问:
Finding linear transformation matrix without much information
I have a linear transformation $F: \mathbb R^3 \to \mathbb R[t]_2$, where $\mathbb R[t]_2$ is the vector space of polynomials up to the second degree.

I also have three different bases:
$$B = \{(1,1,0),(0,1,0),(0,0,1)\}$$
$$C = \{1, 1 + t, 1 + t^2\}$$
$$D = \{1, t, t^2\}$$

The linear transformation matrix $[F]_{B,C}$ has also been given:

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

The problem asks to write the linear transformation matrix of $[F]_{B,D}$.

What I've tried so far:

$$F(1,1,0) = something = 1(1,0,0) + 1(1,1,0) + 0(1,0,1)$$
$$F(0,1,0) = something = 2(1,0,0) + 0(1,1,0) + 1(1,0,1)$$
$$F(0,0,1) = something = (-1)(1,0,0) + (-1)(1,1,0) + 0(1,0,1)$$

So this is basically my attempt to find the linear transformation formula so I can write the matrix. But I'm not sure what to put where I wrote "something" as I'm dealing with a polynomial.

I've also thought that there may be a way to find the $[F]_{B,D}$ matrix using matrices of changing basis, but I could not think how.

Can anyone give me any hint?

答:
From your work so far, it seems as though you have misinterpreted the matrix of the transformation. In particular, we have
$$
[F(0,1,0)]_{C} = (2,0,1) \implies\\
F(0,1,0) = 2 \cdot 1 + 0 \cdot (1 + t) + 1\cdot (1 + t^2)\\
= 3 + t^2 = 3\cdot 1 + 0 \cdot t + 1\cdot t^2.
$$
With that, we see that the coordinate vector of $F(0,1,0)$ with respect to $D$ is given by
$$
[F(0,1,0)]_D = (3,0,1).
$$
So, the second column of $[F]_{B,D}$ is given by $(3,0,1)$. That is, we have
$$
[F]_{B,D} = \pmatrix{? & 3 & ?\\? & 0 & ?\\ ? & 1 & ?}.
$$
The first column is equal to $[F(1,1,0)]_D$, and the third column is equal to $[F(0,0,1)]_D$.


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答: This is only for Group Theory, but I thought Groups and Symmetry by Armstrong is quite a nice little book and serves well for an introduction to Groups. </p>

ContentLicense: CC BY-SA 3.0


问:
Quotient $SO(3)/SO(2)$
To identify an element of $SO(3)$, i.e. a rotation in $\mathbb{R}^3$, we must specify a direction in space, i.e. a straight line through the origin, and a rotation angle.

To specify an element of $SO(2)$ we just need the rotation angle.

Then one might argue that the quotient $SO(3)/SO(2)$ corresponds to the space of directions in $\mathbb{R}^3$, i.e. a projective space. But this is not the case: the quotient is actually $S^2$, the spherical surface.

Why must we not identify antipodal points of $S^2$? Do they not correspond to the same direction?

答: Your assumption that “To identify an element of $SO(3)$ […] we must specify a direction in space, i.e. a straight line through the origin, and a rotation angle” is not correct. To see why, consider, say, the straight line $\Bbb R(0,0,1)$ and the angle $\frac\pi2$. Does this describe an element of $SO(3)$? No; it applies to two elements of $SO(3)$, since there are two distinct rotations with angle $\frac\pi2$ around that line. So, it is indeed natural that $SO(3)/SO(2)$ can be seen as $S^2$.

ContentLicense: CC BY-SA 4.0

Title: Quotient $SO(3)/SO(2)$


问:
In 1000 tosses of a coin, 560 heads appear. Is it reasonable to assume the coin is fair?
So the question is:

In 1000 tosses of a coin, 560 heads appear. Is it reasonable to assume the coin is fair?

Let $Y = \sum_{i=1}^{1000} X_i = $ number of heads. I chose $H_0: p = 0.5$ and $H_1: p > 0.5$.
The $p$-value is:

$$P(Y\geq 560) = \sum_{y=560}^{1000}\binom{1000}{y}(0.5)^y(0.5)^{1000-y}$$
$$= 0.00008252494$$

Then I try to find $\alpha$ using $\frac{Y}{n} = \overline{X}$ as the estimator for $p$:
$$P(Y \geq 560) = P(\overline{X} \geq 0.56)$$
$$= P(\frac{\sqrt{n}(\overline{X} - p_0)}{p(1-p)} \geq \frac{\sqrt{1000}(0.56 - 0.5)}{0.5*0.5}$$
$$= P(Z\geq 7.59)$$

That Z-score seems incredibly high and I don't think they're even included in any Z-tables.

Am I on the right track here or have I missed something?

答:
From the given $p$-value we calculate a $z$-score of
$$Z=\Phi^{-1}(1-0.00008252494)=3.7672\dots$$
The Z-score is not that high, but still high enough (beyond three SDs) that we can reject $H_0$ and say that the coin is biased.


问:
How to find if a sequence is converging or diverging?
The question is as follows: Determine if the sequence:

$$\{1/1,\ 1/3,\ 1/2,\ 1/4,\ 1/3,\ 1/5,\ 1/4,\ 1/6,\ \ldots\}$$

diverges or converges. If it is convergent, find the limit.

I am not sure where to begin, I'd appreciate it if anybody could help me on this one.

Thankyou

答:
The approach of Bérénice is nice and looks quite reasonable, since the first step is to identify the elements of the sequence
\begin{align*}
\begin{array}{ccccccccc}
a_0&\ a_1&\ a_2&\ a_3&\ a_4&\ a_5&\ a_6&\ a_7&\ \cdots\\
\color{blue}{1}&\ \frac{1}{3}&\ \color{blue}{\frac{1}{2}}&\ \frac{1}{4}&\ \color{blue}{\frac{1}{3}}&\ \frac{1}{5}&\ \color{blue}{\frac{1}{4}}&\ \frac{1}{6}&\ \cdots\\
\\
\end{array}
\end{align*}
which typically results after some analysis in a separation of even and odd elements:
\begin{align*}
a_{2n}=\frac{1}{n+1}\qquad\qquad a_{2n+1}=\frac{1}{n+3}\qquad\qquad n\geq 0\tag{1}
\end{align*}

With (1) in mind we can continue along different lines. One of them is to see  the elements are positive and essentially growing as $\frac{1}{n}$, So,  we could think of squeezing the sequence between the constant sequence $(0)_{n\geq 0}$ and a sequence related with $\left(\frac{1}{n+1}\right)_{n\geq 0}$ where we use $n+1$ in the denominator to avoid division by zero when $n=0$.

Starting from (1) we obtain
\begin{align*}
\color{blue}{a_n}&=\frac{1}{\frac{n}{2}+1}\left(\frac{1+(-1)^n}{2}\right)+\frac{1}{\frac{n-1}{2}+3}\left(\frac{1-(-1)^n}{2}\right)\\
&=\frac{1}{n+2}\left(1+(-1)^n\right)+\frac{1}{n+5}\left(1-(-1)^n\right)\\
&\leq \frac{2}{n+2}+\frac{2}{n+5}\\
&\,\,\color{blue}{\leq \frac{4}{n+2}}
\end{align*}

We derive this way an upper bound  $\frac{4}{n+2}$ for $a_n$ and can conclude from
\begin{align*}
0\leq a_n\leq \frac{4}{n+2}\quad\longrightarrow\quad 0
\end{align*}
the sequence $(a_n)_{n\geq 0}$ is a zero-sequence.


答:
First of all you have $P(Y\geq 560)=1-P(Y\leq 559)$. Next we use the normal approximation without the continuity correction factor.

$$1-P(Y\leq 559)\approx 1-\Phi\left(\frac{0.559-0.500}{\sqrt{\frac{ 0.5\cdot 0.5}{1000}}} \right)$$

You missed the root at $p\cdot 1-p$. But the z-score is still high, $z=3.73$. So we assume approximately that $\Phi(3.73)\approx 1$. This leads to $1-P(Y\leq 559)\approx 1-1=0<0.05$

Body:
So the question is:

In 1000 tosses of a coin, 560 heads appear. Is it reasonable to assume the coin is fair?

Let $Y = \sum_{i=1}^{1000} X_i = $ number of heads. I chose $H_0: p = 0.5$ and $H_1: p > 0.5$.
The $p$-value is:

$$P(Y\geq 560) = \sum_{y=560}^{1000}\binom{1000}{y}(0.5)^y(0.5)^{1000-y}$$
$$= 0.00008252494$$

Then I try to find $\alpha$ using $\frac{Y}{n} = \overline{X}$ as the estimator for $p$:
$$P(Y \geq 560) = P(\overline{X} \geq 0.56)$$
$$= P(\frac{\sqrt{n}(\overline{X} - p_0)}{p(1-p)} \geq \frac{\sqrt{1000}(0.56 - 0.5)}{0.5*0.5}$$
$$= P(Z\geq 7.59)$$

That Z-score seems incredibly high and I don't think they're even included in any Z-tables.

Am I on the right track here or have I missed something?


问:
Proving that exponential is growing faster than polynomial
I want to prove this limit:$\lim_{x \to +\infty}(|P(x)|-e^x)=-\infty$.
I can't use Hopital's rule, but only the fact that $a_{n}=(1+\frac{x}{n})^n$ and $b_{n}=(1+\frac{x}{n})^{n+1}$ converge to $e^x$.

答:
Since
$(1+\frac{x}{n})^n
\to e^x
$,
we can show that
$e^x
=\sum_{n=0}^{\infty} \dfrac{x^n}{n!}
$.

Then,
for $x > 0$,
$e^x \gt \dfrac{x^{n+1}}{(n+1)!}
$
so
$\dfrac{x^n}{e^x}
\lt \dfrac{(n+1)!}{x}
\to 0$.


问:
Shannon measure of information dice question
You play a game with a friend where she rolls two $6$-sided dice. You first have to guess the total, and then once she confirms that you have the correct answer you have to guess the combination of dice (the order does not matter, i.e., guessing $6$-$1$ or $1$-$6$ is the same).

How much information is in the distribution of totals?

How many possible combinations are there? What is the SMI for the dice combinations?

If your friend confirms to you that the total is 5, how much information has she given you and how much remains?

答:
The first point is incorrect. You have 11 possible values for the sum, but the are not equiprobable. See eg here.

The second is almost correct. But again, the 21 combinations are not equiprobable. Because the combination $(6,5)$ has twice the probability of the combination $(6,6)$

Even after these corrections, the question might seem slightly ambiguous. Let $X$ be the "full" result (21 possible combinations) and $Y$ be the total sum. Then, "how much information the sum gives" can be understood in two ways. One, in itself; then the answer is $H(Y)$. Two, how much information gives about the full result; then the answer is $H(X)-H(X|Y)=I(X;Y)=H(Y)-H(Y|X)$.  But in this case $H(Y|X)=0$ (knowing the combination one knows the sum), hence fortunately there is no ambiguity.

For the third, one must compute $H(X | Y=5)$


问:
Bernstein set and it proof.
Definition. A set $B\subset\mathbb R$ is called a Bernstein set if both $B$ and $\mathbb R\setminus B$ intersect each perfect set.

I know the construct  the of Bernstein set but I got confused with this construction that I found in a book.

Let $\mathcal F=\{P_\xi\colon \xi<c\}$ be an enumeration of all perfect subsets of $\mathbb R.$ By recursion induction on $\xi<c$ define the sequences $\{a_\xi\colon \xi<c\}$ and $\{b_\xi\colon \xi<c\}$ by choosing in step $\xi<c$ points $a_\xi\neq b_\xi$ from $\mathbb R$ such that $$ a_\xi, b_\xi\in P_\xi\setminus(\{a_\zeta\colon \zeta<\xi\}\cup \{b_\zeta\colon \zeta<\xi\})$$  clearly, this choice is possible. Now define $$B:=\{a_\xi\colon \xi<c\}$$ Then $B$ is Bernstein set.

This is not my proof but I have doubts about this proof is correct. More specifically, I did not see how $B$ will intersect each perfect at continuum many points. Am I missing some thing ?

Any help will appreciated greatly.

答: I think the proof is correct here is why $B$ is Bernstein set. For every prefect set $P$ there is a $\xi<c$ such that $P=P_\xi$. Then $$a_\xi\in P\cap  B$$ and $$b_\xi\in P\cap( \mathbb R\setminus B) $$ Thus $B$ is as needed.

Title: Bernstein set and it proof.


问:
How can I show that $T(\omega) = T(\overline{\omega})$ when $X_{t}(\omega)=X_{t}(\overline{\omega})$ for all $t \in [0,T(\omega)]\cap [0,\infty)$
Consider the stochastic process $(X_{t})_{t\in \mathbb R_{+}}$ on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t}^{X})_{t\in \mathbb R_{+}},\mathbb P)$ where $\mathcal{F}_{t}^{X}=\sigma (X_{s}:0\leq s\leq t) $. Let $T$ be a $(\mathcal{F}_{t}^{X})_{t\in \mathbb R_{+}}-$stopping time such that there is a pair $\omega, \overline{\omega} \in \Omega$ where:

$ X_{t}(\omega)=X_{t}(\overline{\omega})$ for all $t \in [0,T(\omega)]\cap [0,\infty)$

Then show that $T(\omega) = T(\overline{\omega})$

My failed attempt:

Perhaps a contradiction may help. Assume that $T(\omega) < T(\overline{\omega})$, then $\overline{\omega}\notin \{T\leq t\} \in \mathcal{F}_{t}^{X}$ for any $t \leq T(\omega)$, in particular:

$\overline{\omega}\notin \{T\leq T(\omega)\} \in \mathcal{F}_{T(\omega)}^{X}$ but how will this help me? Any ideas/tips?

答:
The $\sigma$-fields $\mathcal F^X_t$ ($t>0$) have a "saturation" property that is the key: If $\omega\in A\in\mathcal F^X_t$ and if $\overline\omega$ is such that $X_s(\omega) = X_s(\overline\omega)$ for all $0\le s\le t$ then also $\overline\omega\in A$.

Now suppose we have $\omega$ and $\overline\omega$ such that $X_s(\omega)=X_s(\overline\omega)$ for all finite $s$ in $[0,T(\omega)]$. Consider first the case  $T(\omega)<+\infty$ and apply the saturation property with $t=T(\omega)$ and $A=\{T=t\}$. The conclusion is that $\overline\omega\in A$ as well; that is, $T(\overline\omega) = t$.

