问:
if $\mathbf x$ is sampled randomly from a hypercube on $R^n$, what is the probability density for $|\mathbf x| = d$
if the vector $\mathbf x$ is sampled randomly from a uniform distribution on $[0, 1]^d$, what is the probability density function for $|\mathbf x|$? Is it easy to scale for $[0, n]^d$?</p>

答:
You want to think about how to calculate the measure of a hypersphere of radius $r$. If we have $n$ dimensions, we are asking for the set $\sum x_i^2 \le r^2$. This is because this set describes all the points with magnitude $r$ or less. If we imagine points distributed uniformly throughout this hypersphere, we will see that the probability of finding a point at radius $r$ is related to the size of the shell with radius $r$.

We can calculate the measure of this hypersphere by doing an iterated integral. For the four dimensional case we have:

$2^4\int_0^r \int ^{\sqrt{r^2-x_1^2}}_0 \int_0 ^\sqrt{r^2-x_1^2-x_2^2}\int_0^\sqrt{r^2-x_1^2-x_2^2-x_3^2} dx_4 dx_3 dx_2 dx_1$

We can easily generalize this to any dimension. Then we note that the actual problem asks for the density inside a hypercube, so we only take the portion of the hypersphere which intersects our hypercube.

Let $M(x) = \text{min}(x,1)$. Then we drop our $2^n$ factor because we are only worried about positive components. Then our CDF$(r)$ is

$\int_0^{M(r)} \int ^{M(\sqrt{r^2-x_1^2})}_0 \int_0 ^{M(\sqrt{r^2-x_1^2-x_2^2})}\int_0^{M(\sqrt{r^2-x_1^2-x_2^2-x_3^2})} dx_4 dx_3 dx_2 dx_1$

For $r \le 1$  we have an easy solution because the $M$s disappear and we end up with the volume of the n-dimensional hypersphere of radius $r$ divided by $2^n$. I will update the solution if I find a simpler form where $r \gt 1$.

Here is a different formulation / generalization to arbitrary dimension:

Let a particle appear with equal likelihood in any part of the unit n-hypercube. For $r \ge 0$ Let $F_n(r)$ be the probability that the particle will appear within $r$ distance of the origin.

Obviously $F_1(r) =\text{min}(1,r)$ because the one-dimensional case is a line segment of length 1. (this is just a silly way of writing the uniform CDF because I don't know how to format latex)

Now let's try to compute $F_n(r)$ in terms of $F_{n-1}$ by imagining that the particle is restricted to a hyperplane where one of the coordinates is fixed to $k$. That is, it must appear in the plane $\{k\} \times [0,1]^{n-1}$. Since we are only allowed to go $r$ distance away from the origin and we used up $k$ distance already, we can only use $\sqrt{r^2 - k^2}$ distance in traveling away from the origin of the sub-cube $[0,1]^{n-1}$.

$F_n(r) = P(\sum{x_i^2} \le r^2)$

$ = \int_0^1P(\sum x_i^2 \le r^2 | x_1 = k)f_{x_1}(k)dk $

$ = \int_0^1P(\sum_{i\ne1} x_i^2 \le r^2 - k^2 | x_1 = k)dk $

$ = \int_0^1P(\sum_{i\ne1} x_i^2 \le r^2 - k^2 | x_1 = k)\mathbb{I}\{k \le r\le 1\}dk $

$ = \int_0^rP(\sum_{i\ne1} x_i^2 \le r^2 - k^2 | x_1 = k)\mathbb{I}\{r\le 1\}dk $

$ = \int_0^rF_{n-1}(\sqrt{r^2-k^2})\mathbb{I}\{r\le 1\}dk $

$ = \int_0^{\text{min}(r,1)}F_{n-1}(\sqrt{r^2-k^2})dk $

So we end up with:

$F_n(r)  = \int_0^{\text{min}(r,1)}F_{n-1}(\sqrt{r^2-k^2})dk$

$F_1(r) =\text{min}(1,r)$

Of course if you expand this formula for $F_4$ you will get the same four dimensional case mentioned above.


问:
Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$
Is it possible to prove that $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ without using a calculator or logarithms (middle school math only recommended)?

My idea for the first one was to use the obvious $5^3<2^7$ and then raise to the power of $50$ to get $5^{150}<2^{350}$. But since $5^2>2^3$, I couldn't use this approach to get the desired result.

Can you please help me find a relatively short proof for these inequalities? Thank you.

答:
Not an answer, but some thoughts too long to put in the comment section.

For $5^{152} < 2^{353}$, probably not, it's really tight. If you look at the continued fraction of $\ln(5)/\ln(2)$:

$$[2;3,9,2,2,4,\ldots]$$

The first approximation is $2+\frac 13 = \frac 73$ which gives you $5^3 < 2^7$. The next one is $2+\frac{1}{3+\frac 19}=\frac{65}{28}$ which yields $5^{28}> 2^{65}$. Or you can have $2+\frac{1}{3+\frac{1}{10}}=\frac{72}{31}$ and $5^{31} < 2^{72}$. So even if this is allowed it still doesn't work because

$$5^{152} = \frac{5^{155}}{5^3} < \frac{2^{360}}{5^3}=2^{353}\frac{2^7}{5^3}$$ but unfortunately $\frac{2^7}{5^3}>1$.

ContentLicense: CC BY-SA 4.0


答:
I don't think so because it is so close.  You can write
$$2^7=\left(1+\frac 3{125}\right)5^3\\
2^{350}=\left(1+\frac 3{125}\right)^{50}5^{150}$$
I asked Alpha for $\left(1+\frac 3{125}\right)^{50}$ and got about $3.2734$ so now you can write
$$2^{353} \approx 8 \cdot 3.2734 \cdot 5^{150}\\
\approx 26\cdot 5^{150}\gt 5^{152}$$
but you need to evaluate the $50^{th}$ power to within $4\%$.  I would say that is outside the reasonable range of hand computation.

Title: Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$

Body:
Is it possible to prove that $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ without using a calculator or logarithms (middle school math only recommended)?

My idea for the first one was to use the obvious $5^3<2^7$ and then raise to the power of $50$ to get $5^{150}<2^{350}$. But since $5^2>2^3$, I couldn't use this approach to get the desired result.

Can you please help me find a relatively short proof for these inequalities? Thank you.


问:
Extended Tonelli's Theorem
Let $f: \mathbb{R}^2\to\mathbb{C}$ be a measurable sign changing function with $f = f^+ - f^-$ such that $f^+,f^-\geq 0$ and:
$$ \int f^+ =\infty $$
$$ \int f^- < \infty $$
Can one still apply Tonelli's theorem in this case?

答: We can apply Tonelli's Theroem to $f^{+}$, Fubini's Theorem to $f^{-}$ and subtract to conclude that $\int f(x,y)dxdy=\int f(x,y)dydx=\infty$.

Title: Extended Tonelli's Theorem


问:
Non-coprime pairs with coprime $\gcd$'s.
Given $m,n,k,l\in\mathbb Z$ such that $a = \gcd(m,n)\neq 1$ and $b = \gcd(k,l)\neq 1$, but $\gcd(a,b)=1$, are there $c,d\in\mathbb Z$ such that
$$
\gcd(cm+dn,\,ck+dl) = 1\,?
$$
I have already tried a lot, but I cannot seem to figure this out. When I denote $m' = m/a$, $n' = n/a$, $k' = k/b$ and $l' = l/b$, then
$$
\gcd(cm+dn,\,ck+dl) = \gcd\big(a(cm'+dn'),\,b(ck'+dl')\big).
$$
Clearly, I can find $c,d$ such that $cm'+dn' = 1$, but then it is not clear whether $a|(ck'+dl')$. Any help is appreciated.

答:
You're on the right track. As you already basically stated,

$$a = \gcd(m, n) \implies m = am', \, n = an', \, \gcd(m', n') = 1 \tag{1}\label{eq1A}$$

$$b = \gcd(k, l) \implies k = bk', \, l = bl', \, \gcd(k', l') = 1 \tag{2}\label{eq2A}$$

With $a \neq 1$, $b \neq 1$ and $\gcd(a, b) = 1$, then there exist integers $c$ and $d$ such that

$$\gcd(cm + dn, \, ck + dl) = 1 \tag{3}\label{eq3A}$$

As you've already done, the left side is equivalent to

$$\gcd(cm + dn, \, ck + dl) = \gcd(a(cm' + dn'), \, b(ck' + dl')) \tag{4}\label{eq4A}$$

If $m' = n' = 0$, then $\gcd(m, n) = \gcd(0, 0) = a$ is not well defined but, if it's to be anything, it should be considered be $0$. However, using the third bullet point in the Properties section of Wikipedia's "Greatest common divisor" article, $1 = \gcd(a, b) = \gcd(0, b) = |b|$, so $b = 1$, but this is not allowed (this may be why this condition was specified in the question). This means $a$ and $b$ are non-zero, plus at least one of $m'$ and $n'$ is non-zero (and, similarly, at least one of $k'$ and $l'$ is non-zero). As such, using Bézout's identity, as you've already stated, there exist integers $c$ and $d$ such that

$$cm' + dn' = 1 \implies \gcd(a(cm' + dn'), \, b(ck' + dl')) = \gcd(a, \, b(ck' + dl')) \tag{5}\label{eq5A}$$

For any particular solution $c = c_{1}$ and $d = d_{1}$, the general solution is, for any integer $e$,

$$c = c_{1} - e(n'), \; d = d_{1} + e(m') \tag{6}\label{eq6A}$$

Using this gives

$$\begin{equation}\begin{aligned}
ck' + dl' & = (c_{1} - e(n'))k' + (d_{1} + e(m'))l' \\
& = c_{1}k' + d_{1}l' + e(m'l' - n'k')
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Let

$$f = \gcd(c_{1}k' + d_{1}l', m'l' - n'k') \tag{8}\label{eq8A}$$

If $f \neq 1$, then there's a prime $p \mid f$. Thus, if $l' \neq 0$, then

$$c_{1}k' \equiv -d_{1}l' \pmod{p} \implies c_{1}k'\color{blue}{m'} \equiv -d_{1}l'\color{blue}{m'} \pmod{p} \tag{9}\label{eq9A}$$

$$n'k' \equiv m'l' \pmod{p} \implies \color{red}{d_{1}}n'k' \equiv \color{red}{d_{1}}m'l' \pmod{p} \tag{10}\label{eq10A}$$

Adding \eqref{eq9A} and \eqref{eq10A} gives

$$c_{1}k'm' + d_{1}n'k' \equiv 0 \pmod{p} \implies k'(c_{1}m' + d_{1}n') \equiv 0 \pmod{p}  \tag{11}\label{eq11A}$$

Since $c = c_{1}$ and $d = d_{1}$ is a particular solution to \eqref{eq5A}, this means $c_{1}m' + d_{1}n' = 1$, so \eqref{eq11A} becomes

$$k' \equiv 0 \pmod{p} \tag{12}\label{eq12A}$$

The first part of \eqref{eq10A} gives $m'l' \equiv 0 \pmod{p}$, but since \eqref{eq2A} gives $\gcd(k', l') = 1$, this means $p \mid \color{blue}{m'}$. Also, the first part of \eqref{eq9A} gives $d_{1}l' \equiv 0 \pmod{p} \implies p \mid \color{red}{d_{1}}$. Thus, $p \mid c_{1}\color{blue}{m'} + \color{red}{d_{1}}n' = 1$, which is not possible. Note if $l' = 0$, then $k' \neq 0$, so you can use a similar procedure to that above to eliminate $k'$ from \eqref{eq9A} and \eqref{eq10A} instead to get $l' \equiv 0 \pmod{p}$, and then continue to end up with the same contradiction.

This means the assumption $f \neq 1$ must be false, so $f = 1$. For simpler algebra, let

$$g = c_{1}k' + d_{1}l', \; h = m'l' - n'k' \tag{13}\label{eq13A}$$

From $f = \gcd(g, h) = 1$, we get

$$j = \gcd(g, a) \implies \gcd(j, h) = 1 \tag{14}\label{eq14A}$$

Using that $g$ and $h$ are coprime, or by doing direct calculations (note I'm not providing the details as this answer is already quite long), then if $g = 0$ then $h = \pm 1$ (so can choose $e = 1$), and if $h = 0$ then $g = \pm 1$. In either case, \eqref{eq3A} is then true.

Otherwise, for $g \neq 0$ and $h \neq 0$, have all of the prime factors of $a$ which divide $j$ be combined into $q$, i.e., have

$$a = qr, \; \gcd(q, j) = j, \; \gcd(q, r) = 1 \implies \gcd(q, h) = 1 \tag{15}\label{eq15A}$$

Thus, $h$ has a multiplicative inverse modulo $q$. Let

$$e \equiv h^{-1}(1 - g) \pmod{q} \implies g + eh \equiv g + h^{-1}(1 - g)h \equiv 1 \pmod{q} \tag{16}\label{eq16A}$$

Due to how $r$ is defined in \eqref{eq15A}, we also have $\gcd(r, g) = 1$ (if not, then any $p \mid r$ means $p \mid a$, so with $p \mid g$ this means $p \mid \gcd(a, g) = j$, so $p \mid q$, which is not allowed). Thus, set

$$e \equiv 0 \pmod{r} \implies \gcd(g + eh, r) = 1 \tag{17}\label{eq17A}$$

Since $\gcd(q, r) = 1$, the Chinese remainder theorem guarantees there's a solution for $e$ using the left sides of \eqref{eq16A} and \eqref{eq17A}. Also, those $2$ equations combined shows (using $a = qr$ from \eqref{eq15A}) that $\gcd(g + eh, q) = \gcd(g + eh, r) = 1 \implies \gcd(g + eh, a) = 1$. Thus, using \eqref{eq13A}, then \eqref{eq7A} becomes $\gcd(ck' + dl', a) = 1$. This means the $\gcd$ in \eqref{eq5A} is $1$, so \eqref{eq3A} is true.


