问:
How often does it happen that the oldest person alive dies?
Today, we are brought the sad news that Europe's oldest woman died. A little over a week ago the oldest person in the U.S. unfortunately died. Yesterday, the Netherlands' oldest man died peacefully. The Gerontology Research Group keeps records: Guinness World Records.

If you live in a country with $N_{\text{country}}$ people, a continent with $N_{\text{continent}}$ people, and a world with $N_{\text{world}}$ people, during a year and on average, how often will you be notified (if you're paying attention to your quality tabloid) of the death of the oldest man/woman/person alive of your country/continent/world? (Note that a death will result in at most one notification.)

Edit, Suggestions (due to comments; thanks!) only:

Ultimately, I am looking for a realistic formula. That means that life tables are certainly allowed (but mind the ending, and note that some of the oldest people are older than the maximum in the tables).

I guess that the Gompertz–Makeham law of mortality, or any plausible model of late-life mortality, is also fair game.

However, these suggestions do not preclude some insight or realistic assumption that might not need any of those things.

Further potentially useful assumptions (with regard to a subpopulation) may include (if reasonable w.r.t. the question):

Time is discrete.

The number of births at any time is constant.

The number of deaths at any time is constant.

The number of births is equal to the number of deaths at any time.

Time is defined in such a way that, at any time, one birth and one death occur.

In the limit to infinite age, the probability of a person of such age dying goes to $1$.

At any time, the oldest person(s) alive is (are) the most likely to die.

At any time, an older person(s) alive is (are) more likely to die than any younger person alive.

(Although such assumptions aren't necessarily realistic, I don't immediately see how they would distort the outcome.)

答: Super-simpler answer: "How often will the oldest living person die?" is just a paraphrase of "How long, typically, will the oldest person (on earth/in country/in any given population) live?" According to one source (http://life-span.findthedata.org/q/101/166/What-is-the-life-expectancy-of-a-100-year-old-man), life expectancy for a man is 2.07 years, for a woman 2.37 years, which is consistent with the 50%-per-year mortality rate mentioned earlier. It seems reasonable to assume that the oldest living person is "very old" in this sense. And since one person will stay the oldest living person until they die, questions of population size are irrelevant. Naturally, there will be variation from population to population, but since this is a probabalistic question anyway, "two and a quarter years" seems a reasonable answer, yes? </p>


问:
Understanding Fourier Transform Theorem in Folland: Theorem 8.35
While reading Folland's Real Analysis textbook, I came across the Fourier Analysis section and became troubled by Theorem 8.35.  It goes as follows:

Focusing on part $a.$ of the theorem, I am confused about the sudden change of supposition where we originally considered $f \in L^1 + L^2$, but then in $a.$ suddenly consider $f \in L^p \, (1 \leq p < \infty)$.  In particular, I have not seen any mention of the Fourier transform if $f \in L^p$ for $p > 2$ (could we use Schwartz functions?).  Furthermore as he begins the proof he lets $f = f_1 + f_2$ where $f_1 \in L^1$ and $f_2 \in L^2$, which to me signifies that he is still using $f \in L^1 + L^2$.  The proof of $a.$ makes sense to me, but only for $f \in L^1 + L^2$, how are we suddenly also considering $f \in L^p \, (1 \leq p < \infty)$?

答: If $f=f_1+f_2$ and $f_j\in L^j,$ then we know that $\hat{f_1}\in L^\infty$ and $\hat{f_2}\in L^2$ by the mapping properties of the Fourier transform. In the second sentence of the statement of the theorem, they assume that $f\in L^1+L^2,$ so such a decomposition is perfectly kosher. They assumed that $f\in L^p$ in part (a) to use theorem 8.14.

ContentLicense: CC BY-SA 4.0


问:
How many regions can be selected in a chess board?
How many regions can be selected in a 8 by 8 chess board?

Definition of region: A region is a set of cells that are all connected together(by edge). i.e. a possible region:

I want to run(or do complexity analysis for) a backtrack algorithm, and need to know the number of possible regions.

答: The value you are looking for is $51016818604894742$ and it is part of OEIS sequence A059525.


问:
Does the computation of the inverse function of a set that is not in the image of $f$ makes sense?
I am studying measure theory, and I was wondering if the computation of the inverse function $f^{-1}$ of a set that is not in the image of $f$ makes sense. The fact is that all the functions that I have seen in my course are defined $(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$. But for example, let us consider the logistic function $f(x)=\frac{1}{1-e^{-x}}$, would that function be defined on $(\mathbb{R},B(\mathbb{R}))\rightarrow ([0,1],B([0,1])$ or in $(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$? In the latter case, what would happened for the sets that are not in the image of $f$?.

答:
Note that here $f^{-1}$ does not necessarily refer to the inverse function. Indeed, $f$ might not admit an inverse. Instead, $f^{-1}$ denotes the pre-image under $f$. I.e.
$$
f^{-1}(A)=\{x\in\Omega|\;f(x)\in A\}
$$
In the case where $A\cap f(\Omega)=\emptyset,$ you'll simply conclude that $f^{-1}(A)=\emptyset$.

Note that if $f$ is injective and $A\subseteq f^{-1}(\Omega)$, then the two notions (pre-images and images under the inverse function) agree and thus, the difference is pedantic at best. However, the pre-image always makes sense for any function and thus, is a better concept in general.


问:
If X is a totally order set with the order topology, show that Cl((a,b))$\subset[a,b]$
My proof goes as follows:
Consider the set Cl(a,b) and an element $x\in Cl(a,b)$, then $a<x<b$. We take an open set containing $x$: $(-\infty,x]$ and $(-\infty,x]\cap(a,b)=(a,x]\implies a\leq x$.

Analogously with the open set $[x,\infty)$ we have $[x,b)\implies x\geq b$.

It follows that $x\in [a,b]$ $\blacksquare$

I feel that my proof is too simple to be correct.

As well I have to show by example that this contention is proper.

答:
$\newcommand{\cl}{\operatorname{cl}}$You get off on the wrong foot right at the beginning: $x\in\cl(a,b)$ does not imply that $a<x<b$, and $(\leftarrow,x]$ is not in general an open set.

The easiest way to show that $\cl(a,b)\subseteq[a,b]$ is to prove the contrapositive: if $x\notin[a,b]$, then $x\notin\cl(a,b)$. If $x\notin[a,b]$, then either $x<a$, or $x>b$. If $x<a$, then $(\leftarrow,a)$ is an open nbhd of $x$ disjoint from $(a,b)$, and if $x>b$, then $(b,\to)$ is an open nbhd of $x$ disjoint from $(a,b)$, and in either case $x\notin\cl(a,b)$.


问:
How to linearize the following LP
I want to minimize $|d_1-d_2|+e1+e2+e3$ where $d_1,d_2,e_1,e_2,e_3>=0$ and $|.|$ denotes the absolute value, for some linear constraints. Is there any way I can linearize the objective function?</p>

答:
Is this homework? If you have the problem

min $|d_1-d_2|+e_1+e_2+e_3$

with the constraints
$d_1,d_2,e_1,e_2,e_3\geq0$

then by introducing a new variable $\lambda$ with $\lambda\geq d_1-d_2\geq -\lambda$ we have the following LP whos optimum is just the same as the above

min $\lambda+e_1+e_2+e_3$

with the constraints
$\lambda - d_1 + d_2\geq 0$,
$\lambda + d_1 - d_2\geq 0$
and
$\lambda,d_1,d_2,e_1,e_2,e_3\geq0$

ContentLicense: CC BY-SA 3.0

Title: How to linearize the following LP


问:
Is it possible to fill a table (64 rows and 6 columns) with zeros and ones so that the conditions are met?
Is it possible to fill a table (64 rows and 6 columns) with zeros and ones so that the following conditions are met:

All binary words of length 6 are written in the rows of the table;

Not a single row in the table can be obtained from any higher row by replacing ones with zeros;

And does the 57th line contain the binary word 011100?

Every help will be appreciated!

答: Hint: think about which of the entries whose relation to 011100 (the target) is forced. For example, you can obtain the target from 011101 by replacing the last one with a zero, so 011101 must be below the target. Conversely, 010100 must above the target, since it is made by replacing the middle $1$ in the target with a zero. What are the all other word whose position relative to the target is known? What does that tell you about the target's position?


问:
Volume of cylinder between planes
Calculate the volume inside the cylinder $x^2+4y^2=4$, and between the two planes $z=12-3x-4y$ and $z=1$.

Converting to cylindrical coordinates gives $$r^2\cos^2\theta +4r^2\sin^2\theta=4\\ z=12-3r\cos\theta-4r\sin\theta\\ z=1$$.

$r$ goes from $0$ to $1$ and $2$. Can i use $0\le r\le 1.5$?

$0\le \theta \le 2\pi$

$1\le z \le 12-3r\cos\theta-4r\sin\theta$

$$\int_{0}^{2\pi}\int_{0}^{1.5}\int_{1}^{12-3r\cos\theta-4r\sin\theta} rdzdrd\theta $$

What have i done wrong?

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}
{\cal V} & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + 4y^{2} < 4}
\bracks{z < 12 - 3x - 4y}}
\\[2mm] &\ \phantom{AAAAA}
\bbox[5px,#ffd]{\bracks{z > 1}\dd x\,\dd y\,\dd z}
\\[5mm] & \stackrel{x/2\ \mapsto\ x}{=}\,\,\,
2\int_{1}^{\infty}\iint_{\mathbb{R}^{2}}\bracks{x^{2} + y^{2} < 1}
\\[2mm] &\ \phantom{AAAAAAAAAAA}
\bracks{z < 12 - 6x - 4y}
\dd x\,\dd y\,\dd z
\end{align}
Lets use Cylindrical Coordinates:
\begin{align}
{\cal V} &
\,\,\,\stackrel{x/2\ \mapsto\ x}{=}\,\,\,
2\int_{1}^{\infty}\iint_{\mathbb{R}^{2}}\
\bracks{0 < \rho < 1}\ \times
\\[2mm] &\ \phantom{\stackrel{x/2\ \mapsto\ x}{=}\,\,\,\,\,\,\,}
\bracks{z < 12 - 6\rho\cos\pars{\theta} - 4\rho\sin\pars{\theta}}
\rho\,\dd \rho\,\dd\theta\,\dd z
\\[5mm] & =
2\int_{1}^{\infty}\int_{0}^{2\pi}\int_{0}^{1}
\\[2mm] &\ \phantom{2 =}
\bracks{z < 12 - 6\rho\braces{\cos\pars{\theta} +
{2 \over 3}\sin\pars{\theta}}}
\rho\,\dd \rho\,\dd\theta\,\dd z
\end{align}
With $\ds{\alpha \equiv \arctan\pars{2 \over 3}}$:
\begin{align}
{\cal V} & =
2\int_{1}^{\infty}\int_{0}^{2\pi}\int_{0}^{1}
\bracks{z < 12 -
2\root{13}\rho\cos\pars{\theta - \alpha}}
\\[2mm] &\
\phantom{AAAAA}\rho\,\dd \rho\,\dd\theta\,\dd z
\end{align}
Indeed, the last integral is $\ds{\alpha}$-independent because the integrand is a periodic function of period $\ds{2\pi}$ and it depends on the difference $\ds{\theta - \alpha}$. Namely,
\begin{align}
{\cal V} & =
2\int_{-\pi}^{\pi}\int_{0}^{1}\int_{1}^{\infty}
\bracks{z < 12 + 2\root{13}\rho\cos\pars{\theta}}
\\[5mm] &\
\phantom{AAAAA}\rho\,\dd z\,\dd \rho\,\dd\theta
\\[5mm] & =
2\int_{-\pi}^{\pi}\int_{0}^{1}\int_{1}^{12 + 2\root{13}\rho\cos\pars{\theta} \color{red}{\ >\ 1}}
\rho\,\dd z\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{0}^{\pi}\int_{0}^{1}
\bracks{11 + 2\root{13}\rho\cos\pars{\theta}}
\rho\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{-\pi/2}^{\pi/2}\int_{0}^{1}
\bracks{11 - 2\root{13}\rho\sin\pars{\theta}}
\rho\,\dd \rho\,\dd\theta
\\[5mm] & =
4\int_{-\pi/2}^{\pi/2}\int_{0}^{1}
11\rho\,\dd \rho\,\dd\theta = \bbx{22\pi}
\approx 69.1150 \\ &
\end{align}

Title: Volume of cylinder between planes


问:
Proving $\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)=n \times 3^{n-1}$ with a combinatoric argument

I want to prove
$$\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)=n \times 3^{n-1}$$
with combinatoric argumentation.

I have tried to understand it by comparing it with similiar problems. One such problem would be to prove the following equation with combinatoric argumentation:

$$\sum_{k=1}^n {{n}\choose{k}}*k = n* 2^{n-1}$$  The proof would be

https://math.stackexchange.com/a/7767/788271

How can i apply that solution to the given problem?

Body:
Playing algebraically.

