问:
$V_\omega$ is countable
Is there an easy way to prove this? I found a book that suggests the injection $h:V_\omega\to\omega$ defined by

$$h(\{x_1,x_2,\dots,x_n\})=2^{h(x_1)}+2^{h(x_2)}+\cdots+2^{h(x_n)},$$

but I hit some snags while proving that the function is well-defined. Are there any other clever ways to prove this fact? (Note: I am making a formal proof, so I need to be able to justify my steps.)

Edit: Let me see if I can make my difficulty more clear. Before I can start the proof by induction, I need a precise statement of the induction hypothesis, and this is what I am trying to hammer out. The hypothesis:

For each $n\in\omega$, there exists a unique bijection $h_n:V_n\to|V_n|$ such that $$h_n(x)=\sum_{m<|V_{n-1}|}\operatorname{if}(h_{n-1}^{-1}(m)\in x,2^m,0).$$
Moreover, the $h_n$ as defined satisfies $h_n\restriction |V_{n-1}|=h_{n-1}$.

Does this hypothesis contain enough to prove the claim? The reason for the odd way of stating the sum is that it avoids the issue in the linked question (we need to show that addition is a well-defined operation over sets), but as a consequence I have to prove that
the $h_n$ is a bijection and not just an injection, and I think I also need the restriction part in order to prove that the union $h=\bigcup_{n\in\omega}h_n$ is the required bijection.

答:
To prove that $h$ is well-defined and injective, it is most natural to proceed by induction on the rank of sets. The rank $\mathsf{rk}(x)$ of a set $x$ is an ordinal, defined recursively as follows (see also MathWorld for a definition for all sets):

$$\mathsf{rk}(x):= \sup\{\mathsf{rk}(y)+1: y \in x\}$$

Lemma: The rank of a set is well-defined.

This lemma can be proved using the Axiom of Foundation, see also this question. Since all suprema involved in computing $\mathsf{rk}$ for $x \in V_\omega$ are finite, the results are also finite ordinals.

The only set of rank zero is of course $\varnothing$; we define/obtain $h(\varnothing) = 0$.

Now given a set $X = \{x_1, \ldots, x_m\}$ of rank $n+1$, we use the Axiom of Replacement for $f(x) = 2^{h(x)}$, well-defined and injective by induction hypothesis.

We now have the set $Y = \{2^{h(x_1)},\ldots, 2^{h(x_m)}\}$; using your earlier question this gives rise to a well-defined number $h(X) = \sum Y$, so $h$ is well-defined on $X$ as well.

To show that $h$ is injective, let $X'$ be such that $\mathsf{rk}(X') \le \mathsf{rk}(X)$. Then we can obtain the sets $\{h(x_1),\ldots, h(x_m)\}$ and $\{h(x'_1),\ldots, h(x'_{m'})\}$ by the Basis Representation Theorem. The Axiom of Replacement can be applied to these sets since $h^{-1}$ is a functional relation, and we recover $X$ and $X'$; injectivity of both the Basis Representation and $h^{-1}$ ensures that $h(X) = h(X')$ implies $X = X'$. Thus $h$ is injective and well-defined on sets of rank at most $n+1$.

In conclusion, $h$ is well-defined on $V_\omega$; its injectivity on $V_\omega$ is trivially verified.

Title: $V_\omega$ is countable


问:
Find curve at the intersection of two level surfaces.
Let C be the curve at the intersection of two level surfaces M(x, y, z) = 5 and N(x,y,z) = 0, passing through a point P (1,1,1). Let $M(x, y, z) = 2x^2 + y^2 + 2z^2$ and $N(x, y, z) = xy-z$. Find curve C in the parametric form <x(t), y(t), z(t)>.

I am quite stuck on this one. I could let $z = xy$ and then plugging that into M I'd have $2x^2 + y^2 + 2x^2y^2 = 5$, and I have no idea what would be good ways to parametrize that.

Another way is to let $y=z/x$, in which case I end up with $2x^4 + (2z^2-5)x^2+z^2 = 0$, but again, I'm quite stuck. Any help would be really appreciated!

Or, if it is impossible to find the curve C, the question I'm working on actually asks for the tangent line to curve C at P if x'(0) = 3. If this makes things easier, how exactly is using this tangent more useful?

答:
It often helps to visualize the situation:

First, eliminate $z$ from your two equations (quite simple... just look at your second equation and plug $z$ into your first equation). Solve that first equation to get $y(x) = \pm \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}}$.  (It is quadratic, so you'll get a + and a - solution.)

Now you have your three-dimensional parameterization:  $(x, y(x), z(x,y)) = (x, \pm \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}}, \pm x \frac{\sqrt{5-2 x^2}}{\sqrt{2 x^2+1}})$.

Note especially that this curve includes the point $(1,1,1)$, as you can verify by simple substitutions of $x=1$.

(Note that $-2 \leq x \leq 2$.)

ContentLicense: CC BY-SA 4.0


问:
Why is $\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\left<(1,2,4)\right>$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$?
I understand that this is isomorphic to either $\mathbb{Z}_4 \times \mathbb{Z}_8$ or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_8$, but I can't think of any way to show that it is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_8$.</p>

答:
In $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, let $v=(1,2,4)$, $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$.

Then $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} = \langle v,e_2,e_3 \rangle$ because $e_1= v-2e_2-4e_3$.

Thus $\langle v,4e_1,4e_2,8e_3 \rangle = \langle v,4e_2,8e_3 \rangle$ and so
$$
\mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_8 /\langle(1,2,4)\rangle
=
\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_1,4e_2,8e_3 \rangle =
\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle v,4e_2,8e_3 \rangle
\cong
\mathbb{Z}_4 \times \mathbb{Z}_8
$$

ContentLicense: CC BY-SA 3.0


问:
Uniform arrival times at a bus stop

40 people arrives at a bus stop at a uniform continuous time between
8:00 and 8:15. The bus arrives at the stop at 8:10, 8:12, and 8:15
with probabilities 0.3, 0.5, and 0.2, respectively and takes everyone
at the stop. Let X be the number of people taking the bus. What is the expectation and variance of X?

Intuitively, I am getting $E(X)=40*(0.3*10/15+0.5*12/15+0.2*15/15)=32$. But how can I get the pdf of X to calculate the expectation and variance?

答:
Let $Y \in \{1, 2, 3\}$ represent the arrival time of the bus, with $Y = 1$ representing 8:10, and $Y = 2$ representing 8:12, and $Y = 3$ representing 8:15.  Then $$\Pr[Y = 1] = 0.3 \\ \Pr[Y = 2] = 0.5 \\ \Pr[Y = 3] = 0.2.$$  We want to compute $\Pr[X = x]$, the unconditional PMF of the number of passengers the bus takes.  To this end, we condition on $Y$.  For instance, $\Pr[X = x \mid Y = 1]$ represents the probability that exactly $x$ people arrive at the stop by 8:10.  Since the arrivals are uniformly distributed from 8:00 to 8:15, each person arriving has a $2/3$ probability of arriving within the first 10 minutes, and each person's arrival is independent of the arrivals of the other people.  Thus, $$X \mid Y = 1 \sim \operatorname{Binomial}(n = 40, p_1 = 2/3)$$ with
$$\Pr[X = x \mid Y = 1] = \binom{40}{x} (2/3)^x (1/3)^{40-x}.$$  Similarly, $$X \mid Y = 2 \sim \operatorname{Binomial}(n = 40, p_2 = 4/5).$$  In the final case, $X \mid Y = 3$ is not random: we have $\Pr[X = 40 \mid Y = 3] = 1$, since everyone is assured to arrive by 8:15.  Then by the law of total probability,
$$\begin{align}
\Pr[X = x] &= \sum_{y=1}^3 \Pr[X = x \mid Y = y]\Pr[Y = y] \\
&= \binom{40}{x} \left( \frac{3}{10} (2/3)^x (1/3)^{40-x} + \frac{1}{2} (4/5)^x (1/5)^{40-x} + \frac{1}{5} \mathbb 1(x = 40) \right). \\
\end{align}$$

This PMF does not need to be calculated in order to compute the expectation and variance.  To get the expectation, we apply the law of total expectation:
$$\begin{align}
\operatorname{E}[X] &= \sum_{y=1}^3 \operatorname{E}[X \mid Y = y]\Pr[Y = y] \\
&= n p_1 \Pr[Y = 1] + n p_2 \Pr[Y = 2] + n \Pr[Y = 3] \\
&= 40((2/3)(3/10) + (4/5)(1/2) + (1/5)) \\
&= 32.
\end{align}$$
Then to compute the variance, we calculate $$\operatorname{E}[X^2] = \sum_{y=1}^3 \operatorname{E}[X^2 \mid Y = y]\Pr[Y = y]$$ in the same manner as above, noting that for a binomial random variable, $\operatorname{E}[X^2] = np(1-p) + n^2 p^2$.  Then we finally use $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$

For the sake of curiosity, the PMF for $X$ is shown in the following table:

$$\begin{array}{c|c}
x & \Pr[X = x] \\
\hline
 0 & 2.467579 \times 10^{-20} \\
 1 & 1.974063 \times 10^{-18} \\
 2 & 7.698847 \times 10^{-17} \\
 3 & 1.950374 \times 10^{-15} \\
 4 & 3.608193 \times 10^{-14} \\
 5 & 5.195798 \times 10^{-13} \\
 6 & 6.061765 \times 10^{-12} \\
 7 & 5.888572 \times 10^{-11} \\
 8 & 4.858073 \times 10^{-10} \\
 9 & 3.454632 \times 10^{-9} \\
 10 & 2.141874 \times 10^{-8} \\
 11 & 1.168298 \times 10^{-7} \\
 12 & 5.646798 \times 10^{-7} \\
 13 & 2.432489 \times 10^{-6} \\
 14 & 9.382629 \times 10^{-6} \\
 15 & 0.0000325276 \\
 16 & 0.000101656 \\
 17 & 0.000287071 \\
 18 & 0.000733841 \\
 19 & 0.00170041 \\
 20 & 0.00357503 \\
 21 & 0.00682546 \\
 22 & 0.0118443 \\
 23 & 0.0187105 \\
 24 & 0.0269928 \\
 25 & 0.0357957 \\
 26 & 0.0441756 \\
 27 & 0.0517703 \\
 28 & 0.0591383 \\
 29 & 0.067258 \\
 30 & 0.076186 \\
 31 & 0.0838148 \\
 32 & 0.0861411 \\
 33 & 0.079579 \\
 34 & 0.0639085 \\
 35 & 0.0432651 \\
 36 & 0.0238812 \\
 37 & 0.0102935 \\
 38 & 0.00324528 \\
 39 & 0.000665157 \\
 40 & 0.200066 \\
\end{array}$$

Title: Uniform arrival times at a bus stop


问:
Finding the Function that Gives Investment Value as a Function of t
I have this question in a course review, and I can't wrap my ahead around it conceptually. Can anyone help and explain?

"When the interest on an investment is compounded continuously, the investment grows at a rate that is proportional to the amount in the account, so that if the amount present is P, then
dP/dt = kP
Where P is in dollars, t is in years, and k is a constant. If $180000

is invested (when t = 0) and the amount in the account after 17 years is $421,136
, find the function that gives the value of the investment as a function of t. (Round your value of k to two decimal places.)"

(a) P =

(b) What is the interest rate on this investment? (Round your answer to the nearest whole number.)

答:
The basic formula (from solving the differential equation) is cited per a source (see, for example, discussion here) as:

${A_t = Pe^{rt}}$

where P is the starting investment (better perhaps expressed as, say, ${A_0}$), ${t}$ is the length of the investment period and ${r}$ is the continuous interest rate.

So, we must solve the following based on the inputs:

${421136 = 180000e^{r17}}$

or:

${421136/180000 = e^{17r}}$

Taking the natural logs:

${ln(421136/180000) = 17r}$

Dividing through by 17 implies a value of r = 0.049999, which rounded to the nearest whole percent is 5%.


问:
Let $a,b\in\Bbb{N}.$ Show that there is a minimum natural number of the form $a-b m,$ where $m \in \Bbb{Z}$.
Let $a,b\in\Bbb{N}.$ Show that there is a minimum natural number of the form $a-b m,$ where $m \in \Bbb{Z}$.

If I define the set $T=\{a-bm \in \Bbb N \mid a,b \in \mathbb{N}, m \in\mathbb{Z}\}$, it's clear that $T\neq\emptyset$ because $1\in T$ and $T\subset\mathbb{N}$. Then by Well-ordering theorem $T$ has a minimum.

Is this right or I have to do something more?

答:
Reading your answer and your comment I think you're misunderstanding an important part of the problem: when you say let $a,b\in \mathbb{N}$ you're fixing these values. For the rest of the problem, $a,b$ are fixed. So the definition of $T$, which is dependent on $a,b$ should actually be:

$$ T_{a,b}= \{ a-bm\in \mathbb{N} | m\in \mathbb{Z}\}$$

Note that in your definition of $T$, it seems that $a,b$ could be variable, and I think that was what made you think $1\in T$, when this is false in general. In the expression $a-bm$ everything is fixed except $m$.

With this in mind, try again the same argument you were trying before, having in mind that in general $1$ doesn't have to be in $T$.


问:
How to demonstrate that there would exist a vertex with degree less than 5? (Cross-check working)
Referring to a past question of mine that can be found here:

How to demonstrate that there would exist a vertex with degree less than $ 5$?

This is the work that I have come up with following the guidance kindly provided by other members.