问:
Integrate ${\sec 4x}$
How do I go about doing this? I try doing it by parts, but it seems to work out wrong:

$\eqalign{
  & \int {\sec 4xdx}   \cr
  & u = \sec 4x  \cr
  & {{du} \over {dx}} = 4\sec 4x\tan 4x  \cr
  & {{dv} \over {dx}} = 1  \cr
  & v = x  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \int {4x\sec 4x\tan 4xdx}   \cr
  & \int {4x\sec 4x\tan 4xdx} :  \cr
  & u = 4x  \cr
  & {{du} \over {dx}} = 4  \cr
  & {{dv} \over {dx}} = \sec 4x\tan 4x  \cr
  & v = {1 \over 4}\sec x  \cr
  & \int {4x\sec 4x\tan 4xdx}  = x\sec 4x - \int {\sec 4x} dx  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $

I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!

EDIT: Is there an easier way to do this?

答:
Let $4x = t$, i.e., $4dx = dt$.
$$I=\int \sec(4x) dx = \dfrac14\int \sec(t) dt = \dfrac14\int \dfrac{dt}{\cos(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{\cos^2(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{1-\sin^2(t)}$$
Now let $\sin(t) = y$, to get $\cos(t) dt = dy$. Hence, we get that
$$I = \dfrac14 \int \dfrac{dy}{1-y^2} = \dfrac18\left(\int\dfrac{dy}{1+y} + \int\dfrac{dy}{1-y}\right) = \dfrac18\left(\log(\vert 1+ y \vert) - \log(\vert 1 - y \vert)\right)+\text{const}$$
This gives us
$$I = \dfrac18 \log \left(\left\vert \dfrac{1+y}{1-y}\right\vert \right) + \text{const} = \dfrac18 \log \left(\left \vert \dfrac{1+\sin(t)}{1-\sin(t)}\right \vert\right)+\text{const} = \color{red}{\dfrac18 \log \left(\left \vert \dfrac{1+\sin(4x)}{1-\sin(4x)}\right \vert\right)+\text{const}}$$

EDIT

You can simplify it further as much as you like.

\begin{align}
\dfrac{1+\sin(4x)}{1-\sin(4x)} & = \dfrac{1+\sin(4x)}{1-\sin(4x)} \times \dfrac{1+\sin(4x)}{1+\sin(4x)} = \dfrac{(1+\sin(4x))^2}{1-\sin^2(4x)}\\
& = \dfrac{(1+\sin(4x))^2}{\cos^2(4x)} = (\sec(4x) + \tan(4x))^2
\end{align}

Hence,
$$\boxed{\color{blue}{I = \dfrac{\log(\vert \sec(4x) + \tan(4x)\vert)}4 + \text{const}}}$$

ContentLicense: CC BY-SA 3.0

Title: Integrate ${\sec 4x}$


问:
Isn't saying ${\mathbb{N}=\{0,1,2,3,...\}}$ with Peano axioms a little circular?
My title probably doesn't explain my worry / concern too well, but it's the best title I could think of. I am researching construction of real numbers for a college project, and as a consequence I am researching the Peano axioms. A couple of sources I have read come to the conclusion that
$$
\{0,1,2,3,...\} \subseteq \mathbb{N}
$$
(where ${1=S(0)}$, ${2=S(1)}$,...) and by the axiom of induction we have
$$
\mathbb{N} \subseteq \{0,1,2,3,...\}
$$
hence
$$
\mathbb{N}=\{0,1,2,3,...\}
$$
(in other words, the Natural Numbers are the set of all possible successors of $0$). The part that makes me feel uneasy is the definition of ${\{0,1,2,3,...\}}$. Doesn't defining this set in the first place use induction in some sense? Like - we start with $0$. Then ${S(0)}$ cannot be $0$, and so we define ${1:=S(0)}$. Then ${S(1)}$ can be neither $0$ or $1$, and so we define ${2:=S(1)}$... I'm not sure how doing this for finitely many elements justifies the existence of the set ${\{0,1,2,3,...\}}$. Does what I'm saying even make sense? Am I overthinking things? Thank you!

EDIT: I should specify, I am following the definition of the axiom of induction as is on Wikipedia (https://en.wikipedia.org/wiki/Peano_axioms)

EDIT 2: source example: http://www2.hawaii.edu/~robertop/Courses/TMP/7_Peano_Axioms.pdf , page $3$

答: The notation $\{0,1,2,3,\cdots\}$ is being used here to symbolize "A set containing $0$, containing $S(n)$ whenever it contains $n$, and containing nothing else." It is true that the Peano axioms do not justify the existence of this set, but that is no surprise, since the Peano axioms do not deal with sets at all! With this line of thought, your two claims become "For every $n\in \mathbb N$, we have $S(n)\in \mathbb N$" and "If a set contains $0$ and $S(n)$ for every $n$ which it contains, then it contains $\mathbb N$", both of which are directly asserted by the Peano axioms. So your sources, at least as you paraphrase them, are saying nothing at all.


问:
Graph Theory - Can two vertices have two distinct edges?
My question is simple:

Can two vertices in a graph have two distinct edges connecting them? Why or why not?

答:
I put this answer up only because OP asked for it in a comment, and also to present the other two terms I've heard of besides the widely used "simple graph".

In a graph, two or more edges connecting two distinct vertices are called parallel edges,
and an edge connecting only one vertex to itself is called a loop.

According to some sources: a graph allowing no loop or parallel edges is a "simple graph", a graph allowing parallel edges but no loops is a "multigraph", and if both parallel edges and loops are allowed the graph is a "pseudograph". [It's not clear to me whether a multigraph must have at least one case of a parallel edge, or whether a pseudograph which happens not to have loops would qualify as a multigraph.]

I also don't know whether there is general agreement about the definitions of multigraph and pseudograph. (Nor do I think it matters much provided a text/article makes the definitions used clear.)


问:
Method of moment, MLE and information matrix
We have $$\mathbb{E}[Y_i| X_i] = β_0 + β_1X_i$$ with the following characteristics. The conditional distribution of $Y_i$ given $X_i$ normal, as follows. $$Y_i| X_i ∼ N(β_0 + β_1X_i, σ^2)$$  The distribution of $X_i$ is fixed in repeated samples.

The error term $ε_i$ is: $ε_i = Y_i − β_0 − β_1X$.

Under the hypotheses above, $ε_i$ follows an unconditional i.i.d. normal distribution. $$ε_i ∼ N(0,σ^2)$$

This model has 3 parameters : $θ = (β_0, β_1, σ^2)$.

What would be the Method of Moments estimator and MLE for this model? Construct the information matrix for $θ = (β_0, β_1, σ^2)$

Here is what I've done so far:
Starting with the MM:

The three population moments are as follows:

1- $\mathbb{E}[y_1-\beta_0-\beta_1x_i]=0$

2- $\mathbb{E}[(y_1-\beta_0-\beta_1x_i)x_i]=0$

3- $\mathbb{E}[ε_i^2]=\sigma^2$

So I generated the sample analogs to the three populations moment conditions:

1- $\frac{1}{N}\sum_{i=1}^{n}(y_1-\hat{\beta_0}-\hat{\beta_1}x_i)=0$

2-$\frac{1}{N}\sum_{i=1}^{n}x_i(y_1-\hat{\beta_0}-\hat{\beta_1}x_i)=0$

3- $\frac{1}{N}\sum_{i=1}^{n}\hat{ε_i^2}=\hat{\sigma^2}$

Now moving on to the MLE:

Based on the $pdf$ of the normal distribution: $$f_{Y_i}(y; β_0, β_1, σ^2)= \frac{1}{\sqrt{
2πσ^2}} e^{{-\frac{1}{2σ^2}}(y-\beta_0-\beta_1x_i)^2}$$

The log likelihood is: $$log(L(β_0, β_1, σ^2)= -\frac{n}{2}log(σ^2)−\frac{1}{2σ^2}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)^2$$

and have found the 3 first derivatives of $\beta_0,\beta_1$ and $\sigma^2$ for the MLE

$\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \beta_0}= \frac{1}{\sigma^2}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)=0$

$\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \beta_1}= \frac{1}{\sigma^2}\sum_{i=1}^{n}x_i(y_1-\beta_0-\beta_1x_i)=0$

$\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \sigma^2}= -\frac{n}{2\sigma^2}+\frac{1}{2(\sigma)^{4}}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)^2=0$

Multiplying throughout by $\frac{2\sigma^4}{N}$ and rearranging the result, gives $$\sigma^2(\beta_0,\beta_1)=\frac{1}{N}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)^2=\frac{1}{N}\sum{e_i^2}$$

I have no idea about the information matrix.

答:
The observed Fisher information matrix is
$$
I(\hat{\theta}) = - \frac{\partial ^ 2}{\partial \theta_i \partial \theta_j} \big|_{\theta = \hat{\theta}}.
$$
For $I(\hat{\theta})_{11} = - \frac{\partial}{\partial \beta_0} \sigma^{-2}\sum (y_i - \beta_0 - \beta_1x_i) = n/\sigma^2\Big|_{\sigma = \hat{\sigma}} = n/\hat{\sigma}^2$.

For $I(\hat{\theta})_{12} = I(\hat{\theta})_{21} =  - \frac{\partial}{\partial \beta_1} \sigma^{-2}\sum (y_i - \beta_0 - \beta_1x_i) = n\bar{x}_n/\sigma^2\Big|_{\sigma = \hat{\sigma}} = n\bar{x}_n/\hat{\sigma}^2$.

For $I(\hat{\theta})_{22} =  - \frac{\partial}{\partial \beta_1} \sigma^{-2}\sum (y_i - \beta_0 - \beta_1x_i)x_i = \sum x_i^2/\sigma^2\Big|_{\sigma = \hat{\sigma}} = \sum x_i^2/\hat{\sigma}^2$.

For $I(\hat{\theta})_{33} =  - \frac{\partial}{\partial \sigma^2} \left( \frac{\partial}{\partial \sigma^2} \ln L(\theta;x_1,...,x_n) \right)=...= \frac{n}{2\hat{\sigma}^4} $.

$I(\hat{\theta})_{13} = I(\hat{\theta})_{23} = 0$.


问:
Compute $\mathbb{E}(\int_{0}^{T} |B_{t}|^{2} dt)$
For $(B_t)_{t\geq0}$ a standard brownian motion and $T$ the first time the standard brownian motion hits $1$ or $-1$ I have to compute $\mathbb{E}(\int_{0}^{T} |B_{t}|^{2} dt)$. My lecturer has given us a hint to first apply Ito's formula to $B_t^4$. There are two things which kind of make sense to me and I'm not sure which is correct if either. First, using the hint and then taking expectation I think we get

$\mathbb{E}(|B_t|^4) = 6\mathbb{E}(\int_{0}^{t} |B_{s}|^{2} ds)$,

then to evaluate this at $t=T$, we get

$\mathbb{E}(|B_T|^4) = 6\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds)$,

but $|B_T| = 1$, so $\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds) = \frac{1}{6}$.

That's the first way that kind of makes sense. The second way is ignoring the hint and using Fubini's theorem to get

$\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds) = \int_{0}^{T} \mathbb{E}(|B_{s}|^{2}) ds = \int_{0}^{T} s ds = \frac{T^2}{2}$.

I'm slightly confused as to whether either of these ways is correct and if not, why not?

答: The second method is wrong. $T$ is a random variable. While $E\int_0^{T}|B_s|^{2}ds$ is a number, $\int_0^{T}E|B_s|^{2}ds$  is a random variable.


问:
Probability that the dot product of two binary vectors is k and sum of vectors equals a and b respectively
I have two binary vectors $x$, and $y$, each with $n$ elements; Each element of $x$ and $y$ belongs to {-1, 1}, and is drawn from uniform random distribution.

How would I compute the probability of $x \cdot y == k$ and $\sum x == a$ and $\sum y == b$? where $k$, $a$, and $b$ are all intergers ranging from $-n$ to $n$.

I am able to compute the probability of $x \cdot y == k$ from combinatorial analysis, assuming (1) $-n \le k \le n$ and (2) $k \text{ mod } 2 == n \text{ mod } 2$, then $P(x \cdot y == k) = $ $n \choose {(n+k)/2}$ $  2^{-n}$.

From here, I tried to derive the probability for $x \cdot y == k$ and $\sum x == a$ and $\sum y == b$, but I was stuck due to coupled relations between $x \cdot y$ and $\sum x$ and $\sum y$.

答:
Let's put
$$
\left\{ \matrix{
  {\bf u} = \left( {1,1, \cdots ,1} \right)^T \quad {\bf u}^T {\bf u} = n \hfill \cr
  {\bf z} = {{{\bf x} + {\bf u}} \over 2}\quad {\bf w}
 = {{{\bf y} + {\bf u}} \over 2} \hfill \cr}  \right.
$$
so that $\bf z , \bf w$ are actual binary vectors.

Then
$$
\eqalign{
  & S_z  = {\bf u}^T {\bf z} = {1 \over 2}S_x  + {n \over 2}\quad S_w  = {1 \over 2}S_y  + {n \over 2}  \cr
  & S_{zw}  = {\bf z} \cdot {\bf w} = {\bf z}^T {\bf w}
 = {1 \over 4}\left( {{\bf x}^T  + {\bf u}^T } \right)\left( {{\bf y} + {\bf u}} \right) =   \cr
  &  = {1 \over 4}\left( {{\bf x}^T {\bf y} + {\bf x}^T {\bf u} + {\bf u}^T {\bf y}
 + {\bf u}^T {\bf u}} \right) =   \cr
  &  = {1 \over 4}\left( {{\bf x}^T {\bf y} + S_x  + S_y  + n} \right)  \cr
  & S_{zw}  = {\bf z}^T {\bf w} = {\rm N}{\rm .}\,{\rm of}\,{\rm coincident}\,{\rm ones} \cr}
$$
where $S_z, S_w$ are the respective number of ones and are integers.

So given two binary strings, with $S_z$ and $S_w$ ones, the number of strings with
$S_{zw}$ coincident ones is readily computed.

We can rearrange $\bf z$ as to have all the $S_z$  ones on the same side (e.g. at the lower indices)
and then consider the combinations of  $\bf w$'s  with a number $0 \le S_{zw} \le S_w$ of coincident ones.

There are totally
$$\binom{n}{S_w}$$
different $\bf w$'s with $S_w$ ones.