$\begin{array}\\
s(n)
&=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)\\
&=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\dfrac{n!k!}{k!(n-k)!l!(k-l)!}l\right)\\
&=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\dfrac{n!l}{(n-k)!l!(k-l)!}\right)\\
&=\sum_{k=0}^{n}\sum_{l=0}^{k}\dfrac{n!l}{(n-k)!l!(k-l)!}\\
&=\sum_{l=0}^{n}\sum_{k=l}^{n}\dfrac{n!l}{(n-k)!l!(k-l)!}
\quad\text{(when in doubt, reverse the order of summation)}\\
&=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=l}^{n}\dfrac{1}{(n-k)!(k-l)!}\\
&=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=0}^{n-l}\dfrac{1}{(n-(k+l))!(k+l-l)!}\\
&=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=0}^{n-l}\dfrac{1}{(n-k-l)!k!}\\
&=\sum_{l=0}^{n}\dfrac{n!l}{l!(n-l)!}\sum_{k=0}^{n-l}\dfrac{(n-l)!}{(n-k-l)!k!}\\
&=\sum_{l=0}^{n}l\binom{n}{l}\sum_{k=0}^{n-l}\binom{n-l}{k}\\
&=\sum_{l=0}^{n}l\binom{n}{l}2^{n-l}\\
&=\sum_{l=0}^{n}(n-l)\binom{n}{n-l}2^{l}\\
&=\sum_{l=0}^{n}n\binom{n}{n-l}2^{l}-\sum_{l=0}^{n}l\binom{n}{l}2^{l}\\
&=n3^n-\sum_{l=0}^{n}l\binom{n}{l}2^{l}\\
&=n3^n-2n3^{n-1}
\qquad (*)\\
&=n \times 3^{n-1}\\
\end{array}
$

(*)
$(1+x)^n
=\sum_{k=0}^n \binom{n}{k}x^k
$
so
$n(1+x)^{n-1}
=\sum_{k=1}^n \binom{n}{k}kx^{k-1}
$
so,
with $x = 2$,
$n3^{n-1}
=\sum_{k=1}^n \binom{n}{k}k2^{k-1}
$
or
$2n3^{n-1}
=\sum_{k=1}^n \binom{n}{k}k2^{k}
$


问:
Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$
$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?

EDIT: The sum should have been infinite.

答:
Let $S = \sum_{k=1}^\infty \left(\frac{1}{3}\right)^k$. Then,
$$ S = \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \ldots = \frac{1}{3} \left[ 1 + \underbrace{\frac{1}{3} + \left(\frac{1}{3}\right)^2  + \ldots}_{=S} \right] =
\frac{1}{3} \left[ 1 + S \right]$$
So, we have $S = \dfrac{1}{3} + \dfrac{1}{3}S \Rightarrow \dfrac{2}{3} S = \dfrac{1}{3} \Rightarrow S = \dfrac{1/3}{2/3}$.


问:
Is my proof for this correct? Prove $2^{n+1} > n^{2}$ for any positive integer n.
Claim: $2^{n+1} >n^{2}$ for any positive integer n.

My proof:

$n = 1 \longrightarrow$

$2^{1+1} > 1^{2} \iff 4 > 1$

Check.

$n=2\longrightarrow$

$2^{2+1} > 2^{2} \iff 8 > 4$

Check.

Induction hypothesis: The claim is true for some positive integer $k$.

Inductive step:

$2^{k+1} > k^2$ (inductive hypothesis)

$2^{k+1} > k^2\ge k\implies 2^{k+1} > k\implies 2^{(k+1)+1} > (k+1)$, because $(k+1)$ is just some positive integer.

This implies that $2^{(k+1) + 1} > (k+1)^{2}$.

So we are done.

Is my proof sound?

答:
I don't understand the final step you make... I think your proof is incorrect.

I do the exercise in this way:

$k=1$ check;

$k=2$ check.

Now I should say that if $2^{k+1}>k^2$ then $2^{k+2}>(k+1)^2$ (where $k\in\mathbb{N}$ and $k>1$) .

Firstly observe that if $k>1$ you have that $2k+1<2^{k+1}$

(you can try this by induction:
It holds for $k=2$ and if $2k+1<2^{k+1}$ with $k>1$ so $2(k+1)+1<2^{k+1}+2\leqslant 2^{k+1}+2^{k+1}=2^{k+2}$).

Now if $2^{k+1}>k^2$, then $(k+1)^2=k^2+2k+1< 2^{k+1}+2^{k+1}=2^{k+2}$. So you are done.


答:
Firstly, prove that $k^2>2k+1$ for sufficiently large $k$. It works for $k\ge 3$.  I wont prove it.

Second, prove the base case of your hypothesis for $k=3$.  We have $2^{3+1} = 16 > 9 = 3^2.$

Third, the inductive step. Assume $2^{k+1} > k^2$ for some $k\in\mathbb{N}$. Then, $2^{(k+1)+1} = 2\cdot 2^{k+1}> 2\cdot k^2=k^2+k^2>k^2+2k+1 = (k+1)^2$.

Then you can show this relationship works for $k=0,1,2$ as well.

Title: Is my proof for this correct? Prove $2^{n+1} > n^{2}$ for any positive integer n.

Body:
Claim: $2^{n+1} >n^{2}$ for any positive integer n.

My proof:

$n = 1 \longrightarrow$

$2^{1+1} > 1^{2} \iff 4 > 1$

Check.

$n=2\longrightarrow$

$2^{2+1} > 2^{2} \iff 8 > 4$

Check.

Induction hypothesis: The claim is true for some positive integer $k$.

Inductive step:

$2^{k+1} > k^2$ (inductive hypothesis)

$2^{k+1} > k^2\ge k\implies 2^{k+1} > k\implies 2^{(k+1)+1} > (k+1)$, because $(k+1)$ is just some positive integer.

This implies that $2^{(k+1) + 1} > (k+1)^{2}$.

So we are done.

Is my proof sound?


问:
Finding the probability function/distribution of Coupon Collector's problem without Stirling Numbers
I'm trying to find the probability mass function for X in the coupon collector's problem which says:

"There are $m \in \mathbb{N} $ different types of coupons, and each coupon
obtained is equally likely to be any one of the $m$ types. Let X be the number of coupons that
need to be collected until the collection contains each of the coupon types. Find the probability mass function of X."

Most if not all of the solutions I've found online seem to use "Stirling numbers of the second kind", which I don't think we're allowed to use without either derivation of the Stirling numbers themselves. Hence, I'm thinking of an approach for finding a closed form for $P(X > k)$, and then determing $P (X = k) = P(X > k) - P(X > k+1)$.

I've found an expression for the above for the case where $m$ is small, say $m=3$, and I obtained that
$$P(X > k) = 3 \cdot \left(\frac{2}{3}\right)^k - 3 \cdot \left(\frac{1}{3}\right)^k$$

But I'm having trouble generalizing this to any m?

So is there some solution to this problem hopefully without using Stirling numbers, whether with the above approach or otherwise? Thank you!

答:
If you want to avoid the Stirling numbers, you will instead need the principle of inclusion exclusion, and a summation.

You had the right idea with looking at $P(X>k)$. The event $\{X>k\}$ is equivalent to saying that at least one coupon is missing after $k$ draws. To find this, first, add up the probabilities of each particular coupon being missing. The result is
$$
m\cdot \bigg(\frac{m-1}{m}\bigg)^k
$$
However, this double counts all the situations where two coupons are missing. THere are $\binom{m}{2}$ pairs of coupons, so we correct for this with
$$
m\cdot \bigg(\frac{m-1}{m}\bigg)^k-\binom{m}2\bigg(\frac{m-2}{m}\bigg)^k
$$
However, now the situations with three missing coupons need to be added back in, and so on. The result works out to be
$$
P(X>k)=\sum_{i=1}^m(-1)^{i+1}\binom{m}i\bigg(\frac{m-i}{m}\bigg)^k
$$


问:
Why is mathematical induction necessary to prove results (eg, commutativity) for natural numbers but not for real numbers?
I've been studying the construction of the natural numbers, and I can't solve my own question, namely

Why is it necessary to use mathematical induction?

Let me clarify this. For example, we know that, for all  $n,m \in \mathbb{N}$,  then $n \cdot m = m \cdot n$. In order to prove this, we use mathematical induction, but when we think about $n,m \in \mathbb{R}$ (real numbers),  in order to prove $n \cdot m= m \cdot n$, we don't need mathematical induction.

Why sometimes in a proof it's enough to take $x \in \mathbb{R}$, arbitrary number, but for natural numbers mathematical induction is necessary?

Anyone can help me, please?

I hope somebody can give me a hint in order to understand this question.

答:
There are two ways to introduce the real numbers.

First way: a set of layers, starting from the natural numbers, then the integers, then the rational numbers, then the real numbers.

Second way: a set of axioms for the real numbers is taken and a structure satisfying them is assumed to exist.

In the first way, we need to prove commutativity of multiplication in the integers, then in the rational numbers, and then in the real numbers. This chain of proofs is based on the proof of commutativity of multiplication in the natural numbers.

In the second way, commutativity of multiplication is taken as an axiom. But we need to embed the natural numbers in the real numbers and this again requires induction and commutativity of multiplication in the natural numbers.


问:
I am not sure on where to start to prove $∀x∈R |x−a|≤1 ⇒ |x+a|≤1 + 2|a|$
I don't know if the epsilon-delta definition of limit will help here. $\forall_{x \in \mathbb{R}}$ we have $|x-a| \leq 1 \implies |x+a| \leq 1 + 2|a|$

答:
Hint:

By the triangle inequality,

$|x+a|\le|x|+|a|,$

and $|x|\le|x-a|+|a|.$


问:
Is it true that for any integer $c_1, c_2$, we have $\gcd(n_1, n_2, n_3) = \gcd(n_1, n_2, c_1 n_1 + c_2 n_2 + n_3)$ for $n_1, n_2, n_3 \in \mathbb N$?
I got this task to check that if $n_1, n_2, n_3 \in \mathbb{N}$ the $\text{gcd}(n_1, n_2, n_3) =\text{gcd}(n_1, n_2, c_1n_1 + c_2n_2 + n_3)$ is valid for any $c_1, c_2 \in \mathbb{Z}$.

I started the idea saying that if:

$\text{gcd}(n_1, n_2, n_3) = k$, only if:

$k|n_1 \land k|n_2 \land k|n_3$ which means that:
$$
\left\{
\begin{array}{c}
n_1=ke \\
n_2=kf \\
n_3=kg
\end{array}
\right. \quad e, f, g\in \mathbb{N}
$$
Now $gcd(n_1, n_2, c_1n_1 + c_2n_2 + n_3) = k$ only if:
$$ c_1n_1 + c_2n_2 + n_3 = kr, \quad r \in \mathbb{Z}
$$
Substituting the values of $n_1, n_2, n_3$:
$$ k(c_1e + c_2f + g) = kr
$$
Which means that it is valid only if:$$c_1e + c_2f + g = r$$
But now I got stucked, I cannot say if this is valid for any $c_1, c_2$, could anyone help?

Body:
I suppose that you have to proof that $gcd(n_1,n_2,n_3)=gcd(n_1,n_2,c_1n_1+c_2n_2+n_3)$, (not $c_3$ at the end... Who is $c_3$?)

$gcd(n_1,n_2,n_3)=k$;

$gcd(n_1,n_2,c_1n_1+c_2n_2+n_3)=p$.

$k$ divides $n_1$,$n_2$ and $n_3$.

So $k$ divides $n_1$,$n_2$,$c_1n_1+c_2n_2+n_3$.

So $k$ divides $p$.

But $p$ divides $n_1$,$n_2$,$c_1n_1+c_2n_2+n_3$. It suffices to show that $p$ divides also $n_3$. This is clear:

$n_1=ps_1$, $n_2=ps_2$, $c_1n_1+c_2n_2+n_3=ps_3$.

So $n_3=ps_3-c_1n_1-c_2n_2=p(s_3-c_1s_1-c_2s_2)$.

So $p$ divides $k$.

So $gcd(n_1,n_2,n_3)=gcd(n_1,n_2,c_1n_1+c_2n_2+n_3)$.


答:
For all $x\in R$

$$|x-a| \leq 1 \implies |x+a|=|x-a+2a|\leq |x-a|+|2a|\leq 1 + 2|a|$$

Title: I am not sure on where to start to prove $∀x∈R |x−a|≤1 ⇒ |x+a|≤1 + 2|a|$

Body: I don't know if the epsilon-delta definition of limit will help here. $\forall_{x \in \mathbb{R}}$ we have $|x-a| \leq 1 \implies |x+a| \leq 1 + 2|a|$


答:
You have to use the following triangle inequality:
\begin{eqnarray}
|x|-|y| \le |x-y| \tag{1}\label{1}
\end{eqnarray}
Suppose $|x-a| \le 1$. Writing:
$$|x-a| = |x-2a+a|$$
call $x+a = z$. Then, by (\ref{1}):
$$|z| -2|a| \le |z-2a| = |x-a| \le 1 \Rightarrow |x+a| \le 1+2|a|$$


问:
Prove that $(-1)^n \text{Laguerre}_n(2) \leq 1$.
I would like to prove the following inequalities on Laguerre polynomials evaluated at point 2:
$$
(-1)^n \text{Laguerre}_n(2) \leq 1
$$
This seems to hold numerically.
I tried to use the recurrence relation between Laguerre polynomials but I was not successful. Any ideas? I also tried expression via Bessel function $J_0$ and then contour integral but it was not successful.

