问:
I am having a difficult time understanding what I am doing wrong with quadratics. $5x^2-15x-4=0$
I think I am missing a key part of solving quadratic equations. When I solve by grouping, I run into an issue of my values being incorrect, even though I am doing the same thing to both sides of the equation. Could someone explain to me why this doesn't work and what foundation I need to revisit? Here is my work:
$$
\begin{align}
5x^2-15x-4&=0\\
5x^2 -15x &= 4\\
5x^2–15x-20&=-16\\
5x^2–20x+5x-20&=-16\\
5x(x-4)+5(x-4)&=-16\\
(5x+5)(x-4)&=-16\\
\implies x=-4.20, -12
\end{align}
$$

答: Your fundamental error is thinking that if $a b = c$ then the two "solutions" are $a = c$ and $b = c$.


问:
Denseness of a particular sequence space

For $a\in\mathbb R$, let $h_a$ be the Hilbert space of sequences defined by
$$ h_a=\left\{(x_n):\sum_{n\in\mathbb Z}(1+n^2)^a|x_n|^2<\infty\right\}$$
and inner product $\langle(x_n),(y_n)\rangle_a=\sum_{n\in\mathbb Z}(1+n^2)^ax_n\overline{y_n}$.

Prove that if $b>a$ then $h_b$ is dense in $h_a$.

Attempt: So far, I have shown that $b>a$ then $h_b\subset h_a$. However, I don't have a clear idea on how to show that $h_b$ is dense in $h_a$. Probably I need to show that for any element in $h_a$ there is a sequence in $h_b$ converging to it; beyond that, I'm stuck.

Any help or guidance on this question would be appreciated greatly. Thank you!

答: Let $(x_n) \in h_a$. Let $y_n=x_n$ for $|n| \leq N$ and $y_n=0$ for $|n| >N$. Then $(y_n) \in h_b$. Now the norm of $(x_n-y_n)$ in $h_a$ is $\sqrt {\sum_{|n|>N} (1+n)^{a}|x_n|^{2}}$. Recall that of  series $\sum c_n$ of non-negative numbers is convergent than $\sum_{|n|>N|} c_n \to 0$ as $ N \to \infty$. Hence, given $\epsilon >0$ we can find $N$ such that $\|(x_n)-(y_n)\| <\epsilon$.

ContentLicense: CC BY-SA 4.0


问:
Is there is any example of continuous bounded function on $\mathbb{R}$ which attains 'either' maximum 'or' minimum on $\mathbb{R}$ but not both.
I need to find example of continuous bounded function on $\mathbb{R}$ which attains 'either' maximum 'or' minimum on $\mathbb{R}$ but not both.

My attempt:

First of all, is such function exists? (How)? If it exists then isn't it will contradict extreme value theorem?

Is $f(x)=\frac{1}{x^2+1}$ for $x\in\mathbb{R}$ works?

答: Let $f(x)=e^{-x}$ for $x \geq 0$ and $f(x)=1$ for $x<0$. Then $f$ is continuous and bounded. Its maximum value $1$ is attined but its infimum  $0$ is not attained.


问:
Partitions of $n$: proving $p(n+2)+ p(n) \geq 2p(n+1)$

For $n \geq 2$ give an alternative description of $p(n) - p(n-1)$ as the number of partitions of $n$ which have a certain property.

I have done that part, it is fine. I have not included it here as I am not accustomed with LaTeX but can put it up if wanted. But then it says

Hence or otherwise prove $p(n+2)+ p(n) \geq 2p(n+1)$.

This is the part I do not know how to do. Any help appreciated thank you

答:
Hint: The stated inequality is equivalent to
$$p(n+2)-p(n+1)\ge p(n+1)-p(n)$$

ContentLicense: CC BY-SA 3.0


问:
Prove the following for $n\geq5$ , where $p_i$ is a prime number
p3n+1 < p1p2....pn

What I did was I was able to see that p2n+1 was always much smaller than p1p2.....pn-1 but  pn+1 was definitely bigger than  pn. So how can I prove that the product of p2n+1 * pn+1 <
(p1p2....pn-1) * pn

答:
Here is a solution with a bit of motivation, so you can solve questions like these on your own next time.

It would be nice to be able to prove this by induction. In this case, we'd multiply
$$ p_{n+1}^3 < p_1 p_2 \dots p_n $$
by $p_{n+1}$ on both sides to get
$$ p_{n+1}^4 < p_1 p_2 \dots p_n p_{n+1} .$$
If we knew that $p_{n+2}^3 < p_{n+1}^4$, then we'd be done.

Fortunately, this is true. This follows from Bertrand's postulate for sufficiently large $p_{n+1}$ (which holds because we have $n \geq 5$).

In fact, I believe this argument can be used to prove
$$ p_{n+1}^k < p_1 p_2 \dots p_n $$
for any nonnegative $k$, for sufficiently large $n$.


Title: Prove the following for $n\geq5$ , where $p_i$ is a prime number

Body:
p3n+1 < p1p2....pn

What I did was I was able to see that p2n+1 was always much smaller than p1p2.....pn-1 but  pn+1 was definitely bigger than  pn. So how can I prove that the product of p2n+1 * pn+1 <
(p1p2....pn-1) * pn

Body:
You're asking to prove

$$p_{n+1}^3 \lt \prod_{i=1}^{n}p_i \tag{1}\label{eq1A}$$

for $n \gt 5$. Note Bertrand's postulate gives

$$p_{n+1} \lt 2p_{n} \lt 4p_{n-1} \lt 8p_{n-2} \tag{2}\label{eq2A}$$

Thus,

$$p_{n+1}^3 \lt (2p_{n})(4p_{n-1})(8p_{n-2}) = 64(p_{n-2}p_{n-1}p_{n}) \tag{3}\label{eq3A}$$

For $n = 5$, \eqref{eq1A} becomes

$$13^3 = 2179 \lt 2(3)(5)(7)(11) = 2310 \tag{4}\label{eq4A}$$

so it's true. For $n = 6$, \eqref{eq1A} becomes

$$17^3 = 4913 \lt 2(3)(5)(7)(11)(13) = 30030 \tag{5}\label{eq5A}$$

so that is also true. For $n \ge 7$,

$$\prod_{i=1}^{n}p_i \ge (p_1 p_2 p_3 p_4)p_{n-2}p_{n-1}p_{n} = 210(p_{n-2}p_{n-1}p_{n}) \tag{6}\label{eq6A}$$

Since $210 \gt 64$, combining \eqref{eq3A} and \eqref{eq6A} gives \eqref{eq1A} is also true for all $n \ge 7$.


问:
Coins in two piles game — where have I gone wrong?
In a certain game, a player can put his money in two piles labelled A and B. A fair coin is tossed. If it lands heads, then all the money in pile A is multiplied by a factor $\alpha$ and returned to you. Likewise, if it lands tails, then all the money in pile B is multiplied by a factor $\beta$ and returned to you. Determine an optimal allocation of money into the two piles if i) $\alpha=2$ and $\beta=0.5$; ii) or $\alpha=2$ and $\beta=1.5$.

My thought process is: suppose we place a fraction $f\in[0,1]$ into pile A and $(1-f)$ into pile B. Then, double our expected returns (so that we can ignore the factor of half) are $f\alpha-(1-f)\beta=(\alpha-\beta)f+\beta$. This is a linear relation between the payoff and $f$. My gut tells me I should not put all my money in the same pile, but the monotonicity implies otherwise. Where have I gone wrong?

答: As long as you have some money, you can play the game again, which changes the expectation from linear to exponential.</p>


问:
Meaning of a function being "defined" on $\mathbb{R}$
I have a somewhat basic question: What does it mean when someone says that a function, say, $f$ is defined on all of $\mathbb{R}$? Does it means that the domain of $f$ is $\mathbb{R}$, or that the range of $f$ is $\mathbb{R}$, or both?

答: It means the domain is $\Bbb R$.  The range can be anything.


问:
Number of arrangements possible
There are $m$ persons and $n$ chairs. Each person needs to maintain social distance between themselves and the person they are sitting next to. Therefore, once a person sits, the person who sits next to them sits at a distance that is in multiples of the integer $k$. We have to find the total number of arrangements possible so that all the people can be seated.

Notes:

1.All the people must get a chair to sit on

2.Two arrangements are different either if the state of a chair is different in both cases or if two different people are seated on it. For example, for chairs $n=10$, person $m=2$, and $k=2$, if the first person sits at the 2nd position, then the second person can sit only at the 4th, 6th, 8th, or 10th position.

For  example, $n=5,m=2,k=2$
no of arrangements possible are 4. (1,3)  (1,5)  (2,4)  (3,5).

Can anyone suggest some approach to me for this combinatorics problem?

答:
Here is how you might go about solving this problem.

One way to do this is to handle the $k$ condition first. For example, when $k = 2$, either all the occupied chairs have even number or odd number. So in general, we can consider $k$ different kinds of seating arrangements: one which has chair numbers $\equiv 1 \mod k$, one with $2 \mod k$, and so on until $0 \mod k$. If $n$ is a multiple of $k$, all of these will be equally large. If not, they will roughly have the same size but be slightly different.

Once we've handled the $k$ condition, this is just a problem about placing $m$ people in either $\lfloor n/k \rfloor$ or $\lceil n/k \rceil$ chairs. This is a standard counting problem.

Title: Number of arrangements possible


问:
Are bounded sequences always strictly less than some fixed number $M$?

Definition: A sequence $(x_n)$ is
   bounded if there exists a number $M \gt 0$ such that $ \left| x_{n}
 \right| \leq M$ for all $n \in
 \mathbb{N}$.

Question: Is there any reason why one can't replace $ \left| x_{n} \right| \leq M$ with the strict inequality $ \left| x_{n} \right| < M$? Since $M$ is a fixed number it seems that $1$ could always be added to make the inequality strict, but maybe I am missing something... If I am right, and the inequality can always be strict, then why didn't they write it that way in the first place?

答:
You are absolutely right, one could also write "there exists $M$ such that $|x_n| < M$ for all $n$"; the two statements are equivalent.  But if they had written it that way, you'd be asking why they didn't write "$x_n \le M$"...

There's a slight advantage to writing $\le$: if the sequence is bounded, then there is always a smallest $M$ such that $|x_n| \le M$ for all $n$.  (Namely, $M = \sup_n |x_n|$.)  But there may not be a smallest $M$ such that $|x_n| < M$ for all $n$.  (Consider the constant sequence $x_n = 0$...)


问:
Closed form of $I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$
I need help with this integral:

$$I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

Where $κ,λ>0$.

Neither Mathematica nor Maple could find a closed form for this integral.

Let $G$ follow a $\Gamma\left(\kappa+1,\lambda\right)$ distribution, i.e. its density can be written as $f_G\left(x\right)=\mathbb{I}_{\mathbb{R}_+^\ast}\left(x\right)\ x^\kappa\ e^{-\ \frac{x}{\lambda}}\frac{1}{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}$.

For fixed values of κ and λ, I can use Monte-Carlo to simulate : $$I=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\int_{0}^{+\infty}{f_G\left(t\right)\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\ \mathbb{E}\left(\sin^2{\left(\frac{G\pi}{2\kappa\lambda}\right)}\right)$$

But I would actually rather have a closed form. Any help or insight will be very much appreciated.

PS : although fluent in english, I mostly study math in french...

Edit : following another user's suggestion, I know have the following :

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\frac{1-\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}}{2}dt}$$

$$I=\frac{1}{2}\left(\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}dt}-\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}dt}\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\left(e^{i\frac{\pi t}{\kappa\lambda}}+e^{-i\frac{\pi t}{\kappa\lambda}}\right)dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}+i\frac{\pi t}{\kappa\lambda}}\ dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}\ -\ i\frac{\pi t}{\kappa\lambda}}\ dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa-i\pi\right)t}{\kappa\lambda}\ }dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa+i\pi\right)t}{\kappa\lambda}\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa-i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa+i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }du}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\left(\frac{\kappa}{\kappa+i\pi}\right)^{1+\kappa}-\left(\frac{\kappa}{\kappa-i\pi}\right)^{1+\kappa}\right)$$

Not sure what to do about the complex numbers I get in the end though...

