问:
Integration U-substitution
I'm having trouble understanding how they go from $du = 3x^2$ to $dx = \frac{1}{3x^2} du$.

I've set $u$ to be $x^3$ and then $du$ is $3x^2 dx$.  Then $3x^2 dx$ becomes $3$.  Then I get a bit lost on how they get to $dx = \frac{1}{3x^2} du$.

$x^3e^{x^4}$

答:
We have $$\int x^3e^{x^4}dx$$
Notice in U-Substitution method we want to eliminate all the $x$s and $dx$ instead have $u$s and $du$. so by substitution $u=x^4$ we achieve this purpose because:

$$u=x^4\quad\quad du=4x^3dx$$
Therefore $x^3dx=\frac14du$ (note that it was not necessary to extract $dx$ like you mentioned in your question):

$$\frac14\int e^u du=\frac14e^u+C$$
Since $u=x^4$ the integral is equal to $\frac14 e^{x^4}+C$.

Title: Integration U-substitution


问:
Two equal circles inscribed in a triangle
Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem  and algebraic manipulations are allowed to be used]

答:
Observe that the side $BC$ is the combination of three segments. One of them is directly related to the radius $r$. The problem is how to get the other two segments through trigonometry.

Notice that this triangle ABC is a combination of two Pythagorean triangles. You can certainly get the values like $\sin$ and $\cos$ for these angles, then use the half-angle formula to express the segments in terms of $r$, then you are done, because you can express $BC$ as a multiple of the radius $r$.

ContentLicense: CC BY-SA 4.0


答:
Hint: Cut the circles vertically in half, then glue the figure back together as shown.

Now two circles merges into one, and the small triangle formed is similar to the original triangle, with a scale factor of $\frac {21-2r}{21}$ (why?)

The original triangle has area $126$ and perimeter $54$, hence its inradius is $\frac {14}3$.

Thus the new triangle has inradius

$$\frac {14}{3} \times \frac{21-2r}{21} = r$$

hence $r = \dfrac {42}{13}$.

Title: Two equal circles inscribed in a triangle

Body: Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem  and algebraic manipulations are allowed to be used]


问:
n unique sums of squares for 2p^(2n) for prime p (1 mod 4) and n natural number >= 1?
After looking at many sums of squares, this is something I am hypothesizing. I'd love to know if there is any work related to this, if it is easy to solve or even if there exist counterexamples.

Conjecture: Let $p$ be some prime number where $p \equiv 1 \bmod 4$. Then, for $n \in \mathbb{N}$ and $n \geq 1$, $2p^{2n}$ will give $n$ unique pairs $a,b \in \mathbb{N}$ such that:

\begin{align*}
    a^2 + b^2 &= 2p^{2n}
\end{align*}

In the image below I have provided 3 different examples along with the $a,b$ sums of squares for more details.

Ex 1: $p=617$ and $n=1$.

Ex 2: $p = 13$ and $n=3$.

Ex 3: $p = 5$ and $n=7$.

link to image with examples described above

答:
None of your resulting triples will be primitive because that requires that the RHS be an odd number. To find them we use Euclid's formula for $\quad A^2+B^2=C^2\quad$ which is
$ \quad A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2\quad$ We solve for $k$ and test a defined range of values to see which, if any yield integers.

\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\quad\text{for}\quad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}

The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

We will use $C$ instead of $C^2:\quad$ $\quad C^2=2p^{2n}\implies C=2p^n$
$$C=2\cdot 617^1=1234
\implies \bigg\lfloor\frac{ 1+\sqrt{1234-1}}{2}\bigg\rfloor = 25
\le m \le \big\lfloor\sqrt{1234-1}\space\big\rfloor=35\\
\quad\text{and we find} \quad m\in\{35\}\implies k\in\{3\}\\
F(35,3)=(1216,210,1234)\quad \\$$

$$C=2\cdot 13^3=4394
\implies \bigg\lfloor\frac{ 1+\sqrt{4394-1}}{2}\bigg\rfloor = 47
\le m \le \big\lfloor\sqrt{4394-1}\space\big\rfloor=66\\
\quad\text{and we find} \qquad m\in\{55,65\}\implies k\in\{37,13\}\\
F(55,37)=(1656,4070,4394)\qquad F(65,13)=(4056,1690,4394)\quad \\$$

$$C=2\cdot 5^7=156250
\implies \bigg\lfloor\frac{ 1+\sqrt{156250-1}}{2}\bigg\rfloor = 280
\le m \le \big\lfloor\sqrt{156250-1}\space\big\rfloor = 395\\
\quad\text{and we find} \quad m\in\{307,325,375,395\}\implies
k\in\{249,225,125,15\}\\
f(307,249)=(32248,152886,156250)\\
 f(325,225)=(55000,146250,156250)\\
 f(375,125)=(125000,93750,156250)\\
 f(395,15)=(155800,11850,156250)$$


问:
I am confused with the def of derivative.
The problem given :

Let $f:\Bbb R\rightarrow \Bbb R$ and $F:\Bbb R^2\rightarrow \Bbb R$ be differentiable and satisfy $F(x,f(x))=0$ and $\displaystyle \frac {dF}{dy}$ is not zero.
Prove that $f'(x)= -\dfrac {dF}{dx}/\frac {dF}{dy}$ where $y=f(x)$

I am so confused here.

Since $F(x,f(x))=0$, I think, $\displaystyle \frac {dF}{dx}$ must be $0$.

But when I see the $f'(x)$, it seems not.

Is there any difference between $\displaystyle \frac {dF(x,f(x))}{dx}$ and $\displaystyle \left.\frac {dF}{dx} \right|_{(x,f(x))}$  ?

答:
Taking $f(x) = y$, note that by the chain rule:

$$0 = \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx}$$

The result you want follows immediately from this observation.

So to recap:

You are correct in supposing that $dF / dx = 0$.  This only works because we are taking the total derivative: it emphatically does not mean that $\partial F / \partial x = 0$.

ContentLicense: CC BY-SA 3.0


问:
Discrete Math logically equivalent
Prove that (p ∨ (q ∧ ¬r)) → p is logically equivalent to p ∨ ¬q ∨ r.

I'm hitting the dead end all the time.

Here's what I have done:

(p ∨ (q ∧ ¬r)) → p

¬(p ∨ (q ∧ ¬r)) ∨ p        Implication

¬(p) ∧ ¬(q ∧ ¬r)) ∨ p      De Morgan’s Laws

¬(p) ∧ (¬q ∨ ¬¬r) ∨ p      De Morgan’s Laws

¬(p) ∧ (¬q ∨ r) ∨ p        Double Negation

then I have no idea  what to continue.

答:
Based on the properties of the logical operators involved, we deduce the proposed equivalence:

\begin{align*}
(p\vee(q\wedge\neg r))\to p & \Longleftrightarrow \neg(p\vee(q\wedge\neg r))\vee p\\\\
& \Longleftrightarrow (\neg p\wedge\neg(q\wedge\neg r))\vee p\\\\
& \Longleftrightarrow (\neg p\wedge(\neg q\vee r))\vee p\\\\
& \Longleftrightarrow (\neg p\vee p)\wedge(p\vee\neg q\vee r)\\\\
& \Longleftrightarrow p\vee\neg q\vee r
\end{align*}

where we have made use of the equivalence
\begin{align*}
P\to Q \Longleftrightarrow \neg P\vee Q
\end{align*}

Hopefully this helps!

Title: Discrete Math logically equivalent


问:
Getting graphs for linear inequalities
I am answering some activity on a book and the instructions is as follows:

Graph the following inequalities and identify at least $3$ points that belong to the solution set of the inequalities.

So I went to get the graphing with the help of Desmos, but then:

This actually confused me. If I were to get $3$ points from the solution set of inequalities, I need to get both sides to intersect.

But this one is parallel, it won't intersect, so how can I get the solution set of inequalities if this is so?

Try it, the 2 inequalities are:

$ y \ge 2x + 3 $

$ y < 2x - 4 $

答:
Your graph correctly shows that there's no intersection for these two inequalities. There are no points that will solve them simultaneously.

We can also demonstrate the same thing algebraically. Given the two inequalities $y\ge 2x+3$ and $y<2x-4$, we can rewrite the intersection as

$$ 2x+3 \le y < 2x-4 . $$

So, any solution must have $ 2x+3 < 2x-4 $. But, notice, if you subtract $2x$ from both sides of this inequality, we have $3 < -4$. This is obviously not true, regardless of what values of $x$ and $y$ we choose, so there is no solution to this pair of inequalities.

Title: Getting graphs for linear inequalities


问:
Prove that sin A/2 * sin B/2 * sin C/2 = r/4R
The other day I came across an identity in the book "Problems from the Book" and it was presented as well known:

$$\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{r}{4R}$$

However I wasn't familiar with thee identity so I tried proving it.