答:
Theorem 8.22. on Szego's Orthogonal polynomials the asymptotics of the Laguerre polynomials (called there Fejer's formula) looks like
$$L_n(x)=\frac{e^{x/2}}{2^{1/4}\sqrt{\pi}\cdot n^{1/4}}cos \left ( 2\sqrt{nx}-\pi /4\right )+O(n^{-3/4}),\, \text{for }x>0.$$
Take $x=2,$ we have then that
$$|L_n(2)|\leq \frac{e}{2^{1/4}\sqrt{\pi}}\left |\frac{cos(2^{3/2}\sqrt n-\pi /4)}{n^{1/4}} \right |+\left |O(n^{-3/4})\right |\leq 1.2896\cdot \frac{1}{n^{1/4}}+\frac{C}{n^{3/4}},$$
for some constant $C.$

Even if the constant is something big, you can check the result for small values of $n.$
Here is a graph using $C=2$ which I would say is big according to simulation:

Also, this is a very good survey about the whole problem


答:
Let $d_1 = \gcd (n_1, n_2, n_3), d_2 = \gcd (n_1, n_2, c_1 n_1 + c_2 n_2 + n_3)$. We want to show $d_1 = d_2$.

I will prove $d_2 | d_1$. Then you can do $d_1 | d_2$ similarly then you are done.

There exists $x_1, x_2, x_3 \in \mathbb{Z}$ such that
$$
x_1 n_1 + x_2 n_2 + x_3 n_3 =d_1.
$$

Therefore
$$
(x_1 - x_3 c_1) n_1 + (x_2 - x_3 c_2)n_2 + x_3 (c_1 n_1 + c_2 n_2 + n_3) =d_1 \Rightarrow d_2 | d_1.
$$

Title: Is it true that for any integer $c_1, c_2$, we have $\gcd(n_1, n_2, n_3) = \gcd(n_1, n_2, c_1 n_1 + c_2 n_2 + n_3)$ for $n_1, n_2, n_3 \in \mathbb N$?

Body:
I got this task to check that if $n_1, n_2, n_3 \in \mathbb{N}$ the $\text{gcd}(n_1, n_2, n_3) =\text{gcd}(n_1, n_2, c_1n_1 + c_2n_2 + n_3)$ is valid for any $c_1, c_2 \in \mathbb{Z}$.

I started the idea saying that if:

$\text{gcd}(n_1, n_2, n_3) = k$, only if:

$k|n_1 \land k|n_2 \land k|n_3$ which means that:
$$
\left\{
\begin{array}{c}
n_1=ke \\
n_2=kf \\
n_3=kg
\end{array}
\right. \quad e, f, g\in \mathbb{N}
$$
Now $gcd(n_1, n_2, c_1n_1 + c_2n_2 + n_3) = k$ only if:
$$ c_1n_1 + c_2n_2 + n_3 = kr, \quad r \in \mathbb{Z}
$$
Substituting the values of $n_1, n_2, n_3$:
$$ k(c_1e + c_2f + g) = kr
$$
Which means that it is valid only if:$$c_1e + c_2f + g = r$$
But now I got stucked, I cannot say if this is valid for any $c_1, c_2$, could anyone help?


问:
How do I find the derivative $f(x) = \int_{x^{13/14}}^{15\pi} \sqrt{q} \cot(q)dq$
How do I find the derivative $f(x) = \int_{x^{13/14}}^{15\pi} \sqrt{q} \cot(q)dq$

The problem I am running into is that $15\pi$ is not in the continuous range of the function $\sqrt{x}\cot(x)$ hence the fundamental theorem of calculus can not be applied. How can I get around this?

答: Your integrand has a non-integrable singularity at multiples of $\pi$, so the integral doesn't converge, and $f(x)$ does not exist.


问:
If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.

If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.

Please prove using $$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
and
$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}$$

Thankyou

答:
Using $\lim_{x\to c}f(x)=f(c)$ to check continuity, in this case, is just the same as checking that you have a composition of continuous functions, which is continuous wherever it is defined.

Since $1-x^2\ge0$ means $|x|<1$ and implies $\sqrt{1-x^2}\le1$, your function is defined (and continuous) in $[-1,1]$.

The derivative can be computed with the chain rule, keeping in mind that the derivative of $x\mapsto\sqrt{x}$ exists only for $x>0$:

$$
f'(x)=\frac{1}{2\sqrt{1-\sqrt{1-x^2}}}\frac{x}{\sqrt{1-x^2}}
$$

You have to be careful where $1-\sqrt{1-x^2}=0$ and $\sqrt{1-x^2}=0$, that is at $x=0$, $x=1$ and $x=-1$, where the formula doesn't apply.

However, since the function is continuous, you can compute the limit of the derivative; it's easy to show that
\begin{align}
\lim_{x\to-1^+}f'(x)&=-\infty\\
\lim_{x\to 1^-}f'(x)&=\infty
\end{align}
so your function is not differentiable in $-1$ and $1$. For the limit at $0$ you can write the derivative as
\begin{align}
f'(x)&=
\frac{1}{2\sqrt{1-x^2}}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}\\
&=
\frac{1}{2\sqrt{1-x^2}}\frac{x\sqrt{1+\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\sqrt{x^2}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\lvert x\rvert}\\
\end{align}
so that
$$
\lim_{x\to0^-}f'(x)=-\frac{\sqrt{2}}{2},\qquad
\lim_{x\to0^+}f'(x)=\frac{\sqrt{2}}{2}
$$
which proves that $f$ is not differentiable at $0$.

Here's the graph


答:
First of all, induction only really works on well-ordered, countable sets: things like $\mathbb{Z}$ and $\mathbb{N}$, where you can always define a "next" number. You can well-order $\mathbb{Q}$, but not in any way that makes induction useful: the most common well-ordering of $\mathbb{Q}$ starts $0,1,-1,2,\frac12,-\frac12,-2,3,\frac13,-\frac13,-3,4,\frac32,\frac23,\frac14,-\frac14,-\frac23,-\frac32,-4$ and continues on like that. (See if you can spot the pattern!) For any rational number, it really doesn't make sense to talk about the "next" rational number, and so induction isn't really useful.

$\mathbb{R}$ has all the issues of $\mathbb{Q}$, but it's not even countable. An induction on $\mathbb{R}$ that considers each real number separately requires all sorts of weirdness like the Axiom of Choice and maybe even some form of Transfinite induction in order to make it work, and it'll be a mess. Of course, there are ways around this, like slicing up $\mathbb{R}$ into countably many pieces, but that's usually just standard induction with intervals.

Incidentally, though, we do need to establish that multiplication on $\mathbb{R}$ is commutative, and the way we do it depends on the way we're constructing $\mathbb{R}$. Sometimes, we poof $\mathbb{R}$ into existence using the magic spell, "the unique Dedekind-complete ordered field up to isomorphism" and then we don't need to do much work; the field axioms establish that the operation we call "multiplication" must be associative. Careful, though; if you're doing this for the first time, you may want to establish that this approach actually does give a unique, well-defined set, and that this set is indeed the "real numbers" that you've grown to know.

The other way to define real numbers is by "building them up" in some way from the rationals (which, incidentally, are defined by "building up" from the integers, which are in turn defined by "building up" from the naturals). There are a few ways of doing this, and some of them make showing commutativity easier than others do, but either way we do have to prove it. It just turns out that induction is usually totally useless, because by the time you've shown that this new set behaves the way you want it to behave (i.e. being Dedekind-complete, closed under certain operations, etc) you've already probably shown that the multiplication operator is commutative. If you're curious about how that's done, here's a paper that goes into depth on two of the most common ways to construct the reals, Dedekind cuts and Cauchy sequences.

The naturals, on the other hand, are far more fundamental; there's really not much that you can "build up" from in a rigorous and meaningful way, and there's really only one commonly accepted definition of the naturals as far as I know. In that definition, commutativity is a theorem, not an axiom. I suppose if you really tried, you could find some contrived definition of the naturals where commutativity of multiplication has to be an axiom, but it's so much simpler to just use Peano and induct.

Title: Why is mathematical induction necessary to prove results (eg, commutativity) for natural numbers but not for real numbers?

Body:
I've been studying the construction of the natural numbers, and I can't solve my own question, namely

Why is it necessary to use mathematical induction?

Let me clarify this. For example, we know that, for all  $n,m \in \mathbb{N}$,  then $n \cdot m = m \cdot n$. In order to prove this, we use mathematical induction, but when we think about $n,m \in \mathbb{R}$ (real numbers),  in order to prove $n \cdot m= m \cdot n$, we don't need mathematical induction.

Why sometimes in a proof it's enough to take $x \in \mathbb{R}$, arbitrary number, but for natural numbers mathematical induction is necessary?

Anyone can help me, please?

I hope somebody can give me a hint in order to understand this question.


问:
Choosing Convergent Sequence From Infinitely Many Matrix Solutions
For some research I'm doing for my PhD, I encountered this situation where I'm pretty sure what I'm doing is correct, but I'd like some confirmation from someone with more knowledge of this stuff. The situation can be described in a very general way as follows:

Suppose we have a 3xN (N > 3) matrix M with any three columns of M being linearly independent, an Nx1 vector U where each entry is strictly positive, and let M*U = P.

Suppose further that we perturb M by $\epsilon$ in some way, to $M_\epsilon$. If $\epsilon$ is small enough, $M_\epsilon$ is still full rank, and thus there are infinitely many solutions $U_\epsilon$ to $M_\epsilon$ * $U_\epsilon$ = P for each $\epsilon$. What I would like to do is say "For every $\epsilon$ > 0, define $U_\epsilon$ to be the unique solution to $M_\epsilon$ * $U_\epsilon$ = P such that ||$U_\epsilon$ - U|| is minimal (and there's no reason to assume that's unique, so I would have to maybe specify a further condition, like minimal under a lexicographical ordering or something?).

Once $U_\epsilon$ is defined in this way, then I want to say that it converges to U as $\epsilon$ goes to 0, and therefore we can choose a small enough $\epsilon$ so that all entries of $U_\epsilon$ are positive (this is the main goal). The main problem I have is that without trying to specify a unique $U_\epsilon$, all I would know is that the $U_\epsilon$'s would converge to some solution of M*X = P, but not necessarily the U that I want, which I need to prove that $U_\epsilon$ can have all positive entries. But I'm not 100% sure the way that I've specified $U_\epsilon$ is reasonable or gives me the result that I want.

Hopefully the way I have explained the problem and what I'm looking for is clear enough, if not just let me know. Figuring this out will be very helpful for my PhD research.

答:
The solutions to $M_\epsilon X = P$ form an affine subspace $S$ of $\mathbb R^n$, and in particular a convex set.  If $\| \cdot \|$ is the usual Euclidean norm, since that norm is strictly convex there is always a unique member of $S$ that minimizes $\|X - U\|$.

EDIT: Since $M$ has rank $3$, there is a set of three columns which are linearly independent and thus form a nonsingular $3 \times 3$ matrix.  WLOG let's say these are the first three columns and write $M$ in block
form as $[B \mid C]$, where $B$ is $3 \times 3$, and correspondingly $U = \pmatrix{V \cr \hline  W}$. Note that $M U = P$ is equivalent
to $V = B^{-1}(P - C W)$.

Now if $M_\epsilon = [B_\epsilon \mid C_\epsilon]$ is sufficiently close to $M$,
$B_\epsilon$ will still be nonsingular and
we have a solution $\tilde{U}_\epsilon = \pmatrix{B_\epsilon^{-1}(P - C_\epsilon W)\cr \hline W}$.
Since inversion is a continuous map on the
nonsingular matrices, $\|\tilde{U}_\epsilon - U\| \to 0$ as $\epsilon \to 0$.  Since $\|U_\epsilon - U\| \le \|\tilde{U}_\epsilon - U\|$, that also goes to $0$ as $\epsilon \to 0$.  In particular, if $\epsilon$ is sufficiently small all entries of $U_\epsilon$ must be strictly positive.


问:
Is $0 \rightarrow SU(n) \rightarrow SU(n+1) \rightarrow S^{2n + 1} \rightarrow 0$ exact?
I recently went to a talk where the speaker mentioned that

$$ 0 \rightarrow SU(n) \rightarrow SU(n+1) \rightarrow S^{2n + 1} \rightarrow 0$$

was exact. I think it's well-known that this sequence is a fibration, but is it an exact sequence too?

For this to the the case we would need to put a group structure on $S^{2n + 1}$ and define a group homomorphism from $SU(n+1)$ to it... but I don't know how to do this. There is a bijection between $S^{2n+1}$ and cosets $SU(n+1)/SU(n)$, but since $SU(n)$ is not normal in $SU(n+1)$, $SU(n+1)/SU(n)$ does not have a group structure. Hence my source of confusion.

Was the speaker correct, or did they mean something else by 'exact'?

答: This is not a standard usage of "exact", and looks like a misstatement.  In particular, $S^{2n+1}$ does not admit any topological group structure if $n>1$.  I imagine they really just meant that it's a fiber sequence.


问:
What is law of random variable $ S_n = X_1 + \cdots + X_n $
We consider a sequence $ (X_n)_{n \leq 1} $ of mutually independent random variables defined on a probability space $ (\Omega, \mathcal{A}, \mathbb{P}) $  with value in $\{-1,1\}$ and such that, for all $ k \geq 1$ $$ \mathbb{P} (X_k = -1) = \mathbb{P} (X_k = 1) = \frac{1}{2} $$ For all $ n \geq 1 $, we set $ S_n = X_1 + \cdots + X_n $

Problem

What is law of random  variable $S_n$?

An idea please.

答: Hint: The Count for values of $+1$ among the sequence of ${(X_k)}_{k=1}^n$ shall be the count for successes among $n$ independent Bernoulli trials with an identical success rate.


