ell}{\ell-1}}}{(1+\frac{n\ell-\ell-d\ell+\ell}{\ell-1})!(\frac{n-1-d+1}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
We'll follow up with some reduction and necessary factoring to get what we need: $$\frac{(n-1)!(\frac{n-(d-1)(\ell-1)}{\ell-2})!(\ell-2)^{2+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
Since this is only the first half of the left side, we need to multiply by $\frac{n}{n-(d-1)(\ell-1)}$. So we now have $$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)=\frac{n}{n-(d-1)(\ell-1)}\frac{(n-1)!(\frac{n-(d-1)(\ell-1)}{\ell-2})!(\ell-2)^{2+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
We'll notice that $n(n-1)!$ is just $n!$, so we can cancel that part out plain and simple.$$\frac{1}{n-(d-1)(\ell-1)}\frac{n!(\frac{n-(d-1)(\ell-1)}{\ell-2})!(\ell-2)^{2+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
This next step took me a few seconds to see but let's notice the converse of that exact statement but for $(\frac{n-(d-1)(\ell-1)}{\ell-2})!$. We'll also bring that $\frac{1}{n-(d-1)(\ell-1)}$ up to the top half. So, now we have $$\frac{n!(\frac{n-(d-1)(\ell-1)}{\ell-2}-1)!(\frac{n-(d-1)(\ell-1)}{\ell-2})\frac{(\ell-2)^{2+\frac{n-d\ell}{\ell-1}}}{n-(d-1)(\ell-1)}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
We can cancel out a $n-(d-1)(\ell-1)$ and a $\ell-2$ to get $$\frac{n!(\frac{n-(d-1)(\ell-1)}{\ell-2}-1)!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
Now for our last step. That middle term, $(\frac{n-(d-1)(\ell-1)}{\ell-2}-1)!$, looks kinda weird with that $-1$ in there. We'll do some un-weirding to get the final product. $$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)=\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
But, is it true for our function that it meets the condition we had? $$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)\leq a(n,d,\ell)$$
Well, we know what the right side is, and we now know what the left side is, so we'll plug them in: $$\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}\leq\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$
To see it more clearly, let me cancel some terms out from both sides: $$1\leq1$$ which is true. Therefore, our original function $$a(n,d,\ell)=\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$ satisfies the condition that $$\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)\leq a(n,d,\ell)$$

The second condition we'll look at is the condition that $0<a(n,d,\ell)<1$ where $n\geq1+\ell(d-1)$. To do this, we need to notice that, if $\ell\geq1$, then  $n\leq1+\frac{n\ell-d\ell}{\ell-1}$ and $d\leq1+\frac{n-d}{\ell-1}$. The proof for this is trivial. We use this fact to our advantage by remembering the (very) crude formulas for our numerator and denominator:$$a(n,d,\ell)=\frac{\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}}{1+\prod_{g=0}^{(1+\frac{n-d}{\ell-1})-1}\frac{(1+\frac{n\ell-d\ell}{\ell-1})-g}{(1+\frac{n\ell-d\ell}{\ell-1})-g-((1+\frac{n-d}{\ell-1})-1-g)(\ell-1)}}$$
It is trivial to see that for any arbitrary $t\geq0$, $$0<\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}\leq\prod_{g=0}^{d+t-1}\frac{n+t-g}{n+t-g-(d+t-1-g)(\ell-1)}$$
This gives us all we need to know. We know that $n\leq1+\frac{n\ell-d\ell}{\ell-1}$ and $d\leq1+\frac{n-d}{\ell-1}$, but we also know that $n\leq n+t$ and $d\leq d+t$ for any $t\geq0$ (not just integers). Then we can safely say that $$\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}\leq\prod_{g=0}^{(1+\frac{n-d}{\ell-1})-1}\frac{(1+\frac{n\ell-d\ell}{\ell-1})-g}{(1+\frac{n\ell-d\ell}{\ell-1})-g-((1+\frac{n-d}{\ell-1})-1-g)(\ell-1)}$$
With that out of the way, we have only proven that $0<a(n,d,\ell)\leq1$, or so it may seem. In our original function $a(n,d,\ell)$, we see in the denominator that we added a one. By adding one to the denominator, we ensure that the numerator is strictly less than the denominator. In math terms, $a\leq b\Rightarrow a<b+1\Rightarrow\frac{a}{b+1}<1$. Just replace $a=\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}$ and $b=\prod_{g=0}^{(1+\frac{n-d}{\ell-1})-1}\frac{(1+\frac{n\ell-d\ell}{\ell-1})-g}{(1+\frac{n\ell-d\ell}{\ell-1})-g-((1+\frac{n-d}{\ell-1})-1-g)(\ell-1)}$ to get back to our original (crude) function $$a(n,d,\ell)=\frac{\prod_{h=0}^{d-1}\frac{n-h}{n-h-(d-1-h)(\ell-1)}}{1+\prod_{g=0}^{(1+\frac{n-d}{\ell-1})-1}\frac{(1+\frac{n\ell-d\ell}{\ell-1})-g}{(1+\frac{n\ell-d\ell}{\ell-1})-g-((1+\frac{n-d}{\ell-1})-1-g)(\ell-1)}}$$
We already simplified that previously, so therefore we have proven that $0<a(n,d,\ell)<1$ given $$a(n,d,\ell)=\frac{n!(\frac{n-d\ell+d+1}{\ell-2})!(\ell-2)^{1+\frac{n-d\ell}{\ell-1}}}{(1+\frac{n\ell-d\ell}{\ell-1})!(\frac{n-d}{(\ell-1)(\ell-2)}-\frac{\ell-2}{\ell-3})!+1}$$

This function meets all your given conditions.

Title: Existence of a particular function


问:
Why the ith term can be represented in a quadratic form as any difference of $2$ adjacent elements of the sequence is a form of linear?
$0<a_i$

($1\le i\le n$)

$$\alpha,\beta:=\text{constants for any } i$$

$$a_{i+1}-a_i=\alpha \cdot i+\beta$$

The textbook claims that $a_i$ can be represented in a quadratic form ($bi^{2}+ci^{1}+d$).

Why this claim is adequate?

I thought below.

$$a_{i+1}=\alpha\cdot i+\beta+a_i$$

$$a_2=\alpha\cdot(1)+\beta+a_1 \text{ (linear form)}$$

As 2nd term is a form of linear so the 3rd one will also be a linear one, and same for any $i$

Hence the any term takes a linear form.However actually the textbook states the ith term takes 2nd degree polynomial.

答:
Note that

$$a_4=(a_4-a_3)+(a_3-a_2)+(a_2-a_1)+a_1\,,$$

because the $a_2$ and $a_3$ terms cancel. We can do this for any $a_n$: $$a_n=\sum_{i=1}^{n-1}(a_{i+1}-a_i)+a_1\,.$$

Now you need only know how to calculate the sum of consecutive integers:

$$\begin{align*}
a_n&=\sum_{i=1}^{n-1}(a_{i+1}-a_i)+a_1\\
&=\sum_{i=1}^{n-1}(\alpha i+\beta)-a_1\\
&=\alpha\sum_{i=1}^{n-1}i+\beta(n-1)-a_1\\
&=\alpha \cdot\frac{n(n-1)}2+\beta(n-1)-a_1\\
&=\frac{\alpha}2n^2+\left(\beta-\frac{\alpha}2\right)n-(\beta+a_1)\,.
\end{align*}$$


问:
$x,y\in \mathbb Z^+$ Show that $P(x)=P(y) \implies x=y$

Let $x,y\in \mathbb Z^+$, we define $P(a)$ as the product of all the divisors of $a$ including $1$ And $a$, Show that $$P(x)=P(y)\implies x=y.$$

Im gonna show my Attempt, which is based on cases.

