问:
An intuitive proof about the expected number of trials it takes for an event to happen
Imagine there is a fair spinner that can land on any number from $1$ through $100$. It is not too difficult to verify that if you spin this spinner $100$ times, then the expected number of $5$'s, for example, is equal to $1$. This can be done using linearity of expectation. Let $X_i$ be an indicator variable that is equal to $1$ if the $i$th spin results in a $5$ and is equal to $0$ if the spinner does not land on a $5$. Then let $X$ equal the number of $5$'s that come up after $100$ spins. We know that $X=\sum X_i$, and so
\begin{align}
\mathbb{E}(X) &= \mathbb{E}(X_1)+\mathbb{E}(X_2)+\mathbb{E}(X_3)+\ldots+\mathbb{E}(X_{100}) \\
&= \underbrace{0.01 + 0.01 + 0.01 + \ldots + 0.01}_{\text{$100$ times}} \\
&= 1 \, .
\end{align}
I find this proof to be fairly easy to understand. It also means that the idea of expectation becomes intuitive to me. However, there is another conception of expectation that I find more difficult to get my head around. Imagine you spin the spinner as many times as are needed to get a $5$. Let's denote the number of spins as $Y$. It turns out that $\mathbb{E}(Y)$, just like $\mathbb{E}(X)$, is equal to $100$. However, I am unable to find a satisfying proof of this fact. I know you can use an infinite series to compute $\mathbb{E}(Y)$, but is there a proof like the one shown above that doesn't require much algebraic manipulation or number crunching? If it is possible to use linearity of expectation, then I would be very interested to see this.

I don't mind if the proof involves infinite series; however, is it possible to improve upon this?

\begin{align}
\sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1} \cdot \frac{1}{100} \cdot n &= \frac{1}{100} \sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1} \cdot n \\[6pt]
&= \frac{1}{100} \sum_{n = 1}^{\infty} (n-1)\left(\frac{99}{100}\right)^{n-1} + \left(\frac{99}{100}\right)^{n-1}\\[6pt]
&= \frac{1}{100} \left(\sum_{n = 1}^{\infty} (n-1)\left(\frac{99}{100}\right)^{n-1} + \sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1}\right)\\[6pt]
&= \frac{1}{100} \left(\frac{99/100}{(1-99/100)^2} + \frac{1}{1-99/100}\right)\\[6pt]
&= \frac{1}{100} \left(9900 + 100\right)\\[6pt]
&= 100
\end{align}

(This proof uses the 'Gabriel's staircase’.)

答:
Well, it's easy enough with a series:

$$\sum_{i=1}^{\infty} i \times \mathcal{P}(\text{it takes exactly $i$ spins to get a $5$}) = \sum_{i=1}^{\infty} i \times \left(\frac{99}{100}\right)^{i-1} \times \frac{1}{100}$$

This is $\frac{1}{100} f'(0.99)$ where $$f(x) = \sum_{i=1}^{\infty} x^i = \frac{1}{1-x} - 1$$
That is readily calculated to be $100$.

I don't know if that's "not much number crunching" - it doesn't seem too onerous to me, though I will certainly admit that it is practically intuition-free!


问:
Let $|f(x)-f(y)| \leqslant (x-y)^2$ for all $x,y\in \mathbb{R}.$ Show that $f$ is a constant.

Let $|f(x)-f(y)| \leqslant (x-y)^2$ for all $x,y\in \mathbb{R}.$ Show that $f$ is a constant.

This seemed quite straightforward just using the definition of the derivative, but I've ran into some weird issue. I have that $$|f(x)-f(y)| \leqslant (x-y)(x+y) \Longrightarrow \frac{|f(x)-f(y)|}{x-y} \leqslant x+y$$

from where I have that $\lim_{x\to y} x+y = 2y$.

If the problem assignment would instead had $|f(x)-f(y)| \leqslant |(x-y)^2|$ I would have gotten $$|f(x)-f(y)| \leqslant |x-y||x+y| \Longrightarrow \frac{|f(x)-f(y)|}{x-y} \leqslant |x-y|$$

from where $\lim_{x\to y} |x-y| = 0$. And $f$ would be a constant. What might I be missing here?

答:
We have
$$
|f(x)-f(y)| \leqslant (x-y)^2
$$
so
$$
\frac{|f(x)-f(y)|}{|x-y|} \leqslant |x-y|
$$
so
$$
\frac{|f(x)-f(y)|}{|x-y|} \to_{y \to x} 0.
$$
Thus $f'=0$ and $f$ is constant.

ContentLicense: CC BY-SA 4.0


问:
Solution verification: when is the irreflexive closure of a relation defined?
I am working on a problem that states the following:

When is it possible to define the irreflexive closure of a relation $R$, that is, a relation that contains $R$, is irreflexive, and is contained in every irreflexive relation that contains $R$?

Attempt:
First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) \notin R$ for all $a \in A$.

Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)\notin S$ for all $a \in A$ and $R \subseteq S$ which implies that $(a,a) \notin R$ for all $a \in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.

Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.

Question: is my approach correct?

答:
Your answer is perfectly correct! After all, since the closure $S$ must be irreflexive, then $(a,a) \not \in S$ $\forall a \in A$, but that in turn enforces the same restrictions on subsets of $S$, namely the relation we're trying to close. Thus the closure only exists for relations which themselves are already irreflexive.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


答: For any $x \in \mathbb{R}$, you have: $|f(x+h) - f(x)| \le ((x+h) - x)^2=h^2=|h|^2\implies \left|\dfrac{f(x+h) - f(x)}{h}\right|\le |h|\implies \displaystyle \lim_{h \to 0} \left|\dfrac{f(x+h) - f(x)}{h}\right| = 0 \implies \displaystyle \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = 0\implies f'(x) = 0\implies f = \text{constant}.$

Body:
Let $|f(x)-f(y)| \leqslant (x-y)^2$ for all $x,y\in \mathbb{R}.$ Show that $f$ is a constant.

This seemed quite straightforward just using the definition of the derivative, but I've ran into some weird issue. I have that $$|f(x)-f(y)| \leqslant (x-y)(x+y) \Longrightarrow \frac{|f(x)-f(y)|}{x-y} \leqslant x+y$$

from where I have that $\lim_{x\to y} x+y = 2y$.

If the problem assignment would instead had $|f(x)-f(y)| \leqslant |(x-y)^2|$ I would have gotten $$|f(x)-f(y)| \leqslant |x-y||x+y| \Longrightarrow \frac{|f(x)-f(y)|}{x-y} \leqslant |x-y|$$

from where $\lim_{x\to y} |x-y| = 0$. And $f$ would be a constant. What might I be missing here?


答: We needn't assume $f$ is differentiable, or even continuous. Just use the triangle inequality:$$|f(y+h)-f(y)|\le\sum_{k=1}^n|f(y+hk/n)-f(y+h(k-1)/n)|\le n(h/n)^2=h^2/n$$for all $n\in\Bbb N$, so $f(y+h)-f(y)=0$.

Title: Let $|f(x)-f(y)| \leqslant (x-y)^2$ for all $x,y\in \mathbb{R}.$ Show that $f$ is a constant.


问:
An Olympiad geometry question on area lemma.
In ABC, E, F, G are points on AB, BC, CA respectively such that $\frac{AE}{EB} = \frac{BF}{FC} = \frac{CG}{GA} = \frac{1}{3}$. K, L, M are the intersection points of the lines AF and CE, BG and AF, CE and BG, respectively. Suppose the area of ABC is 1; find the area of KLM.

I know there is a general formula when dividing in m:n ie. $\frac{(m-n)^2}{m^2+mn+n^2}$

P.S. please no Bary/ Complex

To verify the result, I put $m=n$ and it worked. Also, I tried the area lemma, Ceva's theorem but to no use. I tried this problem for a long time and then realised that Menelaus worked but as this excercise was in the area and ratio section, that felt like 'cheating' as I want to know the author's intended method.

Thanks!

答: The general case of this problem is called Routh's theorem.  It states that if $EB/AE = x$, $FC/BF = y$, $GA/CG = z$, then $$\frac{[KLM]}{[ABC]} = \frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}.$$  Note I have taken the reciprocals so that in your case, $x = y = z = 3$.  The proof of this theorem is given in the Wikipedia link.


问:
An explicit formula of $\|\,f\,\|_0$ where $f \in L^p(S, \Sigma, \mu)$
The following question is from Problem VII 11.32 in Linear Operators edited by Nelson Dunford and Jacob T. Schwartz.

Assume $(S, \Sigma, \mu)$ is a probability space and $f \in L^p(S, \Sigma, \mu)$(where $p > 0$ but not necessarily $\geq 1$). Using Hölder's inequality we can show that $\|\,f\,\|_p$ is a monotone function of $p$ for any $p > 0$. Now show that as $p \downarrow 0, \|\,f\,\|_p \downarrow \exp[\int_S \log\vert\,f(s)\,\vert\mu(ds)]$, which will be called $\|\,f\,\|_0$.
Furthermore, this result can also apply to any element $g \in L^p(X, \mathcal{B}, \mathcal{v})$ where $(X, \mathcal{B}, \mathcal{v})$ is a positive measure space, $p > 0$ and the support of $g$ has finite measure.

If we let $f \in L^p(S, \Sigma, \mu)$ be a simple function, it suffices to consider the sequence $\{\|\,f\,\|_{\frac{1}{n}}\}_{n \in \mathbb{N}}$. Without losing generality, we can assme $f = \sum_{i \leq k}r_i\chi_{A_i}$ where all $r_i$ are real numbers. Then we may need to directly show $$\|\,f\,\|_{\frac{1}{n}} = \exp[n \log(\sum_{i \leq k}r_i^{\frac{1}{n}}\mu(A_i))] \rightarrow \exp[\sum_{i \leq k}(\log r_i)\mu(A_i)] = \prod_{i \leq k}r_i\exp(\mu(A_i))$$ which is not obvious to me ... Any hints about this convergence or new ideas to attack on this problem will be appreciated.

答:
You need two facts to prove the convergence:

$n(t^{1/n}-1) \to \ln t$ for all $t >0$.  To see this note that $t^{1/n}-1=e^{1/n \ln t}-1\sim \frac {\ln t} n $

$n \ln (\sum r_i^{1/n} \mu(A_i))\sim n  (\sum r_i^{1/n} \mu(A_i)-1)=
n (\sum (r_i^{1/n}-1) \mu(A_i))$ since $\sum \mu(A_i)=1$. [I ahve used the fact that $\ln x \sim x-1$ as $x \to 1$.


问:
How to find the following limit?

Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.

答:
You may just use squeezing. In a right neighbourhood od the origin we have $\cos(z)\leq 1$ and $\cos(z)\geq 1-\frac{z^2}{2}$, since $1-\cos(z)=2\sin^2\left(\frac{z}{2}\right)\leq 2\left(\frac{z}{2}\right)^2$. This leads to $1-\frac{t^4}{2}\leq \cos(t^2)\leq 1$ and to
$$ x-\frac{x^5}{10}\leq \int_{0}^{x}\cos(t^2)\,dt \leq x$$
so the wanted limit is clearly one.

Title: How to find the following limit?


问:
Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$
Kazakstan 2012

Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$

prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$

My idea :

$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$

答: We have: $4(ab)^2 = (a+b)^2 \ge 4ab \implies ab \ge 1\implies LHS \ge 2\sqrt{ab} + \dfrac{1}{1+\sqrt{ab}}= f(t), t = \sqrt{ab} \ge 1\implies f'(t) = 2 - \dfrac{1}{(1+t)^2}> 0\implies f(t) \ge f(1) = 2+\dfrac{1}{2} = \dfrac{5}{2} \implies LHS \ge \dfrac{5}{2} = RHS $. Equality occurs when $a = b = 1$.


问:
please prove the following proof related to F distribution.
Suppose $S_1^2$ and $S_2^2$ are two independent unbiased estimate of the common population variance $\sigma^2$ from two random sample of sizes $n_1$ and $n_2$ respectively.

Then show that

$\frac{S_1^2}{S_2^2} $   follows $F$-distribution with $\left(n_1-1\right)$  and  $\left(n_1-1\right)$ degrees of freedom.

答:
HINT:

Note that $$\dfrac{S_1^2}{\sigma^2}\sim \chi^2_{\displaystyle n_1-1} \text{and} \dfrac{S_2^2}{\sigma^2}\sim \chi^2_{\displaystyle n_2-1}$$.

Hence we get $$\displaystyle \dfrac{S_1^2}{\sigma^2} / \dfrac{S_2^2}{\sigma^2} =\dfrac{S_1^2}{S_2^2}\sim F_{\displaystyle n_1-1,n_2-1}$$

ContentLicense: CC BY-SA 3.0


答:
One way to think of it is to think about the first spin.

Case 1 (probability $.01$): First spin is $5$. Then you are done with $Y=1$.

Case 2 (probability $.99$): First spin is not $5$, and you can expect $E[Y]$ additional spins to get your first $5$.