问:
How do you solve for the largest value of x for which the acceleration is zero in this problem?
This is my problem:

A block of mass 1.50 kg is accelerated across a rough surface by a light cord passing over a small pulley shown in the image below. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.140 m above the top of the block. The coefficient of kinetic friction is 0.360.

This is the question it is asking:

Find the largest value of x for which the acceleration is zero.

I have been trying to solve this for hours with no luck, please help.

Image of the problem here

Edit: To further explain my problem and what I have tried:

I know that to find x when acceleration = 0 you set the force of friction equation equal to force giving you Tcos(theta) = mu(mg - Tsin(theta) and then you put it in terms of x. I got (1/m)(mu)(m)(g)[x^2 + h^2]^-1/2 = (1/m)Tx + (1/m)(mu)(h)(T) (h being the height of the pulley). I then substitute in my variables m = 1.5kg, h = 0.14m, mu = 0.36, and T = 10N. I guess what I’m asking is if that equation is correct because it worked for a practice problem with different numbers but not the numbers in the problem where it counts.

答:
You have

$T\cos(\theta) = \mu (mg-T\sin(\theta))$

$\cos(\theta) = \frac{x}{\sqrt{x^2+h^2}}$

$\sin(\theta) = \frac{h}{\sqrt{x^2+h^2}}$

plugging these into the original equation we get:

$T\frac{x}{\sqrt{x^2+h^2}} = \mu (mg-T\frac{h}{\sqrt{x^2+h^2}})$

multiply both sides by $\sqrt{x^2+h^2}$

Is this what you also got?

$Tx = \mu (mg\sqrt{x^2+h^2}-Th)$

Now plug in everything and solve for x. You'll have to isolate the square root. Then square both sides, simplify and use the quadratic formula. What do you get?


问:
Proving a series is not convergent
I have an exercise where I have to show that a series is not convergent. I have tried some convergence tests but I'm unfamiliar with working a product in a series.

Let $a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}$

Now let $c_n = \sum_{k=0}^n a_{n-k}b_k$

Prove that $\sum_{k=0}^\infty c_n$ is not convergent.

As before mentioned. I haven't really worked with series like these. It stated that the Cauchy product formula is wrong here. So I've tried to just calculate the limit using the ratio test, but that did not seem to work.

Is it correct to just write it like this?

$\sum_{k=0}^\infty c_n$ $=$ $\sum_{k=0}^\infty \frac{(-1)^{n-k+k}}{\sqrt{n-k+1}\sqrt{k+1}}$ $=$ $\sum_{k=0}^\infty \frac{(-1)^{n}}{\sqrt{n-k+1}\sqrt{k+1}}$

I'm not sure if I can rewrite $\sqrt{n-k+1}\sqrt{k+1}$ $=$ $\sqrt{(k+1)(-k+n+1)}$ But I'm equally unsure if this is even helps.

Now my question is: Is my work this far even correct or am I making crucial mistakes even just writing these down and if it is, how can I proceed to prove that this series does in fact not converge? (Since the ratio tests was inconclusive)

答:
We'll make use of the following theorems:

$\lim\limits_{n\to\infty}a_n=L$ if and only if $\lim\limits_{n\to\infty}a_{2n}=L$ and $\lim\limits_{n\to\infty}a_{2n+1}=L$.

If $\lim\limits_{n\to\infty}a_n\neq 0$ or does not exist, then $\sum_{n=0}^{\infty}a_n$ is divergent

Combining both theorems, it follows that if either $\lim_{n\to\infty}c_{2n}$ or $\lim_{n\to\infty}c_{2n+1}$ doesn't exist, then $\lim\limits_{n\to\infty}c_{n}$ doesn't exist either and hence $\sum_{n=0}^{\infty}c_n$ is divergent. Therefore, we may assume that $c_{2n}$ and $c_{2n+1}$ are both convergent.

We proceed by first analyzing the behavior of $c_{2n}$.
\begin{align*}
c_{2n} &= \sum_{k=0}^{2n}\frac{(-1)^{2n}}{\sqrt{2n-k+1}\sqrt{k+1}}\\
&= \sum_{k=0}^{2n}\frac{1}{\sqrt{2n+1-k}\sqrt{k+1}}
\end{align*}
Note that for $0\leq k\leq 2n$, we have that $k+1\geq 1$ and $1\leq 2n+1-k\leq 2n+1$, so each of the terms in the sum $\sum_{k=0}^{2n}\frac{1}{\sqrt{2n+1-k}\sqrt{k+1}}$ is positive. It follows that $c_{2n}$ is strictly positive and strictly increasing, so the limit of $c_{2n}$ must be greater than zero.

Now we analyze the behavior of $c_{2n+1}$.
\begin{align*}
c_{2n+1} &= \sum_{k=0}^{2n+1}\frac{(-1)^{2n+1}}{\sqrt{2n+1-k+1}\sqrt{k+1}}\\
&= \sum_{k=0}^{2n+1}\frac{-1}{\sqrt{2n+2-k}\sqrt{k+1}}
\end{align*}
For $0\leq k\leq 2n+1$, we have that $k+1\geq 1$ and $1\leq 2n+2-k\leq 2n+2$, so each of the terms in the sum $\sum_{k=0}^{2n+1}\frac{-1}{\sqrt{2n+2-k}\sqrt{k+1}}$ is strictly negative (the $-1$ present in each of the terms has this effect). It follows that $c_{2n+1}$ is strictly negative and strictly decreasing, so the limit of $c_{2n+1}$ must be less than zero.

We have shown that $c_{2n}$ has a positive limit and $c_{2n+1}$ has a negative one. Since no real number can be both positive and negative, it follows that $\lim_{n\to\infty}c_n$ cannot exist, and thus $\sum_{n=0}^{\infty}c_n$ is divergent.

Title: Proving a series is not convergent


问:
Show that $n$ even $ \land\ f^{n}(b)<0 \implies f$ has strict maxima in $b$
I have already asked this question once, but did not quit understand all the steps in the provided hint. So here goes:

The $n$-th derivative of $f$ is continuous in a neighbourhood around a point $b$

$f'(b)=f''(b)=...=f^{(n-1)}(b)=0$

$f^{(n)}(b)\neq 0 $

From the conditions above I am supposed to come to the following conclusions:

$n$ even $\land\: f^{n}(b)<0 \implies f$ has strict maxima in $b$

$n$ even $\land\: f^{n}(b)>0 \implies f$ has strict minima in $b$

$n$ odd $\implies f$ has no extreme in $b$

My attempt is based on Taylor's formula:

$f(x) = f(b) + \sum_{k=1}^{n-1} \frac {f^{(k)}(b)}{k!} (x-b)^{k} + R_{n}(x) \iff f(x)-f(b)= R_{n}(x)$

$R_{n}(x)$ is the Lagrange remainder $\iff R_{n}(x) = \frac {f^{(n)}(b+c \ (x-b)))}{n!} \ (x-b)^{n}$ , $0<c<1$

So $f(x)-f(b) = \frac {f^{(n)}(b+c \ (x-b)))}{n!} \ (x-b)^{n}$

This is where I get stuck.

In order to move forward, I am supposed to understand that $ \frac {f^{(n)}(b+c \ (x-b)))}{n!} \ (x-b)^{n} = \frac {(x-b)^n}{n!} (f^{(n)} (b) + \epsilon(x)), \epsilon(x)$ approaches $0$ when $x$ approaches $b$.

The only way for me to understand the equality above is by assuming that all continuous functions can be linearized in a (small enough) neighbourhood of b, but I don't know how to prove that or even if it is true.

答:
hint

Yes, as you said, by Taylor-Young formula, we can write for $ x $ close to $ b $,

$$f(x)-f(b)=$$
$$\frac{(x-b)^n}{n!}f^{(n)}(b)\Bigl(1+\epsilon(x)\Bigr)$$

If $n $ is even , then $(x-b)^n\ge 0$.

$1+\epsilon(x)>0$ because $ \epsilon(x) \to 0$ which means that for $ x $ near $ b $, we have
$$-\frac 12 <\epsilon(x)<\frac 12$$
and
$$0<\frac 12 <1+\epsilon(x)$$

You can conclude.
What can you say about the sign of $ f(x)-f(b)$.


问:
Maximum dimensionality of the linear manifold
What is the maximum dimensionality of the linear manifold spanned by:

```<code>a) 20 points in a 40 dimensional space
b) 40 points in a 20 dimensional space
c) 20 points in a 20 dimensional space
</code>```

It has been a long time since I took the linear algebra course, so please correct me if I am wrong.

My answers:

```<code>a) 40
b) 20
c) 20
</code>```

答:
The maximum dimension $d$ of the linear manifold $M$ spanned by $m$ points in a $n$-dimensional vector space $V$ is $d=\min(m,n)$.

Indeed, $M$ is a submanifold of $V$, so $d\le n$, and $M$ is spanned by $n$ points, so $d\le m$, hence $d\le \min(m,n)$. Conversely, since $n=\dim V$ there are $n$ linearly independent vectors in $V$; take $\min(m,n)$ of them (and throw in $m-n$ copies of the zero vector if $m>n$) and then take the span to get a $\min(m,n)$-dimensional manifold $M$ spanned by $m$ points, showing that $d\ge \min(m,n)$.

So the answer to each of your questions should be $20$.


问:
Testing my uniform distribution
i have two vectors (15 elements and 120 elements) of numbers that i generated with Linear congruential generator U[-1,1) in python and now i have to test them for
uniform distribution . I need help choosing a test and running it (i cant do to this with any built-in test, I have to do this by my self with calculations).
Anyone can help me with choosing a test and showing me any solustions?
I think I know how to conduct tests but i need help with choosing it.
I dont know am I right but first i should start with testing mean with t student test ?

Thank you in advance for your help

Mean for 15 elements:  0.037882

Mean for 120 elements  0.06044

SD for 15:   0.502921

SD for 120:  0.593756

答:
(a) Instructions seem vague and it would be easier to
know what approach you are supposed to take if we
knew what topics in statistics you have studied already--especially recently. (b) Also, you do not say specifically what
distribution your LCG is programmed to generate.

One reason we ask you to show your work so far, it that
your work may provide some context, even for vague
problems. Such information is missing in your question.

I will answer briefly assuming in (a) that you know about z -tests and in (b) that your 'random' numbers are
supposed to be indistinguishable from $\mathsf{Unif}(0,1).$
I understand that these guesses may be wrong. If so,
maybe you can revise your question with more context so
that someone else will make better assumptions.

From the Central Limit Theorem one would expect the mean
of $n$ such random variables should be approximately
$\mathsf{Norm}\left(\mu = 1/2, \sigma = \sqrt{\frac{1/12}{n}}\right)$ because the mean and variance of the standard uniform distribution are $1/2$ and $1/12,$ respectively.

Means of samples of 120 properly generated standard uniform random variables are extremely close to normally distributed. Even for samples of $n=15,$ means are nearly normal.

Using z-scores to asses LCG. So you can find the z-score
$z = \frac{\bar X = 1/2}{\frac{1/12}{n}}$ for each
of your samples and reject the null hypothesis that
the LCG is performing properly if $|z| > 1.96.$

For example, your second sample has $z = \frac{0.0604 - .5}{\sqrt{1/12}{120}} \approx -16.7,$ which obviously leads to rejection.

```<code>(0.0604 - 1/2)/sqrt((1/12)/120)
[1] -16.68165
</code>```

Simulated results for means of properly generated
standard uniform samples. The following simulation in R looks at results
for 100,000 samples $n=120$ observations drawn from a well-vetted random number generator.

```<code>set.seed(113)
a = replicate(10^5, mean(runif(120)))
summary(a); sd(a)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.3812  0.4822  0.5000  0.5001  0.5179  0.6185
[1] 0.02637311
</code>```

The 100,000 sample means $\bar X$ (vector a in the code) average about .5 and are never below 0.38 or above 0.52.

So for the second sample, the mean $\bar X = 0.0604$ seems
an essentially impossible value for the mean of 120 actual
observations from a standard uniform distribution.

You might also ponder why the standard deviation 0.593756 of this sample is so far from the anticipated value $\sqrt{\frac{1/12}{120}} = 0.0264.$

Title: Testing my uniform distribution

Body:
(a) Instructions seem vague and it would be easier to
know what approach you are supposed to take if we
knew what topics in statistics you have studied already--especially recently. (b) Also, you do not say specifically what
distribution your LCG is programmed to generate.

One reason we ask you to show your work so far, it that
your work may provide some context, even for vague
problems. Such information is missing in your question.

I will answer briefly assuming in (a) that you know about z -tests and in (b) that your 'random' numbers are
supposed to be indistinguishable from $\mathsf{Unif}(0,1).$
I understand that these guesses may be wrong. If so,
maybe you can revise your question with more context so
that someone else will make better assumptions.

From the Central Limit Theorem one would expect the mean
of $n$ such random variables should be approximately
$\mathsf{Norm}\left(\mu = 1/2, \sigma = \sqrt{\frac{1/12}{n}}\right)$ because the mean and variance of the standard uniform distribution are $1/2$ and $1/12,$ respectively.

Means of samples of 120 properly generated standard uniform random variables are extremely close to normally distributed. Even for samples of $n=15,$ means are nearly normal.