Here is part of my attempt in solving this problem:

I first rewrote $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$ as $\frac{r^3}{AI\cdot BI \cdot CI}$ (where I is the incenter of $\bigtriangleup ABC$). Then I tried using the fact that [ABC] =abc/4R (where [ABC] represents the area of $\bigtriangleup ABC$)which allowed me to rewrite the equation as $\frac{r^2}{AI\cdot BI \cdot CI} = \frac{[ABC]}{abc}$. I tried various things at this point but none of my attempts were successful...

Does anyone have a proof of this identity?

答:
$$LHS = \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} $$

$$= \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-c)(s-a)}{ca}}  \cdot \sqrt{\frac{(s-a)(s-b)}{ab}}$$

$$= \frac{(s-a)(s-b)(s-c)}{abc}$$

Note that $$r = \frac{\Delta}{s}$$
$$R = \frac{abc}{4 \Delta}$$
$$\Delta^2 = s(s-a)(s-b)(s-c)$$
Now $$ RHS = \frac{r}{4R} = \frac{\Delta}{s} \cdot \frac{\Delta}{abc} = \frac{(s-a)(s-b)(s-c)}{abc}$$

Hence LHS = RHS


问:
How to evaluate $\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$
I want to evaluate: $$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx,$$ but I don't see how can I achieve so.

My attempts so far have been rewriting the integral using algebraic identities such as: $$ab=\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{2}\left(a-b\right)^2=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2,$$ that yield other integrals like:
$$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(1-x\right)}{1+x}\:dx,\:\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)}{1+x}\:dx.$$Normally the beta function and expanding terms into series is used to deal with these kind of integrals yet neither can be used because of the upper bound.

What else can be done in order to compute the main integral? Thanks.

Body:
I think that I should expand one part of the integrand as a Taylor series around $x=0$.

For symmetry reasons, I chose
$$\frac{\log (1-x) \log (x+1)}{x+1}=\sum_{n=2}^\infty a_n\,(x^{2n+1}-x^{2n})$$  where the $a_n$'s make the sequence
$$\left\{1,\frac{17}{12},\frac{151}{90},\frac{9413}{5040},\frac{5647}{2800},\frac{444
   811}{207900},\frac{84902267}{37837800},\frac{706664611}{302702400},\cdots\right\}$$ and use
$$I_n=\int_0^{\frac 12}(x^{2n+1}-x^{2n})\,\log(x)\,dx=$$
$$\frac{(7+6 \log (2))+ 2(6+11 \log (2))\,n+4 (1+6 \log (2))\,n^2+8  \log
   (2)\,n^3}{4^{n+2}\,(n+1)^2 \,(2 n+1)^2}$$ and the partial sums
$$S_p=\sum_{k=1}^p a_k\,I_k$$ converge quite fast
$$\left(
\begin{array}{cc}
p & S_p \\
 1 & 0.02803334672 \\
 2 & 0.03276936442 \\
 3 & 0.03366455049 \\
 4 & 0.03384586175 \\
 5 & 0.03388419298 \\
 6 & 0.03389253226 \\
 7 & 0.03389438389 \\
 8 & 0.03389480131 \\
 9 & 0.03389489653 \\
 10 & 0.03389491845 \\
 11 & 0.03389492353 \\
 12 & 0.03389492472 \\
 13 & 0.03389492500 \\
 14 & 0.03389492506 \\
 15 & 0.03389492508
\end{array}
\right)$$

Edit

You could use this for any other value of the upper bound since
$$J_n=\int_0^{t}(x^{2n+1}-x^{2n})\,\log(x)\,dx=$$
$$ t^{2n+1}\Bigg[\frac{1}{(2 n+1)^2}-\frac{t}{4(n+1)^2} +\frac{(2n+1)t-2(n+1) }{2(n+1) (2 n+1) }\log(t)\Bigg]$$


问:
Why modulo 2 can be express as cos function for optimization?
I have been reading the paper Monotonic optimization based decoding for linear codes. In this paper the author show that modulo 2 operation can be rewritten as cos function.

Particularly, an $\left( {n,k} \right)$ with the corresponding generator matrix $G$ of size $n \times k$ is considered

$G = \left[ {\begin{array}{*{20}{c}}
{{g_1}}\\
{{g_2}}\\
{...}\\
{{g_n}}
\end{array}} \right]$ each row is ${g_i} \in {\left\{ {0,1} \right\}^k}$

Let $u, x$ be the column vectors of the input bit sequence/message and the corresponding
encoder output codeword. This will give the relation $x = Gu\left( {\bmod \,\,2} \right)$. The author then show this without any proof or intuition explain

$Gu\left( {\bmod \,\,2} \right) = \frac{{{1_n} - \cos \left( {\pi Gu} \right)}}{2}$ where
$\cos \left( {\pi Gu} \right): = {\left( {\cos \left( {\pi {g_1}u} \right),\cos \left( {\pi {g_2}u} \right),\cos \left( {\pi {g_3}u} \right),...,\cos \left( {\pi {g_n}u} \right)} \right)^T}$

How can modulo 2 be express as cos function (for optimization latter in the continuous space) ?

Please help me understand this

Thank you very much

Body:
$$\frac{1-\cos0}2 = 0\\
\frac{1-\cos\pi}2 = 1\\
\frac{1-\cos2\pi}2 = 0\\
\frac{1-\cos3\pi}2 = 1\\
\vdots$$


问:
Prove that $\sqrt{13}$ is irrational in 3 ways.
I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.

However, I am not sure where to begin with the other two proofs. I thought that this was the only way to prove irrationality of a number. I would really appreciate some help on where to start/look for to start the two other proofs.

答:
Suppose $\sqrt{13} = \dfrac k \ell$ where $k,\ell$ are positive integers and the fraction is in lowest terms.

Then $$\frac{65\ell -18 k}{5k - 18\ell} = \sqrt{13}$$ but here the numerator and denominator are positive integers smaller than $k,\ell$ respectively.


问:
Proof of $\sum_{n=1}^\infty a_nb_n^2c_n^3<\infty$.
Let $a_n \geq 0, b_n \geq 0, c_n \geq 0 \; \forall n \in \mathbb{N}$ and suppose that $\sum_{n=1}^\infty a_n^3b_n^3, \;\sum_{n=1}^\infty b_n^3c_n^3, \; \sum_{n=1}^\infty c_n^6$ are convergent. Prove that the series $\sum_{n=1}^\infty a_nb_n^2c_n^3$ is convergent.

I saw that $(a_n^3b_n^3 \cdot b_n^3c_n^3 \cdot c_n^6)^\frac{1}{3}=a_n b_n^2 c_n^3$ and here I stopped.
I was thinking about using Cauchy-Schwarz inequality, but I don't really know how to apply it (especially having $x_n^3$, not $x_n^2$).

I would be grateful for some hints! :)

答:
With Hölder's inequality:
$$\begin{align}
\sum_{n} a_n b_n^2 c_n^3
&= \sum_{n} a_n b_n \cdot b_n c_n^3\\
&\leq \left(\sum_{n} a_n^3 b_n^3\right)^{1/3}\left(\sum_{n} b_n^{3/2}c_n^{9/2} \right)^{1/3}\\
&= \left(\sum_{n} a_n^3 b_n^3\right)^{1/3}\left(\sum_{n} b_n^{3/2}c_n^{3/2}\cdot c_n^{3} \right)^{1/3}\\
&\leq \left(\sum_{n} a_n^3 b_n^3\right)^{1/3}\left(\sum_{n} b_n^{3}c_n^{3} \right)^{1/6}\left(\sum_{n} c_n^{6} \right)^{1/6} < \infty
\end{align}$$
The first inequality is Hölder's with $p=3,q=3/2$ and the second is Cauchy—Schwarz (i.e., Hölder's with $p=2,q=2$).


问:
Graphing a systems of linear equations?
I'm trying to understand how to graph systems of linear equations.

Say for example we have the following:

$x_1 - 2x_2 = -1$

$-x_1 + 3x_2 = 3$

I'm only used to graphing $x$ and $y$ coordinates, how am I supposed to interpret these?

Thanks in advance!

答:
Hint.

The first equation can be written as
$$
x_2=\frac{1}2(x_1+1)
$$
and the second one:
$$
x_2=\frac13 x_1+1
$$

If you replace $x_1$ and $x_2$ with $x$ and $y$ respectively, you will see something you are familiar with.

Title: Graphing a systems of linear equations?


答:
You could

Treat $x_2$ as $y$

and $x_1$ as $x$

so you are drawing

$x_2=\frac12x_1+\frac12$

$x_2=\frac13x_1+1$

and label the axes appropriately

Title: Graphing a systems of linear equations?

Body:
I'm trying to understand how to graph systems of linear equations.

Say for example we have the following:

$x_1 - 2x_2 = -1$

$-x_1 + 3x_2 = 3$

I'm only used to graphing $x$ and $y$ coordinates, how am I supposed to interpret these?

Thanks in advance!


答:
Just set $x=x_1, y=x_2$, or when drawing the graph, label the axes $x_1$ and $x_2$ instead of $x$ and $y$.
A solution is where the lines cross.

Title: Graphing a systems of linear equations?