Working:

$$\text{Assume that every vertex of the graph have a degree of at least six.}\\
\text{ Using} ∑_{v∈V}deg(v)=2e \text{ for all graphs,}\\
\text{Since } deg(v)≥6 \text{ for all } v∈V \text{ according to the assumption, we have }\\ 2e=∑_{v∈V}deg(v)≥6n \\ \text{would imply } e \geq 3n \\ \text{ yet, according to Euler's theorem } e \leq 3n -6 \\ \text{therefore implying that there is a contradiction.}\\ \text{Now, it can be concluded that the initial assumption cannot be true, and S has at least one vertex with a degree at most five}$$

I was wondering if anyone would be wiling to check my work and provide any correction/advice on how to improve it?

My main issue is that I do not know if this working would be detailed enough.

I do not know if this is allowed on the site.

If not, just let me know and I will delete the question.

Thank you.

答: This looks good, but you should clarify at the very end that the contradiction implies that the assumption is false (as you say), thus there is at at least one vertex with degree $\leq 5$. Also be careful, this result is for having some vertex having degree $\leq 5$, not $<5$.


问:
What are some upper bounds for the number of factors of a number (a proof would also be nice)?
I found that the bound $\sqrt{3x}$ appears to work, but I just want to see if there are any tighter bounds, and if not, what a proof for this would look like.

答:
I will start off with the simplest type, $$ d(n) \leq \sqrt{3 n} $$ and $$ d(n) \leq 48 \left(\frac{n}{2520}\right)^{1/3} $$ and
$$ d(n) \leq 576 \left(\frac{n}{21621600}\right)^{1/4}. $$
The first one has equality only  at $n = 12,$   second only at $n =2520,$ third only at $n= 21621600.$   Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a 1988 paper by J. L. Nicolas in a book called Ramanujan Revisited.

With equality at $n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $d(n) = 2304,$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)} =  n^{ \left( \frac{1.0660186782977...}{\log \log n} \right) }. $$ Full details of the proof appear in J.-L. Nicolas et G. Robin. Majorations explicites pour le nombre de diviseurs de n, Canad. Math. Bull., 26, 1983, 485--492. The next two appear in the dissertation of Robin, are repeated in the 1988 Nicolas survey article indicated.

With equality at a number $n$ near $6.929 \cdot 10^{40},$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}.   $$ Compare this one with Theorem 317 in Hardy and Wright, attributed to Wigert (1907),
$$ \limsup \frac{\log d(n)  \log \log n}{\log n} = \log 2.  $$

With equality at a number $n$ near $3.309 \cdot 10^{135},$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)}   $$

Just to fill in one blank, the special integers $n$ here are "superior highly composite numbers" using Ramanujan's original recipe for prime factorization, which I like to write, with  $ \delta > 0,$ as
$$ N_\delta = \prod_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor  } $$
The first (largest) $\delta$ that assigns an exponent $k$ to a prime $p$ is
$$ \delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log p}.  $$

So $$  N_{1/2} = 12, \;  N_{1/3} = 2520, \;  N_{1/4} = 21621600, $$

$$  N_{0.23} = 6983776800, \;  N_{0.155} \approx 6.929 \cdot 10^{40}, \;  N_{0.1218} \approx 3.309 \cdot 10^{135}.$$


问:
How to show [a,b) is not open with respect to Euclidean topology
Show that, if $a,b\in\mathbb{R}$ and $a < b$, then $[a,b)$ is not open set.

I am using Topology Without Tears by Sydney Morris

My thoughts on it

Observe  $a\in [a,b)$.

Suppose that there exists $c,d\in\mathbb{R}$ such that $c\leq d$ $a\in(c,d)\subseteq[a,b)$

$a\in(c,d)$ implies $c\leq a < d$ and $c\leq 0.5(a+c) < c < b$.

So $0.5(a+c)\notin [a,b)$ and $0.5(a+c)\in (c,d)$.

So $[a,b)\not\subseteq (c,d)$. Contradiction.

So $[a,b)$ is not open.

Basically I am tweaking Morris’s argument.

It is obviously not open since it does not contain the point $b$

Let $I_{n} = [a,b-1/n) \bigcup_{n\in N} I_n = [a,b)$.

答:
You have the exact right idea!

$[a,b)$ is open if and only if each $x \in [a,b)$ has an open interval $(x-\epsilon, x+\epsilon) \subseteq [a,b)$. So to show that $[a,b)$ isn't open, we must find a point $x$ for which that condition fails.

You're exactly right to think $a$ is a good choice of point. We want to show that no matter how small an $\epsilon$ we choose, $(a-\epsilon, a+\epsilon) \not \subseteq [a,b)$.

Of course, this is easy! Since $a - \frac{\epsilon}{2} \in (a-\epsilon, a+\epsilon)$, and yet $a-\frac{\epsilon}{2} \not \in [a,b)$.

So $(a-\epsilon, a+\epsilon) \not \subseteq [a,b)$. As needed.

As a remark, I'm not sure what you mean by "It is obviously no open since it does not contain the point b". Perhaps this was a typo and you meant it is obviously not closed since it doesn't contain $b$, and this would be true. After all, $b$ is in the closure of $[a,b)$ (do you see why?).

As you saw from the argument above (and the argument you outlined), the relevant point for showing $[a,b)$ is not open is $a$, not $b$.

I hope this helps ^_^


答:
I shall prove it for the case of $(0,1]$ and you shall prove this for $(\alpha, \beta]$ using a similar argument.

Consider the interval $(0,1]$, we claim that such set is not open in $\mathbb{R}$ since
$1$ is not an interior point of $(0,1]$. To see this, notice that $\forall \varepsilon>0$, the neighbourhood $(1-\varepsilon, 1+\varepsilon) \nsubseteq(0,1]$ . For instance :
$$1+\frac{\varepsilon}{2} \in(1-\varepsilon, 1+\varepsilon)$$
However :
$$1+\frac{\varepsilon}{2} \notin(0,1]$$

Title: How to show [a,b) is not open with respect to Euclidean topology

Body:
Show that, if $a,b\in\mathbb{R}$ and $a < b$, then $[a,b)$ is not open set.

I am using Topology Without Tears by Sydney Morris

My thoughts on it

Observe  $a\in [a,b)$.

Suppose that there exists $c,d\in\mathbb{R}$ such that $c\leq d$ $a\in(c,d)\subseteq[a,b)$

$a\in(c,d)$ implies $c\leq a < d$ and $c\leq 0.5(a+c) < c < b$.

So $0.5(a+c)\notin [a,b)$ and $0.5(a+c)\in (c,d)$.

So $[a,b)\not\subseteq (c,d)$. Contradiction.

So $[a,b)$ is not open.

Basically I am tweaking Morris’s argument.

It is obviously not open since it does not contain the point $b$

Let $I_{n} = [a,b-1/n) \bigcup_{n\in N} I_n = [a,b)$.


问:
Deducing $t_1 = t_2 \land t_2 = t_3 \Rightarrow t_1 = t_3$ with special axioms of equality.
In Manin's "Mathematical Logic for Mathematicians," the beginning section is dedicated to syntactic properties of various first-order languages. I've found a strong weakness of mine is in close syntactic manipulation, so I'm struggling with the direct deduction proofs.

To clarify, a deduction of $P$ in $\mathcal{E} \subset L \in \mathcal{L}_1$ (first-order languages), written $\mathcal{E} \vdash P$, is a sequence of formulae in $\mathcal{E}$ where either $P$ is in $\mathcal{E}$, or can be obtained from $\mathcal{E}$ via deductive rules. For example, $(P, P \Rightarrow Q, Q)$ is a deduction of $Q$ from $\{P, P \Rightarrow Q\}$.

Other valid rules that have been derived are simple predicate manipulations, such as $\forall x(P) \Rightarrow P(t)$.

Right now I'm a little stumped on a derivation of $t_1 = t_2 \land t_2 = t_3 \Rightarrow t_t = t_3$ using:

The deductive rules of MP and Universal Generalizaton, and Conjunction,

Basic predicate manipulations,

The rules $x=x$ for all variables $x$, and $x=y \Rightarrow \{P(x,x) \Rightarrow P(x,y)\}$ for atomic predicates $P$.

I believe the proof should go something along the lines of:

List out some premises, such as the second axiom in the third bullet.

Create some tautologies that get me from those true premises to $x=y \land y=z \Rightarrow x=z$.

Generalize then instantiate over $t_1, t_2, t_3$.

But I can't figure out (2). If someone could give a hint, I'd be appreciative. I have solutions with more flexible techniques such as proof by contradiction, but I think the intent of these types of exercises is that they're strict deduction.

答: Hint: the trick is this: if you have $t_1 = t_2$ and $t_2 = t_3$, then with $P(x, y)$ defined to be $t_1 = y$, you can use the rule $x = y \Rightarrow P(x, x) \Rightarrow P(x, y)$ to transform $t_1 = t_2$ into $t_1 = t_3$. (The first argument of $P$ is actually irrelevant here.)


问:
Find integers $x,y$ such that $21x + 13y = 1$
Given  Bézout's identity, how do I find the $x,y$, I already performed the euclidean algorithm. So.

21 = 1 * 13 + 8

13 = 1 * 8 + 5

8 = 1 * 5 + 3

5 = 1 * 3 + 2

3 = 1 * 2 + 1

2 = 1 * 1 + 1

1 = 1 * 1 + 0

I am not sure how to find the x,y though I tried starting from 1 = 2 - 1 and then substituting
2 as 3 - 1 and so on but I wasn't getting anywhere I know I need to somehow get 21*some number + 13 * some number but I am confused on how to start.

答:
(Using the extended Euclidean algorithm, not Fibonacci identities)

$1=3-2=3-(5-3)=2\cdot3-5=2\cdot(8-5)-5=2\cdot8-3\cdot5=2\cdot8-3\cdot(13-8)=5\cdot8-3\cdot13=5\cdot(21-13)-3\cdot13=5\cdot21-8\cdot13$


答: Eliminate $2z^2.$ We have $5-2x^2-y^2=2x^2y^2\Rightarrow y^2=\frac{5-2x^2}{1+2x^2}.$ We take the positive square root since in a neighborhood of $(1,1,1),\ y>0.$  Then we may take $(t, \sqrt{\frac{5-2t^2}{1+2t^2}},t\sqrt{\frac{5-2t^2}{1+2t^2}}).$ This paramterization is valid for $t\in [-\sqrt {5/2},\sqrt {5/2}].$

Title: Find curve at the intersection of two level surfaces.

Body:
Let C be the curve at the intersection of two level surfaces M(x, y, z) = 5 and N(x,y,z) = 0, passing through a point P (1,1,1). Let $M(x, y, z) = 2x^2 + y^2 + 2z^2$ and $N(x, y, z) = xy-z$. Find curve C in the parametric form <x(t), y(t), z(t)>.

I am quite stuck on this one. I could let $z = xy$ and then plugging that into M I'd have $2x^2 + y^2 + 2x^2y^2 = 5$, and I have no idea what would be good ways to parametrize that.

Another way is to let $y=z/x$, in which case I end up with $2x^4 + (2z^2-5)x^2+z^2 = 0$, but again, I'm quite stuck. Any help would be really appreciated!

Or, if it is impossible to find the curve C, the question I'm working on actually asks for the tangent line to curve C at P if x'(0) = 3. If this makes things easier, how exactly is using this tangent more useful?


问:
Prove that $A = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : e^{x_1^2 + x_2^2} < 1 + x_3^2 \} \subset \mathbb{R}^3$ is open and connected
I need to prove that set $$A = \{ (x_1, x_2, x_3) \in \mathbb{R}^3 : e^{x_1^2 + x_2^2} < 1 + x_3^2 \} \subset \mathbb{R}^3$$ is open and connected in $\mathbb{R^3}$

So, I did a change of variable $x = x_1^2 + x_2^2$ and $y = x_3^2$ and then it's easy to show, that set $$\tilde{A} = \{ (x, y) \in \mathbb{R^2}_+: e^x < 1 +y \} \subset \mathbb{R}^2 $$ is open and connected in $\mathbb{R^2}$. Does it follow that the set $A \subset \mathbb{R^3}$ is also open and connected?

答: $A$ is open because $f : \mathbb R^3 \to \mathbb R, f(x_1,x_2,x_3) =  e^{x_1^2 + x_2^2} - x_3^2$ is continuous and $A = f^{-1}((-\infty,1))$. It is not connected because the plane $P = \{ (x_1,x_2,x_3) \in \mathbb R^3 \mid x_3 = 0\}$ has empty intersection with $A$ (note that $e^{x_1^2 + x_2^2} \ge e^0 = 1$). But the points $(0,0,\pm1) \in A$ lie on different sides of $P$.


问:
What is the origin of the (V)BODMAS rule?
When finding out the origin of the (V)BODMAS rule, found out this from google

"BODMAS was introduced by Achilles Reselfelt to help in solving mathematical problem involving operational signs. Whenever an operation is introduced in a sum,BODMAS rule is applicable. The BODMAS rule determines the order of operations."

(V)BODMAS Means:

B = brackets
o = of                              (power  and square root of the terms are carried )
d = division                      (operation carried out from left- right )
m = multiplication             (operation carried out from left- right )
a = addition
s = subtraction
v (vinculum) = A horizontal line drawn over a group of terms in a mathematical expression to indicate that they are to be operated on as a single entity by the preceding or following operator.

First do bracket then powers of next do Division and multiplication then addition and subtraction of the sum. Concerned to bracket,if expression contains all the three types of brackets, the first to be disposed of are round bracket followed by curly brackets and square brackets in turn.