Case 1: if  $x,y$ Are primes.
$$x,y\in \mathbb P\implies \cases{P(x)=x\\P(y)=y}\implies (P(x)=P(y) \iff x=y)$$
Where $\mathbb P$ Is the set of prime numbers.

Case 2: if $x,y$ are some powers of prime.
$$\cases{x=p^n\\ y=q^m } \implies \cases{P(x)=1\cdot p\cdot p^2...p^n=p^{\frac{n(n+1)}{2}}\\P(y)= 1\cdot q\cdot q^2...q^m= q^{\frac{m(m+1)}{2}}} $$ So to prove the statement We set $$p^{\frac{n(n+1)}{2}} =q^{\frac{m(m+1)}{2}}$$
The next step is to show that this statement implies $x=y$, I have no idea how to do that. (By the way the next case is to consider $x,y$ As a product of primes.)

答:
Let $\mathcal E(x)$ be the exponents of the prime factorization of $x$, for example $\mathcal E(2^7 \times 3^0 \times 5^3) = [7, 0, 3]$.

Let $X = \mathcal E(n)$ and $Y = \mathcal E(P(n))$.  As noted in Calvin Lin's Answer, $P(n) = n^{\tau(n)/2}$, so $Y_k = X_k \times \tau(n)/2$.  So given $Y$, to determine $\tau$:

$$\tau = (X_0 + 1)(X_1 + 1)\dots(X_{z - 1} + 1) = (2Y_0/\tau + 1)(2Y_1/\tau + 1)\dots(2Y_{z-1}/\tau + 1)$$
$$\tau^{z+1} = (\tau + 2Y_0)(\tau + 2Y_1)(\tau + 2Y_2) \dots (\tau + 2Y_z)$$

By Descartes' Sign Rule, the above only has 1 positive solution for $\tau$ if it has any, so $P$ is injective.


问:
How to prove Riemann sum wrt. any point will give same result (left, right, middle, etc.)
If $f(x)$ is a differentiable function. If I want to find the Riemann sum between points $a$ and $b$.

I'll divide the part of function from $a$ to $b$ into $n$ pieces. So the width of each piece will be $\Delta x=\frac{b-a}{n}$. Let the points that demarcate each part are $x_0,x_1,...,x_n$ such that $x_0=a$ and $x_n=b$, and $x_i=x_{i-1} + \Delta x$. The area of each (rectangular) piece will be $\Delta x f(x_i^*)$, such that $x_i^*$ is a point between $x_i$ and $x_{i+1}$.

I want to prove that $x_i^*$ can be any point between $x_i$ and $x_{i+1}$, and the Riemann sum (i.e. $\sum_{i=0}^n f(x_i^*)\Delta x$) as $n \to \infty$ will give the same result.

Will this kind of proof work?

$x_i\le x_i^*\le x_{i+1}$

$\Rightarrow x_i\le x_i^*\le x_i+\Delta x$

as $n\to \infty$, $\Delta x\to 0$

$\Rightarrow \lim_{\Delta x\to 0} x_i\le \lim_{\Delta x\to 0} x_i^*\le \lim_{\Delta x\to 0} x_i+\Delta x$

$\therefore x_i \le x_i^*\le x_i$

Hence $x_i^*$ can be any point between $x_i$ and $x_{i+1}$

答:
Integrals are defined as the limit of the Riemann sum for any partition with distinguished points, if such a limit exists. So, the question you are asking is actually defined as being true in order for an integral to exist. You can imagine that, if the limit depended on the points $x_i^*$ that you chose, you'd get a different limit for different points and the integral wouldn't be unique.

Perhaps, though, a better way to think about this, with respect to the question you are asking is by looking at
\begin{align*}
s&:=\sum_{i=1}^n\inf_{x\in[x_i+\delta x]}f(x)\Delta x\\
S&:=\sum_{i=1}^n\sup_{x\in[x_i+\delta x]}f(x)\Delta x
\end{align*}
$\int_a^bf(x)dx$ exists only when $\lim s=\lim S$. Since $\inf_{x\in[x_i+\Delta x]f(x)}\leq f(x_i^*)\leq \sup_{x\in[x_i+\Delta x]}f(x)$, it follows that it will work for any values of $x_i^*$ that you choose.

btw, if you're unfamiliar with the notation, you can think of $\inf$ as minimum and $\sup$ as maximum, though it's a little more technial than that.


问:
How do you obtain Variance of the "method of moment estimator" for the beta distribution using delta method?
I'm trying to solve a question where I have to find $V(\hat{\alpha})$ using the delta-method where $V$ is notation for variance and $\hat{\alpha}$ is the method of moment estimator for a beta distribution. To be more specific, I write down the question below:

Suppose $X_1, ..., X_n$ be a random sample from the following distribution,

$$f(x)=\alpha x^{\alpha -1}, 0<x<1\text{ and }\alpha>0.$$

Let $\hat{\alpha}$ be the method of moment estimator. Obtain the $V(\hat{\alpha})$ using the delta-method.

I know that this function is just Beta distribution where $X\sim\mathrm{Beta}(\alpha,1)$. So the method of moment $E(X^k)=\frac{\Gamma(\alpha +1)\Gamma(\alpha +k)}{\Gamma(\alpha)\Gamma(\alpha +k+1)}$, where $\Gamma(\cdot)$ is a Gamma function.

But how do I get the variance of the method of moment using delta method? I am not sure how the usage of the delta method will get me the variance. Thank you.

答:
I believe the delta method only gives you an approximation of the variance, which becomes more accurate as $n$ goes to infinity (because it is based on a Taylor approximation).

The variance I get using the delta method is $\frac{\alpha(\alpha+1)^2}{n(\alpha+2)}.$ Here is how.

The method of moment estimator $\alpha$ solves the equation
$$ \frac{\Gamma(\hat\alpha+1)\Gamma(\hat\alpha+1)}{\Gamma(\hat\alpha)\Gamma(\hat\alpha+2)} = \frac{\hat\alpha}{\hat\alpha+1} = \frac{1}{n}\sum_{i=1}^n X_i=: \hat\mu$$
where I am using the fact that $\Gamma(x+1)=x\Gamma(x)$.
By inverting this we obtain $\hat\alpha = \frac{1}{1-\hat\mu} - 1$. Therefore we are interested in $\mathrm{Var}(\frac{1}{1-\hat\mu})$. We can use the following Taylor approximation, valid if $\hat\mu\approx\mu = \frac{\alpha}{\alpha+1}$ the true mean (which should hold by LLN):
$$
\frac{1}{1-\hat\mu} \approx \frac{1}{1-\mu} + \frac{1}{(1-\mu)^2}(\hat\mu - \mu).
$$

Therefore,
$$
\mathrm{Var}(\hat\alpha) \approx \mathrm{Var}\left(\frac{1}{(1-\mu)^2}(\hat\mu-\mu)\right)
= \frac{1}{(1-\mu)^4}\mathrm{Var}(\hat\mu).
$$

Using the fact that $\mathrm{Var}(\hat \mu)$ is $\frac{1}{n}$ times the variance of a single $\mathrm{Beta}(\alpha, 1)$ variable, which is $\frac{\alpha}{(\alpha+1)^2(\alpha+2)}$, and plugging in $1-\mu = \frac{1}{1+\alpha}$, we obtain the final answer for the delta method approximated variance

$$
\mathrm{Var}(\hat\mu) \approx \frac{\alpha(1+\alpha)^2}{n(\alpha+2)}
$$


问:
Covariance and Random Variables
Suppose $X$ and $Y$ are independent random variables. The mgf for $X$ is $M_{X}(t) = (1-t)^{-1}, \ t<1$. The mgf for $Y$ is $M_{Y}(t) = (1-2t)^{-1}, \ t< 0.5$. Two other random variables $U$ and $V$ are defined by $U = \frac{1}{2}(X+Y)$ and $V = \frac{1}{2}(X-Y)$. Calculate $\text{Cov}(U,V)$.