So $E[Y]=.01\cdot(1)+.99\cdot(1+E[Y])$.

Solving for $E[Y]$ in this equation gives $E[Y]=100$.

Title: An intuitive proof about the expected number of trials it takes for an event to happen

Body:
Imagine there is a fair spinner that can land on any number from $1$ through $100$. It is not too difficult to verify that if you spin this spinner $100$ times, then the expected number of $5$'s, for example, is equal to $1$. This can be done using linearity of expectation. Let $X_i$ be an indicator variable that is equal to $1$ if the $i$th spin results in a $5$ and is equal to $0$ if the spinner does not land on a $5$. Then let $X$ equal the number of $5$'s that come up after $100$ spins. We know that $X=\sum X_i$, and so
\begin{align}
\mathbb{E}(X) &= \mathbb{E}(X_1)+\mathbb{E}(X_2)+\mathbb{E}(X_3)+\ldots+\mathbb{E}(X_{100}) \\
&= \underbrace{0.01 + 0.01 + 0.01 + \ldots + 0.01}_{\text{$100$ times}} \\
&= 1 \, .
\end{align}
I find this proof to be fairly easy to understand. It also means that the idea of expectation becomes intuitive to me. However, there is another conception of expectation that I find more difficult to get my head around. Imagine you spin the spinner as many times as are needed to get a $5$. Let's denote the number of spins as $Y$. It turns out that $\mathbb{E}(Y)$, just like $\mathbb{E}(X)$, is equal to $100$. However, I am unable to find a satisfying proof of this fact. I know you can use an infinite series to compute $\mathbb{E}(Y)$, but is there a proof like the one shown above that doesn't require much algebraic manipulation or number crunching? If it is possible to use linearity of expectation, then I would be very interested to see this.

I don't mind if the proof involves infinite series; however, is it possible to improve upon this?

\begin{align}
\sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1} \cdot \frac{1}{100} \cdot n &= \frac{1}{100} \sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1} \cdot n \\[6pt]
&= \frac{1}{100} \sum_{n = 1}^{\infty} (n-1)\left(\frac{99}{100}\right)^{n-1} + \left(\frac{99}{100}\right)^{n-1}\\[6pt]
&= \frac{1}{100} \left(\sum_{n = 1}^{\infty} (n-1)\left(\frac{99}{100}\right)^{n-1} + \sum_{n = 1}^{\infty} \left(\frac{99}{100}\right)^{n-1}\right)\\[6pt]
&= \frac{1}{100} \left(\frac{99/100}{(1-99/100)^2} + \frac{1}{1-99/100}\right)\\[6pt]
&= \frac{1}{100} \left(9900 + 100\right)\\[6pt]
&= 100
\end{align}

(This proof uses the 'Gabriel's staircase’.)


问:
expected number of flips until $k$ consecutive misses or number of flips finish
I followed the argument given in the answers for this question to find the expected number of misses in a sequence of coin-tosses with success probability $p$. The stopping condition for tossing is either until the $k^{th}$ consecutive miss or until the number of rounds reach $n$, whichever comes first. We can consider tail as a miss and head as hit.

The question in the link differs from my question. In my question stopping happens only when $k$ consecutive miss happens or rounds reach $n$. In the linked question stopping happens when $k$ miss happens.

The probability that $k^{th}$ consecutive miss happen at $i^{th}$ trial can not be simply represent in terms of $i$ and $k$ as the number of hits or misses before the start of the sub-sequence of $k$-consecutive miss is still unknown. There can be many misses and hits before the start of the sub-sequence which results in stopping. Similarly there can be unknown number of misses or hits before reaching the end of round ($n$).

Is it even possible to represent this number in a closed form ? Any help and suggestions ?

答:
When $p=1/2$, the required expectation can be written in terms of the Fibonacci $n$-step numbers. Let $T$ denote the number of trials until $k$ consecutive tails. Then
$$
\mathsf{P}(T>i)=\frac{F_{i+2}^{(k)}}{2^i},
$$
which is the probability that no runs of $k$ consecutive tails occur in $i$ tosses. Therefore,
$$
\mathsf{E}[T\wedge n]=\sum_{i=0}^{\infty}\mathsf{P}(T\wedge n>i)=\sum_{i=0}^{n-1}\mathsf{P}(T\wedge n>i) =\sum_{i=1}^{n}\frac{F_{i+1}^{(k)}}{2^{i-1}}.
$$


问:
Can two random variables $X$ and $Y$ be identically distributed while satisfying $P(X<Y)=1$?
I realise this is not possible if $X$ and $Y$ are discrete with a finite support and I think it is true in general, but I am having trouble with the continuous and discrete-with-infinite-support cases.

Any help would be appreciated.

答:
Step 1

If $U \leq V$ and $U,V$ are bounded above with same  distribution then $U=V$ a.s. To prove this note that $e^{U}$ and $e^{V}$ have the same distribution. Since $e^{U}\leq e^{V}$ and $Ee^{U}=Ee^{V}$ we get $e^{U}=e^{V}$ as and hence $U=V$ a.s

Step 2

Note that for any real number $t$,  $\min \{t,X\} \leq \min \{t, Y\}$ for every real number $t$. Applying Step 1 we see that $\min \{t,X\} =\min \{t, Y\}$ a.s . Letting $t \to \infty$ we get $X=Y$ a.s.


答:
From the hypothesis: $a+b=2ab$
So the claim can be written as: $2ab+\frac{1}{1+\sqrt{ab}}\geqslant 2.5$
Now to make everything based on $ab$, we analyze the hypothesis:
For which real values $x$, there exist $a,b$ such that $ab=x , \frac{1}{a}+\frac{1}{b}=2$?
For solving this put $a$ as our variable. We want to find $a$ such that $\frac{1}{a}+\frac{a}{x}=2$ or equivalently: $2ax=x+a^2$ now if we see this as a polynomial, $a$ exists iff $4x^2-4x\geqslant 0$ which happens iff $x\geqslant 1$ or $0\geqslant x$.

So now we can state the problem as follows:
For any real number $x$ such that $x$ isn't between $0$ and $1$, prove that $2x+\frac{1}{1+\sqrt x }\geqslant 2.5$

Now note that $x$ can't be negative because of the square root and this function is increasing for $x\geqslant 1$ so the minimum value is taken on $x=1$ which completes the proof.

Title: Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$

Body:
Kazakstan 2012

Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$

prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$

My idea :

$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$


问:
Calculating $a^p+b^p$ given $a^2+b^2=c$
If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?</p>

答: The other responses showing you cannot find it just from $a^2+b^2=c$ are correct.  You may be interested in the fact that given $a^2+b^2=c, a+b=d$, you can form $(a^2+b^2)(a+b)=cd=a^3+a^2b+ab^2+b^3$ and  $d^2-c=2ab$, so $cd-\frac{d^3-cd}{2}=a^3+b^3$.  Given as many symmetric polynomials as variables you can find all the higher orders.</p>

Title: Calculating $a^p+b^p$ given $a^2+b^2=c$


问:
Show that $B_n = \{x \in X : |\{j \in \mathbb{N} : x\in A_j\}| \geq n \}$ is measurable.
Let $(X, \Sigma, \mu)$ be a measure space and $\{A_j\} \subseteq \Sigma$ be a countable family of measurable sets. For $n \in \mathbb{N}$, define
$$B_n = \{x \in X : |\{j \in \mathbb{N} : x\in A_j\}| \geq n \},$$
so $B_n$ is the set of points in $X$ in at least $n$ of the $A_j$s. Show that $B_n \in \Sigma$.

I think that this boils down to showing that you can construct $B_n$ via complements, countable unions, or countable intersections of the $A_j$s, but I cannot seem to get an angle on it.

答:
Notice that $$B_n = \cup_{k_1 \ge 1} \cup_{k_2 \ge 1, k_2 \ne k_1} \cup_{k_3 \ge 1, k_3 \ne k_1, k_3 \ne k_2} \ldots \cup_{k_n \ge 1, k_n \ne k_j, 1 \le j \le n-1} A_{k_1} \cap A_{k_2} \cap \ldots \cap  A_{k_n}.$$

As $\cup_{k_n \ge 1, k_n \ne k_j, 1 \le j \le n-1} A_{k_1} \cap A_{k_2} \cap \ldots \cap  A_{k_n}$ is a countable union than it is an element of $\Sigma$. Similarly $B_n$ is an element of $\Sigma$.


问:
Verifying properties of events $A$, $B$ given $P(A)$, $P(B)$, $P(A\cup B)$: where is the error?
Given that:

$P(A) = 0.5$

$P(B) = 0.7$

$P(A \cap B) = 0.3$

I have to choose one option that is true... However they all seem to be false which means I am possibly making a mistake.. The only option that is almost true is B it seems. Any help is appreciated. Here are the options followed by my work so far:

A) $A$ and $B$ are independent

If $A$ and $B$ are independent $P(A \cap B) = (0.5)(0.7) = 0.35$; so this is not true

B) $A$ and $B$ are mutually exclusive

If $A$ and $B$ are mutually exclusive then $P(A \cap B) = 0$; so this is not true

C) $P(A \cup B) = 0.8$

$(0.5 + 0.7) - 0.3 = 0.9$; so this is not true

D) $P(A|B) = 0.6$

$$P(A \mid B) = \frac{P (A \cap B)}{P(B)} = \frac{0.3}{0.7} = 0.42$$ so this is not true

答:
Your explanations are correct.

$A,B$ are independent if and only if $P(A \cap B) = P(A)P(B)$. You've shown this isn't true.

$A,B$ are mutually exclusive if and only if $P(A \cap B) = 0$. This obviously isn't true.

From inclusion-exclusion,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
which you've calculated and shown is $0.9$, not $0.8$.

The conditional probability formula gives
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ and you've calculated this to be (about) $0.42$, not $0.6$.

The only reasonable conclusion is that there is an error in the problem.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue. Posting as Community Wiki since I don't have much to add.


Body:
Calculate $142^{381}$ mod $773$ without a calculator .

Attempt:

$$142^{(3\cdot 127)}=142^{381}$$
By try some number's
$$142^1\equiv142\pmod{773}$$
$$142^2\equiv66\pmod{773}$$
$$142^3\equiv96\pmod{773}$$

Lets check the gcd between $773,142$
$$\gcd(773,142)$$
$$773=142\cdot 5+63$$
$$142=63\cdot 2+16$$
$$63=16\cdot3+15$$
$$16=15\cdot 1+1$$
$$15=1\cdot15+0$$
$$\gcd(773,142)=1$$
How to find the answer from here ?

Body: Easily we verify $\,773\,$ is prime. So $\!\bmod 773\!:\, $ if $\,142 \equiv a^2\,$ then $142^{386}\equiv a^{772}\equiv 1\,$ by Fermat, so $\,142^{381}\equiv 142^{386} 142^{-5}\equiv (142^{-1})^5.\,$ By here $\,142^{-1}\equiv 7^2\,$ so $\,142\equiv 7^{-2}\,$ is indeed a square, so $\,142^{381}\equiv 49^5\equiv 49(82)^2 \equiv 49(-223)\equiv 178.\,$ Total time: a few minutes by hand.


问:
Finding Expected Value of vector length?
We build a vector of digits, such that every time we pick a number randomly from {0,1,2,3,...,9} (For each digit the possibility of picking it is equal this 1/10 and not related to previous choices). We stop building out vector at the first time each digit appears at least once (Which means the last digit appears exactly once or else we can have a shorter vector which is a contradiction) Now I want to calculate Expected value of length of the vector created.

How can I solve this, I tried with indicators but don't think of any specific indicator what will be helpful here.

答:
We can split up the procedure into several waiting times.

Let $T_1$ be the number of draws before seeing the first new number. We have $T_1=1$ since the first draw will give you the first new number.

Let $T_2$ be the number of subsequent draws before seeing the second new number. So for instance, if your sequence is $3,3,3,3,1$, then $T_2=4$ since your first new number is "3", and you had to draw $3,3,3,1$ before seeing the second new number "1".

Let $T_3$ be the number of subsequent draws before seeing the third new number, and so on.

In the end, the total number of draws is $T_1 + T_2 + \cdots + T_{10}$. By linearity of expectation,
$$E[T_1 + T_2 + \cdots + T_{10}] = E[T_1] + E[T_2] + \cdots + E[T_{10}],$$
so it suffices to compute each expectation separately.

Recognize that $T_i$ is a geometric distribution. In a sequence of independent draws of successes (seeing the $i$th new number) and failures (seeing an old number), it counts the number of draws until the first success. Note that a success her happens with probability $1 - \frac{i-1}{n}$ since there are $i-1$ "old" numbers you have already seen so far. You can then refer to your prior knowledge about the expectation of a geometric random variable to compute $E[T_i]$.