Using z-scores to asses LCG. So you can find the z-score
$z = \frac{\bar X = 1/2}{\frac{1/12}{n}}$ for each
of your samples and reject the null hypothesis that
the LCG is performing properly if $|z| > 1.96.$

For example, your second sample has $z = \frac{0.0604 - .5}{\sqrt{1/12}{120}} \approx -16.7,$ which obviously leads to rejection.

(0.0604 - 1/2)/sqrt((1/12)/120)
[1] -16.68165

Simulated results for means of properly generated
standard uniform samples. The following simulation in R looks at results
for 100,000 samples $n=120$ observations drawn from a well-vetted random number generator.

set.seed(113)
a = replicate(10^5, mean(runif(120)))
summary(a); sd(a)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.3812  0.4822  0.5000  0.5001  0.5179  0.6185
[1] 0.02637311

The 100,000 sample means $\bar X$ (vector a in the code) average about .5 and are never below 0.38 or above 0.52.

So for the second sample, the mean $\bar X = 0.0604$ seems
an essentially impossible value for the mean of 120 actual
observations from a standard uniform distribution.

You might also ponder why the standard deviation 0.593756 of this sample is so far from the anticipated value $\sqrt{\frac{1/12}{120}} = 0.0264.$


问:
Equivalent definitions of the second-order/Lorentz cone
I have come across two different definitions of the second-order/Lorentz cone. The first is the standard form where $t$ is a scalar and $\mathbf{y} \in \mathbb{R}^n$.
$$
\mathcal{C}_1 = \bigg\{ \begin{bmatrix}\mathbf{y} \\ t\end{bmatrix} \in \mathbb{R}^{n+1}: t \ge \lVert \mathbf{y}\rVert_2\bigg\}
$$

The second definition is from this paper. Here the cone is defined for $\mathbf{x} \in \mathbb{R}^{n+1}$.
$$
\mathcal{C}_2 = \big\{\mathbf{x} : \mathbf{x}^{\sf T} \mathbf{1} \ge \sqrt{n}\lVert \mathbf{x}\rVert_2 \big\}
$$

Are $\mathcal{C}_1$ and $\mathcal{C}_2$ the same?

答: The axis of the first cone is in the direction $(0,0,\ldots,0,1)$, while the axis of the second cone is in the direction $(1, 1, \ldots, 1)$. There is also some scaling difference in the sharpness of the cone (the factor $\sqrt{n}$).


问:
Finiteness of absolute moments
If $r>0$ then why is it that $E(|X|^r)<\infty$ if and only if $E(|X-a|^r)<\infty$ for every $a$. In the simplified case of a probability distribution describing $X$ then it says that

$$\int |x|^rp(x)dx <\infty \leftrightarrow \int |x-a|^r p(x)dx<\infty$$

But why should that be true?

答:
Everything follows from the inequalities
$$|x+a|^r\leqslant (|x|+|a|)^r\leqslant \max\{1,2^{r-1}\}\cdot(|x|^r+|a|^r),$$
where the second inequality follows from this thread when $r\geqslant 1$ and concavity of $x\mapsto |x|^r$ if $0<r<1$.</p>

Title: Finiteness of absolute moments


问:
Units/zero divisors in $R[x]$

Let $R$ be the set of integers modulo $9$, $\mathbb{Z}_9$, and define $R[x]$ to be the corresponding polynomial ring. Determine whether $x, 3x, 1 + x$, and $1 + 3x$ are units or zero divisors.

My attempt:

$3x$ is a zero divisor since $3x \cdot 3 = 9x = 0$ in mod $9$. $1 + 3x$ is a unit since
\begin{align*}
(1 + 3x)(1 - 3x) = 1 - 3x + 3x + 9x^2 \equiv 1.
\end{align*}

I cannot figure out $x$ and $1 + x$. By trial and error, I believe they are neither units nor zero divisors, but I cannot figure out how to prove it.

REVISED ATTEMPT:

Consider a non-zero polynomial $p(x) = a_0 + a_1 x + \ldots + a_n x^n$ where $a_n \neq 0$ for some $n \geq 0$. Then
\begin{align*}
x p(x) & = x (a_0 + a_1 x + \ldots + a_n x_n) \\
& = a_0 x + a_1 x^2 + \ldots + a_n x^{n+1}.
\end{align*}
Since $a_n \neq 0$, $p(x)$ is a polynomial of degree $n + 1$. Since $n \geq 0$, this implies that $p(x)$ is a polynomial of degree $\geq 1$. Therefore, it cannot equal either $0$ or $1$, so it is neither a zero divisor or a unit.

答:
What about $1+x$? Let's try multiplying by a nonzero polynomial
$$
(1+x)(a_0+a_1x+\dots+a_nx^n)=
a_0+(a_1+a_0)x+(a_2+a_1)x^2+\dots+(a_n+a_{n-1})x^n+a_nx^{n+1}
$$
If $1+x$ is a unit, we should be able to find
$$
a_n=0,\quad a_n+a_{n-1}=0,\quad\dots,\quad a_1+a_0=0,\quad a_0=1
$$
and you immediately see the contradiction.

Can $1+x$ be a zero divisor? Again not, because we'd need $a_0=0$, so also $a_1=0$ and so on up to $a_n$.

What's the difference between $1+x$ (neither a unit nor a zero divisor) and $1+3x$ (a unit)? That $3x$ is nilpotent.

Your attempt for $x$ is good: if $p(x)$ has degree $n\ge0$, then $xp(x)$ has degree $n+1$.

Any time a polynomial $q(x)$ has a unit as its leading term, the degree formula holds $\deg(q(x)p(x))=\deg q(x)+\deg q(x)$. Prove it. This provides a different proof of the fact that $1+x$ is neither a unit nor a zero divisor.

Title: Units/zero divisors in $R[x]$


问:
More edges implies larger eigenvalues of graph Laplacian?
Suppose

$G_1$ and $G_2$ are two connected graphs with the same nodes.

$G_2 \subset G_1$. $G_2$ is the connected subgraph of $G_1$.

I know that $\lambda_2(L(G_2))\leq \lambda_2(L(G_1))$ since $\lambda_2$ represents the connectivity.
($L$ represents the graph Laplacian matrix.)

Can we say the following

$\lambda_i (L(G_2))\leq \lambda_i(G_1)$ for all $i$?

$L(G_1) \succeq L(G_2)$?   (We know graph Laplacian matrix is positive semi-definite.)

I believe 1. implies 2.

If you know any reference of above, please let me know.

Thanks in advance.

Note: By my old and not clear memory, I guess this is true in graph theory. But I forgot where I read it

答:
This is a standard interlacing problem. We will use the well-known interlacing theorem (it is a specific case of Poincaré separation theorem):

If $B$ is a principal submatrix of a symmetric matrix $A$, then the eigenvalues of $B$ interlace the eigenvalues of $A$.

Let $G$ be a graph on $n$ vertices and let $H$ be a spanning subgraph (this is your case as $G_1$ and $G_2$ have same vertex set). I claim that for $i\in\{1,\ldots,n\}$,  $\lambda_i(L(H)) \leq \lambda(L(G))$.

Proof :
Let $N$ be the directed incidence matrix obtained by a random orientation the edges of $G$. Then it is easy to verify that $L = NN^\perp$, and that $NN^\perp$ and $N^\perp N$ have the same eigenvalues.
Removing an edge from $G$ results in removing a column from $N$ and
yields a principal submatrix of $N^\perp N$. Therefore we can apply Cauchy's interlacing theorem and the result holds. $\square$


问:
Graphing $f(2-x)$
Sorry for this very trivial question, but I've become slightly confused by this question. Consider a graph
$y=f(x)$. How would I draw the graph $y=f(2-x)$?

It seems to me that as this is obviously equal to $y=f(-(x-2))$ this should represent the graph being translated $2$ units in the positve $x$ direction and then reflected in the $y$ axis.

Is that true? It doesn't seem to be from the graphs I have plotted using Desmos. If not, please explain why it is incorrect.

Thanks for your help.

EDIT:
I have now slept over my problem and I believe that it lies in the following statement I have been led to believe in class:

The graph of $f(\text{Blah}+a)$ is ALWAYS  a translation of $a$ units of the graph $f(\text{Blah})$ in the negative direction.

More specifically, I thought that the as graph of $f(x+a)$ is a translation of $a$ units of the graph $f(x)$ in the negative direction, then the graph of $f(-x+a)$ is a translation of $a$ units of the graph $f(-x)$ in the negative direction as well. After thinking it over logically however, I now think this is wrong.

This is my reasoning:

Consider $y=f(x+a)$. For a given $y$ value on the $y=f(x+a)$ graph, the $x$ value needed for it must be $a$ smaller than the $x$ value needed if it was just the function $y=f(x)$; hence the graph $y=f(x+a)$ must be the graph of $y=f(x)$ but shifted $a$ units to the negative $x$ direction.

But, if we consider $y=f(-x+a)$: For a given $y$ value on the $y=f(-x+a)$ graph, the $x$ value needed for it must be $a$ bigger than the $x$ value needed if it was just the function $y=f(-x)$; hence the graph $y=f(-x+a)$ must be the graph of $y=f(-x)$ but shifted $a$ units to the positive $x$ direction.

Is my reasoning correct now?  Thanks again for your help.

答:
This is indeed equal to $f(-(x-2))$, but your interpretation of the latter is incorrect.

You identified the correct operations:

Translate 2 units in the positive $x$ direction (replace $x$ with $x-2$).

Reflect in the $y$ axis (replace $x$ with $-x$).

But what order do you have to do these in to get $f(-(x-2))$?

The reasoning added in revision 2 of the question, a few minutes before this edit to my answer, is correct.

Title: Graphing $f(2-x)$


答:
hint

If you know the graph of the curve whose equation is $ y=f(x) $, the graph of $ y=f(-x) $ is the symetric with respect to $ Oy$ axis.

if you know the graph of $ y =g(x)$ , the graph of $ y=g(x-a) $ is got by the translation of vector $ (a,0)$.

Title: Graphing $f(2-x)$

Body:
Sorry for this very trivial question, but I've become slightly confused by this question. Consider a graph
$y=f(x)$. How would I draw the graph $y=f(2-x)$?

It seems to me that as this is obviously equal to $y=f(-(x-2))$ this should represent the graph being translated $2$ units in the positve $x$ direction and then reflected in the $y$ axis.

Is that true? It doesn't seem to be from the graphs I have plotted using Desmos. If not, please explain why it is incorrect.

Thanks for your help.

EDIT:
I have now slept over my problem and I believe that it lies in the following statement I have been led to believe in class:

The graph of $f(\text{Blah}+a)$ is ALWAYS  a translation of $a$ units of the graph $f(\text{Blah})$ in the negative direction.

More specifically, I thought that the as graph of $f(x+a)$ is a translation of $a$ units of the graph $f(x)$ in the negative direction, then the graph of $f(-x+a)$ is a translation of $a$ units of the graph $f(-x)$ in the negative direction as well. After thinking it over logically however, I now think this is wrong.

This is my reasoning:

Consider $y=f(x+a)$. For a given $y$ value on the $y=f(x+a)$ graph, the $x$ value needed for it must be $a$ smaller than the $x$ value needed if it was just the function $y=f(x)$; hence the graph $y=f(x+a)$ must be the graph of $y=f(x)$ but shifted $a$ units to the negative $x$ direction.

But, if we consider $y=f(-x+a)$: For a given $y$ value on the $y=f(-x+a)$ graph, the $x$ value needed for it must be $a$ bigger than the $x$ value needed if it was just the function $y=f(-x)$; hence the graph $y=f(-x+a)$ must be the graph of $y=f(-x)$ but shifted $a$ units to the positive $x$ direction.

Is my reasoning correct now?  Thanks again for your help.


问:
Modulo calculation with inverse and larger divisor
I have this equation $x \equiv (2003^{-1}) \mod 1511$ and am confused with how to try solving it. Since $2003^{-1} < 1511$, you can't use the extended Euclidean algorithm with regards to the inverse. So I'm looking for some tips on where to start with solving this equation. Thanks.

答:
since $gcd(2003,1511)= 1$ it has one and only solution

$\Rightarrow $ $2003x\equiv 1mod1511$

($2003mod1511=492)$$\Rightarrow $

$492x\equiv1mod1511$$\Rightarrow $

find the inverse of $492$ in $mod1511$
and you will see its $777$ so

$777\cdot492x\equiv777mod1511 $ $\Rightarrow $

$x\equiv777mod1511$


答: Denote $g(x)=f(2-x)$ and set $x'=2-x$. What you want is drawing the graph of $g$. Now the points $x$ and $x'$ are symmetric (on the $x$-axis) w.r.t. the point $1$ since $\frac{x+x'}2=1$, and $g(x)=f(x')$. Therefore the graph of $g$ is the symmetric of the graph of $f\,$ w.r.t.  the line $x=1$.

Title: Graphing $f(2-x)$


问:
Prove if a power series is zero on an interval then all coefficients are zero
Suppose the power series $P(x) =  \sum_{n=1}^\infty b_n x^n$  converges for $|x| \leq 1$ and that for some $0<c\le1$ it is given that $$P(x)=0 \quad \forall x \;\text{such that}\;|x| < c$$ Show that $b_n = 0$ for all $n \geq 1$.

What I was able to do was:

$$\lim_{n\to\infty} |b_n x^n|^{1/n} < 1$$ (by the root theorem for convergence)

and as $P(x)=0$ for $|x| < c$, we have:

$$\sum_{n=1}^\infty b_n x^n = 0$$

Now how to get what is asked in the question from these known facts? I am not able to proceed beyond this. Thanks.