So $\text{Cov}(U,V) = E(UV)-E(U)E(V)$. This is the same thing as $\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y))$. Expanding out, we get

$\text{Cov}(\frac{1}{2}(X+Y), \frac{1}{2}(X-Y)) = \frac{1}{4} \text{Var}(X) -\frac{1}{4} \text{Cov}(X,Y)+\frac{1}{4} \text{Cov}(Y,X)-\frac{1}{4} \text{Var}(Y)$ which equals $\frac{1}{4}(\text{Var}(X)-\text{Var}(Y))$.

Is there an easier way to obtain this without going through all this work? Could one use the fact $U+V = X$ and $U-V = Y$?

答:
The method you have is pretty short already. I'd have pulled out the $\frac{1}{4}$ at the start, but otherwise the straightforward way is a lot faster than trying to think up something trickier.  Sometimes straightforward is best.

The moment generating functions indicate that $X$ is exponential with parameter $1$ so its mean is $1$ and its variance is also $1$. $Y$ is exponential with parameter $2$, so its mean is $2$ and variance is $4$.  Those are both special cases of gamma distributed, so I was hoping the theorem about the sum of gammas being gamma would help, but the wrong parameter matches.

We can use $M_{aX}(t) = M_{X}(at)$ to find

$M_{\frac{1}{2}X} = \frac{1}{1-\frac{1}{2}t}$,
$M_{\frac{1}{2}Y} = \frac{1}{1-t}$,
$M_{\frac{-1}{2}Y} = \frac{1}{1+t}$,

giving $M_{U} = \frac{1}{(1-\frac{1}{2}t)(1-t)}$ and $M_{V} = \frac{1}{(1-\frac{1}{2})(1+t)}$

but $U$ and $V$ are not independent so that approach doesn't seem to be fruitful either.

Really I think the answer to your question is that no, there isn't any faster or easier or more elegant way to do it than what you did or what was given in the comment.

ContentLicense: CC BY-SA 2.5

Title: Covariance and Random Variables


问:
discrete math monotonic functions question
i had this question to answer:

how many monotonic function $f:[1...k]\to[1...n]$, $(i>j \implies f(i)\geqslant f(j))$ are there?

my answer is: $\sum_{r_1=1}^n\sum_{r_2=r_1}^n\cdots\sum_{r_k=r_{k-1}}^n1$

is this the right answer?

Thanks in advance,

Yaron

答:
After correction, your iterated summation formula describes the correct number, but it is not really an answer. You are saying there are as many monotonic functions $f[1,k]\to[1,n]$ as there are sequences $r_1,\ldots,r_k$ with $1\leq r_1\leq r_2\leq\cdots\leq r_k\leq n$, which is true because taking $r_i=f(i)$ for $i\in[1,k]$ establishes a bijection.

If you want to know what that number is, you many want to establish a bijection with something that you know how to count in terms of some simple expression in $k$ and $n$ (not an iterated summation). Here is an idea: if you define $s_i=r_i+(i-1)$ for $i\in[1,k]$, then you get a sequence $1\leq s_1< s_2<\cdots< s_k\leq n+k-1$ and the correspondence $(r_1,\ldots,r_k)\mapsto(s_1,\ldots,s_k)$ is a bijection since setting $r_i=s_i-(i-1)$ for $i\in[1,k]$ inverts it. You many know a formula for the number of such sequences $s_1,\ldots,s_k$.

Hint: $\{s_1,\ldots,s_k\}$ is a combination of $k$ elements chosen in the set $[1,n+k-1]$, and any combination of $k$ such elements occurs exactly once in this way.


问:
Prob. 10, Chap. 2, in Royden's REAL ANALYSIS: If, for some $\alpha>0$, $|a-b|\geq\alpha$ for all $a\in A$ and $b\in B$, then ...
Here is Prob. 10, Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Let $A$ and $B$ be bounded sets for which there is an $\alpha > 0$ such that $\lvert a-b \rvert \geq \alpha$ for all $a \in A$ and $b \in B$. Prove that $m^*(A \cup B) = m^*(A) + m^*(B)$.

Of course, we have
$$
m^*(A \cup B) \leq m^*(A) + m^*(B), \tag{1}
$$
by virtue of Proposition 3, Chap. 2, in Royden.

How to prove the equality? In particular, how to prove the inequality $m^*(A) + m^*(B) \leq m^*(A \cup B)$?

For each $a \in A$, we have $B \cap (a-\alpha, a+\alpha ) = \emptyset$, and for each $b \in B$, we have
$A \cap (b-\alpha, b+\alpha) = \emptyset$. Thus of course we also have $A \cap B = \emptyset$.

How to make use of these facts in arriving at our desired equality?

PS:

Based on the comments below, here is my attempt at proving the desired inequality:

If $m^*(A \cup B) = \infty$, then we are done. Su let us assume that $m^*(A \cup B) < \infty$.

If either of $A$ and $B$ is countable, say set $A$, then we have $m^*(A) = 0$, and thus we have
$$
m^*(A) + m^*(B) = m^*(B) \leq m^*(A \cup B),
$$
as required. So let us assume that both the sets $A$ and $B$ are uncountable.

Then by definition $m^*(A \cup B)$ is the infimum of the set of all the sums $\sum_{k=1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty, bounded open intervals covering $A \cup B$.

Let $\left\{ I_k \right\}_{k=1}^\infty$ be any countable collection of non-empty, bounded open intervals covering $A \cup B$. Suppose  $I_k := \left( a_k, b_k \right)$ is any interval in this covering. Then we can sub-divide $I_k$ into $n_k$ disjoint sub-intervals $I_{k, i} := \left( a_{k, i}, b_{k, i} \right)$ such that, for each $i = 1, \ldots, n_k$, we have
(1) $b_{k, i} - a_{k, i} < \alpha$,
(2) no endpoints $a_{k, i}$ and $b_{k, i}$ are in $A \cup B$,  and
(3) $$
a_k = a_{k, 1} < b_{k, 1} = a_{k, 2} < b_{k, 2} = \cdots = a_{k, n_k} < b_{k, n_k} = b_k.
$$
Then of course we have the equality
\begin{align}
\sum_{i = 1}^{n_k} l \left( I_{k, i} \right) &= \sum_{i=1}^{n_k} \left( b_{k, i} - a_{k, i} \right) \\
&= b_{k, n} - a_{k, 1} \\
&= b_k - a_k \\
&=  l \left( I_k \right),
\end{align}
and thus
$$
\sum_{k=1}^\infty l \left( I_k \right) = \sum_{k=1}^\infty \sum_{i = 1}^{n_k} l \left( I_{k, i} \right). \tag{2}
$$

Note that no open interval of length less than $\alpha$ can contain a point of set $A$ and a point of set $B$ at the same time, because each point of $A$ is at a distance of at least $\alpha$ from each point of $B$.