Of these, those which have $S_{zw}$ ones in the first $S_z$ places and the remaining
$n-S_{zw}$ in the last $n-S_z$ places are
$$
N(S_{zw} \;\left| {\,n,S_z ,S_w } \right.)
 = \left( \matrix{  S_z  \cr   S_{zw}  \cr}  \right)
\left( \matrix{  n - S_z  \cr  S_w  - S_{zw}  \cr}  \right)
$$
and the probability will be
$$
p(S_{zw} \;\left| {\,n,S_z ,S_w } \right.)
 = {{\left( \matrix{  S_z  \cr   S_{zw}  \cr}  \right)
\left( \matrix{  n - S_z  \cr   S_w  - S_{zw}  \cr}  \right)}
 \over {\left( \matrix{  n \cr   S_w  \cr}  \right)}}
$$
wihich is a Hypergeometric distribution


问:
$J_0(\gamma)=\sum_{m=-\infty}^\infty(J_m(a)J_m(\beta)e^{im\theta})$ and a.β,γ be triangle edges
How can I proove his relation for a,β,γ be triangle edges and $\gamma^2=a^2+\beta^2-2a\beta cos\theta
$
$$J_0(\gamma)=\sum_{m=-\infty}^\infty(J_m(a)J_m(\beta)e^{im\theta})$$

答:
Use the following integral representation :
$$J_m(\alpha)=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i(\alpha\sin\tau-m\tau)}$$

Substitute that into the RHS to obtain

$$RHS=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i\alpha\sin\tau}\sum_{m=-\infty}^{\infty}J_{m}(\beta)e^{im(\theta-\tau)}$$

which by virtue of the generating function of the Bessel functions can be written as follows:
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i\alpha\sin\tau+i\beta\sin(\theta-\tau)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i(\alpha-\beta\cos\theta)\sin\tau+i\beta\sin\theta\cos\tau}\\= \frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i
\gamma\sin(\tau+\phi_0)}$$
where in the last line we defined an angle $\phi_0$ such that
$$\gamma\cos\phi_0=\alpha-\beta\cos\theta~,~\gamma\sin\phi_0=\beta\sin\theta$$
Since the function $e^{i\gamma\sin\tau}$ is periodic with period $2\pi$, it is true that
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\gamma\sin(\tau+\phi_0)}d\tau=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\gamma\sin\tau}d\tau=J_0(\gamma)$$
which concludes the proof.


问:
prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$

attempt:

$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$

How I can continue from there to apply the Euler's theorem

Body:
By Fermat's little theorem, we have

$$2^{18} \equiv 1 \pmod{19} \implies 2^{36} = (2^{4})^{9} \equiv 1 \pmod{19} \tag{1}\label{eq1A}$$

Since $2^6 = 64 \equiv 1 \pmod{9}$, we thus get $2^{6k} = (2^6)^k \equiv 1 \pmod{9}$, so there's an integer $j$ where $2^{6k} = 9j + 1$. Thus,

$$(2^4)^{2^{6k}} = (2^4)^{9j + 1} = ((2^4)^{9})^{j}(2^4) \equiv 2^4 \pmod{19} \tag{2}\label{eq2A}$$


答: First try $u = 4x$. Then multiply numerator and denominator by $(\tan u + \sec u)$</p>

Title: Integrate ${\sec 4x}$

Body:
How do I go about doing this? I try doing it by parts, but it seems to work out wrong:

$\eqalign{
  & \int {\sec 4xdx}   \cr
  & u = \sec 4x  \cr
  & {{du} \over {dx}} = 4\sec 4x\tan 4x  \cr
  & {{dv} \over {dx}} = 1  \cr
  & v = x  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \int {4x\sec 4x\tan 4xdx}   \cr
  & \int {4x\sec 4x\tan 4xdx} :  \cr
  & u = 4x  \cr
  & {{du} \over {dx}} = 4  \cr
  & {{dv} \over {dx}} = \sec 4x\tan 4x  \cr
  & v = {1 \over 4}\sec x  \cr
  & \int {4x\sec 4x\tan 4xdx}  = x\sec 4x - \int {\sec 4x} dx  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $

I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!

EDIT: Is there an easier way to do this?


问:
Is there a notation for repeatedly applying a function?
There is notation for applying both multiplication and addition a set amount of times (can be infinite): $\sum$, $\prod$ but there is no notation (to my knowledge) of repeating a function. Whenever I need this I use this notation: $R_{n}^{k} f(a) $ where $k$ is the amount of times it is repeated, $a$ is where the function is repeated, and $n$ is the input. For example: $$R_{n=\sqrt{2}}^{\infty} n^{a}=2$$
If anyone knows of the proper notation for this it would be greatly appreciated.

答:
A common notation is $f^k(a)$, so
$$f^0(a)=a, f^1(a)=f(a), f^2(a)=f(f(a)), f^3(a)=f(f(f(a))),\dots.$$
You can see this notation used for $k=-1$ to denote the inverse of a function, e.g. $\sin^{-1}(x):=\arcsin(x)$.

For some functions, this notation is not very good; for example, $\sin^2(x)$ is usually defined as $\sin(x)^2$ instead of $\sin(\sin(x))$, since the former is a more useful function. It's a good idea to define this notation anywhere you might want to use it, and if there could be confusion, use something related like $f^{(k)}(a)$ (which can also be confused with the $k$th derivative of $f$) or $f^{\circ k}(a)$ (with the latter denoting "$f$ composed with itself $k$ times").


问:
Is it true that a mixed Nash equilibrium cannot be strict?
A Nash equilibrium $y  = (y_i, y_{-i}) \in \Delta$ is said to be strict if for all players $i$,

$$u(y_i; y_{-i}) \geq u(x_i; y_{-i}), \forall x_i, y_i \neq x_i$$

Is it true that no mixed equilibrium can be strict? (and only pure strategies can be strict?)

答:
Mixed-strategy Nash equilibria are necessarily always weak, while pure-strategy Nash equilibria can be either strict or weak. We can see this because if there are greater than or equal to two pure strategies that are best responses to the other players strategy then any mixture of them is also a best response. For example take the game,

\begin{array}{|c|c|}
\hline
& H & T \\
\hline
H & (2, 1) & (0, 0) \\ \hline
T & (0, 0) & (1, 2) \\ \hline
\end{array}

the mixed strategy for the row player is $(\frac{2}{3}, \frac{1}{3})$ and $(\frac{1}{3}, \frac{2}{3})$ for the column player, with each player having an expected payoff of $\frac{2}{3}$. Now assuming the column player does not deviate from their equilibrium mixed strategy, every strategy, $(p, 1 - p)$, is a best response for the row player, so they can deviate if they want.


问:
Sequence of Functions in $L^{p}(X)$
Consider the following statement :

[Lemma] : Let $(X, \mathcal{A}, \mu)$ be a measurable space. Then any function in $L^{p}(X)$ is the limit in $L^{p}$ of a sequence of simple functions in $L^{p}$.

How can we prove this?

答:
Let $\left(g_{n}\right)$ be an increasing sequence of simple nonnegative functions that converges (pointwise) to $f^{+}$ and let $\left(h_{n}\right)$ be an increasing sequence of of simple nonnegative functions that converges to $f^{-} .$ Let $\varphi_{n}:=g_{n}-h_{n} .$ Then $\varphi_{n} \rightarrow f$ pointwise and :
$$\left|f-\varphi_{n}\right| \leq|f|+\left|\varphi_{n}\right| \leq |f|+h_{n}+g_{n} \leq 2|f| .$$ Thus, $\left|f-\varphi_{n}\right|^{p} \leq 2^{p}|f|^{p} .$ It follows first that $\varphi_{n} \in L^{p}$ and second that $\left\|f-\varphi_{n}\right\|_{p} \rightarrow 0$ by the Lebesgue dominated convergence theorem.

Title: Sequence of Functions in $L^{p}(X)$


问:
$6$ points in the interior of a square of side length $2$.
Prove that having $6$ points in the interior of a square of side length $2$, we can choose $3$ of them so that the sum of distance between them is less than $3\sqrt{2}$. Is there a simple way?</p>

答:
There is no simple way to prove this, because the statement is false!

Place four points at the corners of the square, and two points at $(x,x)$ and $(2-x,2-x)$, where $x\approx 0.48735$ (though any $x\in[0.44,0.5)$ works).

Then the minimum perimeter of any triangle in the configuration is

$$\frac{6+5\sqrt{2}+\sqrt{22-4\sqrt{2}}}{4}\approx4.2784\ldots\approx3\sqrt{2}+0.03578$$

attained by both the triangles consisting of two adjacent corners and the nearest interior point, and the smaller of the two degenerate triangles lying on the diagonal.

(Because this configuration has all triangles with a strictly larger perimeter, we can move any of the points around by epsilon if we wish to e.g. avoid degenerate triangles, or have all points lie in the interior of the square.)


问:
Cardinality: $\left|\Bbb N^{\{1,2\}}\right| = \left|\Bbb N^{\{1,2,3\}}\right|$
Left $F$ be the set of functions from $\{1,2\}$ to $\Bbb N$ and $G$ be the set of functions from $\{1,2,3\}$ to $\Bbb N$, prove that $\left|F\right| = \left|G\right|$.

Seems intuitive to me that the sets are cardinal equivalent as $\aleph_0^2 = \aleph_0^3$.
I've been trying to write a proof that is not based on cardinal number arithmetic. My approach was to show that $\left|F\right|\le\left|G\right|$ and $\left|G\right|\le\left|F\right|$ then according to Schröder–Bernstein theorm $\Rightarrow \left|F\right|=\left|G\right| \Rightarrow F\sim G$ .

It is clear that $\{1,2\}\subset\{1,2,3\}$ therefore $F\subseteq G$ therefore $\left|F\right|\le\left|G\right|$.

Another explanation is that for every $f\in F$ we can match $g\in G$ such as:
$$g(x) =
\begin{cases}
f(x) + 1 & x\in\{1,2\} \\
0  & x\in \{3\}
\end{cases}$$

I need help to prove the second part.

答: HINT: For $f\in\Bbb N^{\{1,2,3\}}$ define $\hat f\in\Bbb N^{\{1,2\}}$ by setting $\hat f(1)=f(1)$ and $\hat f(2)=2^{f(2)}\cdot3^{f(3)}$. Show that the map $f\mapsto\hat f$ is injective.


答:
You can reduce the exponent $\bmod \varphi(19)=18$, by Euler's theorem.

And now we can apply the Chinese Remainder Theorem:  $4\cdot2^{6k}\equiv4\bmod9$, since $\varphi(9)=6$  (Euler's theorem again),  and of course $4\cdot2^{6k}\equiv0\bmod2$.

So $4\cdot2^{6k}\equiv4\bmod{18}$.

Body:
number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$

attempt:

$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$

How I can continue from there to apply the Euler's theorem


问:
If $N,K\in\mathcal{F}$ are subgroups of $G$ with $K\unlhd G$, then $NK\in\mathcal{F}$.
This is Exercise 4.19 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

The set product of two subsets $S,T$ of a group $G$ is given by $$ST=\{st\in G\mid s\in S, t\in T\}.$$

Since definitions vary, Roman's book defines normal as follows:

Definition: A subgroup $H$ of $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

The Question:

Let $\mathcal{F}$ be a family of groups with the following properties:

If $G\in\mathcal{F}$ and $H\cong G$, then $H\in\mathcal{F}$

If $G\in\mathcal{F}$ and $N\unlhd G$, then $G/N\in\mathcal{F}$.

If $N\unlhd G$ and if $N\in\mathcal{F}$ and $G/N\in\mathcal{F}$, then $G\in\mathcal{F}$.

Prove that if $N\in\mathcal{F}$ and $K\in\mathcal{F}$ are subgroups of $G$ with $K\unlhd G$, then $NK\in\mathcal{F}$.

Thoughts:

Two previous questions of mine seem related and they're from the same book (and even the same set of exercises):

Find a property of groups that is inherited by quotients but not by subgroups.

Find a property of groups that is inherited by subgroups but not by quotients.

A guess of mine is that the isomorphism theorems come into play here, especially the second one (since we're aiming for $NK$), except that there's nothing about $N\cap K$ in $\mathcal{F}$.

This seems like it should have a simple answer.

Perhaps a more modest goal would be to show $G\in\mathcal{F}$ if at all possible; it might even be a crucial lemma.

Please help :)

答:
Note that $NK$ is a subgroup of $G$.

By the second isomorphism theorem we have $$NK/K \cong N/(N \cap K).$$

Since $N \in \mathcal{F}$ and $N \cap K \trianglelefteq N$, property 2 tells us that $N/(N \cap K) \in \mathcal{F}$.

Property 1 now tells us that $NK/K \in \mathcal{F}$.

Finally, since $K \in \mathcal{F}$ and $NK/K \in \mathcal{F}$, property 3 tells us that $NK \in \mathcal{F}$.


问:
Munkres Topology section 50 exercise 8(c), error?
The main problem is to prove the following theorem.

Let $X$ be a $\sigma$-compact Hausdorff space. If every compact subspace of $X$ has topological dimension at most $m$, then so does $X$.

Also, we have a definition.

$X$ is said to be $\sigma$-compact if there is a countable collection of compact subspaces of $X$ whose interiors cover $X$.

Let $\mathbb{A}$ be an open cover of $X$. The strategy is to find an open cover $\mathbb{B}$ of $X$ refining $\mathbb{A}$ that has order at most $m+1$.

The following is part (c), which I can't understand.

Suppose $n \geq 0$ and $\mathbb{B}_n$ is an open covering of $X$ refining $\mathbb{B}_0$ such that $\mathbb{B}_n$ has order at most $m+1$ at points of $X_n$. Choose an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$. Choose $f : \mathbb{C} \to \mathbb{B}_n$ so that $C \subset f(C)$. For $B \in \mathbb{B}_n$, let $D(B)$ be the union of those $C$ for which $f(C) = B$. Let $\mathbb{B}_{n+1}$ consist of all sets $B \in \mathbb{B}_n$ for which $B \cap X_{n-1} \neq \emptyset$, along with all sets $D(B)$ for which $B \in \mathbb{B}_n$ and $B \cap X_{n-1} = \emptyset$. Show that $\mathbb{B}_{n+1}$ is an open covering of $X$ that refines $\mathbb{B}_n$ and has order at most $m+1$ at points of $X_{n+1}$.