答:
Let $f(x):=\sum_{_{j\ge 1}}1_{A_j}(x)$. Since $f$ (from $X$ to $[0,\infty]$) is a measurable function,
$$
B_n=\{x:f(x)\ge n\}\in \Sigma.
$$

Title: Show that $B_n = \{x \in X : |\{j \in \mathbb{N} : x\in A_j\}| \geq n \}$ is measurable.

Body:
Let $(X, \Sigma, \mu)$ be a measure space and $\{A_j\} \subseteq \Sigma$ be a countable family of measurable sets. For $n \in \mathbb{N}$, define
$$B_n = \{x \in X : |\{j \in \mathbb{N} : x\in A_j\}| \geq n \},$$
so $B_n$ is the set of points in $X$ in at least $n$ of the $A_j$s. Show that $B_n \in \Sigma$.

I think that this boils down to showing that you can construct $B_n$ via complements, countable unions, or countable intersections of the $A_j$s, but I cannot seem to get an angle on it.


答:
Let

$$
x = 142^{381} \pmod{773}.
$$

Since $773$ is prime, by Fermat's little theorem we have

$$
142^{772} = 1 \pmod{773}.
$$

Therefore, either

$$
142^{386} = 1 \pmod{773}
$$

or

$$
142^{386} = -1 \pmod{773}.
$$

We can distinguish the two cases using Euler's criterion. To that end, we need to compute the Legendre symbol

$$
\left(\frac{142}{773}\right) = \left(\frac{71}{773}\right)\left(\frac{2}{773}\right)
$$

where we used the fact that Legendre symbol is a completely multiplicative function. Next, we use the law of quadratic reciprocity to find

$$
\left(\frac{71}{773}\right) = \left(\frac{773}{71}\right) = \left(\frac{710 + 63}{71}\right) = \left(\frac{63}{71}\right)
$$

and similarly

$$
\left(\frac{63}{71}\right) = -\left(\frac{71}{63}\right) = -\left(\frac{8}{63}\right) = -\left(\frac{2}{63}\right)\left(\frac{2}{63}\right)\left(\frac{2}{63}\right).
$$

Substituting, we see that

$$
\left(\frac{142}{773}\right) = -\left(\frac{2}{63}\right)\left(\frac{2}{63}\right)\left(\frac{2}{63}\right)\left(\frac{2}{773}\right).
$$

Now, using the property known as the second supplement to the law of quadratic reciprocity

$$
\left(\frac{2}{p}\right) = (-1)^{\frac{p^2-1}{8}}
$$

we find

$$
\left(\frac{2}{773}\right) = -1 \\
\left(\frac{2}{63}\right) = 1.
$$

Therefore,

$$
\left(\frac{142}{773}\right) = 1
$$

and so $142$ is a quadratic residue. Consequently,

$$
142^{386} = 1 \pmod{773}.
$$

Now, substituting $x$ and partial results listed in the question

$$
x \cdot 142^5 = 1 \pmod{773} \\
x \cdot 142^2 \cdot 142^3 = 1 \pmod{773} \\
x \cdot 66 \cdot 96 = 1 \pmod{773} \\
x \cdot 152 = 1 \pmod{773}.
$$

Thus, we see that $x$ is the multiplicative inverse of $152$ modulo $773$. We can find it by computing Bézout's coefficients using Euler's algorithm

$$
152 \cdot 178 + 773 \cdot (-35) = 1
$$

and so we see that

$$
178 \cdot 152 = 1 \pmod{773}.
$$

Therefore,

$$
x = 178 \pmod{773}.
$$

Title: Calculate $142^{381}$ mod $773$ without a calculator .


问:
Relaxation of valuation axioms?
Let $K$ be a field, and $v:K\rightarrow \mathbb{R}$ be a valuation, i.e. a function such that, for all $x,y\in K$: $v(x)\geq 0$, $v(x+y)\leq v(x)+v(y)$ (if $v$ is Archimedian, alternatively $v(x+y)\geq \min(v(x),v(y))$) and $v(xy)=v(x)v(y)$. My question is, are there meaningful objects that come out of relaxing any of these conditions? I would be particularly interested to know if one can relax to a "super" or "sub" valuation satisfying $v(xy)<v(x)v(y)$ or $v(xy)>v(x)v(y)$.

答:
The natural relaxation is to replace $\Bbb{R}$ by any ordered abelian group, for example $$v(\sum_{n,m} c_{n,m}x^ny^m)=\inf_{n,m,c_{n,m}\ne 0} n+m\epsilon, \qquad v(f/g)=v(f)-v(g)$$ on $\Bbb{Q}(x,y)$.

The valuation ring (elements with valuation $\ge 0$) is $$\Bbb{Q}[x,y]_{(x,y)}+x\Bbb{Q}(y)[x]_{(x)}$$

Title: Relaxation of valuation axioms?


问:
How to find a pair of solutions of Linear Diophantine equation such that you minimize this expression?
This is a typical Math coding Problem I encountered here Problem.

So Let me break it to you

Suppose you have two values $a$ and $b$ and to select this  value $a$ one time it takes you $c_1$ cost and it takes $c_2$ cost to use value $b$.

You are to solve this equation for $(x,y)$
 $ax$ + $by$ = $N$ where $N$ is a value given to you

Find such $(x,y)$ that you minimize this expression $x*c_1 + y*c_2$.
 or report if it is not possible to solve

It is a clear cut Linear Diophantine equation and I can easily solve for $(x,y)$ by using Extended GCD But I don't how to do it by minimizing the given expression.

An example for this question is suppose you have $a=3$ $b=4$ and $c_1 = 1$ and $c_2=2$ and $N=43$

 $x=13$ and $y=1$ can do the trick for us  by minimizing  the given expression.

答:
Let $d>0$ be the greatest common divisor of $a$ and $b$. Fix an arbitrary solution $(x_0,y_0)$ of the equation $ax+by=N$. Let $(x,y)$ be any solution of this equation. Then $a(x-x_0)+b(y-y_0)=0$, so $d(x-x_0)|b$, that is $x=x_0+kb/d$ for some integer $k$. Conversely, for any integer $k$, $(x_0+kb/d, y_0-ka/d)$ is a solution of the equation. This solution costs $c_1(x_0+ kb/d)+c_2(y_0-ka/d)$, which is $k(c_1b-c_2a)/d$ than the cost of the solution $(x_0,y_0)$. This follows that a solution of the equation with the minimum cost is:

a solution with minimum $x_1$, when $c_1b>c_2a$,

a solution with    minimum $x_2$, when $c_1b<c_2a$,

any solution of the equation, when    $c_1b=c_2a$.


问:
Example of a continuous function $ f: X\rightarrow Y$ that is not closed in product of X and Y
I need to show f is not a closed subset ( different than saying its not a closed function. And I have been thinking of examples of continuous but not closed subsets. if I defined a function f from R to R , the topology on the domain is standard topology and on the codomain is Cofinite topology( which isnt a hausdorff space) such that f(x)=3. its easy to prove why its continuous but to show why it isn't a closed subset of product of X and Y, I took a point (1,4) that doesn't belong to f but then i kind of got confused of what to do next. If I go on to prove that the complement of f isn't open. I don't know If I am on the right track  or even if my example is correct or not? any examples and simple explanations will be appreciated.</p>

答:
HINT: Your function actually is closed in $\Bbb R\times\Bbb R$ when $\Bbb R$ has the cofinite topology. Use the identity function, $f(x)=x$, instead. Let $p=\langle x,y\rangle$, where $x,y\in\Bbb R$ and $x\ne y$. Let $U$ be any open nbhd of $p$; then there are open sets $V$ and $W$ in $\Bbb R$ such that $p\in V\times W\subseteq U$. Since $\Bbb R$ has the cofinite topology, there are finite sets $F,G\subseteq\Bbb R$ such that $V=\Bbb R\setminus F$ and $W=\Bbb R\setminus G$. Use this to show that $f\cap(V\times W)\ne\varnothing$; this will show that $p\in(\operatorname{cl}f)\setminus f$ and hence that $f$ is not closed.

If you need a further hint, there’s one in the spoiler-protected box:

 Find a point $y\in V\cap W$.


问:
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$

I can see that $x - 1$ and $x^2 + x + 1$ are nonzero and non-units (for any field $F$ the set of all units of $F[x]$ = $F-\{0\}$)?

Added: Actually I would like to show $f(x)|(x-1)(x^2+x+1)~\text{where }\deg f(x)\ge 1~\implies f(x)=(x-1)(x^2+x+1)~or~(x-1)~or~(x^2+x+1).$ Does showing $(x-1)~and~(x^2+x+1)$ irreducible help me anyhow?

答:
For the first polynomial, you might want to see what would happen if you could express
$x-1=p*q$, where p and q are polynomials with smaller degree. For the second, maybe check out the roots of the quadratic.</p>


问:
Understanding when a relation is transitive
I'm having a bit of a problem in understanding when a relation is transitive.

So let's say we have $A=\{a,b,c,d\}$. Do these following constitute a transitive relation? $\{(a,b), (b,c), (c,a)\}$?

I'm just confused since what I know is that for a set to have a transitive relation, $xRy$ and $yRz$ -> $xRz$. So it should be $(a,c)$ instead of $(c,a)$? Or is it the same thing?

答:
As stated in the comments, $(a,c)$ and $(c,a)$ are different things. Transitivity would mean that $(a,b),(b,c) \in R$ (a relation on $A$) implies $(a,c) \in R$. Similarly, it would imply $(c,b),(b,a) \in R \implies (c,a) \in R$. These are ordered pairs, after all, and thus each component must be equal for two ordered pairs to be equal.

Of course, if we had symmetry, it would be a different matter. But that isn't always the case.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
In how many ways $n$ dishes around a round table can be colored with 3 colors with adjacent dishes with different colors

Consider a round table with $n$ places. In each place there is a colored dish. In how many ways the dishes in this table can be colored with 3 colors, in such a way that adjacent dishes have different colors?

Solution: If $P(n)$ is the number of possibilities for a table with $n$ places, it is true that $$P(n)=P(n-1)+2P(n-2).$$By inspection it is easy to show that $P(2)=6$ and $P(3)=6$, and we can find the solution for any $P(n)$ by substitution using the recursion, or by the solution for the previous difference equation given by:
$$P(n)=2^n+2(-1)^n$$.

My problem: I don't understand why $$P(n)=P(n-1)+2P(n-2)$$ is true, or why, directly, $$P(n)=2^n+2(-1)^n$$ is true. Yes, I know how last equation is obtained from the previous equation, but cannot find a reasonable argument why it makes sense from a combinatoric argument. Any help?

答:
Here is a different approach.  Look at the circular pattern of differences between successive colors (in a given order, say: red, green, blue) as you go around the table, so the color sequence R,G,R,B,R... gives rise to the sequence $+1,-1,-1,+1,\ldots$ say, since RG has difference $+1$, GR has difference $-1$, and so on.  The condition that the $n$ differences must satisfy is that they add up to $0 \bmod 3$.  (The cumulative sum of the first $k$ of these numbers tells you the difference in color between the $0$-th  plate and the $k$th; if you follow all the way around the circle the $0$th plate is the $n$th plate, and hence of the same color, so the sum of all $n$ differences should be $0\bmod 3$.)  Let $D_n$ count the number of such difference patterns. The number of colorings is $3D_n$: the starting chair can have any of $3$ colors, then the rest are determined by the difference pattern.

A formula for $D_n$ can be obtained by evaluating an $n$-fold convolution product:   $D_n= f^{*n}(0)$, where $f:\mathbb Z_3\to\mathbb R$ is given by $f(0)=0, f(1)=f(2)=1$. (Convolution with respect to the additive group of integers $\bmod 3$.) Writing $f=\mathbb 1-\delta$, where $\mathbb1$ is the constant function $1$, and $\delta(x)$ is $1$ at $x=0$ and $0$ elsewhere, and using the fact that $\mathbb1*\mathbb1=3 \mathbb1$, we see
\begin{align*}
D_n &= \sum_{k=0}^{n-1} \binom n k (-1)^{n-k}\mathbb1^{*k} (0)+(-1)^n\\
&= \sum_{k=0}^{n-1} \binom n k (-1)^{n-k} 3^{k-1} + (-1)^n\\
&=\frac{ (3-1)^n - (-1)^n}3 + (-1)^n
\end{align*}
so the number of colorings is
$$
3D_n = 2^n +2(-1)^n .$$


问:
Equivalence relations: showing $[b] \subseteq [a]$ while knowing $[a] \subseteq [b]$

Let $R$ be an equivalence relation on $A$ for two elements $a,b \in A$. Show that $[b]\subseteq [a]$.

In a previous section it was already proven that $[a]\subseteq [b]$. Can I just argue that by symmetry it must be true that $[b]\subseteq [a]$?

EDIT: Here's my attempt.

Assume $aRb$. Take any $t\in [a]$. Then $aRt$. By symmetry, $tRa$. Given $[a]\subseteq [b]$ then $t\in [b]$, and so $bRt$. Since $tRa$ and $bRt$, then $[a]=[b]$, therefore it follows $[b]\subseteq [a]$.