Thus we can partition the collection $\left\{ I_k \right\}_{k=1}^\infty = \left\{ \bigcup_{i=1}^{n_k}  I_{k, i} \right\}_{k=1}^\infty$ into two disjoint countable sub-collections  $\left\{ I_k^\prime \right\}_{k=1}^\infty$ and $\left\{ I_k^{\prime\prime} \right\}_{k=1}^\infty$ covering $A$ and $B$, respectively.

We by virtue of (2) above therefore obtain
$$
\begin{align}
m^*(A) + m^*(B) &\leq \sum_{k=1}^\infty l \left( I_k^\prime \right) + \sum_{k=1}^\infty l \left( I_k^{\prime\prime} \right) \\
&= \sum_{k = 1}^\infty l \left( I_k \right).
\end{align}
$$

But $\left\{ I_k \right\}_{k=1}^\infty$ was any covering of $A \cup B$ by a countable collection of non-empty, bounded open intervals. Thus we have shown that the number $m^*(A) + m^*(B)$ is a lower bound for the set of all the sums $\sum_{k=1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty, bounded open intervals covering $A \cup B$.  Thus it follows that
$$
m^*(A) + m^*(B) \leq m^*(A \cup B),
$$
as required.

Is my proof correct in each and every detail? If so, is my presentation clear enough? Or, are there any issues?

Last but not least, is the assumption of boundedness for the sets $A$ and $B$  essential for our conclusion to hold?

答:
Actually, $m^*()$ in this chapter is not only an outer measure but also a Lebesgue outer measure. And more generally, if this is a metric outer measure, it satisfies the property of totally additive of disjoint sets.

By the definition in this book,
$m^*(A)$
is
$$m^*(A)=\inf \left\{\sum_{k=1}^∞ \mathscr l(I_k) \, | \, \mathsf A \subseteq \bigcup_{k=1}^∞ I_k \right\}$$

And due to the definition of the $\inf$, for any arbitrary $\epsilon > 0$, we could find a covering of $A \cup B$, say $\cup_{k=1}^∞ I_k^{A \cup B} $,  which satisfies
$$
\sum_{k=1}^∞ \mathscr l(I_k^{A \cup B})\le m^*(A \cup B) + \epsilon.
$$
And due to this is Lebesgue measure, we could let $\mathscr l(I_k^{A \cup B})\lt\alpha/2$ for any $k$. So none of $I_k^{A \cup B}$ could intersect with $A$ and $B$ at the same time.

Denote the covering of A, $\cup_{\{ I_k^{A \cup B} \cap A \neq \emptyset\}} I_k^{A \cup B}$ as $C_A$, and the covering of B, $\cup_{\{ I_k^{A \cup B} \cap B \neq \emptyset\}} I_k^{A \cup B}$ as $C_B$ .

Because of $\alpha \neq 0$ ,$C_A$ and $C_B$ are disjoint, then we have $$\sum_{k=1}^∞ \mathscr l(A\cap I_k^{A \cup B})  +\sum_{k=1}^∞ \mathscr l(B\cap I_k^{A \cup B}) \le  m^*(A \cup B) + \epsilon.$$

By the definition of $m^*(A)$ and $m^*(B)$, we have
$$
m^*(A) + m^*(B) \le \sum_{k=1}^∞ \mathscr l(A\cap I_k^{A \cup B})  +\sum_{k=1}^∞ \mathscr l(B\cap I_k^{A \cup B}) \le  m^*(A \cup B) + \epsilon
$$

We could let $\epsilon$ limit to $0$, then we have the other direction of the equation.


问:
Arithmetic on infinite formal prime factorizations
By the fundamental theorem of arithmetic, we can identify a natural number $x$ with the sequence $(a_n)$ of exponents in its prime factorization $x=\prod_np_n^{a_n}$, where $p_n$ is the $n$th prime. A sequence obtained in this way will of course have only finitely many nonzero elements.

However, multiplication/$\gcd$/$\operatorname{lcm}$ operations on natural numbers correspond respectively to pointwise addition/$\min$/$\max$ of these sequences, and these operations make sense on arbitrary sequences. So I wonder: How much arithmetic can we do with these sequences, thinking of them as formal infinite products of prime powers?

Specifically:

Is there a semiring structure (without $0$) on $R=(\Bbb Z_{\ge 0})^\Bbb N$ where $\times$ is pointwise addition and $+$ coincides with normal addition on finite natural numbers (represented as described above)?

If so, I'm also interested in whether we can get other familiar properties.

Does $\gcd(a,a+b)=\gcd(a,b)$ hold (where $\gcd$ = pointwise $\min$)?

Is $<$, defined by $a<b\iff\exists c.a+c=b$, a total order?

If we include $0$ and negatives (e.g. by considering equivalence classes of formal differences), does Bézout's lemma hold?

Can we in fact get a (non-standard) model of arithmetic? Are these the hyperintegers or something?

Edit: Commenters have pointed out that the answer to the last bullet is no, since this structure lacks large primes and includes elements with no largest prime divisor (and both of these properties are inconsistent with the axioms of arithmetic). The latter issue prevents this structure from even being a sub-semiring of a model of arithmetic.

答:
The comments supplied some intuition about why this construction doesn't come close to modeling arithmetic or even supporting addition in a "reasonable" way.

For one thing, we always expect $x$ and $x+1$ to have no common prime divisor, but this forces $\prod p_n+1=1$. This is undesirable since $x+a$ should be infinite whenever $x$ is infinite, and $a+b=a$ should never be true if we're trying to represent positive values.

This essentially answers my question, even though I think a subtraction axiom is needed to turn the argument into an actual contradiction.

I briefly considered avoiding this problem by restricting to sequences that have infinitely many $0$s, but this doesn't work since we still get e.g. $\prod p_{2n}+\prod p_{2n-1}=1$. We could try harder by requiring the density of nonzero entries to be small, but I doubt this will work. In fact, "thinning out" the infinite numbers seems like the opposite of what we need, since the infinite products are already all "highly composite" and therefore too sparse for addition to work. As the comments suggest, if the goal is to model arithmetic, we should add nonstandard primes and consider only finite products (edit: bounded or "internal" products) of them.


问:
Type of singularities of $\frac{z}{e^z-1}$
I don't really understand how one can find the type of singularities for a given function. Say if
$$f(z) = \frac{z}{e^z-1}$$
then I know that the singularities are at $z = 2n\pi i$

However, how do I find the type? If I try to write out Laurent series for this by using
$$e^z = 1 + z + z^2/2! + z^3/3! + \cdots $$, then I got a summation in the denominator, which I don't know how to rearrange as a Laurent series.

Can anyone please give me a hint?

答:
Some of key points to decide the type of the singularity

I am not claiming that these points are complete. But certainly these point may help you to decide about the types of singularity of a function at a point.

1:  A singular point $z_0 $is called an isolated singular point of an analytic function $f(z)$ if there exists a deleted $\epsilon$-spherical neighborhood of $z_0$ that contains no singularity. If no such neighborhood can be found, $z_0 $is called a non-isolated singular point. Thus an isolated singular point is a singular point that stands completely by itself, embedded in regular points.

2. An isolated singular point $z_0$ such that $f$ can be defined, or redefined, at $z_0$ in such a way as to be make it analytic at $z_0$ is called removable singularity. A singular point $z_0$ is removable if  $\lim _{z\to z_{0}} f(z)$ exists. For example The singular point $z = 0$ is a removable singularity of $f(z) = \frac{\sin z}{z}$
 since $\lim _{z\to z_{0}} \frac{\sin z}{z} = 1$. Dominic's answer says about when a singular point $z_0$ is a pole of order $k$.