And the following are part (a), (b), (d) of the same question, respectively.

(a). Show that $X = \bigcup X_n$, where $X_n$ is compact and $X_n \subset$ Int $X_{n+1}$ for each $n$. Let $X_0 = \emptyset$.

(b). Find an open covering $\mathbb{B}_0$ of $X$ refining $\mathbb{A}$ such that for each $n$, each element of  $\mathbb{B}_0$ that intersects $X_n$ lies in $X_{n+1}$.

(d). Define $\mathbb{B}$ as follows: Given a set $B$, it belongs to $\mathbb{B}$ if there is an $N$ such that $B \in \mathbb{B}_n$ for all $n \geq N$.

(a), (b) is done, I solved them. But I don't understand what (c) is trying to do. It seems that I should inductively define $\mathbb{B}_n$ such that it refines $\mathbb{B}_0$ and has order at most $m+1$ at points of $X_n$. So at the beginning of (c), $\mathbb{B}_n$ satisfying the condition should be given and I should use it to make a $\mathbb{B}_{n+1}$ that refines $\mathbb{B}_n$ and has order at most $m+1$ at points of $X_{n+1}$. But at the second sentence, the question already assumes that a collection $\mathbb{C}$ satisfying this condition exists('Choose an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$'). It seems to be a nonsense. And I have no idea how to prove that the collection $\mathbb{C}$ satisfying the condition does exist.

答:
Let us start with

Lemma. Let $A \subset X$ be closed. Then $\dim A \le m$ if and only if for each open covering $\mathcal U$ of $X$ there exists an open covering $\mathcal V$ of $X$ which refines $\mathcal U$ and has order at most $m+1$ at all points of $A$. [The latter was not explicitly defined by Munkres, but it means that some point of $A$ lies in $m+1$ elements of $\mathcal V$ and no point of $A$ lies in more than  $m+1$ elements of $\mathcal V$.]

"If part" : Let $\mathcal U'$ be an open covering of $A$. For each $U' \in \mathcal U'$ choose an open $U \subset X$ such that $U \cap A = U'$. These $U$ together with $X \setminus A$ form an open covering $\mathcal U$ of $X$. There exists an open covering $\mathcal V$ of $X$ which refines $\mathcal U$ and has order at most $m+1$ at all points of $A$. Let $\mathcal V'$ be the set of all nonempty $V \cap A$ with $V \in \mathcal V$. This is an open covering of $A$. Since each $V$ admits $U \in \mathcal U$ such that $V \subset U$, the same is true for the intersections: $V \cap A \subset U \cap A$. Since $V \cap A \ne \emptyset$, we cannot have $V \subset X \setminus A$, hence we must have $U \cap A \in \mathcal U'$. Thus $\mathcal V'$ refines $\mathcal U'$. But by construction $\mathcal V'$ has order at most $m+1$.

"Only if part" : Let $\mathcal U$ be an open covering of $X$. Then the set $\mathcal U'$ of all nonempty $U \cap A$ with $U \in \mathcal U$ is an open covering of $A$. There exists an open covering $\mathcal V'$ of $A$ which refines $\mathcal U'$ and has order at most $m+1$. For each $V' \in \mathcal V'$ there exists $U_{V'} \in \mathcal U$ such that $V' \subset U_{V'} \cap A$. Let $V \subset X$ be open such that $V \cap A = V'$ and define $W_{V'} = U_{V'} \cap V$. The set of these $W_{V'}$ together with the set of all nonempty $U \setminus A$ with $U \in \mathcal U$ forms an open covering of $X$ which refines $\mathcal U$. By construction it has order at most $m+1$ at all points of $A$.

In (c) you start with an open covering $\mathbb{B}_n$ of $X$ which refines $\mathbb{B}_0$ and has order at most $m+1$ at points of $X_n$. Since $X_{n+1}$ is compact and thus has $\dim X_{n+1} \le m$, the lemma gives us an open covering $\mathbb{C}$ of $X$ refining $\mathbb{B}_n$ that has order at most $m + 1$ at points of $X_{n+1}$.


问:
How to show this tensor is not decomposable
Let $V$ a vector space over $\mathbb{F}$ with basis $B=\{e_1,\dots, e_n \}$ and $B'=\{e^1,\dots,e^n \}$ adjoint basis in $V^*$, Consider the following tensor:
$$t^{ij}=(i+j) $$
I need to see that is not decomposable. This is the proof of my professor:

$\delta=\sum_{i,j=1}^{n}(i+j)(e_i\otimes e_j), $ suppose $\delta=\left(\sum_{i=1}^{n}x_ie_i\right) \otimes \left(\sum_{i=1}^{n}y_j e_j \right)$. As $n\geq 2$, we have $x_1y_1=2$,$x_1y_2=3$, so $y_1/y_2=2/3$, but $x_2y_1=3$, $x_2y_2=4$ and $y_1/y_2=3/4$ that is a contradiction.

I am not sure why this numbers are related with the fact that is decomposable, tensors are pretty hard to me, can you help me to understand this?

答:
Let $\delta=\sum_{i,j=1}^n(i+j)(e_i\otimes e_j)$, as you have. By definition, a decomposable element of $V\otimes V$ is one of the form $v\otimes w$, for some $v,w\in V$. To show that $\delta$ is not decomposable, we therefore need to show that $\delta\neq v\otimes w$, for any $v,w\in V$. On the other hand, since $B$ is a basis of $V$, every element of $V$ is of the $\sum_{i=1}^nx_ie_i$, for some $x_i\in \mathbb{F}$. Thus it suffices to show that $\delta\neq\left(\sum_{i=1}^nx_ie_i\right)\otimes\left(\sum_{j=1}^ny_je_j\right)$ for any $x_i,y_i\in\mathbb{F}$.

Assume for contradiction therefore that $\delta$ is of that form, for some $x_i$ and $y_i$. We can expand the second expression as follows, using the bilinear and $\mathbb{F}$-balanced properties of the tensor product: \begin{align}
\left(\sum_{i=1}^nx_ie_i\right)&\otimes\left(\sum_{j=1}^ny_je_j\right)=\sum_{i=1}^n\left(x_ie_i\otimes \sum_{j=1}^ny_je_j\right)
\\ &=\sum_{i,j=1}^n(x_ie_i\otimes y_je_j)=\sum_{i,j=1}^n(x_iy_j)(e_i\otimes e_j).
\end{align}
If this equals $\delta=\sum_{i,j=1}^n(i+j)(e_i\otimes e_j)$, then – because $\{e_i\otimes e_j\}_{i,j\leqslant n}$ is a basis for $V\otimes V$ – we must have $x_iy_j=i+j$ for every $i,j\leqslant n$.

In particular, we have

            $\bullet$ $x_1y_1=1+1=2$                         $\bullet$ $x_2y_1=1+2=3$

            $\bullet$ $x_1y_2=2+1=3$                         $\bullet$ $x_2y_2=2+2=4$

This gives  that $$\frac{y_1}{y_2}=\frac{x_1y_1}{x_1y_2}=\frac{2}{3}\ \ \ \ \ \ \ \ \text{ and }\ \ \ \ \ \ \ \ \frac{y_1}{y_2}=\frac{x_2y_1}{x_2y_2}=\frac{3}{4},$$ which gives the desired contradiction.

(In dividing by $3$ and $4$, we are assuming here that $\mathbb{F}$ does not have characteristic $3$ or $2$. I suspect you are probably assuming that $\mathbb{F}$ has characteristic $0$, so I assumed this too. However, do correct me if I'm wrong; the result still holds no matter what $\operatorname{char}(\mathbb{F})$ is, we just need a tiny bit of extra casework in the case where it equals $2$ or $3$.)


问:
Extention of an Amenable Group by an Amenable Group is Amenable
I would like to prove that if $H\subset G$ is a normal amenable subgroup such that $G/H$ is amenable, then $G$ is amenable. The definition of amenability that I'm using is the following:

A group $G$ is amenable if every action of $G$ by homeomorphisms of a compact metric space admits an invariant probability measure.

This definition can be found on Navas's "Groups of Circle Diffeomorphisms". I've tried a lot of different ways but i couldn't prove it, I know there are many equivallent definitions for amenability but I would like (if possible) a proof that only uses this definition.

Here is what I've done so far: If $G$ acts on $(M,d)$ then $G/H$ acts on $M/H$ (the quotient of $M$ by the orbits of $H$), the problem is that this group is not necessarily metric, whe could endow the quotient group with the pseudometric $d'$ given in wikipedia https://en.wikipedia.org/wiki/Metric_space#Quotient_metric_spaces (the topology could be weaker than the quotient topology), and then do another quotient $X=(M/H)/\sim$ where $[x]\sim [y]$ if $d'([x],[y])=0$. Here $X$ is a compact metric space and we could take the action of $G/H$ on $X$ given by ${[g]}({[[x]]})=[[y]]$ if $[[g(x)]]=[[y]]$, since $G/H$ is amenable there exists an invariant probability measure, namely $\nu$. Now the sets $A_{[[x]]}=\lbrace y\in M:[[y]]=[[x]]\rbrace$ are compact and invariant under the action of $H$, so each one has an invariant probability measure namely $\mu_{[[x]]}$ and we could define the probability measure on $M$ as
$$\mu(B)=\int_X \mu_{[[x]]}(B\cap A_{[[x]]})d\nu([[x]]).$$

I don't know if this works in general, i couldn't prove or disprove it, i suppose this doesn't work since there could be some internal shifting of the orbits of $H$ in the sets $A_{[[x]]}$, but I hope this gives you some insight of what I'm trying so far.

I hope I was clear, many thanks in advance.

Something that might help: The space of probability measures on a metric space is compact, so you could use convergence of probability neasures.

答:
I think, the equivalence of Navas' definition and the standard notion of amenablity is called Bogolyubov-Dey theorem. You can find it in many places, see for instance Proposition 3.6 in

Grigorchuk, Rostislav; de la Harpe, Pierre, Amenability and ergodic properties of topological groups: from Bogolyubov onwards, Ceccherini-Silberstein, Tullio (ed.) et al., Groups, graphs and random walks. Selected papers of the workshop, Cortona, Italy, June 2–6, 2014 on the occasion of the 60th birthday of Wolfgang Woess. Cambridge: Cambridge University Press (ISBN 978-1-316-60440-3/pbk; 978-1-316-57657-1/ebook). London Mathematical Society Lecture Note Series 436, 215-249 (2017). ZBL1397.43001.

(Read here for a free version.) Given this result, you can use many of the available proofs of the fact that the class of amenable groups is closed under extensions, e.g. here or one of many other books dealing with amenable groups.

Edit. It is clear from the context of the book that Navas defines amenability (and, for instance, property T) only for groups equipped with discrete topology. It is unfortunate that he never mentions amenability in the context of topological groups (equipped with nondiscrete topology), uses a nonstandard definition of amenability and provides no references (as far as I can tell) for a general textbook treatment of amenable groups (and there are several of these, see references here, at least in the case of locally compact groups which includes discrete groups).


问:
Expected Value of Flips Until HT Consecutively
Suppose you ﬂip a fair coin repeatedly until you see a Heads followed by a
Tails. What is the expected number of coin ﬂips you have to ﬂip?

By manipulating an equation based on the result of the first flip, shown at this link:

http://www.codechef.com/wiki/tutorial-expectation

the answer is 6. This also makes sense intuitively since the expected value of the number flips until HH or TT is 3. But is there a way to tackle this problem by summing a series of probabilities multiplied by the values?

Thank you!

答:
Surprised no one mentioned the absorbing Markov chain, which elegantly solves the problem.
https://en.wikipedia.org/wiki/Absorbing_Markov_chain#String_generation</p>


Title: prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$

Body:
Induction:
\begin{align*}2^{2^{6(k+1)}\cdot 4}=2^{2^{6k+6}\cdot 4}=\left(2^{2^{6k}\cdot 4}\right)^{64}&\equiv (2^4) ^{64}\mod 19\\&\equiv 2^{256}\mod19\end{align*}
By Fermat's little theorem, $2^{18}\equiv 1\mod 19$. Since $256=18*14+4$, then
\begin{align*}2^{2^{6(k+1)}\cdot 4}&\equiv (2^{18})^{14}2^4\mod19\\&\equiv 1^{14}2^4\mod 19\\&\equiv 2^4\mod19\end{align*}


答:
well if $2^{6k}4\equiv m \pmod {18}$ then $2^{2^{6k}4}\equiv 2^m \pmod {19}$.

$2^{6k}4 \equiv 0 \pmod 2$ and $\phi(9) = 6$ so $2^6 \equiv 1 \pmod 9$ so $2^{6k}4\equiv 4 \pmod 9$.  So by Chinese remainder theorem there is one solution $\pmod {18}$ to $2^{6k}4\equiv 0 \pmod 2$ and $6^{6k}4\equiv 4 \pmod 9$.

And that unique solution is $4$

And that's that.

$2^{6k}4 \equiv 4 \pmod{18}$.  So there exists an $M$ so that $2^{6k}4=18M + 4$ so $2^{2^{6k}4}=2^{18M + 4} \equiv 2^4 \pmod{19}$.


问:
proof $x-1$ is always invertible in $\mathbb{Z}_{^3}$.
in $a\in\mathbb{N}$ we know that $x>1$. How can I prove the question above? I see the proof from this question, $x+1$ is always invertible in $\Bbb Z_{x^3}$.

But is the answer the same when the sign is just flipped?

答:
$x^3-1 = (x-1)(x^2+x+1)$ , we know that $(x^2+x+1)$ is never $0$ since $0<1<(1+x+x^2)$.

Later, $x^3 = 1+ (x-1)(x^2+x+1)$, so
$$1+ (x-1)(x^2+x+1) \equiv 0 ~~{ mod}(x^3) $$
$$(x-1)(x^2+x+1) \equiv -1 ~~{ mod}(x^3) $$
$$(x-1)(-x^2-x-1) \equiv 1 ~~{ mod}(x^3) $$
so $x-1$ is always invertible.

A Better one:
Since $x^3 = 1+ (x-1)(x^2+x+1)$, then $$x^3 + (x-1)(-x^2-x-1) = 1$$
so, for Bezout's theorem, $x^3$ and $(x-1)$ are always coprime, and hence, $(x-1)$ is invertible .

