问:
Calculating eigenvectors of a 3x3 symmetric matrix from Cayley–Hamilton theorem?
Let's say I have a 3x3 real symmetric matrix M and I have its three eigenvalues $\alpha_1, \alpha_2, \alpha_3$ and that $\alpha_1 < \alpha_2 < \alpha_3$.  (Notice all eigenvalues are different).

We can deduce that $(M-\alpha_{1}I)(M-\alpha_{2}I)(M-\alpha_{3}I) = 0$ must be true.

Therefore, the three columns of $S = (M-\alpha_{2}I)(M-\alpha_{3}I)$ must all be eigenvectors of $\alpha_1$.

Notice that S must be symmetric because M and I are symmetric, and the sum and product of two symmetric matrixes is also symmetric.  (Not sure if that's relevant.)

My question is about the columns of S.  How many columns of S will be non-zero vectors?  Zero, one, two or three?  Or will it vary depending on M?  and why?

答: If all columns of $S$ are eigenvectors of $M$ associated with $\alpha_1,$ and $S$ is symmetric, then $S=c\,v_1v_1^T$ with a suitable $c\in\mathbb{R}$ (and $v_1$ is one eigenvector of $M$ associated with $\alpha_1.$) This shows that there are exactly as many columns filled with zeroes in $S$ as there are zeros in $v_1.$


问:
How can I solve probability questions using the lebesgue integral?
Here are examples of questions I found in Google. I was curious about whether they can be solved via lebesgue integration and how. Examples:

A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after $n$ selections ($n>1$), the product of the $n$ numbers selected will be divisible by 10.

Basically, I’ve been taking a break off olympiad math and been reading schilling. I was curious about whether I can apply this knowledge.

答:
Here's how you can utilize Markov chains to solve this problem.

Let $S_0$ be the state that the number isn't divisible by $2$ nor $5$, $S_1$ the state the number is divisible by $2$ but not $5$, $S_3$ the state the number is divisible by $5$ but not $2$, and $S_4$ the (absorbing) state the number is divisible by $10$.

This Markov chain generates the following transition matrix $P$: $$P=\begin{pmatrix}\frac{7}{9}&\frac{1}{9}&\frac{1}{9}&0\\ 0&\frac{8}{9}&0&\frac{1}{9}\\ 0&0&\frac{8}{9}&\frac{1}{9}\\ 0&0&0&1\end{pmatrix}$$ The probability the number is divisible by ten after $n$ selections equals $$p_{14}^{(n)}=e_1^TP^n e_4$$

ContentLicense: CC BY-SA 4.0


问:
$x^{n}$ such that n is positive and odd, the range is all real numbers. Prove it.
What I think the answer is : The range is the set of all real numbers because we can have a number $x^1$ that is equal to $x^2$

Specifically, to access even powers of n, all we need do is scale up x of the odd powers.

And since the function is not a quotient or a square root, there are no constraints of the functions domain. Therefore, since the domain is the set of all real numbers, the range is the set of all real numbers as well.

答: The reason why the range of all functions of the type $f: \mathbb{R} \rightarrow \mathbb{R},$ $f(x) = x^n,$ where $n \in 2 \mathbb{N} + 1$ is the entire $\mathbb{R}$ is that obviously all of these mappings are continuous, which can be readily verified and also $\lim_{x \to -\infty} x^n = -\infty,$ $\lim_{x \to \infty} x^n = \infty$ and a simple application of Darboux's property helps you get your conclusion.


问:
Set construction
Have the set $H = \{1, 2, 3, ..., 24\}$, then construct the set:

$$ P(H) \setminus \emptyset = \{\{1\}, \{2\},..., \{24\}, \{1,2\}, ..., \{23,24\}, ..., \{1,2,3,...,24\}\}$$

The set of subsets of $P(H)$ that have cardinality 1 is given by:

$$P(H_1) = \{\{1\}, ..., \{24\}\} $$

Now, have the complement of $P(H_1)$:

$$C = P(H_1)^c = \{\{1,2\}, \{2,3\} ..., \{1,2,3,...,24\}\} $$

Suppose $h$ is a function from N to R, such that: $$h(n) = \frac{n}{60}$$

where $n$ is an arbitrary natural number.

I want to construct a subset of $C$, in which each subset of this set is the smallest subset that satisfies:

$$|s| \geq h(n)$$

For example: For h(30) = 0.5, the set of smallest subsets that satisfies the condition above is:

$$ S = \{\{1,2\}, \{2,3\}, ..., \{23,24\}\} $$

Where each $s \in S$ is the smallest subset of $C$ that satisfies $|s| \geq h(n)$.

My attempt at constructing this set is:

$$ S = \{s | inf\{|h(n) - |s||\}, s \in C, h(n) \in R \} $$

Is it correct?

答:
There are a few things wrong with your expression:

Taking the absolute value of $h(n)-|s|$ is going to prevent you from distinguishing between $h(n)>|s|$ and $h(n)<|s|$.

$\inf\{\dots\}$ is just a number, but the stuff after the $\{s|$ should be complete propositions representing conditions that $s$ must satisfy.

The condition $h(n)\in\Bbb R$ has nothing to do with $s$, so it doesn't make sense here. (We aren't building a set by collecting several values of $h(n)$, we're just using $h(n)$ as a fixed constant.)

I'd approach this in two steps:

Figure out an expression for the size $|s|$ you're looking for. This is the smallest integer that's at least 2 and at least $h(n)$. I'd write this as $\max(2,\lceil h(n)\rceil)$.

Write an expression that selects the subsets of $H$ having that size. I'd write it as $\{s\subseteq H\text{ }|\text{ }|s|=\max(2,\lceil h(n)\rceil)\}$.

Title: Set construction


问:
Solve the equation in natural numbers $x^3+2x+1=y^2$
The Question

Solve the equation in natural numbers:

$x^3+2x+1=y^2,x,y\in N$

The solution is:

$x=1,y=2;x=8,y=23$

My Understanding

The given equation can be written as $x(x^2 +2)=(y-1)(y+1)$, as we're working in $\mathbb{N}$ we've the following cases:

Case 1: $x=y-1$ and $x^2 +2 =y+1$ using these we get $y^2 -3y +2=(y-2)(y-1)=0$ therfore it follows that $(x,y)=(1,2)$ is the only solution in this case.

Case 2: $x=y+1$, and $x^2+2 =y-1$ , using these we get $y^2 +y+4=0$. As the quadratic is irreducible over $\mathbb{Z}$ we can conclude the non existence of solutions in this case.

Therefore we can conclude that the only solution in $\mathbb{N}^{2}$ is $(x,y)=(1,2)$

What about $(8,23)$?

答:
Solving this problem using Elliptic Curves is probably not the best idea.

Examining the solutions to $E:y^2=x^3+2x+1$ over finite fields we see that $E(\mathbb{F}_{11})=16$ and $E(\mathbb{F}_{19})=27$. This mean that $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $16\cdot11^{r_1}$ for some $r_1$ and $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $27\cdot19^{r_2}$ for some $r_2$. The only solution is $|E(\mathbb{Q})_{\mathrm{tors}}|=1$, and so the torsion group is trivial.

As for the rank of the curve, the method of descent applied with $p=2$ shows that $E$ has rank $1$ (this can be done in Magma or Sage).

Combining both results, we see that $E(\mathbb{Q})$ is generated entirely by the free point $(0,1)$ on the curve. The issue is, there is no simple way to reduce this to just the solutions over $\mathbb{Z}$. It defiantly can be done, Magma or Sage can tell us that the the points $(0,\pm1), (1,\pm 2)$ and $(8,\pm 23)$ are the only solutions. If this question is being asked to high schoolers or undergraduates, this is not the way to go about it.


答:
Your notation $2^{H_1}$ is somewhat unfortunate as $2^X$ is standard notation for the power set of $X$ or (essentially equivalently) the set of functions from $X$ to a set with two elements. So the notation suggests there is a set $H_1$ whose power set comprises the singletons in the power set of $H$, which is not true. I will just write $C$ for what you call $C(2^{H_1})$, the set of subsets of $H$ with at least two elements. Here is a set comprehension that gives the set $S_n$ of elements $X$ of $C$ that are minimal subject to the constraint that $|X| \ge h(n)$.

$$S_n  = \{X \in C \mid \forall Y \subseteq X\cdot |Y| \ge h(n) \Leftrightarrow Y = X\}$$

Title: Set construction

Body:
Have the set $H = \{1, 2, 3, ..., 24\}$, then construct the set:

$$ P(H) \setminus \emptyset = \{\{1\}, \{2\},..., \{24\}, \{1,2\}, ..., \{23,24\}, ..., \{1,2,3,...,24\}\}$$

The set of subsets of $P(H)$ that have cardinality 1 is given by:

$$P(H_1) = \{\{1\}, ..., \{24\}\} $$

Now, have the complement of $P(H_1)$:

$$C = P(H_1)^c = \{\{1,2\}, \{2,3\} ..., \{1,2,3,...,24\}\} $$

Suppose $h$ is a function from N to R, such that: $$h(n) = \frac{n}{60}$$

where $n$ is an arbitrary natural number.

I want to construct a subset of $C$, in which each subset of this set is the smallest subset that satisfies:

$$|s| \geq h(n)$$

For example: For h(30) = 0.5, the set of smallest subsets that satisfies the condition above is:

$$ S = \{\{1,2\}, \{2,3\}, ..., \{23,24\}\} $$

Where each $s \in S$ is the smallest subset of $C$ that satisfies $|s| \geq h(n)$.

My attempt at constructing this set is:

$$ S = \{s | inf\{|h(n) - |s||\}, s \in C, h(n) \in R \} $$

Is it correct?


问:
How do you turn a proof of a mathematical statement into a zero-knowledge proof?
I recently watched a video on Numberphile2 in which Avi Wigderson describes how one can prove a graph has a 3-colouring in zero-knowledge and that as 3-colouring is NP-complete, all NP statements have zero-knowledge proofs. He further emphasises that this is constructive; one can turn a proof of an NP statement into a zero-knowledge proof of that statement.

As I already understood the Cook-Levin reduction from NP problems to 3-SAT and the reduction from 3-SAT to 3-colouring, I thought I'd be able to do exactly this with a simple mathematical statement like the infinitude of the primes, but I seem to be missing some steps in the process.

I should first say that my understanding is that the statements which are in NP are precisely the statements which have a second-order formula only involving existential second-order quantification, which certainly seems to cover all the mathematical statements I can think of. For my desired example, one can express the infinitude of the primes with the following second-order formula:

$\exists f \forall x \forall y (x < f(x) \land 1 < f(x) \land (1 \nless y \lor y \nless f(x) \lor y \nmid f(x)))$

It expresses that there exists a function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the output of $f$ is prime and always produces output bigger than its input, for all natural number inputs, therefore there must be infinitely many primes. And so proving the existence of such a function $f$ constitutes a proof of the statement.

Though it's an awkward way of writing it, it does have the advantage that the predicate being quantified over is in Conjunctive Normal Form, which is what is required for the reduction from 3-SAT to 3-colourability. But I don't understand where to go from here, as the existential quantifier isn't quantifying over predicates (having Boolean outputs), the universal quantifiers are quantifying over natural numbers and the predicate involves two binary predicates on natural numbers $<$ and $|$.

How does one reduce the problem of proving a mathematical statement (second-order formula with no second-order universal quantifiers), such as the infinitude of the primes, to the problem of proving a SAT instance has a satisfying assignment? Preferably in a way that is somewhat good at minimising the size of the resultant SAT instance.

答: Instead of converting the statement "the primes are infinite", you can convert the statement "there is a proof that the primes are infinite with less than n steps in deductive system X", for some X. This new statement can be more easily converted to SAT and 3-colorability.</p>


问:
How can an average decrease in numbers show up as an average percentage increase?
In the list below:

the first column shows the population

the second column shows the number by which the population has increased (so the total new population in the first row would be 27312).

the third column shows the percentage increase i.e. column 2 / column 1

I then took the average of the second column and got an average number.
But then I took the average of the third column and got an average percentage increase.

That doesn't make sense to me. My brain is a little fried. I'm trying to get an intuition for why this is.

Surely an average fall in numbers should mean an average fall in
percentage change?

答:
When you average the percentages, you’re giving each of them the same weight in the average. But a $10\%$ change starting at $100,000$ is a change of $10,000$, while a $10\%$ change starting at $10,000$ is a change of only $1000$. That first change has $10$ times the effect on the overall population, even though it’s an identical percentage of its base population. If the first is an increase and the second a decrease, the average of the percentages as $0\%$, but the total population has gone from $110,000$ to $119,000$, an increase of $9000$, or $8.\overline{18}\%$.

For an average of the percentages to be meaningful, it would have to be a weighted average, with each percentage weighted by the fraction of the total population that it affects. In this case the first change affects $\frac{10}{11}$ of the total population of $110,000$, while the second affects only $\frac1{11}$ of it, so the weighted average of the percentages is

$$\frac{10}{11}(10)+\frac1{11}(-10)=\frac{90}{11}=8\frac2{11}=8.\overline{18}\,,$$

which is indeed the percentage by which the total population has changed.


问:
Verification of my proof for irrationality of $\sqrt2$
Here's my proof:

Let us suppose for sake of contradiction, $\sqrt2$ is rational. Then, we can write
$$\sqrt2=\frac{a}{b}$$
for non-zero coprime integers $a$ and $b$.
Squaring our equation,
$$2=\frac{a^2}{b^2}$$
Here, LHS is an integer, but since $a^2$ and $b^2$ are coprime, RHS is not an integer. This is a contradiction, and so our original assumption was wrong. Therefore, $\sqrt2$ is irrational.

Is this valid? Thanks in advance.

答:
You need to point out that $\sqrt 2 =\frac ab$ where $a,b$ are relatively prime that $b \ne 1$.

That's easy enough.  You just have to show there is no integer so that $n^2 = 2$ and there obviously aren't as $|n| \ge 2 \implies n^2 \ge 4$ and $|n| < 2\implies |n| \le 1\implies n^2 \le 1$ so there are no integer square roots of $2$ or $3$.