3. A singular point that is not a pole or removable singularity is called an essential singular point.

4 Isolated essential singularity

Limit points of the sequence of a zeros of the nonzero analytic function $f(z)$ is an isolated singularity. Since it is not a pole it is an isolated essential singularity. For example consider the function $\sin \frac{1}{z}$ it has zeros at $z = \frac{1}{k\pi}$ . Limit points of the zeros is the zeros is the point $z =0$ hence, $z = 0$ is an isolated essential singularity of $f(z)$.

5 . Non -isolated essential singularity

Let $z_1, z_2,\ldots, z_n, \ldots$ be the set of points having limit point $z_0$ in a region $D$. Let $f$ be analytic in $D$ except at the poles  $z_1, z_2,\ldots, z_n, \ldots$, then $z_0$  is also singularity of $f(z)$. The reason is that $f$ is unbounded in the neighborhood of $z_0$. But  $z_0$ is not isolated because it is limit points of the poles. Hence $f(z)$ has a non-isolated singularity at $z_0$. For example consider the function $\frac{1}{\cos(1/z)}$ this function has poles at $k = \frac{2}{(2k+1)\pi}$ the point $z =0$
is the limit point of these poles hence given function has non -isolated essential singularity at $z = 0$.


问:
Odd Periodic Extension to Obtain Fourier Series
I'm asked to obtain the Fourier series for $f(x)=x|x|$ on the interval $[-1,1]$. I'm thinking it's easier to do an odd periodic extension of $x^2$ on $[0,1]$. My book says this is given by $$f(x)=\sum_{n=1}^{\infty}b_{n}\sin{\frac{n\pi}{L}x},$$ with $$b_{n}=\frac{2}{L}\int_{0}^{L}f(x)\sin{\frac{n\pi}{L}x}\ dx.$$ Taking $f(x)=x^2$, I computed \begin{align*}b_{n}&=2\int_0^1x^2\sin{n\pi x}\\&=\frac{2(2-\pi^2n^2)\cos{(\pi n)-4}}{\pi^3n^3}\\&=\frac{2(2-\pi^2n^2)(-1)^n-4}{\pi^3n^3}\end{align*} and so $$\hat{f}(x)=\sum_{n=1}^{\infty}\frac{2(2-\pi^2n^2)(-1)^n-4}{\pi^3n^3}\cos{n\pi x}.$$ But when I plot this up to n=1000 I get nothing  close to $x|x|$. What am I doing wrong?

答:
It should be summing over sine terms rather than cosines:

$$\hat{f}(x)=\sum_{n=1}^{\infty}\frac{2(2-\pi^2n^2)(-1)^n-4}{\pi^3n^3}\color{red}\sin{n\pi x}.$$


问:
combinations question help
Let $N > 10$ be an integer. We have $N$ one-dollar bills that we split between three people. No person can be given less than two one-dollar bills, and everyone gets a whole number of bills (no cents). In how many ways can we do this?

This is how I tried solving the problem. Before splitting, if we give each of the three people two (2) one-dollar bills, then we'll be starting with $N-6$ total bills. Since order doesn't matter, there could be $\displaystyle{{N-6}\choose3}$ ways to do this. Is this right or am I missing something? Any help would be appreciated.

答:
This problem can be viewed as How many integer solutions which each lager than $2$ of the following integer equation
$$ N_1 + N_2 + N_3 = N $$
where the $N_i$ is the number of one-dollar bill that the i-th people get, so what we want is find integer solutions with $N_i \ge 2$ , we can simplify the problem further in the following:
$$ (N_1-2) + (N_2-2) + (N_3-2) = N-6 $$
rewrite as following with denoting $x_i = N_i - 2$
$$ x_1 + x_2 + x_3 = N-6 $$ where $x_i$ is nonegative integer.

what we want to solve is a classic problem , The Number Of Non-negative Integer Solutions Of Equations

Using the proposition $2$ of the above link ,There are $n+r-1\choose r-1$ distinct non-negative integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$ x_1 + x_2 + ... + x_r = n$$

Apply it to this question , we get the answer $N-6 + 3 -1 \choose 3-1 $ and that
is $N-4 \choose 2 $

Title: combinations question help


问:
If $a-\sqrt{2021a} = 1$, find $a-45\sqrt{a}$.
If a is a positive real number such that $$a-\sqrt{2021a} = 1$$ evaluate $$a-45\sqrt{a}$$
I've factored the $2021$ into $43\cdot 47$ and then $(45+2)(45-2)$ so $$a -\sqrt{(45^2-2^2)a} = 1$$ but I can't figure out what to do next.

答:
VIVID's solution is complete. Here is a solution in other words. By rearranging and squaring the equation we have:

$$(a-1)^2 = 2021a = (45+2)(45-2)a = (45^2 - 2^2)a$$
Therefore:
$$45^2a = (a-1)^2 + 4a = (a+1)^2 $$
And recalling that $a$ is positive (and so $a>-1$) we have:
$$45\sqrt a = a+1$$
Therefore:
$$a - 45\sqrt a = -1 $$


问:
Prove that a subspace contains the span
Let vectors $v, w \in \mathbb{F}^n$. If $U$ is a subspace in $\mathbb{F}^n$ and contains $v, w$, then $U$ contains $\operatorname{Span} \{v,w\}.$

--

My attempt: if $U$ contains vectors $v, w$. Then $v+w \in U$ and $av \in U$, $bw \in U$ for some $a, b \in \mathbb{F}$. Since the span is $au+bw$, then $U$ contains the span. I feel like I am missing a step here, but dont know what it is. So, can someone please point it out?

答:
Implement Clayton's suggestion, and also replace "Since the span is ..." with:

"Since the sum $av + bw$ is a fully general element of the span, then applying the previous sentence shows that $U$ contains the span."

Title: Prove that a subspace contains the span


问:
Given a list of 10 random numbers, can you make two sets of any length that add up to each other?
So the question is pretty simple if we have a set of 10 random numbers, how do we prove that we can make 2 subsets of any length such that the sum of numbers in each subset adds up to each other? The numbers in each subset must also not be reused. I am just looking for ways to approach this problem since it has left me confused. I have been trying to use proof by contradiction, but that hasn't worked out.

EDIT: Random numbers from 1 to 100 inclusive.

答: As was said in the comments we have $10^{10}=1024$ subsets and the sums are all in the range $[0,10\times100]$ so by the pidgeonhole principle at least one quantity must be the sum of two distinct subsets.


问:
How do I model the Expected Value in a trial and error until find some value?
The problem I was thinking is: Supose you have m pieces of cake and  n of them are spoiled, I start picking one piece, if it's normal I take it out and continue doing this until I find one spoiled.

What is the expected value of cakes I catch before the spoiled? I thought it could be something geometric but the probabilities change. How can I attack this problem?

答:
EDIT: This answer assumes sampling with replacement. Sampling without replacement is a bit more complicated, but this should lead you in the right direction.

This looks to be a homework question, so I will go through the steps a little more carefully and allow you to finish it.

The first thing we need to ask is "What probability distribution is this?" In our situation, there are 2 values: Spoiled and not spoiled. Many can work in this situation, but since we know that this is like a coin flip with many coins, we can say it's likely a binomial distribution.

We then ask "What is an expected value?" The short answer is that it is the mean of the collected data, so we will treat it like a simple mean. But this question isn't actually about the expected value, we are asking "how many successes until failure", where failure is finding a bad piece. So instead, we ask, what is the probability of 0 successes? Then of 1 success? Then 2? Then 3? Etc.