Note: This observation was taken from one answer of the link attached to question.


问:
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$
I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$

I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2}  \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?

Body:
Note

$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} dx
= \int_{0}^{1} \frac{1+\frac1{x^{2}}}{x^{2}+\frac1{x^2}} dx
=\int_0^1 \frac {d(x-\frac1x)}{(x-\frac1x)^2+2}
= \frac\pi{2\sqrt2}
$$


问:
if $\mathbf x$ is sampled randomly from a hypercube on $R^n$, what is the probability density for $|\mathbf x| = d$
if the vector $\mathbf x$ is sampled randomly from a uniform distribution on $[0, 1]^d$, what is the probability density function for $|\mathbf x|$? Is it easy to scale for $[0, n]^d$?</p>

答:
I'll try to explain and expand mark's comment.

Think for a momnet in two dimensions. The probability density that $|x|=r$ is the probability that a uniformly chosen point lies at a distance $r$ from the origin. This equals the length of an arc of a circle of radius $r$, intersected with the unit box. See the circles in the drawing:

It is therefore clear that for $r<1$ the PDF will be $\frac{\pi r}{2}$. For $r>1$ simple trigonometry shows that this arclength is $\frac{\pi}{2}-2\cos^{-1}\left(\frac{1}{r}\right)$. So we get that for 2D
$$f(r)=\begin{cases}
 \frac{\pi }{2}r & r<1 \\
 \frac{\pi }{2}-2\cos^{-1}\left(\frac{1}{r}\right) & r\geq 1 \\
\end{cases}$$
Which is the result that WimC gave, but simplified.

In a general dimension, finding the hyper-area of the intersection of the $d-1$ sphere with the hypercube is pretty complicated for $r>1$. However, it is quite simple for $r<1$ - the sphere is completely contained in the first hyper-quadrant, so its area is simply it's total area divided by $2^d$. We get that for a general dimension $d$ the PDF is

$$f(x)=\begin{cases}
 \frac{v_{d-1}}{2^d}r^{d-1} & r<1 \\[5mm]
 ??? & r\geq 1 \\
\end{cases}$$

Where $v_{d-1}$ is the area of the unit $d-1$ sphere.


问:
Does this series diverge? If so, any ideas on how to prove it?
I suspect that the series
$$\sum_{n=1}^\infty\left(\frac{\cos(2\pi n/3)}{\ln(1+n)}\right)^3$$
diverges, but having used all the criterion for divergence I can think of, I haven't come up with anything yet. Note that for $n=1,2$ the numerator is $-1/8$, and for $n=3$ it is 1. I suspect it diverges because we don't even have the conditions to apply the Dirichlet criterion, since the partial sums of $\cos(2\pi n/3)$ are not bounded.

Does anyone have an idea?

答:
Indeed, the series diverges. Notice that $\frac1{\ln(1+3n)^3},\frac1{\ln(1+3n+1)^3},\frac1{\ln(1+3n+2)^3}\sim\frac1{\ln(n)^3}$ are the terms of diverging series by Bertrand's criterion. Setting $S_n=\sum_{k=2}^n\frac1{\ln(k)^3}$ we have $\sum_{k=1}^n\frac1{\ln(1+3k)^3}, \sum_{k=1}^n\frac1{\ln(1+3k+1)^3}, \sum_{k=1}^n\frac1{\ln(1+3k+2)^3}\sim S_n$ and so
$$\sum_{k=1}^{3n}\left(\frac{\cos(2\pi k/3)}{\ln(1+k)}\right)^3=S_n-\frac28S_n+o(S_n)=\frac34S_n+o(S_n)$$
which diverges to $+\infty$.


问:
Number of groups of Lie Type of order $n$
As a combined math/programming exercise, I want to make a function that takes in a natural number $n$ and returns the number of simple groups of order $n$. From the classification of simple finite groups, every simple finite group is cyclic, alternating, Lie type or sporadic.

So far,

Cyclic is straightforward, since there always is one cyclic group of order $n$.

Alternating is straightforward, since there is one alternating group of order n if $n=m!/2$ for some $m$, and $0$ otherwise.

Since there are just $26$ sporadic groups, I can do them case by case. The main challenge of these will be storing the size of the monster as an integer.

But how many are there of Lie type of order $n$? Is there a straightforward formula for calculating this?

As a bonus, if I successfully create this function (takes in $n$ and returns the number of simple groups of order $n$) how could I extend it to a function that takes in a number $n$ and returns the number of groups of order $n$?

答:
The orders of the groups of Lie type are polynomials in the size of the underlying field with the degree depending on the rank. You can find formulas at https://en.wikipedia.org/wiki/List_of_finite_simple_groups

So you could calculate through all series and check which once come up for a given order.
Since there are quite a few series, it will be a bit of a slog to go through all cases. Be aware that the group orders are not monotonous in the field order, e.g. $L_2(128)$ is larger than $L_2(157)$.

You might want to check/compare your results with:
http://www.madore.org/~david/math/simplegroups.html


问:
$xy''-(2x^2+1)y'=x^5e^{x^2}$
$xy''-(2x^2+1)y'=x^5e^{x^2}$

Hi there,
needs help with the second order ODE.
I've tried (for the homogenous part) to substitute $\frac{dy}{dx} =u$ to get $ u'-(2x+\frac{1}{x})u=0$

solving for the first non trivial solution

$y_1=C_1 e^{x^2}$
however with $y_2$ and the particular solution i stuck.
asks for help.

答:
$$xy''-(2x^2+1)y'=x^5e^{x^2}$$
$$(xy''-y')-2x^2y'=x^5e^{x^2}$$
$$\dfrac {(xy''-y')}{x^2}-2y'=x^3e^{x^2}$$
$$\left (\dfrac {y'}{x}\right)'-2y'=x^3e^{x^2}$$
Better to substitute $u=x^2$:
$$\left (2 {y'}\right)'-2y'=\dfrac 12ue^{u}$$
$$\left ( {y'e^{-u}}\right)'=\dfrac u4$$
$$y'e^{-u}=\dfrac {u^2}8+c_1$$
$$y'=\dfrac {u^2}8e^u+c_1e^u$$
$$y(u)=\dfrac 18\int  {u^2}e^u \, du+c_1e^u+c_2$$

Edit1

$$\left (\dfrac {y'}{x}\right)'-2y'=x^3e^{x^2}$$
Note that:
$$y'=\dfrac {dy}{dx}=\dfrac {dy}{dx^2}\dfrac {dx^2}{dx}=2x\dfrac {dy}{du}$$
$$\implies \left (\dfrac {y'}{x}\right)=2\dfrac {dy}{du}$$
And:
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d}{dx}\dfrac {dy}{du}$$
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d}{du}\dfrac {dy}{du}\dfrac {du}{dx}$$
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d^2y}{du^2}\dfrac {du}{dx}$$

And $\dfrac {du}{dx}=2x$ since $u=x^2$.

Now we have: $$
4x\dfrac {d^2y}{du^2}-4x\dfrac {dy}{du}=x^3e^{x^2}$$
Divide the equation by $x$ you have:
$$
4\dfrac {d^2y}{du^2}-4\dfrac {dy}{du}=x^2e^{x^2}$$
Substitute $u=x^2$
$$
4\dfrac {d^2y}{du^2}-4\dfrac {dy}{du}=ue^{u}$$
$$
\dfrac {d^2y}{du^2}e^{-u}-\dfrac {dy}{du}e^{-u}=\dfrac u 4$$
So that the differential equation is simply:
$$
\boxed {(y'e^{-u})'=\dfrac u 4}$$
And it's easy to solve it.

Title: $xy''-(2x^2+1)y'=x^5e^{x^2}$


问:
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$
I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$

Is it OK how I want to do?

$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$

I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?

I thinks is not because  if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.

Thanks :)

答:
Your approach is correct: The cosine part keeps oscillating, but it's bounded between -1 and 1, while the sine goes to zero due to its argument going to zero. Their product thus goes to zero too (even when multiplied by 2).

The final remark about bounding the value of sine is valid (in the sense of the inequality being true), but has no relevance -- if you know the sine and cosine are bounded between -1 and 1, you can conclude that their product falls between these bounds too (which is true, but doesn't provide much information about the actual limit or its existence).


答:
As a bonus, if I successfully create this function (takes in n and returns the number of simple groups of order $n$) how could I extend it to a function that takes in a number $n$ and returns the number of groups of order $n$?

This is hopeless, we don't know a simple way to calculate the number of groups of order $n$ even if $n$ is a prime power (in which case the only simple factors are cyclic). It would require you to solve the group extension problem which is quite hard.

To give a concrete example, it's currently not known how many groups there are of order $2^{11}=2048$.

Title: Number of groups of Lie Type of order $n$

Body:
As a combined math/programming exercise, I want to make a function that takes in a natural number $n$ and returns the number of simple groups of order $n$. From the classification of simple finite groups, every simple finite group is cyclic, alternating, Lie type or sporadic.

So far,

Cyclic is straightforward, since there always is one cyclic group of order $n$.

Alternating is straightforward, since there is one alternating group of order n if $n=m!/2$ for some $m$, and $0$ otherwise.

Since there are just $26$ sporadic groups, I can do them case by case. The main challenge of these will be storing the size of the monster as an integer.

But how many are there of Lie type of order $n$? Is there a straightforward formula for calculating this?

As a bonus, if I successfully create this function (takes in $n$ and returns the number of simple groups of order $n$) how could I extend it to a function that takes in a number $n$ and returns the number of groups of order $n$?


问:
Explanation in the proof of De Rham Theorem
With this Lemma in "Basic Algebraic Topology" by Shastri, the author want prove the De Rham Theorem.

Can anyone help me to prove the underline claim?

答:
This claim is false.  For instance, consider the set$$V=(0,1/2)\cup(1/2,2/3)\cup(2/3,3/4)\cup\dots\subset\mathbb{R}.$$  The closure of $V$ is $[0,1]$ which is compact, but $V$ cannot be written as a finite union of convex open sets, since it has infinitely many connected components.  (In higher dimensions, you can similarly construct examples that are connected but whose higher homology groups are not finitely generated, so you cannot even use axiom (iii) to rescue the argument.)

To fix the proof, note that $f^{-1}([i-1/2,i+1/2])$ is compact and thus can be covered by finitely many open balls that are all contained in $U_i$.  The union of these open balls is then a set $U_i'$ which is a finite union of convex open sets, and you can replace $U_i$ with $U_i'$ for the rest of the proof.


问:
For the following question show that locus is a circle.
Two circles touch one another internally at A, and a variable chord PQ of the outer circle touches the inner circle. Show that the locus of the incentres of the triangle APQ is another circle touching the given circles at A.

First I found the equation of the chord which is also the tangent to the smaller circle. Then I thought a lot on how to continue further;  but literally have no idea how to do it. Even if we use the formula of the incetre of a triangle then we have to find the length of the sides of the triangle which is damn tedious.
Please help me with shorter method.

答:
Given two circles internally tangent at $A$, and any chord $PQ$ in the greater circle tangent to the lesser circle, prove that the incenter of triangle $APQ$ lies on a circle which is also internally tangent at $A$.

$B$ is center of the greater circle, $AC$ the diameter of the lesser, and $D$ the point of tangency.

Join $AD$ and extend it to $F$. Join $DC$ and $FB$ intersecting at $J$.

Let $QG$ bisect $\angle AQP$, and draw $GH\parallel DC$.

Since$$\angle JDE=\angle CAD=\angle JFD$$and complements$$\angle DJF=\angle DCA$$then$$\triangle ADC\sim\triangle FDJ\sim\triangle DEJ$$Therefore $\angle JED$ is also right, making $QE=EP$, arc $QF$=arc $PF$, and$$\angle QAF=\angle PAF$$

Hence $G$ is the incenter of $\triangle APQ$.

And since by angle bisection and parallels$$\frac{QA}{QD}=\frac{GA}{GD}=\frac{HA}{HC}$$and point $C$ is fixed, then $H$ is also fixed.

And since $\angle AGH$ is right, it follows that incenter $G$ lies on the circumference of a circle with diameter $AH$ and internally tangent at $A$.


问:
Indexing set for partition
On page 3/(4) of Dummit&Foote's Abstract Algebra (3rd ed.):

(Let $A$ be a nonempty set) A partition of $A$ is any collection $\{A_i\lvert i\in I\}$ of nonempty subsets of $A$ ($I$ some indexing set) such that (a) $A=\bigcup_{i\in I}A_i$, and (b) $A_i \cap A_j=\emptyset$, for all $i,j\in I$ with $i\neq j$. i.e., $A$ is the disjoint union of the sets in the partition.

May I ask, by the word indexing in this context, if the indexing set can contain infinitely many elements? In particular, can it contain uncountably many elements? (The answer to this question is probably "yes")

Furthermore, can the function from $I$ to those nonempty subsets of $A$ (i.e. the process of indexing) not be injective? More importantly, are there cases where we intentionally want it not to be injective? And why? [For this question we shall modify the definition (b) to "$A_i \cap A_j=\emptyset$, for all $A_i,A_j\in A$ with $A_i\neq A_j$"]

Some stranger questions:

Am I allowed to have $I=A$? If so, is it meaningful? (A kind of self-indexing?)

Is that possible, by any chance, to have some partition of some set being unable to be indexed?

答:
In general an indexing set can be any set at all. Yes, one member of a partition can have more than one index, though the default would be an injective map. Yes, it’s possible to index a partition of $A$ by $A$ itself: if $\mathscr{P}$ is a partition of $A$, for each $a\in A$ let $A_a$ be the unique member of $\mathscr{P}$ containing $a$. In this case the indexing is clearly not injective unless $\mathscr{P}$ is the partition of $A$ into singletons. This of course shows that every partition can be indexed. In fact, every partition can be injectively indexed, since it can trivially be indexed by itself: the function is $$f:\mathscr{P}\to\mathscr{P}:P\mapsto P\,,$$ or, in the notation of Dummit & Foote’s definition, $A_P=P$ for each $P\in\mathscr{P}$. This is pretty pointless, but it is technically an indexing of the partition.