The question is that,in other sites(like quora),many have stated that the origin is not known.When the name "Achilles Reselfelt" is googled, am getting some kind of greek god named "Achilles".(You can try and check).

So can anyone point out to me the origin of this word and if you can, please link me to this person if he exists?

Thanks for your time.If any doubt persists please comment.

答:
6÷2(a+b) =  6÷2a+2b (if parentheses are 'ignored')
So, if I now give you the values of a as 1 and b as 2 ..
Then the answer would be 7!
However, by retaining the parentheses it would be  6÷(2a+2b) and give 1.
For me, the parentheses and correct use thereof, becomes rather important in describing the intent. Especially in more complex equations like I=(LR)/(1-(1+r)^-p) to calculate the installment on a bond.</p>


答:
Use the extended Euclidean algorithm: it relies on the fact that  all remainders in the Euclidean algorithm, satisfy a Bézout's identity, and corresponding coefficients can be computed recursively.

\begin{array}{rrrl}
r_i&x_i&y_i &q_i\\ \hline
21&1&0 \\
13&0&1&1 \\ \hline
8&1&-1&1\\5&-1&2&1 \\
3&2&-3&1 \\
2&-3&5&1 \\  \hline
1&\color{red}5&\color{red}{-8}
\end{array}

Title: Find integers $x,y$ such that $21x + 13y = 1$

Body:
Given  Bézout's identity, how do I find the $x,y$, I already performed the euclidean algorithm. So.

21 = 1 * 13 + 8

13 = 1 * 8 + 5

8 = 1 * 5 + 3

5 = 1 * 3 + 2

3 = 1 * 2 + 1

2 = 1 * 1 + 1

1 = 1 * 1 + 0

I am not sure how to find the x,y though I tried starting from 1 = 2 - 1 and then substituting
2 as 3 - 1 and so on but I wasn't getting anywhere I know I need to somehow get 21*some number + 13 * some number but I am confused on how to start.


问:
Constructing a triangle given an angle, the side opposed to this angle, and the median to the given side.
I came across the following problem in my Euclidean Geometry text:

Construct a triangle having given an angle, the side opposed to this angle, and the median to the given side.

I have tried solving this problem using method of loci, with Locus 1 being the circumcircle of the sought for triangle and Locus 2 being the circle of radius of the given  median. However I don't know how to start this construction.

答:
If the given angle is acute: To construct the circumscribed circle, observe that the inscribed angle is half of the central angle (see here for the terms and a picture). So you can start with your given side, construct an isosceles triangle with double the given angle opposite that side, and the newly constructed vertex of this triangle will be the center of the circumscribed circle you are looking for.

If the given angle is obtuse: Use the supplementary angle of the given angle to construct the isosceles triangle. The  new vertex you construct is the center of the circumscribed triangle in this case.

When completing the construction you need to be mindful of which of these two cases you are in: the side you start with divides the circumcircle into two arcs; the third vertex of your triangle will lie on the same side as the center (i.e. on the longer arc) if the angle was acute, and on the opposite side (i.e. on the shorter arc) if it was obtuse.

If the angle just happened to be a right angle, then the center of the circumscribed circle is the midpoint of the given side (and you will have infinite number of solutions if the median length was half the size of the given side, the third vertex  being anywhere on the circumcircle; or no solution if it was not).


问:
What are the extreme points of this polytope?
Let $n, k$ be positive integers with $n\geq k$.

Let $P_{n,k}$ be the set of vectors $x$ in $[0,1]^n$ for which

$$ \sum_{i=1}^n x_i = k $$

$P_{n,k}$ is defined by linear equations, so it is a polytope. What are its corners (extreme points)?

My guess is that the corners are the "$k$-binary" vectors - the vectors with exactly $k$ ones and $n-k$ zeros. To prove this, it is sufficient to prove (I think) that every vector in $P_{n,k}$ is a convex combination of $k$-binary vectors. Is this correct?

Note: when $k=1$, $P_{n,1}$ is just the standard simplex in $\mathbb{R}^n$. Does it have a name when $k > 1$?

答:
From the course notes A Course in Combinatorial Optimization from Alexander Schrijver (= Lex Schrijver) we find

Theorem 2.2. Let $P = \{ x \space | \space Ax \le b \}$ be a polyhedron in
$\mathbb{R}^{n}$ and let $z \in P$. Then $z$ is a vertex of $P$ if and
only if $\text{rank}(A_{z}) = n$.

Where they also explain the terminology about a vertex and what $A_{z}$ means:
$A_z$ is the submatrix of $A$ consisting of those rows $a_i$ of $A$ for which $a_i z=b_i$.

So the question gives this polyhedron

$\begin{equation*}
Ax =
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
-1 & -1 & \cdots & -1 \\
1 &  &  & \\
  &  1 &  & \\
 &  &  \ddots & \\
 &  &  &  1  \\
-1 &  &  & \\
  &  -1 &  & \\
 &  &  \ddots & \\
 &  &  &  -1  \\
\end{bmatrix} x \le  \begin{bmatrix}
k \\
-k\\
1 \\
1\\
 \vdots  \\
1 \\
0 \\
 0 \\
\vdots \\
 0 \\
\end{bmatrix} = b.
\end{equation*}$

It is not difficult to see that a vertex $z$ with $k$ ones and $n - k$ zeros gives raise to a submatrix $A_{z}$ which has rank $n$ so those points are vertices. It is obvious $a_{1}z = k = b_{1}$ and $a_{2}z = -k = b_{2}$ because $z$ has $k$ ones. So you keep those rows. Now an arbitrary $z_{i}$ equals $1$ or $0$. So look at column $i$ of $A$. If $z_{i} = 1$ you keep the row with $1$ in the diagonal, otherwise if $z_{i} = 0$ you keep the row with the $-1$ in the diagonal because $-1 \cdot 0 = 0 = b_{q}$. So you will end up with a matrix $A_{z}$ with $a_{1}$ and $a_{2}$ on top and a diagonal with $\pm 1$ below. Obviously $\text{rank}(A_{z}) = n$, you have $n$ pivots after all. So we conclude $z$ is a vertex.

Any other $z$ in the polyhedron must have a $z_{i}$ with $0 < z_{i} < 1$ so that also implies there is a $z_{j}$ with $i \ne j$ with $0 < z_{j} < 1$, because $z_{1} + \dots + z_{n} = k$ after all. Again $a_{1}z = k = b_{1}$ and $a_{2}z = -k = b_{2}$. However at column $i$ of $A$ you will multiply with $z_{i} \ne 0, 1$ and therefore you drop the row from the matrix $A$ with $1$ on the diagonal but also the row with $-1$ on the diagonal, because at those rows it must equal $1$ or $0$, those are the values in $b$ at the corresponding positions. This applies to column $j$ as well (note $j \ne i$). So from both diagonals (the one with $1$'s and the one with $-1$'s) you will keep at most $n - 2$ rows with a $\pm 1$. However the first two rows are still there, however they can only supply you with at most 1 pivot so $\text{rank}(A_{z}) \le n - 1$ so those $z$ are not vertices.

It is important to understand at column $i$ from $A$ you can pick at most one row from $A$ corresponding to the $b_{q}$ where $b_{q} = 1$ or $b_{q} = 0$. Try $n = 4$ and $k = 2$ for example if you need some extra help understanding it. Concrete examples are good for that.


问:
If A is an $m\times n$ matrix, show that $||A||_2 \le \sqrt{||A||_1*||A||_\infty}$

If $A$ is an $m\times n$ matrix, show that $$\| A \|_2 \le \sqrt{\|A\|_1 \, \| A \|_\infty}$$

I reduced this to:

$$\rho(A^TA) \le||A||1*||A||_\infty$$

I created a matrix for experimenting:

$$A = \begin{bmatrix}a&b\\c&d\end{bmatrix} \\
A^T = \begin{bmatrix}a&c\\b&d\end{bmatrix} \\
A^tA = \begin{bmatrix}a^2+b^2&ac+bd\\ac+bc&c^2+d^2\end{bmatrix}$$

I tried to calculate the eigenvalues but that got me nowhere.

$$-\lambda^2-\lambda(a^2+b^2+c^2+d^2)+(ad-bc)^2 = 0 \Rightarrow(ad-bc)^2 = \lambda(a^2+b^2+c^2+d^2)+\lambda^2$$

As for $||A||_1$ it can be either $a+c$ or $b+d$, and $||A|||_\infty$ can be either $a+b$ or $c+d$. This gives me 4 possible combinations:

$$c^2+cd+a(c+d) \\
a^2+a(b+c)+bc \\
b^2+a(b+d)+db \\
d^2+d(b+c)+bc$$

I suppose I could do:

$$\lambda = \frac{(a^2+b^2+c^2+d^2)\pm \sqrt{(a^2+b^2+c^2+d^2)^2+4*(ad-bc)^2}}{-2}$$

Since that root will always be bigger than $(a^2+b^2+c^2+d^2)$, $\lambda$ will be either negative (because of the -2), in which case it's "proven" because the right hand side of the initial equation is always positive, or... it will be so big that the denominator will approach zero from the negative side, when the sign is negative, because you'll have $(a^2+b^2+c^2+d^2)-((a^2+b^2+c^2+d^2)+ something)$. In that case, the result will be a very small negative value, divided by -2, which is a very small positive value.

Going from there it's "intuitive" that the equation is correct but not rigorous, I suppose.

Help?

答:
The conventional proof is
$$
\|A\|_2^2=\rho(A^\ast A)\le\|A^\ast A\|_\infty\le\|A^\ast\|_\infty\|A\|_\infty=\|A\|_1\|A\|_\infty,
$$
where the first inequality is due to the fact that the spectral radius is the infimum of all submulticative norms, and the second inequality is due to the submultiplicativity of the induced maximum norm.


问:
How can I see that the rank of $\mathrm{SU}(N)$ is $N-1$?
Is there a way to understand intuitively why this must be the case? Or is there a kind of easy-to-see proof of this result? Thank you in advance.</p>

答: It seems easier to work on the level of the Lie algebras, the tangent to a maximal torus being precisely a maximal abelian subalgebra. The Lie algebra of $SU(n)$ is $n\times n$ traceless skew-Hermitian matrices. If they commute, they are simultaneously diagonalizable (by a version of spectral theorem), so in some basis are diagonal. That means (recalling the traceless condition) that the dimension is at most $n-1$. Of course it is then exactly $n-1$.

OwnerDisplayName: Polarized photon


问:
Finding the limit of a function?
The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$

One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer.

答:
You could notice that
$$\begin{array}{rcl} \lim_{t\to\infty} \frac{t^{k+1}}{e^t} &=& \lim_{t\to\infty} \frac{((k+1)t)^{k+1}}{e^{(k+1)t}}\\ &=& (k+1)^{k+1}\lim_{t\to\infty} \left(\frac t{e^t}\right)^{k+1} \end{array}$$

Now you can use continuity of $x\mapsto x^{k+1}$ and you only need to use L'Hôpital once.

Title: Finding the limit of a function?


问:
Determining the number of solutions of a system of linear inequalities.
I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system
$$
\begin{array}{lcl}
x & \leq & y \\
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has 1 unique solution, namely $x=y=\frac{1}{2}$, while the system

$$
\begin{array}{lcl}
x & \geq & y \\
x + y & \leq & 1 \\
x + y & \geq & 1
\end{array}
$$

has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

答:
Firstly, let us present the inequality system in the unified form.
For example,
\begin{cases}
-x+y\ge 0\\
x-y\ge 0\\
-x-y+1\ge0\\
x+y-1\ge0\\
x+3y-2\ge0,\tag1
\end{cases}
$$L_k(x,y,1)\ge 0, \quad k=1,2,\dots,5.$$

Easily to see, that

$L_1+L_2=0,\;$ i.e. the sum of non-negative values equals to zero. Then should $L_1=L_2=0.\;$ Therefore, we have the equation instead the pair of the inequalities.

Similarly $L_3+L_4=0,\;\Rightarrow\;L_3=L_4=0.$

$L_2+2L_3+L_5 = 0,\;$i.e. the positive linear combination of non-negative values equals to zero. Then should overdefined $L_2=L_3=L_5 =0,$ and really we have two independent equalities instead of three inequalities.

$L_5 =L_1+2L_4,\;$ i.e. inequality $(1.5)$ follows from the pair $(1.1),(1.4)$ and can be eliminated.

Finally, we have the system $\;L_1=L_3=0,\;$ with the rank $2$ and the single solution.

In the common case, the Jordan algorithm can be applied additionally. Finally, this leads to the inequality system for 'independent'  unknowns and the set of the linear functions for the others.

Since the system $(1)$ is presented in the homogenius form, then solutions can exist only if its matrix has rank $2$ or less, and any three expressions $\;L_k\;$are linearly dependent. The similar situation takes place also in the common case.


问:
Four Relations on $\{1,2,3\}$

Define four relations on $\{1, 2, 3\}$.

Both transitive and symmetric.

Transitive and not symmetric.

Symmetric but not transitive.

Neither transitive nor symmetric.

I have:

$R= \{(1,2),~(2,1),~(2,2),~(1,1)\}$ Transitive and Symmetric

$R= \{(1,1),~(2,2),~(1,2)\}$ Transitive and Not Symmetric

$R= \{(1,2),~(2,1),~(2,3),~(3,2)\}$ Not Transitive and Symmetric

$R= \{(1,1),~(1,2),~(2,3)\}$ Not Transitive and Not Symmetric

Am I understanding what the question was asking correctly and are my relations correct?