答:
Assume otherwise and let $n$ be moinimal with $b_n\ne 0$.
Then $Q(x)=\sum_{k=0}^\infty b_{n+k}x^k$ converges for $|x|<1$ and we have $P(x)=x^nQ(x)$ and hence also $Q(x)=0$ for $0<|x|<c$. by continuity, also $Q(0)=0$, i.e. $b_n=0$ - contradiction.</p>


问:
Distance to a closed subset of $\mathbb{R}^n$
This has been asked heavily in the past, for instance here and here.

Let $x\in X$ and $\operatorname{dist}(x,A)=\inf\lbrace \rho(x,a): a\in A\rbrace$ for any nonempty subset $A$ of X.

It is a fact that if $A$ is compact, then there exists an $a\in A$ such that $\operatorname{dist}(x,A)=\rho(x,a)$.

If $X=\mathbb{R}^n$ and $A$ is closed, then the conclusion of $b$ holds.

I don't see quite well why I need $\mathbb{R}^n$ besides the obvious. If $A$ is also bounded, then $A$ is compact and there is nothing to prove. If $A$ is closed but unbounded, how can I prove the statement? I'm confused because closed subsets  can be quite different. For instance I can choose $\mathbb{N}$, but also $\cup [0,a], a\geq0$. I think both are closed and unbounded, but one has a nonempty set of limit points while the other not.

答:
Let's say $A$ is closed and $\text{dist}(x,A)=d\ge0$. Now, for each $d+\frac{1}{n}$ there has to be some element $y_n$ in $A$ such that $\rho(x,y_n)<d+\frac{1}{n}$.

Consider the sequence $(y_n)_{n\in\Bbb N}$. We have that $\|y_n\|=\rho(y_n,0)\le\rho(y_n,x)+\rho(x,0)=\rho(x,y_n)+\|x\|<d+\frac{1}{n}+\|x\|\le d+1+\|x\|$, so the sequence is bounded. By Bolzano-Weierstrass theorem (which is valid in $\mathbb{R^n}$) it has a convergent subsequence, let's say $(y_{n_k})$ with limit $y$. Then $\rho(y_{n_k},x)<d+\frac{1}{n_k}\xrightarrow{n_k\to\infty}d$, which means $\rho(y,x)=d$, since $A$ is closed and $(y_{n_k})$ is in $A$, so $y\in A$.


问:
Application of first Borel Cantelli lemma
Q) Let $(X,\mu)$ be a positive measure space. Suppose $f,f_1,f_2,\cdots$ are functions in $L^1(X,\mu)$ such that $\sum_n \int_X |f_n - f|d\mu < \infty$. Show that $f_n$ converges $\mu$-a.e. to $f$.

My attempt: Let $E_n = \{x\in X: f_n(x)\not\to f(x)\}$. If $\mu(\limsup E_n)=0$, then $f_n\to f$ a.e. but I am not sure how to use my definition of $E_n$ and the information given in question: $\sum_n \int_X |f_n - f|d\mu < \infty$?

答: $$\mu \{x \in X: |f_n-f| >\epsilon\}$$ $$ \leq \frac1  {\epsilon}  \int_{\{x \in X: |f_n-f| >\epsilon\}} |f_n-f|d\mu$$ $$ \leq \frac1  {\epsilon}  \int |f_n-f|d\mu$$ Now sum over $n$ and use Borel Cantelli Lemma. You see that almost surely for any $\epsilon >0$, $|f_n-f|\leq \epsilon$ for $n$ suffcienltly large.


问:
Is it true that for any integer $c_1, c_2$, we have $\gcd(n_1, n_2, n_3) = \gcd(n_1, n_2, c_1 n_1 + c_2 n_2 + n_3)$ for $n_1, n_2, n_3 \in \mathbb N$?
I got this task to check that if $n_1, n_2, n_3 \in \mathbb{N}$ the $\text{gcd}(n_1, n_2, n_3) =\text{gcd}(n_1, n_2, c_1n_1 + c_2n_2 + n_3)$ is valid for any $c_1, c_2 \in \mathbb{Z}$.

I started the idea saying that if:

$\text{gcd}(n_1, n_2, n_3) = k$, only if:

$k|n_1 \land k|n_2 \land k|n_3$ which means that:
$$
\left\{
\begin{array}{c}
n_1=ke \\
n_2=kf \\
n_3=kg
\end{array}
\right. \quad e, f, g\in \mathbb{N}
$$
Now $gcd(n_1, n_2, c_1n_1 + c_2n_2 + n_3) = k$ only if:
$$ c_1n_1 + c_2n_2 + n_3 = kr, \quad r \in \mathbb{Z}
$$
Substituting the values of $n_1, n_2, n_3$:
$$ k(c_1e + c_2f + g) = kr
$$
Which means that it is valid only if:$$c_1e + c_2f + g = r$$
But now I got stucked, I cannot say if this is valid for any $c_1, c_2$, could anyone help?

答:
$${\rm RHS} =\:\! \overbrace{((n_1,n_2),\,c_1n_1\!+\!c_2n_2\!+\!n_3) = ((n_1,n_2), \color{#c00}{n_3})}^{\textstyle\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \bmod d\!=\!\color{#0a0}{(n_1,n_2)}\!:\, c_1\color{#0a0}{n_1}\!+\!c_2\color{#0a0}{n_2}\!+\!\color{#c00}{n_3}\equiv \color{#c00}{n_3} } =\:\! \rm LHS\qquad$$

via gcd mod reduction: $\,(d,n,\ldots) = (d,\color{#c00}{\bar n},\ldots)\, $ if $\, n\equiv \color{#c00}{\bar n}\pmod{\!d},\,$ and  gcd associative law.

Similarly a gcd stays the same if we replace any argument by any integer congruent to it mod the gcd of (some of) the other arguments. That's essentially  the Euclidean algorithm reduction step.


问:
Triple Integral Over a Ball
How do I compute the following:

$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz$ where $V = \{x^2 +y^2 +z^2 \leq 1\}$?

答:
Your question as it stands is likely incorrect.  Are you really interested in the cases where $n<0$, for instance??

Assuming $n \geq 1/2$, and using symmetry:

$$8 \int\limits_{x=0}^1 \int\limits_{y=0}^{\sqrt{1-x^2}} \int\limits_{z=0}^{\sqrt{1 - x^2 - y^2}} (x^{2 n} + y^{2 n} + z^{2 n})\ dx\ dy\ dz = \frac{12 \pi }{4 n (n+2)+3}$$,

which is the same as @Quanto's later answer.  (Though somehow mine is downvoted.)

Title: Triple Integral Over a Ball


问:
Counting the Number of Vectors of the Form $\{-1, 1\}^n$
Let the $L_{\infty}$-norm of a vector $\textbf{u} = (u_1,\dots,u_n) \in \mathbb{R}^n$ be defined as $\text{max}_{1 \leq i \leq n} |u_i|$. How can we find the total number of ordered triplets $\langle \textbf{u, v, w} \rangle \in ( \, \{-1, 1\}^n ) \, ^ 3$ of $n-$dimensional vectors with the property $L_{\infty} ( \, \textbf{u+v+w} )\, \leq 1$. (Notice that $\langle \textbf{u, v, w} \rangle \in ( \, \{-1, 1\}^n ) \, ^ 3$ means that each of the three vectors is a sequence of $1$'s and $-1$'s.

I came across this problem in my combinatorics textbook, and I have been unable to solve it for a  long time now. I would appreciate any help on this!

答: Well, for a fixed $i$ you either have $|u_i+v_i+w_i|=3,1$ the way to have just $1$ is you have to choose two to place a sign and the remaining one has other sign. This can be done in $3\times 2=6$ ways by choosing which of the $3$ has $1$ or $-1.$ So the answer is $$6^n.$$ by the multiplication principle.


问:
$2^n-3^m=1 , m,n \in \mathbb N =?$
$2^n-3^m=1 , m,n \in \mathbb N =?$ my questions are:

do m,n exist?

are they finitely many $m,n$?

if there are infinitely many is there a way to describe them all?

Same question about $3^n-2^m=1 $, besides $m=n=1$ are there any others?

PS: This is not from number theory or home work, just personal pondering

答:
$\mathsf{Hint}$: Consider mod 3, then mod 4, then mod 8.

 $\displaystyle\begin{array}{lllllll} \rm 2^n-3^m=1 & \implies & \rm (-1)^n\equiv 1\bmod 3 & \iff &  \rm 2\mid n & \iff & \rm n=2u \\ \rm 4^u-3^m=1 & \implies & \rm (-1)^m\equiv-1\bmod 4 & \iff & \rm 2\nmid m & \iff & \rm m=2v+1 \\ \rm 4^u-3\cdot 9^v=1 & \implies & \rm 4^u\equiv 4\bmod 8 & \iff & \rm u=1 &\implies & \rm n=2,m=1\end{array}$

Title: $2^n-3^m=1 , m,n \in \mathbb N =?$


问:
Find parabola given lines it is differentiable with
I have been given two lines and am asked to connect them with a parabola; the resulting shape needs to be continuous and differentiable (at either point where the parabola meets a line, the instantaneous rates of change need to be the same. The project itself is to "design a rollercoaster track" by getting the formula of the lines/parabola making up the track.)

The parabola is/needs to be in y = ax^2 + bx + c formula. I need to find a, b, and c.

So the line on the left, L1, has a slope of 0.7 and meets the parabola at point P, which is at (0, 0)

The line to the right of the parabola, L2, has a a slope of -1.5 and meets the parabola at point Q.

I know point P and point Q are 40 meters apart, so the x value of Q must be 40. I know by the wording of the question that P is higher than Q.

This is my rough sketch of the graph, just to put an image to my words. (it's rough so ignore the units of measurement) Link to image.

I know about derivatives but I know next to nothing about parabolas. I know my c = 0 because c is the Y intercept and the parabola starts at point P which is the origin. But... I know nothing else. I don't even know where to start. Can I get some hints or something? :(

EDIT: I think I graphed this right with the answer I got and the parabola & the second line don't ever intersect

答:
Use the general form $y=ax^2+bx+c$, and its derivative, $y'=2ax+b$ with the $(x,y)$ coordinates and slopes at the required points. As stated, you know $c=0$. You must find $a$ and $b$.

Solution assuming that $Q(x_Q,y_Q)$ lies on the circle with radius 40 centered at $P(0,0)$:

$f(x)=ax^2+bx$

$f'(x)=2ax+b$

$P$ lies on the parabola and $m_P=0.7$ so $f'(0)=b=0.7$

Now we have

$f(x)=ax^2+0.7x$

$f'(x)=2ax+0.7$

Q lies on both the parabola and the circle. We also know $m_Q=-1.5$.

(1)   $y_Q=ax_Q^2+0.7x_Q$

(2)   $x_Q^2+y_Q^2=40^2$

(3)   $-1.5=2ax_Q+0.7$

This is a system of three independent equations with three unknowns, $a$, $x_Q$, and $y_Q$. Use substitution to first eliminate $x_Q$ and then again to eliminate $y_Q$. Now you can solve for $a$.


问:
Which root systems admit a proper root subsystem with full span?
Let $\Phi$ be an irreducible root system of $\mathrm{rank}(\Phi) = n$. I am allowing the case where $\Phi$ is not reduced.

Say that a subset $\Psi \subset \Phi$ is a root subsystem if it is a root system in its own right. For example, the intersection of $\Phi$ with any proper linear subspace of $\mathrm{span}(\Phi)$ is a root subsytem.

I would like to list all the examples of pairs $(\Phi,\Psi)$ where $\Psi$ is a full rank root subsystem of $\Phi$, namely $\mathrm{span}(\Psi) = \mathrm{span}(\Phi)$.

An obvious family of examples includes $(BC_n,B_n)$ and $(BC_n,C_n)$.
Possibly there are other examples? How  to systematically find all such pairs?

A textbook reference would also be appreciated.

答:
Here is a strategy to find such root subsystems for irreducible (reduced) root systems:

Take the extended Dynkin diagram of the root system you're looking at, and remove one of the vertices (and the edges touching it). You are left with a Dynkin diagram of the same rank as the original one, and this gives a root subsystem.

This gives the following proper root subsystems of full rank:

For type $A_n$: None

For type $B_{n\ge 3}$: $D_i \times B_{n-i}$ (for $ 2\le i \le n$)

For type $C_{n\ge 2}$: $C_i \times C_{n-i}$ (for $ 1\le i \le n-1$)

For type $D_{n\ge 4}$: $D_i \times D_{n-i}$ (for $2 \le i \le n-2$)

For type $E_6$: $A_1 \times A_5$ and $A_2 \times A_2 \times A_2$

For type $E_7$: $A_1 \times D_6$, $A_2 \times A_5$, $A_3 \times A_3 \times A_1$, and $A_7$

For type $E_8$: $A_1 \times E_7$, $A_2 \times E_6$, $A_3 \times D_5$, $A_4 \times A_4$, $A_5 \times A_1 \times A_2$, $A_7 \times A_1$, $D_8$ and $A_8$

For type $F_4$: $A_1 \times C_3$, $A_2 \times A_2$, and $B_4$

For type $G_2$: $A_1 \times A_1$ and $A_2$.