You need to point out that $a,b$ are relatively prime (and $b\ne 1$) that would imply $a^2$ and $b^2$ are relatively prime (and $b^2 \ne 1$).

If we have the fundamental theorem of arithmetice (all integers have unique prime factorizations)  and/or euclid's lemma (if $p$ is prime and $p|nm$ then $p|n$ or $p|m$) that's short work.

$b\ne 1$ so $b^2 \ne 1$ and $b^2$ must have a prime factor and that prime factor, call it $p$ must, by Euclid's, lemma divide $b$.  By transitivity that prime factor must divide $a^2$ and must therefore divide $a$.  But that contradicts $a,b$ are relatively prime.

Then your proof is complete.  As $a^2,b^2$ are relatively prime and $b^2 \ne 1$ then $b^2$ is not itself a factor of $a^2$ and $\frac {a^2}{b^2}$ can not be the integer $2$.

If you don't have the FTA or EL then you have your work cut out for you.  Better to just do the classic.

(That is proving that if $m$ is even/odd then $m^2$ is even/odd and therefore $2 =\frac {a^2}{b^2}$ must imply $a,b$ are both even and thus not relatively prime.)

(This assumes that all rationals can be written as a pair of relatively prime integers.  Although that is not a basic definition and should be proven, I'll allow that one to go assumed.)


问:
Do Groebner bases over rationals lose information about complex solutions?
Given a system of polynomial equations, if I calculate the Groebner basis over the rationals and get {1} then:

My understanding is that I can then conclude there are no rational solutions. Is this correct?

Can I also conclude that there are no complex solutions as well? (Because there are no complex solutions to 1=0? ... or is that being too flippant?)

If they do lose information on complex solutions, under what conditions does this occur? I've made some tests and haven't found a case where information is lost. So it appears, at the very least, there are many cases in which no information is lost. In particular, what if I restrict to systems of equations which are multilinear in the variables? (ie. it could have a fourth order term such as $wxyz$ but there is never a term with a variable raised to second or higher power)

Or is there no simple condition and I just need to recalculate the Groebner basis over the complex polynomial space?

I'm sorry if these questions sound very basic, but this is outside of the math I've learned and I used a CAS to help me solve a problem. So I want to be absolutely sure I am not misusing the tools and making some incorrect assumptions (and Mathematica only calculates Groebner basis over the rationals, yet makes claims about complex systems of polynomials, so I need to understand what I can trust).

Here is a simple test:

```<code>In[1]: GroebnerBasis[{x^2 + y, y - 1}, {x, y}, CoefficientDomain -> Rationals]
Out[1]: {-1 + y, 1 + x^2}
</code>```

And even though there are no rational solutions, it did not reduce to {1}. So at least in this case, it did not lose information about complex solutions.

In the comments, Dietrich Burde suggested adding x^2 + y^2 + 1

```<code>In[2]: GroebnerBasis[{x^2 + y, y - 1, x^2 + y^2 + 1}, {x, y}, CoefficientDomain -> Rationals]
Out[2]: {1}
</code>```

This collapsed to {1}, so this is more directly related to my question. In this case there are no complex solutions either, so no information was lost.