答:
Overall, as already stated, your work is perfectly fine and correct, and you understand what is necessary. However, I would like to present a bit of a visual intuition for understanding these relations and properties, and how your relations meet the desired criteria.

It is easiest to see these properties by means of a directed graph, such as the below one for your first relation:

In a graph of this form, node $x$ points to node $y$ if and only if $(x,y) \in R$ (or, alternatively stated, $xRy$).

In graphs like these, then, your properties can be seen as so:

Reflexivity: Every node has an arrow pointing to itself.

Symmetry: If an arrow goes from $x$ to $y$, there is an arrow going in the opposite direction: $y$ to $x$. Basically, between two distinct nodes, there is either no "loop" or a closed loop.

Transitivity: If node $x$ points to $y$, and $y$ points to $z$, then $x$ points to $z$. Essentially, for any three distinct nodes, if two pairs are connected, the third pair must also be connected, to form a triangle.

So let's look at your examples:

First, your initial $R$:

You can see this is symmetric because there is the closed loop between $1$ and $2$. Neither point to $3$, so that's okay. (We only require that a given path allow you to backtrack between distinct nodes. Stray nodes are okay.)

Transitivity follows from the "self-pointing" arrows. Walking from $1$ to $2$ to $1$, is the same as staying at $1$ (following the loop pointing to and from $1$). Same for the arrow for $2$.

For your second example $R$:

This is not symmetric, since the loop $1 \to 2$ is not closed in the other way.

This is transitive since the only case to consider is $1 \to 1 \to 2$, and since $1 \to 2$, this follows trivially. Perhaps it's easier to see this if we have multiple $2$ and $1$ nodes to avoid the self-pointing arrows and rely more on the triangle analogy:

Perhaps it's clearly with this one why any two sides of a "triangle" in the directed graph always have a third side completed as well.

For your third example $R$:

In this case, we have symmetry since we only have closed loops and no one-way paths.

We don't have transitivity, however, since the top-left side of either triangle ($1 \to 3$ resulting from $3 \to 2 \to 1$, nor $3 \to 1$ resulting from $1 \to 2 \to 3$) is not present. (Some arguments could also be made through the lack of self-pointing arrows -- for instance, $1 \to 2 \to 1$ should imply $1 \to 1$ exists if $R$ is transitive -- but the triangle bit is easiest to visualize.)

For your final relation:

Here, we have no symmetry since there are some one-way paths instead of closed loops. Similarly, transitivity is not present since the completion of this triangle ($3 \to 1$) is not present.

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

Title: Four Relations on $\{1,2,3\}$


问:
CDF of $W=\max\left(\frac{X}{Y},\frac{Y}{X}\right)$
Given a joint probability density function of $X$ and $Y$

$$f(x,y)=\begin{cases} \frac{1}{a^2} & 0<x,y\\ 0, & \text {otherwise} \end{cases}$$

I want to find the cumulative distribution function of $W=\max\left(\frac {X}{Y},\frac {Y}{X}\right)$.

I can find the cdf of $Z=\frac XY$ (is $F(z) =1-\frac {z}{2}$ correct?), and I know how to find the CDF of max of two independent RVs. Is it correct to assume that the CDF of $W$ is equal to the CDF $X>Y$?

答:
Actually, with $Z:=\frac XY$, if $z\geq1$, $\mathbb{P}(Z\leq z) = \mathbb{P}(X\leq zY) = \frac 1{a^2} \int_0^a\int_0^a \chi_{x \leq zy} dydx = \frac 1{a^2}\int_0^a\min(a,zy)dy = \frac 1{a^2}\big( \int_0^\frac azzydy+\int_\frac az^aady\big) = \frac 1{a^2}\big( \frac{a^2}{2z}+a^2-\frac{a^2}z\big) = 1-\frac{1}{2z}$

If $z\leq1$, $\mathbb{P}(Z\leq z) = \mathbb{P}(X\leq zY) = \frac 1{a^2} \int_0^a\int_0^a \chi_{x \leq zy} dydx = \frac1{a^2}\int_0^azydy = \frac z2$

Now , with $w\geq1$, $\mathbb{P}(W\leq w) = \mathbb{P}(\frac1w\leq Z \leq w) = \mathbb{P}(Z\leq w) - \mathbb{P}(Z < \frac1w) = 1-\frac1{2w}-\frac1{2w} = 1-\frac1w$


问:
Is this a Standard Brownian Motion?
B(t) is a standard Brownian motion

Is $t\in[0,+\infty]$$\rightarrow$$\sqrt{t}B(1) $ a Standard Brownian Motion?

Not very sure how to verify if it is a standard Brownian motion, because I am unsure of what $\sqrt{t}$ does to the process

答: $(\sqrt tB_1)_{t\geq 0}$ is obviously not a Brownian motion. Indeed, $t\mapsto \sqrt tB_1$ is a.s. differentiable on $(0,\infty )$, whereas a Brownian motion is not. Nevertheless, $B_t$ and $\sqrt tB_1$ have the same distribution.

Title: Is this a Standard Brownian Motion?


问:
Tricky Integral $\int_{0}^{\pi/2} \left[\frac{\sec^2 \theta}{2}\cdot \exp(\frac{\sec^2 \theta}{2})\right]^{-1} d\theta$ - Brain Slayer
Given the integral
$$
\int_{0}^{\pi/2}
\left[\frac{\sec^{2}\left(\theta\right)}{2}
\exp\left(\frac{\sec^{2}\left(\theta\right)}{2}\right)\right]^{\,-1} \mathrm{d}\theta
$$
How am supposed to evaluate it within few minutes $?$. Is there a quick trick solve the integral $?$.

Should you wish to see my attempt at doing so, then you should know that my solution is analyzed the author's one, because I could not solve it without seeing clues. So, I would like to get some ideas from; I think that I fail to solve this guy because I could not see something...

答:
Note

$$I= \int_{0}^{\pi/2} \left(\frac12{\sec^2 \theta}\>e^{\frac{\sec^2 \theta}{2}}\right)^{-1} d\theta
= J\left(\frac12\right)
$$
where

$$J(a)=2 \int_{0}^{\pi/2} \frac{e^{-a\sec^2\theta}}{\sec^2\theta}d\theta, \>\>\>\>\>\>\>J’(a)= -2\int_{0}^{\pi/2} e^{-a\sec^2\theta}d\theta, \\
J’’(a)=2\int_{0}^{\pi/2} e^{-a\sec^2\theta}\sec^2\theta d\theta
= 2\int_{0}^{\pi/2} e^{-a(1+\tan^2\theta)}d(\tan\theta)=\frac{\sqrt\pi e^{-a}}{\sqrt a}
$$
and $J(\infty)=J’(\infty)=0$. Then
\begin{align}
I &=J\left(\frac12\right)=-\int_{\frac12}^\infty J’(a)da\\
&=\int_{\frac12}^\infty\left( \int_a^\infty J’’(s)ds \right)\>da
=\int_{\frac12}^\infty\left(\int_{\frac12}^s J’’(s) da\right) \> ds\\
&=\int_{\frac12}^\infty (s-\frac12) J’’(s) ds
=\int_{\frac12}^\infty (s-\frac12)\frac{\sqrt\pi e^{-s}}{\sqrt s}ds\\
& =-\sqrt\pi \int_{\frac12}^\infty d(\sqrt s e^{-s} )
=\sqrt{\frac{\pi}{2e}}
\end{align}


问:
How to advance Physics?
I am currently in high school and I am an avid lover of Maths and Physics, I am currently doing my GCSE's and (without sounding arrogant) I don't really enjoy how basic our Physics is, I used to have this problem with Maths too until I picked up a (text)book on Linear Algebra (Linear Algebra: A Modern Introduction) which I'm enjoying very much. The issue is, I don't really know where to start: I am reading books on Physics like "QED by Richard Feynman" which I am really enjoying but I don't feel it's enough. I want to advance my Physics and I want to start applying Maths to Physics (why I posted on Maths.Exchange), I want to be able to pick up more challenging books on this topic and fully understand them. Yet I also really enjoy the "Physics-Physics side of Physics", I find it very interesting and I am wondering: how can I learn more about Physics (both Mathematically and Theoretically). Through experience I have seemed to grow a higher appreciation for learning stuff through books rather than online, so are there any recommendations of books I can pick up? (I mean as in textbooks, or something which I can study from rather than normal books- those which I am already reading). I know Physics is a very vague name for all the fascinating branches there are, but I would be content with anything - personally, I am really into Astrophysics, but I find everything interesting.

Nevertheless, thank you.

答: For studying astrophysics, I would recommend using the textbook "Relativity: Special, General, and Cosmological" by Wolfgang Rindler to learn more about relativity and "Black Holes, White Dwarfs, and Neutron Stars" by Shapiro and Teukolsky to learn more about the current understanding of some of the most interesting stellar objects. However, a warning: both of these are more graduate level textbooks than for the undergraduate.</p>

OwnerDisplayName: anon

LastEditorDisplayName: anon

Title: How to advance Physics?


答:
The cumulative distribution function is $F_W(w)=\mathbb{P}\left(W < w\right)$, which in this case is $\mathbb{P}\left(\max\left(\frac {X}{Y},\frac {Y}{X}\right) < w \right)$. Equivalently, this is $\mathbb{P}\left(\frac {X}{Y} < w \cap \frac{Y}{X} < w \right) = \mathbb{P}\left(\frac {X}{Y} < w \cap \frac{1}{w} < \frac{X}{Y} \right) = \mathbb{P}\left(\frac{1}{w} < \frac{X}{Y} < w \right)$. For $w < 1$, $F_W(w)$ will be $0$ because either $\frac{X}{Y}$ or $\frac{Y}{X}$ must be greater than $0$.

Then from $F_Z(z)$, $F_W(w) = F_Z(w)-F_Z\left(\frac{1}{w}\right)$. $F_Z(z)$ is given by $$\mathbb{P}\left( \frac{X}{Y} < z \right) = \mathbb{P}\left(X < Yz\right)$$

As an integral, this would be $$\int_0^{a} \int_0^a [x<yz]f(x, y)\ \mathrm dx\ \mathrm dy$$ $F_Z(z)$ is then given by $$\begin{cases}
      0 & z \le 0 \\
      \frac{z}{2} & 0\le z \le 1 \\
      1-\frac{1}{2z} & z \ge 1
   \end{cases}$$

Plugging this in for $F_W(w)$ for $w > 1$ yields $$F_W(w) = \left(1-\frac{1}{2w}\right)-\frac{\frac{1}{w}}{2} = 1-\frac{1}{w}$$

Title: CDF of $W=\max\left(\frac{X}{Y},\frac{Y}{X}\right)$

Body:
Given a joint probability density function of $X$ and $Y$

$$f(x,y)=\begin{cases} \frac{1}{a^2} & 0<x,y\\ 0, & \text {otherwise} \end{cases}$$

I want to find the cumulative distribution function of $W=\max\left(\frac {X}{Y},\frac {Y}{X}\right)$.

I can find the cdf of $Z=\frac XY$ (is $F(z) =1-\frac {z}{2}$ correct?), and I know how to find the CDF of max of two independent RVs. Is it correct to assume that the CDF of $W$ is equal to the CDF $X>Y$?


问:
Differentiability on normed spaces
My Advanced Calculus homework have the following problem: define differentiability on normed spaces $E,F$. Show that if $U \subset E$ is open and $f: U \to F$ is differentiable at $x \in U$ then continuity of $f$ at $x$ is equivalent to continuity of the derivative (linear transformation) $f'(x): E \to F$.

Now, the definition of differentiability on normed spaces should be the same as the one on euclidian spaces: $f: U \subset E \to F$ is differentiable at $x \in U$ if there exists a linear transformation $T: E \to F$ such that:
\begin{eqnarray}
f(x+h) = f(x) + T(h) + r(h) \tag{1}\label{1}
\end{eqnarray}
where:
$$\lim_{h\to 0}\frac{r(h)}{||h||} = 0,$$ provided $x+h \in U$.
Now, write $\rho(h) = r(h)/||h||$ for every $h \neq 0$ and set $\rho(0) := 0$. Then, (\ref{1}) becomes:
\begin{eqnarray}
f(x+h) = f(x) + f'(x)(h) + \rho(h)||h|| \tag{2}\label{2}
\end{eqnarray}
where $\lim_{h\to 0}\rho(h)=0$ (I wrote $f'(x)$ instead of $T$). Thus, $\lim_{h\to 0}\rho(h)||h|| = 0$ and we have:
\begin{eqnarray}
f(x+h)-f(x) = f'(x)(h) \tag{3}\label{3}
\end{eqnarray}
Now, if we write $x+h = y$, (\ref{3}) becomes
\begin{eqnarray}
f(y) - f(x) = f'(x)(y)-f'(x)(x) \Rightarrow \lim_{y\to x}f(y) - f(x) = 0 \iff \lim_{y\to x}f'(x)(y)-f'(x)(x) = 0 \tag{4}\label{4}
\end{eqnarray}

Question: Is (\ref{4}) sufficient to prove the second statement of the problem, i.e. that continuity of $f$ at $x$ is equivalent to continuity of the derivative $f'(x): E \to F$?

答: $(3)$ is an equation containing the variables $x$ and $h$. If it were true for all $h$, then subtraction of $(3)$ from the correct equation $(2)$ would give $r(h)= \rho(h)/\lVert h \rVert = 0$ for all $h$. This is true if and only if $f(x + h) = f(x) + T(h)$ which is a very special case ($f$ is an affine map). In other words, $(3)$ and $(4)$ are in general false. This does not matter, you do not need $(4)$. Since $r(h) \to 0$ as $h \to 0$, we see that $\lim_{h \to 0} f(x+h) = f(x)$ if and only if $\lim_{h \to 0} T(h) = 0$. But $\lim_{h \to 0} f(x+h) = f(x)$  means that $f$ is continuous at $x$ and $\lim_{h \to 0} T(h) = 0$ means that $T$ is continuous at $0$ which is equivalent to the continuity of $T$ (since $T$ is linear).