To deal with the possibility that $T(\omega)$ might be $+\infty$, apply the preceding to the stopping time $T_n:=T\wedge n$ ($n=1,2,\ldots$) to conclude that  $T(\overline\omega)\wedge n=T_n(\overline\omega) = T_n(\omega) =n$ for each $n$. Consequently, $T(\overline\omega)=+\infty=T(\omega)$.


问:
Graphing $y= \frac{x^3}{x^2-1}$
I'm having a lot of problems trying to graph this... other than using a graphics calculator!

I know the domain is all real values of $x , x$ does not equal $-1 \text{or} 1$. The point of inflection is easy to find since when $x=0 , y=0.$ How do I find the local maxima and minima?

答:
Let $f(x):=\displaystyle\frac{x^3}{x^2-1}$. $\ $
Find out the following informations:

$\displaystyle\lim_{x\to-1-0} f(x)$, $\ \ \ \displaystyle\lim_{x\to-1+0} f(x)$

$\displaystyle\lim_{x\to +1-0} f(x)$, $\ \ \ \displaystyle\lim_{x\to +1+0} f(x)$

$\displaystyle\lim_{x\to-\infty} f(x)$, $\ \ \ \displaystyle\lim_{x\to +\infty}f(x)$

$f'(x)$, $\ \ f''(x)$

where $f'(x)=0$ -- these might be the local maximum/minimum points.
(Because if a differentiable function has local maximum or minimum at $x$, then its tangent line is horizontal there.)

if $f'(x)>0$, then $f$ is increasing locally around $x$

if $f''(x)>0$ then $f$ is convex locally around $x$ (convex from upwards, i.e. smiley-like)

if $x$ is an inflection point (where $f$ is changing convexity), we necessarily have $f''(x)=0$.

Title: Graphing $y= \frac{x^3}{x^2-1}$


问:
Proof that $\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}$ if $j_1+ j_2+ j_3+\cdots+ j_k = n$
$n, j_1, j_2, j_3, \ldots, j_k \in \mathbb{N}$ are such that: $j_1+ j_2+ j_3+\cdots+ j_k = n$. Prove that $$\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}.$$ I don't know how to do it. Tried induction and $e^{\ln(j_1! j_2! j_3! \cdots j_k!)}$ but non of those seem to be working.

答: The given expression counts the number of ways to order $n$ objects, with $j_1$ objects of type $1$, $j_2$ of type $2$ and so on until $j_k$ objects of type $k$, and there are no other objects or types. Thus the expression must be an integer.


问:
finite sets and countable sets.
Let $ A \neq \emptyset $. If $ A $ is finite and $ B $ is countable, then $ A \times B $ is countable.

My attempt is as follows, since $ A $ is finite and $ B $ is countable then $$ A = \{a_1, a_2, \ldots, a_n \} \quad \quad \quad B = \{b_1, b_2, \ldots, b_n ,\ldots \} $$
So $$ A \times B = \{(a_1, b_1), (a_1, b_2), (a_2, b_1), (a_1, b_3), \ldots (a_1, b_n), (a_2, b_ {n-1} ), \ldots, (a_n, b_1), \ldots \} $$ How could I set a bijection from $ \mathbb N \times \mathbb N $ to $ A \times B $, or is there an error in my argument?

答: The countable union of countable sets is countable, and in your case $$A \times B = \bigcup_{i = 1}^{n} \{a_i\} \times B$$so $A \times B$ is countable.

Title: finite sets and countable sets.


答: For induction on $k$, the base step $k=1$ gives a ratio $\tfrac{n!}{n!}=1$. To go from $k=j$ to $k=j+1$ in the inductive step, replace $\tfrac{1}{j_k^\text{old}!}$ with $\frac{1}{j_k^\text{new}!j_{k+1}!}$, wth $j_k^\text{new}=j_k^\text{old}-j_{k+1}$. This multiplies the ratio by $\frac{j_k^\text{old}!}{j_k^\text{new}!j_{k+1}!}=\binom{j_k^\text{old}}{j_{k+1}}$, which is an integer. In fact, this insight allows us to prove by telescoping product$$\frac{n}{\prod_{j=1}^kj_j!}=\prod_{l=1}^{k-1}\binom{n-\sum_{j=1}^{l-1}j_j}{j_l}.$$

Title: Proof that $\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}$ if $j_1+ j_2+ j_3+\cdots+ j_k = n$

Body: $n, j_1, j_2, j_3, \ldots, j_k \in \mathbb{N}$ are such that: $j_1+ j_2+ j_3+\cdots+ j_k = n$. Prove that $$\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}.$$ I don't know how to do it. Tried induction and $e^{\ln(j_1! j_2! j_3! \cdots j_k!)}$ but non of those seem to be working.


答:
Assuming that binomial coefficients are integers (they are, by a simple combinatorial argument), you can use them with telescoping to write your expression:
\begin{align*}
&\binom{j_1+\cdots +j_k}{j_1}\cdot \binom{j_2+\cdots +j_k}{j_2}\cdots \binom{j_{k-1}+j_k}{j_{k-1}}\cdot \binom{j_k}{j_k}\\
    &= \frac{(j_1+\cdots +j_k)!}{j_1!\cdot (j_2+\cdots +j_k)!}\cdot \frac{(j_2+\cdots +j_k)!}{j_2!\cdot (j_3+\cdots +j_k)!}\cdots \frac{(j_{k-1}+j_k)!}{j_{k-1}!\cdot j_k!}\cdot \frac{j_k!}{j_k!\cdot 0!}\\
    &= \frac{(j_1+j_2+\cdots +j_k)!}{j_1!\cdot j_2!\cdots j_k!}
\end{align*}


问:
Rotations by degrees other than $90, 180,$ and $270$.
Say I have a triangle with vertices $(0,0), (2,4), (4,0)$ that I want to rotate along the origin. Rotation by multiples of $90^{\circ}$ is simple. However, I want to rotate by something a bit more complicated, such as $54^{\circ}$. How do I figure out where the vertices would be then?</p>

答: In the answer to this question, I mentioned the formula for the rotation matrix; one merely takes the product of the rotation matrix with the coordinates (treated as 2-vectors) to get the new rotated coordinates. Note that I gave the matrix for clockwise rotation; for anticlockwise rotation, negate the angle (thus switching the sign of the two sine components).</p>

ContentLicense: CC BY-SA 2.5


问:
Entire sequence $g_{n} \rightarrow g$ with $g_{n}$ having only real zeros
Say we have a sequence of entire functions $g_{n}$, and $g_{n} \rightarrow g$ uniformly on compact sets in $\mathbb{C}$. I want to show that if each $g_{n}$ has only real zeros, then all the zeros of $g$ must also be real.

My best guess was to assume $g$ has a non real zero $w$ and take a compact set $K$ containing no real numbers and containing $w$. Then $g_{n}$ never vanishes on $K$ yet $g$ does. What if we look at the the sequence of functions $\frac{1}{g_{n}}$ analytic on $K$. Is it true $\frac{1}{g_{n}} \rightarrow \frac{1}{g}$ uniformly on compact sets? Because then we would get $\frac{1}{g}$ is analytic in $K$, a contradiction. Thanks for the help!

答: That is not true. Take for example $f_n(z)= \frac{z}{n}$. Then $f_n$ vanishes only at 0 but $f=0$ for all $z$.</p>


问:
Property of normal coordinates
Let $M$ be a Riemannian manifold and $\nabla$ the Levi-Civita conection. I need to prove the following.

Let $B$ be an open ball of radius $r$ in $T_pM$ such that $\left.exp_p\right|_B$ be a diffeomorphism over an open $U\subset M$ and
let $\{u_1,...,u_n\}$ be an orthonormal basis of $T_pM$.

Let $F:T_pM\supset B\rightarrow U\subset M$ given by $F(x_1,...,x_n)=exp_p(x_1u_1+\cdots+x_nu_n)$. Clearly, $(U,F^{-1})$ is a chart. Show that

$$\left(\nabla_{X_j}X_i\right)_p=0$$

where $X_i:=\dfrac{\partial }{\partial x_i}$.

答:
By the definition of your coordinates, $X_i$ is tangent to a geodesic through $p$ so $\left(\nabla_{X_i}X_i\right)_{p}=0$. For the crossed terms we use the polarization trick

$0=\left(\nabla_{X_i+X_j}X_i+X_j\right)_{p}=\left(\nabla_{X_i}X_{i}\right)_{p}+2\left(\nabla_{X_i}X_j\right)_{p}+\left(\nabla_{X_j}X_j\right)_{p}$.

As a side, these are called geodesic normal coordinates and they are really useful to carry out computations in Riemannian geometry.

Title: Property of normal coordinates


问:
Number of distinct quadruples of non-negative integers
I needed to find the number of distinct quadruples $(x_i), i=1,2,3,4,5$ of non-negative integers such that $x_1 \geq 3, x_2 \geq 3, x_4 \geq 8, x_5 \leq 3$, such that $x_1+x_2+x_3+x_4+x_5 \leq 23$. My answer does not agree with the correction. My reasoning is : substitute variables $$y_1=x_1-3 \geq 0,y_2=x_2-3 \geq 0,y_3=x_3,y_4=x_4-8 \geq 0,y_5=x_5,$$ and find the number of tuples satisfying $$\begin{cases}y_1+y_2+y_3+y_4 \leq 9 & (y_5=0)\\y_1+y_2+y_3+y_4 \leq 8 & (y_5=1)\\y_1+y_2+y_3+y_4 \leq 7 & (y_5=2)\\y_1+y_2+y_3+y_4 \leq 6 & (y_5=3)\\\end{cases}.$$ However (and this is where I disagree/don't understand the correction), this is the exact same thing as the number of tuples satisfying $y_1+y_2+y_3+y_4 \leq 9$, as all the former ones are included in this latter one. Hence my solution is $\overset{9}{\underset{k=0}{\sum}}{k+3 \choose 3}=715,$ where it seems that I should have found $1750$. Did I go wrong somewhere ?

答:
If you introduce a nonnegative integer slack variable $y_6$ so that you get four equalities in $y$, you obtain the correct count of
$$\sum_{k=6}^9 \binom{k+4}{4}=1750$$

Alternatively, count the number of nonnegative integer solutions to
$$\sum_{j=1}^6 y_j = 9$$
and subtract the ones that have $y_5 \ge 4$:
$$\sum_{j=1}^6 y_j = 5$$
This approach yields
$$\binom{9+5}{5}-\binom{5+5}{5}=1750$$


问:
Inequality $x - \sqrt{x^2 + 1} > 0$
This is how I solved this inequality:
$$ x - \sqrt{x^2 + 1} > 0$$
$$ x > \sqrt{x^2 + 1} $$
$$x < 0 \implies x^2 < x^2 + 1 \iff 1 > 0 \implies\text{inequality holds for all x}$$
$$x\geq 0 \implies x^2 > x ^2 +1 \iff 1 < 0 \implies \text{inequality doesn't hold }$$

The thing is that this shouldn't hold for any x. Can somebody explain why is my procedure wrong.

答:
The inequality is false.In fact is the oppposite.

Indeed, For any real number $x$, $x^2+1>x^2\ge0$, so (since $f(t)=\sqrt{t}$ is increasing on $[0,+\infty)$) $\sqrt{x^2+1}>\sqrt{x^2}=|x|$. But $|x|\ge x$, so $\sqrt{x^2+1}>x$, whence $$x-\sqrt{x^2+1}<0.$$

Title: Inequality $x - \sqrt{x^2 + 1} > 0$


答:
You need merely a injection from $A\times B$ to $\mathbb N$, so list the elements of $A\times B$ as follows:

$ (a_1,b_1),(a_2,b_1),...,(a_n,b_1),(a_1,b_2),(a_2,b_2),...,(a_n,b_2),...,$

$(a_1,b_m),(a_2,b_m),...,(a_n,b_m),...$

Title: finite sets and countable sets.

Body:
Let $ A \neq \emptyset $. If $ A $ is finite and $ B $ is countable, then $ A \times B $ is countable.

My attempt is as follows, since $ A $ is finite and $ B $ is countable then $$ A = \{a_1, a_2, \ldots, a_n \} \quad \quad \quad B = \{b_1, b_2, \ldots, b_n ,\ldots \} $$
So $$ A \times B = \{(a_1, b_1), (a_1, b_2), (a_2, b_1), (a_1, b_3), \ldots (a_1, b_n), (a_2, b_ {n-1} ), \ldots, (a_n, b_1), \ldots \} $$ How could I set a bijection from $ \mathbb N \times \mathbb N $ to $ A \times B $, or is there an error in my argument?


问:
Ordinal numbers arithmetic $((\omega + 1) + (\omega + 1))^\omega$ Cantor Normalform
is the attempt to bring the term in Cantor Normalform valid, especially how would you rate the step: $$(2^{\omega + 1}+1)^\omega = ((2^{\omega+1}) 2)^\omega$$

The term is $((\omega + 1) + (\omega + 1))^\omega = (\omega2 +1)^{\omega}=(2^{\omega+1}+1)^{\omega}=((2^{\omega+1}) 2)^\omega = (2^{\omega+2})^\omega=2^{(\omega2)\omega)}$

答:
The claimed equalities are true, but not particularly clear imo. I would rather write everything in base $\omega$ and bound the expression:

$$\omega^\omega\le(\omega\cdot2+1)^\omega\le(\omega^2)^\omega=\omega^{2\cdot\omega}=\omega^\omega$$


问:
Flirting Sequences, Convergence (Real Analysis)
Can anyone provide some insight or help into the following problem?

Definition: In a metric space, a sequence ${p_n}$ flirts with $p$ iff for each $\epsilon > 0$ there are $m,n \in \mathbb{N}$ where $m>n$ such that $0<d(p_n, p) < \epsilon$ and $d(p_n,p) < d(p_m,p)$.

Problem: If a sequence ${p_n}$ flirts with each point in a sequence ${q_n}$ and $q_n$ converges to $q$, does ${p_n}$ flirt with $q$? Prove it.

I wrote out the definitions for "${p_n}$ flirts with each point in a sequence ${q_n}$" and for "$q_n$ converges to $q$. I also wrote out the definition for "${p_n}$ flirt with $q$". I've been staring at them, trying to manipulate the first two so that they work for the third definition, but am stuck. I am not even sure how to proceed. Any help in trying to solve this problem will be helpful!

For reference, you can check this post for examples of flirting sequences.

By the way... I think the answer is no, but its more of intuition than anything.