Here is a more involved example, with multilinear equations that only has rational solutions with $w=x=0$, but has complex solutions with all variables non-zero ($x=y=1$, $w=z=i$):
$$
\begin{aligned}
x y + w z &= 0 \\
-x z + y w &= 0 \\
\end{aligned}
$$
And here is what Mathematica calculates for the Groebner basis over the rationals

```<code>In[1]: GroebnerBasis[{x y + w z, -x z + y w}, {w, x, y, z},
 CoefficientDomain -> Rationals]
Out[1]: {x y^2 + x z^2, x y + w z, w y - x z}
</code>```

The non-trivial complex solution is still a solution, so it does not appear any information was lost here either. However the basis is no-longer multi-linear, which suggests multi-linear equations are not somehow special here.

答:
Consider the set of complex solutions to a system of polynomial equations $$S = \{(p_1,\ldots,p_n) \in \mathbb{C}^n \mid f_i(p_1,\ldots,p_n) = 0 \text{ for each } i=1,\ldots,r \}$$ where the $f_i \in \mathbb{Q}[x_1,\ldots,x_n]$ are polynomials in $x_1,\ldots,x_n$ with rational coefficients.

If a Groebner basis of the ideal $I = \langle f_1, \ldots, f_r \rangle \subset \mathbb{Q}[x_1,\ldots,x_n]$ contains $1$, then in particular $1$ is contained in the ideal $I$, which means $1 = \sum_{i=1}^r c_i f_i$ for some polynomials $c_i \in \mathbb{Q}[x_1,\ldots,x_n]$. Any solution $p = (p_1,\ldots,p_n)$ in $S$ is then also a solution to $1 = \sum_{i=1}^r c_i(p)f_i(p) = 0$ (i.e. $1=0$), so there are no solutions: $S = \varnothing$. In the same way there are no rational solutions either ($S \cap \mathbb{Q}^n = \varnothing$).

If (a Groebner basis of) $I$ does not contain a nonzero constant, then the system of equations has a complex solution: $S \neq \varnothing$ (this is Hilbert's weak Nullstellensatz). The question of existence of rational solutions (rational points) is generally more subtle (it is a whole field of study).

When you compute a Groebner basis of $I$ as an ideal in $\mathbb{Q}[x_1,\ldots,x_n]$ (e.g. using Buchberger's algorithm), then effectively you will only do plain arithmetic with the input polynomials $f_i$, like division with remainder, addition, subtraction, multiplication, and division by nonzero constants. In other words, all the arithmetic takes place in $\mathbb{Q}[x_1,\ldots,x_n]$, and there is never any need to go to a bigger ring, or bigger field of coefficients. If you consider $I$ as an ideal in $K[x_1,\ldots,x_n]$, for any field extension $K$ of $\mathbb{Q}$ (in particular for $K=\mathbb{C}$), and you run the same algorithm, then all the steps will be the same and so the end result will be the same (effectively, all computation happens in the subring $\mathbb{Q}[x_1,\ldots,x_n]$). That is, a Groebner basis of $I$ computed in $\mathbb{Q}[x_1,\ldots,x_n]$ will also be a Groebner basis of $I$ in $K[x_1,\ldots,x_n]$, in particular also for $K=\mathbb{C}$.


答: Suppose that $\sqrt{2}$ is rational number then exists $p$, $q$ coprime natural numbers such that $\sqrt{2} = \dfrac{p}{q}$. Implies $2q^2 = p^2$, so $p^2$ is divisible by 2. So on $p\vdots 2$, assume that $p=2p_1$ then $q^2=2p_1^2$, implies $q\vdots 2$, So $\gcd(p, q)\vdots 2$ is contradict $p$, $q$ are coprime.

Title: Verification of my proof for irrationality of $\sqrt2$

Body:
Here's my proof:

Let us suppose for sake of contradiction, $\sqrt2$ is rational. Then, we can write
$$\sqrt2=\frac{a}{b}$$
for non-zero coprime integers $a$ and $b$.
Squaring our equation,
$$2=\frac{a^2}{b^2}$$
Here, LHS is an integer, but since $a^2$ and $b^2$ are coprime, RHS is not an integer. This is a contradiction, and so our original assumption was wrong. Therefore, $\sqrt2$ is irrational.

Is this valid? Thanks in advance.


问:
If for integer $x_1,x_2,x_3,x_4$ holds $x_i\cdot x_j = y_{ij}^2 - 13$ ($i\neq j$), is it possible that $x_1+x_2+x_3=10^6$?
Let $a$, $b$, $c$, $d$ be integers. Given that a product of any two of them equals perfect square minus $13$, is it possible for the sum of some three of them to be equal to $10^6$?

My guess is that it is impossible.
My approach was to use the remainders modulo $5$, as $n^2\equiv0,\pm1\pmod{5}$.
If $a+b+c=10^6$, then $d(a+b+c)=k^2 + l^2 + m^2 - 39 = d \cdot 10^6$ for some $k,l,m$.

It is clear that (with the proper rearrangement of variables), $k^2\equiv l^2 \equiv 0,\; m^2 \equiv -1$ or $k^2\equiv l^2 \equiv -1,\; m^2 \equiv 1$.
One can elaborate a bit further on the possibilities for the remainders, but I can't arrive to any contradiction.

I have also tried solving this for smaller numbers in Wolfram, but it uses to much RAM.

答:
You are incredibly close. Consider the equation $k^2 + l^2 + m^2 -39 = d\cdot 10^6$ modulo $8$.

Note that $n^2 \equiv 0,1,4 \pmod 8$. Hence $k^2 + l^2 + m^2 \not\equiv 7 \pmod 8$. But $39 \equiv 7 \pmod 8$ and $d \cdot 10^6 \equiv 0 \pmod 8$. Hence there are no integer solutions to $k,l,m$.


问:
Suppose $f(x) = ax^3 + bx^2 + cx + d$, with $a, b, c, d ∈ R$
Questions

Find $a,b,c,d$ so the function $f$ is an even function.

Find $a,b,c,d$ so the function $f$ is an odd function.

My understanding:

Even function means $f(x) = f(-x)$.

Odd function means $-f(x) = f(-x)$

We will have 2 equations, one for even function and the other for odd function:

$ax^3+bx^2+cx+d = -ax^3+bx^2-cx+d$

$-(ax^3+bx^2+cx+d) = -ax^3+bx^2-cx+d$

If I cancel out what I can I'll have:

$ax^3 + cx = -ax^3 - cx$ $<=>$ $2(ax^3+cx) = 0$ $<=>$  $ax^3+cx = 0$ $<=>$ $x*(ax^2+c)=0$

$-bx^2-d = bx^2 + d$ $<=>$ $bx^2+d = 0$

Should I not cancel? Any hints to how I should continue? First equation is basically $x = 0$ or $ax^2+c = 0$

答: Indeed, your reasoning is correct so far. In the even case, since $b$ and $d$ canceled out (thus they can actually be anything), you are only left with the equation you mentioned, $x(a x^2 + c) = 0$ for all $x \in \mathbb{R}.$ But this is obviously not true if $a$ and $c$ are not $0$, so we should in fact have $a = c = 0.$ Thus, the general solution for the first part is $f(x) = b x^2 + d.$ Doing the same for the odd case, we get the general solution $f(x) = a x^3 + cx.$


问:
Product on Greatest integer
I want to prove the following equality: $\left[\frac{[x]}{n}\right] = \left[\frac{x}{n} \right]$, if $n \in \mathbb{N}$. I did it:
$$h = \left[\frac{[x]}{n}\right] \rightarrow h\leq \frac{x}{n}< h + 1$$
For definition of greatest integer:
$$h \leq \left[\frac{x}{n} \right] $$
How to prove: $h \geq \left[\frac{x}{n} \right]  $

答: Hint: ${x-[x] \over n}<{1 \over n}$ and ${[x] \over n} \in \mathbb Q$.

Title: Product on Greatest integer


问:
Integral $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}$, may involve beta function
Evaluate the integral: $$\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2} dx$$

I have used substitutions like $x^4 = u$, or $x^{-4} = u$. After many hours, I came up with $\frac{2}{1+x^4} = u$ which reduces the integral to :

$$\int_1^2 \frac{2^{3/4}}{8} \left(\frac{u-1}{2-u}\right)^{3/4} du$$

and then taking $\frac{u-1}{2-u} = z$ gives:

$$\frac{2^{3/4}}{8} \int_0^\infty \frac{z^{3/4} }{(1+z)^2} dz = \color{red}{ \frac{2^{3/4}}{8}\beta(\tfrac{7}{4},\tfrac{1}{4})}$$

Last answer comes from using:

$$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} dx = \beta(m,n)$$

This answer seems to be correct as online integral calculator gives $0.7005...$ as answer.

What can be alternate approaches

答:
Easiest way to solve this is by taking 4 substitutions
$$x^4=u \\
\dfrac 1u=z \\
z-1=t\\
\dfrac t2=v$$
After these the integral will come out in the form of the last identity op has mentioned
I know my formatting and illustration is weak,so kindly someone pls form a answer on basis of this and post it as it may help people searching for the same Q


问:
kinematics: what does the answer mean?
$v = -0.01t^3 + 0.22t^2 - 0.4t$. Find the 2 positive values of t for which the particle is instantaneously at rest. This part has been done: $t= 2$ and $t= 20$.

Find the distance travelled by the particle while its velocity is positive.

We find an expression for the distance $s$ and we just substitute $t=2$ and $t=20$ in it and then substract them together. My question is:

Why do we take these values of t?

答:
Here is a plot of your given velocity function. Notice that the velocity is positive positive between $t=2$ and $t=20$, and so when you find the expression by integrating the velocity function, these two points of time become your limits of the integral. (Integral of the velocity is the displacement. In this case, since velocity is always positive, we are only travelling in one direction, so distance = displacement!)

Just to make this clear,
$$s = \int_2^{20} v(t) dt = \int_2^{20} (-0.01t^3 + 0.22t^2 - 0.4t) dt = [f(t)]_{2}^{20} = f(20) - f(2)$$

where $f = -0.0025 t^4 + 0.0733333 t^3 - 0.2 t^2$

Title: kinematics: what does the answer mean?


问:
Factorize $f(X)=X^4+1$ on $L$.
Let $L= \mathbb{Q}[Y]/(Y^4+1)$. Factorize $f(X)=X^4+1$ on $L$.

My attempt: We know that $\alpha = \overline{X}=X+(X^4+1)$ is one of the root of $f(X)$ in $L$. Now using the relation $\alpha^4+1=0$ and using simple division method, we can factorize $X^4+1 = (X-\alpha)(X^3+\alpha X^2+\alpha^2 X+\alpha^3)$. Is this the final factorization? How do we know that there are not more roots in $L$?

Edit: I have changed the variables in the field $L$ from $X$ to $Y$.

答:
Hint: consider the $\mathbb Q$-algebra map $f \colon \mathbb Q[X] \to \mathbb Q(\xi_8)$, $X \mapsto \xi_8$, where $\xi_8$ is an $8$-th primitive root of unity. Since $\xi_8^8 = 1$, we have $(\xi_8^4)^2 = 1$. But $\xi_8^4\neq 1$, as $\xi_8$ is primitive, so we necessarily obtain $\xi_8^4 = -1$ and thus $\xi_8$ vanishes at $x^4+1$.

On the other hand, a standard fact of cyclotomic polynomials is that

$$
\deg m(\xi_8,\mathbb Q) = \varphi(8) = 4,
$$

so $m(\xi_8,\mathbb Q) = X^4+1$. By construction, this is (a generator of) the kernel of $f$, hence

$$
\mathbb Q[X]/(X^4+1) \simeq \mathbb Q(\xi_8).
$$
Maybe the latter presentation can help you.

 The cyclotomic polynomial has as roots primitive roots of unity. These are characterized as powers of $\xi_8$ with exponent coprime with $8$; hence $T^4+1 = (T-\xi_8)(T-\xi_8^7)(T-\xi_8^3)(T-\xi_8^5)$.

Title: Factorize $f(X)=X^4+1$ on $L$.


问:
Integral closure in an extension ring
I'm trying to prove the following statement:

Let $R$ be an integral domain and let $S\subset R$ be a multiplicative closed subset. If $R\subset R'$ is an extension ring of $R$ with $\overline{R}$ the integral closure of $R$ over $R'$, then $S^{-1}\overline{R}$ is the integral closure of $S^{-1}R$ in $S^{-1}R'$.

My attempt

I want to prove that $S^{-1}\overline{R}=\overline{S^{-1}R}$.

$\subseteq)$ Let $\frac{b}{s}\in S^{-1}\overline{R}$. Since $b\in \overline{R}$, $\exists$ a monic relation $b^n+c_{n-1}b^{n-1}+\cdots+c_0=0$ with $c_0,\cdots,c_{n-1}\in R$ and thus

$$\left(\frac{b}{s}\right)^n+\frac{c_{n-1}}{s}\left(\frac{b}{s}\right)^{n-1}+\cdots+\frac{c_1}{s^{n-1}}\left(\frac{b}{s}\right)+\frac{c_0}{s^n}=0$$

where $\frac{c_{n-i}}{s^i}\in S^{-1}R$, $i=1,\cdots,n$. Then $\frac{b}{s}\in\overline{S^{-1}R}$.

I tried to prove $\supseteq)$ starting in this way:
Let $\frac{b}{s}\in\overline{S^{-1}R}$. Then $\exists$ a monic relation

$$\left(\frac{b}{s}\right)^n+z_{n-1}\left(\frac{b}{s}\right)^{n-1}+\cdots+z_1\left(\frac{b}{s}\right)+z_0=0$$
with $z_0,\cdots,z_{n-1}\in S^{-1}R$.

At this point I don't know how to continue the proof. I am grateful for any ideas you can give me to continue it.

答: Let $b \in \overline{S^{-1} R}$. Then there is some monic polynomial $p \in S^{-1} R[X]$, say of degree $n$, with $p(b) = 0$. Now let $s \in S$ such that $s \cdot p(X) \in R[X]$. Then $s^n \cdot p(X) = q(s X)$ for an appropriate monic polynomial $q \in R[X]$ (write it down). Now $q(sb) = 0$, hence $sb \in \overline R$ and $b = \frac{sb}{s} \in S^{-1} \overline{R}$.

Title: Integral closure in an extension ring


答:
To avoid confusion I will give a new name to the indeterminate:

Factor $f(t)=t^4+1$ over $L$.

You already found the root $\alpha$ and hence the factor $t-\alpha$ of $f$ in $L$. But there are more! Notice that $f$ contains only even powers of $t$, so $-\alpha$ will also be a root. Now by division we get $f(t)=(t-\alpha)(t+\alpha)(t^2+\alpha^2)$. So we are reduced to factor $t^2+\alpha^2$. Notice that if there was an element $i\in L$ such that $i^2=-1$ we could write $t^2+\alpha^2=(t-i\alpha)(t+i\alpha)$. Can you find such an element?

Answer:

 Take $i=\alpha^2$.

Title: Factorize $f(X)=X^4+1$ on $L$.

Body:
Let $L= \mathbb{Q}[Y]/(Y^4+1)$. Factorize $f(X)=X^4+1$ on $L$.

My attempt: We know that $\alpha = \overline{X}=X+(X^4+1)$ is one of the root of $f(X)$ in $L$. Now using the relation $\alpha^4+1=0$ and using simple division method, we can factorize $X^4+1 = (X-\alpha)(X^3+\alpha X^2+\alpha^2 X+\alpha^3)$. Is this the final factorization? How do we know that there are not more roots in $L$?

Edit: I have changed the variables in the field $L$ from $X$ to $Y$.


问:
How to evaluate $\lim_{x\to 0} (1+2x)^{1/x}$
Good night guys! I'm having some trouble with this:

$$\lim_{x\to 0} (1+2x)^{1/x}$$

I know that $\lim_{x\to\infty} (1 + 1/x)^x = e$
but I don't know if i should take $h=1/(2x)$ or $h=1/x$

Can someone please help me? Thanks!

答:
$$\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e,\displaystyle\lim_{x\to-\infty}\left(x+\frac{1}{x}\right)^x=e$$

$(-\infty,0),(0,\infty)\subset\mathbb R-\{0\}=\text{domain}\displaystyle\left(1+\frac{1}{x}\right)^x.$ So $\displaystyle\lim_{x\to0+}(1+x)^\frac{1}{x}=\displaystyle\lim_{x\to0-}(1+x)^\frac{1}{x}=e.$

Now $0$ is a limit point of both $(-\infty,0)$ and $(0,\infty)\implies\displaystyle\lim_{x\to0}(1+x)^\frac{1}{x}=e.$ Let

$$f:\mathbb R-\{0\}\to\mathbb R:x\mapsto2x,\\g:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+x)^\frac{1}{x}$$

$f(\mathbb R-\{0\})\subset\mathbb R-\{0\}\implies gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+2x)^\frac{1}{2x}$ is defined and $\displaystyle\lim_{x\to0}f(x)=0$ being a limit point of $\mathbb R-\{0\}$ and $\displaystyle\lim_{x\to0}g(x)=e,$ we have $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=e.$

Consequently, $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=\displaystyle\lim_{x\to0}\left((1+2x)^\frac{1}{2x}\right)^2=e^2.$

ContentLicense: CC BY-SA 3.0


问:
Probability of $X > \max\{Y,Z\}$ exponentials
Let $X,$ $Y$ and $Z$ be exponentially distributed with $E[X] = 1/\lambda_X$, $E[Y] = 1/\lambda_Y$, $E[Z] = 1/\lambda_Z$ respectively.

How do I calculate $P(X > \max\{Y,Z\})$?