Added: In the first version of this answer, I thought this (with iteration, see below) would be enough, but it is not. For example, the inclusion $D_n \subset C_n$ does not show up yet. In a way, this procedure only "sees" subsystems which contain the highest root, which is always a long root. In the case of BCFG root systems i.e. those with two different root lengths, we also have to look at an extension of the Dynkin diagram with the (negative of the) highest short root, and then erase one of the other vertices. The diagrams look like this:

and we get the following possibilities which have not already been listed:

For type $B_{n\ge 3}$: $B_i \times B_{n-i}$ (for $ 1\le i \le n$)

For type $C_{n\ge 2}$: $D_i \times C_{n-i}$ (for $ 2\le i \le n$)

For type $F_4$: $C_4$, $A_1 \times A_3$, and $B_3 \times A_1$

I have a feeling, but no rigorous proof yet, that this plus the following iteration procedure exhausts all possibilities.

One word about why this procedure with the Dynkin diagram first extended by the negative of the highest, or highest short, root works: Well, what we are doing with such extended Dynkin diagrams is writing down the angle and length relations between some roots. As soon as he resulting diagram is a union of standard (non-extended) Dynkin diagrams, we know by the classification that the roots making up the vertices do generate a root system of the respective type. Now in both cases, when we first extend the Dynkin diagram with the negative -of-highest(-short) root vertex, and then remove a different vertex, what remains is a classical Dynkin diagram; meaning that the roots corresponding to the remaining vertices (all but one of them from the original basis, one the negative of the highest (short) root) will generate such a classical root system. And it is obviously of full rank and contained in our original root system.

Now in a second step, one can iterate the procedure until one arrives at all factors being of type $A$ (note that $A_1 =B_1 =C_1$, $B_2=C_2$, $D_2 = A_1 \times A_1$ and $D_3 =A_3$). E.g. within $F_4$ we further have
$$A_1 \times C_3 \supset A_1\times A_1 \times \underbrace{C_2}_{=B_2} \supset A_1 \times A_1 \times \underbrace{A_1 \times A_1}_{C_1 \times C_1}$$
as well as
$$B_4 \supset D_4 \qquad ( \supset \underbrace{A_1 \times A_1}_{D_2} \times \underbrace{A_1 \times A_1}_{D_2})$$
and also
$$C_4 \supset C_2 \times C_2.$$

Note that the only root systems of rank $n$ which do not contain an $n$-fold product of
$A_1$ (i.e., $n$ mutually orthogonal roots) are $A_{n\ge 2}$, $D_{ n \text{ odd}}$ and $E_6$, which matches well-known results about $-id$ being in the Weyl group or not.

This somewhat matches the (far, far more elaborate) tables at the end of R. W. Carter: Conjugacy Classes in the Weyl Group (Comp. Math. 25 no.1 (1972) p. 1-59), although in those tables, among the "$\Gamma$"'s for the exceptional root systems, only the subsystems from my first approach are listed. I am sure there is a reason for this in that paper.

Finally, to include the non-reduced and the reducible case from here should be easy.


问:
Proof by contradiction - cubic equation has a root in the interval (N,N+1)
The question is you have to prove by contradiction

The equation $x^3-3x^2-2x+3=0$ has a root in the interval $(N,N+1)$ where N is an integer. Prove that there are three possible values of $N$

I don't understand how to get started with this. $(N,N+1)$I think this means that the root will be between two consecutive integers so wont be a whole number so I cant use factor theorem? I am also stuck on what is the initial contradicting assumption is, is it "there are not three possible values of N" ?

Thanks in advance

答:
Let $f(x) = x^{3} - 3x^{2} -2x+3$.

Note that $f(3) = -3$ and $f(4) = 11$, so $f(x)$ has a zero in $(3,4)$ by the Intermediate Value Theorem..

Similarly, $f(0) = 3$ and $f(1)=-1$, so $f(x)$ has a zero in $(0, 1)$. Also, $f(-2) = -13$ and $f(-1) = 1$, so $f(x)$ must have a zero in $(-2,-1)$.

Thus, $\boxed{f(x)\text{ has zeroes in }(-2,-1),(0,1),\text{ and }(3,4).}$


答: Or: Monotone convergence shows that $$\int\sum |f_n-f|=\sum\int|f_n-f|<\infty.$$ Hence $\sum|f_n-f|<\infty$ almost everywhere, so $f_n\to f$ a.e.

Title: Application of first Borel Cantelli lemma

Body:
Q) Let $(X,\mu)$ be a positive measure space. Suppose $f,f_1,f_2,\cdots$ are functions in $L^1(X,\mu)$ such that $\sum_n \int_X |f_n - f|d\mu < \infty$. Show that $f_n$ converges $\mu$-a.e. to $f$.

My attempt: Let $E_n = \{x\in X: f_n(x)\not\to f(x)\}$. If $\mu(\limsup E_n)=0$, then $f_n\to f$ a.e. but I am not sure how to use my definition of $E_n$ and the information given in question: $\sum_n \int_X |f_n - f|d\mu < \infty$?


答:
Due to symmetry

$$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz
= 3\iiint _V z^{2n} dxdydz\\
= 3\int_0^{2\pi}\int_0^\pi \int_0^1 r^{2n}\cos^{2n}\theta \>r^2\sin\theta drd\theta d\phi
 = \frac{12\pi}{(2n+3)(2n+1)}
$$

Title: Triple Integral Over a Ball

Body:
How do I compute the following:

$\iiint _V (x^{2n} + y^{2n} +z^{2n}) dxdydz$ where $V = \{x^2 +y^2 +z^2 \leq 1\}$?


问:
Counting Problem: Briefcases
Suppose you and a friend are presented with nine identically-looking briefcases, three of which have $100 in them while the rest are empty. You both take turns picking briefcases (with the first person choosing 5 and the second choosing 4), and only at the end are the briefcases opened. I have done 1-4 parts of these HW questions but am unsure how to do the last:

If you go first, what is the probability that you pick an empty briefcase on your first turn.

Say you went first and picked an empty briefcase. What is the probability that your friend also picks an empty briefcase on their first turn, given that you also picked an empty one your first turn.

Say you went first and picked a briefcase with 100 dollars. What is the probability that your friend picks an empty briefcase on their first turn, given that you also picked one with 100 dollars on your first turn.

If the game were to end after each person has only one go, show that the probability that you choose a briefcase with 100 dollars is the same regardless of whether you start or not.

If the game were to end once each briefcase was picked, would it be better to start or go second?

EDIT: Goal of the game is to make as little money as possible since 'gambling is bad'.

My attempt:

The answer is the no. of ways you can pick an empty briefcase divided by the no. of ways you can choose a briefcase. This is 6C1 (since there are 6 empty briefcases) divided by 9C1 which is equal to 2/3.

I believe the answer to this is 2/3 (from part 1) multiplied by 5C1 (since now there are 5 empty briefcases remaining and the friend picks one of these) divided by 8C1 (since now there are only 8 remaining cases to choose from). This is approx. 0.4167.

I believe this is $\frac{3C1}{9C1}\cdot\frac{6C1}{8C1}$ since the first person must pick one of the three money briefcases out of a total of 9, while the second person must choose one of the 6 empty briefcases out of the 8 remaining. This is 0.25.

The way I thought to approach this was by considering four different events, where:

Event A: You go first and get 100 dollars, then friend chooses empty.

Event B: You go first and get 100 dollars, then friend also gets 100 dollars.

Event C: Your friend goes first and gets 100 dollars, then you also get 100 dollars.

Event D: Your friend goes first and chooses empty, then you get 100 dollars.

Therefore, if you go first the probability of getting 100 dollars is the probability of event A plus the probability of event B. If you go second, it's the probability of event C plus the probability of event D. Both these expressions are $\frac{3C1\cdot 6C1}{9C1 \cdot 8C1}+\frac{3C1\cdot 2C1}{9C1 \cdot 8C1}$ which is 1/3.

This part I am confused about how to approach, since there are many different sequences possibly of choosing 6 empty briefcase and 3 with 100 dollars.

Any advice would be greatly appreciated.

答:
If you go first, the number of "money" suitcases you get follows a hypergeometric distribution, since your five suitcases are drawn without replacement from a population of $3$ "money" suitcases and $6$ empty ones.

$$P(X=k) = \begin{cases}
\frac{\binom{3}{0} \binom{6}{5}}{\binom{9}{5}} = \frac{6}{\binom{9}{5}} & k=0
\\
\frac{\binom{3}{1} \binom{6}{4}}{\binom{9}{5}} = \frac{45}{\binom{9}{5}}
& k = 1
\\
\frac{\binom{3}{2} \binom{6}{3}}{\binom{9}{5}} = \frac{60}{\binom{9}{5}}
& k = 2
\\
\frac{\binom{3}{3} \binom{6}{2}}{\binom{9}{5}} = \frac{15}{\binom{9}{5}}
& k = 3
\end{cases}$$

If you go second, the corresponding probabilities will be in reverse order. You can see that the player that goes first tends to get more money suitcases.

Title: Counting Problem: Briefcases


问:
Can we see both sides of faces of 4D tesseract from 3D?
To see the whole surface of a cube it is sufficient to have two viewpoints above two opposite vertices of a cube. We can see three faces of a cube from the first viewpoint and other three faces from the second viewpoint.

Suppose each face is marked by numbers from 1 to 6 on outside of faces and from 7 to 12 on inside of faces. Also suppose faces are not transparent.

By rotating the cube we can only see numbers from 1 to 6 from the two viewpoints because faces are not transparent.

Analogously now consider we have a tesseract in 4D.

Its 24 faces are marked with numbers from 1 to 24 on one side of faces and with numbers from 25 to 48 on opposite sides of faces. Faces are not transparent.

Now my question is:

If the tesseract is rotating (in all possible angles of rotation about all three planes) in 4D, can an observer in 3D see all 48 numbers or only 24 numbers?

(Observer has the same two viewpoints as with a cube.)

Seeing the animation on the Wikipedia link, my intuition tells me that an observer can see both sides of the faces even though they were not transparent.

答:
The "faces" of a 3-cube (i.e., the things you get by setting one coordinate to $\pm 1$) are cubes, and there are eight of them. They each have an "inside" and an "outside" and of the 16 numbered sides, you can see only 8 from your two exterior viewpoints.

The 2D-components of a 3-cube are sort of analogous to edges of a 2-cube in 3-space. There are indeed 24 of those. But they don't have an "inside" and an outside, just as a line-segment in 3-space doesn't have "sides".

But just as the edges of a cube in 3-space, if you think of them as very thin tubes, have 1/4 inside the cube and 3/4 outside, so you could label them with two different colors, the 2D-components of a 3-cube in 4-space can also be labelled similarly, and indeed, you can "see" only 24 of the 48 labels. By "See", I mean that you can pick a point on one of the labelled parts of the tube, and connect it by a line segment to the viewpoint, with no portion of the interior of the line segment intersecting the 3-cube.

It's often easiest to understand these things by working up from lower dimensions. Let's start with a square (a "1-cube in 2-space"). 4 vertices (which are 0-dimensional --- they're just points) and 4 edges (each 1-dimensional, a line-segment). The line-segments could be labelled with an "inside" and an "outside", but if we draw the top row of this figure:

you can see what I've done: I've colored the edges blue, the vertices red, and I've made a smaller and larger copy of the square, and filled in the "corners" with red arcs. These arcs are kind of "fattened up" versions of the vertices. And now, where we used to have one edge with two sides, we've got two parallel edges; where we used to have a vertex (where it was hard to talk about "sides") we've got two parallel arcs, serving as proxies for the "inside" and "outside". From the exterior of the square, we can see the four "outside" blue edges and red arcs, but none of the "inside" ones.

When we move to a cube (a 2-cube in 3-space) we have
6 vertices (zero-dimensional)
12 edges (1-dimensional)
6 faces (2-dimensional)

I've drawn only three of the faces, to save myself time and drawing problems. Again I "blow up" the shape and make two copies of each top-dimensional thing; I've drawn the "inside" copy in light blue. Now for each edge in the left-most picture, I've drawn a quarter-tube in green, and for each vertex in the left-most picture, I've ... skipped drawing a 1/8-sphere that would fill in the little triangular hole you see. But in this "fattened" version, you can see the outside of the vertex, of three edges, and of three faces. Actually, if I'd drawn them, you'd see part of the outside of 6 more edges. So 3 of the edges are visible ONLY from this vantage point, 3 are visible ONLY from the opposite vantage point, and 6 are partly visible from both. But NONE of the interior edges are visible, etc.

When you get to 4-space, there are

12 vertices
24 "edges"
24 "faces"
12 "solid" faces (3-dimensional things)

And you can blow up things in much the same way, and you'll find that the outsides of all these things are visible (given two opposing viewpoints on opposite ends of a main diagonal of the hypercube), but that the "insides" are not.


问:
Using Induction to prove a squared summation statement
I have the following problem:

Let $p_{1}, . . . , p_n$ be positive real numbers such that $\sum_{k=1}^{n} p_k = 1$.

Prove:

$\sum_{k=1}^{n} p_k^2 \geq \frac{1}{n}$ and

$\sum_{k=1}^{n} \frac{1}{p_k} \geq n^2$.

I am not sure how to approach this problem.