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
I \equiv &\bbox[5px,#ffd]{\int_{0}^{\infty}t^{\kappa} \expo{-t/\lambda}\sin^{2}\pars{\pi t \over 2\kappa\lambda}\,\dd t}
\\[5mm] & =
\left.\lambda^{\kappa + 1}
\int_{0}^{\infty}t^{\kappa} \expo{-t}
\sin^{2}\pars{\alpha t \over 2}\,\dd t
\,\right\vert_{\ \color{red}{\alpha\ =\ \pi/\kappa}}
\\[5mm] & =
{1 \over 2}\,\lambda^{\kappa + 1}\
\overbrace{\int_{0}^{\infty}t^{\kappa}
\expo{-t}\dd t}
^{\ds{\Gamma\pars{\kappa + 1}}}
\\[2mm] &\
-{1 \over 2}\,\lambda^{\kappa + 1}
\underbrace{\int_{0}^{\infty}t^{\kappa}
\expo{-t}\cos\pars{\alpha t}\,\dd t}
_{\ds{\cal J}}\label{1}\tag{1}
\end{align}

$\ds{\large{\cal J}\ \mbox{Evaluation:}}$
\begin{align}
{\cal J} & \equiv
\int_{0}^{\infty}t^{\kappa} \expo{-t}
\cos\pars{\alpha t}\,\dd t =
\Re\int_{0}^{\infty}t^{\kappa} \expo{-\pars{1 + \ic\alpha}t}\,\dd t
\end{align}
Note that
$$
\expo{-\pars{1 + \ic\alpha}t} =
\sum_{n = 0}^{\infty}{\bracks{-\pars{1 + \ic\alpha}t}^{n} \over n!} =
\sum_{n = 0}^{\infty}\color{red}{\pars{1 + \ic\alpha}^{n}}\,{\pars{-t}^{n} \over n!}
$$

In order to evaluate $\ds{\cal J}$, I'll use Ramanujan's Master Theorem:
\begin{align}
{\cal J} & =
\Re\bracks{\Gamma\pars{\kappa + 1}\pars{1 + \ic\alpha}^{-\kappa - 1}\,}
\\[5mm] & =
\Gamma\pars{\kappa + 1}
\pars{1 + \alpha^{2}}^{-\kappa/2 - 1/2}\,\,\, =\
{\Gamma\pars{\kappa + 1} \over
\pars{1 + \alpha^{2}}^{\kappa/2 + 1/2}}
\\[5mm] & =
\kappa^{\kappa + 1}\
{\Gamma\pars{\kappa + 1} \over
\pars{\kappa^{2} + \pi^{2}}^{\kappa/2 + 1/2}}
\quad\mbox{with}\quad\alpha = {\pi \over \kappa}
\label{2}\tag{2}
\end{align}

With (\ref{1}) and (\ref{2}):
\begin{align}
I \equiv &\bbox[5px,#ffd]{\int_{0}^{\infty}t^{\kappa} \expo{-t/\lambda}\sin^{2}\pars{\pi t \over 2\kappa\lambda}\,\dd t}
\\[5mm] & =
\bbx{{1 \over 2}\,\lambda^{\kappa + 1}\
\Gamma\pars{\kappa + 1}
\bracks{%
1 -  {\kappa^{\kappa + 1} \over
\pars{\kappa^{2} + \pi^{2}}^{\kappa/2 + 1/2}}}}
\\ &
\end{align}


答:
What is your definition of optimal?  Unless otherwise specified, we take it as expected value.  In that case, put all your money in the pile where it is multiplied the most.  See this  Your intuition that going broke is infinitely bad is wrong.

Note that as defined betting on pile A is break-even and betting on pile B is losing in both cases.  If you are not going to put everything in pile A you should not play at all.

Title: Coins in two piles game — where have I gone wrong?

Body:
In a certain game, a player can put his money in two piles labelled A and B. A fair coin is tossed. If it lands heads, then all the money in pile A is multiplied by a factor $\alpha$ and returned to you. Likewise, if it lands tails, then all the money in pile B is multiplied by a factor $\beta$ and returned to you. Determine an optimal allocation of money into the two piles if i) $\alpha=2$ and $\beta=0.5$; ii) or $\alpha=2$ and $\beta=1.5$.

My thought process is: suppose we place a fraction $f\in[0,1]$ into pile A and $(1-f)$ into pile B. Then, double our expected returns (so that we can ignore the factor of half) are $f\alpha-(1-f)\beta=(\alpha-\beta)f+\beta$. This is a linear relation between the payoff and $f$. My gut tells me I should not put all my money in the same pile, but the monotonicity implies otherwise. Where have I gone wrong?


问:
Prove the generating function for the Bessel function.
The generating function states that  $$e^\left (\frac12 \left (t - \frac1t \right ) x\right )= \sum_{n = -\infty}^{\infty} J_n(x) t^n.$$ However, I do not know where the $n = -\infty$ came from in

.

Can someone point me in the right direction?

答:
A typical term after the multiplication is:

$$\frac{(-1)^k(\frac{x}{2})^{m+k}}{m!k!}z^{m-k}$$

The author then collects like-terms by 'level' $n=m-k$.


问:
Partial Derivative Leads to Contradiction
Starting from the same equation $g(x,y,c) = 0$, I partially differentiate with regards to c and end up in a contradiction. However, if I manipulate the equation a bit I do not. I will post the entire solution, which starts from an ODE.
$$(3y-a)^2*y\prime^2 - 4y = 0$$
$$y\prime = +(\frac{4y}{3y-a})^{1/2}, or\ y\prime = -(\frac{4y}{3y-a})^{1/2}$$
$$solving\ both\ ODEs: y^{3/2} - ay^{1/2} = \pm*x + c \quad (1)$$
$$(1)\rightarrow y^{3/2} - ay^{1/2}  \mp*x - c = 0 \quad let\ g_1(x,y,c) = y^{3/2} - ay^{1/2}  \mp*x - c = 0 \quad (2)$$
$$(1)\rightarrow (y^{3/2} - ay^{1/2})^2 = (\pm*x + c)^2 $$
$$\Leftrightarrow y*(y^3-a)^2 = (\pm*x + c)^2 \quad let\ g_2(x,y,c) = y*(y^3-a)^2 - (\pm*x + c)^2 = 0 \quad (3)$$

Now if we try and diferentiate both  $g_1$ and $g_2$ partialy with regards to c we end up with the following:

$$(2) \rightarrow -1 = 0$$
$$(3) \rightarrow c = \mp x$$

So my question really is, why can I not partially differentiate (2)? I know from where I got this question that (3) is the correct way to go, but why is that?

答:
There is no contradiction. The equation (3) holds the information of equation (2) but unadvisedly expanded your solution domain. In other words, when you have chosen to use exponentiation, $ -(y^{3/2} - ay^{1/2}) = (\pm*x + c) $ joined $ +(y^{3/2} - ay^{1/2}) = (\pm*x + c) $ as a "possible" solution.

Do the same thing in the context of simpler equations like $x=2$ and you will end up with $x=\pm\,2$, for example.


问:
Completely stuck at integrating $\int x\sqrt{\cos x}\,dx$
I am currently learning Calculus and I just started studying Integrals. I want to find the general indefinite integral $$\int x\sqrt{\cos x}\,dx$$

I have tried using some substitution such as letting $u=\cos x$ or $u=\sqrt x$, etc. but it seems like none of them is effective here.

I have also tried using online integral calculators to solve this question, but all I got is something like "steps are currently not supported for this problem."

Since even an integral calculator couldn't handle this, I wondered if this question can be solved. Therefore I came back to MathSE to seek help.

Is it solvable? If so, then how? Thanks in advance.

答:
As said, at the price of the "monster" given by @Raffaele in comments, you can compute it.

What you could also do (but at the price of another infinite summation, is to use
$$\sqrt{\cos(x)}=t \implies x=\cos ^{-1}\left(t^2\right)\implies dx=-\frac{2 t}{\sqrt{1-t^4}}\,dt$$ to make
$$I=-2\int \frac{t^2}{\sqrt{1-t^4}}\, \cos ^{-1}\left(t^2\right)\,dt$$

Now, using
$$\frac{t^2}{\sqrt{1-t^4}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, t^{4 n+2}$$ we face the problem of
$$J_n=\int t^{4 n+2}\, \cos ^{-1}\left(t^2\right)\,dt$$ which are
$$J_n=\frac{t^{4 n+3} \left(2 t^2 \,
   _2F_1\left(\frac{1}{2},\frac{4n+5}{4};\frac{4n+9}{4};t^4\right)+(4 n+5) \cos
   ^{-1}\left(t^2\right)\right)}{(4 n+3) (4 n+5)}$$
But, these integrals express also in terms of elliptic integrals of the first kind. Their general form is
$$J_n= t \sqrt{1-t^4}P_{n}(t^4)+\frac {t^{4n+3}}{4n+3} \, \cos ^{-1}\left(t^2\right)+a_n\, F\left(\left.\sin ^{-1}(t)\right|-1\right)$$

Concerning the $a_n$'s, they form the sequence
$$\left\{\frac{2}{9},\frac{10}{147},\frac{30}{847},\frac{26}{1155},\frac{442}{27797},
   \frac{1326}{110561},\frac{11050}{1168101},\frac{320450}{41575743},\cdots\right\}$$ which do not seem to be known in $OEIS$.

However, in a private discussion, @Raymond Manzoni did identify the sequence
$$a_n=\frac{2}{3 (4 n+3)}\prod_{k=1}^n \frac{4 k+1}{4 k+3}=\frac{2}{3 (4 n+3)}\frac{\Gamma \left(\frac{7}{4}\right) \Gamma \left(n+\frac{5}{4}\right)}{\Gamma
   \left(\frac{5}{4}\right) \Gamma \left(n+\frac{7}{4}\right)}$$

I think that the above would be interesting from a computing point of view.

Edit

If we are concerned by
$$I=\int_0^{\frac \pi 2} x\sqrt{\cos (x)}\,dx=\frac{4}{15} \,
   _3F_2\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};1\right)$$

Now, using

$$I=-2\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, J_n$$ Integrated between $0$ and $1$, the first two terms of the expressions for $J_n$ are $0$ and we are left with
$$J_n=a_n\, K(-1)=\frac{2\, \sqrt{\pi } \,\Gamma \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma   \left(n+\frac{3}{4}\right)}$$

$$I=-4 \sqrt{\pi }\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\,\frac{   \Gamma
   \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-\frac{\sqrt{\pi }\, \Gamma \left(\frac{1}{4}\right)}{9 \,\Gamma
   \left(\frac{3}{4}\right)}\,
   _3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right
   )$$

Update

All of the above is so complex that, may be, approximations could be used.

For example, using the $1,400$ years od approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for}\qquad -\frac \pi 2 \leq x\leq\frac \pi 2 $$ in this range
$$\int x\sqrt{\cos (x)}\,dx \sim \frac{1}{2} \sqrt{\left(\pi ^2-4 x^2\right) \left(x^2+\pi ^2\right)}+\frac{5}{4}
   \pi ^2 \sin ^{-1}\left(\frac{2 \sqrt{x^2+\pi ^2}}{\sqrt{5} \pi }\right)$$ which, integrated between $0$ and $\frac \pi 2$ would give
$$\frac{5 \pi ^3}{8}-\frac{\pi ^2}{4}  \left(2+5 \sin
   ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\approx 0.785221$$ while the exact value is $\approx 0.784608$.


Title: I am having a difficult time understanding what I am doing wrong with quadratics. $5x^2-15x-4=0$

Body:
I think I am missing a key part of solving quadratic equations. When I solve by grouping, I run into an issue of my values being incorrect, even though I am doing the same thing to both sides of the equation. Could someone explain to me why this doesn't work and what foundation I need to revisit? Here is my work:
$$
\begin{align}
5x^2-15x-4&=0\\
5x^2 -15x &= 4\\
5x^2–15x-20&=-16\\
5x^2–20x+5x-20&=-16\\
5x(x-4)+5(x-4)&=-16\\
(5x+5)(x-4)&=-16\\
\implies x=-4.20, -12
\end{align}
$$

Body:
Are you trying to complete the square in your problem?

$5x^2-15x-4 = 0$

Then you'd need to do this:

$5x^2-15x = 4$

Then divide by 5:

$x^2-3x=\frac{4}{5}$

Now complete the square:

$x^2-3x + (\frac{-3}{2})^2 = \frac{4}{5} + (\frac{-3}{2})^2$

$(x-\frac{3}{2})^2 = \frac{4}{5}+\frac{9}{4}$

$(x-\frac{3}{2})^2 = \frac{61}{20}$

$(x-\frac{3}{2})^2 = (\sqrt{\frac{61}{20}})^2$

$(x-\frac{3}{2})^2 - (\sqrt{\frac{61}{20}})^2 = 0$

Now you can factor this using $a^2-b^2 = (a-b)(a+b)$

$(x-\frac{3}{2}-\sqrt{\frac{61}{20}})(x-\frac{3}{2} + \sqrt{\frac{61}{20}}) = 0$

Now you can use that ab=0 means a=0 or b=0.

So

$x-\frac{3}{2}-\sqrt{\frac{61}{20}}=0$

or

$x-\frac{3}{2}+\sqrt{\frac{61}{20}}=0$

So solving these two gives

$x = \frac{3}{2} + \frac{1}{2}\sqrt{\frac{61}{5}} \approx3.2464$

or

$x = \frac{3}{2} - \frac{1}{2}\sqrt{\frac{61}{5}} \approx0.2464$

Does this help at all?

I could also have done it like this:

$5x^2-15x = 4$

$5(x^2-3x) = 4$

$5(x^2-3x+(\frac{-3}{2})^2) - 5(\frac{-3}{2})^2 = 4$

$5(x-\frac{3}{2})^2 - \frac{45}{4} = 4$

$5(x-\frac{3}{2})^2 = \frac{61}{4}$

$(x-\frac{3}{2})^2 = \frac{61}{20}$

Then take the square root of both sides. That gives

$x-\frac{3}{2} = \sqrt{\frac{61}{20}}$ or

$x-\frac{3}{2} = -\sqrt{\frac{61}{20}}$

which gives the same two solutions.


问:
A question regarding random variables
Consider the average $Y$ of $n$ independent random variables, each uniformly distributed on $[a,b]$. Does that mean $Y= \sum_{i=1}^{n}Y_{i}$ or $Y= \dfrac{\sum_{i=1}^{n}Y_{i}}{n}$?

Thank you for your time.

答: Average $\implies$ the second one.</p>

Title: A question regarding random variables

Body: Average $\implies$ the second one.


答:
Let us follow your approach: till $(5x+5)(x-4) = -16$, all your statements are correct.  Now you still need to find which values of $x$ satisfy this, and those are the roots.

However none of the values you mention satisfy - for e.g. taking $x = -12$, this statement reads $(-60+5)(-12-4) = -55\cdot -16 = 880 \neq -16$.

If the RHS of $-16$ factors as $a\times b$ for some $a, b$, and you can find a solution for $x$ which simultaneously satisfies $5x+5=a, x-4=b$, indeed you have a solution.  However such factors aren't obvious and in general finding them may be tougher than solving the original quadratic through other methods.  Further, you may get only one value of $x$ through this approach, for the other root, you will have to keep checking other possible factorisations.