答:
This is easier than you think.
Since $p_n$ flirts with each $q_n$ then you have that $d(p_n,q_n)\to 0$.
Because you are given $q_n\to q$, we have $||q_n-q||$ converges to 0. By the triangle inequality you obtain that
$d(p_n,q)\le d(p_n,q_n)+  ||q_n-q||$ therefore $d(p_n,q)<\epsilon$ right away. Hence the sequence $p_n$ actually converges to $q$.

Finally the fact $d(p_n,q)<d(p_m,q)$ comes from the fact that the sequence converges hence the distance is getting smaller and smaller, therefore
We can esasily find an $n,m$ with $d(p_n,q)<d(p_m,q)$ by arguing that as $n$ approaches to $\infty$, by the Archimedean Principle, such $n$ exists.

This was easier than we thought!  I love this stuff!


问:
Find volume using triple integration with cylindrical coordinates
Find volume of $E$ using triple integration and cylindrical coordinates, when $E$ is bounded by $$x^2+y^2=x,\quad y=0,\quad y=x,\quad z=0,\quad z=\sqrt{x^2+y^2}$$
I know that in cylindrical coordinates $$x=r\cos\varphi,\quad y=r\sin\varphi,\quad z=h,\quad \text{ where } r\geq0 \text{ and } 0\leq\varphi\leq2\pi$$
but I'm very confused how to set up this integral. Cylindrical coordinates are quite new for me and it's hard to understand, how to make this conversion. So I would be grateful if anyone can help me with this.

答:
I imagine the projection of your figure on $Oxy$ as upper segment cropped from circle by line $y=x$. So, firstly let's consider volume in Cartesian coordinates
$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{x}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$
So to find borders for cylindrical coordinates we need to solve inequalities

$$\begin{array}{}
0 \leqslant r \cos \phi \leqslant \frac{1}{2} & \\
r \cos \phi \leqslant r \sin \phi \leqslant \sqrt{\frac{1}{4} -\left(r \cos \phi-\frac{1}{2}\right)^2 } & \\
0 \leqslant z \leqslant r
 \end{array}$$
From 1-st and 2nd line inequalities we have $0 \leqslant \cos \phi \leqslant \sin \phi$, so we obtain $\phi \in \left[\frac{\pi}{4},\frac{\pi}{2}  \right]$ and $0 \leqslant r \leqslant \cos \phi$

$$\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$

Addition:
In case where we consider "sector" of circle i.e. part of circle between lines $y=x$ and $y=0$, then volume in Cartesian coordinates will be

$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{x}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx+\int\limits_{\frac{1}{2}}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$

Good news in this case is, that we can calculate this volume again using the founded volume in previous , circle segment, case  subtracting it from the volume over the entire semicircle, which is

$$\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx=\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$
so we obtain
$$\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi = \int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$


问:
Niceness of the projection of a closed subscheme of affine space?
Let $k$ be an algebraically closed field, and suppose $C\subseteq \mathbb{A}^{n+m}_k$ is a closed subscheme. What can we say about the image under the projection $\pi: \mathbb{A}^{n+m}_k\rightarrow \mathbb{A}^n_k$? We have the very general result from Chevalley's Theorem that $\pi(C)$ is a constructible set. Can we say more? For instance does it have a natural scheme structure?

$\pi(C)$ is not necessarily closed, for instance when $C=(xy-1)\subseteq\mathbb{A}^2_k$. However in this case we do have that $\pi(C)$ is locally closed rather than being a finite union of locally closed subsets. Does this hold in general? The examples I can think of where the image of a nice subscheme is properly a constructible subset don't involve a morphism as nice as the projection from an affine space.

Note that it doesn't seem that the image under $\pi$ of a locally closed set is locally closed. For instance take the standard "constructible but not locally closed set" example in $\mathbb{A}^2_k$ which is the complement of the $x$-axis, but including the origin (i.e. $\left(\mathbb{A}^2_k\setminus (y)\right) \cup (x,y)$). This is the image under $\pi:\mathbb{A}^3_k\rightarrow\mathbb{A}^2_k$ of $(z)\setminus (z,y)$ union the line $(t,t,1)$ (i.e. $((x-y),(z-1))$).

答:
I think $\pi(C)$ is not always locally closed.

For instance, take $C=V(x-yz)$ in $\mathbb{A}_k^3$, and $\pi:\mathbb{A}_k^3\rightarrow \mathbb{A}_k^2$ is defined by $\pi(x,y,z)=(x,y)$, then $\pi(C)=\left(\mathbb{A}_k^2\setminus (y)\right)\cup(x,y)$, which is exactly the standard "constructible but not locally closed set" in your question.


问:
What happens when you multiply a cycle index notation?
I recently was introduced to this lovely combinatorial object https://en.wikipedia.org/wiki/Cycle_index#Applications. Now I'm curious if there are any operations we can do with it? Like if we square it or multiply it with another does the outcome have any relationship to the original two? Can we take derivatives? Can we take sums? Or is this just purely an organizational object with no capicity to be manipulated algebraically?</p>

答:
The cycle index polynomial is an object attached to a permutation group, meaning an embedding $G \hookrightarrow S_n$ of a group into a symmetric group. Given two such permutation groups $G \hookrightarrow S_n, H \hookrightarrow S_m$, the product of their cycle index polynomials $Z(G) Z(H)$ is the cycle index polynomial $Z(G \times H)$ for the permutation group given by the embedding

$$G \times H \hookrightarrow S_n \times S_m \hookrightarrow S_{n+m}.$$

I'm not aware of any corresponding interpretation for sums. Derivatives correspond to generating functions for the number of cycles of a given length, and evaluating them with all variables set to $1$ tells you the average / expected number of such cycles. For example, using the exponential formula

$$\sum_{n \ge 0} Z(S_n) t^n = \exp \left( \sum_{k \ge 1} z_k \frac{t^k}{k} \right)$$

and taking derivatives with respect to each cycle variable $z_k$ you can compute (and there are other ways to do this computation) that the expected number of $k$-cycles in a random permutation in $S_n$ is $\frac{1}{k}$ (for $n \ge k$). By taking further derivatives you can compute higher moments also, and eventually learn that the number of $k$-cycles is asymptotically (as $n \to \infty$) Poisson distributed with rate $\lambda = \frac{1}{k}$. See this blog post for the calculation.


答:
The proposed inequality is equivalent to
\begin{align*}
x > \sqrt{x^{2} + 1}
\end{align*}

If $x \leq 0$, clearly there is no solution because the LHS is nonpositive and the RHS is positive.

Let us assume that $x > 0$. In this case, due to the AM-GM inequality, one has that
\begin{align*}
x > \sqrt{x^{2} + 1} > 2|x| = 2x \Rightarrow x < 0
\end{align*}
which contradicts our assumption.

Thus the proposed inequality has no solution.

Hopefully this helps!

Title: Inequality $x - \sqrt{x^2 + 1} > 0$

Body:
This is how I solved this inequality:
$$ x - \sqrt{x^2 + 1} > 0$$
$$ x > \sqrt{x^2 + 1} $$
$$x < 0 \implies x^2 < x^2 + 1 \iff 1 > 0 \implies\text{inequality holds for all x}$$
$$x\geq 0 \implies x^2 > x ^2 +1 \iff 1 < 0 \implies \text{inequality doesn't hold }$$

The thing is that this shouldn't hold for any x. Can somebody explain why is my procedure wrong.


问:
Evaluate $\int_{-\pi/2}^{\pi/2} (1+e^{2i\phi})^{\alpha} (1+e^{-2i\phi})^{\beta} \, \mathrm{d}\phi$
I should evaluate:

$$ \int_{-\pi/2}^{\pi/2} (1+e^{2i\phi})^{\alpha} (1+e^{-2i\phi})^{\beta} \, \mathrm{d}\phi $$

by using the binomial theorem and the identity:

$${}_2F_1 \left(\begin{array}{c}a , b \\ c \end{array};x\right) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_{0}^{1} t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a} \, \mathrm{d}t$$

So first using binomial theorem I get:

\begin{align*}
&\int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \binom{\alpha}{k} e^{2i\phi k} \sum_{k=0}^{\beta} \binom{\beta}{k} e^{-2i\phi k} \, \mathrm{d}\phi \\
&= \int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \sum_{l=0}^{\beta} \binom{\alpha}{k} e^{2i\phi k} \binom{\beta}{l} e^{-2i\phi l} \, \mathrm{d}\phi \\
&= \int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \sum_{l=0}^{\beta}  \binom{\alpha}{k} \binom{\beta}{l} e^{2i\phi(k-l)} \, \mathrm{d}\phi
\end{align*}

But from here I don't know how to proceed or rather how to use the identity. Any hints?

答: If $\beta$ is a non-negative integer, with $z=e^{2i\phi}$ this becomes$$\oint_{|z|=1}(1+z)^{\alpha+\beta}\frac{dz}{2iz^{\beta+1}}=\pi[z^\beta](1+z)^{\alpha+\beta}=\pi\binom{\alpha+\beta}{\beta}=\frac{\pi\Gamma(\alpha+\beta+1)}{\Gamma(\alpha+1)\Gamma(\beta+1)}.$$Update: @Iridescent has pointed out how we can generalize to complex $\beta$. The integral is $2^{\alpha+\beta-1}\int_0^{\pi/2}\cos^{\alpha+\beta}\phi\cos[(\alpha-\beta)\phi]d\phi$, since the integrand's imaginary part integrates to $0$ on $[-\tfrac{\pi}{2},\,\tfrac{\pi}{2}]$. An old question proves this is indeed $\tfrac{\pi\Gamma(\alpha+\beta+1)}{\Gamma(\alpha+1)\Gamma(\beta+1)}$.


问:
Galois action on $K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell$
Let $K/\mathbb{Q}$ be a finite extension and let $\ell$ be a prime number.

For each prime $\lambda$ of $K$ lying over $\ell,$ choose an embedding $\sigma_\lambda:K\hookrightarrow K_\lambda$ of $K$ into the $\lambda$-adic completion of $K.$

There is a natural isomorphism
$$K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell \longrightarrow \prod_{\lambda\mid\ell} K_\lambda\,; \; \; \; x\otimes y \longmapsto (\sigma_\lambda(x)y \, : \, \lambda\mid\ell).$$

If we now assume, in addition, that $K/\mathbb{Q}$ is Galois with group $G,$ then $G$ has a natural action on $K\otimes_\mathbb{Q} \mathbb{Q}_\ell$ (acting on the first factor).

Is there a "nice" description of resulting action on the product $\prod_{\lambda\mid\ell} K_\lambda$?

答: For example, for unramified $\ell$, the Galois group permutes the product factors, with decomposition groups being the isotropy groups. (We know that decomposition groups surject to local Galois groups...)


问:
complete the table by providing an example of a binary operation $*$ defined on $\{a , b ,c\}$
I have a problem with one of my questions. The question is:

complete the table by providing an example of a binary operation $*$ defined on $\{a , b ,c\}$
such that $*$ is commutative and has the identity element $c$.
\begin{array}{c|ccc}
\ast & a & b & c \\
\hline
a & a & ? & ? \\
b & ? & b & ? \\
c & ? & ? & c \\
\end{array}

(I need the letters in place of the ?)

I don't understand what they want me to do ???

答: $c$ is the identity and the operation is commutative so $$a\cdot c=c\cdot a=a$$ Similarly $$b\cdot c=c\cdot b=b$$ It seems to me that $a\cdot b$ can be $a$ or $b$.</p>


问:
Show $\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$
I am tasked with the following:

Show the following inequality for $\lvert b \rvert < \frac{\lvert a \rvert}{2}$ where $a,b \in \Bbb R$:
$$\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$$

I'm a bit at a loss as to where to even start with this one. I'm mostly looking for a little nudge.

答:
WLOG assume $a>0$ then $|b|<\frac{a}{2}$
$$
\frac{1}{|a-b|} = \frac{1}{a-b}<\frac{1}{a-\frac{a}{2}}=\frac 2a=\frac{2}{|a|}.\blacksquare
$$


问:
If $\gcd (a,n) = 1$, how can we prove that $\gcd(a + kn, n) = 1$?
If $\gcd (a,n) = 1$, then how can we prove that $\gcd(a + kn, n) = 1$, where $k$ is any integer?

I feel like I should start with linear combinations theorem, but I'm quite lost. Any help would be greatly appreciated!

答: If some prime $p|n$, $p|a+kn$, then $p|kn$ and...


问:
Bessel function, characteristic function, semicircle distribution
I am trying to prove that $\dfrac{1}{x} J_1(2x)$, where $J_n$ is the Bessel function of order n, is the characteristic function of the semicircle distribution, i.e. $σ(x)=\dfrac{1}{2π}\sqrt{4-x^2}
 \textbf{1}_{|x|<2} $. Basically, I would like to calculate the integral

$$\dfrac{1}{2π} \displaystyle\int_{\mathbb{R}}{\dfrac{1}{t}J(2t)}e^{-itx} dt $$

So, I do not want to calculate the characteristic equation of $σ$, I want to find $σ$ when I am given the characteristic function. I have tried some integral forms of $J_1$ but can't seem to find what I am looking for. Any hints?

答:
$$J_1(z) = \dfrac{1}{iπ} \int_{0}^π e^{iz\cos θ}\cos θ  dθ $$

Now, we want to calculate the integral shown above to find the density function corresponding to the characteristic function $ \dfrac{1}{x} J_1(2x) $.

\begin{align*}
f(x) &= \dfrac{1}{2π} \int_{\mathbb{R}} φ(t) e^{-itx} dt = \dfrac{1}{2π} \int_{\mathbb{R}} \dfrac{1}{t} J_1\left(2t\right) e^{-itx} dt \\[7pt]
&= \dfrac{1}{2iπ^2} \int_{\mathbb{R}}\int_{0}^π \dfrac{1}{t} e^{2it\cos θ }\cos θ e^{-itx} dθ dt  \\[7pt]
&=  \dfrac{1}{2iπ^2} \int_{0}^π  \cos θ  \underbrace{ \int_{\mathbb{R}}  \dfrac{1}{t}  e^{it(2\cos θ-x)} dt }_{-iπ \text{sgn}(2cosθ-x) } dθ \\[7pt]
&= - \dfrac{1}{2π} \int_{0}^π \cos θ \text{sgn}(2\cos θ-x)  dθ
\end{align*}

If $ x > 2 $ then $ \text{sgn} (2 \cos θ-x)  = -1$, and if  $ x < -2 $ then $ \text{sgn} (2 \cos θ-x)  = 1$ and so, since $ \int_{0}^π \cos θ dθ  = 0$, $ f(x) = 0$ as well. Now, let $ -2\le x\le 2 $. If we set $ u=2\cos θ-x  $, then

\begin{align*}
&-\dfrac{1}{2π}\int_{-2-x}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}  \text{sgn}(u)  du \\[7pt]
&= \dfrac{1}{2π}\int_{-2-x}^{0} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}  du-\dfrac{1}{2π}\int_{0}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}}   du\\[7pt]
   &= \dfrac{1}{2π}  \left(    2 \int_{-1}^{\frac{x}{2}}  \dfrac{y}{\sqrt{1-y^2}}  dy - 2\int_{\frac{x}{2}}^{1}  \dfrac{y}{\sqrt{1-y^2}}   dy   \right)\\[7pt]
&=\dfrac{1}{2π} \left(     2 \sqrt{1-y^2} \bigg\rvert_{-1}^{x/2}  -  2 \sqrt{1-y^2} \bigg\rvert_{x/2}^{1}    \right) = \dfrac{1}{2π} \sqrt{4-x^2}
\end{align*}


问:
Graphing $y= \frac{x^3}{x^2-1}$
I'm having a lot of problems trying to graph this... other than using a graphics calculator!