I dislike that definition, because, as some of what I’ve written may suggest, the indexing is completely unnecessary. A partition $\mathscr{P}$ of a non-empty set $A$ is simply a family of pairwise disjoint, non-empty subsets of $A$ such that $\bigcup\mathscr{P}=A$. Any surjective function $f$ from a set $I$ to $\mathscr{P}$ can then be used to index $\mathscr{P}$: for each $i\in I$ we have $P_i=f(i)$.

Title: Indexing set for partition


答: It doesn't pass the divergence test.  Use L'hopital on the general term:  $|\dfrac{-2\pi/3\sin(2\pi k/3)}{1/(1+k)}|\to\infty$.

Title: Does this series diverge? If so, any ideas on how to prove it?

Body:
I suspect that the series
$$\sum_{n=1}^\infty\left(\frac{\cos(2\pi n/3)}{\ln(1+n)}\right)^3$$
diverges, but having used all the criterion for divergence I can think of, I haven't come up with anything yet. Note that for $n=1,2$ the numerator is $-1/8$, and for $n=3$ it is 1. I suspect it diverges because we don't even have the conditions to apply the Dirichlet criterion, since the partial sums of $\cos(2\pi n/3)$ are not bounded.

Does anyone have an idea?


问:
Why does $A^TAw = A^Tb$ describe the least square solution to a multivariate system?
Note that this builds off a previous question, found here.

Suppose I have a system of equations of the form:

$$Wv_1 + Xv_2 + Yv_3 + Zv_4 = b$$

Where $v_1$, $v_2$, $v_3$, $v_4$, and $b$ are vectors. I want to determine the least squares solution for unknowns $W,X,Y,Z$. As I understand it, the solutions are the vector $w$ in:

$$A^{T}Aw = A^{T}b$$

Where $A$ is the matrix who's columns are $v_1$, $v_2$, $v_3$, $v_4$.

It isn't clear to me, however, why this is. Why is $w$ the least-squares solution, mathematically and intuitively? I've tried to do some research online, however I haven't found much (and what I have found uses extensively terminology with which I'm unfamiliar— unfortunately, this isn't a subject I'm particularly familiar with).

答:
If you have $Ax =b$ then the least squares solution can be taken from the normal equations which are given by multiplying by the transpose on each side. This makes a square matrix.

$$ A^TAx = A^Tb$$

Now the least squares problem is

$$ f(x) =  \|  b - Ax \|_{2}^{2} = (b-Ax)^T (b-Ax) = b^Tb - x^TA^Tb - b^TAx  + x^TA^TAx$$

this comes from how $\| \cdot \|$ works. $ b - Ax = r$ which is the residual and $\| r \|_{2}^{2} = rr^{T}$ so they are attempting to minimize the residual. You can distribute $(b-Ax)^T(b-Ax)$ and you'll get that above.

How is it achieved? If $f(x)$ is a global minimum then $ \nabla f(x) = 0$ and

$$ \nabla (x^TA^Tb) = A^Tb \\ \nabla(b^TAx) = A^Tb \\ \nabla (x^TA^TAx) = 2A^TAx$$

which gives us
$$ \nabla f(x) = 2A^TAx -  2A^Tb$$

so we solve that for 0


问:
Scale rectangles so they have same height and don't exceed a total width?
I have three rectangles of different sizes side by side.

I want to scale them all (maintaining their aspect ratio) so they have the same height and don't exceed a total width.

I know I could find the optimal height by trial and error, but I'd like to find a formula that will work no matter what the sizes of the rectangles.

Can anyone point me in the right direction?

答:
First, scale them all to the same height (tallest, first, some constant).

Second, add the (new) widths and divide the sum by the desired total width.

Third, scale them all by the inverse of this factor.


问:
Simple algebraic step in establishing estimation of variance of errors, $e_{i}$ i.e $\sigma^{2}$ for variables in linear regression
This is something I definitely know how to do, but is slipping my mind at the moment. I'm reading through Rice's Mathematical Statistics and the following is the proof to establish an estimate for the errors of the independent variables of a linear regression (You can ignore the majority of the proof it is only the last line I'm stuck on):

I understand the proof of $E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2}) = (n-p) \sigma^{2}$$. What I'm having trouble figuring out is how the

$$s^{2} = \frac{||\textbf{Y} - \hat{\textbf{Y}} ||^{2}}{(n-p)}$$

Came about, specifically what was on the other side to allow us to divide?

$$\text{?something?} = E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2}) = (n-p) \sigma^{2}$$

I ask because if we're taking the expectation of $E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2})$, then the only thing I see appearing on the other side would be:

$$\sum_{i = 1}^{n}E(\textbf{Y} - \hat{\textbf{Y}})^2$$

Which to me doesn't amount to anything special. Since what I want is

$$||\textbf{Y} - \hat{\textbf{Y}} ||^{2}$$
without the expectation operator.

答:
$s^2$ is an unbiased estimator of $\sigma^2$ (i.e., its expectation equals the parameter of interest) because
$$
\mathsf{E}s^2=\mathsf{E}\left[\frac{\|\bf{Y}-\hat{\bf{Y}}\|^2}{n-p}\right]=\frac{(n-p)\sigma^2}{n-p}=\sigma^2.
$$


问:
Finding the n'th $(n\geq2)$ derivative of the equation $\frac{x^3}{2x^2 - x - 3}$
I tried spotting some pattern after derivating it a few times, but I couldn't find any. Maybe I should use the Leibniz formula somehow?</p>

答:
Polynomial long division gives:

$$\frac{x^3}{2x^2 - x - 3} = \frac{1}{2}x+\frac{1}{4}+\frac{\frac{7}{4}x+\frac{3}{4}}{2x^2-x-3}$$

and now use partial fractions on the last term:

$$\frac{1}{4} \frac{7x+3}{(2x-3)(x+1)} = \frac{1}{4} \left( \frac{A}{2x-3} + \frac{B}{x+1} \right)$$
$$A(x+1) + B(2x-3) = 7x+3$$
$$x = -1 \Rightarrow B = \frac{4}{5} \ ; \ x = \frac{3}{2} \Rightarrow A = \frac{27}{5}$$

and therefore:
$$\frac{x^3}{2x^2 - x - 3} = \frac{1}{2}x+\frac{1}{4}+\frac{1}{4} \left(\frac{27/5}{2x-3} +  \frac{4/5}{x+1} \right)$$

Treat the first derivative as a special case. All other derivatives can be found using the power rule and chain rule.


问:
Boolean Algebra: Demonstrate that the pentagon lattice is non-distributive
I just started learning Boolean Algebra and have this homework question

Demonstrate that the pentagon lattice is non-distributive

I know this is non-distributive because $b$ complements $a$ and $c$. So when I check if distributive law holds for $ a \lor (b \land c) = (a \lor b) \land (a \lor c)$.  I get

$$ a \lor (b \land c) = (a \lor b) \land (a \lor c)$$
$$ a \lor 0 = (a \lor b) \land (a \lor c)$$
$$ a = (a \lor b) \land (a \lor c)$$
$$ a = 1 \land c $$
$$ a = c $$

Since $ a \neq c$ the pentagon is not distributive. But why is it if I use the other law of distribution that I don't get the same answer? I would think they both would fail if the lattice is non-distributive. What am I doing wrong here?

$$ a \land (b \lor c) = (a \land b) \lor (a \land c)$$
$$ a \land 1 = (a \land b) \lor (a \land c)$$
$$ a = 0 \lor a$$
$$ a = a $$

答:
This is very simple. If try to find the complement of 'b' you will get 'a' and 'c'. And any element in a distributve lattice should have atmost one complement. Distributive property comes from set theory where every element has atmost one complement and if lattice has an element with more than one complement then the lattice will not follow distributive property.

How to find complement?
ans: If any element two elements 'a' and 'b' such that a∨b = upper bound of lattice and a∧b = lower bound of lattice, a and b are complements of each other.

what are upper bound and lower bound?
Ans:  upper bound : an element to which every element relates.
Lower bound :this element relates to every element.


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

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

答:
This is because many phenomena in nature can be described by well-defined fields. When we look at the boundaries of some surface enclosing the fields, it tells us everything we need to know.

Take a conservative field, such as an electric or gravitational field.  If we want to know how much energy needs to be added or removed to move something from one spot in the field to another, we do not have to look at the path. We just look at the initial and final location, and the potential energy equation for the field will give us the answer.

The endpoints of a path are boundaries in a 1D space, but the idea extends to surfaces bounded by curves and so on.

This is not so strange. For instance, to subtract two numbers, like 42 - 17, we do not have to look at all the numbers in between, like 37, right? 37 cannot "do" anything that will wreck the subtraction which is determined only by the values at the boundary of the interval.


问:
Does every topology have singletons?
I have seen many topology proofs where if $X$ is the space in question, they say things like for some $x \in X$...., but how do we know the space even contains the singleton $\{x\}$ to be doing that?

答: If $X$ is non-empty, it contains elements $x \in X$.

Title: Does every topology have singletons?


答: $x \in X $ does not imply the topology on $X$ contains $\{x\}$.

Title: Does every topology have singletons?

Body: I have seen many topology proofs where if $X$ is the space in question, they say things like for some $x \in X$...., but how do we know the space even contains the singleton $\{x\}$ to be doing that?


答:
Consider $A\in\mathbb{R^{m\times n}}$ and $b\in\mathbb{R^{m}}$ The least squares problem defined as seeking $x\in\mathbb{R^{n}}$ so that $\|b-A x\|_{2}=\min _{y \in \mathbb{R}^{n}}\|b-A y\|_{2}$ is equivalent to minimizing the function $\Phi(x)=\|b-A x\|_{2}^{2}$ over $\mathbb{R}^{n}$ such
that
$$
\begin{aligned}
\Phi(.) &: \quad \mathbb{R}^{n} \rightarrow \mathbb{R}^{+} \cup\{0\} \\
\Phi(x) &=\|b-A x\|_{2}^{2}=(b-A x)^{T}(b-A x)=b^{T} b-b^{T}(A x)-(A x)^{T} b+(A x)^{T}(A x) \\
&=\|b\|_{2}^{2}-2(A x)^{T} b+x^{T} A^{T} A x \\
&=\|b\|_{2}^{2}+2 \Psi(x)
\end{aligned}
$$
where $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^{m}$ are given. Thus $,\|b\|_{2}^{2}$ is a constant and minimizing $\Phi(x)$ is equivalent to minimizing
$$
\Psi(x)-\frac{1}{2} x^{T} A^{T} A x-x^{T} A^{T} b
$$
To prove this equivalence, we note that : \begin{aligned}
\Psi(x) &=& \frac{1}{2}(A x)^{T}(A x)-x^{T}\left(A^{T} b\right)=\frac{1}{2} y^{T} y-x^{T} c & & \text { where } y=A x \\
&=& \frac{1}{2} \sum_{i=1}^{m} y_{i}^{2}-\sum_{j=1}^{n} c_{j} x_{j} \\
&=& \frac{1}{2} \sum_{i=1}^{m}\left(\sum_{j=1}^{n} A(i, j) x_{j}\right)^{2}-\sum_{j=1}^{n} c_{j} x_{j} & & \text { since } y_{i}=\sum_{j=1}^{n} A(i, j) x_{j}
\end{aligned}
Thus, $\Psi(x)=\Psi\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a quadratic function of the unknowns $x_{1}, x_{2}, \cdots, x_{n} .$ To find the minimum of $\Psi(x)$
it should be such that :
$$
\nabla \Psi(x)=\left[\begin{array}{c}
\frac{\partial \Psi}{\partial x_{1}}(x) \\
\frac{\partial \Psi}{\partial x_{2}}(x) \\
\vdots \\
\frac{\partial \Psi}{\partial x_{n}}(x)
\end{array}\right]=0 $$ $$ \begin{aligned}
\Longleftrightarrow\frac{\partial \Psi}{\partial x_{k}}(x) &=\sum_{i=1}^{m} A(i, k)\left(\sum_{j=1}^{n} A(i, j) x_{j}\right)-c_{k}=0, \quad \text { for } \quad k=1,2, \cdots, n \\
&=\sum_{i=1}^{m} A(i, k)(A(i,:) x)-c_{k}=0 \\
&=\sum_{i=1}^{m} A^{T}(k, i) y_{i}-c_{k}=0 \\
&=A^{T}(k,:) y-c_{k}=0
\end{aligned} $$

$$
$$

$$
\therefore\left\{\begin{array}{l}
A^{T}(1,:) y=c_{1} \\
A^{T}(2,:) y=c_{2} \\
\vdots \\
A^{T}(n,:) y=c_{n}
\end{array} \quad \Longleftrightarrow \quad A^{T} y=c \quad \Longleftrightarrow \quad A^{T} A x=A^{T} b\right.$$

Now likewise, We will show that if $x$ solves the normalized system $A^{T} A x=A^{T} b,$ then it solves the Least squares problem.
For any $y \in \mathbb{R}^{n}$ we have that
$$
\begin{aligned}
\Psi(x)-\Psi(y) &=\frac{1}{2} x^{T} A^{T} A x-x^{T}\left(A^{T} b\right)-\frac{1}{2} y^{T} A^{T} A y+y^{T}\left(A^{T} b\right) \\
&=\frac{1}{2} x^{T} A^{T} A x-x^{T}\left(A^{T} A x\right)-\frac{1}{2} y^{T} A^{T} A y+y^{T}\left(A^{T} A x\right) \\
&=-\frac{1}{2} x^{T} A^{T} A x-\frac{1}{2} y^{T} A^{T} A y+y^{T} A^{T} A x \\
&=-\frac{1}{2}(x-y)^{T} A^{T} A(x-y) \\
&=-\frac{1}{2}\|A(x-y)\|_{2}^{2} \leqslant 0 \\
\therefore \Psi(x) & \leqslant \Psi(y) \quad \text { for any } y \in \mathbb{R}^{n}
\end{aligned}
$$
Thus the existence of solution/s to the least squares problem reduces to studying the existence of solution/s to the
system
$$
A^{T} A x=A^{T} b
$$
where $G=A^{T} A$ is an $n \times n$ symmetric positive semi-definite and $\operatorname{rank}\left(A^{T} A\right)=\operatorname{rank}(A)=r \leqslant \min (m, n)$

Title: Why does $A^TAw = A^Tb$ describe the least square solution to a multivariate system?