Title: Differentiability on normed spaces


问:
Cardinality restrictions of topological spaces
For every natural number $n \geq 2$, is there a set $X$ and a topology $T$ on $X$ such that $T$ has cardinality $n$? Or are there forbidden cardinalities? What exactly is the subset of $\mathbb{N}$ which contain precisely those natural numbers which can be cardinalities of finite topologies?

答:
For $n\in\Bbb Z^+$ let $U_n=\{1,2,\ldots,n\}$, and let $$\tau_n=\{\varnothing\}\cup\{U_k:k\le n\}\;$$ then $\tau_n$ is a $T_0$ topology on $U_n$ with $n+1$ members.

A finite $T_1$ topology on a set $X$, however, must have cardinality a power of $2$: each singleton is closed, so $X$ must be finite, and then every subset of $X$ is closed, so every subset of $X$ is open.


问:
Surface bundle with flat connection and monodromy representation
I saw the following statement:

An $S_g$-bundle $p: E\to B$ admits a flat connection if and only if the
monodromy representation $\rho:\pi_1(B)\to\text{Mod}(S_g)$ lifts to $\rho':\pi_1(B)\to\text{Diff}^+(S_g)$.

I have no idea why it is true, is there a reference or can you give proof?

答:
Suppose that you have smooth connected manifolds $B$, $F$ and a representation $\phi: \pi_1(B)\to Diff(F)$. This defines the associated bundle
$$
E=B\times_\phi F
$$
over $B$ together with a flat connection $\nabla$ on $E$. Namely, start with the product $\tilde E=\tilde{B}\times F$, where $\tilde{B}$ is the universal covering space of $B$. The fundamental group $\pi=\pi(B)$ acts on the product space diagonally: by covering transformations on $\tilde B$ and via the representation $\phi$ on $F$. This product action is free and properly discontinuous, hence, you form the quotient $E= \tilde E/\pi$. The projection $\tilde E\to \tilde B$ descends to a bundle map $p: E\to B$. At the same time, the product foliation $\{x\}\times F, x\in \tilde B$ on $\tilde E$ descends to a foliation on $E$ which is transversal to fibers of $p$. The tangent distribution of that foliation is your flat connection $\nabla$ on $p: E\to B$.

Conversely, given a flat bundle $p: E\to B$ (i.e. a bundle with a flat connection), you have its holonomy homomorphism $\pi_1(B)\to Diff(F)$.

My favorite reference for this staff is

Steenrod, Norman, The topology of fibre bundles., Princeton Landmarks in Mathematics. Princeton, NJ: Princeton University Press. viii, 229 p. (1999). ZBL0942.55002.

For surface bundles, check also Morita's book

Morita, Shigeyuki, Geometry of characteristic classes. Transl. from the Japanese by the author, Translations of Mathematical Monographs. Iwanami Series in Modern Mathematics. 199. Providence, RI: American Mathematical Society (AMS). xiii, 185 p. (2001). ZBL0976.57026.


问:
Is there a universal math proof verification program than can verify every finite purported math proof?
For some axiom systems, we can verify candidate proofs. For example Mizar.

Is there a universal language we can write proofs in, so that the axioms are included in the proof, along with every single step in complete detail, so that some universal verifier (Turing machine) can decide its correctness? Assume proofs are finite in length.

(I believe this is similar to asking "Can every math proof be written in Metamath?")

Basically instead of hard-coding the axiom system into the verifier, I want the axioms as part of the input to the verifier, for all possible axiom systems.

Turing machines have Universal Turing machines that compute all other turing machines.

Turing machines and math proofs seem to be closely related (See Curry-Howard correspondence and the time- and space- hierarchies from Computer Science showing up in higher-order-logic via Descriptive Complexity).

So there may be some analogous universal math axiom system that can describe all other systems.

答:
There's some discussion in the comments about what exactly the question means, so I'll try to give an answer that covers a variety of possible meanings.  Here's some helpful terminology.  A proof system is:

complete if every statement that is true under all interpretations can be proved in the system.

sound if every statement that can be proved in the system is true under all interpretations.

effective if there is an algorithm that can correctly detect whether a given string of symbols is a correct proof.

Now whether or not there is proof system like you're asking for depends on exactly what you want, and which logical framework you want to work in.  For example, first-order logic does have a complete, sound, effective proof system (in fact, there are several).  On the other hand, it follows from the incompleteness theorems that second-order logic (with full semantics) does not have a complete, sound, effective proof system.

I feel like I should note that getting a sound and effective proof system is trivial - just declare that no string of symbols is a proof. Completeness, or at least something like it, is thus highly desirable.

I also think I should point out that despite having a complete, sound, effective proof system, there is not an algorithm for determining whether or not a given statement is a theorem from some axioms, but there is an algorithm which, given a statement and some reasonably-presented list of axioms, will return TRUE if the statement can be proved from the axioms, and otherwise will run forever.  The terminology here is that first-order logic is semidecidable.


答:
You can also use the series representation $e^t = \sum_{n=0}^\infty \frac{t^n}{n!}$ to observe that for any given $k \in \mathbb{N}$ we have $e^t > t^{k+2}$ for $t > 1$. Hence
$$
\frac{t^{k+1}}{e^t} < \frac{t^{k+1}}{t^{k+2}} = \frac{1}{t} \xrightarrow{t \to \infty} 0.
$$</p>

Title: Finding the limit of a function?

Body:
The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$

One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer.


答:
If you fix on some logic, e.g., first-order logic and as is usual require the axiom system to be a recursive set, then you can design a machine that takes as input a purported proof $\Pi$ and a code $\cal A$ for a function that decides whether a formula is an axiom and tests whether $\Pi$ is a valid proof from the axioms in $\cal A$. Generalising, you could come up with a fancier machine that took as input a purported proof in any reasonable deductive sstem and a function that decomposes the proof into steps and decides whether each step is valid. Systems along these lines have been implemented and are called logical frameworks. Metamath is one such system. However, I don't think any such system enjoys any very interesting universal property: there is no useful sense in which they describe the systems that they model.

However, all this depends on our logic being tractable to computation. In the logic whose language is the language of natural number arithmetic and whose axioms are all true statements about the natural numbers, there is no possibility of providing a verifier.

Title: Is there a universal math proof verification program than can verify every finite purported math proof?

Body:
For some axiom systems, we can verify candidate proofs. For example Mizar.

Is there a universal language we can write proofs in, so that the axioms are included in the proof, along with every single step in complete detail, so that some universal verifier (Turing machine) can decide its correctness? Assume proofs are finite in length.

(I believe this is similar to asking "Can every math proof be written in Metamath?")

Basically instead of hard-coding the axiom system into the verifier, I want the axioms as part of the input to the verifier, for all possible axiom systems.

Turing machines have Universal Turing machines that compute all other turing machines.

Turing machines and math proofs seem to be closely related (See Curry-Howard correspondence and the time- and space- hierarchies from Computer Science showing up in higher-order-logic via Descriptive Complexity).

So there may be some analogous universal math axiom system that can describe all other systems.


问:
Principal value of complex exponential
Here you can see that Cauchy's principle value ($PV$) of the complex exponential is zero, i.e.,

$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = 0
\tag1 $$

I'm trying to prove this a follows:

$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = \lim_{T\rightarrow +\infty}\int_{-T}^{+T} e^{iax}dx = \frac{1}{a} \lim_{T\rightarrow +\infty} \left[ \sin(T) - \sin(-T)\right] \tag2
$$

In principle, one would write $\sin(T) - \sin(-T) = 2\sin(T)$ and would realize that the limit does not exist so the only thing I can try to prove the result in Eq. (1) is to work out both sines separatedly and assume that

$$
\lim_{T\rightarrow +\infty} \sin(T) = \lim_{T\rightarrow +\infty}\sin(-T) \tag3
$$

so finally you arrive to Eq. (1). Nevertheless, this is not very convincing from my point of view because  RHS limit is undetermined as well as LHS, so you should not assume Eq. (3) is right. Then, what's the right approach?

答:
Note that

$$\int_{-\infty}^{\infty} e^{iax} \, dx=\delta(a)$$

which vanishes when $a\ne 0$.

Please also refer to post 1 and post 2 for your further interests.

In Wolfram or Mathematica, may the machine assume the integrand is evaluable:

Title: Principal value of complex exponential


问:
Surface integral over an ellipsoid
I am trying to determine the outward flux integral of the vector field $$F(x,y,z) = \frac {(x-y)i + (x+y)j + zk}{(x^2+y^2+4z^2)^{3/2}}$$
across the ellipsoid $x^2 + y^2 + 4z^2=4$.

Could I have a hint for this? I can't use the divergence theorem because of the singularity, so I am trying to evaluate it directly. I have replaced the denominator with 8 but I'm not sure what to do next.

答:
Parametrizing is unnecessary after a simplification. The integrand of the surface integral simplifies to

$$F(x,y,z) = \frac{(x-y,x+y,z)}{8}$$

by plugging in $x^2+y^2+4z^2=4$. This is a vector field divergence theorem applies to:

$$\nabla\cdot F = \frac{3}{8}$$

which means the integral evaluates to $\frac{3}{8}\operatorname{vol}(x^2+y^2+4z^2\leq 4) = 2\pi$

Title: Surface integral over an ellipsoid


问:
$B\otimes_A A/\mathfrak{m}$ a domain with $B$ not a domain
I want to find an example of where $A$ is a dedekind domain, $B$ is a flat algebra over $A$, $m$ is some maximal ideal of $A$ with $B\otimes_A A/\mathfrak{m}$ a domain but $B$ is not a domain.

Specifically could you hint me towards how you would come up with one of these. This is my thought process so far:
We want $B$ to have some zero divisors, $ab=0$. But since we want $B\otimes_A A/\mathfrak{m}$ a domain we do not want $(a\otimes 1)$ to be non zero since then $(a\otimes 1)(b\otimes 1)=0$ and then $B\otimes_A A/\mathfrak{m}$ is not a domain.
But how to try an construct a ring so that $(a\otimes 1)$ is zero? My best guess would be to have all the zero divisors in $B$ to be elements in $\mathfrak{m}$. But I don't see how we construct this?

答: Consider $A=\Bbb Z$, $\mathfrak{m}=(p)$ for some prime number and $B=\Bbb Z \times\Bbb Q$. This is a flat algebra. We have $B \otimes_A A/\mathfrak{m} \cong \Bbb Z/(p)$ which is a field, but $B$ is not a domain.


问:
Intuition behind spectrum of an operator
I am trying to understand better the concept of the spectrum of an operator, and it gives me some troubles, I would like to know if there is an intuitive idea behind this, for example, a part of the spectrum are the eigenvalues (I think this is the point spectrum), and they have a very strong "geometric" interpretation.

But, what happens with the elements in the continuous spectrum or the residual spectrum? And we can also define something like the spectral radius ($r_\sigma (T)=\sup \{|\lambda|\colon \lambda \in \sigma (T) \}$), all this stuff can be somehow interpreted? Or how should I approach them to understand them in a better way?

Also I am not sure how to "think" the spectrum when I see linear transformations related to matrices. If you can help me to see this, it would be very helpful.

Thanks.

答:
This is not a complete answer, but since you are looking for intuition, it may be helpful.

Some background: Historically, the spectrum originated in quantum mechanics, and in particular in the energy states of atoms.  In this context, there is an operator $H$ (typically on a space like $L^2(\mathbb{R})$) whose spectrum describes the possible states of the atom.  There are two kinds of states: "bound states" are those where all the particles of the atom stay close together (they are "bound"), whereas "scattering states" are those where the particles fly apart.  The only states which can stay constant in time are bound states; the scattering states necessarily entail changes to the atom in time.

Mathematically, the bound states correspond to eigenvalues of $H$ and the scattering states correspond to the rest of the spectrum.  The bound states have corresponding eigenvectors in $L^2$, which typically look something like $e^{-kx^2}$, and the atom is allowed to be in such a state.  On the other hand, the scattering states have pseudo-eigenvectors which are not in $L^2$ and typically look something like $e^{ikx}$.  Though the scattering states do not live in $L^2$, it is possible to form combinations of them which do live in $L^2$, e.g. by a Fourier transform.  So the atom cannot live in a single scattering state, but it can live in a combination of them, which can be arbitrarily close (in a suitable sense) to a pure scattering state.

Here are some concrete examples:

Consider the "multiply by x" operator $\hat{x}$ on $L^2(\mathbb{R})$.  For any $f\in L^2(\mathbb{R})$ and any $x_0\in\mathbb{R}$ we have $\hat{x}f (x_0) = x_0 f(x_0)$.  This operator has no eigenvalues, but its spectrum is the entire real line.  Intuitively, a delta function $\delta(x-x_0)$ is a pseudo-eigenvector to $\hat{x}$ with pseudo-eigenvalue $x_0$, and morally we have $\hat{x} \delta(x-x_0) = x_0 \delta(x-x_0)$.

Similarly, the Laplacian $\nabla^2$ has pseudo-eigenvectors $e^{ikx}$ for any $k\in \mathbb{R}$, corresponding to pseudo-eigenvalues $-k^2$, because $\nabla^2 e^{ikx} = -k^2 e^{ikx}$.  So the entire non-positive real line is in the spectrum of $\nabla^2$.