问:
Compute $\lim_{n\to \infty} \sup A_{n}$ and $\lim_{n\to \infty} \inf A_{n}$, where $A_{n} = [0,a_{n}]$
I want to compute the limit (if exists) of the sequence of sets defined by $A_{n} = [0,a_{n}]$. Where $\{a_{n}\}$ is a sequence of nonnegative numbers that converges to $a\geq 0$.

I know that I have to compute $ \lim_{n\to \infty } \inf A_{n} $ and  $ \lim_{n\to \infty } \sup A_{n} $.

Here's my attempt:

$ \lim_{n\to \infty } \inf A_{n} := \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_{k} $.

Let $T_{n} := \{a_{n},a_{n+1},... \} $

I proved that $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $, and note that $\{ \inf T_{n} \}_{n \geq 1}$ is a non decreasing sequence.

Then $\lim_{n\to \infty } \inf A_{n} = \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_{k} = \cup_{n=1}^{\infty} [0,\inf T_{n}] = [0,\sup_{n\geq 1}\{\inf T_{n}\}]  $

since $ \lim_{n\to \infty} a_{n} = a \implies \lim_{n \to \infty} \inf a_{n} :=\sup_{n\geq 1}\{\inf Tn\} = a $. Then

$\lim_{n\to \infty } \inf A_{n} = [0,a]$

Now, I have a problem when I try to compute
$ \lim_{n\to \infty } \sup A_{n}$

$ \lim_{n\to \infty } \sup A_{n} := \cap_{n=1}^{\infty} \cup_{k=n}^{\infty} A_{k} $.

I tried to prove that $\cup_{k=n}^{\infty} A_{k} = [0,\sup{T_{n}}]$ but I was not able to prove that $ [0,\sup{T_{n}}] \subset \cup_{k=n}^{\infty} A_{k}  $.

Proof that $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $

$\subset:$ Let $x \in \cap_{k=n}^{\infty} A_{k} \implies \forall k \geq n \ \  x \in A_{k} \implies \forall k \geq n \ \  0\leq x \leq a_{k}  \implies x \leq \inf T_{n} \implies x \in [0,\inf T_{n}] $. Then $\cap_{k=n}^{\infty} A_{k} \subset [0,\inf T_{n}] $

$\supset:$ Let $x \in [0,\inf T_{n}] \implies 0 \leq x \leq \inf T_{n} \leq a_{k} \ \ \ \forall k \geq n \implies x \in \cap_{k=n}^{\infty} [0,a_{k}] = \cap_{k=n}^{\infty} A_{k} $. Then $\cap_{k=n}^{\infty} A_{k} \supset [0,\inf T_{n}] $

which implies $\cap_{k=n}^{\infty} A_{k} = [0,\inf T_{n}] $

Questions:

What do you think about the computation of $\lim_{n\to \infty} \inf A_{n} $, is it correct?

Do you know how to compute $\lim_{n\to \infty} \sup A_{n}$?

Thanks in advance.

答: Take $\epsilon>0$. By definition of limit, there exists $n_{\epsilon}: k>n_{\epsilon} \rightarrow |a_k-a|<\epsilon$. We can then say $[0,a-\epsilon] \subset \cap_{k=n_{\epsilon}}^{\infty} A_{k} $. Since we can take $\epsilon$ arbitrarily small, it follows that $[0,a) \subset \lim_{n\to \infty } \inf A_{n} $. The limsup will be either $[0,a)$ or $[0,a]$; if there are an infinite $a_n$ less than $a$, then limsup =$[0,a)$, otherwise it is $[0,a]$.


答:
The dimensions of the triangle were chosen to make this a little easier than it first appears.  The triangle can be divided to two Pythagorean triples..

The slope of the line from the B to the center of the closer circle is the same as the slope of the line from B to the incenter of the 5,12,13 right triangle.

There are a few ways to find this, but I like this one.

$a+b = 12\\
a+c = 13\\
b+c = 5$

Solving this system: $a = 10, b = 2, c = 3$

The slope is $\frac 23$
Doing the same to the $12,16,20$ triangle gives a slope of $-\frac 13$

$\frac 32 r + 2r + 3r = 21\\
r = \frac {42}{13}$


问:
Show that $\int_1^\infty x^n\sin(x+x^2)dx$ diverges, $n\ge1$.
How do I show that this improper integral diverges? I know certainly it is not always true that if $\int_a^\infty f(x)dx$ converges, then $f(x)\to 0$. My intuition tells me this improper integral diverges because $x^n*\sin(x+x^2)$ oscillates between infinity and -infinity as $x\to \infty$. I am wondering if there is a nice way to prove it is divergent.

答:
Due to the fact that $x$ goes to infinity I think it is fair to approximate it as the following:
$$\int_1^\infty x^n\sin(x^2)dx$$
Now we could split this up into regions, we know that $\sin(x)\ge0$ for $2k\pi\le x\le(2k+1)\pi$ where $k$ in an integer, so for the positive values of the $\sin$ function we have:
$$0\le\sum_{k=0}^\infty\int_\limits\sqrt{2k\pi}^{\sqrt{(2k+1)\pi}}x^n\sin(x^2)dx\le\sum_{k=0}^\infty\int\limits_\sqrt{2k\pi}^\sqrt{(2k+1)\pi}x^ndx$$
Although I feel like to use this we could do with a much better lower limit than zero, so just a suggestion


问:
Difficulty with Equivalence Proof
I'm having difficulty proving that p ↔ q ≡ (p → q) ∧ (q → p). I've tried breaking it down from both sides, but nothing is clicking yet. So far, I believe my best attempt was with p ↔ q and goes something like this:

p ↔ q

(p ∧ q) ∨ (¬p ∧ ¬q) (Law of Equality)

(p ∧ q) ∨ ¬(p ∧ q) (DeMorgan 2)

¬(p ∧ q) ∨ (p ∧ q) (Symmetry)

(p ∨ q) → (p ∧ q) (Implication)

From here I'm stuck as to where I should go next or even if I'm in the right direction in the first place. Any help and advice is greatly appreciated!

答:
The second step gives you a disjunction of conjunctions.  You want a conjunction of conditionals -- and conditionals are equivalent to certain disjunctions.

So the next step(s) should be distribution to obtain a conjunction of disjunctions.

$${p\leftrightarrow q\\(p\wedge q)\vee(\neg p\wedge\neg q)\\(p\lor(\neg p\wedge \neg q))\wedge(q\vee(\neg p\wedge\neg q))\\(p\vee\neg p)\wedge(p\vee\neg q)\wedge(q\vee\neg p)\wedge(q\vee\neg q)\\~~\vdots\\(p\to q)\wedge(q\to p)}$$

Title: Difficulty with Equivalence Proof


问:
The smallest non-zero singular value of AB
For a matrix $A$, let $σ_{min}(A)$ denote the smallest $\textbf{non-zero}$ singular value of A.

I saw some materials (e.g. https://pdfs.semanticscholar.org/e74f/89ac85239811b295787619e73c99fc867724.pdf)
says that
$$σ_{i}(AB)≥σ_{i}(A)σ_{n}(B),~~~ i=1,\cdots,n$$

Here $A$ and $B$ are rectangle matrix with dimension, say, $p\times n$ and $n\times m$. This holds for even zero singular value. So if $B$ is not full rank, then $\sigma_n(B)$ is $0$ and the above bound is trivial.

My question is if the following inequality holds:
$$σ_{min}(AB)≥σ_{min}(A)σ_{min}(B)$$

Thanks.

答:
Let $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times p}$. Let $\sigma_*(A)$ be the smallest non-zero singular value of $A$ if $A \neq 0$, and zero if $A = 0$. I use $\sigma_*$ to avoid confusion with the smallest singular value.

Counter-example for a general lower bound

Let $A = \lceil 1, 0 \rfloor$ (a diagonal matrix). Then $\sigma_*(A) = 1$.

Let $B = \lceil 1, 2 \rfloor$. Then $AB = \lceil 1, 0 \rfloor$, and $1 = \sigma_*(A) \sigma_*(B) = \sigma_*(AB) = 1$.

Let $B = \lceil 2, 1 \rfloor$. Then $AB = \lceil 2, 0 \rfloor$, and $1 = \sigma_*(A) \sigma_*(B) < \sigma_*(AB) = 2$.

Let $B = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$. Then $AB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, and $\sqrt{2} = \sigma_*(A) \sigma_*(B) > \sigma_*(AB) = 1$.

Hence a general lower (or upper) bound which works for every matrix does not exist. Additional assumptions are needed.

Relaxed sufficient conditions

The accepted answer by user1551 gives sufficient conditions for the stated lower bound to hold: $A$ is column-independent and $B \neq 0$, or $B$ is row-independent and $A \neq 0$.