Body:
Note that this builds off a previous question, found here.

Suppose I have a system of equations of the form:

$$Wv_1 + Xv_2 + Yv_3 + Zv_4 = b$$

Where $v_1$, $v_2$, $v_3$, $v_4$, and $b$ are vectors. I want to determine the least squares solution for unknowns $W,X,Y,Z$. As I understand it, the solutions are the vector $w$ in:

$$A^{T}Aw = A^{T}b$$

Where $A$ is the matrix who's columns are $v_1$, $v_2$, $v_3$, $v_4$.

It isn't clear to me, however, why this is. Why is $w$ the least-squares solution, mathematically and intuitively? I've tried to do some research online, however I haven't found much (and what I have found uses extensively terminology with which I'm unfamiliar— unfortunately, this isn't a subject I'm particularly familiar with).


答: In the beginning of most topology books, it states that we always take $ X \ne \emptyset $.

Title: Does every topology have singletons?


答:
We are looking for an estimator $s^2$ for $\sigma^2$ which is unbiased that means:

$$
E[s^2] = \sigma^2
$$

if we define $s^{2} = \frac{||\textbf{Y} - \hat{\textbf{Y}} ||^{2}}{(n-p)}$ by the theorem A we get:

$$
\begin{aligned}
E[s^{2}] &= \frac{E[||\textbf{Y} - \hat{\textbf{Y}} ||^{2}]}{(n-p)} \\
&= \frac{(n-p)}{(n-p)} \sigma^{2} = \sigma^2
\end{aligned}
$$

This proves that, in fact, $s^2$ is unbiased.

In general terms, we have a model that depends on the value of some parameter $\theta$ (in this case: $\sigma^2$; once we have $\sigma^2$ we can fit our model to the data). An unbiased estimate $\hat{\theta}$ (in this case: $s^2$) is a value extracted from data which represents the information that $\theta$ is supposed to show us. It represents that parameter in such a way, that in average (i.e. expected value) it's equal to our parameter, that means:
$$
\text{Bias}(\theta, \hat{\theta}) = 0 \iff E[\hat{\theta}] = \theta \left( \text{ i.e.} \iff E[s^2] = \sigma^2 \right)
$$
There's no general strategy to find $\hat{\theta}$, so we explore the model and try to find some pattern, that allows us to declare such a $\hat{\theta}$. In our case, that pattern is given by theorem A, because we have an expected value on one side and our parameter, namely $\sigma^2$, explicitly on the other side. Given that structure, we propose the functional form for $s^2$ given by the book.

Title: Simple algebraic step in establishing estimation of variance of errors, $e_{i}$ i.e $\sigma^{2}$ for variables in linear regression

Body:
This is something I definitely know how to do, but is slipping my mind at the moment. I'm reading through Rice's Mathematical Statistics and the following is the proof to establish an estimate for the errors of the independent variables of a linear regression (You can ignore the majority of the proof it is only the last line I'm stuck on):

I understand the proof of $E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2}) = (n-p) \sigma^{2}$$. What I'm having trouble figuring out is how the

$$s^{2} = \frac{||\textbf{Y} - \hat{\textbf{Y}} ||^{2}}{(n-p)}$$

Came about, specifically what was on the other side to allow us to divide?

$$\text{?something?} = E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2}) = (n-p) \sigma^{2}$$

I ask because if we're taking the expectation of $E(||\textbf{Y} - \hat{\textbf{Y}} ||^{2})$, then the only thing I see appearing on the other side would be:

$$\sum_{i = 1}^{n}E(\textbf{Y} - \hat{\textbf{Y}})^2$$

Which to me doesn't amount to anything special. Since what I want is

$$||\textbf{Y} - \hat{\textbf{Y}} ||^{2}$$
without the expectation operator.


问:
Elliptic Surfaces: a naive question
Ground field $\Bbb{C}$. Algebraic category. Smooth surfaces.

Let $S$ be a minimal elliptic surface $p:S\rightarrow C$ the elliptic fibration (general fiber = elliptic curve). Suppose the $m$-canonical system is non-empty and let $D\in \lvert m K \rvert$.

Why can we say that $D.F=0$, where $F$ is a fiber of $p$ ?

It appears to be very obvious, but not for me. Can't it be positive?

Enlighten me, s'il vous plait !

答: Recall that by adjunction, $2g-2=F^2+K.F$ where $g$ is the genus of $F.$ But in this case, $g=1,$ and $F^2=0,$ so we end up with $F.K=0.$ This implies in particular that any $D\sim mK$ also satisfies $D.F=0.$</p>

Title: Elliptic Surfaces: a naive question


Title: What is inverse of a matrix whose diagonal elements are all zero.

Body:
I am a science researcher and I got a problem to find the generic inverse of the following matrix:

$$
A_n = \left(\begin{array}{ccc}
        0 & a_2 & a_3 & ... & a_{n-1} & a_n \\
      a_1 &   0 & a_3 & ... & a_{n-1} & a_n \\
      a_1 & a_2 &   0 & ... & a_{n-1} & a_n \\
      ... & ... & ... & ... & ...     & ... \\
      ... & ... & ... & ... & ...     & ... \\
      a_1 & a_2 & a_3 & ... &       0 & a_n \\
      a_1 & a_2 & a_3 & ... & a_{n-1} &   0 \\
      \end{array}\right)
$$

I figured out that for n=2,3

$$
A_2^{-1} = \left(\begin{array}{ccc}
        0 & a_2^{-1}  \\
      a_1^{-1} &   0  \\
      \end{array}\right)
$$

$$
A_3^{-1} = \frac{1}{2} \left(\begin{array}{ccc}
        -a_1^{-1} &  a_1^{-1} &  a_1^{-1} \\
         a_2^{-1} & -a_2^{-1} &  a_2^{-1} \\
         a_3^{-1} &  a_3^{-1} & -a_3^{-1} \\
      \end{array}\right)
$$

but can we extend it to a general case?
Can anyone help?? Thanks!

[later]

It looks like it is

$$
m_{ij} =  -\frac{n-2}{n-1} a_{i}^{-1}\ (i=j)
$$
$$
       = \frac{1}{n-1} a_{i}^{-1}(else)
$$

Body:
Your matrix can be written as $D_n + u_n \, v_n^T$:

$$u_n = \left[ \begin{matrix}  1 & ... & 1 \end{matrix} \right]^T$$

$$v_n = \left[ \begin{matrix}  a_1 & ... & a_n \end{matrix} \right]^T$$

$$D_n = - \text{diag}(v_n) = - \left[ \begin{matrix}
a_1 & ... & 0 \\
\vdots & ... & \vdots \\
0 & ... & a_n \\
 \end{matrix} \right]$$

Which reminds me of the Sherman–Morrison formula:

$$\left(A + u \, v^T\right)^{-1} = A^{-1} - {A^{-1} \, u \, v^T \, A^{-1} \over 1 + v^T \, A^{-1} \, u}$$

[EDIT]

Your pattern seems to be right. Defining $w_n = \left[ \begin{matrix}  a_1^{-1} & ... & a_n^{-1} \end{matrix} \right]^T$:

$$ $$

$$1 + v^T \, A^{-1} \, u = 1-n$$

$$A^{-1} \, u = - w_n$$

$$ v^T \, A^{-1} = - u_n^T$$

$$\left(A + u \, v^T\right)^{-1} = - \text{diag}(w_n) - { w_n \, u_n^T \over 1-n}$$

$$ \left(A + u \, v^T\right)^{-1} =
- \left[ \begin{matrix}
a_1^{-1} & ... & 0 \\
\vdots & ... & \vdots \\
0 & ... & a_n^{-1} \\
 \end{matrix} \right]
-
\frac{1}{1-n}
\left[ \begin{matrix}
a_1^{-1} & ... & a_1^{-1} \\
\vdots & ... & \vdots \\
a_n^{-1} & ... & a_n^{-1} \\
 \end{matrix} \right]
$$

$$ $$

$$
m_{ij} = -a_{i}^{-1} -\frac{a_{i}^{-1}}{1-n} = \frac{n-2}{1-n} \, a_{i}^{-1} = - \frac{n-2}{n-1} \, a_{i}^{-1} \, (i=j)
$$
$$
 = -\frac{a_{i}^{-1}}{1-n} = \frac{1}{n-1} \, a_{i}^{-1} \, (else) \hphantom{aaaaaaaaaaaaaaa}
$$