In both of these examples, the problem is that the "eigenvectors" are not in the domain of the operator--they live in some generalized space (either distributions in the first case, or non-square-integrable functions in the second).  I'm not sure if this is always the case, but it is certainly typical that the part of the spectrum of an operator which is not eigenvalues corresponds to generalized eigenvectors which live in some larger space.

So summarily, my intuition for the part of the spectrum which is not proper eigenvalues is that it corresponds to some pseudo-eigenvector which does not properly live in the domain of your operator, but is still somehow "there" as some kind of complicated limiting object, e.g. a distribution or non-normalizable function.


问:
Reverse implication of Hilbert's Basis Theorem
I have been looking over multiple proof regarding Hilbert's basis theorem. Namely, if $R$ is a commutative noetherian ring, then $R[x]$ is also notherian.

I understand, for the most part, the construction of this proof. My question is, can Hilbert's basis theorem be written as an if and only if statement? In other words, is it true that if $R[x]$ is noetherian then $R$ is also noetherian? If it can be, then I have an idea on how to prove it, but I would like some feedback.

Proof:
Let $R[x]$ be noetherian. Let $$\phi: R[x] \rightarrow R$$ be the homomorphism of rings where $$\phi(f) = f(0).$$
It follows then that $\phi(R) = R$. In other words, $\phi$ is onto. Now, it is not hard to check that the image of a noetherian ring is noetherian. Thus $R$ is noetherian. $\square$

My question is, is my proof is correct? Furthermore, if it is correct, how come a lot of standard algebra texts don't state this theorem as an if and only if?

答: Your proof is correct! I think the reason algebra texts don't state Hilbert's Basis Theorem as a biconditional is that this direction is much more obvious (and less often useful). I also have one nitpick: you should say "the homomorphism" instead of "a homomorphism" because you are clearly singling out a unique map.</p>


问:
Correlation between three variables question
I was asked this question regarding correlation recently, and although it seems intuitive, I still haven't worked out the answer satisfactorily. I hope you can help me out with this seemingly simple question.

Suppose I have three random variables $A$, $B$, $C$. Is it possible to have these three relationships satisfied?
$$
\mathrm{corr}[A,B] = 0.9
$$
$$
\mathrm{corr}[B,C] = 0.8
$$
$$
\mathrm{corr}[A,C] = 0.1
$$
My intuition is that it is not possible, although I can't see right now how I can prove this conclusively.

答:
As a follow up to one of the other answers, I'd like to finish the solution to make it clear. I find this approach very nice (Borat accent).

In order to generate the correlation matrix, you can leave one of the correlations as $\rho$, I'm leaving $\rho_{AC}$ here.
\begin{bmatrix}
1 & 0.9 & \rho\\
0.9 & 1 & 0.8\\
\rho & 0.8 & 1
\end{bmatrix}

Now, if you calculate the determinant, you'll get the bound for $\rho$. Determinant calculation would result in the following equation (note that the matrix above is a correlation matrix and has to be at least semidefinite):

$$
-0.45 + 0.72\rho + 0.72\rho - \rho^2 \ge 0
$$

making it nicer, it will look like this:
$$
\rho^2 - 1.44\rho + 0.45 \le 0
$$

which is correct if $ 0.458 \le \rho \le 0.981 $. This means that the correlation between $A$ and $C$ should be in this range, which isn't.


问:
Statistics, Poisson distribution + hypotheses
Hey, I find this problem quite difficult to handle,I would be grateful to anyone who could at least lead on the path of solving it.

The random variable X has a Poisson distribution with mean µ. The value of µ is
known to be either 1 or 2 so the following hypotheses are set up.

$H_0$ : µ = 1;   $H_1$ : µ = 2

A random sample $x_1$ , $x_2$ , ..., $x_{10}$ of 10 observations is taken from    the distribution of X and the following critical region is defined.

\begin{eqnarray*}
\sum_{i\in 1}^{10}x_i \geqslant 15
\end{eqnarray*}

Determine the probability of a Type 1 and Type 2 errors.

答:
1-poissoncdf(10, 14)= 0.08346 is the correct answer

But only one issue, Why is it P(Y≥15)=1−P(Y<14), isn't this a discrete distribution? Why are we using cdf?


问:
How can I show that if 2 power series are conditionally convergent then it can happen that $\sum_{k=1}^{\infty}c_k$ diverges?
Assume that $\sum_{k=1}^{\infty}a_k =A \in R$ and $\sum_{k=1}^{\infty}b_k =B \in R$ with partial sums $r_n =\sum_{k=1}^{n}a_k$ and $s_n=\sum_{k=1}^{n}b_k.$ If one wants to sum the double series $\sum_{k=1}^{\infty}\sum_{l=1}^{\infty}a_kb_l$ the sum may depend on the order of the summation. If we sum over square we obtain

$\lim_{n \to \infty}\sum_{k=1}^{\infty}\sum_{l=1}^{\infty}a_kb_l = \lim_{n \to \infty} r_ns_n =AB$

If the series $\sum_{k=1}^{\infty}a_k$ and $\sum_{k=1}^{\infty}b_k$ are absolutely convergent, then all orders of summation five the same result, i.e. AB. If we sum over the triangles we obtain

$\lim_{n \to \infty}\sum_{k=1}^{\infty}c_k \quad$ where $c_k =\sum_{l=1}^{k-1}a_lb_{k-l}.$

How can I show that if the series $\sum_{k=1}^{\infty}a_k$ and $\sum_{k=1}^{\infty}b_k$ are conditionally convergent then it can happen that $\sum_{k=1}^{\infty}c_k$ diverges? Allegedly I can use the following hint: consider $a_k=b_k=(-1)^{k+1}/\sqrt{k}.$

答: Hint: $\sqrt{l}\sqrt{k-l} = \sqrt{kl - l^2} \leqslant k$ and $|c_k| \not\to 0$ as $k \to \infty$.


问:
Calculate $\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$, whose indefinite integral is of non-elementary function

Calculate $$\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$$

I know an approach is using the change of variables $t=-u$, so we have the value for the integral $\frac{1}{3}$. How can I calculate that integral using other approach?

I thought about this  problem when observing this indefinite integral indefinite integral.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{\left.{1 \over \expo{t} + 1}\,
\right\vert_{\,t\ \not=\ 0} = \Theta\pars{-t} +
{\on{sgn}\pars{t} \over \expo{\verts{t}} + 1}}$ and equal to $\ds{1 \over 2}$ when $\ds{t = 0}$. $\ds{\Theta}$ is the Heaviside Theta or/and Step Function.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{-1}^{1}{t^{2} \over 1 + \expo{t}}\,\dd t} =
\int_{-1}^{0}t^{2}\,\dd t = \bbx{1 \over 3} \\ &
\end{align}


问:
use central limit theorem to find sample size
a weekly wage is a random variable X, $σ^2$ = 64, $µ$ = 75, p ≥ 0.9

use CLT to find the minimum sample size such that with probability at least 0.9, the average wage of this group is between 70 and 80.

so far I have:

P(70 ≤ X ≤ 80) ≥ 0.9 and I worked with this until I got to

P(-5 ≤ (X-µ) ≤ 5) ≥ 0.9

is this correct so far? if so where do I go from here?

edit

following on, I get

P(X ≤ |40√n|) ≥ 0.9

but I am unsure on how to solve this

答:
Let $X$ be the sample mean. Using CLT, we have that $X$ is approximately a Gaussian random variable with mean equal to $\mu = 75$ and variance equal to $\sigma^2/n = 64/n$. We need to find minimum $n$ such that $\Pr(70 \leq X \leq 80) \geq 0.9$. Thus, we have,

\begin{align*}
\Pr(70 \leq X \leq 80) & \geq 0.9 \\
\implies \Pr(-5 \leq X - 75 \leq 5) & \geq 0.9 \\
\implies \Pr\left(-\frac{5\sqrt{n}}{8} \leq \frac{X - 75}{8/\sqrt{n}} \leq \frac{5\sqrt{n}}{8}\right) & \geq 0.9 \\
\end{align*}

Since $X \sim \mathcal{N}(75, 8/\sqrt{n})$, $(X - 75)/(8/\sqrt{n}) \sim \mathcal{N}(0,1)$, a standard normal random variable, $Z$. Hence, we need to find $n$ such that
\begin{align*}
\Pr(-0.625 \sqrt{n} \leq Z \leq 0.625 \sqrt{n}) & \geq 0.9 \\
\implies 1 - 2 \Phi(-0.625 \sqrt{n}) & \geq 0.9 \\
\implies  \Phi(-0.625 \sqrt{n}) & \leq 0.05 \\
\implies n & \geq \left(\frac{\Phi^{-1}(0.05)}{0.625} \right)^2.
\end{align*}

Using standard tables, we obtain $n \geq 6.9262$.


问:
Is the ascending union of contractible spaces contractible
Let $\{Y_i\}_{i \in \mathbb{N}}$ be a collection of subspaces of $X$ such that each $Y_i$ is contractible and $Y_{i} \subseteq Y_{i+1}$. Is $\bigcup_{i\in \mathbb{N}}Y_i \subseteq X$ also contractible?</p>

答:
No. Consider the half-open arcs
$$Y_n=\{e^{i\theta}\mid \theta\in[0,2\pi-\tfrac{1}{n})\}$$
on the circle $X$.</p>


问:
Prove $s<\infty\to S<\infty \to I<\infty$

Let $p\geq 1$ and $f\geq 0$ Setting

$I:=\int_X|f|^pd\mu$

$s=\sum_{n\in\mathbb{Z}}2^n\mu(\{x:2^n<f(x)\leq2^{n+1}\})^{\frac{1}{p}}$

$S=\sum_{n\in\mathbb{Z}}2^n\mu(\{x:2^n<f(x)\})^{\frac{1}{p}}$

Prove $s<\infty\to S<\infty \to I<\infty$

I've tried to set $A_n:=\{x:2^n<f(x)\leq2^{n+1}\}$,$B_n:=\{x:2^n<f(x)\}$ to prove that $S<\infty$ and from chebyshev inequality we know $2^n\mu(B_n)=2^n\mu(\{x\in B_N:|f|>2^n\})^{1/p}\leq(\int_{B_n}|f|^p)^{\frac{1}{p}}$

And got stuck here:

$\sum_{n\in\mathbb{Z}}2^n\mu(B_n)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(\int_{B_n}|f|^p)^{\frac{1}{p}}=\sum_{n\in\mathbb{Z}}(\int_{\uplus_{k=n}^{\infty}A_k}|f|^p)^{\frac{1}{p}}=\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}(\int_{A_k}|f|^p)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}(\int_{A_k}(2^{k+1})^p)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}2^{k+1}\mu(A_k)^{\frac{1}{p}}$

For if we asumme that $S<\infty$ and trying to prove that  $I<\infty$ I've tried :

$\int_X|f|^{\frac{1}{p}}=\int_{\uplus A_n}|f|^{\frac{1}{p}}\underset{(1)}{=}\sum_{n\in\mathbb{Z}}\int_{A_n}|f|^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(2^{n+1})^{\frac{1}{p}}\mu(A_n)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(2^{n+1})^{\frac{1}{p}}\mu(B_n)^{\frac{1}{p}}$

And I'm not sure about the if (1) is correct aswell.

答:
To show that $s<\infty$ implies $S<\infty$ for $p>1$ set
$a_n:=\mu\big(\{x:2^n<f(x)\leq2^{n+1}\}\big)$,
$b_n:=\mu\big(\{x:2^n<f(x)\}\big)$, $q=p/(1-p)$ and
$M_n=\sup\{a_n\bigm| m\geq n\}$. Then,

$$b_n=\sum_{m\geq n}^\infty a_n
%
=\sum_{m\geq n}^\infty a_n^{1/p}  a_n^{1/q}
%
\leq M_n^{1/q}\sum_{m\geq n}^\infty a_n^{1/p}
$$

so that

\begin{align*}
  \sum_{n=-k}^k2^nb_n^{1/p}& =
%
\sum_{n=-k}^k2^nM_n^{1/pq}\Big\{\sum_{m\geq n}^\infty a_m^{1/p}\Big\}^{1/p}
%
  \leq\sum_{n=-k}^k\Big\{\sum_{m\geq n}^\infty 2^n a_m^{1/p}\Big\}^{1/q}
  \Big\{\sum_{m\geq n}^\infty 2^n a_m^{1/p}\Big\}^{1/p} \\
%
& =\sum_{n=-k}^k\sum_{m\geq n}^\infty 2^n a_m^{1/p}
%
 <\sum_{m=-\infty}^\infty 2^{m+1} a_m^{1/p}=2s.
%
\end{align*}
Thus $s<\infty$ implies $S<\infty$ for $p>1$. If $p=1$ a similar
calculations shows the implication.

To see that $S<\infty$ implies $\int_X |f|^p\mathrm{d}\mu<\infty$, let
$B_n:=\{x:2^n<f(x)\}$ and let $\mathcal{X}_{B_n}$ denote the
characteristic function of $B_n$. Note that

\begin{align*}
  \int_{B_{-k}\setminus B_k }|f|^p\mathrm{d}\mu & =
%
\sum_{m=-k}^{k-1}\int_{B_{m}\setminus B_{m+1} }|f|^p\mathrm{d}\mu =
%
\sum_{m=-k}^{k-1}\int_{X}\big(\mathcal{X}_{B_{m}}-
%
\mathcal{X}_{B_{m+1}}\big)|f|^p\mathrm{d}\mu\\
%
  & \leq \sum_{m=-k}^{k-1}2^{p(m+1)}\int_{X}\big(\mathcal{X}_{B_{m}}-
%
\mathcal{X}_{B_{m+1}}\big)\mathrm{d}\mu
%
   \leq \sum_{m=-k}^{k-1}2^{p(m+1)}\int_{X}\mathcal{X}_{B_{m}}
%
\mathrm{d}\mu\\ & =2^p\sum_{m=-k}^{k-1}\Big[2^{m}b_m^{1/p}\Big]^p,
%
%
\end{align*}

which converges by the comparison test if $S<\infty$.