I'll note that the lower bound holds with the following relaxed condition:

$$\mathrm{range}(BB^T) \subset \mathrm{range}(A^T) \lor \mathrm{range}(A^T A) \subset \mathrm{range}(B).$$

Note that:

when $A$ is column-independent, $\mathrm{range}(A^T) = \mathbb{R}^n$,

when $B$ is row-independent, $\mathrm{range}(B) = \mathbb{R}^n$,

$\mathrm{range}(A^T) = \mathrm{null}(A)^{\perp}$,

$\mathrm{range}((AB)^T) = \mathrm{range}(B^T A^T) \subset \mathrm{range}(B^T)$,

$\mathrm{null}(B) \cap \mathrm{range}(B^T) = \mathrm{null}(B) \cap \mathrm{null}(B)^{\perp} = \{0\}$.

Suppose $A = 0$ or $B = 0$. Then $\sigma_*(AB) = \sigma_*(0) = 0 = \sigma_*(A) \sigma_*(B)$, and the lower bound holds.

Suppose $A \neq 0$, $B \neq 0$, and $\mathrm{range}(BB^T) \subset \mathrm{range}(A^T)$. Then

$$
\begin{aligned}
{} & \sigma_*(AB) \\
= & \min \left\{||AB x|| : x \in \mathrm{range}((AB)^T) \land ||x|| = 1 \right\} \\
\geq & \min \left\{||AB x|| : x \in \mathrm{range}(B^T) \land ||x|| = 1 \right\} \\
= & \min \left\{\left\lVert A \frac{Bx}{||Bx||}\right\rVert ||Bx|| : x \in \mathrm{range}(B^T) \land ||x|| = 1 \right\} \\
\geq & \min \left\{\left\lVert A \frac{Bx}{||Bx||}\right\rVert : x \in \mathrm{range}(B^T) \land ||x|| = 1 \right\} \\
\cdot \; & \min \left\{||Bx|| : x \in \mathrm{range}(B^T) \land ||x|| = 1 \right\} \\
\end{aligned}
$$

For the first factor,

$$
\begin{aligned}
{} & \min \left\{||A \frac{Bx}{||Bx||}|| : x \in \mathrm{range}(B^T) \land ||x|| = 1 \right\} \\
= & \min \left\{||A \frac{Bx}{||Bx||}|| : x \in \mathrm{range}(B^T) \setminus \{0\} \right\} \\
= & \min \left\{||A y|| : y \in \mathrm{range}(BB^T) \land ||y|| = 1 \right\} \\
\geq & \min \left\{||A y|| : y \in \mathrm{range}(A^T) \land ||y|| = 1 \right\} \\
= & \sigma_*(A).
\end{aligned}
$$

Therefore

$$\sigma_*(AB) \geq \sigma_*(A) \sigma_*(B).$$

Suppose $A \neq 0$, $B \neq 0$, and $\mathrm{range}(A^T A) \subset \mathrm{range}(B)$. By the previous case

$$\sigma_*(AB) = \sigma_*((AB)^T) = \sigma_*(B^T A^T) \geq \sigma_*(B^T) \sigma_*(A^T) = \sigma_*(A) \sigma_*(B).$$

Application

I'll now describe an interesting application for the bound. Consider $||(AB)^+||$, where $||\cdot||$ denotes the operator 2-norm, and $+$ denotes the Moore-Penrose inverse. It is not difficult to show (e.g. by employing singular value decomposition) that $||A^+|| = \sigma_*(A)^+$. Supposing now the relaxed sufficient conditions, we get the nice upper bound

$$||(AB)^+|| \leq ||A^+|| ||B^+||.$$

Note that this result is not trivial, since in general $(AB)^+$ does not equal $B^+ A^+$.

Open question

Is the relaxed condition also necessary?


问:
Question about simplifying $(a+b+c)(ab+bc+ac)=abc$
To simplify this expression, according to math110's answer, we can write LHS as:

$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc$$
then $abc$ cancels and we get: $(a+b)(b+c)(a+c)=0$

But my question is, how we can recognize that we have this equality? I have no clue about it. is there any way to quickly realize we can write $(a+b+c)(ab+bc+ac)$ as $(a+b)(b+c)(a+c)+abc$ ?

答:
The original equation is

$$\frac1a + \frac1b+ \frac1c = \frac1{a+b+c}$$

We see that $a=-b, b=-c, c=-a$ are trivial solutions to the equation. Thus

$$abc(a+b+c)\left(\frac1a + \frac1b+ \frac1c - \frac1{a+b+c}\right)$$

must contain $(a+b)(b+c)(c+a)$ as a factor; indeed it is equal to $(a+b)(b+c)(c+a)$.


问:
Uniform convergence of subsequence of function
I have this exercise:

Let $f_n:\mathbb{R}\to\mathbb{R}, n\in\mathbb{N}$ a sequence of continuous functions that punctualy converge to $f:\mathbb{R}\to\mathbb{R}$.
I have to prove that if there exists a sequence in $\mathbb{R}$ ,$(r_n)$, such that $(r_n)\to r$ but $f_n(r_n)\not\to f(r)$ then $(f_n)$ does not converge uniformly to $f$ in $\mathbb{R}$.

Any hints would be great.

答:
Assume $(f_n)$ converges uniformly to $f$. We'll prove that $f_n(r_n)\to f(r)$.

Given $\epsilon>0$, use uniform convergence to find an $m$ such that $|f_m(x)-f(x)|<\epsilon$ for all $x$, for all $n\ge m$. Then use the continuity of $f_m$ to show that for sufficiently large $n$, we have $|f_m(r_n)-f_m(r)|<\epsilon$.

Applying the triangle inequality along the path $f(r)\to f_m(r)\to f_m(r_n)\to f(r_n)\to f_n(r_n)$, we see that for sufficiently large $n$, $|f(r)-f_n(r_n)|<4\epsilon$.


问:
Two equal circles inscribed in a triangle
Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem  and algebraic manipulations are allowed to be used]

答:
Let $a,b, c$ be the sides and note that

\begin{align}
a = 2r + r(\cot \frac{B}2 +\cot \frac{C}2)=2r + r\frac{\cos\frac A2}{\sin\frac B2 \sin\frac C2}\\
= 2r +r\frac{2R\sin A}{4R\sin\frac A2\sin\frac B2 \sin\frac C2}
= 2r +r \frac{a}{r_i}
\end{align}
with $R$ and $r_i$
the circumradius and inradius, respectively. Then,
$$ r = \frac{a}{2+\frac a{r_i}} = \frac{a}{2+\frac {a(a+b+c)}{2K}}
$$
where $K$ is the area of the triangle. Apply the Heron’s area formula to get
$$r= \frac a{ 2+\frac{a\sqrt{s}}{\sqrt{(s-a)(s-b)(s-c)}}},\>\>\>\>\>
s=\frac{a+b+c}2$$
Substitute $a=21$, $b=20$ and $c=13$ to obtain $r=\frac{42}{13}$.


问:
Factorials and Divisibility
I'm having trouble getting started on the following:

Given $n_1, n_2, ..., n_k \in$ $\Bbb N$, show that $n_1!\cdot n_2!\cdot\cdot\cdot n_k! |(n_1+n_2+...+ n_k)!$

I thought about a proof by induction on $k$ but feel that induction shouldn't be needed.
Ideas?

答:
The ratio
$$\frac{(n_1+n_2+\cdots+n_k)!}{n_1!n_2!\cdots n_k!}$$
is a multinomial coefficient. It is the number of words of length $n_1+\cdots+n_k$ with $n_1$ occurrences of the "letter" $a_1$, $n_2$ occurrences of $a_2$, and so on. Since our fraction counts something, it must be an integer.

Title: Factorials and Divisibility


问:
Summing this series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$
I wish to sum the series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$.

I notice that by writing $\displaystyle\sum_{n=1}^{\infty}(2n-1)\dfrac{1}{2^n}$, $\dfrac{1}{2^n}$ is a geometric series and can be summed easily. But the next step I think I did wrong is

$$2n-1\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^n}$$

I don't think I can pull out $2n-1$ like a constant because that will affect the sum. What should I have done. The series converges to $3$ according Wolfram Alpha.

Body:
Write $$\sum_{n=1}^\infty \frac {2n-1}{2^n} = \frac12\sum_{n=1}^\infty(2n-1)\left(\frac 1{\sqrt 2}\right)^{2n-2}$$

Notice that

$$\sum_{n=1}^\infty (2n-1)x^{2n-2} = \frac d{dx}\sum_{n=1}^\infty x^{2n-1} = \frac d{dx}\left(\frac {x}{1-x^2}\right) = \frac {1+x^2}{(1-x^2)^2}$$ for $|x|<1$. Hence:

$$\frac12\sum_{n=1}^\infty(2n-1)\left(\frac 1{\sqrt 2}\right)^{2n-2} = \frac12\left(\frac {1+1/2}{(1-1/2)^2}\right) = 3$$


答:
Imagine you have $k$ different letters. The first letter is available $n_1$ times, the second one $n_2$ times etc.
Now show that the number of words which you can form out of these $n_1 + n_2 + \ldots + n_k$ letters is $\frac{(n_1 + n_2 + \ldots + n_k)!}{n_1!\cdot\ldots\cdot n_k!}$. In particular, this number is an integer, so $(n_1 + n_2 + \ldots + n_k)!$ is divisible by $n_1!\cdot\ldots\cdot n_k!$.</p>

Title: Factorials and Divisibility

Body:
I'm having trouble getting started on the following:

Given $n_1, n_2, ..., n_k \in$ $\Bbb N$, show that $n_1!\cdot n_2!\cdot\cdot\cdot n_k! |(n_1+n_2+...+ n_k)!$

I thought about a proof by induction on $k$ but feel that induction shouldn't be needed.
Ideas?