I have A, B, C and D for 3 cash boxes in a queueing system. Consider D the one waiting. I want to find the probability D is the last one to leave. The service times of the C.Boxes are indep. Exp with mean 5 (same mean for c.box 1, 2 and 3) That's how I'm trying to solve this: P(D last one) = P(T1 > max{T2,T3})*P(T1 < min{T2,T3}) + P(T2 > max{T1,T3})*P(T2 < min{T1,T3}) + P(T3 > max{T1,T2})*P(T3 < min{T1,T2})

And then:
$$(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4 = = 3(\frac{1}{2})^4 = 3\frac{1}{16} = \frac{3}{16}$$

答:
The beginning of this answer addresses the original question of finding $P(X > \max\{Y, Z\})$. OP added more context to the question afterward; I address that at the end of this answer.

[Assuming independence of $X,Y,Z$.]

Note that
$P(X > \max\{Y, Z\}) = P(X > Y> Z) + P(X > Z > Y)$.

\begin{align}
P(X>Y>Z)
&= \int_0^\infty \lambda_Z e^{-\lambda_Z z}\int_z^\infty \lambda_Y e^{-\lambda_Y y}\int _y^\infty \lambda_X e^{-\lambda_X x} \, dx \, dy \, dz
\\
&= \int_0^\infty \lambda_Z e^{-\lambda_Z z}\int_z^\infty \lambda_Y e^{-\lambda_Y y} e^{-\lambda_X y} \, dy \, dz
\\
&= \int_0^\infty \lambda_Z e^{-\lambda_Z z} \frac{\lambda_Y}{\lambda_X+ \lambda_Y} e^{-(\lambda_X + \lambda_Y) z}\, dz
\\
&= \frac{\lambda_Y}{\lambda_X + \lambda_Y} \cdot \frac{\lambda_Z}{\lambda_X + \lambda_Y + \lambda_Z}.
\end{align}
The other term $P(X>Z>Y)$ can be computed similarly.

Verification with BruceET's simulation:
$$\frac{\lambda_Y}{\lambda_X + \lambda_Y} \cdot \frac{\lambda_Z}{\lambda_X + \lambda_Y + \lambda_Z} + \frac{\lambda_Z}{\lambda_X + \lambda_Z} \cdot \frac{\lambda_Y}{\lambda_X + \lambda_Y + \lambda_Z}
= \frac{1/3}{7/12} \frac{1/2}{13/12} + \frac{1/2}{3/4} \frac{1/3}{13/12}=\frac{128}{273} \approx 0.46886$$

Afterthought:

After looking at the expression for $P(X>Y>Z)$ and noting that $\min\{X,Y\} \sim \text{Exponential}(\lambda_X + \lambda_Y)$, one might be tempted to do the following:

\begin{align}
P(X>Y>Z)
&= P(Y<X) P(Z < \min\{X,Y\} \mid Y<X)
\\
&\overset{?}{=} P(Y<X) P(Z < \min\{X,Y\})
\\
&= \frac{\lambda_Y}{\lambda_X + \lambda_Y} \cdot \frac{\lambda_Z}{\lambda_X + \lambda_Y + \lambda_Z}.
\end{align}
However, verifying the step marked "?" takes some care. In general, the distribution of $\min\{X,Y\}$ does depend on the event $X<Y$, but in the special case of independent exponential random variables, $\min\{X,Y\}$ does not depend on this event. This can be verified explicitly with integrals as follows.
$$P(\min\{X,Y\} > t \mid Y<X) = \frac{P(X>Y>t)}{P(X>Y)} = \frac{\frac{\lambda_Y}{\lambda_X+\lambda_Y} e^{-(\lambda_X + \lambda_Y) t}}{\frac{\lambda_Y}{\lambda_X+\lambda_Y}} = e^{-(\lambda_X + \lambda_Y) t} = P(\min\{X,Y\}>t)$$
(I think one could alternatively justify this with a memorylessness argument.)

Update after OP added context:

Stacker's simulations suggest that the answer to the original question "When is D the last to leave?" is $1/3$. Your work almost gets you there except you plug in the wrong numbers. You can use the above result to show that $P(T_1 > \max\{T_2, T_3\}) = \frac{1}{3}$. The other term $P(T_1 < \min\{T_2, T_3\})$ is also $1/3$ if you note that $\min\{T_2, T_3\}$ is exponential with mean $5/2$.

But because $T_1,T_2,T_3$ are identically distributed, you can use symmetry to immediately note that both of these are $1/3$: each of the three boxes is equally likely to end first or end last. So you can skip the complexity of the subproblem of computing $P(X > \max \{Y,Z\})$ altogether.

Plugging in $1/3$ into each term in your expression gives $\frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{3}$ which matches Stacker's simulations.

Also, arriving at the $P(T_1 > \max\{T_2, T_3\}) P(T_1 < \min\{T_2, T_3\}) + \cdots$ expression in your attempt involves a memorylessness argument that should not be glossed over (i.e., given A finishes first, the additional time B needs to finish is also exponential with the same mean, etc.), but I'll give you the benefit of the doubt.


问:
How is this binomial coefficient be written in this way?
$$\begin{pmatrix}
    p\\
    n\end{pmatrix}=\frac{p!}{n!(p-n)!} = \frac{p(p-1)(p - 2) \dots (p-(n-1))}{n!}$$

I was reviewing and going over the binomial theorem. While doing that, I came across this binomial coefficient and began to feel confused about how it simplifies to the result on the right-hand side. What is the process behind this simplification?

答: Simplify $p !$ with $(p - n)!:$ $$\frac{p!}{(p-n)!} = \frac{(p - n)! (p - n + 1) \dots (p - 1) p}{(p - n)!} = (p - n + 1) \dots (p - 1) p,$$ which yields the result.


问:
How to compute/estimate the probability of the mean value of x results, if each one of the results has its own probability
I use a neural network to classify the sentiment of some articles per day. Possible results are [1,2,3,4,5] (1=very negative, ..., 5 = very positive). Using one article as input the network gives me as output, a number 1-5, describing the sentiment and a the corresponding probability (a softmax probability), e.g. sentiment = 2, probability = 0.65.

If I calculate the mean sentiment of today's articles, is there a way to assign to this mean (the average of the articles sentiments) a probability that will take into account the softmax probabilities assigned to each article?

答:
Assume that you have n articles, $X_1,...,X_n$ with mean sentiment $\bar X$. A probability that takes into account the softmax probabilities is

$$\sum_{x_i\in\{1,2,3,4,5\}\\\bar x_i=\bar X} P(X_1=x_1)*P(X_2=x_2)*...*P(X_n=x_n)$$

It is probably better to allow some leeway for the average, with l being a number between 0 and 4, 0 being the most exact (giving you the smallest probability and equivalent to the expression above) and 4 being the most forgiving (giving you a probability of 1).

$$\sum_{x_i\in\{1,2,3,4,5\}\\|\bar x_i-\bar X|\le l} P(X_1=x_1)*P(X_2=x_2)*...*P(X_n=x_n)$$


问:
Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem
How do I find:

$$ \lim_{x\to0}\frac{1-\cos(x)}{x} $$

Using the squeeze theorem.  Particularly, how would I arrive at its bounding functions?

If possible, please try not to use derivatives.

Body:
For any $x$ such that $0 < |x| \leq \pi/2$,
$$
0 \le \bigg|\frac{{1 - \cos x}}{x}\bigg| = \frac{{1 - \cos |x|}}{{|x|}} = \frac{{\int_0^{|x|} {\sin t \,dt} }}{{|x|}} \le \frac{{\int_0^{|x|} {\sin |x| \, dt} }}{{|x|}} = \sin |x|.
$$


问:
How to expand integral of variance for continuous random variable?
Variance of continuous random variable: $$E[(X-\mu_X)^2]=\int_x (x-\mu_X)^2 f_X(x)dx $$

I should expand into two parts, but I don't know how to move from here and I doubt it's even correct:

$$\int_x (x-\mu_X)^2 f_X (x)dx = \int_x(x^2-2x\mu_x+\mu^2_X)f_X dx$$

答: Well, yes, this is perfectly fine. If you want to continue computing, you could do something like this: $$\int x^2 f_X(x) dx - 2 \mu_X \int x f_X(x) dx + \mu_X^2 \int f_X(x) dx = \mathbb{E}[X^2] - 2 \mu_X \mathbb{E}[X] + \mu_X^2 = \mathbb{E}[X^2] - 2 (\mathbb{E}[X])^2 + (\mathbb{E}[X])^2 = \mathbb{E}[X^2] - (\mathbb{E}[X])^2,$$ where I only used the definitions of $\mu_X, \, f_X$ and the law of the unconscious statistician. I hope this is what you meant.


Title: How is this binomial coefficient be written in this way?

Body:
$$\begin{pmatrix}
    p\\
    n\end{pmatrix}=\frac{p!}{n!(p-n)!} = \frac{p(p-1)(p - 2) \dots (p-(n-1))}{n!}$$

I was reviewing and going over the binomial theorem. While doing that, I came across this binomial coefficient and began to feel confused about how it simplifies to the result on the right-hand side. What is the process behind this simplification?

Body:
See, leaving out the n! in the denominator for now

$$\begin{split}\frac{p!}{(p-n)!}&=\frac{p*(p-1)*(p-2)*...*(p-n+1)*(p-n)*(p-n-1)*...*2*1}{(p-n)*(p-n-1)*...*2*1}\\
&=p*(p-1)*(p-2)*...*(p-(n-2))*(p-(n-1))\end{split}$$

Divide by the original n! to achieve

$$\frac{p*(p-1)*(p-2)*...*(p-(n-2))*(p-(n-1))}{n!}$$

Basically, it truncates p! to the first p - (p-(n-1)) + 1 terms


问:
Cauchy's Mean-Value Theorem for three or more functions
Cauchy's mean-value theorem is itself a generalization of the "normal" mean value theorem. My question is, does this generalization go one? If there are three functions $f,g,h: [a,b]\to \mathbb{R}$, which are differentiable on (a,b) and continuous on [a,b]. Also $g'(x)\neq 0 \; \forall x \in (a,b)$. Is then true? $$\implies \exists \xi \in (a,b):\quad \frac{f(b)-f(a)}{g(b)-g(a)} \cdot \frac{h(b)-h(a)}{b-a}=\frac{f'(\xi)}{g'(\xi)} \cdot h'(\xi)$$

答:
Take
$$f(x)=g(x)=h(x)=x$$
the conditions are satisfied

we will get $$h(\xi)=\xi=1$$


问:
A Dedekind Domain if and only if localizations at all nonzero primes are DVRs
I am trying to understand a part of a proof given by Neukirch in his Algebraic Number Theory text. The highlighted part below is what I'm confused by.

In particular, why do we choose $\pi\in\mathfrak{m}-\mathfrak{m}^2$, and not just any nonzero element of $\mathfrak{m}$? Also, why does $\mathfrak{m}^n=(\pi^n)$ imply that $\mathcal{O}_p$ is a PID? Couldn't there be an ideal of $\mathcal{O}$ that isn't necessarily a power of $\mathfrak{m}$?

答:
We choose $\pi\in \mathfrak{m}\setminus \mathfrak{m}^2$ because this implies $\mathfrak{m}=(\pi)$. To see this, factor $(\pi)$ into non-zero prime ideals (possible since we are in a Dedekind domain). Since $\mathfrak{m}$ is the only non-zero prime ideal we necessarily have $(\pi)=\mathfrak{m}^n$ for some $n\geq 0$. As $\pi\in \mathfrak{m}$ we have $n\geq 1$ and since $\pi\notin\mathfrak{m}^2$ we also have $n<2$, hence $n=1$, i.e. $(\pi)=\mathfrak{m}$. If we wouldn't require $\pi\notin\mathfrak{m}^2$, we might have $(\pi)=\mathfrak{m}^n$ with $n>1$ which is not what we want.

By the same argument we see that any non-zero ideal of $\mathcal{O}_P$ is a product of primes and hence a power of $\mathfrak{m}$, therefore principal (actually for this we don't really need the Dedekind-property anymore: any noetherian local domain whose maximal ideal is principal is already principal ideal domain)


问:
A basic probability doubt on independence
Let $X_1$ and $X_2$ be two i.i.d continuous random variable. I need to find the probability that $P(X_1 < X_2)$. I know how to formally find the probability by integrating over appropriate region and how we use the "identically distributed" and "independent" property here.

Now,consider the intuitive argument for this problem. $X_1$ and $X_2$ both have the same distribution. So, any one is equally likely to be the smallest. So, is n't identically distributed sufficient condition ? Where is "independence" needed here ?

答: Suppose as an extreme example that $X_1=X_2$. </p>


问:
Can a theory prove (schematically) its axioms relativized to a set?
In $\mathsf{ZFC}$, a somewhat cheating way to buy "transitive models" without cost in consistency is to add to the language a constant symbol $M$ and add to $\mathsf{ZFC}$ the axioms stating that $M$ is countable transitive, and for each axiom of $\mathsf{ZFC}$, add its relativization to $M$. This allows us to "pretend" that we have a transitive model of $\mathsf{ZFC}$ (the catch is that this is a theorem schema).

I wonder if adding a constant symbol is necessary. In other words, can there be some theory $T$ (extending $\mathsf{ZFC}$) in the language $\{\in\}$, such that T can prove there is a set $M$, and $T$ proves each of its own axioms relativized to $M$?

答:
$\mathsf{ZFC}$ already has this property. Specifically:

The reflection theorem shows that every model of $\mathsf{ZFC}$ - even of $\mathsf{ZFC+\neg Con(ZFC)}$ - contains a model of $\mathsf{ZFC}$. (This model might not internally be a model of $\mathsf{ZFC}$, which is why this isn't an obvious absurdity.)

The fact that satisfiability is absolute to $L$ then lets us pick out a specific model, via the $L$-ordering.

The details are as follows:

Working in $\mathsf{ZFC}$, fix some standard enumeration of the $\mathsf{ZFC}$ axioms and let $T_0$ be the largest initial segment of that enumeration which is consistent. Classically of course we have (under the usual assumptions) that $T_0=\mathsf{ZFC}$; meanwhile, note that $\mathsf{ZFC}$ proves  "$T_0$ is consistent" (trivially) as well as "$\varphi\in T_0$" for each $\varphi\in\mathsf{ZFC}$ (by the reflection theorem).

Now since $\mathsf{ZFC}$ proves that $T_0$ is consistent, we can - in $\mathsf{ZFC}$ - consider $M=$ the least constructible set model of $T_0$, where "least" refers to the standard ordering on $L$. This is provably in $\mathsf{ZFC}$ a model of $T_0$, and so for each $\mathsf{ZFC}$-axiom $\varphi$ we have $\mathsf{ZFC}\vdash M\models\varphi$ as desired.


问:
Why is the inverse of this function negative?
$$f(x) = 2x^2 - 12x + 7 \,\,\,,x\le3$$
$$f^{-1}(x) = - \sqrt{(x+11)/2} + 3$$

why is the square root negative? how to know when it's supposed to be negative or positive? Could you please explain the reasoning and logic?

答:
"Completing the square" produces

$$f(x) \ = \ 2x^2 - 12x + 7 \ = \ 2 \ · \ ( x^2 \ - \ 6x \ + \ 9 ) \ - \ 11 \ = \ 2 \ · \ ( x \ - \ 3)^2 \ - \ 11 \ \ , $$

so restricting the domain to $ \ x \ \le \ 3 \  $ uses only the "left half" of the parabola, which is the graph of a one-to-one function and so will permit the construction of an inverse function.  The function is not negative:  you are just finding the solution from the quadratic equation that use the "negative" square-root.  (The solution with the "positive" square-root corresponds to the "right half" of the parabola.)


问:
Ideal Class Group of $ \mathbb{Q}(\sqrt[3]{3}) $
I'm trying to compute the ideal class group of $\mathbb Q(\sqrt[3]{3})$, and I would like to know if my calculations are right and if I could improve my arguments.

Let $K=\mathbb Q (\sqrt[3]{3}),$ its ring of integers $\mathcal O_K$ is $ \mathbb Z[\sqrt[3]{3}]$, since Disc$(1,\sqrt[3]{3},\sqrt[3]{3^2})=-3^5 \quad $ and $f(x)=x^3-3 \ $ is Eisenstein for $3$.

The Minkowski bound is $\frac{4}{\pi}\cdot\frac{6}{27}\cdot \sqrt{3^5}<5$. Thus we need to check ideals of norm less than $5$. Set $a=\sqrt[3]{3}$ to facilitate notation.

The only ideal of norm 1 is $\mathcal{O}_K$

Every ideal contains its norm, thus ideals of norm $2$ are prime factors of $(2). \quad $ $ (2)=(a-1)(a^2+a+1) \ $, since $2=3-1=a^3-1=(a-1)(a^2+a+1). N(a-1)=2, N(a^2+a+1)=4 $.

$(3)=(a)^3$ and $(a)$ is principal.

Since $(2)=(a-1)(a^2+a+1),$ ideals of norm 4 are:
$(a-1)^2, \ (a^2+a+1).$

All of the ideals above are principal, thus the ideal class group is trivial and $\mathcal O_K$ is a PID.

答:
Your computations are correct and they seem quite optimal. Here is a verification in SageMath:

```<code>sage: x = polygen(QQ)
sage: f = x^3 - 3
sage: f.is_irreducible()
True
sage: K.<a> = NumberField(f); K
Number Field in a with defining polynomial x^3 - 3
sage: OK = K.ring_of_integers(); OK
Maximal Order in Number Field in a with defining polynomial x^3 - 3
sage: OK.basis()
[1, a, a^2]
sage: OK.discriminant().factor()
-1 * 3^5
sage: M = K.minkowski_bound(); M
8*sqrt(3)/pi
sage: bool(M == 4/pi * 6/27 * sqrt(3^5))
True
sage: bool(M < 5)
True
sage: OK.ideal(2).factor()
(Fractional ideal (a^2 + a + 1)) * (Fractional ideal (a - 1))
sage: OK.ideal(3).factor()
(Fractional ideal (a))^3
sage: K.class_number()
1
sage: OK.class_group()
Class group of order 1 of Number Field in a with defining polynomial x^3 - 3
</code>```

Of course you meant to say $\mathcal{O}_K$ is a PID. ($K$ is also always a PID.)

Body:
Your computations are correct and they seem quite optimal. Here is a verification in SageMath:

sage: x = polygen(QQ)
sage: f = x^3 - 3
sage: f.is_irreducible()
True
sage: K.<a> = NumberField(f); K
Number Field in a with defining polynomial x^3 - 3
sage: OK = K.ring_of_integers(); OK
Maximal Order in Number Field in a with defining polynomial x^3 - 3
sage: OK.basis()
[1, a, a^2]
sage: OK.discriminant().factor()
-1 * 3^5
sage: M = K.minkowski_bound(); M
8*sqrt(3)/pi
sage: bool(M == 4/pi * 6/27 * sqrt(3^5))
True
sage: bool(M < 5)
True
sage: OK.ideal(2).factor()
(Fractional ideal (a^2 + a + 1)) * (Fractional ideal (a - 1))
sage: OK.ideal(3).factor()
(Fractional ideal (a))^3
sage: K.class_number()
1
sage: OK.class_group()
Class group of order 1 of Number Field in a with defining polynomial x^3 - 3

Of course you meant to say $\mathcal{O}_K$ is a PID. ($K$ is also always a PID.)


问:
Degree of a polynomial as smooth map - clarification
I would like to a clarification of the following question. I cannot understand the last three lines, where we conclude that the exponents $n_1=n_2=...n_k=1$, which then concludes our proof. Cannot understand why the condition of the jacobian implies the one on the $n_s$. Any help is appreciated! Thanks!

答:
Since $q$ is a regular value, then the derivative of the polynomial at every root is different from $0$.

Note that the only possibility to have the derivative different from $0$ at every $x_j$ for all $j=1,...,k$ is that $n_1=...=n_k=1$, since in this case the derivative does not equal $0$.

By the condition that the sum of all the $n$'s is equal to $d$, you easily get the result recalling that for holomorphic proper functions $deg(F)=\#F^{-1}(q)$.


问:
Calculating eigenvectors of a 3x3 symmetric matrix from Cayley–Hamilton theorem?
Let's say I have a 3x3 real symmetric matrix M and I have its three eigenvalues $\alpha_1, \alpha_2, \alpha_3$ and that $\alpha_1 < \alpha_2 < \alpha_3$.  (Notice all eigenvalues are different).

We can deduce that $(M-\alpha_{1}I)(M-\alpha_{2}I)(M-\alpha_{3}I) = 0$ must be true.

Therefore, the three columns of $S = (M-\alpha_{2}I)(M-\alpha_{3}I)$ must all be eigenvectors of $\alpha_1$.

Notice that S must be symmetric because M and I are symmetric, and the sum and product of two symmetric matrixes is also symmetric.  (Not sure if that's relevant.)

My question is about the columns of S.  How many columns of S will be non-zero vectors?  Zero, one, two or three?  Or will it vary depending on M?  and why?

答: In general,$S$ could have all of its entries non-zero.  Since $\alpha_1$ is an eigenvalue with only one eigenvector (after all, $\alpha_2$ and $\alpha_3$ also have associated eigenvectors), all of the columns of $S$ are pairwise linearly dependent and the determinant of $S$ is zero.