问:
A Diophantine Equation Related to the Markoff Numbers
Consider the following Diophantine equation.
$$5(p^2+q^2+r^2+s^2+t^2)^2-7(p^4+q^4+r^4+s^4+t^4)=90pqrst$$
This equation was discussed at The Diophantine equation $5(p^2+q^2+r^2+s^2+t^2)^2 - 7(p^4+q^4+r^4+s^4+t^4) = 90pqrst$

The author was interested about the source, not the solution. I found the problem quite interesting. I wrote a code which generated all the solutions $1 \leq p \leq q \leq r \leq s \leq t \leq 500$.

```<code>p q r s t
1 1 1 1 1
1 1 1 1 2
1 1 1 2 2
1 1 1 2 4
1 1 2 3 3
1 1 2 3 6
1 2 4 7 7
1 2 7 7 17
1 2 34 47 127
1 17 42 63 124
2 7 19 47 59
2 13 39 97 99
3 3 6 42 87
3 3 27 87 189
6 21 138 417 483
11 11 11 154 286
</code>```

According to the magazine mentioned in the link, this equation has a relation to the Markov numbers (https://en.wikipedia.org/wiki/Markov_number).

I wonder if there are similar tree like structure for the solutions of this problem.

答:
The answer depends on the precise meaning of your
"similar tree like structure". Consider a very general situation.
Fix a symmetric polynomial $\,P(x_1,\dots,x_n).\,$ Define
$$ T:=\{(x_1,\dots,x_n)\in\mathbb{Z}^n:P(x_1,\dots,x_n)=0\}.\tag{1}$$
Define a graph $\,G\,$ where the vertices $\,V\,$ are equivalence
classes of $\,T\,$ where two elements of $\,T\,$ are regarded as
equivalent if they are permutations of each other. Define two
vertices as adjacent if they are equal up to permutation except for
at most one component.
Because $\,P\,$ is a polynomial, there can be at most $\,n-1\,$
vertices adjacent to a given vertex. If there are no cycles, then
the graph is a tree. In the case of the Markov numbers,
$\,P(x,y,z) = x^2+y^2+z^2-3xyz.\,$
In the case in your question,
$$ P(p,q,r,s,t)=5(p^2+q^2+r^2+s^2+t^2)^2-7(p^4+q^4+r^4+s^4+t^4)-90pqrst. \tag{2}$$
It is probably true that there are no cycles but I have no proof
for that, even if the numbers are restricted to positive integers.
Thus, the solutions form a tree like structure.


问:
Prove functions in L1 space are Uniformly Continuous
Let $f \in L^1(R)$ and define $g(x)=\int_{(-\infty,x)} f d\mu$ where $\mu$ is Lebesgue measurable. Show that $g$ is uniformly continuous.

答:
$f \in L^{1}(\mathbb{R})$ implies $|f(t)| \leq M$ a.e x.

Let $E = \{ t \in \mathbb{R} : |f(t)| \leq M\}$, then $\mu (\mathbb{R}\backslash E) = 0$.

Suppose $\epsilon > 0$, let $\delta < \frac{\epsilon}{M}$.

Then $|g(x) - g(y)| \leq \int_{ \mathbb{R}} |f(t)| \cdot \chi_{[x, y]}(t) d\mu = \int_{E} |f(t)| \chi_{[x, y]}(t)d\mu $

$\leq M \mu( E \cap [x, y]) \leq M |x - y| < \epsilon$ for all $x, y \in \mathbb{R}$ for which $|x -y| < \delta$.


问:
Integral $\int_0^\infty \frac{\ln(x)}{1-x^2} \; dx$ using residue theorem
I want to compute the integral
$$
I = \int_0^\infty \frac{\ln(x)}{1-x^2} \; dx.
$$
Here is my attempt: First, in order to extend the integral to the interval $(-\infty,\infty)$ I write
$$
I = \frac{1}{2}\int_0^\infty \frac{\ln(x^2)}{1-x^2} \; dx = \frac{1}{4}\int_{-\infty}^\infty \frac{\ln(x^2)}{1-x^2} \; dx.
$$
Now, let $R>2$ and $0 < \epsilon< 1/2$ and define
\begin{align*}
C_0 &= \{Re^{it} \, : \, 0\leq t\leq \pi\}\\
C_1 &= \{\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_2 &= \{-1+\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_3 &= \{1+\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_4 &= [-R,R]\setminus(]-1-\epsilon,-1+\epsilon[\cup ]-\epsilon,\epsilon[\cup]1-\epsilon,1+\epsilon[),
\end{align*}
and write
$$
C = C_0 \cup C_1 \cup C_2\cup C_4 \cup C_4.
$$
The function $f(z) = \frac{\ln(z^2)}{1-z^2}$ is holomorphic, where $\ln$ denotes a branch of the logarithm. Then, we have that
$$
\int_C f(z)\; dz = 0,
$$
hence
$$
\int_{C_4} f(z)\; dz = -\sum_{j=0}^3 \int_{C_j} f(z)\; dz,
$$
and
$$
\int_{-\infty}^\infty \frac{\ln(x^2)}{1-x^2} \; dx = \lim_{\begin{subarray}{c} \epsilon\to 0 \\ R\to\infty  \end{subarray}} -\sum_{j=0}^3 \int_{C_j} f(z)\; dz.
$$
My problem (if the above is correct) is that I'm unable to compute the integrals over the curves $C_j$. I think that the integral over the curve $C_0$ tends to $0$ as $R\to\infty$ but I cannot have information about the remaining ones.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Integrate
$\ds{\bbox[5px,#ffd]{\oint_{\cal C}
{\ln\pars{z} \over 1 - z^{2}}\dd z}}$ over a quarter circle -of radius $\ds{R \to \infty}$- in the complex plane first quadrant: There isn't any pole "inside" $\ds{\cal C}$:

\begin{align}
0 & = \oint_{\cal C}
{\ln\pars{z} \over 1 - z^{2}}\dd z =
\int_{0}^{\infty}{\ln\pars{x} \over 1 - x^{2}}\dd x
\\[2mm] + &\
\overbrace{\left.\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}\
{\ln\pars{z} \over 1 - z^{2}}\dd z\,\right\vert_{\,z\ =\ R\exp\pars{\ic\theta} \atop R \to \infty}}^{\ds{=\ 0}}\
\quad \quad+\
\int_{\infty}^{0}{\ln\pars{y} + \pi\ic/2 \over 1
+ y^{2}}\,\ic\dd y
\\[5mm] & =
\int_{0}^{\infty}{\ln\pars{x} \over 1 - x^{2}}\dd x -
\ic\int_{0}^{\infty}{\ln\pars{y} \over 1 + y^{2}}\,\dd y +
{\pi \over 2}\
\underbrace{\int_{0}^{\infty}{\dd y \over 1 + y^{2}}}_{\ds{\pi \over 2}}
\end{align}

Take $\ds{\Re}$eal $\ds{\cal P}$art in both sides
$\ds{\implies
\bbx{\int_{0}^{\infty}{\ln\pars{x} \over 1 - x^{2}}\dd x = -\,{\pi^{2} \over 4}}}$.


答:
Note
$$\frac1{1+e^t} = \frac12 -\frac12\tanh\frac t2$$
and the odd function $\tanh\frac t2$ does not survives the integration over $(-1,1)$. Thus, the integral reduces to
$\int_{-1}^{1}\frac{t^{2}}2 dt=\frac13$.

Title: Calculate $\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$, whose indefinite integral is of non-elementary function

Body:
Calculate $$\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$$

I know an approach is using the change of variables $t=-u$, so we have the value for the integral $\frac{1}{3}$. How can I calculate that integral using other approach?

I thought about this  problem when observing this indefinite integral indefinite integral.


问:
Not sure how to apply pigeon hole beyond most simple example
You have a jar of 20 pennies, 8 nickels, 10 dimes, and 15 quarters. How many coins must you pull to be certain you have 12 of the same kind?

I understand that only the pennies and the quarters are eligible.
This can't be as simple as $${20 \choose 12} + {15 \choose 12}$$ I'm not sure if I need to divide it by the total number of ways to pick 12 out of the 53 total coins?

That is a lot harder for me to think about than "How many coins to guarantee two of the same denomination"? Which is clearly pigeon hole principle and the answer is the ceiling of n/2 = 2 which is 3

答:
You're over-complicating. You know that just the pennies and quarters are eligible. So just imagine a scenario where you have $20$ pennies, $15$ quarters, and nothing else. How many coins do you need to ensure there are 12 of one type?

The answer should be obvious: $23$. You can have $22=11+11$, but once you reach $23=2(11)+1$ you must have $12$ pigeons in one "hole," so to speak.

Hopefully, this is enough of a hint to solve the problem you're asking. :-)


答:
This is neither a combinatorics question nor (arguably) a pigeonhole question.  Neither the 8 nickels nor the 10 dimes help getting 12 of the same kind of coin.  Therefore, for starters, you must select all 18 of those coins.  Then, you have to select (worst case) an additional 22 coins, which could be 11 quarters and 11 pennies.  Then the next coin that you pick will have to be either a satisfying quarter or a satisfying penny.

Final Answer:

$$8 + 10 + 11 + 11 + 1.$$

Title: Not sure how to apply pigeon hole beyond most simple example

Body:
You have a jar of 20 pennies, 8 nickels, 10 dimes, and 15 quarters. How many coins must you pull to be certain you have 12 of the same kind?

I understand that only the pennies and the quarters are eligible.
This can't be as simple as $${20 \choose 12} + {15 \choose 12}$$ I'm not sure if I need to divide it by the total number of ways to pick 12 out of the 53 total coins?

That is a lot harder for me to think about than "How many coins to guarantee two of the same denomination"? Which is clearly pigeon hole principle and the answer is the ceiling of n/2 = 2 which is 3


问:
Calculation of probability that at least 2 out of n people have the same birthday leads to paradox
Usually, the way the probability that at least 1 pair out of n people have the same birthday is calculated is by calculating the probability of the converse statement - the probability that no pair has the same birthday. I attempted to calculate the probability the straightforward way.

I use the facts that the probability of $A \lor B$ occurring is $P(A) + P(B)$. And the fact that the number of ways of picking an unordered pair from a set of $m$ elements is $\frac{(m-1)m}{2}$. The probability that a given pair of people have a birthday on the same day is $\frac{1}{365}$. Let our $n$-element set (of people, say) be the following: $\{X_1,X_2,...,X_n\}$. And let the relation $\sim$ of $X_a$ and $X_b$ mean that the people $X_a$ and $X_b$ have the same birthday. The probability that we're after is the following:
\begin{equation}
P[(X_1 \sim X_2) \lor (X_1 \sim X_3) \lor \cdots \lor (X_1 \sim X_n) \lor (X_2 \sim X_3) \lor \cdots \lor (X_{n-1} \sim X_{n})]
\end{equation}
In other words, all possible pairs OR-ed with each other. However, using the OR fact that I mentioned earlier, we can split it up into separate probabilities:
\begin{equation}
P(X_1\sim X_2) + P(X_1 \sim X_3) + \cdots
\end{equation}
Since there are $\frac{(n-1)n}{2}$ pairs and the probablity of each pair having the same birthday is $\frac{1}{365}$, the total probability should be $\frac{n(n-1)}{2}\frac{1}{365}$

However, take 40 people. By the formula I just calculated, the probability should be $\frac{39*40}{2}\frac{1}{365} \approx 2.1$, which is more than 100%... But there exist arrangements of 40 people in 365 days where none have the same birthday, so the porbability should definitely be less than 100%

Also this doesn't seem to match the formula you get the other way:
\begin{equation}
\frac{365!}{365^n(365-n)!}
\end{equation}

Essentially, my question is, where is the mistake in the above calculation? On a side note, this seems like a pretty nice false-proof.

答:
People having different birthdays are dependent events, not independent events.  For example, consider the case of 3 people.

If Person-1 and Person-2 have different birthdays, and Person-1 and Person-3 have different birthdays, then it becomes slightly more likely than normal that Person-2 and Person-3 have the same birthday.

By the way...

After reading Brian Scott's comment, I think that his insight is more relevant than my answer.


问:
how is a first variation of a functional calculated?
given the functional:
$$ F(\phi)=\frac{1}{2} \langle L\phi,\phi \rangle - \frac{1}{2} \langle \phi,f \rangle - \frac{1}{2} \langle f,\phi \rangle $$
where the differential equation with operator $L$ and function $\phi$:
$$ L\phi=f $$
and $ \langle F,G \rangle $ denotes the inner product of two functions.

how is its first variation calculated?
$$ \delta F=\frac{1}{2} \langle L\delta\phi,\phi \rangle + \frac{1}{2} \langle L\phi,\delta\phi \rangle - \frac{1}{2} \langle \delta\phi,f \rangle - \frac{1}{2} \langle f,\delta\phi \rangle $$

答:
Assuming that $L$ is a linear operator, we have
$$
\frac{d}{dt} \langle L (\phi+t\eta), \phi+t\eta \rangle|_{t=0}= \\
\frac{d}{dt} \langle L \phi, \phi \rangle+t(\langle L \phi, \eta \rangle+ \langle L \eta, \phi \rangle) +t^2 \langle L \eta, \eta  \rangle |_{t=0}= \\
\langle L \phi, \eta \rangle+ \langle L \eta, \phi \rangle
$$
At this point, you are already at your whit`s end - if you knew your operator (e.g. $L=- \Delta$), you could probably perform an integration by parts. Adapting your notation, this means that the first term of $\delta F$ is (with the factor of $\frac{1}{2}$ in front)
$$
\frac{1}{2}(\langle L \phi, \delta \phi \rangle+ \langle L \delta\phi, \phi \rangle)
$$
for the linear terms, we have
$$
\frac{d}{dt} \langle \phi+t\eta, f \rangle|_{t=0}=\frac{d}{dt} \langle \phi,f \rangle + t \langle \eta, f \rangle|_{t=0}=\langle\eta,f\rangle
$$
Switching back to your notation, we have with the factor of $\frac{1}{2}$ in front
$$
\frac{1}{2}\langle\delta \phi,f\rangle
$$
The term $\langle f, \phi \rangle$ is taken care of in the same manner. Adding all terms give you the desired variation $\delta F$. However, I wanna note that I computed a Gateaux derivative of a functional (also called "variation" sometimes) and if you knew how $F$ depended on $\phi, \nabla \phi$ explicitly, i.e. $F=F(\phi, \nabla \phi)$, you would arrive at the weak form of the Euler-Lagrange equations.
On another note, you still have to take care of the domain of your differential operator $L$, which is often NOT the whole $L^2$ space. I omitted this part, as it would require more informations on $L$.


答:
The probabilities are not disjoint. For example, if a collection $3$ people all share a birthday, then we have $X_1\sim X_2$ and $X_2\sim X_3$ and $X_1\sim X_3$. Your calculation triple-counts that case, and it sextuple-counts a case where there are $4$ people with a common birthday, etc.

This means that we can't just say $P(A\lor B\lor C\lor \cdots)=P(A)+P(B)+P(C)+\cdots$, because this will (as you notice) lead to an overestimate of the probability. If you're doing this, you need to essentially add and subtract off the probabilities that multiple people share birthdays, utilizing the inclusion-exclusion principle. This leads to complicated computations that don't scale well, so counting the complement is considerably easier.

Title: Calculation of probability that at least 2 out of n people have the same birthday leads to paradox

Body:
Usually, the way the probability that at least 1 pair out of n people have the same birthday is calculated is by calculating the probability of the converse statement - the probability that no pair has the same birthday. I attempted to calculate the probability the straightforward way.

I use the facts that the probability of $A \lor B$ occurring is $P(A) + P(B)$. And the fact that the number of ways of picking an unordered pair from a set of $m$ elements is $\frac{(m-1)m}{2}$. The probability that a given pair of people have a birthday on the same day is $\frac{1}{365}$. Let our $n$-element set (of people, say) be the following: $\{X_1,X_2,...,X_n\}$. And let the relation $\sim$ of $X_a$ and $X_b$ mean that the people $X_a$ and $X_b$ have the same birthday. The probability that we're after is the following:
\begin{equation}
P[(X_1 \sim X_2) \lor (X_1 \sim X_3) \lor \cdots \lor (X_1 \sim X_n) \lor (X_2 \sim X_3) \lor \cdots \lor (X_{n-1} \sim X_{n})]
\end{equation}
In other words, all possible pairs OR-ed with each other. However, using the OR fact that I mentioned earlier, we can split it up into separate probabilities:
\begin{equation}
P(X_1\sim X_2) + P(X_1 \sim X_3) + \cdots
\end{equation}
Since there are $\frac{(n-1)n}{2}$ pairs and the probablity of each pair having the same birthday is $\frac{1}{365}$, the total probability should be $\frac{n(n-1)}{2}\frac{1}{365}$

However, take 40 people. By the formula I just calculated, the probability should be $\frac{39*40}{2}\frac{1}{365} \approx 2.1$, which is more than 100%... But there exist arrangements of 40 people in 365 days where none have the same birthday, so the porbability should definitely be less than 100%

Also this doesn't seem to match the formula you get the other way:
\begin{equation}
\frac{365!}{365^n(365-n)!}
\end{equation}

Essentially, my question is, where is the mistake in the above calculation? On a side note, this seems like a pretty nice false-proof.


问:
Suppose $G$ is a finite simple group whose order is divisible by $p^2$. Show there is no subgroup with index $p$.

Suppose $G$ is a finite simple group such that $p^2$ divides $|G|$ for some prime $p$. Show there is no subgroup of $G$ of index $p$.

We can write the order of $G$ as $|G|=p^2m$. I want to use the third Sylow theorem, but there is no restriction that $p$ does not divide $m$. So, I rewrote the order as $|G|=p^nm_1$ where $n\geq2$ and $p$ does not divide $m_1$. But, I don't know how to convert this into a proof showing the index of a subgroup can't be $p$.

答: If $G$ has a subgroup $H$ of index $p$, left multiplication on cosets $G/H$ gives a non-trivial homomorphism $G \to S_p$. This must be injective as $G$ is simple. But $p^2$ doesn't divide $p!$ the order of $S_p$.


问:
On a Property of Perron Projection
For any positive matrix $A$, let $M=\frac{A}{\rho(A)}$. $G$ is the Perron projection of $M$.

I happen to find for any positive vector $v>0$, there is always
$\left| \Vert v\Vert_1-\Vert vG\Vert_1 \right|\geq \left| \Vert vM\Vert_1-\Vert vG \Vert_1 \right|.$
But have no ideas why...

Thanks for any advice!

Title: On a Property of Perron Projection

Body:
This is not true. Here is a random counterexample. Let
\begin{aligned}
&A=\pmatrix{2&1&2\\ 2&2&1\\ 1&2&1},
\ v=\pmatrix{2\\ 1\\ 1},\\
&M=\pmatrix{
0.42457&0.21229&0.42457\\
0.42457&0.42457&0.21229\\
0.21229&0.42457&0.21229},
\ G=\pmatrix{
0.37543&0.35924&0.29917\\
0.39235&0.37543&0.31265\\
0.31265&0.29917&0.24914}.
\end{aligned}
Then
\begin{aligned}
&|\|v^T\|_1-\|v^TG\|_1| - |\|v^TM\|_1-\|v^TG\|_1|\\
=\,&|4-4.00906|-|4.03344-4.00906|\\
=\,&0.00906-0.02438\\
<\,&0.
\end{aligned}
However, I don't find any counterexamples when $A$ is $2\times2$.


问:
How can I plot my curve, so it can be seen that it is a spiral?
I want to plot the parametric curve $ [\exp(-1/(1-t) )\cos(1/(1-t)) , \, \exp(-1/(1-t)) \sin(1/(1-t)) ]$ for $ t \in [0,1] $ in Maple or Matlab. It should look like a spiral which winds infinitley often around $(0,0)$, but when I plot it, all I can get is an arc which goes to $ (0,0) $ and does not look like a spiral. Is there a way to plot this curve, to see the behaviour of a spiral?

答:
In Maple,