The only case where you can independently set any one the factors in LHS to equal the RHS to find roots, is when the RHS is $0$, for obvious reasons.  This is what you seem to have done, but except when RHS is $0$, this may not solve the original problem. Another e.g. where you may find this working is $(2x-3)(3-x)=1$, due to the coincidence of the factorisation $1^2=1$, even so that gives you only one of the roots.

Hence this approach isn't useful in most cases, and others have suggested valid approaches such as completing the square, or using the quadratic formula, both of which works always.


问:
Given 2 matrices, find T
Trying to find $T$ given $C=T^{-1}BT$

$B = A(A^TA)^{-1}A^T \in \mathbb{R}^{m \times m}$ given $A=\begin{bmatrix}A_1 \\ A_2\end{bmatrix}$ with $A_1 \in \mathbb{R}^{n \times n} $ is invertible and $A_2 \in \mathbb{R}^{k \times n}$ where $k = m-n$.

$C=\begin{bmatrix}I_{n \times n} & 0_{n \times k}\\ 0_{k \times n} & 0_{k \times k}\end{bmatrix}$

I got to this point:
$$C=\begin{bmatrix}A_1A_1^{-1} & 0 \\ 0 & 0\end{bmatrix}=T^{-1}\begin{bmatrix}A_1 \\ A_2\end{bmatrix}(A_1A_1+A_2A_2)^{-1}\begin{bmatrix}A_1 & A_2\end{bmatrix}T$$

I somehow need to find a $T$ in terms of $A_1$ and $A_2$ that will cancel out all of the $A_2$s and leave me with $A_1A_1^{-1}=I$. How can I proceed?

Thanks!

答: $B$ is the orthogonal projection onto the column space of $A$. Its null space is the orthogonal complement to the column space of $A$. If you take $T=\pmatrix{A_1&-(A_2A_1^{-1})^T\\ A_2&I_k}$, its first $n$ columns are precisely the columns of $A$ and its last $k$ columns form a basis of the null space of $B$. Hence $T$ is invertible, $BT=TC$ and $C=T^{-1}BT$.

Title: Given 2 matrices, find T


问:
Expectation of Computational Complexity for a dense matrix linear system
Given a linear system: $A \in \mathcal{R}^{n \times n}$

$$
A = \begin{bmatrix}
n & 1 & 1 & \cdots & 1 \\
1 & n & 1 & \cdots & 1 \\
1 & 1 & n & \cdots & 1 \\
\cdots & & &  \dots &  n \\
\end{bmatrix}
$$
which has elements $n's$ on the diagonal and $1's$ everywhere, and $b = [1,2,...,n]^{T}$. Consider this linear system:

$$
Ax = b
$$

I am trying to find the expectation of execution time for this linear system with LU algorithm solving linear system and Gauss-Seidel method. For LU algorithm, I figured out using computational cost for LU is $W = \frac{2}{3} n^{3} +  2n^{2} + O(n)$ deducing the ratio is near to $8$ when $n$ size is scaled by double its size. I tried to run the Gauss Seidel in python for problem sizes $n =250,500,1000,2000,4000$ the number of convergence iterations are $18,18,19,20,20$ respectively. However, for the method Gauss-Seidel, how can I find the execution time as a scale asymptotically as a function of $n$ ? I was given a hint that for number of iterations as a function of $n$ it is very specific for this type of matrix $A$.

答:
The Gauss-Sidel iterations are $x_{k+1} = L^{-1}(b-Ux_k)$. If your initial guess is $x_0 = 0$, then we have $$x_m = \left(\sum_{k = 0}^{m-1}(-L^{-1}U)^k\right)L^{-1}b.$$ This converges when $\rho(L^{-1}U) < 1$, and the error is $$x_{\infty}-x_m = \left(\sum_{k = m}^{\infty}(-L^{-1}U)^k\right)L^{-1}b = (-L^{-1}U)^{m}(I+L^{-1}U)^{-1}L^{-1}b,$$ which has norm $$\|x_{\infty}-x_m\| = \left|(-L^{-1}U)^{m}(I+L^{-1}U)^{-1}L^{-1}b\right\| \le \|L^{-1}U\|^m \cdot \|(I+L^{-1}U)^{-1}L^{-1}b\|.$$

Using this equation, you can figure out what value of $m$ makes the norm small enough to meet the convergence criteria.


问:
If a measure $\mu$ arises from the restriction of an outer measure $\mu^{*}$, is it true that $\mu$ is saturated?
A measure $\mu$ on a measurable space $(X, \mathcal{M})$ is called saturated if given a subset $E$ of $X$ the condition $E \cap A \in \mathcal{M}$ for every $A \in \mathcal{M}$ with $\mu(A) < \infty$ implies that $E \in \mathcal{M}$ (that is if any locally measurable subset of $X$ is measurable). The Wikipedia article on saturated measures states that "measures arising as the restriction of outer measures are saturated".

I know how to prove this result when the outer measure itself arises from a pre-measure (a nice proof can be found in the answer to this question), but not every outer measure is generated by a pre-measure so I'd like to know if the previous assertion is actually true and, if it is, how can I prove the result in the general case, that is how can I show that any measure $\mu$ resulting from the restriction of an outer measure $\mu^{*}$ is saturated?

I wasn't really sure about the validity of the result so I tried to construct a counterexample by considering the set $X = \{0,1\}$ and the outer measure $\mu^{*} : \mathcal{P}(X) \rightarrow [0, \infty]$ given by $\mu^{*}(\emptyset)=0$, $\mu^{*}(\{0\}) = 2$, $\mu^{*}(\{ 1 \})=2$ and $\mu^{*}(X)=3$ which is not generated by a pre-measure and has $\emptyset$ and $X$ as the only two $\mu^{*}$-measurable subsets of $X$. The problem is that the measure induced by this outer measure $\mu^{*}$ is saturated (since $\mu^{*}(X) is finite), so if a counterexample does exist it would be nice to know it.

If every measure obtained by the restriction of an outer measure is actually saturated, I'd like to know a proof of this result and my attempt was to consider the measure space $(X, \mathcal{M}^{*}, \overline{\mu})$ where $\mathcal{M}^{*}$ is the $\sigma$-algebra on $X$ consisting of the $\mu^{*}$-measurable subsets of $X$ and $\overline{\mu}$ is the restriction of the outer measure $\mu^{*}$ to $\mathcal{M}^{*}$ and then considering the outer measure, say $\mu^{+}$, induced by the measure $\overline{\mu}$. Since the outer measure $\mu^{+}$ is induced by the measure $\overline{\mu}$, we know that the measure obtained by restricting $\mu^{+}$ to the collection of $\mu^{+}$-measurable subsets of $X$, say $\hat{\mu}$, is a saturated measure and I believe I would be able to finish the proof if $\mathcal{M}^{*}$ is equal to the $\sigma$-algebra of $\mu^{+}$-measurable subsets of $X$ and $\mu^{+}= \mu^{*}$ (but I think this is only true if the original outer measure $\mu^{*}$ is induced by a pre-measure which leads us back to the original problem of proving the desired result when we drop this assumption).

Any hints or ideas would be greatly appreciated and thank you all in advance for your answers.

答:
OK I think I found a counterexample which shows the result is false in the general case (when the outer measure is not induced by a pre-measure). If $\mu^{*} : \mathcal{P}(\mathbb{R}) \rightarrow [0, \infty]$ is the outer measure defined by $\mu^{*}(\emptyset) = 0$, $\mu^{*}(A) = 1$ if $A$ is countable and nonempty, and $\mu^{*}(A) = \infty$ if $A$ is uncountable.

It is easy to see that $\mu^{*}$ is an outer measure on $\mathcal{P}(\mathbb{R})$.

Now, if $E$ is a proper nonempty subset of $\mathbb{R}$, $x \in E$ and $y \in \mathbb{R} \setminus E$, then $\{x,y\}$ is clearly a nonempty countable subset of $\mathbb{R}$, $\{x,y\} \cap E = \{x\}$, and $\{x,y\} \cap E^{c} = \{y\}$, so that $\{x,y\} \cap E$, and $\{x,y\} \cap E^{c}$ are nonempty countable subsets of $\mathbb{R}$ as well. From the definition of $\mu^{*}$ we obtain that $\mu^{*}(\{x,y\}) = 1 < 1+1 = \mu^{*}(\{x,y\} \cap E) + \mu^{*}(\{x,y\} \cap E^{c})$ and consequently $E$ is not a $\mu^{*}$-measurable subset of $\mathbb{R}$.

The above argument shows that $\emptyset$ and $\mathbb{R}$ are the two only $\mu^{*}$-measurable subsets of $\mathbb{R}$, that is $\mathcal{M}^{*} = \{ \emptyset , \mathbb{R} \}$ (where $\mathcal{M}^{*}$ denotes the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $\mathbb{R}$). But then any proper nonempty subset $F$ of $\mathbb{R}$ is locally measurable in the measure space $(X, \mathcal{M}^{*}, \mu^{*}|_{\mathcal{M}^{*}})$ since $\emptyset$ is the only element of $\mathcal{M}^{*}$ with finite measure $\mu^{*}|_{\mathcal{M}^{*}}$ and $F \cap \emptyset = \emptyset \in \mathcal{M}^{*}$. Since said nonempty proper subset $F$ of $\mathbb{R}$ is not $\mu^{*}$-measurable we conclude that $\mu^{*}|_{\mathcal{M}^{*}}$ is not a saturated measure.


答:
$\text{The sample $\color{red}{\text{mean}}/\text{average}$  always $\color{blue}{\text{means}}$}$ $$Y = \dfrac{\sum_{i=1}^n Y_i}n$$
$\text{irrespective of the distribution.}$</p>

Title: A question regarding random variables

Body:
Consider the average $Y$ of $n$ independent random variables, each uniformly distributed on $[a,b]$. Does that mean $Y= \sum_{i=1}^{n}Y_{i}$ or $Y= \dfrac{\sum_{i=1}^{n}Y_{i}}{n}$?

Thank you for your time.


问:
For which $u$ does the derivative $f'(u,0)$ of $f(x,y)=x^3/(x^2+y^2)$ exist?
Let $f: \mathbb{R}^2\to \mathbb{R}$ be defined by setting

$$f(x,y)=\begin{cases}\frac{x^3}{x^2+y^2} & (x,y)\neq (0,0)\\0 & (x,y)=(0,0)\end{cases}$$
(a) For which vectors $u\ne 0$ does $f'(0; u)$ exist? Evaluate it when it
exists.

(b) Do $D_1f$ and $D_2f$ exist at $0$?

(c) Is $f$ differentiable at $0$?

(d) Is $f$ continuous at $0$?

I have thought a lot about this problem:

For (a), be $u\neq 0, u:=(h,k)$, so $\lim_{t\to 0}\frac{f(0+tu)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{(th)^3}{(th)^2+(tk)^2}\frac{1}{t}=\frac{h^3}{h^2+k^2}$, then $f'(0,u)$ exists for all $u\neq 0$.

For (b), $\lim_{t\to 0}\frac{f(0+te_1)-f(0)}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=\lim_{t\to 0}\frac{t^3}{t^3}=\lim_{t\to 0}1=1$ so $D_1f(0,0)=1$ and $\lim_{t\to 0}\frac{f(0+te_2)-f(0)}{t}=\lim_{t\to 0}0=0$ so $D_2f(0,0)=0$

I am having problems solving (c) and (d), could someone help me please? Thank you very much.

答:
Is my solution for (c) ok?

(c) Assume that $f$ is differentiable at $\mathbf{0}$.

Let $\mathbf{u}=(1,1)$.

$f'(\mathbf{0};\mathbf{u})=Df(\mathbf{0})\cdot\mathbf{u}=\begin{bmatrix} D_1f(\mathbf{0})&D_2f(\mathbf{0})\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=1$.

On the other hand, from (a), $f'(\mathbf{0};\mathbf{u})=\frac{1}{2}$.

This is a contradiction.

So, $f$ is not differentiable at $\mathbf{0}$.

(d) Let $\mathbf{h}=(h_1,h_2)$.

$|f(h_1,h_2)-f(0,0)|=|\frac{h_1^3}{h_1^2+h_2^2}|\leq|h_1|\frac{|h_1^2|}{|h_1^2+h_2^2|}\leq|h_1|\frac{|h_1^2+h_2^2|}{|h_1^2+h_2^2|}\leq|h_1|\leq\max\{|h_1|,|h_2|\}=|\mathbf{h}|.$

So, $f$ is continuous at $\mathbf{0}$.


问:
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .

What I Tried: Here is a picture:-

Let the perpendicular bisector of $BC$ pass through $BC$ at $E$ .

Then I first noticed that $\Delta BDE \cong \Delta CDE$ from $(SAS)$ congruency.

This gives the required information in the diagram, as well as we have $BD = 7$ .

Now :- $$\Delta ABD \sim \Delta ACB$$
$$\rightarrow \frac{AD}{AB} = \frac{7}{BC} = \frac{AB}{AC}$$
So let $AD = k$ , $AB = m$ , $BE = EC = n$ . We have :-
$$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$

答:
Note:- As others mentioned, this question was unsolvable before because there can be infinitely many such triangles. I also realized that is true because I missed a small detail that $AD = 9$. Sorry for the inconvenience. I also figured out how to solve this now.

I had figured out that :-
$$\frac{k}{m} = \frac{7}{2n} = \frac{m}{(7+k)}$$
With the new information this becomes :-
$$\frac{9}{m} = \frac{7}{2n} = \frac{m}{16}$$

From here, I get $m = 12$ , so I have the lengths of all the sides of $\Delta ABC$.