答:
The proof is mostly superfluous, since it does effectively follow by a symmetric argument. You have that $[a] \subseteq [b]$ for all pairs $(a,b) \in R$. Taking the ordered pair $(a',b') := (b,a)$ will then have $[a'] = [b] \subseteq [a] = [b']$. But we can always go from a more axiomatic perspective, as you and your text did. (And there's no harm in fully justifying things.)

This would be a more basic set-inclusion proof though, and I think you overcomplicate matters just a bit. Here's how I would do it... Assume $aRb$; thus $bRa$ by symmetry. We want to show $[b] \subseteq [a]$, i.e. $\forall x \in [b]$, it follows $x \in [a]$. Observe:

\begin{align*}
x \in [b]
&\iff xRb \tag{def.} \\
&\implies xRa \tag{transitivity} \\
&\iff x \in [a]\tag{def.}
\end{align*}

The first is just definition of membership of an equivalence class represented by $b$

The second is transitivity applied to the facts that $xRb$ and $bRa$

The third is again definition of membership of equivalence classes

Thus, $[b] \subseteq [a]$.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
inverse laplace transform of $\frac{1}{s(e^s+1)}$
How to compute this inverse Laplace transform ?

$$\displaystyle{ \mathcal{L^{-1}} \left\{  \frac{1}{s(\exp(s)+1)} \right\} }$$

Thanks.

答: Google the Square wave function $\mathbf{sqw}(x)=(-1)^{\mathbf{floor}(x)}$ to find out it is a periodic function which is obviously piecewise and by using the L.T. rules we can see that $$\mathcal{L}\{\mathbf{sqw}(x)\}=\frac{1}{s}\tanh(s/2)=\frac{e^s-1}{s(e^s+1)}$$ So since $$\mathcal{L}\{1\}=\frac{1}{s}$$ I think you can do the rest.</p>


问:
Help understanding a basic complex analysis argument
I'm reading about some basic complex analysis results from the Supplementary Notes by Joe Schwartz in order to understand Weierstrass uniformization. I was wondering if someone can help me understand the step-by-step logic behind the following paragraph (on page 7 of the notes linked above) starting at "If $f(0)\neq 0$, then we are done. If $f(0) = 0$, then...":

I know and understand all the results described in the notes up until that point, so feel free to quote them without explicit reference in your explanation. Thanks!

答:
The result that is being proved here is that 'locally a holomorphic function looks like $z^n$ for some $n$'. This idea is made precise as the following lemma.

$\textbf{Lemma}$ Let $f$ be a function holomorphic in a neighborhood $\Delta $of $0.$ Then $f(z)=z^ng(z)$ for some non-negative integer $n$ and a holomorphic function $g$ such that $g(0)\neq 0$ (and therefore $g(z)\neq 0$ in some neighrhood $\Delta'$ of $0$).

The proof of the above lemma is what you are looking at. And, obviously if $f(0)\neq 0$ then you can take $n=0$ and $g(z)=f(z)$ itself, so you are done. When $f(0)=0,$ the first thing to observe is that $\lim_{z\to 0}\frac{f(z)}{z}=f'(0).$ In particular this measn $\frac{f(z)}{z}$ is bounded on $U\setminus\{0\}$ for some neighborhood $U$ of $0.$

After this observation define a new function
$$f_1(z)=\begin{cases}
\frac{f(z)}{z}, & z\neq 0\\
f'(0), & z=0\end{cases}.$$
Clearly $f_1$ is holormophic except possibly at $0.$ But the Riemann removable singularity theorem implies that $f_1$ is holomorphic because $f_1$ is bounded in some neighborhood of $0.$ Obviously, we have $f(z)=zf_1(z).$ Now you repeat the same argument untill you get $f(z)=z^nf_n(z)$ for some $f_n$ such that $f_n(0)\neq 0.$ Only thing you need to be careful is to argue that such an $n$ must exist. But that is easy. If not, it follows that $f^{(k)}=0$ for all $k\ge 0.$ Power series expansion around $0$ therefore vanishes identically.

I should however also add that if you already know that an analytic function is locally represented by a power series, then the proof is very simple. Just let $n$ be the smallest integer such that $f^{(n)}\neq 0.$ Then the power series of $f,$ locally looks like $\sum_{i\ge n}a_nz^n=z^n\left(\sum_{i=0}^{\infty}a_{n+i}z^i\right).$ Take $g(z)=\sum_{i=0}^{\infty}a_{n+i}z^i$ and you are done.


问:
Find all integers $x$ such that $x ^{86}\equiv 6\pmod{29}$
Find all integers $x$ such that $x ^{86}\equiv 6\pmod{29}$

By Fermat’s Little Theorem

$x ^{28}\equiv 1\pmod{29}$

I am stuck from here any idea?

答:
I am continue from end of your attempt.

By Fermat’s Little Theorem,

$$x^{28}\equiv 1 \pmod{29}$$

Thus,
$$x^{86}\equiv x^2\pmod{29}$$

So, we only need to solve

$$x^{2}\equiv 6\pmod{29}$$

This is the same as

$$x^{2}\equiv 64\pmod{29}$$

which means that
$$x^{2}-64\equiv (x − 8)(x + 8)\equiv 0 \pmod{29}$$

Thus,
$$x\equiv 8\pmod{29}$$
$$x\equiv 21\pmod{29}$$

Are the solution's


问:
How to find midpoint of an Arc on a 3D Plane
I'm not sure if this is even possible but here it goes:

Let's assume we have a 3D Circle with 2 points on it
A (1, 5, 2) B (5, 2, 1), and the circle's center O (0, 0, 0)

Is it possible to find C based on the conditions that the distance from C to the center is the radius and distances from C to A and B are equal?

答: The simplest way in my view would be to take the midpoint of $AB$ and scale it to a point on the sphere (ie, a point whose distance from $O$ is the radius).


答:
You know that point $C$ is between $A$ and $B$, in the $AOB$ plane. Using $C=(x,y,z)$ you know that $$\vec{AO}\times\vec{CO}=\vec{CO}\times\vec{BO}$$
This accounts both for the fact that $C$ is in the $AOB$ plane, and also that the angle is the same. It yields a system with three unknown and three equations, but only two are independent. You can get the last one by using $|AO|=|CO|$, or $$x^2+y^2+z^2=1^2+5^2+2^2$$

Title: How to find midpoint of an Arc on a 3D Plane

Body:
I'm not sure if this is even possible but here it goes:

Let's assume we have a 3D Circle with 2 points on it
A (1, 5, 2) B (5, 2, 1), and the circle's center O (0, 0, 0)

Is it possible to find C based on the conditions that the distance from C to the center is the radius and distances from C to A and B are equal?


答: Fermat's Little Theorem gives $x^{28}\equiv 1\mod 29$ and therefore $1\equiv x^{3(28)}\equiv x^{84}$. Thus the congruence $x^{86}\equiv 6\mod29$ is equivalent to the congruence $x^2\equiv 6\mod29$ whose solution set is the union of the residue classes $8+29\Bbb Z$ and $-8+29\Bbb Z$.

Body:
Find all integers $x$ such that $x ^{86}\equiv 6\pmod{29}$

By Fermat’s Little Theorem

$x ^{28}\equiv 1\pmod{29}$

I am stuck from here any idea?


问:
$\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$
If $\{X_n\}$ is a sequence of identically distributed r.v.'s with finite mean, then

$$\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$$

The inequality

$$\frac{1}{n}E(\max_{1\le j\le n} |X_j|) \le \frac{1}{n}E(|X_1| + \cdots + |X_n|)=E(|X_1|)$$

suggests that the result has something to do with the "graphs" of the $|X_j|$ overlapping so that $E(|X_j|)=\int_0^\infty |X_j| P(d\omega)$ isn't fully counted $n$ times in the value of $E(\max_{1\le j\le n} |X_j|)$. But how do we make this rigorous? Notice there is no mention of the $X_j$ being independent.

答:
For a positive $R$, define $$X_j^R:=|X_j|\chi_{\{|X_j|\leqslant R\}}.$$
Recall that
$$E[\max_{1\leqslant k\leqslant n}|Y_k]\leqslant n\max_{1\leqslant k\leqslant n}E[Y_k],$$
hence
$$\frac 1nE\left[\max_{1\leqslant k\leqslant n}|X_k|\right]\leqslant \max_{1\leqslant k\leqslant n}E[|X_k|\chi_{\{|X_k|>R\}}]+\frac Rn=E[|X_1\chi_{\{|X_1|>R\}}|]+\frac Rn.$$
Taking the $\limsup_{n\to+\infty}$, and using the fact that $R$ is arbitrary we can conclude.</p>


问:
Probability that a 2-d random variable falls into an area equals the double integral of its PDF?
Let $x:\Omega\to\mathbb{R}^2$ be a 2-dimensional r.v., and $f(p,q)$ be its probability density function.

For what kind of $D\subseteq\mathbb{R}^2$, $\mathrm{Pr}[x\in D]=\iint_D f(p,q)\mathrm{d}p\mathrm{d}q$ ?

I couldn't find a proof for this from the definition of r.v., even in the most simple case of $D=\{(x,y)|x,y\geq 0,x+y\leq1\}$ (perhaps it's just me being stupid). (okay, now I see that this can be proven using the definition of integration, by splitting D into rectangles.but can this be applied to general D or is there a more elegant proof?)

答:
This is a question for measure theory; I will use the basics of that subject freely in the answer.  It may be possible to redo the proof using Riemann integration if $f$ is Riemann integrable (and attention is restricted to Riemann integrable sets $D$), but I haven't tried writing the details yet.

Let $\mu$ be the law of $x$.  In other words, $\mu$ is the Borel probability measure on $\mathbb{R}^{2}$ defined by
\begin{align*}
\mu(D) = \mathbb{P}\{x \in D\}.
\end{align*}

We know that if $D$ is a rectangle of the form $[a,b) \times [c,d)$, then $\mu(D) = \int_{D} f(p,q) \, dp \, dq$.

I claim that this determines $\mu$, that is, that $\mu(D) = \int_{D} f(p,q) \, dp \, dq$ for each Borel set $D$.

To see this, let $\mathscr{B}$ be the Borel subsets of $\mathbb{R}^{2}$.  First, I claim that $\mathscr{B} = \sigma(\mathcal{D})$, where $\mathcal{D} = \{[a,b) \times [c,d) \, \mid \, a,b, c, d \in \mathbb{R}\}$.  To see this, notice that if $\mathcal{O} \subseteq \mathbb{R}^{2}$ is open, then we can find open rectangles $\{[a_{n},b_{n}) \times [c_{n},d_{n}) \, \mid \, n \in \mathbb{N}\}$ with $\{(a_{n},b_{n},c_{n},d_{n}) \, \mid \, n \in \mathbb{N}\} \subseteq \mathbb{Q}^{4}$ such that $\mathcal{O} = \bigcup_{n = 1}^{\infty} [a_{n},b_{n}) \times [c_{n},d_{n})$.  (Here I'm using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$.)  Hence $\sigma(\mathcal{D})$ contains all the open sets, which implies it contains $\mathscr{B}$.  Conversely, $\mathscr{B} \supseteq \mathcal{D}$ so $\mathscr{B}$ contains $\sigma(\mathcal{D})$.

Next, note that the following statement is an elementary consequence of Dynkin's $\pi-\lambda$ theorem:

If $\mu_{1},\mu_{2}$ are two Borel probability measures on a
measurable space $(\mathbb{R}^{2},\mathscr{B})$, then $\mu_{1} =
 \mu_{2}$ if and only if $\mu_{1}(D) = \mu_{2}(D)$ independently of the
choice of $D \in \mathcal{D}$.

Here I am using the fact that $\mathcal{D}$ is a $\pi$-system (finite intersections of rectangles are rectangles again) and the set $\{D \in \mathscr{B} \, \mid \, \mu_{1}(D) = \mu_{2}(D)\}$ is a $\lambda$-system (since $\mu_{1}(\mathbb{R}^{2}) = \mu_{2}(\mathbb{R}^{2}) = 1$).

Applying this with $\mu_{1} = \mu$ and $\mu_{2}(D) = \int_{D} f(p,q) \, dp \, dq$ and $\mathcal{A}$ the smallest algebra containing $\mathcal{D}$, we conclude that $\mu(D) = \int_{D} f(p,q) \, dp \, dq$ for all Borel sets $D$.