问:
Diophantine Equation with 3 Variables example
$15x+72y+21z=9 (1)$

$15+72y = gcd(15,72).r1$

(1) can be written as :

```<code>15x+72y=gcd(15,72).r1    (2)

gcd(15,72).r1 + 21z = 9  (3)
</code>```

$15x+72y=gcd(15,72).r1$

Calculating $GCD(15,72)$ gives:

$72 = 4*15 + 12$

$15 = 1*12 + 3$

$12 = 4*3 + 0$

from (2) and (3)
$15x+72y=3.r1 (4)$

$3.r1 + 21z = 9$ (5)

Then applying the Extended Euclidean Algorithm:

$3 = (1 * 15) + (-1 * 12) = (-1 * 72) + (5 * 15)$

A particular solution is $x,y = (5,-1)$ multiply by r1 the $(5r1,-r1)$

general solution $(x,y)=(5r1+24m, -r1-5m)$

$3.r1 + 21z = 9 gcd (3,21) = 3 | 9$ so the is solution $(r1,z) = (3+7n, -n)$

$x=5r1+24m = 5(3+7n)+24m=15+35n+24m$

$y=-r1-5m=-3-7n-5m$

$z=-n$

$(x,y,z) = (15+35n+24m, -3-7n-5m, -n)$

I provide this exercise as an example for diophantine equations with 3 variables however please somebody check for mistakes

答:
I'm not sure what you are after with modular arithmetic but the equation has integer solutions that can be found by other means... assuming a solution in integers is what you seek.

Experimenting with values in a spreadsheet, we can let $x$ be any integer and notice that $y,z$ must jump in respective multiples of $7$ and $24$ if the equation is to sum to a consistent value. Further experimentation locates the constants in their separate equations so that the original sums to zero.

$$15x+72y+21z=9\implies 5x+24y+7z-3=0$$

$$\text{Let}\qquad y = 7 n + 3 x + 1\qquad z = -24 n - 11 x - 3\qquad n,x \in Z$$
Here are some example triples of $(x,y,z)$ as $n$ and $x$ vary from $-2$ to $2$

For $n=-2$
$$\cdots\quad (-2,-19,67)\quad
(-1,-16,56)\quad
(0,-13,45)\quad
(1,-10,34)\quad
(2,-7,23)\quad\cdots$$

For $n=-1$
$$
\cdots\quad (-2,-12,43)\quad
(-1,-9,32)\quad
(0,-6,21)\quad
(1,-3,10)\quad
(2,0,-1)\quad\cdots$$

For $n=0$
$$\cdots\quad (-2,-5,19)\quad
(-1,-2,8)\quad
(0,1,-3)\quad
(1,4,-14)\quad
(2,7,-25)\quad\cdots$$

For $n=1$
$$
\cdots\quad (-2,2,-5)\quad
(-1,5,-16)\quad
(0,8,-27)\quad
(1,11,-38)\quad
(2,14,-49)\quad\cdots
$$

For $n=2$

$$\cdots\quad (-2,9,-29)\quad
(-1,12,-40)\quad
(0,15,-51)\quad
(1,18,-62)\quad
(2,21,-73)\quad\cdots$$


问:
Function of a martingale
Let $X_n$ be a martingale such that $|X_n-X_{n-1}|\leq 1$. How would one bound something like $E[e^{\lambda(X_n-X_{n-1})}|\mathcal{F_{n-1}}]$ where $\mathcal{F}_{n}$ is the standard filtration and $\lambda$ a real parameter? I can tell that an $e^{-\lambda X_{n-1}}$ is measurable and can be taken out, but wouldn't know how to proceed. Any hint would do.

答:
Yes, you are right that $e^{-\lambda X_{n-1}}$ is $\mathcal{F}_{n-1}$ measurable, but you do not need to use that fact.

The bound can be found as a direct consequence of Hoeffding's lemma,

$$\mathbb{E}[e^{\lambda(X_n-X_{n-1})}|\mathcal{F}_{n-1}] \leq e^{\lambda^2/2}$$

Note that the standard form of the Hoeffding's lemma applies to total expectation (not to conditional expectation), but the result holds when the bounds for $|X_n-X_{n-1}|$ are themselves measurable by the conditioning variable (as is the case here). See this for reference.

Title: Function of a martingale


问:
Double limit question
Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous function where $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$. Suppose that

$(x_{n},y_{n}) \rightarrow (x,y)$ in $\mathbb{R}^{2}$,

$(x_{n},y_{n}) \in  X \times Y$, and

$\lim _{y_{n} \rightarrow y} f(x',y_{n}) = \infty$ for all $x' \in X$.

Is it true that there exists an $N>0$ such that $f(x_{n},y_{n}) > 0 $ for all $n \geq N$?
Seems intuitive but I am a bit stuck.

答: I think it is not true. Suppose f(x,y)=-x+ 1/|y-1|. let y=1, yn=1+1/n, xn=2n.</p>

Title: Double limit question


问:
How to show that something has the same probability distribution function as something else?
I understand the logic behind what I'm about to ask, but I'm not sure exactly how to write it mathematically. I have a random variable $X$ with probability density function:

$$
f_X(x)=
\begin{cases}
x^{-2} & x\ge 1\\
0 & x<1
\end{cases}
$$

and therefore a probability distribution function:

$$
F_X(x)=
\begin{cases}
1-\frac{1}{x} & x\ge 1\\
0 & x<1
\end{cases}
$$

Then I need to show that if $$ is uniformly distribution between $0$ and $1$ (i.e. $∼(0,1)$) and
$=(1−)^{−1}$, then $$ has the same distribution function as $X$.

I get that $0<\frac{1}{x}<1$, but I'm not sure how to use this to show the above - let alone how to show it in a mathematically sound way.

答:
Hints:

If $y=\frac{1}{1-w}$, then can you find $w$ in terms of $y$? Let's call this $w=g(y)$

Is $y$ increasing or decreasing when $w$ is positive and increasing?

What is $\mathbb P(W \le w)$ in terms of $w$ when $w \le 0$ or $0 \lt w \lt 1$ or $1 \le w$?

What is $\mathbb P(W \le g(y))$  in terms of $g(y)$? Over what ranges of $y$?

What is $\mathbb P(Y \le y)$ in terms of $y$? Over what ranges of $y$?

Does this match $F_X(x)$?


问:
Why is this eigenvector wrong?
To make this simple I'm going to skip to the part of the problem I'm having trouble with.
I'm trying to find the eigenvector for the eigenvalue $\lambda = 1$, I've done the math and verified this part after putting the matrix in reduced row echelon form we have a $3\times 3$ matrix:
$$\begin{pmatrix}0 & 2 & 1 \\
1 & -3 & 0 \\
0 & 0 & 0\end{pmatrix}$$
Therefore,
$v_1 - 3 v_2 = 0$ and $2 v_2 + v_3 = 0$.

This means that $v_1 = 3 v_2$ and $v_3 = -2 v_2$.

This is where I get confused so bear with me.
Because we can define both $v_1$ and $v_3$ with $v_2$, I set $v_2 = t$.
$$[v_1, v_2, v_3] = t[\ ]$$
I think about the problem this way: in order to get $v_1$, what do I multiply "$t$" by? We established earlier that $v_1 = 3 v_2$ or $v_1 = 3 t$, therefore we multiply $t$ by $3$ to get $v_1$, so the first value is $3$.

We repeat. In order to get $v_2$, what do I multiply $t$ by? We established earlier that $v_2 = t$. so the next value is $1$.

In order to get $v_3$, what do we multiply $t$ by? We established earlier that $v_3 = -2 v_2$, therefore $v_3 = -2 t$, so the last value is $-2$.

$$[v_1, v_2, v_3] = t[3, 1, -2]$$

because $3t = v_1$, $t = v_2$, $-2t = v_3$.

I've read my book, read explanations online, and watched video after video (Khan Academy and Krista King Math), yet I still don't understand how to do this.
Apparently the answer is $[-3, -1, 2]$. I'm not sure how the signs flipped, but they did and everywhere I look I can't find an explanation.

答:
Your solution is correct. In fact, $(3,1,-2)$ is still an element of the subspace (eigenspace) generated by $(-3,-1,2)$.

If $v$ is an eigenvector of $A$ for some eigenvalue $\lambda$, i.e. $Av=\lambda v$, then $-v$ is an eigenvector as well, just as any multiple $\mu v$ (with $\mu\in\mathbb R$, or whatever field you're working on):
$$ Av=\lambda v \ \implies A(\mu v)=\mu (Av)=\mu (\lambda v)= \lambda (\mu v). $$

Title: Why is this eigenvector wrong?


问:
integral of heaviside function, with variable inside heaviside part
I don't understand how the integral of a heaviside function works if there is a variable inside the heaviside part.

$$\int_{0}^{t}\tau^2H(t-\tau-1)d\tau$$

答:
The Heaviside function is telling you where to integrate. $$H(t-\tau-1) = \chi_{[0,+\infty]}(t-\tau-1) = \chi_{[-\infty,t-1]}(\tau)$$
Since $t-\tau-1\geq 0 \iff \tau \leq t-1$. Then the integral in question is:
$$\int_0^t\tau^2H(t-\tau-1)d\tau = \int_0^{t-1}\tau^2d\tau$$
which can be solved easily.


问:
What are the eigenvalues of $(a_m+a_n)^2$?
Let $a_m\in \mathbb{R}$ and consider the matrix defined by $A_{mn} = (a_m+a_n)^2$. Using python and inputting a random vector $a = (a_n:a_n\in \mathbb{R})$ of various length and various values, the eigenvalues of $A$ always seem to consist of 1 negative value and 2 positive values. All the other eigenvalues are $0$.

Based on this, it seems that there should be a formula which determines the 3 nontrivial eigenvalues of $A$ (and possibly all the eigenvectors of $A$), but I can't think of a rigorous proof.

EDIT: If $a_n$ had some "symmetry" so that $\sum a_n =\sum a_n^3=0$, then it's easy to see that $a_n$  is an eigenvector and that $a_n^2 \pm c$ are eigenvectors where
$$
c^2 = \frac{1}{N}\sum a_n^4
$$
However, the statement seems to be more general. Also, I'm not sure why all the other eigenvalues are 0.

答:
First, the matrix definitely has rank at most $3$, since it is the sum of three matrices
$$
A = \begin{pmatrix}
a_1^2 & a_1^2 & \cdots & a_1^2 \\
a_2^2 & a_2^2 & \cdots & a_2^2 \\
\vdots & \vdots & \ddots & \vdots \\
a_n^2 & a_n^2 & \cdots & a_n^2
\end{pmatrix}
+2
\begin{pmatrix}
a_1a_1 & a_1a_2 & \cdots & a_1a_n \\
a_2a_1 & a_2a_2 & \cdots & a_2a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_na_1 & a_na_2 & \cdots & a_na_n
\end{pmatrix}
+
\begin{pmatrix}
a_1^2 & a_2^2 & \cdots & a_n^2 \\
a_1^2 & a_2^2 & \cdots & a_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
a_1^2 & a_2^2 & \cdots & a_n^2
\end{pmatrix}
$$
each of which has rank at most $1$.

Let's assume some genericity going forward, which is necessary to avoid more degenerate outcomes (like the zero matrix, for instance). I'll assume that $a_1\ne0$ and that $a_1,a_2,a_3$ are distinct.

Since it is a symmetric matrix, $A$ is positive (or negative) semidefinite if and only if all of its leading principal minors are nonnegative (or nonpositive). Note that the first leading principal minor is simply the positive quantity $a_1^2$, which rules out the possibility that $A$ is negative semidefinite. On the other hand, (my computer tells me that) the third leading principal minor is the negative quantity $-2(a_1-a_2)^2(a_1-a_3)^2(a_2-a_3)^2$, which rules out the possibility that $A$ is positive semidefinite.

That gets us down to two possibilities: one negative and two positive eigenvalues (what the OP observes), or one positive and two negative eigenvalues. I don't know how to finish from here. I feel that the fact that all the $3\times3$ minors are nonpositive (following the same pattern as the third leading principal minor) might be relevant to the fact that the product of the three nonzero eigenvalues being negative, but I can't make the exact connection.


问:
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$
I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$

Is it OK how I want to do?

$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$

I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?

I thinks is not because  if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.

Thanks :)

答:
A simpler argument is to write $f(x)=\sqrt{x+1}-\sqrt{x}$ and note that since $\lim\limits_{x\to \infty}f(x)=0$ and $\sin$ is uniformly continuous, we have
$$\lim\limits_{x\to \infty}\sin(\sqrt{x}+f(x))-\sin(\sqrt{x})=0.$$

Why does this work? Let $g$ be a uniformly continuous function, $f$ a function such that $\lim\limits_{x\to\infty}f(x)=0$ and $h$ an arbitrary function. Then
$$\lim\limits_{x\to\infty} g(h(x)+ f(x))-g(h(x))=0$$
as for any $\epsilon>0$ we can choose some $\delta>0$ such that $|a-b|<\delta\implies |g(a)-g(b)|<\epsilon$, so choosing $x_0$ such that $x>x_0\implies |f(x)|<\delta$ gives us that for $x>x_0$, $|g(h(x)+f(x))-g(h(x))|<\epsilon$, so the limit as $x\to \infty$ is $0$.


问:
Uniform convergence to show an integral
I struggled with the following question, and would like to know how to do it correctly.

The exponential function is defined by $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$.  Assuming that $\lim_{N\to\infty} \sum_{n=0}^N \frac{x^n}{n!}$ converges uniformly on every interval of the form $[-M,M]$, show that $\int_0^x e^t\ dt = e^x - 1$.

My strategy that I believe was incorrect was to say since $f_n \rightarrow f$ is uniformly convergent then $\int f_n=\int f$

Then I said because of this, $\lim_{n\to\infty}\sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^n}{n!}dt=\int_{0}^{x}e^{x}dx=e^{x}-1$ but I realize I have not really "shown" anything.

Well, what is the correct way to solve this problem?

答:
$\int_0^x\sum_{n=0}^N \frac{t^n}{n!}dt=\sum_{n=0}^N \frac{x^{n+1}}{(n+1)!}=\sum_{n=0}^{N+1}\frac{x^n}{n!}-1\to e^x-1$

The convergence is uniform on any finite interval.

Title: Uniform convergence to show an integral


问:
Given $dy/dt = 6y/39 -(y^2)/2002$ and the point $(0,14)$, what is the original equation in "$y=\dots$" form?
This is how I tried to solve it:
$$\frac{dy}{dt}=\frac{6y}{39}-\frac{y^2}{2002}$$
I multiplied and divided:
$$\frac{dy}{\frac{6y}{39}-\frac{y^2}{2002}}=dt$$
I simplified:
$$\frac{dy}{\frac{2y}{13}-\frac{y^2}{2002}}=dt$$
I found a common denominator:
$$\frac{dy}{\frac{308y-y^2}{2002}}=dt$$
I divided the fractions:
$$\frac{2002dy}{308y-y^2}=dt$$
I completed the square:
$$\frac{2002dy}{308y-y^2-154^2+154^2}=dt$$
$$\frac{2002dy}{-(y^2-308y+154^2)+154^2}=dt$$
$$\frac{2002dy}{-(y-154)^2+154^2}=dt$$
Now I am temporarily looking at only the left side of the equation to find the integral:
$$\int\frac{2002}{-(y-154)^2+154^2}dy$$
I substituted u:
$$2002\int\frac{1}{(-1)(u)^2+154^2}du$$
At this point I was very confused so I tried to use arctan when I probably wasn't allowed to. I used the following formula:
$$\int\frac{b}{b^2+a^2x^2}dx=\arctan(\frac{ax}{b})+c$$
My integral:
$$\frac{2002}{154}\int\frac{154}{(-1)(u)^2+154^2}du$$
$$\frac{2002}{154}\arctan(\frac{iu}{154})$$
$$13\arctan(\frac{i(y-154)}{154})$$
The integral of $1dt$ is $t$ so I got this equation:
$$13\arctan(\frac{i(y-154)}{154})=t+c$$
I then solved for y:
$$\frac{i(y-154)}{154}=\tan(\frac{t+c}{13})$$
$$y=\frac{154\tan(\frac{t+c}{13})}{i}+154$$
I then solved for c by substituting the initial point (0,14) into the equation:
$$14=\frac{154\tan(\frac{0+c}{13})}{i}+154$$
$$\frac{-140i}{154}=\tan(\frac{c}{13})$$
$$\arctan(\frac{-10i}{11})=\frac{c}{13}$$
$$c=13\arctan(\frac{-10i}{11})$$
And I substituted the found $c$ value into the equation:
$$y=\frac{154\tan(\frac{t+13\arctan(\frac{-10i}{11})}{13})}{i}+154$$
Is this technically correct or did do I something wrong?
If I did do something wrong, what did I do wrong?
What would be an alternative method of solving this?

Body:
Note
$$\frac{2002}{308y-y^2}=\frac{2002}{y(308y-y)}=\frac{2002}{308}\bigg(\frac{1}{y}+\frac{1}{308-y}\bigg)=\frac{1001}{154}\bigg(\frac{1}{y}+\frac{1}{308-y}\bigg)$$
and hence
$$\frac{2002dy}{308y-y^2}=dt$$
becomes
$$\frac{1001}{154}\bigg(\frac{1}{y}+\frac{1}{308-y}\bigg)dy=dt.$$
Integrating both sides gives
$$ \frac{1001}{154}\bigg(\ln|y|-\ln|308-y|\bigg)=t+C.$$
So
$$ \ln\bigg|\frac{y}{308-y}\bigg|=\frac{154}{1001}t+C' $$
or
$$ \frac{y}{308-y}=ke^{\frac{154}{1001}t}. $$
From this, you can solve for $y$ to get the answer.


问:
Let $u_1,u_2,u_3,u_4$ be four linearly independent vectors in $R^n$. For what value of $a$ is the set linearly dependent?
Let $u_1,u_2,u_3,u_4$ be four linearly independent vectors in $R^n$. For what value of $a$ is the set {${u_1+4u_2, u_2+3u_3, u_3+3u_4, u_4+au_1}$}
linearly dependent?

My method for solving this question was to write the terms as a vector equation equal to 0, expand the equation, and collect like terms until I was left with an equation of the following form. The $c$ terms are scalars multiplying the vectors:

$$u_1(c_1+c_4a) +u_2(4c_1+c_2)+u_3(3c_2+c_3)+u_4(3c_3+c_4)$$

I then set the four $c$ expressions equal to zero, and solved between them to determine that $a=1/36$ makes the set linearly dependent.

What I don't understand is how setting the scalar terms equal to zero, which I believe indicates the trivial solution and thus linear independence, returns the value of $a$ that will cause linear dependence.

答: The two sets differ by the application $\begin{pmatrix}1&4&0&0\\0&1&3&0\\0&0&1&3\\a&0&0&1\end{pmatrix}$.  The linear independence of the first set implies independence of the second precisely when the transformation is invertible, or, when the determinant is nonzero.  We get $\rm{det}=1-a(4)(3)(3)=1-36a=0$ when $a=1/36$.


问:
Simple example of independent events fails to satisfy
Simplest example is tossing a coin.

Event A: Tossing a coin results in heads

Event B: Tossing a coin results in heads again.

These are independent events i.e. A doesn't influence the B.

$A=\{H\}$

$B=\{H\}$

$A∩B=\{H\}$

$P(A) = P(B) = P(A\cap B) = \frac{1}{2}$

So, here clearly,

$P(A)(B) \neq P(A \cap B)$

Am I writing this in wrong way?

答:
I'm afraid that what you wrote is not correct. Note that $A$ and $B$ are separate events that happen to have the same probability of occurring. Here's how we can proceed
$$P(A\cap B)=P(A\vert B)P(B)$$
Since $A$ and $B$ are independent events, we have
$$P(A\cap B)=P(A)P(B)$$
$$=\frac12\cdot\frac12=\frac14$$
I hope this helps you understand.