Now from Heron's Formula I have :-
$$[\Delta ABD] = \sqrt{s(s-a)(s-b)(s-c)}$$
$$\rightarrow \sqrt{14 * 2 * 5 * 7}$$
$$\rightarrow \sqrt{980} = 14\sqrt{5}$$
So we have $a = 14$ as our solution.


答:
The expression $g(x,y,c)=0$ does not specify the function. It specifies the value which the function must have (from its range of possible values) to solve the DE, but that is not the same thing. The function $g(x,y,c)$ will have value $0$ for particular values of the three quantities $x$, $y$, and $c$, and the equation $g(x,y,c)=0$ can be used to find relationships between the values of $x$, $y$ and $c$ Which solve it.

The function $g$ is not identically zero; it takes other values for suitable values of the three variables. To partially differentiate a function with respect to a variable requires changing that variable without changing the other variables. However the condition $g=0$ contradicts the requirement of changing only one variable at a time so doing a simple partial derivative violates this condition. In other words the process of performing the partial differentiation assumes that there are values of $x$, $y$ and $c$ for which $g=0$ is false.

We can instead do a more complex process where we take into account the changes to the other variables. In your question, if you change $c$ and keep $y$ constant then $x$ cannot be constant (in fact $x=\pm c$), so you must include the partial derivative $\frac{\partial x}{\partial c}$ to account for changes in $x$.

Title: Partial Derivative Leads to Contradiction

Body:
Starting from the same equation $g(x,y,c) = 0$, I partially differentiate with regards to c and end up in a contradiction. However, if I manipulate the equation a bit I do not. I will post the entire solution, which starts from an ODE.
$$(3y-a)^2*y\prime^2 - 4y = 0$$
$$y\prime = +(\frac{4y}{3y-a})^{1/2}, or\ y\prime = -(\frac{4y}{3y-a})^{1/2}$$
$$solving\ both\ ODEs: y^{3/2} - ay^{1/2} = \pm*x + c \quad (1)$$
$$(1)\rightarrow y^{3/2} - ay^{1/2}  \mp*x - c = 0 \quad let\ g_1(x,y,c) = y^{3/2} - ay^{1/2}  \mp*x - c = 0 \quad (2)$$
$$(1)\rightarrow (y^{3/2} - ay^{1/2})^2 = (\pm*x + c)^2 $$
$$\Leftrightarrow y*(y^3-a)^2 = (\pm*x + c)^2 \quad let\ g_2(x,y,c) = y*(y^3-a)^2 - (\pm*x + c)^2 = 0 \quad (3)$$

Now if we try and diferentiate both  $g_1$ and $g_2$ partialy with regards to c we end up with the following:

$$(2) \rightarrow -1 = 0$$
$$(3) \rightarrow c = \mp x$$

So my question really is, why can I not partially differentiate (2)? I know from where I got this question that (3) is the correct way to go, but why is that?


问:
Continued Fraction for $\log(\frac{1+\alpha z}{1+z})$
There is a continued fraction for
$$\log\left(\frac{1+z}{1-z}\right)$$
given by
$$\cfrac{2z}{1-\cfrac{1z^2}{3-\cfrac{4z^2}{\cdots}}}.$$

This was known even to Gauss.
See Handbook of Continued Fractions for Special Functions by Annie A.M. Cuyt et al pg 196
or Analytic theory of Continued fractions by H.S. Wall pg 343.

Is there a continued fraction of the same form for the more general
$$\log\left(\frac{1+\alpha z}{1-z}\right)?$$

答:
There is
$$
\log \left(\frac{a z+1}{1-z}\right)-\frac{(a+1) z}{1-\frac{\frac{1-a}{2} z}{1-\frac{\frac{(a+1)^2}{6 (1-a)} z}{1-\frac{\frac{a^5-a^4-8 a^3-8 a^2-a+1}{3(-a^4-2 a^3+2 a+1)} z}{1-\ldots}}}}.
$$
Any formal power series can be expanded into a cfrac of the form $a_0+\operatorname{K}\limits_{n=0}^\infty\frac{a_nz}{1}$, where coefficients $a_n$ can be determined through Hankel matrices composed of coefficients of the formal power series. But one can not always expect a regular pattern in the coefficients $a_n$.


问:
Subgroup of $S_8 $ of order $9$
This particular question was asked in my abstract algebra quiz and I was unable to solve it and I am looking for help here.

Let $H$ be a subgroup of $S_8 $ such that order of $H$ is $9$, then $H$ is  :

A) cyclic

B) not abelian

C) abelian but not cyclic

D) if $H$ is abelian then it must be cyclic

$H$ cant be cyclic as then it will have an element of order $9 $.  Also, by cauchy theorem, there exists an element of order $3$. Disjoint cycles always commute. But how to use that condition here?

But I am unable to think about other options, as i am not really good while analysizing permutation groups. Can you please tell how to approach these kind of problems in permutation groups?

答:
Suppose $H$ is cyclic. Then there exists an element of order $9$ in $S_8$. But $S_8$ does not have any element of order $9$ because a permutation of order $9$ should contain a $9$-cycle. So (A) and (D) are not correct.

Note that for any prime $p$, group of order $p^2$ is always abelian. (Refer here).  So (B) is not correct and the only correct answer is (C).

Title: Subgroup of $S_8 $ of order $9$


问:
Is $\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}$ true?

Prove/disprove $$\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}.\tag 1$$

As far as I can tell, this is true, although it seems to converge very slowly.

I came up with a proof but I don't know if it's valid.

Let $$J=\int_0^\pi \frac{xdx}{1+\sin x}.$$
On one hand, we have
$$\frac1{1+\sin x}=\sum_{n\ge0}(-1)^n\sin(x)^n,$$
so that
$$J=\sum_{n\ge0}(-1)^n p_n,\tag 2$$
where
$$
\begin{align}
p_n&=\int_0^\pi x\sin(x)^ndx\\
&=\int_\pi^0 -(\pi-x)\sin(\pi-x)^ndx\\
&=\pi\int_0^\pi\sin(x)^ndx-p_n\\
\Rightarrow p_n&=\frac\pi2\int_0^\pi\sin(x)^ndx.
\end{align}
$$
And since $\sin(x)=\sin(\pi-x)$, $$p_n=\pi\int_0^{\pi/2}\sin(x)^ndx=\frac{\pi^{3/2}}{2}\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}.\tag 3$$

On the other hand, we have $1+\sin x=2\sin^2(\tfrac{x}2-\tfrac\pi4)$, so that
$$\begin{align}
J&=\frac12\int_0^\pi\frac{xdx}{\sin^2(\tfrac{x}2-\tfrac\pi4)}\\
&=2\int_{\pi/4}^{3\pi/4}\frac{tdt}{\sin^2t}-\frac\pi2\int_{\pi/4}^{3\pi/4}\frac{dt}{\sin^2 t}\\
&=2\left(\ln\sin x-x\cot x\right)\bigg|_{\pi/4}^{3\pi/4}-\frac\pi2\left(-\cot x\right)\bigg|_{\pi/4}^{3\pi/4}\\
&=2\pi-\frac\pi2\cdot2=\pi.
\end{align}$$
Then from $(2)$ and $(3)$, we have
$$\frac{\pi^{3/2}}{2}\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\pi,$$
which is equivalent to $(1)$. $\square$

Can you come up with any other proofs to $(1)$? Thanks!

Edit (11/12/2020):

Here is a proof that the interchange of the sum and integral in $(2)$ is valid.

The partial sums $$S_M(x)=\sum_{n=0}^M(-1)^n\sin(x)^n$$
form a uniformly convergent sequence of functions for $x$ in $[0,\pi/2)$ or $(\pi/2,\pi]$, and they converge to the limit $$\lim_{M\to\infty}S_M(x)=\frac1{1+\sin x},\qquad x\in[0,\pi]\setminus\{\pi/2\}.$$
Choose $\varepsilon>0$ and notice that
$$J=\int_{0}^{\pi}\frac{xdx}{1+\sin x}=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\int_0^{\pi/2-\varepsilon}\frac{xdx}{1+\sin x}+\int_{\pi/2+\varepsilon}^\pi\frac{xdx}{1+\sin x}.$$
The sums $S_M(x)$ converge uniformly as $M\to\infty$ when $x\in[0,\pi/2-\varepsilon]\cup[\pi/2+\varepsilon,\pi]$, so we can interchange the sum and integral to get
$$J=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\sum_{n\ge0}(-1)^n(a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)),$$
where
$$\begin{align}
a_n(t)&=\int_0^t x\sin(x)^ndx\\
b_n(t)&=\int_t^\pi x\sin(x)^ndx.
\end{align}$$
We have $a_n(t)+b_n(t)=p_n$ for all $t\in[0,\pi]$. As $\varepsilon$ approaches $0$, $\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}$ approaches $0$.

And since $a_n(t), b_n(t)$ are continuous, $a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)$ approaches $a_n(\pi/2)+b_n(\pi/2)=p_n$, so that
$$J=\sum_{n\ge0}(-1)^np_n$$
as desired. $\square$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(\bracks{n+1}/2) \over \Gamma(n/2 + 1)} =
{2 \over \root{\pi}}} \approx 1.1284:\ {\Large ?}}$

\begin{align}
&\bbox[5px,#ffd]{\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(\bracks{n+1}/2) \over \Gamma(n/2 + 1)}}
\\[5mm] = &\
{1 \over \root{\pi}}\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(n/2 + 1/2)\Gamma\pars{1/2} \over
\Gamma(n/2 + 1)}
\\[5mm] = &\
{1 \over \root{\pi}}\sum_{n\ \geq\ 0}\pars{-1}^{n}
\int_{0}^{1}t^{n/2 - 1/2}\,\,\,
\pars{1 - t}^{-1/2}\,\,\dd t
\\[5mm] = &\
{1 \over \root{\pi}}\int_{0}^{1}{1 \over \root{t}\root{1 - t}}\sum_{n\ \geq\ 0}\pars{-\root{t}}^{n}\,\dd t
\\[5mm] = &\
{1 \over \root{\pi}}\int_{0}^{1}{1 \over \root{t}\root{1 - t}}{1 \over 1 + \root{t}}\,\dd t
\\[5mm] \stackrel{t\ \mapsto\ t^{2}}{=} &\
{2 \over \root{\pi}}\int_{0}^{1}{1 \over
\root{1 - t^{2}}}{1 \over 1 + t}\,\dd t
\\[5mm] \stackrel{t\ \mapsto\ \sin\pars{\theta}}{=} &\
{2 \over \root{\pi}}\int_{0}^{\pi/2}
{\dd\theta \over 1 + \sin\pars{\theta}}
\\[5mm] = &\
{2 \over \root{\pi}}\int_{0}^{\pi/2}
\bracks{\sec^{2}\pars{\theta} - \sec\pars{\theta}\tan\pars{\theta}}\dd\theta
\\[5mm] = &\
{2 \over \root{\pi}}\
\underbrace{\bracks{\sin\pars{\theta} - 1 \over \cos\pars{\theta}}_{0}^{\pars{\pi/2}^{\,-}}}
_{\ds{=\ 1}}\ =\
\bbx{2 \over \root{\pi}} \approx 1.1284 \\ &
\end{align}


问:
Focus of rectangular hyperbola touching a given ellipse $\frac{x^2}{36}+\frac{y^2}{18}=1$
Consider an ellipse $\frac{x^2}{36}+\frac{y^2}{18}=1$

There is a hyperbola whose one asymptote is the major axis of a given ellipse. If the eccentricity of the given ellipse and hyperbola are reciprocal to each other, both have the same center and both touch each other in the first and third quadrant. The focus of hyperbola is at?

Clearly, it will be a rectangular hyperbola say $xy=k$ for $k>0$, and if hyperbola and ellipse touch each other we get biquadratic equation $x^4-36x^2+2k^2=0$. I can consider $D=0$ for tangency, but as it is quadratic in $x^2$ it may have another condition when one root is positive and another root is negative. That's where I am stuck. Please help!

答:
E: $$x^2/36+y^2/18=1 \implies y'=-\frac{x}{2y}$$
RK: $$xy=k^2 \implies y'=-y/x$$
Equating the two derivatives (slopes), we get $2y^2=x^2$ Putting this in RH
we get $x= 2^{1/4}k, y= 2^{-1/4} k$ in the first quadrant. Using these in E we get
$\frac{k\sqrt{2}}{36}+\frac{k}{18\sqrt{2}}=1 \implies k=3 (2)^{1/4}$. So the required Two RH are $$xy= \pm 9 \sqrt{2}.$$


问:
Quick way to solve computational congruences
The specific problem at hand is $$34x \equiv 60 \bmod{98}$$
I reduced to get $$17x \equiv 30 \bmod{49}$$
and from this I have
$$17x \equiv 30 \bmod{7}$$
which is easy to solve and yields $x \equiv 3 \bmod{7}$. How can I use this fact to help me solve the more complex equation? I know the solution will be of the form $7k + 3$, but how can I utilize this fact?</p>

答:
This may be more than you need to know, but Hensel's Lemma is the big tool here.  You have $f(x)=17x-30$, which you solved modulo 7, i.e. $f(3)\equiv 0\pmod{7}$.  Because $f'(3)=17\not\equiv 0\pmod{7}$, there will be a unique solution modulo $7^2$.  That solution will be $3+t7$, where $t$ is the unique solution to $f(3)/7+f'(3)t\equiv 0\pmod{7}$.  In your case, $f(3)=21$ and $f'(3)=17$ so you want to solve $3+17t\equiv 0\pmod{7}$.  This has solution $t=6$, so $3+6\cdot 7=45$ is the unique solution mod $49$.