问:
Showing that $\mathscr B$ is a tautology iff $\lnot \mathscr B'$ is a tautology.
I have the following logic question about duality:(Introduction to mathmatical Logic by elliot mendelson , exercise $1.30$)

If $\mathscr B$ is a statement form involving only $\lnot$ , $\land$, and $\lor$, and $\mathscr B'$ results from  $\mathscr B$  by replacing each $\land$ by $\lor$ and each $\lor$ by $\land$ , show that $\mathscr B$ is a tautology if and only if $\lnot \mathscr B'$ is a tautology.

But I think there is a counterexample if  $\mathscr B = A$ , then  $\mathscr B' = A$ .If  $\mathscr B$  is a tautology , then $\lnot  \mathscr B' = \lnot A$ will be a contradictory and this contradicts the thing we want to prove.I am confused with this.Is there a flaw in my reasoning?

答:
As Mauro ALLEGRANZA points out, your counterexample does not work, since if $A$ is a propositional variable, it can not be a tautology: just take a valuation for which it is mapped to $0$.

I will prove a slightly stronger result by induction: for a valuation $\nu$, define $\nu^{*}$ as the valuation such that, for any propositional variable $A$, $\nu^{*}(A)=1$ if, and only if, $\nu(A)=0$. Then I state that
$$\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}'),$$
for any formula $\mathfrak{B}$.

Suppose, first, that the formula $\mathfrak{B}$ has no connectives between $\vee$ and $\wedge$: since we are considering only formulas with $\vee$, $\wedge$ and $\neg$ as connectives, we must have $\mathfrak{B}=\neg\cdots\neg A$, for a propositional variable $A$ and some number of negations applied to $A$. In that case, $$\mathfrak{B}'=\mathfrak{B}\quad\text{and so}\quad\neg\mathfrak{B}'=\neg\mathfrak{B},$$ and one easily sees (apply induction in the number of negations of $\mathfrak{B}$ if you must) that $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$.

Suppose now that the result holds for formulas with, at most, $n-1$ ($n\geq 1$) connectives in $\{\vee, \wedge\}$, and take $\mathfrak{B}$ with $n$ of these connectives; we can write
$$\mathfrak{B}=\neg\cdots\neg(\mathfrak{C}\#\mathfrak{D})=\neg^{m}(\mathfrak{C}\#\mathfrak{D}),$$
for some $\#\in\{\vee, \wedge\}$ and some number $m$ of negations (possibly zero) applied to $\mathfrak{C}\#\mathfrak{D}$. Then we have that $\mathfrak{B}'=\neg^{m}(\mathfrak{C}'\#'\mathfrak{D}')$, where $\vee'=\wedge$ and $\wedge'=\vee$. Since $\mathfrak{C}$ and $\mathfrak{D}$ have fewer connectives among $\{\vee, \wedge\}$ than $\mathfrak{B}$, the induction hypothesis applies to them and then
$$\nu^{*}(\neg\mathfrak{B}')=\nu^{*}\Big(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}')\Big)=\nu^{*}\Big(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}')\Big)=-^{m}\Big(\nu^{*}(\neg\mathfrak{C}')\#\nu^{*}(\neg\mathfrak{D}')\Big)=$$
$$-^{m}\Big(\nu(\mathfrak{C})\#\nu(\mathfrak{D})\Big)=-^{m}\nu(\mathfrak{C}\#\mathfrak{D})=\nu\Big(\neg^{m}(\mathfrak{C}\#\mathfrak{D})\Big)=\nu(\mathfrak{B}),$$
where: by "$-$" I mean the complement on the two-valued Boolean algebra (so $-0=1$ and $-1=0$); and to show $\nu^{*}(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}'))=\nu^{*}(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}'))$ I used De Morgan's law. This finishes our proof.

Finally, you prove the result you want to prove by using this as lemma: if $\mathfrak{B}$ is a tautology, by noticing that $\nu^{**}=\nu$, for any valuation $\nu$ one has $\nu(\neg\mathfrak{B}')=\nu^{*}(\mathfrak{B})$, which equals $1$ since $\mathfrak{B}$ is a tautology.

Reciprocally, if $\neg\mathfrak{B}'$ is a tautology, for any valuation $\nu$ one has $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$, which is $1$ since $\neg\mathfrak{B}'$ is a tautology.


问:
Prove that for any two rational numbers $r$ and $s$ with $r ≠ 0$ , $\frac{1}{r} −s$ is rational
I have the following to problem

Prove that for any two rational numbers $r$ and $s$ with $r ≠ 0$ , $\frac{1}{r} −s$ is rational

My proof process is as follows (may be incorrect):

Suppose $r$ and $s$ are rational numbers.

By definition of rational, they can be expressed as a quotient of two
  integers with a nonzero denominator.

Then $\frac{1}{r}-s$ is equivalent to $\frac{(1- sr)}{r}$

Not sure where to go after this step

答:
Problems of this level, I'd so the proof is to find two actual integers $n,m$ so that $\frac 1r - s = \frac nm$.

First we are told $r$ is rational and not $0$ so there are integers $a,b$ so that $r = \frac ab$ (and $a\ne 0$).  And we are told $s$ is rational so there are integers $c,d$ so that $s = \frac cd$.

So $\frac 1r - s = \frac 1{\frac ab} -\frac cd$.

Show that we can write that as a fraction with an integer in the numerator and an integer in the denominator.

  $\frac 1{\frac ab} -\frac cd = \frac ba -\frac cd = \frac {bd}{ad} -\frac {ca}{da}= \frac {bd-ca}{ad}$.

 And integers are closed under multiplication so $bd,ca, da$ are all integers.  And integers are closed under subtraction so $bd-ca$ is an integer.  So as $bd-ca$ is an integer and $da$ is an integer (not equal to $0$... as $d\ne 0$ and $a \ne 0$) the $\frac {bd-ca}{da}$ is a rational number.


问:
Show that the sequence defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}$ converges
Show that the sequence $(x_n)$ defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}, x_0=3,x_1=2$ converges. Here i wil try to prove that $(x_n)$ converges considering all details. So. i would like to have a critical feedback on my proof and would like to know if i missed something or did much more than it should be.. Here $\mathbf{R}^*$ will denote $\mathbf{R}\ \ \not \ \{0\}$.

Consider a function $f(x)=\sqrt[3]{x}$. First, we show that it's defined on $\mathbf{R}$ and then we will show that $f(x)$ is an increasing function on $\mathbf{R}$.

We remark that $f(x)$ is differntiable on $\mathbf{R}^*$. Let $a \in \mathbf{R}^*$. By definition of differentiable function the $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ msut exist:

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\frac{1}{3\sqrt[3]{a^2}}$ and so it exists $\forall a \in \mathbf{R}^*$ as we chose an arbitrary $a$. So $f(x)$ is differentiable on $\mathbf{R}^*$ and so is continious on $\mathbf{R}^*$.

We show now that $f(x)$ is continious on $0$. To show that we pass by the definition :

$\forall \varepsilon>0 \ \exists \delta>0 \ \forall x \in \mathbf{R}$: $|x|<\delta \implies |f(x)|=|\sqrt[3]{x}|<\varepsilon$

To satisfy the definition, one can take $\delta=\varepsilon^3$, so it holds and we can conclude that $f(x)$ is continious on $\mathbf{R}$.

Let $a,b \in \mathbf{R}$ such that $a<b$. As $f(x)$ is continious on $\mathbf{R}$ we can write:

$a<b \iff \sqrt[3]{a}<\sqrt[3]{b} \iff f(a)<f(b)$

As $a,b$ are arbitrary numbers (satisfying $a<b$), we can conclude that $f(x)$ is strictly incrasing function.

Now, as all preliminary results are proven, we can show that $(x_n)$ is a convergent sequence. For that, we show that $(x_n)$ is monotonic and bounded.

$\bullet$ $(x_n)$ is decreasing: Suppose $x_{n+1}<x_n$. The previous inequality holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We show nowthat it holds for $n+1$:

$x_{n+2}<x_{n+1} \iff \sqrt[3]{x_{n+1}+x_n}<\sqrt[3]{x_n+x_{n-1}}$. But, as $f(x)=\sqrt[3]{x}$ is strictly increasing function and by induction hypothesis $x_{n+1}<x_n$, we can conclude that $x_{n+2}<x_{n+1}$ and so $x_{n+1}<x_n$ holds $\forall n\in\mathbf{N}^*$. So $(x_n)$ is decreasing.

$\bullet$ $(x_n)$ is bounded below: Suppose $x_n>1$. It holds for $n=1$ so we can suppose that it works up to a certain $n\in \mathbf{N}^*$. We want to show now that t holds for $n+1$:

$x_{n+1}>1 \iff \underbrace{x_{n}}_{>1}+\underbrace{x_{n-1}}_{>1}>1$ which is true, because $(x_n)$ is decreasing. So we conclude that $x_n>1 \ \forall n\in \mathbf{N}^*$.

So, we have shown that $(x_n)$ converges.

答:
$$x_{n+1}=\sqrt[3]{x_{n}+x_{n-1}};\;x_0=3;\;x_1=2$$
$(x_n)$ is a decreasing sequence

Proof. (by strong induction)

for $n=1$

$x_2=\sqrt[3]{x_0+x_1}=\sqrt[3]{5}<x_1$

Now suppose that it is true for all integers $1,2,\ldots, n$,  and let's prove it for $(n+1)$

$$x_{n+2}=\sqrt[3]{x_{n+1}+x_{n}}$$

$$x_{n+2}^3=x_{n+1}+x_n<x_n+x_{n-1}=x_{n+1}^3$$
thus $x_{n+2}<x_{n+1}$. Proved.

The sequence is decreasing and is bounded above by $x_n\le 3$

$x_n>1$ for any $n$.

$x_1=2>1$. Suppose $x_n>1$ for $0,1,2,\ldots,n$.

$x_{n+1}=\sqrt{x_n+x_{n+1}}>\sqrt[3]{1+1}>1$. Proved.

Thus the sequence is bounded above and below $1<x_n<3$,  therefore it converges

$x_n\to x$ as $n\to\infty$

To find the limit let's use the definition

$$x=\sqrt[3]{x+x}\to x^3=2x\to x=\sqrt 2$$

The sequence converges to $\sqrt 2$.


问:
Descent on elliptic curves over a quadratic field extension
I've heard it's possible to find the rank / bound the rank of an elliptic curve using the descent via 2-isogeny method, but over a quadratic field extension (eg. over $\mathbb{Q}(\sqrt{-3})$).
Does anyone know if there are any proofs or examples in any books/online of how this method works?

Secondly, I also believe I can do this using Magma - does anyone know the Magma commands for calculating bounds on the rank of an elliptic curve using 2-isogeny over a quadratic field extension?

Many thanks for your help.

答:
This appears in Chapter X of Silverman The Arithmetic of Elliptic Curves as Example 4.8 and Proposition 4.9 (where he works over a general number field).

If one wishes to use magma all of this is implemented as SelmerGroup in magma. Beware, to do all of this over a general number field Magma uses some very expensive class and unit group calculations, so things can get hairy fast.