```<code>ee := [exp(-1/(1-t))*cos(1/(1-t)),
       exp(-1/(1-t))*sin(1/(1-t))]:

plot([ee[], t=0..1]);
</code>```

You could also make an interactive app, to illustrate zooming in on that. For example,

```<code>ee := [exp(-1/(1-t))*cos(1/(1-t)),exp(-1/(1-t))*sin(1/(1-t))]:
ff := eval(ee,t=a):
Explore(plots:-display(
    plots:-textplot([ff[], sprintf("t = %.3g",a)],
                    align={piecewise(sin(1/(1-a))<0,'below','above')}),
    plots:-textplot([0, 0, sprintf("t = 1.00",a)],
                    align={piecewise(sin(1/(1-a))<0,'below','above'),
                           piecewise(sin(1/(1-a))<0,'left','right')}),
    plots:-pointplot([ff,[0,0]],color=blue,symbol=solidcircle,symbolsize=6),
    plot([ee[], t=0 .. 99*a/100]),
    plot([ee[], t=a..1.0]),
    plot([ee[], t=99*a/100..a])
    ,scaling=unconstrained, axes=box
    , size=[600,500], gridlines
    , title=typeset([x,y]=ee,"\n\n t = ",0.0..1.00,"\n")
    ,labels=["",cat("",seq(" ",i=5-log[10](1/4*exp(1/(1-a)))..0,-1))]
    ,view=map(u->map(`*`,u,(4*exp(-1/(1-a)))),[-0.3..0.3,-0.3..0.3])
         ),
     parameters=[[a=0.0 .. 0.95, width=400, label=""]]):
</code>```

If I'm lucky then that could be viewed (with a movable slider) as spiral zoom on the Maple Cloud.

Body:
In Maple,

ee := [exp(-1/(1-t))*cos(1/(1-t)),
       exp(-1/(1-t))*sin(1/(1-t))]:

plot([ee[], t=0..1]);

You could also make an interactive app, to illustrate zooming in on that. For example,

ee := [exp(-1/(1-t))*cos(1/(1-t)),exp(-1/(1-t))*sin(1/(1-t))]:
ff := eval(ee,t=a):
Explore(plots:-display(
    plots:-textplot([ff[], sprintf("t = %.3g",a)],
                    align={piecewise(sin(1/(1-a))<0,'below','above')}),
    plots:-textplot([0, 0, sprintf("t = 1.00",a)],
                    align={piecewise(sin(1/(1-a))<0,'below','above'),
                           piecewise(sin(1/(1-a))<0,'left','right')}),
    plots:-pointplot([ff,[0,0]],color=blue,symbol=solidcircle,symbolsize=6),
    plot([ee[], t=0 .. 99*a/100]),
    plot([ee[], t=a..1.0]),
    plot([ee[], t=99*a/100..a])
    ,scaling=unconstrained, axes=box
    , size=[600,500], gridlines
    , title=typeset([x,y]=ee,"\n\n t = ",0.0..1.00,"\n")
    ,labels=["",cat("",seq(" ",i=5-log[10](1/4*exp(1/(1-a)))..0,-1))]
    ,view=map(u->map(`*`,u,(4*exp(-1/(1-a)))),[-0.3..0.3,-0.3..0.3])
         ),
     parameters=[[a=0.0 .. 0.95, width=400, label=""]]):

If I'm lucky then that could be viewed (with a movable slider) as spiral zoom on the Maple Cloud.


问:
The Diophantine equation $5(p^2+q^2+r^2+s^2+t^2)^2 - 7(p^4+q^4+r^4+s^4+t^4) = 90pqrst$
I came across the Diophantine equation $5(p^2+q^2+r^2+s^2+t^2)^2 - 7(p^4+q^4+r^4+s^4+t^4) = 90pqrst$ many years ago in the 'Numbers Count' column of the March 1986 issue of 'Personal Computer World' magazine. The column was about Markoff Numbers and Markoff Triples, and simply described this equation as 'a related Diophantine equation'.

It presumably wasn't made up out of thin air, but I've never been able to find any earlier references to it, where it comes from, or if it is indeed related to Markoff numbers. Does anyone recognise it, or have any ideas as to where it might have come from?

(Note: a bit of googling will find the equation in 'Surfing on the Ocean of Numbers' by Henry Ibstedt, but this was in response to the PCW column, not prior to it.)

Edit 2020-12-14: I'm not looking for solutions to this equation, just it's origins.

答: You can find it in Hirzebruch and Zagier, "The Atiyah-Singer theorem and elementary number theory" (pg. 159, eqn. (7)).</p>


问:
what is the sum of this?$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $
What is the sum of $$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $$</p>

答:
For sufficiently large $N$:

$$\sum_{n=1}^N \frac{1}{n} = \log{N} + \gamma + \frac{1}{2 N}+O\left(\frac{1}{N^2}\right)$$

Plugin $N=2013$ and subtract 1.  Note that $\gamma  \approx 0.577216$.  This approximation for this value of $N$ is good to about 3 significant figures.


问:
Ratio of Volume of Standard Simplex to its Dual
In the two-dimensional case, the ratio of the area of an equilateral triangle whose vertices are the mid-points of the edges of another equilateral triangle is 1/4th.  For a tetrahedron and its dual, the ratio is known to be 1/27.

In the general case, is it $\frac{1}{n^{n}}$?  By general case, I mean the standard simplex $S$ in $\mathbb{R}^n$ and the simplex whose vertices are the Euclidean barycenters of the faces of $S$.

答:
The barycenter of an $n$-simplex is the average of all the vertices: $O=\frac{1}{n+1}(V_1+\ldots+V_{n+1})$. The barycenter of the face opposite $V_i$ is the average of all the vertices except $V_i$: $W_i=\frac{1}{n}[(V_1+\ldots+V_{n+1})-V_i]=\frac{n+1}{n}O-\frac{1}{n} V_i$. So $O=\frac{n}{n+1}W_i+\frac{1}{n+1}V_i$. This means that $O$ lies on the segment $V_iW_i$ and divides it in ratio $n:1$ (this is a generalization of the fact that the point of intersection of the medians divides each median in the ratio of $2:1$). Therefore the "dual" simplex is obtained from the original one by a similarity transform centered at $O$ and with ratio $-\frac{1}{n}$. Since we are in $n$ dimensions, this multiplies the volume by $\frac{1}{n^n}$.

In fact this is true for $n$-volume any $n$-simplex in arbitrary $\mathbb{R}^m$, with same proof.