If you wanted to solve it modulo $7^3$, you will again get a unique solution because $17$ remains nonzero modulo $7$.  The answer will be $45+t49$, where $t$ is the unique solution to $f(45)/49+17t\equiv 0\pmod{7}$.  That is, $15+17t\equiv 0\pmod{7}$, which has solution $t=2$.  Hence $45+2\cdot 49=143$ is the unique solution mod $7^3=343$.


问:
Unable to find example / contradiction of this question involving infinite order groups , non -abelian groups
This question was asked in my abstract algebra quiz and I was completely clueless in proving / contradicting any of the options and I am looking for help.

Question : Which of the following statements are right:

A There exists a group of infinite order whose every subgroup is of finite order except itself.

B There exists a non cyclic group of infinite order whose every subgroup is cyclic except itself.

C There exists a non abelian group such that that for all n $\in \mathbb{N}$ there exists an element of oirder n.

D there exists a non abelian group such that for all n $\in \mathbb{N}$- {0}, there are infinite number of elements of order n.

For abelian groups I tried $\mathbb{Z} ,+$ and $\mathbb{Q} ,+$ and there subgroups but no progress could be made . I have no stratergy or examples to prove / contradict any of the options although I have solved many problems in group theory.

Can you please guide?

答:
For A and B, take $\Bbb Z[\frac12]/\Bbb Z$.

For the answers to C and D, here are the suggested example groups - however, it depends on tiny details of the problem statement whether they actually work:

For C, take $(\Bbb Q/\Bbb Z)\times S_3$.

For D take a product of infinitely many copies of the C example.

A,B: Let $H$ be a proper group of $\Bbb Z[\frac12]/\Bbb Z$.
If $\frac k{2^n}+\Bbb Z\in H$ where wlog $k$ is odd, then by Bezout, there exist  $u,v\in\Bbb Z$ with $uk+2^nv=1$, so $\frac1{2^n}+\Bbb Z=\frac {u\cdot k}{2^n}+\Bbb Z\in H$ and then clearly all $\frac \ell{2^m}+\Bbb Z$ with $m\le n$ are $\in H$. Hence for any proper subgroup, the denominators are bounded and if $2^N$ is the maximal denominator of an element of $H$, we see by the preceding argument that $H=\langle\frac1{2^N}+\Bbb Z \rangle$. In other words, every proper subgroup of the (non-cyclic) infinite group $\Bbb Z[\frac12]/\Bbb Z$ is a finite cyclic group.

C: The group is non-abelian because it has a subgroup isomorphic to $S_3$. And for $n\in\Bbb N$, the element $(\frac1n+\Bbb Z,e)$ has order $n$. Note: This assume that $0\notin \Bbb N$, the truth of which may depend on the author. If in your context, $0$ is an element of $\Bbb N$, then of course no such group can exist - because the order of a group element is always a positive integer (or $\infty$) and is never $0$.

D: It is clear from C that we can obtain an element of order $n$ by picking one in an arbitrary factor. Note: For $n=1$ this will always produce the neutral element - which is of course the only element of order $1$ in any group. So if we correct the problem statement to read "... for all $n\in\Bbb N-\{0,1\}$ ...", then the group specified above is an example. But if the problem statement is exactly as posted, there cannot exist such a group because there is always only one element of order $1$.


问:
Which convex shape requires the longest slice to break in half?
Suppose you are given a convex unit area region of the plane, and you wish to minimize the length of a line segment that cuts the shape into two regions of equal area. Which shape results maximizes this length, i.e., is the hardest to slice into two equal regions. Is it the circle?

For generalizations, does anything change if we allow the slice to be any curve? What if we want to chop it into $n$ equal regions for $n>2$? What if we are minimizing the area of planes that slice higher dimensional regions in half? I'm curious in any generalizations along these lines.

答:
Edit: This question is discussed at length in Chords halving the area of a planar convex set (Grüne et al., 2007), along with the question of the longest possible splitting chord. However, the paper (while establishing many other bounds) leaves the relation between area and minimal halving distance open, except to note that the $\sqrt[4]{3}$ upper bound is not tight.

Notably, the circle is not optimal, and any (non-circular) convex Zindler curve serves as a counterexample. The aforementioned paper cites a 2005 preprint On geometric dilation of closed curves, graphs and point sets (Dumitrescu et al., 2005) for this fact, where perusing Section 7 we can see that the "rounded triangle" $C_\triangle$ is conjectured to have maximal cutting length $h$ for a fixed area, at
$$A(C_\triangle)=\sqrt{3}\left(\log(3) - \frac{\log(3)^2}{8} - \frac12\right)h^2$$
so $h\approx 1.135547\sqrt{A}$.

The curve $C_\triangle$ is pictured below, from Dumitrescu et al.'s paper. (The central region $M$ is not a hole, but the locus of the midpoints of area-splitting chords; we have $A(Z)=\frac\pi4h^2-2A(M)$ for any Zindler curve $Z$.)

My original answer, written before locating the above papers, is given below.

Let our convex set be $C$. Note that such a cut is at most the width $w$ of the shape, because if we place $C$ between two parallel lines, then some cut orthogonal to those lines will cut $C$ in half, and its length must be at most the distance between the lines. (This argument extends to the same question for dividing $C$ into regions of equal perimeter, or for that matter most natural measures of "size" of a convex set.)

We can then use the fact that $w^2\le \sqrt{3}A$ among bounded convex planar sets; see for instance Inequalities For Convex Sets (Scott and Awyong, 2000), which cites Yaglom and Boltyanski's book Convex Figures. This gives us a minimal cut width of at most $\sqrt[4]{3}\approx 1.316$.

Note that since the extremal convex set for this inequality is the equilateral triangle, which has a cut width of $\sqrt{2}/3^{1/4}\approx 1.075$, it does not in fact surpass the circle's $\frac{2}{\sqrt{\pi}}\approx 1.128$, but at least provides a reasonable upper bound.

As an unrelated piece of evidence, I believe I can get a cut of length around $1.1038$ on the Releaux triangle of unit area, so that particular candidate (worth checking for any question about convex sets) seems to not quite beat the circle.


问:
Does the series $\sum\ln(1+\frac{1}{n\ln(n)})$ Converge?
As the title states I wanna prove this series converges or not however I noted that if we fan transform $\ln(1+\frac{1}{n\ln(n)})$ into some known limit identity and use Comparison test I would be done

答:
For large $n$:

$\log (1+\dfrac{1}{n\log n})\sim$
$\dfrac{1}{n\log n}.$

$\displaystyle{\sum }\dfrac{1}{n\log n}$ diverges (Cauchy condensation test).


问:
Conjecturing when $M$ is good

We say an integer $M>1$ is good if whenever $n^n \equiv 1 \mod M$ then we also have  $n \equiv 1 \mod M$ and bad otherwise, for any integer $n\ge 2$ . Prove that  all odd $M$  are bad. Find all good $M$ .

My progress: First taking example, we get Among $M \in \{2,3,4,5\}$ $  , 2,4$ are good.

Now for all odd $M$ , consider $(M-1)^{M-1}$ , note that $(M-1)^{M-1}\equiv -1^{M-1}\equiv 1 \mod M-1$ , since $M-1$ is even .But $M-1 \equiv -1 \mod M $. This proves the first part.

For second part I thought that it would be true for all even M, but then $9^9\equiv 1 \mod 14$ . However I got $2,4,6,8,10,12$ good.

Also $M=42$ is good too . Here's the proof showing $M=42$ is good:

If $n^n\equiv 1\mod 42 \implies n^n\equiv 1\mod 2 \implies n$ is odd . Also we have $n^n\equiv 1\mod 6 \implies n\equiv 1\mod 6$. Now for $n^n \equiv 1 \mod 7 \implies n \equiv \text{1 or 6} \mod 7 $. If $n=6 \mod 7 $ , then by CRT $n\equiv 13 \mod 42$ which is not possible as $13^{13} \equiv 13 \mod 42 $ ( Thanks! @Ross Millikan for correction )

I couldn't formulate a conjecture about when $M$ is $good$. Any hints? Thanks in advance!

答:
This is the complete solution.

Theorem: Let $\phi(M)$ be the Euler totient value of $M$. Then $M$ is good if and only if $\operatorname{sqrf}(\phi(M))\mid M$, where $\operatorname{sqrf}(\cdot)$ denotes the squarefree part.

For the “if” part, this follows because given $n\not\equiv 1\pmod M$ but coprime to $M$, then $\operatorname{ord}_M(n)>1$; however,
$$n^n\not\equiv 1\pmod M\,,$$
because otherwise, we would have that $\operatorname{ord}_M(n)\mid n$ whereas $\gcd(n,\phi(M))=1$.

For the “only if” part, suppose $p\mid\phi(M)$ but $p\nmid M$. Thanks to Cauchy’s Theorem (see https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)) we can find a natural number $m$ such that $\operatorname{ord}_M(m)=p$. Now let $v:= \operatorname{ord}_M(p)$ and define $n:=p^vm$. Clearly $n\not\equiv 1\pmod M$; however,  $$n^n=(p^vm)^{p^vm}= p^{v(p^vm)}m^{p^vm}\equiv 1\pmod M\,,$$
which completes the proof.

Title: Conjecturing when $M$ is good


问:
Group $G=\{f : \mathbb{R} \to \mathbb{R}\ |\ f \text{ is a bijective function}\}$ and elements of order $4$ and $9$

This question was asked in my abstract algebra quiz and I was unable to completely solve it.

Question : Let $G=\{f : \mathbb{R} \to \mathbb{R}\  |\ f \text{ is a bijective function}\}$ and $\circ$ be operation of composition of functions , then:

A) $G$ is abelian

B) $(G,\circ)$ has an element of order $9$.

C) there are uncountable elements of order $4$.

I proved A wrong . But I think that B can be right but there is no intution for C .

Can you please give a rigorious argument for B and C .

Thanks!!

答:
(B) Let $f$ be the map such that $f:1\mapsto 2\mapsto 3\mapsto 4\mapsto 5\mapsto 6\mapsto 7\mapsto 8\mapsto 9\mapsto 1$, and for every other $x$, $f:x\mapsto x$. The $f$ has order $9$.

(C) For each $\alpha\in\mathbb{R}$ let
$f_\alpha: \alpha+1\mapsto\alpha+2\mapsto\alpha+3\mapsto\alpha+4\mapsto\alpha+1$ and for every other $x$, $f_\alpha:x\mapsto x$. Then each $f_\alpha$ has order $4$, and they are distinct.


答:
The more general formulation of your problem is as follows: let $A$ be an infinite set and let $\Sigma(A)$ be the symmetric group on $A$, in other words the group of all permutations of $A$ with respect to composition (of maps). In your particular case $A=\mathbb{R}$.

As $A$ is infinite, for any $n \in \mathbb{N}^{\times}=\mathbb{N} \setminus \{0\}$ there exists an injection $\varphi \colon [1, n] \to A$ -- where by $[r, s]$ I am referring to the natural interval between $r$ and $s$ -- and you can consider the cyclic permutation $(\varphi(k);)_{1 \leqslant k \leqslant n}$, which will of course be of order $n$. Thus $\Sigma(A)$ contains elements of any finite order.

Furthermore, as it is not difficult to show that that the collection of subsets of $A$ whose cardinality is $n$ has itself cardinality equal to $|A|$, it is easy to argue that the subset of all cyclic permutations of $A$ of order $n$ has cardinality $|A|$. Since in your case $|\mathbb{R}|=2^{\aleph_0}$, you have uncountably many elements of any fixed order $n \geqslant 2$.

This is a claim that requires a bit of method in order to demonstrate, but it can be shown that the total number of elements of order $n \geqslant 2$ in $\Sigma(A)$ is actually equal to the order $|\Sigma(A)|=2^{|A|}$ of the full symmetric group on $A$.

Title: Group $G=\{f : \mathbb{R} \to \mathbb{R}\ |\ f \text{ is a bijective function}\}$ and elements of order $4$ and $9$

Body:
This question was asked in my abstract algebra quiz and I was unable to completely solve it.

Question : Let $G=\{f : \mathbb{R} \to \mathbb{R}\  |\ f \text{ is a bijective function}\}$ and $\circ$ be operation of composition of functions , then:

A) $G$ is abelian

B) $(G,\circ)$ has an element of order $9$.

C) there are uncountable elements of order $4$.

I proved A wrong . But I think that B can be right but there is no intution for C .

Can you please give a rigorious argument for B and C .

Thanks!!


问:
Is there a HSP-like theorem for algebras that can be axiomatized by a finite number of equations?
It is a famous classic theorem that classes of algebras which can be axiomatized by equations are precisely those which are closed under homomorphisms, subalgebras, and products. Is there a corresponding theorem for those classes of algebras which can be axiomatized by finitely many equations?</p>

答:
Here's a weak negative observation:

Since any infinite equational theory is the union of its finite subtheories, we dually have that the variety corresponding to an infinite equational theory is the intersection of the family of varieties corresponding to its finite subtheories. Basically, this implies that any 'simple closure property' holding of each of the varieties in the latter family will also hold a fortiori of the variety of the whole infinite equational theory. In particular, suppose $\mathbb{F}$ is any monotone map between classes of isomorphism types of algebras (such as the $\mathbb{H}$, $\mathbb{S}$, and $\mathbb{P}$ of the original HSP theorem). Then if every finitely axiomatizable variety is closed with respect to $\mathbb{F}$, every variety whatsoever is closed with respect to $\mathbb{F}$.

Of course this doesn't rule out more intricate conditions of higher quantifier complexity than simply "is closed with respect to [thing]," but at that point we are straying a bit from the "flavor" of the original HSP theorem.