I know the domain is all real values of $x , x$ does not equal $-1 \text{or} 1$. The point of inflection is easy to find since when $x=0 , y=0.$ How do I find the local maxima and minima?

答:
Even without calculus, we can make a rough sketch.  $$y=\frac{x^3}{(x+1)(x-1)}$$

This function is odd, so it's rotationally symmetric about the origin.  It is positive on $(1,+\infty)$, negative on $(0,1)$, positive on $(-1,0)$, and negative on $(-\infty, -1)$.  It blows up as it approaches $\pm 1$, so $\lim_{x\rightarrow 1+}y=+\infty$, while $\lim_{x\rightarrow 1-}y=-\infty$.  Similarly $\lim_{x\rightarrow -1+}y=+\infty$, while $\lim_{x\rightarrow -1-}y=-\infty$.  The degree of the numerator is greater than the degree of the denominator (by 1), so $\lim_{x\rightarrow \infty}y=\infty$, and $\lim_{x\rightarrow -\infty}y=-\infty$.  Hence one should expect a local minimum somewhere in $(1,\infty)$ and a local maximum somewhere in $(-\infty,-1)$.  The only place it crosses the $x$-axis is where the numerator is zero, namely $x=0$.

To compute exactly what the minima are, and to determine inflection points, you'll need to take the derivative though.  Note that $x=0$ may or may not be an inflection point; the correct test is given by @Berci's answer.

Title: Graphing $y= \frac{x^3}{x^2-1}$


问:
Proof by induction - is this correct?
Proof by induction that: For all $n\in \mathbb{N}$, $7^{2n}+ 2^{(2n+1)}$ is a multiple of $3$.

I think I got pretty far but I don't know if it's correct/how I should continue. My working:

Base case: show that $n=1$ holds: $7^2 + 2^3 = 57$ and $3|57$ so $n=1$ holds.

Assume that $n=k$ holds: $7^{2k}+2^{(2k+1)}$.

Prove that $n=k+1$ holds: $7^{(2k+2)} + 2^{(2k+3)}$

I rearranged this so it's in the same form as $n=k$ and got $7^2 \cdot 7^{2k} + 2^2 \cdot 2^{(2k+1)}$.

I then simplified and rearranged this to $4 \cdot 7^2k + 4 \cdot 2^{(2k+1)} + 45 \cdot 7^{2k}$.

Taking out a multiple of $4$ gives $4(7^{2k} +2^{2k+1}) + 45 \cdot 7^{2k}$ and since $(7^{2k} +2^{2k+1})$ is a multiple of $3$, I let it equal $3m$ so it's $4(3m) + 45 \cdot 7^{2k}$.

Finally, I took out a multiple of $3$ to get $3(4m + 15 \cdot 7^{2k})$ which is a multiple of $3$, hence the statement holds by induction.

Is my proof completely correct? Was there an easier way I could've done this?

答:
Your proof is correct but too verbose. Why not just write down
$$
7^{2k+2}+2^{2k+3} = 49(7^{2k})+4(2^{2k+1})=45(7^{2k})+4(7^{2k}+2^{2k+1})
$$
and you are done.

Title: Proof by induction - is this correct?


答:
1)$x \le 0;$ the inequality is false.

2)$x>0;$

Assume the inequality is true.

Multiply both sides by $x+\sqrt{x^2+1} >0.$

Then

$(x-\sqrt{x^2+1})(x+\sqrt{x^2+1})>0;$

$x^2-x^2-1>0;$

$-1>0,$ a contradiction.

Title: Inequality $x - \sqrt{x^2 + 1} > 0$


问:
Prove that following limit exists

Suppose that $x_n$ is a sequence of real numbers that converges to $1$ as $n\to\infty$. Using the definition, prove that following limit exists.
$$\frac{x^2_n − e}{x_n}→ 1 − e \quad\text{ as }\quad n \to\infty.$$

Here is my attempt, Is there anything wrong here?

By hypothesis, given $\epsilon>0$ there is an $N_1\in\mathbb{N}$ such that $n\geq N
_1$ implies $|x_n-1|<\frac{\epsilon}{5}$.

$$|\frac{x_n^2-e}{x_n}-(1-e)|=|x_n-\frac{e}{x_n}-1+e|=|(x_n-1)+e(\frac{x_n-1}{x_n})|$$

since $e<3$ and $\epsilon=\frac{5}{4}$ choose $N_2$ sutch that $n≥N_2$  implies $|x_n-1|<\frac{1}{4}$, (i.e
$x_n>\frac{3}{4})$ set $N=\max(N_1,N_2)$

$$|\frac{x_n^2-e}{x_n}-(1-e)|=|x_n-1||1+\frac{e}{x_n}|<|x_n-1|(1+\frac{3}{x_n})<5|x_n-1|<\epsilon$$

for all $n≥N$.

Thank you!!!

答:
That is not correct. You cannot decide that $\varepsilon=\frac54$; it can be any number greater than $0$.

You can do it as follows. Take $\varepsilon>0$. Now, take $N_1\in\Bbb N$ such that $n\geqslant N_1\implies|x_n-1|<\frac\varepsilon2$. Then, take $N_2\in\Bbb N$ such that $n\geqslant N_2\implies|x_n-1|<\frac12$ and note that this implies that $|x_n|>\frac12$. Then, take $N_3\in\Bbb N$ such that $n\geqslant N_3\implies|x_n-1|<\frac1{4e}\varepsilon$. Now, if $N=\max\{N_1,N_2,N_3\}$, then\begin{align}n\geqslant N\implies\left|\frac{x_n^{\,2}-e}{x_n}-(1-e)\right|&=\left|x_n-1+e\left(\frac{x_n-1}{x_n}\right)\right|\\&\leqslant|x_n-1|+2e|x_n-1|\text{ (because $\frac1{|x_n|}<2$)}\\&\leqslant\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}

Title: Prove that following limit exists


问:
A problem connecting Hilbert spaces and normed linear spaces
The problem is the following:

Let $X$ be a normed linear space and $Y$ be a Hilbert space.

Let $ A : X \rightarrow Y$ be a linear operator.

For $y \in Y \, $, let $ \, S_y = \{ x \in X : \| Ax − y \| \le \|Au − y \|, \, \forall \, u \in X \} $

Show that $S_y$ is nonempty if and only if $ \, y \in R(A) + R(A)^{\bot}$ .

In addition, if $X$ is Hilbert and $N(A)$ is a closed subspace, show that for every $y \in R(A) + R(A)^{\bot}$,
there is a unique $x_y \in S_y$ such that
$ \| x_y \| = inf \{ \| x \|: x \in S_y \}$.

My attempt:

I guess that if I prove that $S_y$ is a convex subset of $Y$, and as $Y$ is Hilbert, I can say that $ \exists ! \, x_y \in S_y $, as required in the second part.
For the first part. I'm not quite sure how to go about it.

答:
Assume that $y \in R(A)\dotplus R(A)^\perp$, so it can be written as $y = Ax+z$ for some $x \in X$ and $z \in R(A)^\perp$.

Now for any $u \in X$ we have
$$\|Ax-y\|^2 = \|-z\|^2 \le \|Au\|^2+\|-z\|^2 = \|Au-z\|^2$$
since $Au \perp -z$. Therefore $x \in S_y$ so $S_y$ is nonempty.

Conversely, assume $y \notin R(A)\dotplus R(A)^\perp$. Then since $\overline{R(A)}\dotplus R(A)^\perp = Y$, it has to be $y\in (\overline{R(A)}\setminus R(A))\dotplus R(A)^\perp$. We can therefore write $y = a+b$ with $a \in \overline{R(A)}\setminus R(A)$ and $b \in R(A)^\perp$. Pick a sequence $(x_n)_n$ in $X$ such that $Ax_n \to a$.

Assume that $x \in S_y$. Since $Ax \ne a$, we have $\|Ax-a\|>0$ so there exists $n\in\Bbb{N}$ such that $\|Ax_n-a\|<\|Ax-a\|$. Therefore
$$\|Ax_n - y\|^2 = \|Ax_n - a - b\|^2 = \|Ax_n-a\|^2+\|-b\|^2 \\
\le \|Ax-a\|^2+\|-b\|^2 = \|Ax - a - b\|^2 = \|Ax - y\|^2.$$
since $Ax_n-a, Ax-a \perp b$. This is a contradiction. Therefore $S_y = \emptyset$.

For the second part, we need to show that $S_y$ is a closed, convex subset of $Y$. Assume $(x_n)_n$ is a sequence in $S_y$ such that $x_n \to x_0 \in X$.

If we write $y = Ax+z$ with $x \in X$ and $z \in R(A)^\perp$, then (as @Berci suggested) for every $v \in S_y$ it holds $Av = Ax$. Indeed,
$$\|Av-Ax\|^2+\|-z\|^2 = \|Av-y\|^2 \le \|Ax-y\|^2 = \|-z\|^2$$
since $Av-Ax \perp -z$ and therefore $\|Av-Ax\| =0$.

Now we also have $Ax_n = Ax$ and hence $(x_n-x)_n$ is a sequence in $N(A)$. Since $N(A)$ is closed, we get that the limit $x_0-x$ is also in $N(A)$. Therefore $Ax_0 = Ax$. The first part of the proof showed $x \in S_y$ so for every $u \in X$ we have
$$\|Ax_0-y\| =\|Ax-y\| \le \|Au-y\|$$
and hence $x_0 \in S_y$.

For convexity let $x,x' \in S_y$ and let $\alpha, \beta \ge 0$ such that $\alpha+\beta = 1$. For any $u \in X$ we have
\begin{align}
\|A(\alpha x+\beta x') -y\| &= \|\alpha(Ax-y)+\beta(Ax'-y)\| \\
&\le \alpha \|Ax-y\| + \beta\|Ax'-y\| \\
&\le \alpha\|Au-y\|+\beta\|Au-y\| \\
&= \|Au-y\|
\end{align}
so $\alpha x + \beta x' \in S_y$. Hence $S_y$ is convex. Now use the fact that every closed, convex, nonempty subset of a Hilbert space has a unique element of minimal norm.


问:
Prove that there exists integers $r,s$ such that $r^k\equiv a \pmod{p_1}$ and $s^k \equiv a \pmod{p_2}$
Given that $k$ is a positive integer and $p_1,p_2$ are co primes such that for some integers $q,a$ and $r,s$
$$q^k \equiv a \pmod{p_1p_2} \iff r^k \equiv a \pmod{p_1} \:\text{ and }\:s^k\equiv a \pmod{p_2}$$

My try:
I tried to prove from right to left first.

$$r^k=mp_1+a$$ and $$s^k=np_2+a$$

Now by Euclid's algorithm we have $m=m'p_2+r_2$ and $n=n'p_1+r_1$

Thus we get $$r^k=m'p_1p_2+p_1r_2+a$$ and $$s^k=n'p_1p_2+r_1p_2+a$$

I got stuck here.

答:
For the right to left case, since $p_1$ and $p_2$ are relatively prime, as Bill Dubuque's comment indicates, you can just use the Chinese remainder theorem to confirm there's a solution for $q \equiv r \pmod{p_1}$ and $q \equiv s \pmod{p_2}$. Alternatively, Bézout's identity states there exist integers $x_1$ and $y_1$ such that

$$x_1 p_1 + y_1 p_2 = 1 \tag{1}\label{eq1A}$$

Multiply both sides by $r - s$ and, for simpler algebra, set $x_2 = x_1(r - s)$ and $y_2 = y_1(r - s)$. This gives

$$\begin{equation}\begin{aligned}
x_2 p_1 + y_2 p_2 & = r - s \\
s + x_2 p_1 & = r - y_2 p_2
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Let $q = s + x_2 p_1 = r - y_2 p_2$. Thus, $q \equiv s \pmod{p_1} \implies q^k \equiv s^k \equiv a \pmod{p_1}$ and $q \equiv r \pmod{p_2} \implies q^k \equiv r^k \equiv a \pmod{p_2}$. Since $p_1$ and $p_2$ are coprime, this means

$$q^k \equiv a \pmod{p_1p_2} \tag{3}\label{eq3A}$$

For the left to right case, just choose $r = s = q$.


答:
Since you asked for an easier way (and assuming you must use induction), consider using modular arithmetic:

For the base case for $n=1$ we have $7^{2n}+ 2^{2n+1}\equiv 1^{2n}+-1^{2n+1} \equiv 0 \pmod 3$

Then $7^{2n+2}+2^{2n+3}=7^{2}\cdot7^{2n}+4\cdot2^{2n+1}\equiv7^{2n+2}+2^{2n+1}\equiv0\pmod 3$ by hypothesis.

Although, this is a bit contrived since in this case the same can be done without needing to check the base case separately.

Title: Proof by induction - is this correct?

Body:
Proof by induction that: For all $n\in \mathbb{N}$, $7^{2n}+ 2^{(2n+1)}$ is a multiple of $3$.

I think I got pretty far but I don't know if it's correct/how I should continue. My working:

Base case: show that $n=1$ holds: $7^2 + 2^3 = 57$ and $3|57$ so $n=1$ holds.

Assume that $n=k$ holds: $7^{2k}+2^{(2k+1)}$.