问:
If $[(p\rightarrow\ q)\land (\lnot p \rightarrow \lnot r)]\rightarrow (\lnot q \rightarrow r)$ is false how to find its equivalent in other words?
The problem is as follows:

Assume the statement which is written below is false,

```<code>If Gaston drinks milk then he takes his vitamins, and if he
doesn't drink milk then he doesn't eat; therefore if Gaston
doesn't take his vitamins, then he eats.
</code>```

Find which of the statements in the alternatives is true.

$\begin{array}{ll}
1.&\textrm{Gaston drinks milk}\\
2.&\textrm{Gaston takes his vitamins}\\
3.&\textrm{Gaston eats}\\
4.&\textrm{Gaston eats or he doesn't drink milk}\\
\end{array}$

How exactly should I match the right one here?.

What I did in my attempt to solve this problem was to formulate the complex statement into terms of propositional logic as follows:

$\textrm{Gaston drinks milk = p}$

$\textrm{he takes his vitamins = q}$

$\textrm{he does eat = r}$

Then this is reformulated as follows:

$$[(p\rightarrow\ q)\land (\lnot p \rightarrow \lnot r)]\rightarrow (\lnot q \rightarrow r)$$

But what to do from here?. How exactly can this be simplified?. Is my intepretation correct?. Can someone help me here?. Please try to be the most detailed as possible so I can understand this part.

答:
If you make a truth table, you’ll find that the statement is false if and only if $p,q$, and $r$ are all false. In other words, that statement is false if and only if $\neg p\land\neg q\land\neg r$ is true. Your four choices of answer are $p$, $q$, $r$, and $r\lor\neg p$. The first three are clearly false, since they contradict $\neg p\land\neg q\land\neg r$. And $\neg p\land\neg q\land\neg r$ clearly implies $\neg p$, which implies $r\lor\neg p$, so $r\lor\neg p$ is true, and the correct answer is the fourth one.

If you want to proceed algebraically, note that saying that

$$\big((p\to q)\land(\neg p\to\neg r)\big)\to(\neg q\to r)$$

is false is the same as saying that

$$\big((p\to q)\land(\neg p\to\neg r)\big)\land\neg(\neg q\to r)$$

is true. And $\neg(\neg q\to r)\equiv\neg q\land\neg r$, so we’re assuming that

$$(p\to q)\land(\neg p\to\neg r)\land\neg q\land\neg r\tag{1}$$

is true. Now $p\to q\equiv\neg p\lor q$, and

$$(\neg p\lor q)\land\neg q\equiv(\neg p\land\neg q)\lor(q\land\neg q)\equiv\neg p\land\neg q\,,$$

so $(1)$ is equivalent to

$$(\neg p\to\neg r)\land\neg p\land\neg q\land\neg r\,.\tag{2}$$

Finally, $\neg p\to\neg r\equiv p\lor\neg r$, and $(p\lor\neg r)\land\neg p\equiv\neg r\land\neg p$, so $(2)$ is equivalent to $\neg r\land\neg p\land\neg q\land\neg r$ and hence to $\neg p\land\neg q\land\neg r$.


问:
Could not understand the physical meaning?

Let's say a plane is defined using normal vector, i.e., $\vec{n_0}$ and a point, i.e., $p_0$ in the plane as shown in the above figure. So we can determine the corresponding equation of the plane as $ax+by+cz+d =0$ where $d = -p_0^T\vec{n_0}$

The following two expressions I count not understand what do they represent

$\frac{-\vec{n_0}\cdot d}{|\vec{n_0}|^2}$

Let assume there is another plane is given by ($\vec{n_1},p_1$), then compute $ \vec{n_1} \times \vec{n_0} \times \vec{n_1}$

These are I encounter when I try to understand how the reference plane intersects with a set of other planes. I could not share the original work due to copyright issue. Any idea on this to understand what actually those represent really appreciated.

答:
$\dfrac{-\vec{n_0}}{\|\vec{n_0}\|}$ is the unit vector pointed in the opposite direction as $\vec{n_0}$. And $\dfrac d{\|\vec{n_0}\|}$ is the distance from the origin to the plane $\vec{n_0}\cdot \vec v = d$. Thus their product
$$\dfrac{-d\vec{n_0}}{\|\vec{n_0}\|^2}$$ would be the vector pointing from the projection of the origin onto the plane back to the origin itself.

For any two non-parallel vectors $\vec v, \vec w$, $\vec v \times \vec w$ is a vector perpendicular to both, with length $\|\vec v\|\|\vec w\|\sin \theta$, where $\theta$ is the angle between the two. So

$\|\vec{n_1} \times \vec{n_0} \times \vec{n_1}\| = \|\vec n_0\|\|\vec n_1\|^2\sin \theta$.

$\vec{n_1} \times \vec{n_0} \times \vec{n_1}$ is perpendicular to $\vec{n_0} \times \vec{n_1}$, and therefore must be in the same plane as the origin, $\vec{n_0}$ and $\vec{n_1}$. It is also perpendicular to $\vec{n_1}$. Some playing around with the RH rule should also convince you that in the plane of $\vec{n_0}$ and $\vec{n_1}$, it lies on the same side of the line through $\vec{n_1}$ as $\vec{n_0}$. In fact, it is the projection of $\vec{n_0}$ onto the line perpendicular to $\vec{n_1}$ in the common plane of the two, multiplied by the the length of $\vec{n_1}$.

Note that in the equation of a plane $\vec n \cdot \vec r = D$, if we divide by the $\|\vec n\|$, we get $$\dfrac{\vec n}{\|\vec n\|}\cdot \vec r = \dfrac{D}{\|\vec n\|}$$ which is another equation of the same time, except that $\vec n$ is replaced with the vector $\dfrac{\vec n}{\|\vec n\|}$, which has length $1$, and the constant is replaced with another constant $\dfrac{D}{\|\vec n\|}$. That constant turns out to be exactly the distance from the plane to the origin, so it is more meaningful that $D$ itself.

For this reason we generally prefer to always write plane equations with the normal being a unit vector. And if we make that restriction the two interpretations become more simple. If $\vec{n_0}$ is a unit vector, then $d\vec{n_0}$ is the projection of the origin onto the plane. And $-d\vec{n_0}$ is its opposite.

When $\|\vec{n_0}\| = \|\vec{n_1}\| = 1$, then $\vec{n_1} \times \vec{n_0} \times \vec{n_1}$ is exactly the projection of $\vec{n_0}$ onto the line perpendicular to $\vec{n_1}$ in their common plane.


问:
Find all positive integers $n$ such that
$n^6 + 6n^5 + 4n^4 - n^2 - n + 2$ is prime.
Find all positive integers $n$ such that
$n^6 + 6n^5 + 4n^4 - n^2 - n + 2$ is prime.

Any idea to start? How to factorize them out to get some ideas?

答:
Well, I just wrote a program on c++ and it seems like there could be a lot :)
There are two alternatives:

You got the wrong polynomial;

Your teacher is better than my pc with calculations.

Here it is the code I used:

```<code> #include <iostream>
#include <cmath>
using namespace std;
int test_prime(long long int x) {
    int d;
    float ths = sqrt(x);
    for(d=2; d<= ths; d++) {
        if(x%d == 0) {
            return 0;
        }
    }
    return 1;
}

long long int poly(long long int n) {
    return pow(n,6)+6*pow(n,5)+4*pow(n,4)-pow(n,2)-n+2;
}

int main()
{
    int n,N;
    cout <<"enter the maximum N you want to test" << endl;
    cin >> N;
    cout << "Here it is the list of n that produce prime numbers:" << endl;
    for(n=0; n<=N;n++) {
        if( test_prime(poly(n)) == 1 ) {
           cout << n << " produces " << poly(n) << endl;
       }
    }
    return 0;
}
</code>```