问:
Independence intuition
Toss two fair dice. There are $36$ outcomes in the sample space $\Omega$, each with probability $\frac{1}{36}$. Let:

$A$ be the event '$4$ on first die'.

$B$ be the event 'sum of numbers is $7$'.

$C$ be the event 'sum of numbers is $8$'.

It says here $A$ and $B$ are independent. I don't understand why this is the case. What is the intuition behind this? Can someone offer an explanation to me that doesn't involve using the definition of $\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)$?

My understanding is informally, an event is independent if the occurrence of one does not affect the probability of the other and vise versa. So if $A$ occurs, wouldn't that affect the probability of $B$? Since if I were to roll a $4$ on the first die, the sample space will be reduced and hence the probability of 'sum of numbers is $7$' will be affected?

It also says $A$ and $C$ are not independent and $B$ and $C$ are not independent. Why?

I think this is because I'm confusing independence with conditional probability?

答:
Here's an analogous example from my previous post:

Pairwise independence can be characterised as:

Let $P(A)\neq0.$

Events $A$ and $B$ are independent iff. the knowledge that $A$ happens doesn't change $B$'s probability.

When considering whether events $X$ and $Y$ are independent, the correct intuition is to ask whether the knowledge that $X$ happens changes the probability of $Y$—not whether it restricts how $Y$ can happen.

Consider this experiment: flip two fair coins, letting $H_1$ be the
event that the first coin lands on Heads, and $X$ be the event that
the coins land on different sides. Then $$ P\left(H_1 \cap
X\right)=\frac14=P(H_1)\:P(X);$$ i.e., Events $H_1$ and $X$ are, by definition, independent. Despite this, the knowledge that $H_1$ happens has reduced the number of ways that $X$ can happen: one of $X$'s outcomes, $TH,$ is no longer possible.

Title: Independence intuition


问:
How to calculate $\int_{-\infty}^{+\infty}\Gamma(x+yi)\Gamma(x-yi) \, dy$?
I tried. Kilbas says that $\int_{-\infty}^{+\infty} \Gamma(x+yi) \Gamma(x-yi) \, dy=(2\pi)^{3/2}$$_2F_1(1/2,1/2,1/2;-1)$. In this case, to the function $_2F(a,b.c;z)$, we have $z=-1 \ (|z|=1)$ and a conditionally convergence if $-1<\Re(c-a-b)\leq 0$.

There is a better way to solve this integral? Thanks for helping!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
I'll use an Identity related to Gamma Function Integration:
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
\Gamma\pars{x + y\ic}\Gamma\pars{x - y\ic}\,\dd y}
\\[5mm] = &
2\pi\bracks{{1 \over 2\pi}\int_{-\infty}^{\infty}
\verts{\Gamma\pars{x + y\ic}}^{2}
\expo{\pars{2b - \pi}y}\,\dd y}
_{\ b\ =\ \color{red}{\pi/2}}
\\[5mm] = &\
2\pi\braces{\Gamma\pars{2x} \over \bracks{2\sin\pars{\color{red}{\pi/2}}}^{\,2x}} =
\bbx{2^{1 - 2x}\,\,\pi\,\Gamma\pars{2x}} \\ &
\end{align}
The above link enforces the conditions $\ds{x > 0}$ and
$\ds{b \in \pars{0,\pi}}$.


问:
Finding an affine cover of a certain surface in $\mathbb{A}_k^4$ (11.2.D vakil)
In 11.2.D in http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf, a surface $S\subset\mathbb{A}_k^4$ is defined as cut out by the equations $wz-xy=0$, $wy-x^2=0$, and $xz-y^2=0$. In computing the dimension of the surface, I feel like we need to find an affine open cover and then compute the dimension of each piece of the cover (so that 11.1.B can be applied). One can easily choose pieces $D(w)$, $D(x)$, $D(z)$, $D(y)$, but these seem not to cover the point $(x,y,z,w)$. For this remaining point, what affine open should be chosen? I feel like something like $D(x-1)$ could be used, but this feels a bit contrived.

Is there an easier affine to use, and is this even the right approach to the problem?

答:
Your proposed cover is fine: for varieties over fields, if $V$ a closed subset and $U$ an open subset of some ambient space, we have that $\dim V \leq \max(\dim V\cap U, \dim U^c)$. Why? Write $V$ as a finite union of irreducible components $V_i$. Then $\dim V = \max_i \dim V_i$, and either:

$V_i\cap U$ is a dense open subset of $V_i$ which has the same dimension

$V_i\subset U^c$, so $\dim V_i\leq \dim U^c$.

Combining the two, we have the stated result.

You could also prove directly that your variety is the image of $\Bbb A^2\to\Bbb A^4$ under the map $(s,t)\mapsto (s^3,s^2t,st^2,t^3)$. (Really, what's going on here is you're computing the dimension of the cone on the twisted cubic.)


问:
In $\Delta ABC$ has area $120$ with $AB = 50, AC= 10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively.

In $\Delta ABC$ has area $120$  with $AB = 50, AC=  10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively. The angle bisector of $\angle BAC$ intersects $DE$ and $BC$ at $F,G$ respectively. Find the area of quadrilateral $FDBG$.

What I Tried: Here is a picture :-

All the results which I got are in the picture with the appropriate use of Midpoint Theorem as well as Angle Bisector Theorem. Since we are given the area, I can use it to find $BC$ and hence find $x$ . But how am I going to find the area of the trapezium $BDFG$ ?  Also, finding $BC$ using Heron's formula will be a bit difficult algebra, I am wanting for some easy methods here.

Can anyone help? Thank You.

答:
$$[FDBG] = (1-\tfrac{1}{4})[ABG]= \dfrac{3}{4} \cdot \dfrac{5}{6} [ABC]   $$

Since $[ADF] = \tfrac{1}{4}[ABG]$. And $\triangle ABG$ has same height but base $5/6$ th of $\triangle ABC$

 So $$[FDBG] = 75$$


答:
Say $\triangle AFE = a$

Then $\triangle ADF = 5a$ (same height as $\triangle AFE$ but base is $5$ times).

$\triangle AGC = 4a$ (both base and height are double of $\triangle AFE$).

$\triangle ABG = 5 \times \triangle AGC = 20a$ ($\triangle AFE$ and $\triangle AGC$ share the same height but base ratio is $5:1$).

Adding them up $\triangle AGC + \triangle ABG = 4a + 20a = 120 \implies a = 5$.

So area of quadrilateral $= \triangle ABG - \triangle ADF = 20a - 5a = 75$.

Body:
In $\Delta ABC$ has area $120$  with $AB = 50, AC=  10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively. The angle bisector of $\angle BAC$ intersects $DE$ and $BC$ at $F,G$ respectively. Find the area of quadrilateral $FDBG$.

What I Tried: Here is a picture :-

All the results which I got are in the picture with the appropriate use of Midpoint Theorem as well as Angle Bisector Theorem. Since we are given the area, I can use it to find $BC$ and hence find $x$ . But how am I going to find the area of the trapezium $BDFG$ ?  Also, finding $BC$ using Heron's formula will be a bit difficult algebra, I am wanting for some easy methods here.

Can anyone help? Thank You.


问:
Formal logic statements symbolic representation.
If the horse is fresh, then the knight will win. $H -> K$
A fresh horse is a necessary condition for the knight to win. $K -> H$

I think the first sentence should have same notation. I don't understand why are different.

答: $p \implies q$ is not the same as $q \implies p$. $q \implies p$ is the converse of $p\implies q$. In other words they are not logically equivalent. Make a truth table and you’ll see why.


答:
We have
$$
\begin{aligned}{}
[FDBC]
&=[ABG]-[ADF]=[ABG]-\frac 14[ABG]=\frac 34[ABG]\\
&=\frac 34\cdot\frac {BG}{BC}[ABC]=
\frac 34\cdot\frac {AB}{AB+AC}[ABC]
\\
&=\frac 34\cdot\frac{50}{50+10}\cdot 120
\ .
\end{aligned}
$$

Title: In $\Delta ABC$ has area $120$ with $AB = 50, AC= 10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively.


问:
Assume $G$ is an Abelian group containing an element of order greater than $2$, then provide a nonidentity automorphism in $G$.
Assume $G$ is an Abelian group containing an element of order greater than $2$, then provide a nonidentity automorphism in $G$.

Let $\Psi:G \to G$ be a mapping given by $g \mapsto g^{\text{ord}(g_1)}$ and let  $g_1  \in G$ be the element with order greater than $2$ then for $g,h \in G$
$$\Psi(gh)=(gh)^{\text{ord}(g_1)}=g^{\text{ord}(g_1)}h^{\text{ord}(g_1)}=\Psi(g)\Psi(h)$$

And this is indeed true since $G$ is Abelian.

Now we need to show that the kernel of the mapping contains only the identity element of $G$ denoted by $e$ but I think the mapping is not the one that is useful since $\Psi(g_1)=g_1^{\text{ord}(g_1)}=e$ which follows that $g_1 \in \text{ker}(\Psi)$ and $g_1$ is not necessarily identical to $e$ and hence the mapping is not necessarily an injection.

So what mapping will be useful and how can I finish the answer?

答:
The mapping $\phi: g \mapsto g^{-1}$ is the desired automorphism. It is not equal to identity because $\phi(g_1)\not= g_1$.

Note that in the non-Abelian case, $\phi$ is not a morphism in $G$. It is a isomorphism from $G$ to its opposite group in this case.


问:
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$
I'm trying to solve this recurrence relation:

$$
a_n = \begin{cases}
    0 & \mbox{for } n = 0 \\
    5 & \mbox{for } n = 1 \\
    6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
    \end{cases}
$$

I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$

(I'm not sure if that's right)

and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)

I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$

but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)

Body:
$$a_0 =0,a_1=5,a_n=6a_{n-1} - 5a_{n-2} + 1, n > 1$$

$$f(x)=\sum_{n=0}^{\infty}a_nx^n=5x+\sum_{n=2}^{\infty}a_nx^n=$$

$$=5x+\sum_{n=2}^{\infty}(6a_{n-1} - 5a_{n-2} + 1)x^n=$$

$$=5x+6\sum_{n=2}^{\infty}a_{n-1}x^n-5\sum_{n=2}^{\infty}a_{n-2}x^n+\sum_{n=2}^{\infty}x^n=$$

$$=5x+6x\sum_{n=2}^{\infty}a_{n-1}x^{n-1}-5x^2\sum_{n=2}^{\infty}a_{n-2}x^{n-2}+\frac{x^2}{1-x}=$$

$$=5x+6xf(x)-5x^2f(x)+\frac{x^2}{1-x}$$ from above follow that g.f. is

$$f(x)=\frac{5x+\frac{x^2}{1-x}}{5x^2-6x+1}=\frac{5x-4x^2}{(1-x)(5x^2-5x-(x-1))}=$$
$$=\frac{x(5-4x)}{(1-x)(5x(x-1)-(x-1))}=\frac{x(5-4x)}{(1-x)(x-1)(5x-1)}$$


问:
Finding the independent term of an expansion
The question and the answer are provided below; however, I have no idea what strategy was implemented to achieve that answer. I would like to know what is happening to solve this question and what each of the lines serves to do. In other words, Just walk me through it, thanks!

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\bracks{x^{0}}
\pars{{1 \over x^{2}} - 3x}^{3}
\pars{{1 \over 2x} + x^{2}}^{3}}
\\[5mm] = &\
\bracks{x^{0}}
{1 \over x^{6}}\pars{1 - 3x^{3}}^{3}
{1 \over 8x^{3}}\pars{1 + 2x^{3}}^{3}
\\[5mm] = &\
{1 \over 8}\bracks{x^{9}}\pars{1 - 3x^{3}}^{3}
\pars{1 + 2x^{3}}^{3}
\\[5mm] = &\
{1 \over 8}\bracks{x^{9}}
\pars{1 - 9x^{3} + 27x^{6} - 27x^{9}}
\pars{1 + 2x^{3}}^{3}
\\[5mm] = &\
{1 \over 8}\
\overbrace{\bracks{x^{9}}\pars{1 + 2x^{3}}^{3}}
^{\ds{2^{3}{3 \choose 3}}}\ -\
{9 \over 8}\
\overbrace{\bracks{x^{6}}\pars{1 + 2x^{3}}^{3}}
^{\ds{2^{2}{3 \choose 2}}}
\\[2mm] - &\
{27 \over 8}
\underbrace{\bracks{x^{3}}\pars{1 + 2x^{3}}^{3}}
_{\ds{2^{1}{3 \choose 1}}}\ -\
{27 \over 8}\
\underbrace{\bracks{x^{0}}\pars{1 + 2x^{3}}^{3}}
_{\ds{2^{0}{3 \choose 0}}}
\\[5mm] = &\
1 - {27 \over 2} + {81 \over 4} -{27 \over 8} =
\bbx{35 \over 8} \\ &
\end{align}


问:
Polygon inscribed in a circle

There is a picture of an inscribed n-side polygon in a circle above. I have a task as follow:

If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve this for n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)

Everything what comes to my mind is θ = 2π/n, but I'm pretty sure, that is not correct answer. Could you help me to deal with it please?

答:
Given that the inscribed polygon is regular, we can say that the angle subtended between any 2 line segments (originating from adjacent corners) at the center of the circle will be equal and since there will be $n$ unique angles totaling up to $2π$, each of them will be equal to $\frac{2π}{n}$.

Your answer is correct and  $θ=\frac{2π}{n}$.