答:
Here is one way:
$$
\sum_{n=1}^\infty \frac{2n-1}{2^n}=\sum_{n=1}^\infty \frac{2n}{2^n}-\sum_{n=1}^\infty\frac{1}{2^n}.
$$
The second sum is geometric and converges to $1$. We just need to deal with the first. To do this, use the relation
$$
\sum_{n=0}^\infty z^n=\frac{1}{1-z}
$$
valid for $\lvert z\rvert <1$ and differentiate to get
$$
\sum_{n=1}^\infty nz^{n-1}=\frac{1}{(1-z)^2}.
$$
Next, evaluate at $z=\frac{1}{2}$ to get
$$
\sum_{n=1}^\infty \frac{2n}{2^n}=\frac{1}{(1/2)^2}=4.
$$
So, the result is $3$.

Title: Summing this series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$

Body:
I wish to sum the series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$.

I notice that by writing $\displaystyle\sum_{n=1}^{\infty}(2n-1)\dfrac{1}{2^n}$, $\dfrac{1}{2^n}$ is a geometric series and can be summed easily. But the next step I think I did wrong is

$$2n-1\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^n}$$

I don't think I can pull out $2n-1$ like a constant because that will affect the sum. What should I have done. The series converges to $3$ according Wolfram Alpha.


答:
Assume $\{f_n\} \longrightarrow f$ uniformly. Then $f$ is continuous, and using triangle inequality,

\begin{eqnarray*}
|f_n(r_n)-f(r)| & \leq & |f_n(r_n)-f(r_n)|+|f(r_n)-f(r)|\\
 & \leq & \sup_{x\in \mathbb{R}}|f_n(x)-f(x)|+|f(r_n)-f(r)|
\end{eqnarray*}
Now take $n \longrightarrow \infty$.

Title: Uniform convergence of subsequence of function

Body:
I have this exercise:

Let $f_n:\mathbb{R}\to\mathbb{R}, n\in\mathbb{N}$ a sequence of continuous functions that punctualy converge to $f:\mathbb{R}\to\mathbb{R}$.
I have to prove that if there exists a sequence in $\mathbb{R}$ ,$(r_n)$, such that $(r_n)\to r$ but $f_n(r_n)\not\to f(r)$ then $(f_n)$ does not converge uniformly to $f$ in $\mathbb{R}$.

Any hints would be great.


答:
The area of the whole triangle can be found using Heron's formula: the semiperimeter $s$ is $\frac{1}{2}(13+20+21) = 27$, and $A = \sqrt{27(27-13)(27-20)(27-21)} = 126$. The altitude of the triangle from the $21$ cm base is thus $\frac{126}{1/2 \cdot 21} = 12$.

Now rearrange the area of the triangle to form three smaller triangles with bases $13, 20$ and $21-2r$ and heights of $r$, and a right-angled trapezoid in the centre. This has area $\frac{1}{2}r(13+20+21-2r) + \frac{1}{2}(2r)(12+r)$.

Equating the two forms of the area, we have $r(54 - 2r) + 2r(12+r) = 252$, or $-2r^2+54r+24r+2r^2=252$. Hence $r = \boxed{\frac{42}{13}}$.


答:
I would suggest to show the equivalence starting with the expression on the right-hand side.

Using another standard notation with $+$ for $\vee$, $\cdot$ for $\wedge$, $0=F$, $1 = T$ and $p'$ for $\neg p$, you get

\begin{eqnarray*}(p\rightarrow q)(q\rightarrow p)
& = & (p'+q)(q'+p)\\
& \stackrel{distributivity}{=} &  p'q' + p'p+qq'+qp \\
& \stackrel{p'p=q'q=0}{=} &  p'q'+0 + qp \\
& \stackrel{a+0 = a}{=} & p'q'+pq\\
& = & p\leftrightarrow q
\end{eqnarray*}

Title: Difficulty with Equivalence Proof

Body:
I'm having difficulty proving that p ↔ q ≡ (p → q) ∧ (q → p). I've tried breaking it down from both sides, but nothing is clicking yet. So far, I believe my best attempt was with p ↔ q and goes something like this:

p ↔ q

(p ∧ q) ∨ (¬p ∧ ¬q) (Law of Equality)

(p ∧ q) ∨ ¬(p ∧ q) (DeMorgan 2)

¬(p ∧ q) ∨ (p ∧ q) (Symmetry)

(p ∨ q) → (p ∧ q) (Implication)

From here I'm stuck as to where I should go next or even if I'm in the right direction in the first place. Any help and advice is greatly appreciated!


答:
Here is an alternative way to solve this question without calculus--

As you have mentioned, $\frac{1}{2^n}$ is indeed a geometric series.

We know that $2n-1$ is an arithmatic series.

Proof--   $2n-1-(2(n-1)-1)=2$, which is a constant. hence we get first term $t_1=1$ and common difference $d=2$

As we have A.P. in the numerator and G.P in the denominator (r<1), it is an infinite convergent Arithmetic-(o)-geometric series, A.G.P, which (according to the question) converges.

The formula for sum of  infinite AGP  where r is less than unity ($r=\frac{1}{2}$)--
$t_0+(t_1)r+(t_2)r^2+...$ is

$S_{\infty}=\frac{t_0}{1-r}+\frac{\left(r\cdot d\right)}{\left(1-r\right)^2}$

but as we do not want $t_0$ in our required sum, can calculate $t_0=-1$, calculate the entire sum using the formula and add 1 to get the desired answer as 3


问:
Sorting balls into groups
Given 26 balls, 8 of them red and 18 of them blue, and 3 distinct, labeled groups to sort them in, how many different combinations can we make that satisfy the following conditions:

All balls are used

Each group has at least one of each color

There are strictly more blue than red balls in each group

So I think I've simplified the problem a bit by ensuring some of the conditions are satisfied by reducing the number of balls to work with to 5 reds and 12 blues (since there has to be at least 1 red and 2 blues in each group). The part that's tripping me up is how I can count all the combinations while ensuring there are more blues than reds.

答:
Remove the five remaining red balls and five blue balls and substitute purple balls for them.  Now use stars and bars twice to ascertain that the five purple balls and seven blue balls can be distributed in $\binom{7}{5} \times \binom{9}{7}=756$ ways.

These are in $1$-$1$ correspondence with acceptable distributions of red and blue balls, as you can see by splitting each purple ball into a red ball and a blue ball.

Title: Sorting balls into groups


问:
Antisymmetry about the points 1/2

The first (trigonometric) integral gives 0, because the integrand in the interval (0, 1) is antisym- metric about the point 1/2.

What does this exactly mean for the following equation?

Equation

答:
$$I = \int_0^1 \pi\sin^2(\pi t)\cos(\pi t)dt$$

Using the substitution $t = 1-u$, we have

$$I = \int_0^1\pi \sin^2(\pi-\pi u)\cos(\pi-\pi u)du = -I$$

$$\implies 2I=0 \implies I=0$$

Title: Antisymmetry about the points 1/2


问:
If a connected open set is evenly covered, then its preimage is uniquely partitioned into slices
This is from Topology by Munkres:

Let $p:E \to B$ be a covering map. Suppose $U$ is a open set of $B$ that is evenly covered by $p$. Show that if $U$ is connected, then the partition of $p^{-1}(U)$ into slices is unique.

What I've tried so far

I proved that if $\{V_\alpha\}$ is a slice then each $V_\alpha$ is connected, but I don't know what to do next.

答:
Show that the slices are the connected components of $p^{-1}(U)$. It follows immediately from this that the partition of $p^{-1}(U)$ into slices is unique.

(You should consider an example where $U$ is not connected, to see why the partition is not unique in that case, by the way...)


问:
Help with integral: Show that $\int_0^{\infty} e^{-y^2/4} |y| dy = 2$
According to Wolfram Alpha, https://www.wolframalpha.com/input/?i=%7Cy%7C+exp%28%7B-%5Cfrac%7B1%7D%7B4%7Dy%5E2%7D%29+,

$\int_0^{\infty} e^{-y^2/4} |y| dy = 2$.

Any ideas to show this?  Partial Integration is not possible because $e^{-y^2/4}$ has no primitive.

Is there any intelligent substitution? Please help.