Thus, we need a formula for the probability of each. For the binomial, this is
$$P = {w\choose k} p^k (1-p)^{n=k}$$

To make it easier though, we can instead ask, what is the probability of each result? Using S for success (not rotten) and F for failure (rotten) we see

F - $ n/m$

SF - $(1-n/m)(n/m)$

SSF - $(1-n/m)^2(n/m)$

SSSF - $(1-n/m)^3(n/m)$

And so on. We thus multiply by each value of input and get
$N = n/m + 2 * (1-n/m)(n/m) + 3 * (1-n/m)^2(n/m) + 4 * (1-n/m)^3(n/m) + ...$

This should slowly bring you to a right answer if you do this a total of (n-m) times, so you find the probability for each result. This actually tells you the total number of pulls including failures, so subtract 1 from the exponents for just the successes.


答:
Not an actual answer, but too long for a comment:

When a sequence (e.g., $0, 2, 4, 6,\dots$) is described by the recurrence relation $$a_k=a_{k-1}+d\quad(k\geq1),$$ the terms are increasing at a constant “velocity” $d$, so the sequence is called an arithmetic sequence with a common difference $d,$ and its closed form turns out to be $$a_n=a+(n-1)d\quad(n\geq0).$$

When a sequence (e.g. $0, 3, 8, 15, 24,\dots$) is described by the recurrence relation $$a_k=a_{k-1}+dk+c\quad(k\geq1),$$ the terms are increasing at a constant “acceleration” $d$, and its closed form turns out to be $$a_n=\frac d2 n^2+Xn+Y\quad(n\geq0),$$ where X and Y are constants depending on $a_0, c$ and $d$ (refer to @BrianMScott's answer), so the sequence is called a quadratic sequence with a second common difference $d.$

(Here, there's a constant difference between the difference between successive terms, i.e., the difference between successive terms is changing at a constant rate.)

Title: Why the ith term can be represented in a quadratic form as any difference of $2$ adjacent elements of the sequence is a form of linear?

Body:
$0<a_i$

($1\le i\le n$)

$$\alpha,\beta:=\text{constants for any } i$$

$$a_{i+1}-a_i=\alpha \cdot i+\beta$$

The textbook claims that $a_i$ can be represented in a quadratic form ($bi^{2}+ci^{1}+d$).

Why this claim is adequate?

I thought below.

$$a_{i+1}=\alpha\cdot i+\beta+a_i$$

$$a_2=\alpha\cdot(1)+\beta+a_1 \text{ (linear form)}$$

As 2nd term is a form of linear so the 3rd one will also be a linear one, and same for any $i$

Hence the any term takes a linear form.However actually the textbook states the ith term takes 2nd degree polynomial.


问:
How many times can quadric kiss cosine at given point?
Let a quadric $ax^2+2bxy+cy^2+dx+ey+f=0$ touches the plot of $y=\cos(x)$ at the point $(0,1)$ with multiplicity $n$. What is the maximum possible value of $n$? Recall that a joint point $P$ of parametrized curves $r_1(t)$ and $r_2(t)$ is said to be a tangent point of multiplicity $n$ if the first $n$ derivatives of $r_1(t)$ and  $r_2(t)$ at the point $P$ are equal.

PS. That was asked as one of the problems of a student mathematical olimpiad in Russia. The author gives general directions and the answer $n=5$.

答:
By symmetry the quadric we are looking for has an apex at $(0,1)$, whereby a parabola $y=1- p x^2$ obviously won't do. Therefore the equation of this quadric will be of the form
$a x^2+ c y^2 +e y + f=0$ with $c\ne0$. Solving for $y$ and taking care of $y(0)=1$ we obtain an expression of the form
$$y(x)=1\pm\bigl(\sqrt{q^2 + r x^2} -q\bigr)$$
with $q\geq0$ and $r$ to be determined such that $$y^{(k)}(0)=\cos^{(k)}(0)\qquad(0\leq k\leq n)\tag{1}$$ and $n$ as large as possible. Doing the computation one finds that the hyperbolic arc
$$y(x)=4-\sqrt{9+3x^2}$$
and no other quadric satisfies $(1)$ with $n=5$. Since  for this arc one has
$$y(x)=1-{x^2\over2}+{x^4\over24}-{x^6\over144}+\ {\rm higher\ terms}$$
it follows that $(1)$ cannot be satisfied with $n=6$. Therefore $n=5$ is indeed the maximal order of kissing between a quadric and the cosine curve at $(0,1)$.</p>


问:
Proving pullback of projection to quotient manifold is always injective on de Rham cohomology
I found the following result on Wikipedia:

We can also find explicit generators for the de Rham cohomology of the torus directly using differential forms. Given a quotient manifold ${\displaystyle \pi :X\to X/G}$ and a differential form ${\displaystyle \omega \in \Omega ^{k}(X)}$ we can say that $\omega $ is ${\displaystyle G}$-invariant if given any diffeomorphism induced by ${\displaystyle G}$, ${\displaystyle \cdot g:X\to X}$ we have ${\displaystyle (\cdot g)^{*}(\omega )=\omega }$. In particular, the pullback of any form on ${\displaystyle X/G}$ is ${\displaystyle G}$-invariant. Also, the pullback is an injective morphism.

which states that the pullback $\pi^*: X/G \to X$ is injective. I was able to prove this but only in the case that the group $G$ is finite (it's somewhat of an "averaging" argument). But I can't see how to prove it in this general case (the "averaging" argument breaks down since a sum of infinite terms is not defined). I'd really appreciate some help on this!

答:
When $G$ is compact, take a basis of left-invariant $1$-forms, wedge them to get a left-invariant volume form $dV(g)$, and given any $k$-form $\omega$, the averaged form
$$\tilde\omega = \frac 1{\text{vol}(G)}\int_G L_g^*\omega\,dV(g)$$
is $G$-invariant. What's more, you can check that $\omega$ is closed/exact if and only if $\tilde\omega$ is closed/exact. So there's a unique invariant form in every cohomology class of $M$.

(Without compactness, the integral won't make sense, and nor will the volume.)


问:
How do I show $f$ is one-one on $\mathbb{R}^2$?
Let $f$ be a differentiable map (but not necessarily continuous differentiable) with component functions $f_1$ and $f_2$, that is $f(x_1,x_2)=(f_1(x_1,x_2),f_2(x_1,x_2))$ for all $(x_1,x_2) \in \mathbb{R}^2$. For all $(x_1,x_2) \in \mathbb{R}^2$, one has

$|\frac{\partial f_1}{\partial x_1}-2|+ |\frac{\partial f_1}{\partial x_2}|+|\frac{\partial f_2}{\partial x_1}|+|\frac{\partial f_2}{\partial x_2}-2| \leq \frac{1}{2}$.

Prove that $f$ is one-one on $\mathbb{R}^2$?

I cannot use inverse function theorem to prove this question. Can anyone suggest some direction to solve this question?

答:
Assume not, then there are $x_1, x_2, R, \theta$ so that $f(x_1, x_2)=f(x_1+R\cos\theta, x_2+R\sin\theta)$. Let $F: {\mathbb R}\to {\mathbb R}$ be defined by
$$
F(t)=\cos\theta\cdot f_1(x_1+t\cos\theta, x_2+t\sin\theta)+\sin\theta\cdot f_2(x_1+t\cos\theta, x_2+t\sin\theta).
$$
So $F(0)=F(R)$. Compute
$$
F'(t)=\cos^2\theta \cdot \frac{\partial f_1}{\partial x_1}+
\sin^2\theta \cdot \frac{\partial f_2}{\partial x_2}+
\sin\theta\cos\theta \cdot \frac{\partial f_1}{\partial x_2}+
\sin\theta\cos\theta \cdot \frac{\partial f_2}{\partial x_1},
$$
then
$$
F'(t)\geq \cos^2\theta \cdot \frac 32+
\sin^2\theta \cdot \frac 32-
|\sin\theta\cos\theta |\cdot \frac 12-
|\sin\theta\cos\theta |\cdot \frac 12\geq \frac 32-\frac 12-\frac 12=\frac 12.
$$
So $F$ is strictly increasing, contradicts the fact $F(0)=F(R)$.