Title: Polygon inscribed in a circle


问:
Defining a group by means of translations
Let $R$ be a non empty set and $T$ be a subset of functions $R\to R$ (we call them translations) which satisfies the following:

the identity map $t(x)=x$ is in $T$

given $x,y\in R$ there exists one and only one function $t_{xy}\in T$ such that $t_{xy}(x)=y$.

if $t,s\in T$ then $t\circ s\in T$.

Now fix any element $0\in R$ and define a addition operation as
$$
 x + y = t_{0y}(x)
$$
where $t_{0y}$ is the only map in $T$ such that $t_{0y}(0)=y$.

I'm able to prove that $(R,0,+)$ is a group. Is it commutative?

答:
I don’t think so.

Let $G$ be an nonabelian group and denote the group operation by $\bullet$. The set of left-multiplications $T=\{L_g:G\rightarrow G, h \mapsto g\bullet h \mid g \in G\}$ is a set of translations.

Indeed $\operatorname{id} = L_e \in T$, $L_g,L_h \in T \implies L_g\circ L_h= L_{g\bullet h} \in T$ and given $x,y\in G$ we find that $L_{y\bullet x^{-1}}$ defines a (unique by evaluating at $e$) translation.

Now for $x,y\in G$ we find that $x+y = L_y(x) = y\bullet x$ and $y+x = L_x(y) = x \bullet y$ won’t coincide in general, as we chose $G$ to be nonabelian.


问:
Linear equations including semi-positive definite matrix and vectors of standard bases
Suppose $A \in \mathbb{R}^{n \times n}$ is a positive semi-definite matrix. Can we solve the following for $x \in \mathbb{R}^n$?

\begin{equation}
\begin{split}
A(x-a) + \lambda_1 e_i+\lambda_2e_j&=0\\
e_ix&=0,\,\,\,\text{or }(x_i=0)\\
e_jx&=0,\,\,\,\text{or }(x_j=0)
\end{split}
\end{equation}
where $\lambda_1, \lambda_2 \in \mathbb{R}$, $a \in \mathbb{R}^n$, $e_i$ is the i-th column of the identity matrix, and $i,j$ are fixed and given indices as well as $a$.
I have solved for the case where $e_i, e_j \in R(A)$ but cannot solve it when either one of $e_i$ or $e_j $ is not in $R(A)$ or both are not in range of $A$.

Proof for the case when $e_i, e_j \in R(A)$:

Since $e_i, e_j \in R(A)$, there exist $y_i, y_j$ such that $Ay_i=e_i$ and $Ay_j=e_j$. Let $b$ be in null space of A. Then, one can check the following is a solution:

$$
x=a -\lambda_1(y_i+b)-\lambda_2(y_j+b)
$$

where $\lambda_1, \lambda_2$ are obtained form the following:

$$
\begin{bmatrix}
e_{i_1}^T\\
e_{i_2}^T
\end{bmatrix}
\begin{bmatrix}
y_1+b & y_2+b
\end{bmatrix}
\begin{bmatrix}
\lambda_1\\
\lambda_2
\end{bmatrix}
=
\begin{bmatrix}
a_{i}\\
a_j
\end{bmatrix}
$$

How can one solve for the following cases:

$e_i\in R(A)$ and $e_j\notin R(A)$ (or $e_i\notin R(A)$ and $e_j\in R(A)$)

$e_i, e_j\notin R(A)$

答:
There are three cases depending on how large is the intersection $M:=R(A)\cap\mathrm{Span}(e_i,e_j)$.

Case 1. $\dim M=2$, i.e., $e_i,e_j\in R(A)$. This is covered by the asker.

Case 2. $\dim M=1$, i.e., there are $\alpha_i,\alpha_j\in\mathbb{R}$ and $y\in\mathbb{R}^n$, such that $$Ay=\alpha_ie_i+\alpha_je_j$$ Any other vector in $M$ is a multiple of this since $M$ is one-dimensional. Then the first condition is equivalent to $$\exists\lambda\in\mathbb{R},\quad A(x-a)=-\lambda Ay$$ $$\therefore\quad\exists b\in N(A),\quad x=a-\lambda y+b$$ Thus the required solution lies in $a+N(A)+\mathrm{Span}(y)$. The other two orthogonality conditions then imply $x=a-\lambda y+b\in e_i^\perp\cap e_j^\perp$. This is the intersection of a displaced subspace $a+N(A)+\mathrm{Span}(y)$ and an $(n-2)$-dimensional subspace $\mathrm{Span}(e_i,e_j)^\perp$. This intersection is empty if they are 'parallel', i.e. $$N(A)+\mathrm{Span}(y)\subseteq\mathrm{Span}(e_i,e_j)^\perp$$ $$\iff e_i,e_j\in(N(A)+\mathrm{Span}(y))^\perp$$ $$\iff e_i^TAz=0, e_i^Ty=0, e_j^TAz=0, e_j^Ty=0, \forall z$$ If this is not the case then there exist $b\in N(A)$, $\lambda\in\mathbb{R}$ such that $$0=e_i^Tx=e_i^T(a+b-\lambda y)$$ $$0=e_j^Tx=e_j^T(a+b-\lambda y)$$ These imply $\phi(a+b)=0$, where $\phi:=(e_j\cdot y)e_i^T-(e_i\cdot y)e_j^T$. So take any $$b\in -a+N(\phi),\qquad \lambda=e_i\cdot(a+b)/e_i\cdot y$$

Case 3. $\dim M=0$, i.e., no linear combination of $e_i,e_j$ is in $R(A)$ except $0$. So $\lambda_1=0=\lambda_2$ and $$A(x-a)=0\implies x=a+b,\quad \exists b\in N(A)$$ Again, similarly to case 2., the orthogonality conditions require an intersection of $a+N(A)$ with $\mathrm{Span}(e_i,e_j)^\perp$. This is possible iff $N(A)$ is not 'parallel' to $e_i^\perp,e_j^\perp$, that is, $$e_i,e_j\notin N(A)^\perp$$ To get $b$ take any solution of $$\begin{bmatrix}e_i^T\\e_j^T\end{bmatrix}b=-\begin{bmatrix}e_i\cdot a\\e_j\cdot a\end{bmatrix}$$


问:
Lebesgue integral on a limited interval
Let $(R,B(R),m)$ a Lebesgue measurable space on $R$,

$\forall n\in N$ denote: $f_n(x)=arctan(n|x|)$.

Show: $lim_{n\to \infty} \int_{[-1,1]} f_n dm =π$

My trial:

Notice that $f_n$ is a positive inceasing sequence for all $x\neq 0$ and $n\in N$.
So by the convergence monotone theorem of Lebesgue, we get:
Denote: $f(x)=lim_{n\to \infty} f_n = π/2$ so,

$lim_{n\to \infty} \int_{[-1,1]} f_n dm= \int_{[-1,1]} f dm= \int_{[-1,1]} π/2= π$

Is that right?

答:
The inequality $\arctan (n|x|) \leq \tan (n|x|)$ is not valid.

The result follows immediately from Dominated Convergence Theorem since $\arctan (n|x|) \to \frac  {\pi} 2$ for all $x \neq 0$ and $\arctan $ is a bounded function.

Title: Lebesgue integral on a limited interval


问:
The last 2 digits of $7^{7^{7^7}}$
What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows  $...43$.</p>

答:
Let's check the pattern of the last two digits of powers of $7$.
$$
7^1\to07\\
7^2\to49\\
7^3\to43\\
7^4\to01\\
7^5\to07
$$
and it loops. Thus we need only find out what that exponent is modulo $4$. Can you proceed from here?</p>

Title: The last 2 digits of $7^{7^{7^7}}$


问:
How do I prove that $\Vert f \Vert=\vert f(0)\vert+\int_{0}^1 \vert f'(x)\vert\,dx$ defines a norm on $X=C^1[0, 1]$?
I don't understand how should I go about checking the first condition ($\Vert f\Vert=0$ iff $f=0$) for a norm.

答:
$$\|f\|=0\iff|f(0)|+\int_0^1|f'(x)|\,\mathrm d x=0\iff |f(0)|=0\quad \text{and}\quad \int_0^1|f'|=0.$$

Since $f'$ is continuous on $[0,1]$, $$\int_0^1|f'|=0\implies f'(x)=0,$$
for all $x\in [0,1]$. Therefore $f$ is constant on $[0,1]$. Since $f(0)=0$, we have that $f(x)=0$ for all $x\in [0,1]$.


问:
The sum of digits in a 2-digit number

The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number.

My attempt:

Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at unit's place. According to question:

$$x+y=8$$

I could not figure out the other condition. Please help. Thanks in advance.

答:
Solving by observation and little mathematics;

let the number be
$$\overline{xy}:x,y\in [1,9]  \text{ and } x,y \in \text{Integers}$$

Given, $x+y=8$ and $\overline{xy} +9=\overline{zz}$

We know that,  $\overline{zz} \in \{11,22,33,44,55,66,77,88,99 \} $

So, $(\overline{zz}-9=\overline{xy})\in \{11-9,22-9,33-9,44-9,55-9,66-9,77-9,88-9,99-9 \} $

or, $(\overline{zz}-9=\overline{xy})\in \{13,24,35,55-9,66-9,77-9,88-9,99-9 \} $

We can see our answer.

Title: The sum of digits in a 2-digit number


问:
Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $
Prove that for $a,b\in [-1,1]$:

$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$

Body:
By Cauchy-Schwarz

$$(a\sqrt{1-b^2}+b\sqrt{1-a^2})^2 \leq (a^2+b^2)(1-b^2+1-a^2)=2(a^2+b^2)-(a^2+b^2)^2$$
$$=1-(1-a^2-b^2)^2\leq 1$$


问:
If $A^3$ diagonalizable, then $A$ diagonalizable
I can prove this statement wrong by considering $A\neq\mathbf O$ and $A$ not diagonalizable while $A^3=\mathbf O$.

But what's wrong with my intuitive proof:

$A^3=PDP^{-1}$, and we just consider $A=PD^{1/3}P^{-1}$, it seems that everything is satisfied. And furthermore, we can also show uniqueness by considering in certain eigenspace $V_\lambda$, $A^3=\lambda I$ so the eigenvalue of $A$ must be $\sqrt[3]\lambda$.

Could anyone help me to figure out what's wrong here? And why does the converse hold?

答:
we can also show uniqueness

We cannot, because given a matrix $B$, there may be several matrices such that $A^3=B$. So you finding the one which is diagonalizable simultaneously with $B$ does not mean much.


问:
Is it correct to conclude about the values of limits using comparison test?
I have two infinite sums $\sum_{i=0}^{\infty}a_i$ and $\sum_{i=0}^{\infty}b_i$ such that $0 \le a_i \le b_i, \forall i$ ($a_i,b_i$ are real). And the sum $\sum_{i=0}^{\infty}b_i$ converges, ie. $\lim_{n \to \infty}\sum_{i=0}^{n}b_i = L$ for some finite $L$.

Then by using comparison test I can say that $\sum_{i=0}^{\infty}a_i$ also converges. Let it converge to a value $M$.

Then would it be correct to say $M \le L$ ?

I can see that from the partial sums
$$
A_n = \sum_{i=0}^{n}a_i \\
B_n = \sum_{i=0}^{n}b_i \\
A_n \le B_n, \forall n
$$

and thus $M \le L$ should hold. Is this reasoning valid ? It might be a trivial question but I wanted to be sure.

答: Of course you can say that. At the end of the day, infinite sums are just limits of partial sums. And we have that for limits $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$, if $a_n < b_n$, then $a <b$.


问:
Distance between two circles with known size and intersection area
Caution - biologist at work

I am trying to plot some circles and want to work out how far apart my circles should be with a target in mind. The ditance between the centre points of the two circles is $d$. The first circle has a radius of $r_1$ = 5 cm and the second has a radius of $r_2$ = 4 cm. The area of these circles is 78.54 cm$^2$ and 50.27 cm$^2$ respectively, making the joint area $a_{1,2}$ = 128.81 cm$^2$. I want 30% of that area, $0.3 \times 128.81$ cm$^2$ to be intersection area (area shared by the two circles). Therefore I have a known pair of radii and a known intersection area, and an unknown distance. By trial and error with various values of $d$ in the equation

$$ a_{1,2} = r_1^2 cos^{-1} \frac{d^2 + r_1^2 - r_2^2}{2dr_1} + r_2^2 cos^{-1} \frac{d^2 + r_2^2 - r_1^2}{2dr_2} - \frac{1}{2} \sqrt{(-d+r_1+r_2)(d+r_1-r_2)(d-r_1+r_2)(d+r_1+r_2)} $$

I find that, to get 30% of the total area into the intersection area, when the radii are 5 and 4, $d$ needs to be approximately 2.71. However, is there a way to solve $d$ from known radii and intersection area? I've been searching for hours and come up short - the internet is dominated by solutions to find the intersection are from known radii and distance.

FYI, I'm programming this in R - bonus kudos for answers providing accompanying script!

```<code>r1 <- 5
r2 <- 4
a1 <- pi*r1^2
a2 <- pi*r2^2
aJ <- 0.3 * (a1+a2)

intersectionArea <- function(r1, r2, d){
  (r1^2 * acos((d^2 + r1^2 - r2^2)/(2*d*r1))) +
  (r2^2 * acos((d^2 + r2^2 - r1^2)/(2*d*r2))) -
  (0.5 * sqrt((-d+r1+r2)*(d+r1-r2)*(d-r1+r2)*(d+r1+r2)))
}

aJ/intersectionArea(r1,r2, 2.709)
</code>```

答:
Here is the Python script that effectively solves your problem (sorry, I don't do R). It's simple enough than it can be translated into any other programming language. The solution is obtained iteratively but it converges quite quickly (range of possible solutions is cut by half in each iteration).