Trying to prove the first statement: I tried squaring the summation term $\sum_{k=1}^{n} p_k$ and got $\sum_{k=1}^{n} p_k^2 + \sum_{k\neq j}^{n} p_k p_j$

So $\sum_{k=1}^{n} p_k^2 = 1-\sum_{k\neq j}^{n} p_k p_j$

Subbing into the statement I'm trying to prove: $1-\sum_{k\neq j}^{n} p_k p_j \geq \frac{1}{n}$

I can't prove this as I don't know how many terms are in $\sum_{k\neq j}^{n} p_k p_j$, and am not sure how to figure it out.

Any help is much appreciated. Thank you!

答:
The Cauchy-Schwarz Inequality: Let $A=(\sum_{k=1}^np_k^2)(\sum_{k=1}^nq_k^2).$ Let $B= (\sum_{k=1}^n p_kq_k)^2.$ Let $C=\sum \sum_{1\le j<k\le n}(p_jq_k-p_kq_j)^2.$

Then $A\ge B$ because $A-B=C\ge 0.$

In your Q let every $q_k=1.$

Geometrically, in $\Bbb R^n,$ if $O$ is the origin and $(p_1,...,p_n)=P\ne O\ne Q=(q_1,...,q_n)$ then $A=|OP|^2\cdot |OQ|^2$ and $B=|OP|^2\cdot |OQ|^2\cos^2 POQ$  and $C=|OP|^2\cdot|OQ|^2\sin^2 POQ.$


答:
Hint:

When $n$ is odd, $2^n-1=(3-1)^n-1$ cannot be divided by $3$.

When $n$ is even, say $n=2k$, then $2^n-1=(2^k-1)(2^k+1)$. Note that $2^k-1$ and $2^k+1$ are coprime to each other.

Title: $2^n-3^m=1 , m,n \in \mathbb N =?$

Body:
$2^n-3^m=1 , m,n \in \mathbb N =?$ my questions are:

do m,n exist?

are they finitely many $m,n$?

if there are infinitely many is there a way to describe them all?

Same question about $3^n-2^m=1 $, besides $m=n=1$ are there any others?

PS: This is not from number theory or home work, just personal pondering


答:
Your answer in #2 is wrong.

Say you went first and picked an empty briefcase. What is the probability that your friend also picks an empty briefcase on their first turn, given that you also picked an empty one your first turn.

You are required to assume that you have already gone first and picked an
empty briefcase.  This means that there are 8 briefcases left of which only
5 are empty.

Answer is

$$\frac{5}{8}.$$

For the same reason as above, the answer to #3 is

$$\frac{6}{8}.$$

#4 is tricky and can be attacked in two different ways.

The pedestrian approach is that

your chance of briefcase with money, if you go first is $\frac{3}{9} = \frac{1}{3}.$

if you go second there are two possibilities - either your friend
picked an empty briefcase, and then you picked one with money, or your friend picked a briefcase
with money and then you also picked one with money.

The combined probabilities here are
$\left(\frac{6}{9} \times \frac{3}{8}\right) ~+~
\left(\frac{3}{9} \times \frac{2}{8}\right) ~=~ \frac{24}{72} = \frac{1}{3}.$

The more elegant analysis is to reason that picking a briefcase is mathematically
equivalent to being handed a briefcase at random.  If a moderator picks out
two briefcases at random, one for you, and one for your friend, the chance that
your briefcase contains money is unaffected by whether the moderator hands
you your briefcase before or after the moderator hands your friend his briefcase.

Edit

I misinterpreted

If the game were to end once each briefcase was picked, would it be better to start or go second?

If you go first, you are going to be given 5 briefcases and your friend will only be given 4 briefcases.  This is equivalent to pretending that there are 10 briefcases, instead of 9, and that the 10th briefcase is forced to be empty.

Therefore, if you go first, you will have 5 briefcases, any of which may have money.  If you go second, one of your briefcases (the imaginary 10th briefcase) is forced to be empty.

Therefore, the person who goes second can expect to get less money.

Title: Counting Problem: Briefcases

Body:
Suppose you and a friend are presented with nine identically-looking briefcases, three of which have $100 in them while the rest are empty. You both take turns picking briefcases (with the first person choosing 5 and the second choosing 4), and only at the end are the briefcases opened. I have done 1-4 parts of these HW questions but am unsure how to do the last:

If you go first, what is the probability that you pick an empty briefcase on your first turn.

Say you went first and picked an empty briefcase. What is the probability that your friend also picks an empty briefcase on their first turn, given that you also picked an empty one your first turn.

Say you went first and picked a briefcase with 100 dollars. What is the probability that your friend picks an empty briefcase on their first turn, given that you also picked one with 100 dollars on your first turn.

If the game were to end after each person has only one go, show that the probability that you choose a briefcase with 100 dollars is the same regardless of whether you start or not.

If the game were to end once each briefcase was picked, would it be better to start or go second?

EDIT: Goal of the game is to make as little money as possible since 'gambling is bad'.

My attempt:

The answer is the no. of ways you can pick an empty briefcase divided by the no. of ways you can choose a briefcase. This is 6C1 (since there are 6 empty briefcases) divided by 9C1 which is equal to 2/3.

I believe the answer to this is 2/3 (from part 1) multiplied by 5C1 (since now there are 5 empty briefcases remaining and the friend picks one of these) divided by 8C1 (since now there are only 8 remaining cases to choose from). This is approx. 0.4167.

I believe this is $\frac{3C1}{9C1}\cdot\frac{6C1}{8C1}$ since the first person must pick one of the three money briefcases out of a total of 9, while the second person must choose one of the 6 empty briefcases out of the 8 remaining. This is 0.25.

The way I thought to approach this was by considering four different events, where:

Event A: You go first and get 100 dollars, then friend chooses empty.

Event B: You go first and get 100 dollars, then friend also gets 100 dollars.

Event C: Your friend goes first and gets 100 dollars, then you also get 100 dollars.

Event D: Your friend goes first and chooses empty, then you get 100 dollars.

Therefore, if you go first the probability of getting 100 dollars is the probability of event A plus the probability of event B. If you go second, it's the probability of event C plus the probability of event D. Both these expressions are $\frac{3C1\cdot 6C1}{9C1 \cdot 8C1}+\frac{3C1\cdot 2C1}{9C1 \cdot 8C1}$ which is 1/3.

This part I am confused about how to approach, since there are many different sequences possibly of choosing 6 empty briefcase and 3 with 100 dollars.

Any advice would be greatly appreciated.


问:
Show that $ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1 $

If $f(x, y, z) = 0$ and $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}  \neq 0$, show that

$$ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1  $$

I understand that posting questions without showing your work is frowned upon. I just couldn't come up with anything.

答:
$$\begin{array}{l}given:\left\{ \begin{array}{l}f\left( {x,y,z} \right) = 0\\z = z\left( {x,y} \right)\end{array} \right.\\\left\{ \begin{array}{l}dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy\\\left\{ {dx = \frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right.\\ = \frac{{\partial z}}{{\partial x}}\left( {\frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right) + \frac{{\partial z}}{{\partial y}}dy\\ = \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right)dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\\\left\{ {\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right) = 1} \right.\\ = dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\end{array} \right.\\ \Downarrow \\\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}} = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} =  - \frac{{\partial z}}{{\partial y}}\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}}\frac{{\partial y}}{{\partial z}} =  - 1\end{array}$$

This is a note from a physics teacher. (Not my work, I rearrange it.)

I dont guarantee this is correct, but I think it is.


问:
Odds of Choosing Items out of a bag
Let's say there is a bag filled with a myriad of balls, labeled 1 through 7. There are an infinite number of each ball (i.e. odds don't diminish when one is removed). Edit A better phrasing than infinite-balls would be balls being replaced. /Edit

I get to draw 12 balls out of the bag.

What are the chances I'd end up with at least 1 of each number (and obviously some repeat pulls)? What are the chances I'd end up with all-but-1 of the 7 numbers (i.e. $\frac{6}{7}$)? Or $\frac{5}{7}$, or any lower combination?

(I'm not particularly gifted with Combinatorics. I was playing with smaller values, like 3-draws of 2-different numbers having $2^3$ possible options and $\frac{6}{2^3}$ of them fulfilling the condition. I tried 3-draws of 3-different numbers having $\frac{6}{3^3}$ fulfilling the condition. I tried 4 draws of 4-draws of 4-different numbers and got $\frac{24}{4^4}$ fulfilling the condition. Needless to say, I couldn't see a readily repeatable pattern to use on higher values :-( )

答:
You do need Stirling numbers (second kind) for this. The Stirling number $S(n,k)$ denotes the number of ways to partition an $n$ element set into $k$ non-empty subsets. To make discussion easier, let me replace the bad by a $7$-sided dice, and drawing from the bag by throwing the dice. These are equivalent for our purposes.

Let us first consider the chances of getting each number at least once. The total space will consist of all possible $12$-element sequences of $1,2,\dots,7$ - there are $7^{12}$ of these. How many of these correspond to the event we are interested in? Well, each such event gives a partition of $1,2,\dots,12$ into $7$ non-empty sets (namely, the set of throws when different numbers came out). Each such partition corresponds to precisely $7!$ events (given a partition, there are that many ways to assign numbers $1,2,\dots 7$ to different subsets in this partition). Hence, there are the total of $7! \cdot S(12,7)$ sequences of throws with each result occurring at least once. The probability you are looking for is:
$$ \frac{7! \cdot S(12,7)}{7^{12}}$$
Mathematica tells me this is: $\frac{9218880}{40353607} \simeq 0.22... $

For the second question of getting $6$ out of $7$ possible results, note that any such combination of throws corresponds to a partition of $1,2,\dots,12$ into $6$ subsets, and for  each partition there are $7!$ combinations of throws. Hence the probability is
$$ \frac{7! \cdot S(12,6)}{7^{12}}$$
Note that for $5$ out of $7$, the same goes through, except now each partition corresponds to ${7 \choose 2} \cdot 5! = \frac{7!}{2!}$ combinations of throws. I hope you should be able to generalise it without much problem.

P.S. The reason why you couldn't see a pattern is probably that there are Stirling numbers involved. The reason why they deserve a separate name is that they aren't easily expressible in terms of simpler expressions, so a pattern would not be obvious.

Title: Odds of Choosing Items out of a bag


问:
extension of a trace on the positive cone of a $*$-algebra
Let $A$ be a $*$-algebra. Denote $A^+$ by the positive self-adjoint elements in $A$. If $\tau$ is a tracial state of $A^+$, can we construct a tracial state $\tilde{\tau}$ on $A$ such that $\tilde{\tau}|_{A}=\tau$?

答:
$\newcommand{\Asa}{A_{\text{sa}}}\newcommand{\tt}{\tilde \tau }$There are few strong results on positivity that apply to  general $^*$-algebras, so let me restrict to C*-algebras
here.  Among the crucial tools at our disposal is  the fact that every self-adjoint element decomposes  as a
difference os positive elements.

Definition.  Given a C$^*$-algebra $A$, a   state  on $A_+$ is a function $\tau :A_+\to {\mathbb R}_+$ that is additive ($\tau (a+b)=\tau (a)+\tau (b)$)
and   positively homogeneous $(\tau (\lambda a)=\lambda \tau (a)$ for $\lambda \geq 0$).  If moreover $\tau (a^*a)=\tau (aa^*)$ for all $a$ in $A$, then $\tau $ is called a tracial
state.

Theorem.  Any state $\tau $ on $A_+$ admits a unique linear extension $\tt$ to $A$.  If $\tau $ is moreover a tracial state, then $\tt$ is a trace.

Proof. Denote by $\Asa$ the set of all self-adjoint elements in $A$.   For each $a$ in $\Asa$, write $a=b-c$, with
$b,c\in  A_+$, and set
$$
  \varphi (a) = \tau (b)-\tau (c).
  $$
If $b'$ and $c'$ satisfy the same properties as above then
$$
  b-c=b'-c' \quad\Rightarrow\quad
  b+c'=b'+c  \quad\Rightarrow\quad
  \tau (b)-\tau (c) =    \tau (b')-\tau (c'),
  $$
so we see tht $\varphi $ is well defined and it is easy to see that it becomes a real-linear functional
$$
  \varphi :\Asa\to \mathbb R.
  $$

Now,  given any  $a$ in $A$,  observe that   both
$$
  \text{Re}(a):= {a+a^*\over 2}, \quad \text {and} \quad   \text{Im}(a):= {a-a^*\over 2i}
  $$
lie in $\Asa$, so we may define
$$
  \tt (a) = \varphi (\text{Re}(a)) + i \varphi (\text{Im}(a)),
  $$
and one may then prove that   $\tt $ is a complex linear functional on $A$ extending $\tau $.  The uniqueness of
$\tt $ is clear.

Assuming that $\tau $ is a tracial state, we need to show that $\tt (ab)= \tt (ba)$, for every $a$ and $b$ in $A$.

For this, observe that if $u$ is a unitary element in the multiplier of $A$, and $a\in A_+$, then
$$
  \tt(u^*au) =   \tt(u^*a^{1/2}a^{1/2}u) =  \tt(a^{1/2}uu^*a^{1/2})=  \tt(a).
  $$
By linearity we therefore also have that
$$
  \tt(u^*au) =    \tt(a),
  $$
for every element $a$ in $A$. It $u$ is again assumed to be a unitary  multiplier, then
$$
  \tt(ua) =    \tt(u^*(ua)u) =   \tt(au).
  $$
Any element $b$ in $A$ may be written as the linear combination of four unitary multipliers,  so it follows that
$$
  \tt(ba) = \tt(ab).
  $$
QED.