It returns quite many primes, and after that it goes out of memory:

```<code>0 produces 2
1 produces 11
5 produces 36847
13 produces 7168631
17 produces 32990491
19 produces 62423381
21 produces 111048191
25 produces 304296227
27 produces 475638941
33 produces 1531022891
35 produces 2159398117
53 produces 24705093151
57 produces 37948820291
63 produces 68541129281
69 produces 117393014831
71 produces 139027301641
79 produces 261705587921
107 produces 1585407762541
115 produces 2434441786037
123 produces 3632659864061
143 produces 8911441858321
151 produces 12327007462361
153 produces 13332932524451
159 produces 16770104782241
181 produces 36331706896031
191 produces 50081719101121
195 produces 56677858561157
207 produces 80960042697041
233 produces 164137834805491
235 produces 172737657543917
253 produces 268490472955751
277 produces 461537975400191
289 produces 594746103578411
291 produces 619785617090021
297 produces 700235587063091
299 produces 728911546022701
303 produces 789205631796401
321 produces 1114524100418891
349 produces 1838101582558151
361 produces 2250169250734451
379 produces 3010708332794621
395 produces 3856033675042957
411 produces 4890522937999901
417 produces 5333723407128491
429 produces 6320988884994071
433 produces 6682102616188091
437 produces 7060246828863391
</code>```

NB: For the non-programmer guys like me, remember you can copy and paste the code for example here, then click on "run" above and use the program in the window below. Enjoy!

Body:
Well, I just wrote a program on c++ and it seems like there could be a lot :)
There are two alternatives:

You got the wrong polynomial;

Your teacher is better than my pc with calculations.

Here it is the code I used:

 #include <iostream>
#include <cmath>
using namespace std;
int test_prime(long long int x) {
    int d;
    float ths = sqrt(x);
    for(d=2; d<= ths; d++) {
        if(x%d == 0) {
            return 0;
        }
    }
    return 1;
}

long long int poly(long long int n) {
    return pow(n,6)+6*pow(n,5)+4*pow(n,4)-pow(n,2)-n+2;
}

int main()
{
    int n,N;
    cout <<"enter the maximum N you want to test" << endl;
    cin >> N;
    cout << "Here it is the list of n that produce prime numbers:" << endl;
    for(n=0; n<=N;n++) {
        if( test_prime(poly(n)) == 1 ) {
           cout << n << " produces " << poly(n) << endl;
       }
    }
    return 0;
}

It returns quite many primes, and after that it goes out of memory:

0 produces 2
1 produces 11
5 produces 36847
13 produces 7168631
17 produces 32990491
19 produces 62423381
21 produces 111048191
25 produces 304296227
27 produces 475638941
33 produces 1531022891
35 produces 2159398117
53 produces 24705093151
57 produces 37948820291
63 produces 68541129281
69 produces 117393014831
71 produces 139027301641
79 produces 261705587921
107 produces 1585407762541
115 produces 2434441786037
123 produces 3632659864061
143 produces 8911441858321
151 produces 12327007462361
153 produces 13332932524451
159 produces 16770104782241
181 produces 36331706896031
191 produces 50081719101121
195 produces 56677858561157
207 produces 80960042697041
233 produces 164137834805491
235 produces 172737657543917
253 produces 268490472955751
277 produces 461537975400191
289 produces 594746103578411
291 produces 619785617090021
297 produces 700235587063091
299 produces 728911546022701
303 produces 789205631796401
321 produces 1114524100418891
349 produces 1838101582558151
361 produces 2250169250734451
379 produces 3010708332794621
395 produces 3856033675042957
411 produces 4890522937999901
417 produces 5333723407128491
429 produces 6320988884994071
433 produces 6682102616188091
437 produces 7060246828863391

NB: For the non-programmer guys like me, remember you can copy and paste the code for example here, then click on "run" above and use the program in the window below. Enjoy!


问:
Example of two star shaped polygons on the plane such that every edge of the first polygon is intersected by every edge of the second one internally
This question was asked in my computational geometry exam yesterday. The definition of a star shaped polygon was given as: A polygon is called star-shaped if there is a point inside it such that the whole boundary is visible from that point.

Since every edge of the first polygon must intersect every edge of the second, I first arranged the edges in a grid. I then connected them in a shark teeth fashion. But I couldn't manage to close the ends while still maintaining the intersection criteria. Any ideas on how this could be done?

答:
Don't mind if I answer my own question :)

The trick is to literally "close the ends". And pick  fewer lines, it'll be simpler to depict.

As you can see, the polygons are star shaped and the edges mutually fully intersect.


问:
Defining ordered pairs using only first-order logic
The usual definition of an ordered pair $\langle x,y \rangle$ is $\{\{x\},\{x,y\}\}$, but how may one define that using purely first-order logic contructs (variables, logical operations, quantifiers) and the $\in$ relation?

After reading the wiki, which defines the properties:

$x$ is the first coordinate of $p$: $\forall A[A \in p \rightarrow x \in A]$

$x$ is the second coordinate of $p$: $\exists A[A \in p \land x \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow x \notin A \lor x \notin B)]$

I arrived at the following definition:

$$
\langle x,y \rangle = p \leftrightarrow \forall A[A \in p \rightarrow x \in A] \land \exists A[A \in p \land y \in A] \land \forall A \forall B[A \in p \land B \in p \rightarrow (A \neq B \rightarrow y \notin A \lor y \notin B)]
$$

However, this definition would also consider $\{\{x\},\{x,y\},\{x,z\}\}$ or $\{\{x\},\{x,y,z\}\}$ as representing that ordered pair, and as such, it is not unique.

So, how may one define ordered pairs using only first-order logic such that only the set $\{\{x\},\{x,y\}\}$ represents the ordered pair $\langle x,y \rangle$?

答:
$p= \langle x, y \rangle \iff \forall z (z \in p \iff \exists w_1 \exists w_2 ((z=w_1 \lor z=w_2) \land ((\forall t_1(t_1 \in w_1 \iff t_1=x)) \land \forall t_2(t_2 \in w_2 \iff (t_2 = x \lor t_2 = y)))).$

Translating this to English, $w_1$ and $w_2$ are the two elements of the set comprising our ordered pair, with $t_1$ as the element of $w_1$ (which we're forcing to be $x$) and $t_2$ as the elements of $w_2$ (which we're forcing to be $x$ or $y$).


问:
Why is there so little overlap between calculus and statistics?
I was an applied math major and took several calculus-based courses, including differential equations and real analysis, as well as an introductory course sequence in probability and statistics. Both calculus and statistics are extremely useful and broadly applicable to the natural and social sciences, but it seems like their tools and methods are completely different. Are the "statistical" and "calculus-based" approaches to solving problems fundamentally different, or is there actually significant overlap between these fields that is not evident to someone who only learned about these fields at the undergraduate and introductory levels?

It seems like one of the major goals of a calculus-based approach to understanding a system is to come up with a differential equation to model the data. And it seems like the statistical approach seeks to obtain causal and inferential relationships between variables. So it seems as if one of the primary goals of both fields is to come up with quantitative relationships between variables being observed in a system. So why are their methods and tools so different?

Also, certain sciences use calculus more than statistics (like physics) whereas others use statistics more than calculus (biology), and others use both extensively (economics). What about a scientific discipline makes  either analytical or statistical methods more useful?

答:
It depends on the quantities that you measure or calculate. Most of the measurements are discrete and are processed using statistical methods. This is field independent. Then, when you model the underlying process, one usually has continuous variables (for example some time evolution). Those are described by calculus.

In the undergraduate physics, most of the time one looks at the models. The experiments were done long time ago. Everyone knows that the gravity next to the surface of the Earth is described by a downward constant acceleration. Then you use calculus to describe projectile motion.

In biology, there is much less detailed knowledge about some underlying mechanism. To try to tease out the correlations, one will therefore use some statistical methods first. Once you have a model, you write the differential equations, solve them to predict other outcomes. But this is also true in the current physics research.


问:
Equating Acceleration Terms for a Parametric Surface
Suppose we have a smooth 2D surface embedded in 3D Euclidean space defined parametrically by,
$$
r : \mathbb{R^2} \to \mathbb{R}^3\ ,\ \ \ r(x,y) := \begin{bmatrix} \bar{x}(x,y) \\ \bar{y}(x,y) \\ \bar{z}(x,y) \end{bmatrix}
$$

We also have its Jacobian and higher partial derivatives:
$$
J(x,y) := \begin{bmatrix} \frac{\partial r}{\partial x}(x,y) & \frac{\partial r}{\partial y}(x,y) \end{bmatrix} \in \mathbb{R}^{3 \times 2}
$$

Consider a 1D trajectory through the parametric space:
$$
q : \mathbb{R} \to \mathbb{R}^2\ ,\ \ \ q(t) := \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}
$$

By chain-rule (and over-dot notation) the ambient velocity is,
\begin{align}
\frac{dr}{dt} &= \frac{dr}{dq} \frac{dq}{dt}\\
\dot{r} &= J \dot{q}
\end{align}

By product-rule the ambient acceleration is,
$$
\ddot{r} = J \ddot{q} + \dot{J} \dot{q}
$$

A similar two-term expression for the acceleration can be derived by expressing the velocity as a product of its magnitude $v := ||\dot{r}||$ and direction $\hat{\tau} := \dot{r} / v$.
\begin{align}
\dot{r} &= v\hat{\tau}\\
\implies\ \ddot{r} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}
\end{align}

Obviously we can equate these $\ddot{r}$ expansions, but I am wondering about whether the following red/blue terms are individually equatable:
$$
\ddot{r} = \color{red}{J \ddot{q}} + \color{blue}{\dot{J} \dot{q}} = \color{red}{\dot{v}\hat{\tau}} + \color{blue}{v\dot{\hat{\tau}}}
$$

The term $\color{red}{J \ddot{q}}$ is always in the tangent plane because it is a linear combination of tangent vectors:
$$
\color{red}{J \ddot{q}} = \ddot{x} \frac{\partial r}{\partial x} + \ddot{y} \frac{\partial r}{\partial y}
$$

The same is true for $\color{red}{\dot{v}\hat{\tau}}$ since it is in the direction of the velocity. This inclines me to write a stronger pair of equalities:
\begin{gather}
\color{red}{J \ddot{q} \overset{?}{=} \dot{v}\hat{\tau}} \tag{1}\\
\color{blue}{\dot{J} \dot{q} \overset{?}{=} v\dot{\hat{\tau}}} \tag{2}
\end{gather}

To actually prove this, I think I would need to show that $\color{blue}{\dot{J} \dot{q}}$ and $\color{blue}{v\dot{\hat{\tau}}}$ are orthogonal to the tangent plane. Edit: nevermind, that isn't true, but also isn't necessary.

I have been unable to show this in general (though I might just be missing a fact that could allow me to simplify things). How should I conclude on conjectures (1) and (2)? And if they are false, can you shed some light on the geometry of the relationship between the two different $\ddot{r}$ expansions?

Thanks!

答:
The conjectures are false. To see why, first lets equate our expressions for $\dot{r}$.
$$
\dot{r} = J\dot{q} = v\hat{\tau}
$$

Non-degeneracy of the surface-coordinates imposes that $\frac{\partial r}{\partial x}$ is linearly independent of $\frac{\partial r}{\partial y}$. Therefore, $J$ is full-column-rank and admits a left-pseudoinverse $\ \tilde{J} := (J^\intercal J)^{-1} J^\intercal\ $ such that $\ \tilde{J}J = I$.
$$
\dot{q} = v\tilde{J}\hat{\tau}
$$

Taking another $t$-derivative we have,
$$
\ddot{q} = \dot{v}\tilde{J}\hat{\tau} + v\dot{\tilde{J}}\hat{\tau} + v\tilde{J}\dot{\hat{\tau}}
$$

Lets simplify this with the following definition:
\begin{gather}
h := \tilde{J}\hat{\tau}\ \implies\ \dot{h} = \dot{\tilde{J}}\hat{\tau} + \tilde{J}\dot{\hat{\tau}}\\[4pt]
\therefore\ \ddot{q} = \dot{v}\tilde{J}\hat{\tau} + v\dot{h}
\end{gather}

In general, $J\tilde{J} \neq I$. However, consider that $\tilde{J}$ acts to orthogonally project ambient vectors onto the local tangent-plane. Therefore, vectors that already live in the tangent-plane like $\hat{\tau}$ have the property $J\tilde{J}\hat{\tau} = \hat{\tau}$. (This isn't the case for $\dot{\hat{\tau}}$ which can have a normal component). Thus, left-multiplying our last equation by $J$ yields the corrected red equation:
$$
\color{red}{J\ddot{q} = \dot{v}\hat{\tau} + vJ\dot{h}}
$$

Substituting this into the original acceleration equation yields the corrected blue equation:
\begin{align}
J\ddot{q} + \dot{J}\dot{q} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}\\[4pt]
\dot{v}\hat{\tau} + vJ\dot{h} + \dot{J}\dot{q} &= \dot{v}\hat{\tau} + v\dot{\hat{\tau}}\\[4pt]
\color{blue}{\dot{J}\dot{q}} &\color{blue}{=} \color{blue}{v\dot{\hat{\tau}} - vJ\dot{h}}\\[4pt]
\end{align}

Since in general $v \neq 0$ and $J \neq 0$, the question of whether the original conjectures were true is a question of whether $\dot{h} \overset{?}{=} 0$. Examples can show this is false in general, but there is insight to be had now that we've isolated the correction.

One can geometrically interpret $h = \tilde{J}\hat{\tau} = \dot{q}/v$ as the 2D "heading" vector that an ant on the surface could express its direction of travel with, all in the tangent-plane basis. Whenever the ant turns or the tangent-plane basis-vectors change interior angle, $\dot{h}$ will be nonzero and the original conjectures won't hold. These ideas are put to use here.


答:
As far as I know, all mathematics fields are deeply connected to each other.
For example, Riemann hypothesis can be "reduced" to a "simple" arithmetic problem:
M(n)=O(sqrt(n))
where M(n) is the arithmetic mean of mu(1), mu(2), ... , mu(n), O is just big-O and sqrt(n) is a square root of n.

Here's a discussion in Russian:
Habr.com

And here's the original article.

Title: Why is there so little overlap between calculus and statistics?