After your advices:
Thanks to the substitution $g(v)= e^{-v/4}$, $v(y)=y^2$ and $\partial_y v(y)= 2y$ we get

$$\int_0^{\infty} e^{-y^2/4} y dy = \frac{1}{2}\int_0^{\infty} e^{-y^2/4} 2y dy = \frac{1}{2}\int_0^{\infty} g(v(y)) \partial_y v(y) dy = \frac{1}{2}\int_{v(0)}^{v(\infty)} g(t)  dt\\ = \frac{1}{2}\int_{0}^{\infty} e^{-t/4}  dt.$$

Now we can simply use the primitive $F(x)= -4e^{-x/4}$ of $f(x)= e^{-x/4}$ and get
$$\frac{1}{2}\int_{0}^{\infty} e^{-t/4}  dt= \frac{1}{2}(\lim_{b\rightarrow \infty}F(b)- F(0)) = \frac{1}{2}(\lim_{b\rightarrow \infty}-4e^{-b/4} - -4e^{-0/4}) = 2$$.

Thanks guys!

答:
Since the integral limits are $0 \to \infty$, we have $|y| = y$

$$I = \int_0^\infty \exp(-\frac{y^2}{4})ydy$$

Let $v = y^2$

$$\implies I = \frac{1}{2}\int_0^\infty \exp(-\frac{v}{4})dv$$

Can you proceed?


问:
MLE and Profile Likelihood for a Weibull-Like Distribution
Consider a random sample $X_1,...,X_n$ from a distribution with PDF $$f(x;\alpha,\beta)=\frac{\alpha}{\beta^\alpha}x^{\alpha-1}, 0\leq x\leq\beta$$ and parameters $\alpha,\beta>0$. Write down the likelihood and maximize it for any fixed $\alpha$ with respect to $\beta$. Write down the resulting profile likelihood.

Here is what I have so far, but the result of my computation leaves me uncertain:

Observe that $l(\alpha,\beta;x)=n\ln{\alpha}-n\alpha\ln{\beta}+(\alpha-1)\sum_{x=1}^n\ln{x_i}$.

It follows that $\frac{\partial{l}}{\partial{\beta}}=\frac{-n\alpha}{\beta}=0$.

Thus, for any fixed $\alpha$, $\hat{\beta}=0$ and $l^*(\alpha)=l(\alpha,\hat{\beta)}$ (but $ln{0}$ is undefined?).

答:
Let's calculate profile likelihood for $\beta$

$$L(\beta)\propto \frac{1}{\beta^{n \alpha}}\cdot\mathbb{1}_{[x_{(n)};+\infty)}(\beta)$$

It is evident that L is strictly decreasing thus

$$\hat{\beta}=x_{(n)}=max(\mathbf{x})$$

If you want to get the estimator of $\alpha$ too ( not requested) you can get the constraint log-likelihood

$$l(\alpha)=n \log\alpha- n \alpha\log x_{(n)}+(\alpha-1)\Sigma_i x_i$$

... and you can easy conclude


问:
Finding Mirror Images of Points on a Plane
I'm having trouble with finding the coordinate: "find the coordinates of the mirror image of $A(3, 1, 2)$ when reflected in the plane $x+2y+z=1$."

I've tried multiple methods such as assuming that it is perpendicular and then using the normal to work it out, or trying to find the vector equation of the line and then working it out that way, but I can't seem to get the answer of $(1, -3, 0)$.

Any help is greatly appreciated!! Thanks!!

答:
I'll tell you an easy way out

For any point like $A(3,1,2)$ taken as $(x_1,y_1,z_1)$
And plane like $x+2y+z-1=0$ taken as $ax +by+cz+d=0$
Its mirror image will be $$(x-x1)/a=(y-y1)/b=(z-z1)/c=-2(ax1+y1+cz1+d)/a^2+b^2+c^2$$
which in this case would be $$(x-3)/1=(y-1)/2=(z-2)/1=-2(1(3)+2(1) + 1(2) -1/[(1)^2+(2)^2+(1)^2]$$
Now you just have to solve for $x,y,z$. And there you go :)
(Applicable for every point)


问:
Eigenvector and eigenvalue for power of matrix, will eigenvectors remain the same?
I understand that if you raise a matrix $ A $ to let's say a power of 5, assuming that $ A$ is a 3 x 3 matrix and has 3 distinct eigenvalues, each of the eigenvalues will also be raised to the power of 5.

My question is, will the eigenvectors of the new $A^5$ still remain the same as A?

答:
Yes.  Say $\vec v$ is an eigenvector of $A$; i.e., $A\vec v=\lambda \vec v$.

Then $A^5\vec v=A^4 A\vec v=A^4 \lambda \vec v=\lambda A^4 \vec v=\lambda A^3 A\vec v=\lambda A^3 \lambda\vec v=\lambda^2 A^3\vec v$

$=\lambda^2A^2A\vec v=\lambda^2A^2\lambda\vec v=\lambda^3A^2\vec v=\lambda^3 A A\vec v=\lambda^3A \lambda\vec v=\lambda^4A\vec v=\lambda^5\vec v.$

So $\vec v$ is an eigenvector of $A^5$ with eigenvalue $\lambda^5$.


问:
Flipping a function across $y = -x$
Is there a general rule of flipping a function, for example $$f(x)=\frac{2x-3}{7x-5}$$ across the line $y = -x$ on the Cartesian plane? I’ve read some articles about it but it doesn’t very help because I’m ver poor at math. When you answer can you make it as simple as possible? Thanks!

答: When you flip across the line $y=-x$, the $y$ coordinate of all points obviously becomes $-x$. By rearranging the equation, we can see that the $x$ coordinate becomes $-y$. This means you can just do that substitution to get the new graph. For example, your $y=\frac{2x-3}{7x-5}$ becomes $-x=\frac{-2y-3}{-7y-5}$. Then, you can of course always simplify the equation.

Title: Flipping a function across $y = -x$


问:
Inverse of transformation matrix
I am preparing for a computer 3D graphics test and have a sample question which I am unable to solve.

The question is as follows:

For the following 3D transfromation matrix M, find its inverse. Note that M is a composite matrix built from fundamental geometric affine transformations only. Show the initial transformation sequence of M, invert it, and write down the final inverted matrix of M.

$M =\begin{pmatrix}0&0&1&5\\0&3&0&3\\-1&0&0&2\\0&0&0&1\end{pmatrix} $

I only know basic linear algebra and I don't think it is the purpose to just invert the matrix but to use the information in the question to solve this.

Can anyone help?

Thanks

答:
I know this is old, but the inverse of a transformation matrix is just the inverse of the matrix. For a transformation matrix $M$ which transforms some vector $\mathbf a$ to position $\mathbf v$, then to get a matrix which transforms some vector $\mathbf v$ to $\mathbf a$ we just multiply by $M^{-1}$

$M\cdot \mathbf a = \mathbf v \\
M^{-1} \cdot M \cdot \mathbf a = M^{-1} \cdot \mathbf v \\
\mathbf a = M^{-1} \cdot \mathbf v$

Title: Inverse of transformation matrix


问:
How to find a lower bound on a smallest solution of $\ln( \frac{x}{a} ) +\frac{1}{x}=0$.
The equaiton
\begin{align}
 \ln( \frac{x}{a} ) +\frac{1}{x}=0
\end{align}
has two solutions for a sufficiently large $a$ (e.g., $a>3$).
I want a lower bound on the smaller of the two solutions.

I have found an upper bound on the largest of the two solutions as follows:
\begin{align}
0= \ln( \frac{x}{a} ) +\frac{1}{x}\le  1- \frac{a}{x}  +\frac{1}{x}
\end{align}
where we have used a lower boung $\ln(x) \ge 1-\frac{1}{x}$.  This implies that $x_2<a-1$.

However, I have no idea how to find a lower bound on the smallest solution.

答:
The function$$f(x)=\log \left(\frac{x}{a}\right)+\frac{1}{x}$$ has two solutions as soon as $a >e$.

As said in comments and answers, the smallest solution is given by
$$x=-\frac{1}{W_{-1}\left(-\frac{1}{a}\right)}$$

For the secondary branch, there exist nice bounds
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) < -1 - \sqrt{2u} - \tfrac{2}{3}u$$ which make for your case
$$-\log (a)- \sqrt{2(\log (a)-1)}< W_{-1}\left(-\frac{1}{a}\right)$$
$$ W_{-1}\left(-\frac{1}{a}\right)< -\frac {2\log(a)+1}3 - \sqrt{2(\log (a)-1)}$$

For $a=e^2$,the smallest solution is then bounded by
$$1-\frac{1}{\sqrt{2}} \sim 0.292893 < x <\frac{3}{7} \left(5-3 \sqrt{2}\right)\sim 0.324583$$ while the exact solution is $\sim 0.317844$.