Prove that $n=k+1$ holds: $7^{(2k+2)} + 2^{(2k+3)}$

I rearranged this so it's in the same form as $n=k$ and got $7^2 \cdot 7^{2k} + 2^2 \cdot 2^{(2k+1)}$.

I then simplified and rearranged this to $4 \cdot 7^2k + 4 \cdot 2^{(2k+1)} + 45 \cdot 7^{2k}$.

Taking out a multiple of $4$ gives $4(7^{2k} +2^{2k+1}) + 45 \cdot 7^{2k}$ and since $(7^{2k} +2^{2k+1})$ is a multiple of $3$, I let it equal $3m$ so it's $4(3m) + 45 \cdot 7^{2k}$.

Finally, I took out a multiple of $3$ to get $3(4m + 15 \cdot 7^{2k})$ which is a multiple of $3$, hence the statement holds by induction.

Is my proof completely correct? Was there an easier way I could've done this?


问:
Proof summary by induction
The question is : " proof for any N that -"
$$\sum _{k=1}^n\:\left(-1\right)^{k+1}\cdot \left(2k-1\right)=\left(-1\right)^{n+1}\cdot n$$
I`ve reached to
$$\left(-1\right)^{n+1}\cdot n+\left(-1\right)^{n+2}\cdot \left(2n+1\right)=\left(-1\right)^{n+2}\cdot \left(n+1\right)$$
tried to manipulate the equation few times and failed  while trying to reach from left side to right side.

would be glad if someone can show a hint how to continue / solve it .
Thanks

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 1}^{n}x^{2k - 1} =
{x^{2n + 1} - x \over x^{2} - 1}}$.

Derive, both members, respect of $\ds{x}$:
$$
\sum_{k = 1}^{n}x^{2k - 2}\,\pars{2k - 1} =
{\pars{2n - 1}x^{2n + 2} - \pars{2n + 1}x^{2n} + x^{2} + 1 \over \pars{x^{2} - 1}^{2}}
$$

Set $\ds{x = \ic}$:
$$
\sum_{k = 1}^{n}\pars{-1}^{k + 1}\,\pars{2k - 1} =
\bbx{\pars{-1}^{n + 1}\,\,n} \\
$$

Title: Proof summary by induction


答:
We use Euler's critrion to find r and s:

If $a \equiv t^2 \mod(p_1)$ then:

$a^{\frac{p_1-1}2} \equiv 1 \mod (p_1)$

So : $a^{\frac{p_1-1}{2}+1} \equiv a\mod(p_1)$

Suppose:

$\frac{p_1-1}{2}+1=\alpha \cdot \beta$

then $r$ can  be $a^{\alpha}$ and $k= \beta$ and we have:

$r^k \equiv a \mod p_1$

Similarly  suppose:

$\frac{p_2-1}{2}+1=\gamma \cdot \beta$

and $s=a^{\gamma}$ then $k=\beta$ and we have:

$s^k \equiv a \mod p_2$

provided $a \equiv m^2 \mod(p_2)$

Title: Prove that there exists integers $r,s$ such that $r^k\equiv a \pmod{p_1}$ and $s^k \equiv a \pmod{p_2}$

Body:
Given that $k$ is a positive integer and $p_1,p_2$ are co primes such that for some integers $q,a$ and $r,s$
$$q^k \equiv a \pmod{p_1p_2} \iff r^k \equiv a \pmod{p_1} \:\text{ and }\:s^k\equiv a \pmod{p_2}$$

My try:
I tried to prove from right to left first.

$$r^k=mp_1+a$$ and $$s^k=np_2+a$$

Now by Euclid's algorithm we have $m=m'p_2+r_2$ and $n=n'p_1+r_1$

Thus we get $$r^k=m'p_1p_2+p_1r_2+a$$ and $$s^k=n'p_1p_2+r_1p_2+a$$

I got stuck here.


问:
Upper bound of first return probability of Markov chain

I only have problem about (b).

My efforts:

Consider the special case in which all transition probabilities are 1. Then the first return probabilities are $f^n_{jj}=1$ when $n=r$ and $f^n_{jj}=0$ when $n\neq r$. Since the problem requires $n>r$, any $\alpha$ works for this case.
$$\begin{bmatrix}0&1&0&0
\\\\
0&0&1&0
\\\\
0&0&0&1
\\\\
1&0&0&0\end{bmatrix}$$

Next consider the case in which not every transition probability is 1. Then at least one of them is less than 1. If that less-than-1 probability is only on row $j$ of the transition matrix, then the result is similar to the all-1 case and any $\alpha$ works.
$$\begin{bmatrix}p_j&q_j&0&0
\\\\
0&0&1&0
\\\\
0&0&0&1
\\\\
1&0&0&0\end{bmatrix}$$

Since the problem requires $n>r$, the state must not be in $j$ for $n\leq r$. Suppose the state is in $i$ when $n=r$. Then we have $0<p^r_{ji}<1$. If no $p_{ik}$ is one, then we can simply take $\alpha$ to be the largest of these $p_{ik}$ and the $r$th root of $p^r_{ji}$. But if there is one, I don't know what to do. Since the 1 increases the number of transitions and also the desired $\alpha$.

答:
Let $f^k_{jj}$ be the first time return probability in $k$ steps. Since $j$ is recurrent, $\sum^\infty_{k=1}f^k_{jj}=1$.

Let $p_n$ = Pr(return after $n$). We have the following relationships of events:

{return after $n+1$} $\subset$ {return after $n$}

{return after $n$} = {return at $n+1$} $\cup$ {return after $n+1$}

Let $a_n=\sum^n_{k=1}f^k_{jj}$.

Pr(return after $n+1$)

= Pr(return after $n+1$|after $n$)Pr(after $n$)
\begin{equation}
=\frac{\sum_{k>n+1}f^k_{jj}}{\sum_{k>n}f^k_{jj}}p_n
=\frac{1-\sum_{k=1}{n+1}f^k_{jj}}{1-\sum^n_{k=1}f^k_{jj}}p_n
=\frac{1-a_{n+1}}{1-a_n}p_n=\frac{1-a_n-f^{k+1}_{jj}}{1-a_n}p_n
=(1-\frac{f^{k+1}_{jj}}{1-a_n})p_n
\end{equation}

Let $\alpha$ = max{sup$_{n>r}(1-\frac{f^{k+1}_{jj}}{1-a_n}),\sqrt[r]{p_r}$}.


问:
Unique property of differentiable functions
Recently I was going through the chain-rule derivative proof in Stewart Calculus-

It's a great approach to avoid the objections of $\Delta$$u$$=0$. But I don't understand why do we need to prove that $\epsilon$ is a continuous function of$\Delta$$x$. The book claims it to be a property of a differentiable function, but what is the significance of this property and how are we helped here by using it? Also, does it really help in some later stages of calculus? If so, then how? It'd be really helpful if someone pointed out why it was necessary to use $\epsilon$ as a continuous function here.

答:
Notice that here we are using that $y=f(x)$ and increment of $y$ is defined by
$$
\Delta y=f(a+\Delta x)-f(a).
$$
By the definition of derivative
$$
f'(a)=\lim\limits_{\Delta x \to 0}\frac{\Delta y}{\Delta x}.
$$
Defining $\epsilon$ as
$$
\epsilon=\frac{\Delta y}{\Delta x}-f'(a)
$$
for $\Delta x \neq 0$ we have that
$$
\lim\limits_{\Delta x \to 0} \epsilon=\lim\limits_{\Delta x \to 0}\frac{\Delta y}{\Delta x}-f'(a)=0.
$$
Define then $\epsilon$ to be $0$ when $\Delta x=0$. If we consider $\epsilon=\epsilon(\Delta x)$ as a function of $\Delta x$ we have that $\epsilon$ is continuous. In fact, notice that the only point that continuity might fail is at $\Delta x=0$ but as we defined $\epsilon(\Delta x)$
$$
\lim\limits_{\Delta x \to 0} \epsilon=0=\epsilon(0).
$$
Hence $\epsilon$ is in fact continuous with respect to $\Delta x$. This property is what allows you to write equations 8 and 9 where $\epsilon_1 \to 0$ as $\Delta x \to 0$ and $\epsilon_2 \to 0$ as $\Delta u \to 0$.

Notice that what this property states is that if $f$ is differentiable at $a$ then you can find a continuous function on $\Delta x$ here denoted by $\epsilon$ such that
$$
f(a+\Delta x)-f(a)=f'(a) \Delta x +\epsilon \Delta x
$$
where $\epsilon \to 0$ as $\Delta x \to 0$. Notice that the converse is also true. If there is a function $\epsilon$ of $\Delta x$ and a constant $b$ such that
$$
f(a+\Delta x)-f(a)=b \Delta x +\epsilon \Delta x
$$
where $\epsilon \to 0$ as $\Delta x \to 0$ then $f$ is differentiable at $a$ and $f'(a)=b$. In fact, we have that for $\Delta x \neq 0$
$$
\frac{f(a+\Delta x)-f(a)}{\Delta x}=b+\epsilon
$$
and
$$
\lim\limits_{\Delta x \to 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}=\lim\limits_{\Delta x \to 0}b+\epsilon=b+\lim\limits_{\Delta x \to 0}\epsilon=b+0=b.
$$
Hence by definition $f$ is differentiable at $a$ and $f'(a)=b$.


问:
Homomorphism between $Q_8$ and $S_4$
Prove that there does not exist a homomorphism $\phi :Q_8 \rightarrow S_4$ such that $\phi (i) = (1 2 3 4), \phi (j) = (1 2 4 3) $.

Using homomorphism properties, we obtain $\phi(i) \phi(j) = \phi(ij) = \phi(k) = (1 2 3 4)(1 2 4 3) = (1 3 2)$, then $\phi(j) \phi(k) = \phi(jk) = \phi(i) = (1 2 4 3)(1 3 2) = (3 4)\neq (1 3 2)$, hence $\phi$ is not even a well-defined function. Is what I did right? I am not completely convinced. Thanks!

答:
In fact, it is not much harder to show that there is no homomorphism $\varphi:Q_8\to S_4$ such that $\varphi(i)=(1234)$.
(In other words, we don't even need the hypothesis on $\varphi(j)$.)

Here is a sketch of the proof. Since $S_4$ does not contain a subgroup isomorphic to $Q_8$ (such a subgroup would be a Sylow $2$-subgroup, but those are dihedral), it follows that $\varphi$ cannot be injective. So the kernel of $\varphi$ is non-trivial. Now, $Q_8$ has a unique minimal normal subgroup, namely $\langle -1\rangle$, so this must be contained in the kernel. So  $\varphi(i)^2=\varphi(i^2)=\varphi(-1)$ is the identity.

Title: Homomorphism between $Q_8$ and $S_4$


问:
Integrating $f(x,y,z)$ within the region of a cylinder and a hyperboloid in cartesian coordinates
I'm confident with solving this question in cylindrical coordinates (for which there are few answers) but I'm doubtful about cartesian coordinates and can't seem to figure out the limits. $z$ needs to be the outer integral in the below question:

Right now, I've figured that:

$z$ must vary from $-1$ to $1$ (since it describes the radius of the cylinder).

If we make $x$ the inner integral and dependent on $y$ we could say that it varies from $-2y^2$ to $(1-2y^2)$ (obtained by substituting the cylinder equation and adding in $a=0$ & $a=1$

I'm not sure how to evaluate the middle integral and how exactly there are 2 regions over which we must integrate.

答:
The best way solve problems like this is to try and visualize the solid $V$ in your question. Notice the surface $z=\sqrt{a^2+x^2+y^2}$ is precisely surface formed by rotating the hyperbola $z=\sqrt{a^2+x^2}$ in the $xz-$plane around the $z-$ axis. On the other hand, the cylinder $y^2+z^2=1$ is the surface formed by taking a circle of radius $1$ in the $xz-$plane and translating it across the $x-$axis. You should first notice that $V$ is bounded below by the hyperbloid and above by the cylinder. Moreover, projection of the intersection of these two surfaces onto the $xy-$plane is the ellipse $x^2+2y^2=1-a^2$. So, one way to express this integral would be as $$\int_{-\sqrt{1-a^2}}^\sqrt{1-a^2} \int_{-\sqrt{\frac{1-a^2-x^2}{2}}}^\sqrt{\frac{1-a^2-x^2}{2}} \int_{\sqrt{a^2+x^2+y^2}}^\sqrt{1-y^2}f(x,y,z)dzdydx$$ Try to envision the solid $V$ to see how we can also express this integral as the sum $A+B$ where
$$A=\int_a^{\sqrt{\frac{a^2+1}{2}}} \int_{-\sqrt{z^2-a^2}}^{\sqrt{z^2-a^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdydz$$ $$B=\int_{\sqrt{\frac{a^2+1}{2}}}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdydz$$ or even better as $$\int_{-\sqrt{\frac{1-a^2}{2}}}^{\sqrt{\frac{1-a^2}{2}}} \int_{\sqrt{y^2+a^2}}^{\sqrt{1-y^2}} \int_{-\sqrt{z^2-y^2-a^2}}^{\sqrt{z^2-y^2-a^2}}f(x,y,z)dxdzdy$$


问:
Finding expected value of operating units
Given $N$ operating units, every unit consists $k$ elements.

If one element is missing from one operating unit, it stops working.
The units operate independently.

When $m$ elements are removed from the system, what will be the expected value of the number of operating units?

So, we have to choose $m$ element out of $k$ groups.

If we choose so that $\,m=k\,$, the result can be minimum $\,0\,$ or maximum $\,N-1\,$ operating unit.

In case $\,m<k\,$, it is minimum $\,1\,$ or maximum $\,N-1\,$ operating units possible.

If $\,m>k\,$ it can be minimum $\,0\,$ or maximum $\,N-2\,$ operating units.

How should I continue the solution?

答:
As I read the question, it seems that there are a total of $Nk$ elements in clusters of $k$ elements in a unit, and a total of $m$ elements fail.