As an example:

```<code>K := QuadraticField(-3);

E := EllipticCurve([0,0,0,1,0]);
E := BaseChange(E, K);

phi := TwoIsogeny(E![0,0]);
phi_hat := DualIsogeny(phi); E_hat := Image(phi);

S := SelmerGroup(phi);
S_hat := SelmerGroup(phi_hat);
</code>```

Body:
This appears in Chapter X of Silverman The Arithmetic of Elliptic Curves as Example 4.8 and Proposition 4.9 (where he works over a general number field).

If one wishes to use magma all of this is implemented as SelmerGroup in magma. Beware, to do all of this over a general number field Magma uses some very expensive class and unit group calculations, so things can get hairy fast.

As an example:

K := QuadraticField(-3);

E := EllipticCurve([0,0,0,1,0]);
E := BaseChange(E, K);

phi := TwoIsogeny(E![0,0]);
phi_hat := DualIsogeny(phi); E_hat := Image(phi);

S := SelmerGroup(phi);
S_hat := SelmerGroup(phi_hat);


问:
Why does FTOC apply here, to find the derivative of $\int_{\sin(x)}^{\pi} \frac{t}{\cos(t)} dt$
If we let $F(x) = \int_{\sin x}^{\pi} \frac{t}{\cos t} dt$, and $f(t) = \frac{t}{\cos t}$

My question is why does FTOC apply here? Here is why I think it does not, please correct me if I am wrong:

For F'(x) to be calculated using FTOC, a condition to use FTOC is that f(t) must be continuous over the interval where we are integrating the function. However, since the lower bound on F(x) is sin(x), which ranges from [-1, 1], we know that the interval would always include the point $\frac{\pi}{2}$. At $f(\frac{\pi}{2})$, the function is not defined (it is a vertical asymptote) as $\cos(\frac{\pi}{2})$ is 0. So how does FTOC apply here?

答:
$ f $ is defined for $ t $ such that

$$t\ne \frac{\pi}{2}+k\pi$$
or
$$t\in \Bbb R -\{...,-1.57,157,....\}$$

For $ x\in \Bbb R $, we have

$$[0,\sin(x)]\subset [-1,1]$$

and

$$\frac{\pi}{2}\approx 1.57\notin [-1,1]$$

$ f $ is continuous at $ [-1,1] $ and consequently at $ [0,\sin(x)]$


问:
How can I solve probability questions using the lebesgue integral?
Here are examples of questions I found in Google. I was curious about whether they can be solved via lebesgue integration and how. Examples:

A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after $n$ selections ($n>1$), the product of the $n$ numbers selected will be divisible by 10.

Basically, I’ve been taking a break off olympiad math and been reading schilling. I was curious about whether I can apply this knowledge.

答:
Here is a solution by standard method.

The resulting number is not divisible by 10 if one of the numbers $2,5$ (or both of them) were never selected. The probability that this happens after $n$ selections is
$$
P_n=2\left(\frac9{10}\right)^n-\left(\frac8{10}\right)^n,
$$
and the probability that the resulting number will be divisible by $10$ is, respectively,
$$
1-P_n.
$$


问:
An Useful Result in Measure Theory
Let $X$ be a borelian, finite measure space, WLOG, $\mu(X)=a$.

Are there any additional hypotheses that guarantee the following assertion is true?

" For any real number $0\leq b <a$, there exists a subset $X_1$ of $X$ such that $\mu(X_1)=b$"

Thanks in advance.

答:
If the measure space is complete, then it contains all sets of measure zero. Since the empty set is a subset of every set, then you can always find a subset $X_1 = \emptyset \subset X$ such that your condition holds.

A less trivial example has to do with non-atomic measures. I refer you to the nice links given by angryavian.

Title: An Useful Result in Measure Theory


问:
Find the limits for the following triple integral
The problem goes as follows: $$\iiint_{\mathrm{E}}(x^2-z^2)\, dx\,dy\,dz, $$where $\mathrm{E}$ is defined by $x, y,z \ge 0$
and $x+y+z \le 1$.
I'm have difficulties finding the limits in order to solve the triple integral. Do I use spherical or cylindrical coordinates?
How do I know which of the two to use for future problems? Or does it not matter?

Edit: I have not been able to calculate the integral, any help would be appreciated.

答:
Spherical or cylindrical ? Obviously not !

$$E=\{(x,y,z)\in \mathbb R^3\mid x\in [0,1], y\in [0,1-x], z\in [0,1-x-y]\}.$$


问:
limit of $\cos^n(\frac{\pi}{n})$
Pretty self explanatory - what is a nice way to show that
$(\cos(\frac{\pi}{n}))^n\to 1$ as $n\to\infty$, where $n$ are integers?

答:
I am going to calculate the limit of the original poster with neither series nor De L’Hospital’s rule, but only notable limits.

$\lim\limits_{x\to0}\dfrac{\log(\cos x)}{x^2}=$

$=-\lim\limits_{x\to0}\dfrac{\log\big[1+(\cos x-1)\big]}{(\cos x-1)}\cdot\lim\limits_{x\to0}\dfrac{1-\cos x}{x^2}=$

$=-1\cdot\dfrac12=-\dfrac12\;.$

By letting $\;x=\dfrac{\pi}n\;\big(n\in\mathbb{N}\land n\ge3\big)\;,\;$ it follows that

$\lim\limits_{n\to\infty}\dfrac{n^2}{\pi^2}\log\left(\cos\dfrac{\pi}n\right)=-\dfrac12\;,$

$\begin{align}
\lim\limits_{n\to\infty}n\log\left(\cos\dfrac{\pi}n\right)&=\lim\limits_{n\to\infty}\dfrac{\pi^2}n\cdot\lim\limits_{n\to\infty} \dfrac{n^2}{\pi^2}\log\left(\cos\dfrac{\pi}n\right)=\\
&=0\cdot\left(-\dfrac12\right)=0\;.
\end{align}$

Consequently,

$\lim\limits_{n\to\infty}\left(\cos\dfrac{\pi}n\right)^n=\lim\limits_{n\to\infty}e^{n\log\left(\cos\frac{\pi}n\right)}=e^0=1\;.$

Title: limit of $\cos^n(\frac{\pi}{n})$


答: As Cameron Williams mentioned, non-atomic measures satisfy this. The proof is due to Sierpinski and uses Zorn's Lemma; see this Mathoverflow discussion and this Math.SE answer.</p>

Title: An Useful Result in Measure Theory

Body:
Let $X$ be a borelian, finite measure space, WLOG, $\mu(X)=a$.

Are there any additional hypotheses that guarantee the following assertion is true?

" For any real number $0\leq b <a$, there exists a subset $X_1$ of $X$ such that $\mu(X_1)=b$"

Thanks in advance.


问:
The period of the given function?
I have this function for $\{\alpha,\beta,x\}>0$ and reals.
$$  \frac{\cos \left(\alpha +\frac{\beta }{2} \right)x+ \Big(5 \cos  (\alpha -\beta )x+7 \cos \beta  x-1\Big)\cos  \frac{\beta}{2}x }{\cos  \frac{\alpha}{2} x\; \cos  \frac{\beta -\alpha}{2} x }  $$
How to determine the period of this function?

答:
You really wouldn't want to ignore the parts where it's not defined. :))) Anyhow, doing just that, note that it would be enough to find a $T > 0$ such that $\cos(\frac{\alpha}{2} x) = \cos(\frac{\alpha}{2} (x + T)),$ $\cos(\frac{\beta - \alpha}{2} x) = \cos(\frac{\beta - \alpha}{2} (x + T)),$ $\cos((\beta - \alpha) x) = \cos((\beta - \alpha) (x + T)),$ $\cos(\beta x) = \cos(\beta (x + T)),$ $\cos(\frac{\beta}{2} x) = \cos(\frac{\beta}{2} (x + T))$ and $\cos((\alpha + \frac{\beta}{2}) x) = \cos((\alpha + \frac{\beta}{2}) (x + T)).$ Note that some of these equations are redundant. Taking into consideration the periodicity of $\cos,$ we deduce that $$\begin{cases} \frac{\alpha}{2} T = 2 k \pi,\\ \frac{\beta}{2} T = 2 l \pi, \end{cases}$$ for some non-zero $k, l \in \mathbb{Z}.$ Note how the fact that all the other arguments become multiples of $2\pi$ follows immediately if only these two conditions are satisfied. Thus, we get that $T = \frac{4 k \pi}{\alpha} = \frac{4 l \pi}{\beta}$ for some non-zero $k, l \in \mathbb{Z}.$

We remark that a necessary condition for this to be possible is that the fraction $\frac{\beta}{\alpha} = \frac{l}{k}$ is rational. This is also sufficient, since if $\beta = \frac{l}{k} \alpha,$ then $T := \frac{4 k \pi}{\alpha}$ is a period for the function in question. To get the smallest period provided by this line of reasoning, you compute $\frac{\beta}{\alpha}$ and simplify it as much as you can, getting a rational $\frac{l}{k}$ with $(l, k) = 1$ and then simply define $T:= \frac{4 k \pi}{\alpha}.$

Is this THE period of the function though? It may not be, the expression of the function is pretty gnarly, but maybe it can in fact be simplified to something simple, a function whose period is easily computed. Anyhow, this is more of a brute force algorithm. Nevertheless, a period does indeed appear.

Title: The period of the given function?


答: Hint. The condition $x, y, z\ge 0$ restricts you to the first octant. Then the equation $x+y+z=1$ defines the plane which intersects the axes at the points $(0, 0, 1), (0, 1, 0)$ and $(1, 0, 0).$ Finally, the inequality $x+y+z\le 1$ says you are interested in the finite region bounded by the first octant and the plane. Can you find your limits now?

Title: Find the limits for the following triple integral

Body:
The problem goes as follows: $$\iiint_{\mathrm{E}}(x^2-z^2)\, dx\,dy\,dz, $$where $\mathrm{E}$ is defined by $x, y,z \ge 0$
and $x+y+z \le 1$.
I'm have difficulties finding the limits in order to solve the triple integral. Do I use spherical or cylindrical coordinates?
How do I know which of the two to use for future problems? Or does it not matter?

Edit: I have not been able to calculate the integral, any help would be appreciated.


问:
Intuition of Addition Formula for Sine and Cosine
The proof of two angles for sine function is derived using $$\sin(A+B)=\sin A\cos B+\sin B\cos A$$ and $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ for cosine function. I know how to derive both of the proofs using acute angles which can be seen here http://en.wikibooks.org/wiki/Trigonometry/Addition_Formula_for_Cosines but pretty sure those who have taken trig know what I'm talking about. So I know how to derive and prove both of the two-angle functions using the acute angles, but what I am completely confused about is where those triangles came from. So for proving the two-angle cosine function, we look at two acute angles, $A$ and $B$, where $A+B<90$ and keep on expanding. So my question is, where did those two triangles come from and what is the intuition behind having two acute triangles on top of each other? </p>

答:
My physics professor was from Germany, and he said High School students were taught to just rederive trig identities any time they needed them using complex "algebra". That is, you can write out:
\begin{align}
\cos(x+y)+i\sin(x+y)
 & = e^{i(x+y)}= e^{ix}e^{iy}\\
 & = (\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\
 & = \cos(x)\cos(y)-\sin(x)\sin(y)+i(\cos(x)\sin(y)+\sin(x)\cos(y))
\end{align}

pretty quickly, and get the desired identity by considering real or imaginary part of equation. For instance, try this for $e^{i2x}=e^{ix}e^{ix}$.


问:
Calculate $\iint\limits_D (x+3y)*(-2x+y)-x-3y dxdy$ where D is the rectangle with corners at (4, 1), (5, 3), (-5, 4) and (-4,6)
I am trying to calculate the double integral $\iint\limits_D (x+3y)*(-2x+y)-x-3y dxdy$ where D is the rectangle with corners at (4, 1), (5, 3), (-5, 4) and (-4,6).

I started by sketching the area and realized that I probably have to divide the integration into two areas so that D=D_1 + D_2. I also calculated the linear equations connecting the points because I'm thinking that these lines are going to be the upper and lower bounds for the integrals, is that right? However I am really new to double integrals so I'm thinking this might be wrong or a very complicated way of solving this. I also don't get a volume when setting these linear equations as my bounds, I just integrate and switch x for y and vice versa, which does not seem right. Does anyone have any input? If so, I'd be very grateful. You can see my calculations and reasoning below:

[
[

Thanks in advance!

/Nick

答:
If points of the rectangle are $A(4,1), B(5,3), C(-4,6), D(-5,4)$, please note that once you find the slope of $AB$, $CD$ will have the same slope. Similarly once you find for $BC$, you know it for $DA$. Also, when writing equations of lines note that two adjacent sides of rectangle go through a common point. These are small things but just speeds things up for you a bit.

$AB: (y-1) = 2(x-4), DA: (y-1) = - \frac{1}{3}(x-4)$
$CD: (y-6) = 2(x+4), BC: (y-6) = -\frac{1}{3}(x+4)$

Simplifying, we have a rectangle bound by the below $4$ lines,

$-7 \leq - 2x + y \leq 14, 7 \leq x + 3y \leq 14$

Now if you want to do this integral without change of variable, it is $3$ integrals. But if we look at the integrand and the bounds, it is pretty clear that we do change of variable as $u = - 2x + y, v = x + 3y$.

The Jacobian of transformation, $|J| = \frac{1}{7}$ and here are our new bounds -

$- 7 \leq u \leq 14, \ 7 \leq v \leq 14$.

Integrand $(x+3y) (-2x+y)-x-3y$ becomes $(uv - v)$

The integral becomes

$I = \displaystyle \frac{1}{7} \int_{7}^{14} \int_{-7}^{14} (uv - v) \ du \ dv$


问:
The predictive distribution
I'm currently studying a bit of Bayesian statistics, and I got my self a bit stuck on the following result regarding the predictive distribution for $\mathbf{Y}|\mathbf{X = x}$.

My book states
$$
f(\mathbf{y}|\mathbf{x}) = \int_{\theta\in\Omega} f(\mathbf{y},\theta|\mathbf{x})d\theta
= \int_{\theta\in\Omega} f(\mathbf{y}|\theta, \mathbf{x}) f(\theta|\mathbf{x})d\theta
= \int_{\theta\in\Omega} f(\mathbf{y}|\theta) f(\theta|\mathbf{x})d\theta
$$

I'm unsure how we get
$$
f(\mathbf{y},\theta|\mathbf{x}) = f(\mathbf{y}|\theta, \mathbf{x}) f(\theta|\mathbf{x})
$$
I've attempted to use
$$
f(\mathbf{x_a},\mathbf{x_b}) = f_{\mathbf{x_b}|\mathbf{x_a}}(\mathbf{x_b}|\mathbf{x_a})f_\mathbf{x_a}(\mathbf{x_a})
$$
but unsuccessfully. I've also tried
$$
f(\mathbf{y},\theta) = f_{\mathbf{Y|\theta}}(\mathbf{y}|\theta)f(\theta)
$$
but in unsure how to make it all conditional on $\mathbf{X}$. So yeah, I'm tying knots on my self a bit. Additionally, I'm also unsure about the step $f(\mathbf{y}|\theta, \mathbf{x}) = f(\mathbf{y}|\theta)$. I'm guessing this is because $\mathbf{Y}$ and $\mathbf{X}$ are independent. So, we have $f(\mathbf{y}|\theta, \mathbf{x}) = f(\mathbf{y}|\theta)f(\mathbf{x})$, why why is $f(\mathbf{x}) = 1$?

I'm sure this is quite straightforward, since they offered no explanation of this in the book, and I'm probably just confusing my self a bit. But, I lost the better part of an hour trying on it already, and I think I might be fixating on a wrong approach.

答:
$f(y, \theta \mid x) = f(y \mid \theta, x) f(\theta, x)$ is a special case of $f(y,\theta) = f(y \mid \theta) f(\theta)$ where everything is conditional on $x$.  If this is still not convincing to you, just go back to the original equation $f(y, \theta \mid x) = f(y \mid \theta, x) f(\theta, x)$ and plug in $f(y, \theta \mid x) = f(y,\theta,x) / f(x)$ and $f(y \mid \theta, x) = f(y,\theta,x)/f(\theta, x)$ to explicitly verify this.

$f(y \mid \theta, x) = f(y \mid \theta)$ is not true in general, but is true if $x$ and $y$ are conditionally independent given $\theta$. In fact this equation is sometimes taken as the definition of conditional independence; an equivalent definition is $f(y,x \mid \theta) = f(y \mid \theta) f(x \mid \theta)$.

Title: The predictive distribution


问:
An inequality involving recurrence sequence $b_{n+1}=\frac{b_{n}^{2}+2b_{n}}{b_{n}^{2}+2b_{n}+2}$ with $b_{1}=1$ and still unsolved

Given a recursion $b_{n+ 1}= \dfrac{b_{n}^{2}+ 2b_{n}}{b_{n}^{2}+ 2b_{n}+ 2}$ with $b_{1}= 1.$ Prove that
$$\left | \frac{2}{n}- \frac{2\ln n}{n^{2}}- b_{n} \right |\leq\frac{1}{n^{2}}$$
Source: AoPS/@Ji_Chen (still unsolved)

I think this problem related to the hyperbol functions, how can we find $b_{n}$ or such a relationship between $b_{n}$ and $n.$ Professor Ji always inspires me by beautiful asymptotics, at most, I only used his result to show $\lim b_{n}= 0$ not to use Laurent's series, hope you find more observations involving the recurrence sequence above, I love it.

答:
This is not meant as a complete proof of the inequality above (I've got no motivation to solve a problem without the complete source or information about the importance of exactly that recursion), just a few thoughts how that might be accomplished:

With $r_n=b^{-1}_n$, the recursion becomes (after some elementary algebra)
$$r_{n+1}-r_n=\frac12+\frac1{4\,r_n+2}\tag{1}$$ with $r_1=1$. Now $r_1>0$ implies that all $r_n>0$, so we have $$r_{n+1}-r_n>\frac12,$$ and this, together with $r_1=1$, gives
$$r_n\ge\frac{n+1}2.\tag{2}$$
Plugging (2) into (1), we obtain
$$r_{n+1}-r_n\le\frac12+\frac1{2(n+2)},$$ and by summing up,
$$r_n\le\frac{n+H_{n+1}}2-\frac14,\tag{3}$$ where $$H_n=\sum^n_{k=1}\frac1k$$ is the $n$-th harmonic number. Since $H_n-\ln n\to\gamma$ as $n\to\infty$, the RHS of (3) already gives the first two terms $$\frac2n-\frac{\ln n}{n^2}$$ of the asymptotics of $b_n=r^{-1}_n$, as a lower bound. For the upper bound, and the inequality above, one would have to plug (3) into (1), again, and do some rather technical estimates.

BTW, for $b_1\in(-2,0)$, $b_n$ converges to $-1$.


问:
How was this simplification done?

\begin{align*} \begin{pmatrix}
     -2\\
     n\end{pmatrix}&=\frac{(-2)(-3)(-4)\dots(-2-n+1)}{n!}\\[5pt] &= (-1)^n\frac{2\times3\times4\times\dots n(n+1)}{n!}\\[5pt]
 &=(-1)^n(n+1)\\[5pt] \end{align*}

While reviewing the binomial theorem, I came across this example problem and I feel like this simplification was not clearly explained. It seems like -1 was factored out but what happens to the expression it is multiplied by. How is it written in such a way? Also, why is $-1$ raised to the power of $n$...$(-1)^n$. I do not find this intuitive and felt like a step was skipped although that might not be the case.

Overall, I just want to understand what the process is here.

Title: How was this simplification done?

Body:
I like using the product notation
so everything is explicit.

$\begin{array}\\
\binom{-2}{n}
&=\dfrac{\prod_{k=0}^{n-1} (-2-k)}{n!}\\
&=\dfrac{(-1)^n\prod_{k=0}^{n-1} (2+k)}{n!}\\
&=\dfrac{(-1)^n\prod_{k=2}^{n+1} k}{n!}\\
&=\dfrac{(-1)^n\prod_{k=1}^{n+1} k}{n!}
\qquad\text{(adds the term with } k=1)\\
&=\dfrac{(-1)^n(n+1)!}{n!}\\
&=(-1)^n(n+1)\\
\end{array}
$

More generally,
if $m$ is a positive integer,

$\begin{array}\\
\binom{-m}{n}
&=\dfrac{\prod_{k=0}^{n-1} (-m-k)}{n!}\\
&=\dfrac{(-1)^n\prod_{k=0}^{n-1} (m+k)}{n!}\\
&=\dfrac{(-1)^n\prod_{k=m}^{n+m-1} k}{n!}\\
&=\dfrac{(-1)^n\prod_{k=1}^{n+m-1} k}{n!\prod_{k=1}^{m-1}k}
\qquad\text{(adds the term with } k=1 \text{ to } m-1)\\
&=\dfrac{(-1)^n(n+m-1)!}{n!(m-1)!}\\
&=(-1)^n\binom{n+m-1}{m-1}\\
\end{array}
$


问:
Logic question with premises
Put the following syllogism into the standard form, which uses a horizontal line to separate premises and conclusion. Moreover, tell whether it is valid or not by showing its Venn Diagram.

Bill didn't go to work this morning because he wore a sweater, and he never wears a sweater to work.

As far I know, to indicate absence of elements, we use shade, whereas to indicate existence of an element belonging to a class, we mark "$x$" inside the circle. But, I could not associate this rule with given premises. Can anyone suggest their solutions with correct premise-conclusion and Venn diagram?

Note: What is the name of this subject, and where can I read about these concepts?

答:
As pointed out in comments, we need to write this in a slightly different way to make it better fitted to be written as a syllogism. So, let me read the statement as

"Bill wore a sweater today;

everyday that Bill wears a sweater, he does not go to work;

Bill did not go to work today."

If $t$ stands for "today", $S(x)$ stands for "Bill wore a sweater at day $x$" and $W(x)$ stands for "Bill went to work at day $x$", this gives us the syllogism (sorry for the simply awful appearance...).

$$S(t)$$
$$\forall x\Big(S(x)\rightarrow \neg W(x)\Big)$$
$$\rule{4cm}{0.4pt}$$
$$\neg W(t)$$

Now, to see this as a Venn diagram, start by writing down your propositions (again, sorry for the awful drawings, in a blackboard this would be easier...):

$S(t)$ means that there is an element in $S$, but we don't know if it is also inside $W$, so we place it in the frontier, as this:

Now $\forall x(S(x)\rightarrow\neg W(x))$ implies that every element $x$ which satisfies $S$ (that is, $S(x)$), does not satisfy $W$ (that is, $\neg W(x)$); in other words, no element of $S$ is inside $W$, and we can shade the area they share as empty:

Combining both diagrams, we know the element $t$, we know to be in $S$, can't be also in $W$, given $S\cap W$ is empty, and so must be in $S$ and outside $W$, as below:

The conclusion is that $t$ is not in $W$, that is, $\neg W(t)$ is true (that is, Bill did not go to work today), and so the argument is, in fact, valid.

Title: Logic question with premises


问:
Complex analysis for generating functions
Looking at "Generatingfunctionology" by Wilf, it seems that complex analysis is quite useful when it comes to finding asymptotic approximations. However, what prerequisites from complex analysis are actually needed? I've seen that Cauchy's theorem is used quite often, but is there anything apart from it that is needed?</p>

答:
A great, user-friendly introduction to all the necessary prerequisites we need from complex analysis together with a wealth of applications is presented in part B of Analytic Combinatorics by P. Flajolet and R. Sedgewick.

Each of the chapters IV - VIII

IV:  Complex Analysis, rational and meromorphic asymptotics

V:       Applications of rational and meromorphic asymptotics

VI: Singularity Analysis of generating functions

VII: Applications of singularity analysis

VIII:    Saddle-point asymptotics

starts with explanatory sections and cites resp. derives the theorems needed to understand the given applications.


问:
Probability of $X > \max\{Y,Z\}$ exponentials
Let $X,$ $Y$ and $Z$ be exponentially distributed with $E[X] = 1/\lambda_X$, $E[Y] = 1/\lambda_Y$, $E[Z] = 1/\lambda_Z$ respectively.

How do I calculate $P(X > \max\{Y,Z\})$?

I have A, B, C and D for 3 cash boxes in a queueing system. Consider D the one waiting. I want to find the probability D is the last one to leave. The service times of the C.Boxes are indep. Exp with mean 5 (same mean for c.box 1, 2 and 3) That's how I'm trying to solve this: P(D last one) = P(T1 > max{T2,T3})*P(T1 < min{T2,T3}) + P(T2 > max{T1,T3})*P(T2 < min{T1,T3}) + P(T3 > max{T1,T2})*P(T3 < min{T1,T2})

And then:
$$(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4 = = 3(\frac{1}{2})^4 = 3\frac{1}{16} = \frac{3}{16}$$

答:
Simulation

Here is a simulation in R that you can use to see if D is the last to leave given that there are 3 people ahead of him. You can set the rate parameters for the three boxes using rat1, rat2, rat3 for the three boxes. Here I have used 1/5, 1/5, 1/5 as you stated in the problem description.