问:
Formula to project a vector onto a plane
I have a reference plane formed by $3$ points in $\mathbb{R}^3$ – $A, B$ and $C$. I have a $4$th point, $D$. I would like to project the vector $\vec{BD}$ onto the reference plane as well as project vector $\vec{BD}$ onto the plane orthogonal to the reference plane at vector $\vec{AB}$. Ultimately, I need the angle between $\vec{AB}$ and $\vec{BD}$ both when the vectors are projected on to the reference plane as well as the orthogonal plane. I have completed tutorials on projecting a vector onto a line in $\mathbb{R}^2$ but haven't figured out how to translate that to $\mathbb{R}^3$...

Please note the diagram only shows the reference plane as parallel to the $xy$ plane for the sake of convenience. In my examples, the reference plane could be at any orientation. I am using $3$D coordinate data from an electromagnetic motion tracker and the reference plane will be constantly moving. I understand the cross product of the two vectors $\vec{AB} \times \vec{BC}$ results in the normal vector to their plane. I have $2$ different methods to calculate that but am a little lost once I get to this point. I have seen both unit vector notation and column vector notation but am confused by moving between the different styles. It would be most helpful if you could tell me the formal name of the notation/equations you use. I know the scalar equation of a plane through point $(a,b,c)$ with normal $\hat{n} = [n_1, n_2, n_3]$ is:

$$
n_1(x-a) + n_2(y-b) +n_3(z-c) = 0
$$
and the standard linear equation definition is:
$$
Ax + By + Cz = D
$$
but I could use some tips on when the equation is $=D$ and when it is $=0$ as well as any additional equations for a plane and in which circumstances the different forms are appropriate. I hope I've made sense here. Thanks for any help you can provide.

答:
Let the displacement of $\mathbf{\vec{{v}_{0}{v}_{1}}}$ be a directed vector from $\mathbf{\vec{v}_{0}}$ to $\mathbf{\vec{v}_{1}}$ defined as
$\left(\mathbf{\vec{v}_{1}}-\mathbf{\vec{v}_{0}}\right)$.

$$\mathtt{\operatorname{Let }}\mathbf{V=\left\{\vec{a},\vec{b}\right\}};$$

"Let $\operatorname{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)$ be the orthogonalization of $\mathbf{\vec{a}}$ compared to $\mathbf{\vec{b}}$ about $\mathbf{\vec{0}}$."

$
\mathbf{\forall\vec{a}\forall\vec{b}}\left(\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)=\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\implies\mathbf{\forall\vec{v}}\left(\pm\left(\mathbf{\vec{a}}\times\mathbf{\vec{b}}\right)\cdot\mathbf{\vec{v}}=\mathrm{0}\land\mp\left(\mathbf{\vec{b}}\times\mathbf{\vec{a}}\right)\cdot\mathbf{\vec{v}}=\mathrm{0}\land\mathbf{\vec{v}}\in\mathbf{V}\implies\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\cdot\mathbf{\vec{v}}=0\land\left(\mathbf{\vec{v}}\in\mathbb{R}^3\implies\text{Ortho}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\in\mathbb{R}^3\right)\right)\right)
$

"Let $\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)$ be the (linear) projection of $\mathbf{\vec{a}}$ onto $\mathbf{\vec{b}}$ about $\mathbf{\vec{0}}$."

$
\mathbf{\forall\vec{a}\forall\vec{b}}\left(\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)=\frac{\mathbf{\vec{a}}\cdot\mathbf{\vec{b}}}{\mathbf{\vec{b}}\cdot\mathbf{\vec{b}}}\mathbf{\vec{b}}\land\mathbf{\vec{b}}\not=\mathbf{\vec{0}}\land\frac{\pi}{2}=
\arccos\left(\frac{\left(\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\right)\cdot\left(\mathbf{\vec{b}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\right)}{\Vert\mathbf{\vec{a}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\Vert \Vert\mathbf{\vec{b}}-\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\Vert}\right)\land\left(\mathbf{\vec{b}}\in\mathbb{R}^n\land\mathbf{\vec{a}}\in\mathbb{R}^n\land\mathbf{\vec{0}}\in\mathbb{R}^n\implies\text{Proj}_{\mathbf{\vec{b}}}\left(\mathbf{\vec{a}}\right)\in\mathbb{R}^n\right)\right)
$

The formula to project a vector onto a plane.

"Let $\vec{F}\left(\mathbf{\vec{v}},i,\mathbf{\vec{X}}_{i,j}\right)$ be the orthogonalization of $\mathbf{\vec{v}}$ compared to $\mathbf{\vec{X}}_{i,1}$ about $\mathbf{\vec{X}}_{i,2}$."

$
\therefore\forall \mathbf{\vec{v}}\exists\mathbf{\vec{X}}\left(\mathbf{\vec{X}}\in\left\{\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)},\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}\right\}\implies\vec{F}\left(\mathbf{\vec{v}},i,\mathbf{\vec{X}}_{i,j}\right)=\mathbf{\vec{v}}-\text{Proj}_{\left(\mathbf{\vec{X}}_{i,1}-\mathbf{\vec{X}}_{i,2}\right)}\left(\mathbf{\vec{v}}-\mathbf{\vec{X}}_{i,2}\right)\right)
$

the traced codomain of $\vec{F}$ of a traced random $\mathbf{\vec{v}}$ about one instance of $\mathbf{\vec{X}}$.

"A reference plane formed by 3 points in $\mathbb{R}^3$."

Let plane $\mathbf{P}$ be a binary tuple of a normal constructor $\vec{\text{C}}$ using planer point(s) $\mathbf{\vec{p}}\in\mathbf{V}$:

$\mathbf{P}=
\left(\vec{\text{C}}\left(\mathbf{\vec{p}}\right),\mathbf{\vec{p}}\right)\iff\left(\mathbf{\vec{n}}=\vec{\text{C}}\left(\mathbf{\vec{p}}\right)\implies
\mathbf{P}\overset{\text{def}}{=}\forall\mathbf{\vec{v}}\left(\left(\mathbf{\vec{n}}-\mathbf{\vec{0}}\right)\cdot\left(\mathbf{\vec{v}}-\mathbf{\vec{p}}\right)=0\right)\right);$

$\mathtt{\text{Let }}\mathbf{V}=\left\{\vec{A},\vec{B},\vec{C}\right\};$

Let $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$ is an array data structure that contains all possible implementations of planes using all vectors in $\mathbf{V}$ given $\mathbf{P}$.
$$
\mathtt{\text{Let }}\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}=
\begin{bmatrix}
\left(\frac{(\vec{A}-\vec{C}) \times (\vec{B}-\vec{C})}{\Vert(\vec{A}-\vec{C}) \times (\vec{B}-\vec{C})\Vert}+\vec{C}, \vec{C}\right) \\
\left(\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\
\left(\frac{(\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})}{\Vert(\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \\
\left(\frac{(\vec{B}-\vec{C}) \times (\vec{A}-\vec{C})}{\Vert(\vec{B}-\vec{C}) \times (\vec{A}-\vec{C})\Vert}+\vec{C}, \vec{C}\right) \\
\left(\frac{(\vec{C}-\vec{B}) \times (\vec{A}-\vec{B})}{\Vert(\vec{C}-\vec{B}) \times (\vec{A}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\
\left(\frac{(\vec{C}-\vec{A}) \times (\vec{B}-\vec{A})}{\Vert(\vec{C}-\vec{A}) \times (\vec{B}-\vec{A})\Vert}+\vec{A}, \vec{A}\right)
\end{bmatrix};
$$

Example:
$$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2}=\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,j}=\left(\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right);$$
$$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}=\frac{(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})}{\Vert(\vec{A}-\vec{B}) \times (\vec{C}-\vec{B})\Vert}+\vec{B};$$

$$\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,2}=\vec{B};$$

"The plane orthogonal to the reference plane at $\vec{AB}$."