Body:
I was an applied math major and took several calculus-based courses, including differential equations and real analysis, as well as an introductory course sequence in probability and statistics. Both calculus and statistics are extremely useful and broadly applicable to the natural and social sciences, but it seems like their tools and methods are completely different. Are the "statistical" and "calculus-based" approaches to solving problems fundamentally different, or is there actually significant overlap between these fields that is not evident to someone who only learned about these fields at the undergraduate and introductory levels?

It seems like one of the major goals of a calculus-based approach to understanding a system is to come up with a differential equation to model the data. And it seems like the statistical approach seeks to obtain causal and inferential relationships between variables. So it seems as if one of the primary goals of both fields is to come up with quantitative relationships between variables being observed in a system. So why are their methods and tools so different?

Also, certain sciences use calculus more than statistics (like physics) whereas others use statistics more than calculus (biology), and others use both extensively (economics). What about a scientific discipline makes  either analytical or statistical methods more useful?


问:
Comparing cardinalities between infinite sets
What is the relation of the cardinalities of the following two sets A and B, where A = the set of all subsets of the set of all even natural numbers and B = the set of all ﬁnite subsets of the set of all rational numbers?

My reasoning:

B is countable because the set of finite subset of a countably infinite set is countable. Therefore, we have that B as the same cardinality of the set of natural numbers. Furthermore, the set of all even natural numbers has the same cardinality of the set of natural numbers, as I guess we can find a bijection among them. But by Cantor's theorem as |P(X)|>|X|, therefore A as a greater cardinality of the set of natural numbers, hence, |A|>|B|.

Is this reasoning correct? If not, what should I change? On top of that, do you have any tips on how to write it formally, in a proper proof?

Thank you in advance.

答:
Your reasoning is correct. How to write it up nicely depends on what results you already have available to use. It might be enough to write something like this:

Let $E$ be the set of all even natural numbers. Then $$|A|=|\wp(E)|>|E|=|\Bbb N|=|B|\,,$$ where the inequality is from Cantor’s theorem, and the last two equalities are known results.


问:
Group acting on trees with relations $abc=1$ and $cba=1$
Suppose $G$ is a finitely presented group that acts on a tree $T$ by isometries, and let $a,b,c\in G$ with
relations $abc=1$ and $cba=1$. If two of $a,b,c$ are hyperbolic, does this imply the third element is also hyperbolic? Or any additional conditions that we need to make the implication hold?

答:
Assuming $a$ is hyperbolic, and letting $b=a^{-1}$, it follows that $b$ is hyperbolic too. The relation $ab=c^{-1}$ tells you that $c=1$ which is not hyperbolic.

To say something more general, let $a$ be hyperbolic with axis $A_a \subset T$, and $b$ be hyperbolic with axis $A_b \subset T$.

Suppose that $\alpha = A_a \cap A_b$ is a segment of positive length $L = \text{length}(\alpha)$ (which might be infinite). If the translation lengths of $a$ and $b$ are equal to each other and both less than $L$, and if $a$ and $b$ move points on $\alpha$ in opposite directions, then $ab=c^{-1}$ will fix a point on $\alpha$ and therefore $c$ will not be hyperbolic.

On the other hand suppose $A_a \cap A_b$ is either a single point or the empty set. In this case $ab=c^{-1}$ is hyperbolic and therefore $c$ is hyperbolic.

If you want to read more about this stuff I suggest the paper of Culler and Morgan, "Group actions on $\mathbb R$-trees".


问:
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$
How to show that $x-1$ and $x^2+x+1$ are irreducible over $\mathbb R[x]?$

I can see that $x - 1$ and $x^2 + x + 1$ are nonzero and non-units (for any field $F$ the set of all units of $F[x]$ = $F-\{0\}$)?

Added: Actually I would like to show $f(x)|(x-1)(x^2+x+1)~\text{where }\deg f(x)\ge 1~\implies f(x)=(x-1)(x^2+x+1)~or~(x-1)~or~(x^2+x+1).$ Does showing $(x-1)~and~(x^2+x+1)$ irreducible help me anyhow?

答:
Take $f\in F[x]$ with $F$ field. It is immediate to show that $f$ is irreducible in $F[x]$ if and only if it has no divisors of degree $k$ with $0<k<\deg (f)$ (just unfold the definition).

Now, from this carachterization it is obvious that $x-1$ is an irreducible polynomial (over any field indeed). About $g:=x^{2}+x+1$, suppose it is not irreducible; then there must be a polynomial $ax+b\in \mathbb{R}[x]$ ($a\neq 0$) such that $ax+b\mid g$, i.e $-b/a$ must be a root for $g$. But $g$ has no roots in $\mathbb{R}$ and we are done.


问:
For which $n$ is the symmetric group $S_n$ a subgroup of the special orthogonal group $SO(3)$?
For which $n$ is the symmetric group $S_n$ a subgroup of the special orthogonal group $SO(3)$? For example, this holds for $n≤4$, however I don't know if it holds for $n=5$ or what happens for larger $n$.

答:
This question can be attacked either as a question about the finite subgroups of $SO(3)$ or as a question about the representation theory of $S_n$. Each of these approaches will prove that $S_n \hookrightarrow SO(3)$ iff $n \le 4$.

Finite subgroups: It's a classical result that the complete list of finite subgroups of $SO(3)$ is the following:

the cyclic groups $C_n$,

the dihedral groups $D_n$,

the tetrahedral group $A_4$,

the octahedral group $S_4$, or

the icosahedral group $A_5$.

$S_4$ appears on this list but $S_n$ does not for $n \ge 5$.

Representation theory: The symmetric group $S_n$ is known to have the property  that the lowest-dimensional faithful representation has dimension $n-1$ for $n \ge 5$, as reuns points out in the comments.


问:
If $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$.
My book states the following: if $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$. But I don't understand this. It is not as if $f'$ is given to be continuous, rather $f$ is given continuous and differentiable at $x=c$. So how can we possibly comment about the existence of $f'$ in the neighborhood of $c$ too? Note that f is defined on an open interval I and c belongs to I.

答:
Let $\;g(x):\left]-\infty,+\infty\right[\to\mathbb{R}\;$ be the Darboux function defined as follows

$g(x)=\sum\limits_{n=1}^{\infty}\dfrac1{n!}\sin\bigl((n+1)!x\bigr)\quad\forall x\in\left]-\infty,+\infty\right[\;.$

$g(x)\;$ is continuous everywhere on $\;\left]-\infty,+\infty\right[\;$ and nowhere differentiable.

Let $\;c\;$ be a real number that is $\;c\in\mathbb{R}\;.$

Let $\;f(x):\left]-\infty,+\infty\right[\to\mathbb{R}\;$ be the function defined as

$f(x)=(x-c)^2g(x)\quad\forall x\in\left]-\infty,+\infty\right[\;,\;$

then $\;f(x)\;$ is continuous everywhere on $\;\left]-\infty,+\infty\right[\;,\;$ differentiable only at $\;c\;$ and $\;f’(c)=0\;,\;$ indeed

$\begin{align}
f’(c)&=\lim\limits_{x\to c}\dfrac{f(x)-f(c)}{x-c}=\lim\limits_{x\to c}\dfrac{(x-c)^2g(x)}{x-c}=\\
&=\lim\limits_{x\to c}(x-c)g(x)=0\;.
\end{align}$

Since $\;f(x)\;$ is differentiable only at the point $\;c\;,\;$ there does not exist any $\;h>0\;$ such that $\;f(x)\;$ is differentiable on the interval $\;\left]c-h,c+h\right[\;.$

Hence, it is wrong what your book states about continuous functions differentiable at a point $\;c\;$ such that $\;f’(c)=0\;.$

It is wrong even if the function is continuous everywhere, not only at the point $\;c\;,\;$ that is why my counterexample is different from those proposed by other users.


问:
Reversible Markov chain: aperiodicity and strictly positive self-loop probabilities, simplified proof?
Aperiodicity for Markov chains ensures that powers of self-loop probabilities $[P^{k}]_{ii}$ remain strictly positive from a certain power k onward (for the transition matrix $P$)

The proof relies on generalizing Bezout's lemma to the following result:

additively closed set lemma: If $A \subseteq Z^{+}$ is a coprime set of positive integers (GCD of its elements is 1) closed under addition, then there exists a positive integer N such that $A \supseteq [N, +\infty]$.

The proof of the additively closed set lemma uses Bezout's lemma (sketch given only)

One can show that $A$ must have a finite coprime subset $B$ (details omitted).

By Bezout, there exist integers $k_1, ..., k_n$ such that $k_1b_1 + \ldots, k_M b_M = 1$. Without loss of generality, assume that $k_1, \ldots, k_L > 0$ and $k_{L+1}, \ldots, k_M < 0$.

Let $a_1 = k_1 b_1 + \ldots k_Lb_L$ and $a_2 = -(k_{L+1} b_{L+1} + \ldots k_Mb_M)$. Clearly, both belong to $B$ and $a_1 - a_2 = 1$.

If $a_2 = 0$ then the result is trivial.

Otherwise, use Eucledian division and assume that $N \geq a_{2}^{2}$ to obtain that $N = la_2 + r$ for $0 \leq r < a_2$. From this we can deduce that N must belong to by rewriting $N$ to $N = (l-r)a_2 + ra1$ (relying on $a_1 - a_2 = 1)$. It can be shown that $(l-r) \geq 0$ thus $N \in A$ from $N \geq a_2^2$ onward.

One then applies the additively closed set lemma to the set $A_i = \{n \geq 1|\, [P^n]_{ii} >0\}$ for any state $i$.

The question is the following: for reversible markov chains, one can demonstrate that probabilities of self loop paths are the same, irrespective of the direction followed on the path. For instance a self-loop path $v_1 \rightarrow \ldots \rightarrow v_{j - 1} \rightarrow  v_1$ with probability $p_{12}p_{23} \ldots p_{(j-1)1}$ has the same probability as the reversed self-loop path $v_1 \rightarrow v_{j-1} \rightarrow \ldots \rightarrow v_2 \rightarrow v_1$, i.e. for reversible markov chains: $$p_{12}p_{23} \ldots p_{(j-1)1} = p_{1(j-1)} \ldots p_{32}p_{21}$$.

So, for reversible markov chains the notion of a positive direction on a self-loop and a negative direction could be used (with both directions having the same probability of being run through). This seems to indicate that instead of the generalized lemma discussed above, a more direct application of Bezout might be used to show that powers $[P^{k}]_{ii}$ remain strictly positive from a certain point on. The negative coefficients in Bezout would in this context correspond to running in the "negative" direction of a self-loop. Is this intuition valid? I.e. does the result of positive self-loop powers follow more easily for reversible markov chains?

More precisely: in the above proof, the use of $a_1 = k_1 b_1 + \ldots k_Lb_L$ and $a_2 = -(k_{L+1} b_{L+1} + \ldots k_Mb_M)$, where $a_1 - a_2 =1$ (via Bezout), may not need a further application of Eucledian division. Perhaps a more direct application of interpreting $a_1$ as following a series of positive self-loops and $a_2$ as following a series of negative self-loops can be used for reversible Markov chains?

For reversible markov chains the probabilities for reversed self-loops are the same as the probability of following the original self-loops. Hence if one is positive the other must be too. Could the "negative" self-loops used in Bezout, simply be reversed to yield a proof that the powers  $[P^{k}]_{ii}$ remain strictly positive from a certain point on?

Is this type of simplified proof given somewhere? (reference?) or is there an easy way to use this intuition and get the same result, known as part of the folklore on reversible markov chains? Perhaps a result obtaining a sharper estimate than $N \geq a_{2}^{2}$?

答:
For reversible chains, the additively closed set $A$ of return times contains $2$ (by reversibility) and hence all even positive integers. Now $A$ also contains some odd number $2k+1$ by aperiodicity,
hence $A$ contains all integers in $[2k+1, \infty)$.


问:
How to find the inverse tangent of a complex number's rectangular form as it relates to pi?
One of the hints on KhanAcademy appears to skip how they found the inverse tangent as a fraction of pi over a number.

Every time I type this into a calculator it comes up only as a decimal, and I'm not sure how you could find it with pi inside the fraction unless you already memorized the number to 9 digits or something.

The main problem:

Express $z_1 = 3\sqrt{3} - 9i$ in polar form.

Express your answer in exact terms, using radians, where you rangle is between 0 and $2\pi$ radians, inclusive.

The part that appears to skip how to find the answer as a fraction with pi involved:

Finding $\theta$

Using the formula, we have:

$\theta = \tan^{-1}(\frac{b}{a})$

$\theta = \tan^{-1}(\frac{-9}{3\sqrt{3}})$

$\theta = -\frac{\pi}{3}$

答:
I am assuming you have the requisite knowledge of radians and the unit circle in this answer.

Your question just seems to be a question on why $\arctan({-\sqrt{3}}) = -\frac{\pi}{3}$. The definition of $\arctan$ is just the inverse of the tangent function. If you know how to take the tangent of an angle, then you know the arctangent. If I want the value of $\arctan(x)$, then the arctangent function just asks "What ANGLE do I need to make such that the tangent of that angle is $x$?"

In your case, you want $\arctan({-\sqrt{3}}).$ On the unit circle, what ANGLE do I need to make from the axis such that the tangent of that angle is $-\sqrt{3}$? The answer? $-\frac{\pi}{3}$. And you can verify that $\tan(\frac{-\pi}{3}) = -\sqrt{3}$.

This is assuming you have the requisite knowledge of the unit circle to deduce that $\tan({\frac{-\pi}{3}}) = -\sqrt{3},$ though. Let me know if you have any questions.


问:
Does a finitely generated faithful module over an Artinian ring contain a regular element?
In the text

Nicholson -- Introduction to Abstract Algebra, 4th Ed (2012)

the claim of exercise $8(b)$ of exercise set $11.1$ is:

If $R$ is a left artinian ring with $1\ne 0$, and $M$ is a finitely generated left $R$-module such that $\text{ann}(M)=0$, then $M$ has a submodule isomorphic to $R$.

But in my answer to

$\;\;\;\;\;$
How to show $\operatorname{ann}(M) = \operatorname{ann}(X)$.

I gave a counterexample to the above claim.