```<code>nit=10000
rat1=1/5
rat2=1/5
rat3=1/5
a=rexp(nit, rat1)
ad=a+rexp(nit, rat1)
b=rexp(nit, rat2)
bd=b+rexp(nit, rat2)
c=rexp(nit, rat3)
cd=c+rexp(nit, rat3)
coun=0
for (i in 1:nit) {
  # a is the first to leave & d is last to leave
  if (((a[i]<min(b[i],c[i])) & ad[i]>(max(b[i], c[i]))) | ((b[i]<min(a[i],c[i])) & bd[i]>(max(a[i], c[i]))) | ((c[i]<min(a[i],b[i])) & cd[i]>(max(a[i], b[i])))) {
    coun=coun+1
  }
}
print(coun/nit)
</code>```

There's about a 33%-34% chance that D is the last one to leave.

-out

```<code>0.3363
0.3431
0.339
</code>```

Calculation

We want to find $P(\infty>D+A>C>B>A>0)$ and then multiply this by 6 (switching the order of B and C and then allowing B or C to be the first one to finish etc.). These limits can be expressed as $A>0,B>A,C>B,D>C-A$. We directly compute the integral

$$\begin{split}\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\int_{b}^{\infty}\frac{e^{-\frac c 5}} 5\int_{c-a}^{\infty}\frac{e^{-\frac d 5}} 5\text{dd dc db da}
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\int_{b}^{\infty}\frac{e^{-\frac c 5}} 5\left(e^{\frac a 5-\frac c 5}\right)\text{ dc db da}\\
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\left(\frac{e^{\frac a 5-\frac {2b} 5}}2\right)\text{ db da}\\
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \frac{e^{-\frac{2a}5}}{6}\text{ da}\\
&=\frac{1}{18}\end{split}$$

Multiply this by 6 to get $\frac 1 3$ as the probability that D is last. Surprisingly, this is the same as if there were no person in front of D.

Comment

I think your formula works, but you should be careful not to overload $T_1$ to represent two different things, because $P(T_1>\max(T_2,T_3))*P(T_1<\min(T_2,T_3))$ is just 0, it cannot be greater than the max and less than the min. You could write

$$P(\text{D last one}) = P(T_1' > \max(T_2',T_3'))*P(T_1 < \min(T_2,T_3)) + P(T_2' > \max(T_1',T_3'))*P(T_2 < \min(T_1,T_3)) + P(T_3' > \max(T_1',T_2'))*P(T_3 < \min(T_1,T_2))$$

with the primes indicating after the first customer finishes (involving memorylessness).

Body:
Simulation

Here is a simulation in R that you can use to see if D is the last to leave given that there are 3 people ahead of him. You can set the rate parameters for the three boxes using rat1, rat2, rat3 for the three boxes. Here I have used 1/5, 1/5, 1/5 as you stated in the problem description.

nit=10000
rat1=1/5
rat2=1/5
rat3=1/5
a=rexp(nit, rat1)
ad=a+rexp(nit, rat1)
b=rexp(nit, rat2)
bd=b+rexp(nit, rat2)
c=rexp(nit, rat3)
cd=c+rexp(nit, rat3)
coun=0
for (i in 1:nit) {
  # a is the first to leave & d is last to leave
  if (((a[i]<min(b[i],c[i])) & ad[i]>(max(b[i], c[i]))) | ((b[i]<min(a[i],c[i])) & bd[i]>(max(a[i], c[i]))) | ((c[i]<min(a[i],b[i])) & cd[i]>(max(a[i], b[i])))) {
    coun=coun+1
  }
}
print(coun/nit)

There's about a 33%-34% chance that D is the last one to leave.

-out

0.3363
0.3431
0.339

Calculation

We want to find $P(\infty>D+A>C>B>A>0)$ and then multiply this by 6 (switching the order of B and C and then allowing B or C to be the first one to finish etc.). These limits can be expressed as $A>0,B>A,C>B,D>C-A$. We directly compute the integral

$$\begin{split}\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\int_{b}^{\infty}\frac{e^{-\frac c 5}} 5\int_{c-a}^{\infty}\frac{e^{-\frac d 5}} 5\text{dd dc db da}
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\int_{b}^{\infty}\frac{e^{-\frac c 5}} 5\left(e^{\frac a 5-\frac c 5}\right)\text{ dc db da}\\
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \int_{a}^{\infty}\frac{e^{-\frac b 5}} 5\left(\frac{e^{\frac a 5-\frac {2b} 5}}2\right)\text{ db da}\\
&=\int_{0}^{\infty}\frac{e^{-\frac a 5}} 5 \frac{e^{-\frac{2a}5}}{6}\text{ da}\\
&=\frac{1}{18}\end{split}$$

Multiply this by 6 to get $\frac 1 3$ as the probability that D is last. Surprisingly, this is the same as if there were no person in front of D.

Comment

I think your formula works, but you should be careful not to overload $T_1$ to represent two different things, because $P(T_1>\max(T_2,T_3))*P(T_1<\min(T_2,T_3))$ is just 0, it cannot be greater than the max and less than the min. You could write

$$P(\text{D last one}) = P(T_1' > \max(T_2',T_3'))*P(T_1 < \min(T_2,T_3)) + P(T_2' > \max(T_1',T_3'))*P(T_2 < \min(T_1,T_3)) + P(T_3' > \max(T_1',T_2'))*P(T_3 < \min(T_1,T_2))$$

with the primes indicating after the first customer finishes (involving memorylessness).


问:
Evaluating $\int \arcsin(\sqrt{x+1}-\sqrt{x})\,dx$
I found this integral here: https://www.12000.org/my_notes/ten_hard_integrals/index.htm

Sites like https://www.integral-calculator.com/, which I often find very helpful, did not successfully calculate the integral.

I have found a potential answer, that seems to be correct.

Using the formula

$$\int f(x)\,dx=x \cdot f(x)-\left[{\int f^{-1}(y)\,dy}\right]_{y=f(x)},$$

I calculated :

$$\int \arcsin(\sqrt{x+1}-\sqrt{x})\,dx = x \cdot \arcsin(\sqrt{x+1}-\sqrt{x})+\frac{3}{8}\arcsin(\sqrt{x+1}-\sqrt{x})+\frac{1}{8}\sqrt{\sqrt{x(x+1)}-x}\cdot[(1+2\sqrt{2})\sqrt{x+1}-(1-2\sqrt{2})\sqrt{x}]$$

My question is: How else could I integrate this function? Is there an easier way?
I tried to substitute $x=tan(u)^2$ but it seems that the resulting integral is an extremely long combination of sines and cosines.

Thank you for your help

Additional notes on my reasoning:

The inverse function of $\arcsin(\sqrt{x+1}-\sqrt{x})$ is $\left(\frac{\sin^2 x-1}{2\sin x}\right)^2$

Integrating this function can be done by expanding the square :

$$\int \left(\frac{\sin^2 y-1}{2\sin y}\right)^2 dy=\int \frac{1}{4}(\sin^2 y-2+\csc^2 y)\,dy$$

After having calculated that integral, we replace y by $\arcsin(\sqrt{x+1}-\sqrt{x})$. After some simplification work, I get the result mentioned above.

答:
I just wanted to add on. I am still working on this problem. I was thinking that we could build of the identities:

$$\sin(a-b) = \sin(a)cos(b) - \cos(a)\sin(b)$$
$$a-b = \arcsin(\sin(a)\cos(b) - \cos(a)\sin(b))$$.

We could find two functions that we can integrate, $a$ and $b$, where $\sin(a)\cos(b) = \sqrt{x+1}$ and $\cos(a)\sin(b) = \sqrt{x}.$ I was thinking probably letting $a$ and $b$ being some inverse trig functions and proceeding from there.


问:
Am I doing the proper steps for this differential equation?
I don't need the answer, I just need to know if I am plugging in the numbers right.

Suppose $f(x)= e^{2x}.$ We know that if $h$ is small, then $f'(x) \simeq \frac{f(x + h) - f(x)}{h}.$ Find the approximative value of $f'(1)$ with $h=0.01$.

So, I am plugging in the data like this:

$$(e(1 + 0.01)^2-e^{2(1)})/0.01.$$

Then, I find the answer by finding the derivative of the formula above, right?

答:
First of all, welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. It will come in handy later, I promise you!

Secondly, you should plug in the following: $$f'(1) \simeq \frac{f(1.01) - f(1)}{0.01} = \frac{e^{2.02} - e^2}{0.01} \simeq 14.92688.$$ This is not yet very close to the actual value of $f'(1) = 2 e^2 \simeq 14.7781,$ but making $h$ smaller and smaller will get you there- this is basically the definition of the derivative.


答:
If $f(x)=e^{2x}$, then by definition
$$
f'(1) = \lim_{h \to 0}\frac{f(1+h)-f(1)}{h}=\lim_{h \to 0}\frac{e^{2(1+h)}-e^2}{h}
$$
If you want to approximate $f'(1)$, then it is valid to plug in $h=0.01$, so that
$$
f'(1) \approx \frac{e^{2(1+0.01)}-e^2}{0.01}
$$
The accuracy of this approximation depends on the value of $h$ (which is not a dummy variable as it is in the limiting expression).

Title: Am I doing the proper steps for this differential equation?

Body:
I don't need the answer, I just need to know if I am plugging in the numbers right.

Suppose $f(x)= e^{2x}.$ We know that if $h$ is small, then $f'(x) \simeq \frac{f(x + h) - f(x)}{h}.$ Find the approximative value of $f'(1)$ with $h=0.01$.

So, I am plugging in the data like this:

$$(e(1 + 0.01)^2-e^{2(1)})/0.01.$$

Then, I find the answer by finding the derivative of the formula above, right?


问:
Proving the excess of two split triangles is the same as the original
The excess of a triangle is the number of degrees over $180$ a triangle has in spherical geometry.

Given spherical $\triangle ABC$ that is split into two triangles by an arc from the vertex going to the opposite side. Show that the excess of $\triangle ABC=$ the excess of $\triangle ABD$+ the excess of $\triangle BDC$.

I drew this picture:

Essentially it seems that $\angle ABC=\angle ABD+\angle DB$C.

I therefore need the excess defined by the two "bottom angles" of each smaller triangle to be the same as the excess defined by $\angle BAC$ and $\angle BCD$.

Because each of the smaller triangles only has one angle that is different than those of the larger triangle, the excess from the two angles formed by $d$ needs to be half of what is contributed by $\angle BAD$ and $\angle BCD$. Is this true?

答:
Let

$\alpha$, $\beta$, $\gamma$ be the interior angles of the larger triangle at $A$, $B$, $C$, respectively,

$\beta_1$, $\beta_2$ be the angles of the smaller triangles at $B$, and

$\delta_1$, $\delta_2$ be the angles of the smaller triangles at $D$.

Then $\beta_1 + \beta_2 = \beta$ and $\delta_1 + \delta_2 = 180$ degrees. Therefore the sum  of excesses of the smaller triangles is
$$
 (\alpha + \beta_1 + \delta_1 - 180) + (\gamma+\delta_2 + \beta_2 - 180)
 = \alpha + \beta + \gamma - 180
$$
and therefore equal to the excess of the larger triangle.


问:
Question about logarithmic derivative
In my complex analysis class, my professor mentioned something about how if $f$ and $g$ are functions defined on some domain $D\subset\mathbb{C}$, if they have identical logarithmic derivatives, then they are off by a multiplication by a constant. My question is, how can we show this given the weird behavior of $\log$ on the complex plane?

Namely, before even getting to the proof, how can we be sure that we can define $\log f$ in a way such that we can take its derivative? The function $\log$ must have a branch cut somewhere and there is no guarantee (at least for what I see) of being able to always pick a branch cut such that the values of $f$ dodge the line of discontinuity entirely. If there is a point on such a line, then I would assume that $(\log f)'$ is not even defined there. Is there a way around this that I am missing?

答:
The logaritmic derivative of a holomorphic function $f$ is the meromorphic function $\frac{f'}{f}$, with singularities at the zeros of $f$. Now, to prove the result.

One way to prove the result is as follows:

if $f\equiv 0$, the result is trivial.
So, suppose $f\not\equiv 0$
Now, let $z_0\in D$ be such that $f(z_0)\neq 0$ and there exists a disc $B(z_0,\varepsilon)\subset D$ on which $f\neq 0$.
As they have the same logarithmic derivative, $g\not= 0$ on $B(z_0,\varepsilon)$.
As $B$ is simply connected, we can define both $\ln(f)$ and $\ln(g)$ on $B$
(explicitly as $\int_{z_0}^z\frac{1}{f}$ over some path contained in  $B$: this is independent from the path by Cauchy's theorem).
Having the same logarithmic derivative, we have $\ln(f)=\ln(g)+c$ on $B$. This implies $f=e^c g$ on $B$. The identity principle does the rest, as $\frac{f}{g}=c$ on $B$ implies $\frac{f}{g}=c$ on all the domain (one can also prove the result withouth the identity principle, by considering the constant $c$ as a function of $z_0$ and proving it is a continuous locally constant function, thus constant on all of $D$)

Title: Question about logarithmic derivative


问:
Show that the area vectors for a general tetrahedron sum to zero
Using vector addition and multiplication, it is possible to show that the sum of the area vectors for a general closed tetrahedron in $\mathbb{R}^3$ (3-space) is zero.

Hint: start by writing down three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$ and derive relationships for the other sides in terms of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

答:
So for a general tetrahedron, we can derive that three of the sides are described by the vectors: $\vec{a}$, $\vec{b}$ and $\vec{c}$.

We can then write the area vector – using the outward pointing convention – for the first side as: $$\vec{A}_{ab} = \frac{1}{2}\vec{a}\times\vec{b}$$

Similarly, two of the remaining three area vectors can be written: $$\vec{A}_{bc} = \frac{1}{2}\vec{b}\times\vec{c}$$
$$\vec{A}_{ca} = \frac{1}{2}\vec{c} \times \vec{a}$$

The final area can be found using the fact that the vectors that define the sides define a closed figure and thus must sum to zero as suggested in the first hint by Blue:  $$\vec{A}_{(c-a)(b-a)}=\frac{1}{2}(\vec{c}-\vec{a})\times(\vec{b}-\vec{a})$$

Summing all of these we find that: $$ \frac{1}{2}\vec{a}\times\vec{b} + \frac{1}{2}\vec{b}\times\vec{c}+\frac{1}{2}\vec{c} \times \vec{a}+\frac{1}{2}(\vec{c}-\vec{a})\times(\vec{b}-\vec{a}) = \frac{1}{2}\left(\vec{a}\times\vec{b} + \vec{b}\times\vec{c}+\vec{c} \times \vec{a}+(\vec{c}-\vec{a})\times(\vec{b}-\vec{a})\right) = \vec{0}$$

by applying the properties of distribution, simplifying the cross products and canceling.


问:
Minimum value of $ \lceil{an}\rceil-an $, where a is irrational.
Given the function:

$$ f(n) = \lceil{a \cdot n}\rceil - a \cdot n,  n \in \mathbb{N} $$

Find the minimum value of $f$ in terms of $n$, if $a$ is irrational.

We know that if $n$ was real then the minimum value of $f$ would be $0$ for $n_{k}=\frac{k}{a}, k \in \mathbb{N}$. When $n$ is natural though, the function's minimum goes to $0$ while $n$ approaches the $\frac{1}{a}$ in integer form. For example, if $a=\sqrt{2}$ then $\frac{1}{a}=0.707106...$ and when $n$ approaches $70,707,7071,...$ or $\frac{10^{k}}{a}, k \in \mathbb{N}$ then $f$ goes to $0$.

I tend to believe that $f(n) \geq \frac{1}{10n} $ but I would prefer to find a tighter lower bound, proof for the latter, or maybe even a lower bound depending on the irrational number $a$.

Thank you in advance.

答: There is a pretty standard result showing that $(\{a n\})_{n \in \mathbb{N}} \subseteq [0, 1]$ is dense if $a \in \mathbb{R} \setminus \mathbb{Q}$ (you may find it here, for example: reference). Note that $\lceil a n \rceil - a n = \lceil a n \rceil - \lfloor a n \rfloor - \{a n \} = 1 - \{a n\}.$ Since $(\{a n\})_{n \in \mathbb{N}} \subseteq [0, 1]$ is dense, we deduce that the infimum of this expression is $0.$


问:
If $f$ is differentiable at $c$ then $F$ is differentiable at $c$
I've be having trouble with this question and am looking for some advice.

Let $a,b \in \mathbb{R},  a  <  b$.   Let $f$ be  an  integrable  function  on  $[a, b]$,  let $c \in(a, b)$, and let
$$F(x) =\int_0^x \ f(t)\,dt.$$ Prove or disprove the following:

If $f$ is differentiable at $c$ then $F$ is differentiable at $c$.

答: By the first fundamental theorem of calculus, $F'(c)=f(c)$. So in fact we don't even need $f$ to be differentiable at $c$; it is enough for $f$ to be continuous at $c$.


问:
Computing pullback of valuation (order)
I try to do problem from chapter 2 number 2 from [Silverman] "The Arithmetic of elliptic curves" .

(http://www.pdmi.ras.ru/~lowdimma/BSD/Silverman-Arithmetic_of_EC.pdf exercise 2.2)

Let $\phi : C_1 \rightarrow C_2$ be a nonconstant map of smooth curves, let $f \in \overline{K}(C_2)^*$, and $P \in C_1$. Prove that $$\mbox{ord}_P(\phi^*f) = e_\phi(P)\ \mbox{ord}_{\phi(P)}(f)$$
where $e_\phi(P) := \ \mbox{ord}_P(\phi^*t_{\phi(P)}), t_{\phi(P)}$ is a uniformizer at $\phi(P).$

Attempt : It perhaps about unwrap the definitions, but I cannot see how the extra $e_\phi(P)$ comes from.

Let $f \in \overline{K}(C_2)^* = \overline{K}(C_2)\setminus\{0\}.$ Let $m = \ \mbox{ord}_P(\phi^*f) = \ \mbox{ord}_P(f \circ \phi)$ and $n = \ \mbox{ord}_{\phi(P)}(f)$.

Then by definition, $$f \circ \phi \in M^m_P \rightarrow f\circ \phi = \sum_{i=1}^k h_{i1}h_{i2}...h_{im}$$ and $$f \in M^n_{\phi(P)} \rightarrow f = \sum_{i=1}^l g_{i1}g_{i2}...g_{in}$$ where $h_{ij}(P) = 0 = g_{st}(\phi(P))$ for all $i, j, s, t.$

It does not clear where from definition, $e_\phi(P)$ should factor out.

Guess I might instead try to look at $g := f^{e_\phi(P)}$. So I now want $\mbox{ord}_P(\phi^*f) = \ \mbox{ord}_{\phi(P)}(g)$ (compute two function under ord).

Still why they are equal ? Perhaps more tools on ord will help, but [Silverman] only provide definitions, so it still not so clear.

答:
You can write $f=gt_{\phi(P)}^{\alpha}$ for some $g \in \overline{K}(C_2)$ defined at $\phi(P)$ and equal to a nonzero scalar $\lambda$ that point and $\alpha$ is an integer.

Then $\phi^*g$ is a function on $C_1$ defined at $P$ and its value at $P$ is $\lambda \neq 0$, so it has vanishing order zero at $P$. Now, the vanishing order of $\phi^*t_{\phi(P)}^{\alpha}$ is by definition $e_{\phi}(P)\alpha$, and thus the vanishing order of $\phi^*f=\phi^*g \phi^*t_{\phi(P)}^{\alpha}$ at $P$ is $e_{\phi}(P)\alpha$, and $\alpha$ is the vanishing order of $f$ at $\phi(P)$, QED.

Title: Computing pullback of valuation (order)


问:
Example where $\lim_{x\to 0}f(x^2)$ exists but $\lim_{x\to 0}f(x)$ does not.
Can somebody give me an example where $$\lim_{x\to 0}f(x^2)$$ exists but $$\lim_{x\to 0}f(x)$$ does not?</p>

答: $\lim\limits_{x \to 0} \sqrt{x}$  would be a good choice because the domain of f(x) is not a punctured neighbourhood of zero. Whereas it is trivial that $\lim\limits_{x \to 0} {x}$$ exists