Let $\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}$ inherit the definition of $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$, but redefines the normal constructor $\vec{\text{C}}\left(\mathbf{\vec{p}}\right)$ implemented in $\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}$.
$$
\mathtt{\text{Let }}\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)} =
\begin{bmatrix}
\left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}-\vec{B}\right)\times(\vec{A}-\vec{B})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{2,1}-\vec{B}\right)\times(\vec{A}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\
\left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{3,1}-\vec{A}\right)\times(\vec{B}-\vec{A})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{3,1}-\vec{A}\right)\times(\vec{B}-\vec{A})\Vert}+\vec{A}, \vec{A}\right) \\
\left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{5,1}-\vec{B}\right)\times(\vec{C}-\vec{B})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{5,1}-\vec{B}\right)\times(\vec{C}-\vec{B})\Vert}+\vec{B}, \vec{B}\right) \\
\left(\frac{\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{6,1}-\vec{A}\right)\times(\vec{C}-\vec{A})}{\Vert\left(\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{6,1}-\vec{A}\right)\times(\vec{C}-\vec{A})\Vert}+\vec{A}, \vec{A}\right)
\end{bmatrix};
$$

Project $\vec{BD}$ onto the reference plane for any $i \in [1,6]$ at $R\left(i\right)$:

$$
\mathtt{\text{Let }}R\left(i\right)=\begin{Bmatrix}
\vec{D}_r=\vec{F}\left(\vec{D},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{B}_r=\vec{F}\left(\vec{B},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{A}_r=\vec{F}\left(\vec{A},i,\mathbf{\vec{R}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{BD}_r=\vec{D}_r-\vec{B}_r, \\
\vec{AB}_r=\vec{B}_r-\vec{A}_r
\end{Bmatrix};
$$

Project $\vec{BD}$ onto the orthogonal plane for any $i \in [1,4]$ at $O\left(i\right)$:

$$
\mathtt{\text{Let }}O\left(i\right)=\begin{Bmatrix}
\vec{D}_{o}=\vec{F}\left(\vec{D},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{B}_{o}=\vec{F}\left(\vec{B},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{A}_{o}=\vec{F}\left(\vec{A},i,\mathbf{\vec{O}_{\mathbf{V}}\left(P\right)}_{i,j}\right), \\
\vec{BD}_{o}=\vec{D}_{o}-\vec{B}_{o}, \\
\vec{AB}_{o}=\vec{B}_{o}-\vec{A}_{o}
\end{Bmatrix};
$$

The angle between the projection of $\vec{AB}$ and $\vec{BD}$.

while projected onto the orthogonal plane:
$\theta_{A_{o},D_{o}}=\arccos\left(\frac{\vec{AB}_{o}\cdot\vec{BD}_{o}}{\Vert\vec{AB}_{o}\Vert\Vert\vec{BD}_{o}\Vert}\right)\implies\theta_{a_{o},d_{o}}=
\arccos\left(\frac{\left(\vec{B}_{o}−\vec{A}_{o}\right)\cdot\left(\vec{D}_{o}−\vec{B}_{o}\right)}{\Vert\vec{B}_{o}-\vec{A}_{o}\Vert\Vert\vec{D}_{o}−\vec{B}_{o}\Vert}\right);$

while projected onto the reference plane:
$
\theta_{A_{r},D_{r}}=\arccos\left(\frac{\vec{AB}_{r}\cdot\vec{BD}_{r}}{\Vert\vec{AB}_{r}\Vert\Vert\vec{BD}_{r}\Vert}\right)\implies\theta_{a_{r},d_{r}}=
\arccos\left(\frac{\left(\vec{B}_{r}−\vec{A}_{r}\right)\cdot\left(\vec{D}_{r}−\vec{B}_{r}\right)}{\Vert\vec{B}_{r}-\vec{A}_{r}\Vert\Vert\vec{D}_{r}−\vec{B}_{r}\Vert}\right);
$


问:
How to prove Riemann sum wrt. any point will give same result (left, right, middle, etc.)
If $f(x)$ is a differentiable function. If I want to find the Riemann sum between points $a$ and $b$.

I'll divide the part of function from $a$ to $b$ into $n$ pieces. So the width of each piece will be $\Delta x=\frac{b-a}{n}$. Let the points that demarcate each part are $x_0,x_1,...,x_n$ such that $x_0=a$ and $x_n=b$, and $x_i=x_{i-1} + \Delta x$. The area of each (rectangular) piece will be $\Delta x f(x_i^*)$, such that $x_i^*$ is a point between $x_i$ and $x_{i+1}$.

I want to prove that $x_i^*$ can be any point between $x_i$ and $x_{i+1}$, and the Riemann sum (i.e. $\sum_{i=0}^n f(x_i^*)\Delta x$) as $n \to \infty$ will give the same result.

Will this kind of proof work?

$x_i\le x_i^*\le x_{i+1}$

$\Rightarrow x_i\le x_i^*\le x_i+\Delta x$

as $n\to \infty$, $\Delta x\to 0$

$\Rightarrow \lim_{\Delta x\to 0} x_i\le \lim_{\Delta x\to 0} x_i^*\le \lim_{\Delta x\to 0} x_i+\Delta x$

$\therefore x_i \le x_i^*\le x_i$

Hence $x_i^*$ can be any point between $x_i$ and $x_{i+1}$

答:
The riemann sum is defined as

$$\lim_{n\rightarrow \infty}\sum_{i=1}^n f(t_i+h_i)(t_{i+1}-t_i),$$

where $t_i$ is a dense partition of $[a,b]$, and $t_i+h_i\in [t_{i},t_{i+1}]$.

We use the fact that any function that is continuous is also uniformly continuous in an interval classically. This means for any $t$ and any $\epsilon$ there is $\delta$ such that for all $x$

$$|f(x)-f(t)|<\epsilon, |x-t|<\delta,$$

and the delta works for all $t$. Hence we reduce the intervals until each is within $\delta$ of each other, which happens since the partition is dense. Then

$$\lim_{n\rightarrow \infty}\sum_{i=1}^n f(t_i+h_i)(t_{i+1}-t_i),$$

$$<\lim_{n\rightarrow \infty}\sum_{i=1}^n (f(x_i)+\epsilon) (t_{i+1}-t_i)=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i) (t_{i+1}-t_i)+(b-a)\epsilon,$$

$$>\lim_{n\rightarrow \infty}\sum_{i=1}^n (f(x_i)-\epsilon) (t_{i+1}-t_i)=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i) (t_{i+1}-t_i)-(b-a)\epsilon$$

As $\epsilon$ was arbitrarily chosen, this should suffice. I think uniform continuity should be enough.