I wonder if the claim could be repaired by assuming as an additional hypothesis that $R$ is commutative.

Question:

If $R$ is a commutative artinian ring with $1\ne 0$, and $M$ is a finitely generated $R$-module such that $\text{ann}(M)=0$, must $M$ have a submodule isomorphic to $R$?

Two special cases: The answer is "yes" if

$R$ is a field.$\\[4pt]$

$R$ is finite.

That's as far as I've got.

答:
Wow, that is an unfortunate mistake you found.

For semiperfect rings, there is a notion of the basic module of the ring, which is a faithful module that captures a lot about the ring. This is a theorem:

If $R$ is a semiperfect ring, then right basic module is a summand of any generator of Mod-$R$.

For a commutative Artinian ring, the basic module is just $R$ itself. But the missing ingredient, as you see, is that a faithful module need not be a generator of Mod-$R$.  A ring for which every faithful f.g. module is a generator of Mod-$R$ is called finitely pseudo-Frobenius. So the best you can say, I think is

If $R$ is a commutative, semiperfect, finitely pseudo-Frobenius ring, then $R$ is a summand of every f.g. faithful module.

The nicest class for which this is all true is commutative quasi-Frobenius rings.


问:
Proof of van Handel "Probability in High Dimension" Theorem 2.3
I have a question about the proof of Theorem 2.3 in van Handel's Probability in High Dimension (page 15). At the bottom of page 15, the book writes $$E[E[f(X_1,\dots,X_n)\mid X_1,\dots,X_{k-1},X_{k+1},\dots,X_n]\mid X_1,\dots,X_{k-1}]=E[E[f(X_1,\dots,X_n)\mid X_1,\dots,X_{k-1},X_{k+1},\dots,X_n]\mid X_1,\dots,X_{k}]$$ where $X_1,\dots,X_n$ are independent. I was unable to understand why this is true. Thank you!

答:
Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $X_{1},X_{2},\ldots,X_{n}$
be indpendent random variables. Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$
be a bounded Borel function (Note that boundedness can be relaxed.
Here, we impose this condition so that conditional expectation makes
sense). Fix an integer $k$ with $1<k<n$. Let $\mathcal{G}_{1}=\sigma(X_{1},X_{2},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})$.
Observe that $X_{1},X_{2},\ldots,X_{k-1},X_{k+1},\ldots,X_{n}$ are
$\mathcal{G}_{1}$-measurable while $X_{k}$ and $\mathcal{G}_{1}$
are independent. Therefore, Independent Lemma is applicable. Define
a Borel function $\varphi_{1}:\mathbb{R}^{n-1}\rightarrow\mathbb{R}$
by $\varphi_{1}(x_{1},\ldots,x_{k-1},x_{k+1},\ldots,x_{n})=E\left[f(x_{1},\ldots,x_{k-1},X_{k},x_{k+1},\ldots,x_{n})\right]$.
By Independent Lemma, we have that $E\left[f(X_{1},\ldots,X_{n})\mid\mathcal{G}_{1}\right]=\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})$.

Let $\mathcal{G}_{2}=\sigma(X_{1},X_{2},\ldots,X_{k})$. Note that
$X_{1},X_{2},\ldots,X_{k-1}$ are $\mathcal{G}_{2}$-measurable while
$X_{k+1},X_{k+2},\ldots,X_{n}$ and $\mathcal{G}_{2}$ are independent.
Define a Borel function $\varphi_{2}:\mathbb{R}^{k-1}\rightarrow\mathbb{R}$
by $\varphi_{2}(x_{1},x_{2},\ldots,x_{k-1})=E\left[\varphi_{1}(x_{1},\ldots,x_{k-1},X_{k+1},\ldots,X_{n})\right].$
By Independent Lemma, we have that $E\left[\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})\mid\mathcal{G}_{2}\right]=\varphi_{2}(X_{1},\ldots,X_{k-1})$.

Let $\mathcal{G}_{3}=\sigma(X_{1},X_{2},\ldots,X_{k-1})$. Clearly
$\mathcal{G}_{3}\subseteq\mathcal{G}_{2}$.
Denote $Y=f(X_{1},\ldots,X_{n})$. By tower property of conditional
expectation, we have that
\begin{eqnarray*}
 &  & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{3}\right]\\
 & = & E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right].
\end{eqnarray*}
However,
\begin{eqnarray*}
 &  & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\\
 & = & E\left[\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})\mid\mathcal{G}_{2}\right]\\
 & = & \varphi_{2}(X_{1},\ldots,X_{k-1}),
\end{eqnarray*}
which is $\mathcal{G}_{3}$-measurable. Taking $\mathcal{G}_{3}$-conditional
expectation on both sides yields $E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right]=\varphi_{2}(X_{1},\ldots,X_{k-1}).$

Hence,
\begin{eqnarray*}
 &  & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{3}\right]\\
 & = & E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right]\\
 & = & \varphi_{2}(X_{1},\ldots,X_{k-1})\\
 & = & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right].
\end{eqnarray*}


问:
Asymptotics of $\sum_{m=1}^{n} n \mod m$
Let $\displaystyle a_n = \sum_{m=1}^{n} (n \mod m)$ where $0 \le n \mod m \le m-1$.

It appears that $a_n \sim Cn^{\alpha}$ with $\alpha = 2$ and $C \approx .177448$. Is this true, and if so why?

答:
This is OEIS A004125. Your asymptotics are true and $C=1-\frac{\pi^2}{12}$.

This arises because
$$\sum_{m=1}^n(n\bmod m)=n^2-\sum_{m=1}^n m\lfloor n/m\rfloor$$
and Walfisz showed the asymptotics of the latter to be $\frac{\pi^2}{12}n^2$ (see OEIS A024916 or here).


答:
You start with a solution to the equation $x^2 + x + 1 = 0$ and correctly show that it is also a solution to the equation $x^3 = 1$. You then observe that the only real solution to the second equation is $1$ (this is true).

You then conclude that $x = 1$ is a solution to the first equation (this is not true, there may be no real solutions to your original equation, as is in fact the case).

Analogy: if you started with $x^2 = -1$ and squared both sides, you would get $x^4 = 1$, but you can't conclude that $x = 1$.

Title: What's wrong with this proof of $3=0$ starting from $x^2+x+1=0$?

Body:
What am I missing?

$$ x^2 + x + 1 = 0$$
Then
$$ x^2 + x = -1$$
$$ x(x+1) = -1$$
$$ x(-x^2) = -1$$
$$x^3 = 1 $$
$$x = 1 $$
but
$$(1)^2 + (1) + 1 = 3 $$

So

$$ 3 = 0$$


答:
If you are assuming $x$ is a real number, then all of these steps are valid, and prove that the assumption $x^2+x+1=0$ gives a contradiction.  In other words, this is just a proof that the equation $x^2+x+1=0$ has no real solutions.

If you allow $x$ to be a complex number, then $x^2+x+1=0$ does have solutions, but the step where you go from $x^3=1$ to $x=1$ is wrong: there are complex numbers besides $1$ whose cubes are $1$.  (This step is valid for real numbers, since the function $x\mapsto x^3$ from $\mathbb{R}$ to $\mathbb{R}$ is strictly increasing, so it can take the value $1$ at most once.)


答:
$x^3=1$ does not imply $x=1$. By doing "algebra" you can introduce false solutions which need to be checked.

$x^3=1 \iff (x-1)(x^2+x+1)=0$

You can see your original equation from here, with an additional solution added.


答:
Nothing wrong: it is true that if $x^2+x+1=0$ and $x\in\Bbb R$, then $x=1$. Keeping all the pieces together $$x^2+x+1=0\Leftrightarrow\begin{cases}x^2+x=-1\\ x+1=-x^2\end{cases}\Leftrightarrow \begin{cases}-x^3=-1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ 2=-1\end{cases}$$

So the equation has no solutions in  $\Bbb R$.

Of course, we may argue that there is some degree of circularity in proving that $x^2+x+1$ has no real roots by using the fact that $x^3-1$ has exactly one root in $\Bbb R$...


问:
Help me prove this inequality :
How would I go about proving this?

$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$

Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!

答:
And here's the proof why $(1+1/x)^x$ is concave for $x\le 1$ (the case $x\ge 1$ is already proven by @TCL).

We study the second derivative with respect to $x$, it is equal to (thanks to TCL) to

$\displaystyle \left(1+ \frac 1x\right)^x\left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$;

First of all, $\ln (1+1/x)\ge\frac{1}{1+x}$; we say that $y=1/x\ge 1$ and then study $(\ln (1+y)-\frac{y}{1+y})$ - this function is positive in $y=1$ and it's derivative with respect to $y$ is positive for $y\ge 1$, hence the function itself is always positive; therefore $\sqrt{\left( \ln (1+1/x)-\frac{1}{1+x}\right)^2}= \ln (1+1/x)-\frac{1}{1+x}$.

Now we study the sign of $\displaystyle  \left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$; it's equivalent to study the sign of
$\displaystyle    \ln (1+1/x)-\frac{1}{1+x}  -\frac{1}{\sqrt x(1+x) } $. Once again, we change $y=1/x\ge 1$:

$\displaystyle    \ln (1+y)-\frac{y}{1+y}  -\frac{y\sqrt y}{ (1+y) } $.

In $y=1$ the last expression is negative (we have concavity for $x\ge 1$, after all); its derivative for $y\ge 1$ is (after several manipulations)

$\displaystyle 0.5(1+y)^{-2}(2y -3\sqrt y - y^{3/2})\le 0$,

which ensures that $    \ln (1+1/x)-\frac{1}{1+x}  -\frac{1}{\sqrt x(1+x) } $ remains negative for $y\ge 1$; therefore, $(1+1/x)^x$ is concave for $x\le 1$. QED

Title: Help me prove this inequality :


问:
Upper bound on the number of edges of a ten vertexes graph no $C_4, C_3$ subgraph.
We are given a graph $G$ with ten vertices. Any three or four vertices of $G$ don't form a cycle. What is the maximum number of edges $G$ could have?

I know that if the graph is planar and triangle free, it would have 16 edges. It seems that between any 4 vertices, there can be at most 3 edges.

Maybe there is a formula and can be proven by induction on number of vertex. I'm not sure.

答:
The Petersen graph, which contains no cycle of length $3$ or $4$, has $10$ vertices and $15$ edges. I will show that $15$ is the maximum, and is attained (among graphs with $10$ vertices and no $3$-cycles or $4$-cycles) only by the Petersen graph.

Suppose $G$ is a graph with $10$ vertices and at least $15$ edges which contains no cycle of length $3$ or $4$. An acyclic graph on $10$ vertices has at most $9$ edges, so $G$ must contain a cycle. Let $k$ be the minimum length of a cycle in $G$, and let $C$ be a cycle in $G$ of length $k$; so $5\le k\le10$. Let $D=V(G)\setminus V(C)$.

Now, if a vertex $v$ in $D$ were joined to two vertices in $C$, then $v$ would be in a cycle of length $\lt k$, contradicting the minimality of $k$. Therefore, each vertex in $D$ is joined to at most one vertex in $C$. Since the cycle $C$ has $k$ edges (and no chords), there are at most $k+(10-k)=10$ edges with at least one endpoint in $V(C)$, so there must be at least $5$ edges joining two vertices in $D$. Hence $\binom{|D|}2\ge5$, i.e., $|D|\ge4$. However, a graph on $4$ vertices with no cycle of length $3$ or $4$ is acyclic, so it has at most $3$ edges. Therefore the only possibility is that $k=5$ and $|D|=5$.

Since an acyclic graph on $5$ vertices has at most $4$ edges, $D$ must contain a cycle, necessarily of length $5$, and that cycle can have no chords. Thus there are just $5$ edges joining two vertices in $D$, and $G$ has just $15$ edges all told.

We have shown that a graph, which has $10$ vertices and $15$ edges and contains no cycle of length $3$ or $4$, must consist of two vertex-disjoint $5$-cycles and a matching between them. It is easy to see that such a graph, if it contains no cycles of length $4$, must be the Petersen graph.

Alternatively, let $f(n)$ denote the maximum number of edges in a graph with $n$ vertices and no cycles of length $3$ or $4$. For $n\ge3$ the inequality
$$f(n)\le\left\lfloor\frac{nf(n-1)}{n-2}\right\rfloor\tag1$$
can be seen by counting in two ways the pairs $(v,e)$ where $v\in V(G)$ and $e\in E(G-v)$. Then, after verifying $f(6)=6$, we can use $(1)$ to show that $f(7)\le8$, $f(8)\le10$, $f(9)\le12$, and $f(10)\le15$.


问:
Defining a set by induction under ZFC
The definition of a set by induction is used in first order logic to define a couple of sets and plenty other sets can be defined in this way, for example the set of all powers of 2.

Let $X$ be the smallest set such that

$0 \in X$,

if $x \in X$ then $2x \in X$.

This of course does not prove anything about the existence of the set $X$ and the only possible way I can see to assert the existence is to add an axiom that says so. But this feels unpleasant because is not ZFC anymore.

In general all the definitions i saw (logic and structure van dalen) match the following criteria

Let $X$ be the smallest set such that

$x_0, x_1, \ldots \in X$

If $A_0, A_1, \ldots, A_{k_1} \in X$ and $\phi_1(p_1, p_2, \ldots, p_{l_1})$ then $op_1(A_0, A_1, \ldots, A_{k_1}, p_1, p_2, \ldots, p_{l_1}) \in X$

If $A_0, A_1, \ldots, A_{k_2} \in X$ and $\phi_2(p_1, p_2, \ldots, p_{l_2})$ then $op_2(A_0, A_1, \ldots, A_{k_2}, p_1, p_2, \ldots, p_{l_2}) \in X$

$\vdots$

If $A_0, A_1, \ldots, A_{k_m} \in X$ and $\phi_m(p_1, p_2, \ldots, p_{l_m})$ then $op_m(A_0, A_1, \ldots, A_{k_m}, p_1, p_2, \ldots, p_{l_1}) \in X$

where $op_i$ is not a function but a notation, like $n^+ := n \cup \{n\}$ used in the axiom of infinity, $\phi_i$ is a sentence and $k_i$, $i$ or any number is not a member of $\omega$ but the concept of number used informally like the one from the pairing axiom "for any two sets there exists ...".

In van dalen for example each one of these sets has a induction principle theorem and a primitive recursion theorem with a generalized proof, similar to what we have in $w$, in fact $w$ follows this criteria.

Is there a way to get around this under ZFC (in general) or am I missing something very simple?

EDIT

Another example

Let $[Int]$ be the smallest set such that

$nil \in [Int]$

For every $n \in \omega$ if some $x \in [Int]$ then $n :: x \in [Int]$

(Haskell like lists of integers)

```<code>[1, 2, 3] = 1 :: (2 :: (3 :: nil))
</code>```

答:
(I'm going to work with $1$ in place of $0$ here, since otherwise things are a bit trivial.)

There are various ways to approach this in $\mathsf{ZFC}$ (or similar).

I personally like a "from-above" definition - such definitions are extremely powerful, and - while less concrete - once understood are very easy to work with. First, recognizing that we're living in the context of the natural numbers, we introduce the operation $$f: \mathcal{P}(\omega)\rightarrow\mathcal{P}(\omega): A\mapsto A\cup \{1\}\cup\{2n: n\in A\}.$$ This $f$ can be produced by separation (+ powerset/infinity of course). Now say that a set $U\subseteq\omega$ is an $f$-fixed point if it satisfies $f(U)=U$. We can then consider the least fixed point of $f$, namely the set $$X=\bigcap_{f(U)=U}U=\{n: \forall U\subseteq\omega(f(U)=U\implies n\in U)\}.$$ Again, this exists via separation (and we can show that it satisfies $f(X)=X$ easily). This is exactly the set we want.

Title: Defining a set by induction under ZFC