P(an element in unit $i$ fails) = $\frac{m}{Nk}$

P(an element in unit $i$ does not fail) = $1- \frac{m}{Nk}$

Let $X$ be an indicator random variable that is = $1$ if none of the $k$ elements in that unit fail, and $0$ otherwise

$P(X_i) = (1- \frac{m}{Nk})^k = Y, say $

The expected value of an indicator variable is the same as the Pr indicated by it, so $\mathbb{E}(X_i) = Y$

and by the linearity of expectations over $N$ units, $\mathbb{E}(X) = NY$


答: You have to fill out the Cayley table such that the operation is commutative with identity element 'c'. There is more than one way to accomplish that, for example the operation a*b=max{a,b} with (a,b,c) = (2,1,0) will satisfy the preconditions given in the table.</p>

Body:
I have a problem with one of my questions. The question is:

complete the table by providing an example of a binary operation $*$ defined on $\{a , b ,c\}$
such that $*$ is commutative and has the identity element $c$.
\begin{array}{c|ccc}
\ast & a & b & c \\
\hline
a & a & ? & ? \\
b & ? & b & ? \\
c & ? & ? & c \\
\end{array}

(I need the letters in place of the ?)

I don't understand what they want me to do ???


问:
Bernstein set and it proof.
Definition. A set $B\subset\mathbb R$ is called a Bernstein set if both $B$ and $\mathbb R\setminus B$ intersect each perfect set.

I know the construct  the of Bernstein set but I got confused with this construction that I found in a book.

Let $\mathcal F=\{P_\xi\colon \xi<c\}$ be an enumeration of all perfect subsets of $\mathbb R.$ By recursion induction on $\xi<c$ define the sequences $\{a_\xi\colon \xi<c\}$ and $\{b_\xi\colon \xi<c\}$ by choosing in step $\xi<c$ points $a_\xi\neq b_\xi$ from $\mathbb R$ such that $$ a_\xi, b_\xi\in P_\xi\setminus(\{a_\zeta\colon \zeta<\xi\}\cup \{b_\zeta\colon \zeta<\xi\})$$  clearly, this choice is possible. Now define $$B:=\{a_\xi\colon \xi<c\}$$ Then $B$ is Bernstein set.

This is not my proof but I have doubts about this proof is correct. More specifically, I did not see how $B$ will intersect each perfect at continuum many points. Am I missing some thing ?

Any help will appreciated greatly.

答:
That is in fact the standard construction of a Bernstein set. I’m not sure why you are concerned about whether $B$ intersects each perfect set in continuum many points, since that is not part of the definition of a Bernstein set. As you say in your answer, it’s clear that $B\cap P$ and $B\setminus P$ are non-empty for each perfect subset $P$ of $\Bbb R$, and by definition that makes $B$ a Bernstein set.

However, it is true that if $P$ is a perfect subset of $\Bbb R$, then $|B\cap P|=2^\omega$. One way to see this is to note that $P$ necessarily contains a Cantor set $C$, and every Cantor set contains $2^\omega$ distinct subsets that are Cantor sets.1 Each of these Cantor sets is a perfect subset of $\Bbb R$, so each of them is one of the sets $P_\xi$ in the construction of $B$. $B$ contains a point $a_\xi$ from each of these sets, so $B$ contains $2^\omega$ points of $C$ and hence $2^\omega$ points of $P$.

1 For each $n\in\omega$ let $X_n=\{0,1\}$ with the discrete topology, and let $X=\prod_nX_n$; then $X$ is homeomorphic to $C$. For each infinite $S\subseteq\omega$ let

$$X_S=\{x\in X:x_n=0\text{ for all }n\in\omega\setminus S\}\,;$$

$X_S$ is clearly homeomorphic to $\prod_{n\in S}X_n$ and hence to $X$, so $X_S$ is a Cantor set. And $\omega$ has $2^\omega$ infinite subsets $S$, so $X$, and therefore $C$, contains $2^\omega$ distinct Cantor sets $X_S$.

Title: Bernstein set and it proof.


问:
does Euler's handshaking lemma apply to directed graphs
My textbook and wikipedia both specify that Euler's handshaking lemma applies to undirected graphs. But it seem to work for directed as well. So why does it not apply to directed graphs?</p>

答:
It does apply to directed graphs actually, but not in the way stated for undirected graphs. Because in directed graphs, we have in-degree and out-degree unlike a single degree definition in undirected graphs. But still, one can prove that
$$\sum_{v \in V(G)}d_{in}(v) = \sum_{v \in V(G)}d_{out}(v) = |E(G)|$$
Thus it still holds that
$$\sum_{v \in V(G)}\big(d_{in}(v)+d_{out}(v)\big) = 2|E(G)|$$
But I should also note that in directed graphs, we call what we call normally edge as arc. So, $|E(G)|$ is actually the number of arcs.


问:
Dirac Mass and Joint Distribution.
Take two spaces $X_1$ $X_2$ and consider $\mu\in\mathcal{P}(X_1\times X_2)$ i.e the space of Borel probability measures. Let $\mu_1,\mu_2$ be Borel probability measures on $X_1,X_2$, with $\mu_1$ a dirac mass.

Is it true that if $\mu$ has marginals $\mu_1,\mu_2$ then $\mu=\mu_1\times \mu_2$ i.e the product measure? If so how do I prove this result?

答:
Since $\mu_i$ is a Dirac mass for each $i = 1, 2$, we can find $x_i \in X_i$ such that $\mu_i = \delta_{x_i}$. Now if we write

$$ E_1 = (X_1\setminus\{x_1\})\times X_2, \qquad E_2 = X_1 \times (X_2\setminus\{x_2\}), $$

then we have

$$\mu(E_1) = \mu_1(X_1\setminus\{x_1\}) = 0 \qquad\text{and}\qquad \mu(E_2) = \mu_2(X_2\setminus\{x_2\}) = 0. $$

So it follows that

$$ \mu((X_1\times X_2)\setminus\{(x_1,x_2)\}) = \mu(E_1 \cup E_2) = 0. $$

This tells that

$$ \mu = \delta_{(x_1,x_2)} = \delta_{x_1} \otimes \delta_{x_2}. $$

Title: Dirac Mass and Joint Distribution.


问:
Show that the following language is decidable by finding the algorithm for the finite automaton
Given the language $K$ = $\{<M>: M$ is a finite automaton on the alphabet {0,1}) and $L(M)$ contains at least one word of the form $0^k1^l$ with $k,l\geq 0$}. In other words, describe an algorithm with the entry $M$ ( a finite automaton)  and that decide after a finite number of steps if $M$ accept at least one word of the form described above.

So this is my attempt but i am really not sure if my algorithm is good.

we can see that the accepted words can be : 1, 0 , 001, 011, ..000001 , 011111.., 1111111.., 000000.., the empty word can't be in this i think since $l$ would be equal to $k$ (unless i am wrong)

This was my idea about what the finite automaton (it's not required but helpful for automaton) :

And my algorithm would look something like this:

Input : < M >

If q0 is a final state and if a loop of $0$ exists around q0 return true

Verify that in $M$ if from q0 we can read a 1 and follow a path that leads us to a final state that can have a loop of 1s, if it exists return true.

Else return false.

Thanks a lot for your help.

答:
I think you've got it all wrong, you don't have to construct a finite automata but you are given an encoding of a finite automata you have to decide whether the given automata accept the string of type 0*1*.
The algorithm should be something like this:

Convert the given FA to minimal Deterministic Finite Automata.

If minimal DFA contains n states, then generate all strings of language from 0*1* of length upto n.

Run the strings on a simulated DFA, if any one of the string is accepted the FA $\in$ L, else not.

The algorithm gives both yes and no answer given an encoding of an FA, so it is Turing decidable.
The algorithm is correct as we know that if a string of length n of type 0*1* is not accepted on n-state DFA then and string of length > n will never be accepted from the pumping lemma of regular languages.


问:
Inequality $x - \sqrt{x^2 + 1} > 0$
This is how I solved this inequality:
$$ x - \sqrt{x^2 + 1} > 0$$
$$ x > \sqrt{x^2 + 1} $$
$$x < 0 \implies x^2 < x^2 + 1 \iff 1 > 0 \implies\text{inequality holds for all x}$$
$$x\geq 0 \implies x^2 > x ^2 +1 \iff 1 < 0 \implies \text{inequality doesn't hold }$$

The thing is that this shouldn't hold for any x. Can somebody explain why is my procedure wrong.

答:
$$x < 0 \implies x^2 < x^2 + 1 \iff 1 > 0 \implies\text{inequality holds for all x}$$
I want to try and show why this conclusion of yours is wrong.  Let's see if we can go back to the starting inequality from this.

But first notice that a small square shape has a small side and a large square shape has a large side. This is same as saying  $\sqrt{t}$  function is increasing.

Since $x^2$ and $x^2+1$ are nonnegative,

$1\gt 0 \implies x^2 \lt x^2+1 \implies \sqrt{x^2} \lt \sqrt{x^2+1}\implies |x| \lt \sqrt{x^2+1} \implies x\lt \sqrt{x^2+1}$.

This is not the starting inequality. You got it wrong because you're falsely assuming that squaring both sides of an inequality flips the inequality depending on the polarity of sides. Please look at @player3236 comment at the bottom of your question.

Title: Inequality $x - \sqrt{x^2 + 1} > 0$


问:
Is the set of points having the same distance from a closed set connected?
This question is related to this (Connectivity of this set: $A=\{x: \operatorname{dist}(x,F)=\epsilon\}?$). Suppose $F$ is a closed set in $\mathbb{R}^n$ for $n>1$. Suppose $F$ and its complement are connected. The counterexample given in the previous post prove that for a given $\epsilon>0$ we can not say that the set
$$A_\epsilon=\{x\in\mathbb{R}^n: \text{dist}(x,F)=\epsilon\}$$ is connected. How about the following question: For every $\delta>0$, there exists $0<\epsilon<\delta$ such that $A_\epsilon$ is connected? Notice that the counterexample given in the previous post does not  give a negative answer to this question.

答: Consider the counterexample from the previous post. Now place similar "cut annuli" one next to the other connected with a strip. Have the length of the "cut" be a sequence that goes to 0. I haven't carefully proved it, but it seems that the complement of this set is open, so it is closed. Both the set and its complement are connected, but for all sufficiently small $ \epsilon > 0 $ there will be a component of $ A_\epsilon $ trapped inside one of the "cut annuli" (actually there will be infinitely many such components).


问:
Trying to show $|a|\geq |b|-|b-a|$ for $a,b \in \mathbb{R}$
I'm currently solving a question that boils down to showing this inequality. The only tool I've got in my belt to show this the triangle inequality essentially.

My first step is $|a|=|a-b+b|$, so that I can apply the triangle inequality. I've tried many manipulations on $|a-b+b|$ to prove my inequality but I cannot seem to land on the right step.

Any help would be greatly appreciated!

答:
By the triangle inequality, $|a|+|b-a|\ge|a+b-a|=|b|$.

Now subtract $|b-a|$ from both sides.


问:
Modulus Cancellation Law
I'm trying to understand the proof for cancellation law in modulus which states that:

```<code>    ak = bk mod m

<=>     m | (a-b)k
                    since (k,m) = 1
<=>     m | (a-b)

<=>     a = b mod m
</code>```

However, I don't understand why:

if $\,\gcd(k,m) = 1,\;$ then $\,m\mid(a-b)k \;\iff\; m\mid(a-b)$$

Can anybody help me please? :)

答: Hint: Use prime factorization. Since $m$ and $k$ have no prime in common and $m|(a-b)k$ then all primes in $m$ must appear in $a-b$.</p>

Title: Modulus Cancellation Law


问:
How many simple graphs with 5 vertices of degree 2 do we have?
does anyone know any formula to calculate number of simple graphs with n specific number of vertices that each one of vertices have a degree of m(like 2 or four or etc)

note : suppose we n = 5(number of vertices) and m = 2(degree for  each one of vertices)

How many simple graphs do we have with that information?

答: Hint: if every vertex has degree $2$, your graph is the union of disjoint cycles.


问:
Show that $4\mid f_n$ if and only if $6\mid n$.
We define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n = f_{n−1} + f_{n−2}$. Show that $4\mid f_n$ if and only if $6\mid n$.

I have been trying to make a copy of the following solution for the exercise "$ 3 \mid f_n $ iff $ 4 \mid n $".

But so far I have not found anything satisfactory. In this question $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$ they answer it by redefining the fibonacci sequence, but I wonder if it can be done as I am trying.
By the way, what they do here $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$  I don't understand very well.

答:
To expand on this answer, what they've done is considered the sequence modulo $4$. So you have the following sequence:

\begin{align}
f_0 &= f_1 = 1\\
f_{n+1} &= f_n + f_{n-1} \bmod 4
\end{align}

(Here I use $\bmod 4$ as a function which returns the remainder in $0, \ldots, 3$.)

Thus, your question is now equivalent to showing that $f_n = 0$ iff $6 \mid n$.

Now, what they note is that the given sequence is periodic with period $6$.

Proving this is not difficult since you can see that $f_1 = f_7$ and $f_2 = f_8$. Now, if two consecutive terms repeat, then everything from that point on must too. (Simply because we have a recursion in terms of the previous two terms.)

Since $f_6 = 0$, we get that $f_{6n} = 0$ for all $n \ge 1$ showing that $n \mid 6 \implies f_n = 0$.

For the converse, note that each of $f_1, \ldots, f_5$ is non-zero and hence, so is each of $f_{6n+1}, \ldots, f_{6n+5}$ for each $n \ge 0$.

Thus, we are done.


问:
tolerance dimensions arithmetic or how to add or subtract values with probability distributions?
This is a follow up to this question. Imagin two physical objects A and B:

```
   <img src="https://i.stack.imgur.com/YMgvV.png" width="500">
```

with nominal heights of $l_A$ and $l_B$. These parts are being mass-produced so their heights have a probability distribution (e.g., Natural/Gaussian distributions), with standard deviations of $\sigma_A$ and $\sigma_B$. Now I want to know what are the expected probability distributions if the blow measurements:

```
   <img src="https://i.stack.imgur.com/Q7lF8.png" width="500">
```

In other words, what are the standard deviations of the values $l_1$ and $l_2$ from their nominal values $l_A + l_B$ and $l_A - l_B$? Is it correct to assume $\sigma_1 = \sigma_2 = \sigma_A + \sigma_B$?

P.S. Dear moderators, I have no clue if I have used the right terms or if I have chosen the right tags here. Your help in that regard would be highly appreciated.

答:
I was slightly wrong. The correct value for the standard deviations is:

$$\sigma_1 = \sigma_2 = \sqrt{\sigma_A^2 + \sigma_B^2}$$

Reference


问:
How to solve this reccurence with g(n)=-9n*2^(n-1)
$f(n+1)=9f(n)+5f(n-1)-9n\times2^{n-1}$

How to solve this? I have problem with $g(n)=-9n\times2^{n-1}$ I can start and calculate $\Delta$ and $x_1$ and $x_2$ but when I get $s_n=c_1\times 1^n+c_2\times5^n + g(n)$. I don't know what after to do.

答: First let us get rid of the non-homogeneous member $9n2^{n-1}$. For this look for a solution of a recurrence which has a form $f_1(n)=(an+b)2^{n-1}$. It is easy to see that we can pick $a=\tfrac 9{19}$  and $b=a^2$. Then a general solution of the recurrence is a sum of $f_1$ and a general solution of a recurrence $g(n+1)=9g(n)+5g(n-1)$. The latter is a linear recurrence with constant coefficients. Its characteristic polynomial $\lambda^2=9\lambda+5$  has distinct roots $\lambda_{1,2}=\tfrac 12\left(9\pm\sqrt{101}\right)$, so it has a general solution $C_1\lambda_1^n+ C_2\lambda_2^n$ for some constants $C_1$ and $C_2$.


问:
Why do elliptical copula densities contain $x_1$ and $x_2$, but Archimedean copula densities contain $u_1$ and $u_2$?
$$c\left(u_{1}, u_{2}\right)=\frac{1}{\sqrt{1-\rho_{12}^{2}}} \exp \left\{-\frac{\rho_{12}^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho_{12} x_{1} x_{2}}{2\left(1-\rho_{12}^{2}\right)}\right\}$$

is the copula density of the Gaussian copula, which is an elliptical copula, where $\rho_{12}$ is the parameter of the copula, $x_{1}=\Phi^{-1}\left(u_{1}\right), x_{2}=$ $\Phi^{-1}\left(u_{2}\right)$ and $\Phi^{-1}(\cdot)$ is the inverse of the standard univariate Gaussian distribution function.

$$
c\left(u_{1}, u_{2}\right)=\left(1+\delta_{12}\right)\left(u_{1} \cdot u_{2}\right)^{-1-\delta_{12}} \times\left(u_{1}^{-\delta_{12}}+u_{2}^{-\delta_{12}}-1\right)^{-1 / \delta_{12}-2}
$$
is the copula density of the Clayton copula, which is an Archimedean copula,
where $0<\delta_{12}<\infty$ is a parameter controlling the dependence. Perfect dependence is obtained when $\delta_{12} \rightarrow \infty,$ while $\delta_{12} \rightarrow 0$ implies independence.

Question

As you can see the elliptical copula density, despite being a function of $x_1$ and $x_2$, has the actual random variables $u_1$ and $u_2$ on the right hand-side, whereas the Archimedean copula densities do not have $x_1$ and $x_2$ on the right hand side, but instead what we would expect from the function: $u_1$ and $u_2$. This property seems to be a shared between all elliptical versus Archimedean copula densities. Why?

Source of formulas: Aas et al 2009 "Pair-copula constructions of multiple dependence"

答:
It is not the case that one copula density has random variables as arguments and the other does not. A copula density is just a joint PDF  where the underlying random variables have uniform marginal distributions.

In general, for  a probability space $\Omega$ and random variables $Y_1: \Omega \to [0,1]$ and $Y_2: \Omega \to [0,1]$ the joint PDF $f$ is a function mapping the range $[0,1]\times[0,1]$ of $(Y_1,Y_2)$ into $\mathbb{R}$, that is $f:[0,1] \times [0,1] \to \mathbb{R}$, such that for real numbers $y_1,y_2 \in [0,1]$

$$F(y_1,y_2) =\mathbb{P}(\{\omega \in \Omega |  Y_1(\omega) \leqslant y_1\} \cap \{\omega \in \Omega | Y_2(\omega) \leqslant y_2\}) = \int_0^{y_1} \int_0^{y_2} f(t_1,t_2)\, dt_1 \, dt_2$$

With regard to the copula density $c(u_1,u_2)$, the arguments $u_1$ and $u_2$ are always real numbers in $[0,1]$.  This is a joint PDF for random variables $U_1$ and $U_2$ with uniform marginal distributions.

What may have been confusing is that the notation for the Gaussian copula does not show the explicit dependence on $u_1$ and $u_2$ on the RHS.  The inverse of the normal CDF $\Phi^{-1}$ is a mapping from $[0,1]$ to $\mathbb{R}$ and the intermediate variables $x_1$ and $x_2$  are images $x_1 = \Phi^{-1}(u_1)$ and $x_2 = \Phi^{-1}(u_2)$.

The Gaussian copula density can be written more clearly as

$$c\left(u_{1}, u_{2}\right)=\frac{1}{\sqrt{1-\rho_{12}^{2}}} \exp \left\{-\frac{\rho_{12}^{2}\left([\Phi^{-1}(u_1)]^{2}+[\Phi^{-1}(u_2)]^{2}\right)-2 \rho_{12} \Phi^{-1}(u_1)\Phi^{-1}(u_2)}{2\left(1-\rho_{12}^{2}\right)}\right\}$$

A two-dimensional copula is a function $C:[0,1] \times [0,1] \to [0,1]$ with the following properties. For all $u_1,u_2 \in [0,1]$,

$$C(u_1,0) = 0, \,C(0,u_2) = 0,\\C(u_1,1) = u_1, \,C(1,u_2) = u_2,$$

and for all $0 \leqslant u_1 \leqslant u_1' \leqslant 1$ and  $0 \leqslant u_2 \leqslant u_2' \leqslant 1$,

$$C(u_1',u_2') - C(u_1',u_2) - C(u_1,u_2') + C(u_1,u_2) \geqslant 0$$

Archimedean copulas are a just general class of such functions -- with no relationship to the normal distribution function $\Phi$.

In general, suppose we have random variables $X$ and $Y$ with joint CDF $F_{XY}$ and marginal CDFs $F_X$ and $F_Y$. we can obtain the copula associated with this joint distribution as the function $C: [0,1] \times [0,1] \to [0,1]$ given by the composition

$$\tag{*}C(u_1,u_2) = F_{XY}(F_X^{-1}(u_1),F_Y^{-1}(u_2))$$

As long as we start with a valid joint CDF, the function defined by (*) will always meet the requirements of a copula specified above.

A special case is the bivariate Gaussian copula.  Letting $\Phi$ denote the univariate normal CDF and $\Phi_2$ denote the bivariate normal CDF, the bivariate Gaussian copula is given by

$$C(u_1,u_2) = \Phi_2(\Phi^{-1}(u_1), \Phi^{-1}(u_2))$$


答:
As you have asked for, this can be done using a similar argument to the one in your post:

If you repeatedly apply the recurrence relation, then $f_{n} = 8f_{n-5}+5f_{n-6}$ so that when you use strong induction on some set of elements less than $n$ in the same way by supposing $6|n$, then $6|n-6$ and by hypothesis $4|f_{n-6}$ and by the equation above, $4|f_{n}$.

If $6\nmid n$, then $6\nmid (n-6)$ and $4\nmid f_{n-6}$.
If $4|f_{n}$, then $4|8f_{n-5}+5f_{n-6}\implies4|5f_{n-6}\implies4|f_{n-6}$, a contradiction. Hence, $4\nmid f_{n}$. In this way, both implications have been shown.

For the second implication, I was unable to use the exact same argument as above since, in this case, $f_{n}$ was not able to reduced in such a convenient way.

Body:
We define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n = f_{n−1} + f_{n−2}$. Show that $4\mid f_n$ if and only if $6\mid n$.

I have been trying to make a copy of the following solution for the exercise "$ 3 \mid f_n $ iff $ 4 \mid n $".

But so far I have not found anything satisfactory. In this question $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$ they answer it by redefining the fibonacci sequence, but I wonder if it can be done as I am trying.
By the way, what they do here $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$  I don't understand very well.


问:
Calculating $P(Y>1)$, Y follows an random uniform distribution on $[0,X]$ and X follows an exp. distribution.
I need to calculate $P(Y>1)$ with Y being uniformly distributed on the random interval $[0,X]$ and X being exponentially distributed with parameter $\lambda = 1$.

So I need to condition on X to find $P(Y>1)$ but I do not know to continue. Appreciate some help

答:
The conditional density of $Y$ given $X=x$ is
$$
f_{Y\mid X=x} (y\mid x) = \frac1x\cdot1_{(0,x)}(y).
$$
The joint density of $(X,Y)$ is then:
\begin{align}
f_{X,Y}(x,y) &= f_{Y\mid X=x} (y\mid x)f_X(x)\\
&= \frac1x\lambda e^{-\lambda x}\cdot1_{(0,x)}(y)\cdot 1_{(0,\infty)}(x).
\end{align}
We compute the marginal density of $Y$ by integrating the joint density over the range of $X$:
\begin{align}
f_Y(y) &= \int_{\mathbb R} f_{X,Y}(x,y)\ \mathsf dx\\
&= \int_y^\infty \frac1x\lambda e^{-\lambda x}\ \mathsf dx\\
&= \lambda\cdot\Gamma(0,\lambda y),
\end{align}
where $\Gamma(\cdot,\cdot)$ denotes the upper incomplete gamma function, i.e.
$$
\Gamma(0,\lambda y) = \int_{\lambda y}^\infty t^{-1}e^{-t}\ \mathsf dt.
$$
It follows that
\begin{align}
\mathbb P(Y>1) &= \int_{(1,\infty)} f_Y(y)\ \mathsf dy\\
&= \int_1^\infty \lambda \int_{\lambda y}^\infty t^{-1}e^{-t}\ \mathsf dt\ \mathsf dy\\
&= e^{-\lambda} - \lambda\cdot\Gamma[0,\lambda].
\end{align}
When $\lambda=1$, this is
$$
e^{-1} - \Gamma[0,1] \approx 0.148496.
$$


问:
Topologies and manifolds
This question might seem philosophical a bit:

in a standard manifolds introductory course. when one talks about open , closed sets in $\mathbb{R}^n$ it's always the standard euclidian topology that we think about. Is there a more general framework in which one defines manifolds in different topologies in a way that generalize the standard setting that is closer to the physical world of course.

Thank you in advance for your answers

答:
The topology for $\mathbb{R}^{n}$ is very nice. If you try to replace this topological space by some random topological space $X$ you will have to come to terms with whatever goofy properties the topology for $X$ has. Suppose we call this new type of object $M$ an $X$-manifold. There seems to be two paths to proceed

Show that if $X$ has some nice property then some statement about $M$ is true.

Define all properties of $M$ in terms of the properties of a general topological space.

In the first case you end up butterfly-collecting. For each nice property you might get a theorem. In the second case the results you get will upon your ability to define manifold-like properties in terms that make sense for an arbitrary topological space. An example of the latter case might be the transition from $\delta$ and $\epsilon$ proofs to point set topology or locales. In the first transition one eliminates the metric; in the second transition one eliminates the use of points. In certain instances the reintroduction of a metric or points, respectively, may prove beneficial.

Title: Topologies and manifolds


问:
Improving related estimates
There are three underlying quantities $x$, $y$, and $a$, where $x$ and $y$ are vectors, and $a$ is a scalar. They are related by $x = ay$.

We get noisy observations, $x_0,y_0$. We want to find $a$, and better estimates of $x,y$. Any suggestions on how to do it?

I could think of one possible idea:

minimize: $||x-ay|| + ||x - x_0|| + ||y-y_0||$.

Can anyone direct me to a paper/ book which discusses such ideas?

答:
I don't think there's any way to get an estimate of all three unknowns $(a, x_0, y_0)$ given only one data point. Trivially, $a = x_0 / y_0$ is a zero-error solution.

However, if you have multiple data points, you can use the method of "Total Least Squares" to find $a$.

Title: Improving related estimates


问:
Show that if $x \geq y$, then $(e^{\frac{x}{2}}-e^{\frac{y}{2}})^2 \leq \frac{(x-y)^2}{8}(e^x+e^y)$
I tried to naively fix $y$ and then study the sign of

$$f:x \mapsto \frac{(x-y)^2}{8}(e^x+e^y)-(e^{\frac{x}{2}}-e^{\frac{y}{2}})^2 $$

but it gets ugly pretty fast.

答:
$\newcommand{\d}{\mathrm{d}}$
$\newcommand{\e}{\mathrm{e}}$

Let $f$ be a nonnegative convex function such as $f(t)=\e^{t/2}$. Then the Hermite-Hadamard inequality says that

$$\int_y^xf(t)\d t \leq (x-y)\left(\frac{f(y)+f(x)}{2}\right)$$

and the generalized mean inequality says that
$$\frac{f(y)+f(x)}{2}\leq \sqrt{\frac{f(y)^2+f(x)^2}{2}}$$

so that
$$\left(\int_y^xf(t)\d t\right)^2\leq (x-y)^2\left(\frac{f(y)^2+f(x)^2}{2}\right)\text{.}$$


问:
Find the smallest positive integer $n$ such that $\tau(n)=12$
Find the smallest positive integer $n$ such that $\tau(n)=12$

I have used the following equation below

$$\tau(n)= (k_1+1)(k_2+1)\dots(k_n+1) \text{ where } n = p_1^{k_1}p_2^{k_2}\dots p_n^{k_n}$$

To work out that the smallest integer would be $2^{11}$ since $2$ is the smallest prime. So the answer is $2048$.

However, I have since been told the correct answer is $60$ but I have no idea how to work this out, can anyone explain it? Thanks

答:
First factor $12$ to figure out what the $k_i$ should be.

As $12 = 2^2 \times 3$, we could have

$$12 = 6\times 2 = 4\times 3 = 3\times 2\times 2$$

so there are four cases to check (for $k_1 \ge k_2 \ge k_3$):

Case 1: $k_1= 11$.

Case 2: $k_1 = 5, k_2 = 1$.

Case 3: $k_1 = 3, k_2 = 2$.

Case 4: $k_1 = 2, k_2=k_3=1$.

To keep $n$ small, we choose $p_1 < p_2 < p_3$. Hence $p_1=2, p_2 = 3, p_3= 5$ should give the minimum in each case. Now compare the values of $n = p_1^{k_1}p_2^{k_2}p_3^{k_3}$ to verify that $60$ is indeed the smallest such $n$.