So, a first lower bound is
$$\color{red}{x_0=\frac 1 {A+ \sqrt{2(A-1)} }}\quad \text{where}\qquad \color{red}{A=\log(a)}$$

However, this can be improved a lot since $f(x_0) >0$ and $f''(x_0)>0$. So, by Darboux theorem, the solution will be reached without any overshoot of the solution and better bounds will be given by
$$\color{red}{x_{n+1}=x_n+\frac{x_n \left(x_n\log \left(\frac{x_n}{a}\right)+1\right)}{1-x_n}}$$ For example
$$\color{red}{x_1=\frac{\left(A+2  \sqrt{2(A-1)}-1\right)-\log \left(A+
   \sqrt{2(A-1)}\right)}{\left(A+ \sqrt{2(A-1)}-1\right) \left(A+   \sqrt{2(A-1)}\right)}}$$

For the working case $(a=e^2)$, the left bounds are sucessively
$(n=0,1,2)$
$$\{0.2928932188,0.3154911206,0.3178230036\}$$


问:
Is there any designer-friendly parametric curve, for which $\frac{\text{arclength}}{\text{distance}^2}$ can be calculated?
If I could find an $f(t)$ in the $xy$ plane

where $\displaystyle\int\frac{\sqrt{f'_x(t)^2+f'_y(t)^2}}{f_x(t)^2+f_y(t)^2}\, dt$ is known (basically $\displaystyle\frac{\text{arclength}}{\text{distance}^2}$ integrated)

I have tried to use both quadratic and cubic bezier curves, but the results got waay too messy for my realtime application. And I gave up when I found a full explanation of how to find the arc length of bezier curves (I found that it's quite difficult or impossible).

My idea now is that I could design my curve with some other kinds of parameters, so that my calculation will be possible. Does such kind of curve exist??

答:
Circular arcs. And you can string these together smoothly to get free-form “spline” curves. Look up biarc curves to find out more.
https://en.wikipedia.org/wiki/Biarc

Another option is a so-called Pythagorean hodograph curve. These are specifically designed to ensure that the square root term in your integrand simplifies nicely. Rida Farouki wrote an entire book about these creatures:
https://faculty.engineering.ucdavis.edu/farouki/wp-content/uploads/sites/41/2013/02/Introduction-to-PH-curves.pdf


问:
Vanishing of Dirichlet Series
Suppose the function
$\sum_{n=1}^{\infty}{a_{n}n^{-s}}$ is $0$ on some open set $U\subset\mathbb{C}$. (Can assume the sum converges absolutely on $U$.)

Is it true that $a_{n}=0$ for all $n$?

(This feels like it should be true, but we aren't sure how to see it. As a trivial observation, taking derivatives gives us $\sum_{n=1}^{\infty}{\ln(n^{-1})^{k}a_{n}n^{-s}}=0$ for all $k\in\mathbb{N}$)

答:
This is not a trivial problem at all. However, Apostol's Intoduction to Analytic Number Theory - Chapter 11 gives many interesting results.

Theorem(s). For Dirichlet series, $D$, there are numbers $a,c\in\mathbb{C}\cup\{\pm\infty\}$ such that:

$Re(c)\le Re(a)$.

$|Re(a)-Re(c)|\le 1$.

$D$ converges absolutely at every complex number to the right of $a$ (i.e. $Re(a)\le Re(z)$).

$D$ converges conditionally at every complex number to the right of $c$.

The convergences is uniform on every compact subset in the half plane to the right of $c$. Thus, $D$ is equal to some analytic function on the half plane to the right of $c$.

If $D(p_n)=0$, where $Re(p_n)\to+\infty$, then the coefficients of the Dirichlet series are the zero coefficients. That is, non-zero Dirichlet series must have some right half plane where the Dirichlet series is never zero.

These statements allow us to answer your question. We have a NO then YES.

NO we can't assume that if a Dirichlet series converges then the series converges absolutely. However, if it does converge to zero on some $U$, then the function $D$ must converge to the analytic zero function. But that means that $D$ agrees with zero on the whole right half plane to the right of $U$. But that means that the Dirichlet coefficients of $D$ are zeros. Such $D$ converges absolutely and uniformly in the whole complex plane. So, YES the convergence on $U$ will be absolute.

Title: Vanishing of Dirichlet Series


问:
What is the formula for E[Y|X=x] if you already have fx(x) and f(x,y)?
I just need the formula.  Thanks in advance.</p>

答:
The conditional density $f(y\mid x) = f(x,y)/f(x)$.

Then $\Bbb E[Y\mid X=x] = \int_\Bbb R y\,f(y\mid x)\,\mathrm dy$


问:
Proof that $\int_X K(\cdot,y)f(y) \, d\mu(y)$ is measurable if $K, f\in L^2$?
Let $(X,\mathscr A,\mu)$ be a $\sigma$-finite measure space, $K\in L^2(X^2,\mathscr A \otimes \mathscr A,\mu\otimes\mu),$ and $f\in L^2(X,\mathscr A,\mu).$

I am trying to prove that $F(x)=\int_X K(x,y)f(y) \, d\mu(y)$ is $\mathscr A$-measurable. The book I am reading claims it is, without proof.*

Here is my current progress:

If we can show that $\int_{X^2}|K(x,y)|\,|f(y)|\,d(\mu\otimes\mu)<\infty,$ then Fubini's theorem immediately guarantees that $F$ is measurable. Using Fubini's theorem on $K$ gives $|K(x,\cdot)|^2\in L^1$ a.e., which is equivalent to $K(x,\cdot)\in L^2$ a.e. So Cauchy-Schwartz gives:

$$\int_X |K(x,y)| |f(y)|\, d\mu(y)\leq \| K(x,\cdot)\|_2 \|f\|_2\qquad \text{a.e.}$$

In the comments, Kavi Rama Murthy suggests squaring both sides of this inequality and integrating in $x$. This gives

$$\int_X\left(\int_X |K(x,y)| |f(y)|\, d\mu(y)\right)^2 d\mu(x)\leq \| K(x,\cdot)\|_2^2 \|f\|_2^2.$$

So perhaps the left side here simplifies to the expression $\int_{X^2}|K||f|d(\mu\otimes\mu)?$

*V. Moretti, Spectral Theory and Quantum Mechanics. Example 4.16 (4).

答:
Some versions of Fubini's theorem include the following. Let $(X,\mathcal{F}_X,\mu)$ and $(Y,\mathcal{F}_Y,\nu)$ are $\sigma-$finite. If $g:X\times Y\to\mathbb{R}$ is $\mathcal{F}_X\times\mathcal{F}_Y-$measurable and $g \in L^1(\mu\times\nu)$, then
$$x\to \int_{Y}g(x,y)d\nu(y) $$
is not just $\mathcal{F}_X-$measurable, but is also a function in $L^1(\mu)$. In your example you have $g(x,y)=K(x,y)f(y)$.

I actually don't think that $g$ will necessarily be integrable with respect to the product measure, but I think we can get the result using $\sigma-$finiteness. Let $\{A_n\}$ be an increasing sequence of sets of finite measure which expand to $X$. Then consider the functions $g_n(x,y)=1_{A_n}(x)K(x,y)f(y)$ and call $F_n(x)=\int_X 1_{A_n}(x)K(x,y)f(y)d\mu(y)=F(x)1_{A_n}(x)$. Then for any $n$ we have that
\begin{align*}\int_X\int_X |g_n(x,y)|d\mu(y)d\mu(x) &=\int_X\int_X |1_{A_n}(x)K(x,y)f(y)|d\mu(y)d\mu(x)\\
&= \int_{A_n}\int_X |K(x,y)f(y)|d\mu(y)d\mu(x)\\
&\leq \int_{A_n}||K(x,\cdot)||_2 ||f||_2 d\mu(x)\\
&= ||f||_2\int_{A_n}||K(x,\cdot)||_2 d\mu(x)
\end{align*}
Now the function $x\to ||K(x,\cdot)||_2^2$ is measurable by Fubini's Theorem and the assumption that $K\in L^2(X^2)$. Since $z\to\sqrt{z}$ is a Borel function, the composition $x\to \sqrt{||K(x,\cdot)||_2^2}=||K(x,\cdot)||_2$ is measurable. Hence the set $B=\{x: ||K(x,\cdot)||_2\leq 1\}$ is measurable. Hence
\begin{align*}
\int_{A_n}||K(x,\cdot)||_2 d\mu(x) &= \int_{A_n\cap B}||K(x,\cdot)||_2 d\mu(x) + \int_{A_n\cap B^c}||K(x,\cdot)||_2 d\mu(x)\\
&\leq  \int_{A_n\cap B}1 d\mu(x) + \int_{A_n\cap B^c}||K(x,\cdot)||_2^2 d\mu(x)\\
&= \mu(A_n\cap B) + \int_{A_n\cap B^c}\int_X |K(x,y)|^2 d\mu(y)d\mu(x)\\
&\leq \mu(A_n\cap B) +||K||_2
\end{align*}
Now both terms on the right are finite since $\mu(A_n)<\infty$ and $K\in L^2$ by assumption. This then shows that $\int_X\int_X |g_n(x,y)|d\mu(y)d\mu(x) < \infty$ and by Fubini's Theorem we may conclude that $F_n(x)=F(x)1_{A_n}(x)$ is a measurable function. Well $F_n(x)\to F(x)$ almost everywhere, so $F$ is the pointwise limit a.e. of measurable functions, hence $F$ is a measurable function. Do note that $F$ may not be integrable.