```<code>import math

def area(r1, r2, d):
    s1 = r1**2 * math.acos((d**2 - r2**2 + r1**2)/(2 * d * r1))
    s2 = r2**2 * math.acos((d**2 - r1**2 + r2**2)/(2 * d * r2))
    s3 = math.sqrt((-d + r1 + r2) * (d + r1 - r2) * (d + r2 - r1) * (d + r1 + r2)) / 2
    return s1 + s2 - s3

def solve(r1, r2, targetArea, error):
    d1 = r1 + r2
    a1 = area(r1, r2, d1)
    d2 = abs(r1 - r2)
    a2 = area(r1, r2, d2)
    while True:
        d = (d1 + d2) / 2
        a = area(r1, r2, d)
        if abs(a - targetArea) <= error:
            return d
        if a > targetArea:
            d2 = d
            a2 = a
        else:
            d1 = d
            a1 = a

# input data
r1 = 4
r2 = 5
a = 0.3 * 128.81
error = 0.00001

# find d for the given input
d = solve(r1, r2, a, error)
print("The result is d=%.5f" % d)

# calculate area for the calculated value of d
a2 = area(r1, r2, d)

# print for verification
print("For r1=%.5f, r2=%.5f, d=%.5f the area is %.5f and the target area is %.5f"
      % (r1, r2, d, a2, a))
</code>```

In the hart of this script is the following function:

```<code>def solve(r1, r2, targetArea, error)
</code>```

Simply, it takes 4 parameters: two radii, intersection area and the accuracy that you want to achieve (0.00001 or 0.01 or any other small value).

The output of the script is:

```<code>The result is d=2.70874
For r1=4.00000, r2=5.00000, d=2.70874 the area is 38.64300 and the target area is 38.64300
</code>```

Body:
Here is the Python script that effectively solves your problem (sorry, I don't do R). It's simple enough than it can be translated into any other programming language. The solution is obtained iteratively but it converges quite quickly (range of possible solutions is cut by half in each iteration).

import math

def area(r1, r2, d):
    s1 = r1**2 * math.acos((d**2 - r2**2 + r1**2)/(2 * d * r1))
    s2 = r2**2 * math.acos((d**2 - r1**2 + r2**2)/(2 * d * r2))
    s3 = math.sqrt((-d + r1 + r2) * (d + r1 - r2) * (d + r2 - r1) * (d + r1 + r2)) / 2
    return s1 + s2 - s3

def solve(r1, r2, targetArea, error):
    d1 = r1 + r2
    a1 = area(r1, r2, d1)
    d2 = abs(r1 - r2)
    a2 = area(r1, r2, d2)
    while True:
        d = (d1 + d2) / 2
        a = area(r1, r2, d)
        if abs(a - targetArea) <= error:
            return d
        if a > targetArea:
            d2 = d
            a2 = a
        else:
            d1 = d
            a1 = a

# input data
r1 = 4
r2 = 5
a = 0.3 * 128.81
error = 0.00001

# find d for the given input
d = solve(r1, r2, a, error)
print("The result is d=%.5f" % d)

# calculate area for the calculated value of d
a2 = area(r1, r2, d)

# print for verification
print("For r1=%.5f, r2=%.5f, d=%.5f the area is %.5f and the target area is %.5f"
      % (r1, r2, d, a2, a))

In the hart of this script is the following function:

def solve(r1, r2, targetArea, error)

Simply, it takes 4 parameters: two radii, intersection area and the accuracy that you want to achieve (0.00001 or 0.01 or any other small value).

The output of the script is:

The result is d=2.70874
For r1=4.00000, r2=5.00000, d=2.70874 the area is 38.64300 and the target area is 38.64300


问:
Notation minimum of a column vector
I'd like to know the notation to express the minimum of a column vector.

Is this notation correct?

\begin{equation}
\min
\left[\matrix{
   \left|b_{n}-b_{n+1}\right| \cr
   \left|\left(b_{n}+360\right)-b_{n+1}\right| \cr
   \left|b_{n}-\left(b_{n+1}+360\right)\right| \cr
}\right]
\end{equation}

If not, how would i have to write it down?

答:
You shouldn't need it to be a column vector unless you refer to it elsewhere as being a column vector, in which case I'd give it a name $\mathbf v$ and write $\min_i |v_i|$ or whatever.

If this is the only place it is used, just use the normal $\min\{\cdot,\cdot,\cdot\}$.

It's all personal taste really, I just instinctively feel the above are more natural.

Title: Notation minimum of a column vector


问:
What's the English term for this algebra operation?
I didn't study maths in English so apologies for this trivial question.

How do you say in English when you have a formula 3n + 3 and you want someone to convert it into 3(n+1)
I only know about "rearrange" but that's usually about equality, like U=I*R can be rearranged to U/I = R etc.
Isn't there a more explicit word for this?

For example if I expect 4a - 4b to be the final result from 2a - 4b + 2a and someone also does this 4(a - b), what would I say?

Replace the end of sentences (those are my guesses):

"I want you to just add polynomials, you didn't have to <segregate the 4>?"

"I want you to just add polynomials, don't worry about <inverse of redistribution>?"

答:
The operation $$ab + ac = a(b+c)$$ is often called " factoring out the common factor ", where the common factor is $a$ here.

" I wanted you to add polynomials, you didn't have to factor out 4 "

" I wanted you to add polynomials, don't worry about factoring out the common factor "

The operation $$a(b+c) = ab + ac$$
is often called " expanding the brackets ".

" If your answer is already factorized, do not expand the brackets "


问:
Find an equation for the plane through the point which is perpendicular to the vector.
The point in which the equation must go through is =(−1,−2,6)
The vector is from the origin to point P. How would I do this?</p>

答:
The desired plane passes through the point $P(-1, -2, 6)$ and has a normal vector $OP=(-1, -2, 6)$, where $O$ is the origin.
Let $A(x, y, z)$ be an arbitrary point on the plane. Then any vector $PA = (x+1, y+2, z-6)$ on the plane must be perpendicular to the normal vector $OP=(-1, -2, 6)$. So inner product of $PA$ and $OP$ must be zero.
$$
<PA, OP> = -1.(x+1) + [-2.(y+2)] + 6.(z-6) = -x -2y +6z -41=0
$$
Hence the equation of the plan is given by
$$
-x -2y +6z = 41
$$


问:
Number of arrangements possible
There are $m$ persons and $n$ chairs. Each person needs to maintain social distance between themselves and the person they are sitting next to. Therefore, once a person sits, the person who sits next to them sits at a distance that is in multiples of the integer $k$. We have to find the total number of arrangements possible so that all the people can be seated.

Notes:

1.All the people must get a chair to sit on

2.Two arrangements are different either if the state of a chair is different in both cases or if two different people are seated on it. For example, for chairs $n=10$, person $m=2$, and $k=2$, if the first person sits at the 2nd position, then the second person can sit only at the 4th, 6th, 8th, or 10th position.

For  example, $n=5,m=2,k=2$
no of arrangements possible are 4. (1,3)  (1,5)  (2,4)  (3,5).

Can anyone suggest some approach to me for this combinatorics problem?

答:
OK as per your clarification, people are seated in a row. Based on your question and your example, the next person can sit on the k$th$ chair i.e there are at least $(k-1)$ chairs between any two neighbors on the table.

There are $(m-1)$ places between people so the first condition is $n \geq (m-1) (k-1) + m$.

Now to solve this, here is an approach: we have $m!$ ways to sit $m$ people in a row. The next step is equivalent of finding ways to place remaining $(n - m)$ chairs between $m$ people or at two ends such that there are at least $k-1$ chairs between any two neighbors. This can be found using stars and bars method. Place $(k-1)$ chairs each in $(m-1)$ places between people first (there is only one way to do that) and then the rest $(n - mk + k - 1)$ chairs in $(m + 1)$ places. $[(m - 1$ places between them plus two at the ends$]$.

So altogether it comes to $m! \times {n+m+k-mk-1 \choose m}$ arrangements.

Title: Number of arrangements possible


Title: The last 2 digits of $7^{7^{7^7}}$

Body: What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows  $...43$.</p>

Body:
$7^2=49=50-1$

$\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod {100}$

Alternatively, $100=4\cdot25, 7^2\equiv1\pmod 4$

and $7^2\equiv-1\pmod{25}\implies 7^4\equiv(-1)^2\pmod{25}\equiv1$

$\implies 7^{\text{lcm}(2,4)}\equiv1\pmod {4\cdot25}$ as $(25,4)=1\implies 7^4\equiv1\pmod {100}$

So, we need to determine $7^{7^7}\pmod 4$

As $7^7$ is odd, and $7\equiv-1\pmod4\implies 7^{7^7}\equiv-1\pmod4\equiv3$

$\implies 7^{7^7}=4n+3$ for some integer $n>0$

So, $7^{7^{7^7}}=7^{4n+3}\equiv (7^4)^n\cdot7^3\pmod{100}\equiv1^n\cdot343$


问:
Nice limit: $\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$
Find

$$L=\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

Not much experienced with limits of this kind, I have tried to reduce this limit to a form without exponent:

$$\ln L=\lim_{x \to 0}\ln\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$

$$\ln L=\lim_{x \to 0}\frac{\ln\arcsin x - \ln x}{x^2}$$

...and then I tried to apply L'Hospital to numerator and denominator. However, the solution becomes a complete mess and you can repeat derivation as many times as you want without ever reaching a conclusion.

I also tried to introduce substitution $x=\sin y$ which leads to:

$$L=\lim_{y \to 0}\left(\frac{y}{\sin y}\right)^{\frac1{\sin^2 y}}$$

This makes manipulation slightly easier but the same approach (L'Hospital after log) is still unable to generate a solution.

答:
Use Taylor series (one piece at the time)
$$y=\left(\frac{\sin ^{-1}(x)}{x}\right)^{\frac{1}{x^2}}\implies \log(y)=\frac{1}{x^2}\log\left(\frac{\sin ^{-1}(x)}{x}\right)$$
$$\frac{\sin ^{-1}(x)}{x}=1+\frac{x^2}{6}+\frac{3 x^4}{40}+O\left(x^6\right)$$
$$\log\left(\frac{\sin ^{-1}(x)}{x}\right)=\frac{x^2}{6}+\frac{11 x^4}{180}+O\left(x^6\right)$$
$$\log(y)=\frac{1}{6}+\frac{11 x^2}{180}+O\left(x^4\right)$$
$$y=e^{\log(y)}=\sqrt[6]{e}+\frac{11}{180} \sqrt[6]{e} x^2+O\left(x^4\right)$$ which shows the limit and how it is approached.


问:
why there's not a fourier transform to the function $x(t) = -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$
I know that the Laplace transform of a signal is

$X(s)= \frac{1}{s^2+6s+8}$

$ ROC =  $ {s|-4 < Re{s} < -2}

therefore $x(t) =  -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

$u(t)$ is the heaviside function.

I need to find its Fourier transform (or to prove why there is not such transform).

by wolfram alpha I see that the Fourier transform doesn't exist for this signal.
Someone here can explain why ?

Thank you :)

答: to check if a fourier transform exists from laplace transform you need to have the line $jw$ inside the ROC (equivalently $0 \in ROC$) which is clearly not the case in your question.


问:
If $\lim A_n$ exists, then $P(\lim A_n) = \lim P(A_n)$
Given a sequence $\{A_n\}$ of events with $\lim A_n = A$, I would like to show that $P(\lim A_n) = \lim P(A_n)$.

$\lim A_n = A$ implies that $\limsup A_n = A$. Thus $\bigcap \bigcup_{k=n}^{\infty}A_k=A$

Let $B_n=\bigcup_{k=n}^{\infty}A_k$ for every $n$. Then $B_n$ is non increasing.
We have:

$P(\lim A_n) = P(A) = P(\bigcap \bigcup_{k=n}^{\infty}A_k)=P(\bigcap B_n) = \lim P(B_n)$ since $B_n$ non-increasing.
Thus $P(\lim A_n) = \lim P(\bigcup_{k=n}^{\infty}A_k))$.

How do I go from here to $\lim P(A_n)$?

Any hep would be greatly appreciated.

答:
TL;DR The other answers seem to reinvent the wheel. Just use Fatou's Lemma

$$P(\liminf A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\limsup A_n)$$

Note to justify: I really mean that I expect that at this point that the student knows the inequalities above collectively known Fatou's Lemma (and reverse Fatou's Lemma and the regular $\limsup$ and $\liminf$ of real numbers in calculus). Maybe the student also knows integration version of Fatou's Lemma in measure theory, but I don't expect the student to know Fatou's Lemma in this sense (and thus of course in the expectation sense. Consequently further, I don't expect the student to use or even know $E[1_A]=P(A)$).

Edit: For further justification, see below.

Actually, what we first have to prove is that $\lim P(A_n)$ exists, i.e. $\liminf P(A_n) = \limsup P(A_n)$ and then we define this common value as $\lim P(A_n)$. Then

Fatou's lemma says

$$P(\liminf A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\limsup A_n)$$

Now, saying '$\lim A_n$ exists' means '$\liminf A_n = \limsup A_n$' and then we define this common value as $\lim A_n$.

Then

$$P(\lim A_n) \le \liminf P(A_n) \le \limsup P(A_n) \le P(\lim A_n)$$

Hence, $\liminf P(A_n)  = \limsup P(A_n)$, i.e. $\lim P(A_n)$ exists

Finally, $P(\lim A_n) = \liminf P(A_n)  = \limsup P(A_n) =: \lim P(A_n)$

Edit: For the Note above and the comment of Gabriel Romon:

See for yourself that in David Williams' Probability with Martingales, you can prove Fatou's Lemma without any knowledge of measure-theoretic (Lebesgue) integration.