问:
Integral calculations with exponentials ($\int_{-\infty}^{+\infty}e^{ax^2+bx+c-1}$)
I have a probability distribution function (pdf) $f(x)$ wich has the following form:

$$f(x)=e^{ax^2+bx+c-1}$$

Let's assume that the pdf has a mean value $\mu$ and a variance value $\sigma^2$.

I want to prove that this pdf is the Gaussian function. So, I want to show that:

$$f(x)=\frac{1}{\sqrt{2\pi}\sigma}exp\left \{\frac{-(x-\mu)^2}{2\sigma^2}\right \}$$

I consider finding the parameters $a$, $b$ and $c$ by making use of the following equations:
$$\int_{-\infty}^{+\infty}f(x)=1$$
$$\int_{-\infty}^{+\infty}xf(x)=\mu$$
$$\int_{-\infty}^{+\infty}x^2f(x)=\sigma^2 + \mu^2$$

However, I find diffuculty on calculating these integrals since I am not very familiar with such exponential integrations that have a polynomial exponent with a 2nd or higher degree.

For instance, how do I have to approach the first one ($\int_{-\infty}^{+\infty}f(x)=1$)? Do I have to make use of the Gaussian integral somehow? ($\int_{-\infty}^{+\infty}e^{-x^2}=\sqrt{\pi}$). It seems to me that the integral is divergent.

For the other two equations do I have to make use of integration by parts and if so how do I have to approach it?

Any help is appreciated!Thanks in advance!

答:
$$f(x)=ax^2+bx+c+1=\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2+\left(c+1-\frac{b^2}{4a}\right)$$
now if we have:
$$\int_{-\infty}^\infty e^{f(x)}dx$$
we can use the substitution:
$$u=\sqrt{a}x+\frac{b}{2\sqrt{a}}\Rightarrow du=\sqrt{a}dx\Rightarrow dx=\frac{du}{\sqrt{a}}$$ and our integral becomes:
$$\int_{-\infty}^\infty\frac{1}{\sqrt{a}}e^{u^2+(c+1-b^2/4a)}du=\frac{\exp\left(c+1-\frac{b^2}{4a}\right)}{\sqrt{a}}\int_{-\infty}^\infty e^{u^2}du$$
now this last integral is definately divergent and is easy to show, I have however laid it out like this incase you meant for the polynomial to be negative which would lead you to the gaussian integral :)


问:
Expected distance between random points on hyperplanes
Let $\mathbf{p}$ be a point on a unit $n$-sphere $S_n$ and let $\mathcal{H}_n$ be the set all hyperplanes passing through the center of $S_n$.

Question: What is the simplest way to calculate the expected Euclidean distance $\mathbb{E}[d]$ between $\mathbf{p}$ and a hyperplane selected uniformly at random from $\mathcal {H}_n$?

Should we calculate

$$\mathbb{E}[d]=\mathbb{E}\left[\frac{|x_n|}{\sqrt{\sum_{i=1}^n x_i^2}}\right]$$

where $x_1, x_2, \ldots, x_n$ are the values taken by the i.i.d. random variables $X_1, X_2, \ldots, X_n \sim \mathcal{N}(0,1)$?

答:
Your method would yield the right answer, but it's more complicated than it needs to be. From this answer, to generate a random hyperplane passing through the origin, first pick a random point from the surface of the $n$-sphere. The hyperplane would then be defined by $\langle x,u\rangle=0$.

WLOG, let $\mathbf{p} = (1, \underbrace{0, ..., 0}_{n-1})$. The distance from the hyperplane defined by $\langle x,u\rangle=0$ to $\mathbf{p}$ would then be $\left| \langle u, \mathbf{p}\rangle \right|$. This is then simply $|u_1|$. This is the integral of $|x_1|$ over the $n$-sphere, divided by the "surface area" of the sphere. From this, this is given by $$\frac{\frac{2\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}}{\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}} = \frac{\Gamma\left( \frac{n}{2}-1
 \right)}{\Gamma\left(\frac{n-1}{2}
 \right)\sqrt{\pi}}$$


问:
Show these affirmations using the context-free pumping lemma properties
The question :

Given $G$ a context-free grammar of Chomsky normal form with $k$ symbols. We know that the language $L(G)$ satisfy the pumping lemma for $ p = 2^k + 1$

I have those two questions to proove:

1. Show that for each word $w$ in $L(G)$ such as $|w| \geq p$, a word $w'$ exist in $L(G)$ such as $|w'|\lt|w|$
EDIT THIS ONE IS ANSWERED ( SEE COMMENTS, 2 AND 3 LEFT)

In other words, if a  word w exist in the language generated by the grammar that has a superior length than the pumping length then an other word w' exist and has a smaller length than w but is in the language.

2. Show that if $L(G) \neq \emptyset$ then a word x exist where $x$ is in $L(G)$ such as $|x| \leq p$

3. Show that this language is decidable $NV_{GNC}$ = {< G >: where G is a context-free grammar in Chomsky normal form and $L(G) \neq \emptyset $ }

This is my proof for 3 ( feel free to correct me if this doesn't work.)

If we have a grammar in Chomsky normal form, H, then it takes H exactly 2n − 1 steps
to generate an input of length n. (Unless n = 0, in which case it requires 1 step.) Thus, if G is
in Chomsky-normal form, we can design a turing machine that will generate all derivations in G
of length 2|w| − 1. Since there are finitely many rules, this can be done in finite time and space.
At that point, we only need to check all of our derivations against w; if any one matches, we
accept; if none do, we reject.

Note : to answer 2 you can use 1 and to answer 3 you can use 1,2.

I am really not sure what step i need to do to proove that, if anyone could help it would be appreciated.

答:
HINT for 1: Apply the pumping lemma to $w$ and pump down; i.e., in this version of the pumping lemma let $n=0$.

HINT for 2: If not, let $w$ be a word in $L(G)$ of minimal length, and apply 1 to get a contradiction.


问:
solving $1+\frac{1}{x} \gt 0$
In solving a larger problem, I ran into the following inequality which I must solve:

$$ 1+\frac{1}{x} \gt 0.$$

Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.

答:
There are two cases.  If $x>0$ we may multiply both sides by $x$ and the inequality is unchanged, i.e. $x+(1/x)x>0(x)$, which simplifies to $x+1>0$ or $x>-1$.  This case is the combination of $x>0$ and $x>-1$, which is $x>0$.

If $x<0$ we multiply both sides by $x$ and the inequality reverses, i.e. $x+(1/x)x<0(x)$, which simplifies to $x+1<0$ or $x<-1$.  This case is the combination of $x<0$ and $x<-1$, which is $x<-1$.

Title: solving $1+\frac{1}{x} \gt 0$


问:
$T\subset X \times Y$ is the graph of the function if and only if for each $a\in X$ ,$T\cap (\lbrace a \rbrace \times Y )$ is a singleton

Let $f:X \longrightarrow Y$ and considere their graphic $G_{f}=\lbrace (x,f(x)) \mid x\in X\rbrace$.Prove that a subset $T\subset X \times Y$  is the graph of the function if and only if for each $a\in X$ ,$T\cap (\lbrace a \rbrace \times Y )$ is a singleton

My attempmt

First assume that  $T\subset X \times Y$  is the graph of the function, if $|T\cap (\lbrace a \rbrace \times Y )|=\lbrace \emptyset \rbrace $ then it is a singleton. Now suppose that $|T\cap (\lbrace a \rbrace \times Y )|>1$ then $|T\cap (\lbrace a \rbrace \times Y )|=2$ and let $l_1,l_2\in T\cap (\lbrace a \rbrace \times Y )$with $l_1 \neq l_2$,then  by hypotesis $l_1=(x_1,f(x_1))$ for $x_1\in X$ and $l_2=(x_2,f(x_2))$ for $x_2\in X$.Since $l_1,l_2\in \lbrace a \rbrace \times Y$  then $x_1=x_2=a$ then $f(x_1)=f(x_2)$.And $l_1=l_2$ Which is a contradiction with our suppose.And therefore $|T\cap (\lbrace a \rbrace \times Y )|=1$ this is a singleton.

For the converse supposethat for each $a\in X$ ,$T\cap (\lbrace a \rbrace \times Y )$ is a singleton, we should prove that $T\subset X \times Y$ is a graphic.
Suppose that $T\subset X \times Y$ is not  a graphic.Then for some $l_1\in T$ we had
$l_1=(x_1,y_1)$ such that $y_1 \neq (x_1)$ and therefore
for $a=x_1$ $T\cap(\lbrace a \rbrace \times Y)=\lbrace (x_1,y_1),(x_1,f(x_1))\rbrace$ which is not a singleton.
Therefore $T$ is a graphic.
Is my proof right or i do a mistake, any feedback was very helpful

答:
Suppose that $T$ is the graph of a function, that is, suppose that there exists $g : X \to Y$ such that $T = \{(x,y) \in X \times Y : y = g(x)\}$. Now, take $a \in X$ and observe that $T \cap (\{a\} \times Y)$ is non-empty since $(a,g(a)) \in T \cap (\{a\} \times Y)$. We claim that $T \cap (\{a\} \times Y) = \{(a,g(a))\}$. Indeed, if $(x,y) \in T \cap (\{a\} \times Y)$ then $(x,y) \in T$ implies that $y = g(x)$, but since $(x,y) \in \{a\} \times Y$, we have $(x,y) = (a,g(a))$.

Now suppose that $T \cap (\{a\} \times Y)$ is a singleton for each $a \in X$. Thus, if for each $a \in X$ we let $g(a)$ be the unique element in $Y$ such that $T \cap (\{a\} \times Y) = \{(a,g(a)\}$, it follows that $g : X \to Y$ and that $T$ is the graph of $g$.


问:
Who is the winner in the zero-sum game?
I have a zero-sum game as described in the table below:

Using minimax = maximin approach I get that the first player should play A with $p^*=\frac{5}{12}$ and B with $p^*=\frac{7}{12}$.

The second player should then mix A and B with $q^*=\frac{1}{2}$.

I found the value of the game to be: $v=0.5$.

The question is asking to determine the winner. And since, the game is zero-sum, I know that
one player wins and the other loses (what is one player's gain is the
other's loss etc.). But how does this apply to this situation?
Basically, every time they play the same action, player 2 wins and
when they play different actions - player 1 wins. But how to
determine the overall winner? I think it is based on the value of the
game. My guess is that when the value of the game, v is positive,
then the first player is the winner? Is that correct?

$$\begin{array}{c|c|c|c|}
 & \text{A} & \text{B} & \text{min} \\ \hline
\text{A} & -3 & 4 & -3\\ \hline
\text{B} & 3 & -2 & -2 \\ \hline
\text{max} & 3 & 4 \\ \hline
\end{array}$$

答: There is no such thing as an "overall winner".  Value $v=0.5$, means that (when the two players use the specified "optimal" strategies) then the first player's winnings average $0.5$ per play.  Because of the random nature of the play, he wins some and loses some.  If one player does not use the optimal strategy, then that $0.5$ will change in the other player's favor.

Title: Who is the winner in the zero-sum game?


问:
State a vector in $\mathbb{R}^3$ that does not belong to the image space $f(V)$
The problem

Given the image space $$f(V)=\text{span}\{(1,0,2),(2,-1,0) \} $$
state and argue a vector in $\mathbb{R}^3$ that does not belong to $f(V)$

The instructors gave the following solution to the problem.

"For example, the vector $u=\begin{bmatrix}1 \\0 \\0 \end{bmatrix}$ does not belong to the image space, because you cannot arrive at $u$ through any linear combination of the spanning vectors, $v_1=\begin{bmatrix}1 \\0 \\2 \end{bmatrix}$ and $v_2 = \begin{bmatrix}2 \\-1 \\0 \end{bmatrix}$."

My question

Is another valid answer to this question that the zero vector, $v=\begin{bmatrix}0 \\0\\0 \end{bmatrix}$ doesn't belong to the image space $f(V)$ either, because it belongs to the kernel $\text{ker}(f)$?

I just think this solution is too easy and "straight forward".

答: Nope: clearly, $\begin{bmatrix}0\\0\\0\end{bmatrix} = 0v_1 + 0v_2 \in f(V)$.


问:
Galois action on $K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell$
Let $K/\mathbb{Q}$ be a finite extension and let $\ell$ be a prime number.

For each prime $\lambda$ of $K$ lying over $\ell,$ choose an embedding $\sigma_\lambda:K\hookrightarrow K_\lambda$ of $K$ into the $\lambda$-adic completion of $K.$

There is a natural isomorphism
$$K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell \longrightarrow \prod_{\lambda\mid\ell} K_\lambda\,; \; \; \; x\otimes y \longmapsto (\sigma_\lambda(x)y \, : \, \lambda\mid\ell).$$

If we now assume, in addition, that $K/\mathbb{Q}$ is Galois with group $G,$ then $G$ has a natural action on $K\otimes_\mathbb{Q} \mathbb{Q}_\ell$ (acting on the first factor).

Is there a "nice" description of resulting action on the product $\prod_{\lambda\mid\ell} K_\lambda$?

答:
$G=Gal(K/Q), D(P_1) = \{ \sigma\in G, \sigma(P_1)=P_1\}$
$$\ell O_K=\prod_{j=1}^g P_j^e, \qquad G= \bigcup_{j=1}^g \sigma_j D(P_1), \qquad P_j = \sigma_j(P_1)$$
Take $\pi_1\in P_1,\not \in P_1^2$ and for $j\ne 1$, $\pi_1\equiv 1\bmod P_j$. Then $$(1-\pi_1)^{\ell^n}\equiv 0\bmod P_j^{\ell^n}, \qquad (1-\pi_1)^{\ell^n}\equiv 1\bmod P_1^{\ell^n}$$
Let $$r_1 = \lim_{n\to \infty} ( (1-\pi_1)\otimes 1)^{\ell^n} \in K\otimes_\Bbb{Q}\Bbb{Q}_\ell, \qquad {\sum_{j=1}^g  \sigma_j(r_1)=1,  \sigma_j(r_1)^2= \sigma_j(r_1), \\ \sigma_i(r_1)\sigma_j(r_1)=0}$$

Take a normal basis $$K_{P_1} = \sum_{m=1}^{ef} h_m(b) \Bbb{Q}_\ell$$
Where the $h_m$ are the elements of $D(P_1)$ and $b$ is an element of $K$.

$\sigma_j(b)$ generates a normal basis of $K_{P_j}$.

Then
$$K\otimes_\Bbb{Q}\Bbb{Q}_\ell =\sum_{j=1}^g \underbrace{\sigma_j(r_1) \sum_{m=1}^{ef} \sigma_jh_m \sigma_j^{-1}(\sigma_j(b)\otimes 1) \Bbb{Q}_\ell}_{\cong K_{P_j}}$$
Because $r_1$ is fixed by $D(P_1)$ the Galois action on the RHS is clear.