问:
Show sequence of functions converges in metric space
For each positive integer $k$ define the function $f_k:[0,1] \rightarrow \mathbb{R}$ by $f_k(x)=\cos (\frac{x}{k})$ for $x$ in $[0,1]$. Show that the sequence of functions $\{f_k\}$ converges in the metric space $C([0,1],\mathbb{R})$.

To do this, would I just need to show that the sequence converges uniformly?It converges pointwise to $1$. Let $\epsilon>0$ Since $|\cos (\frac{x}{k})-1|\leq 1$ for all $k \in \mathbb{N}$ now I have to choose a $K$ in $\mathbb{N}$ such that $k \geq K \implies |\cos(\frac{x}{k})-1|< \epsilon$ for all $x \in [0,1]$. Now I am struggling to figure out what to do next, since I cannot just choose an $N$ that depends on $\epsilon$. Can I just choose $K>1$? Any help on how to proceed here?

Body: $|\cos y-1| =2 \sin^{2} (\frac y 2) \leq \frac {y^{2}} 2$.


问:
Probability: Conditioning equations
I saw in many different context that any probability result that is true for unconditional probability remains true if everything is conditioned on some event. For instance, consider the following equation:
$$P(B\cap C)=P(B)P(C|B).$$ Then by conditioning both sides on $A$, we get $$P(B\cap C|A)=P(B|A)P(C|B\cap A).$$
However, it is possible to find counterexamples as given here.

So I was wondering when this conditioning the both sides of equality operation is valid. I am, in particular, interested   in the following equality: $$P(X_4, X_3, X_2, X_1)=P(X_4 \mid X_3, X_2, X_1)\cdot \mathrm P(X_3 \mid X_2, X_1)\cdot \mathrm P(X_2 \mid X_1)\cdot \mathrm P(X_1).$$
Is it possible to condition both sides on $A$ and get $$P(X_4, X_3, X_2, X_1|A)=P(X_4 \mid X_3, X_2, X_1,A)\cdot \mathrm P(X_3 \mid X_2, X_1,A)\cdot \mathrm P(X_2 \mid X_1,A)\cdot \mathrm P(X_1|A)$$
here?

Many thanks in advance.

答:
If $P(A) \neq 0$ then we always have $P(B\cap C|A)=P(B|A)P(C|B\cap A)$.

Here is how to prove it.

$$ P(B \cap C \cap A) =P(B \cap C | A) P(A)  \tag{1}$$
\begin{align*} P(B \cap C \cap A) & = P(C \cap B \cap A) = \\& = P(C |B\cap A) P(B\cap A) = \\ & = P(C |B\cap A) P(B | A) P(A) \tag{2}
\end{align*}

From $(1)$ and $(2)$,

$$  P(B \cap C | A) P(A)  = P(C |B\cap A) P(B | A) P(A) $$

Since $P(A) \neq 0$, we have

$$  P(B \cap C | A)   = P(C |B\cap A) P(B | A) $$

So, if $P(A) \neq 0$, by repeatedly applying the equality we have just proved, we have

$$ \mathrm P(X_4, X_3, X_2, X_1|A)= \mathrm P(X_4 \mid X_3, X_2, X_1,A)\cdot  \mathrm P(X_3 \mid X_2, X_1,A)\cdot \mathrm P(X_2 \mid X_1,A)\cdot \mathrm P(X_1|A)$$

Title: Probability: Conditioning equations


问:
Putting balls into urns
We have two urns A and B. Urn A has 4 red balls and 3 blue balls, and urn B is empty. Randomly draw two balls $X_1$ and $X_2$ from urn A without replacement and place them into urn B. Then, draw two balls $Y_1$ and $Y_2$ from urn B with replacement.

Given that $Y_1$ and $Y_2$ are both blue, what is the probability that $X_1$ and $X_2$ are different colors?

So to start this problem, I labeled the event "$X_1$ and $X_2$ are different colors" as event A, and labeled the event "$Y_1$ and $Y_2$ are both blue" as event B. So what I want to find is $P(A \mid B)$.

Using the definition of conditional probability,  we have that $$P(A \mid B)= \frac{P(A \cap B)}{P(B)}$$ and $$P(A \cap B)=P(B \mid A)P(A)$$ I computed $P(B \mid A)=\frac{1}{4}$ and $P(A)=\frac{4}{7}$, giving me $P(A \cap B)=\frac{1}{7}.$

Then using the law of total probability, we have that $$P(B)=P(B \mid A)P(A)+P(B \mid A^c)P(A^c)$$ which is where I am running into trouble. How can I calculate $P(B \mid A^c)$, or in plain English, the probability that $Y_1$ and $Y_2$ are both blue given that $X_1$ and $X_2$ are the same?

答: There are $\binom42=6$ ways to draw two red balls from urn $1$ and $\binom32=3$ ways to draw two blue balls.  So the probability that both balls are blue, given that they are the same color, is $$\frac3{6+3}=\frac13$$

Title: Putting balls into urns


问:
prove the existence of $c\in \mathbb C$ such as $\mathcal F(g)=c\delta + {1\over2i\pi}V_p({\mathcal F(f)\over x}) $

prove that it exist $c\in \mathbb C$ such as $\mathcal F(g)=c\delta +
{1\over2i\pi}V_p({\mathcal F(f)\over x}) $

with $\mathcal F$ fourier transform , $f\in \mathbb D(\mathbb R)$ , $\delta$ dirac distribution , $g\in \mathbb D'(R)$ and :

$$ g(x)=\int_{-\infty}^xf(t)dt $$

$$ \left\langle V_p\left({u \over x}\right),\phi\right\rangle = \lim_{\epsilon\to0^+} \int_{|x|>\epsilon}{u(x)\phi(x)\over x} dx $$

i couldn't start anything here as i've never encountered this type of problems before , anything  will be appreciated !

答:
I may have some mistake in my work, but this is what apparently seems to appear, so you have to prove for any test function $\phi $

$$ \int_\mathbb{R}\hat{g}\phi dx = c\phi(0) + \lim_{\epsilon \rightarrow 0} \frac{1}{2\pi i}\int_{|x|>\epsilon} \frac{\hat{f}(x)\phi(x)}{x}dx  $$
Now $ g'(x) = f(x) $, so we have $\hat{f}(x) = 2\pi ix\hat{g}(x) $, substituting this we have
$$ \int_\mathbb{R}\hat{g}\phi dx = c\phi(0) + \lim_{\epsilon \rightarrow 0}  \int_{|x|>\epsilon}\hat{g}(x)\phi(x)dx  $$ This equivalently is
$$  \lim_{\epsilon \rightarrow 0}  \int_{|x|\leq\epsilon}\hat{g}(x)\phi(x)dx = c\phi(0) $$


问:
How to get collection of points inside polygon without holes?
I have 2D polygon (integer vertices) without holes. I would like to have a collection of all the points (integer coordinates) inside polygon. Can I do it faster that checking all points within polygon's boundary and applying https://math.stackexchange.com/a/3441442/725387 for each of them? Can I apply any precomputation on polygon?</p>

答:
Since you want to implement this on a computer, here's a very simple approach, in Mathematica:

Define the region:

```<code>RR = Polygon[{{0, 0}, {5, 1}, {7, 3},
           {6, 8}, {4, 9}, {1, 4},
           {0, 0}}];

Graphics[
     {FaceForm[None], EdgeForm[Blue], RR},
     Axes -> Automatic]
</code>```

Here are candidate points that span the region:

```<code>pts = Flatten[Table[{i, j}, {i, 0, 10}, {j, 0, 10}], 1];
</code>```

Select the points inside the region (including boundary):

```<code>insidepts= Select[pts, RegionMember[RR]]
</code>```

{{0, 0}, {1, 1}, {1, 2}, {1, 3}, {1, 4}, {2, 1}, {2, 2}, {2, 3}, {2,
4}, {2, 5}, {3, 1}, {3, 2}, {3, 3}, {3, 4}, {3, 5}, {3, 6}, {3,
7}, {4, 1}, {4, 2}, {4, 3}, {4, 4}, {4, 5}, {4, 6}, {4, 7}, {4,
8}, {4, 9}, {5, 1}, {5, 2}, {5, 3}, {5, 4}, {5, 5}, {5, 6}, {5,
7}, {5, 8}, {6, 2}, {6, 3}, {6, 4}, {6, 5}, {6, 6}, {6, 7}, {6,
8}, {7, 3}}

Show them:

```<code>Graphics[{Red, PointSize[0.03], Point[insidepts]}]
</code>```

Note that there is no restriction on the region being convex, or not having holes...  you just have to be careful defining such regions.

You seem concerned about timing, but this is blazingly fast on a Mac laptop..... 0.0032 seconds.  For the entire code on $10^6$ points:  5.52 seconds.

Body:
Since you want to implement this on a computer, here's a very simple approach, in Mathematica:

Define the region:

RR = Polygon[{{0, 0}, {5, 1}, {7, 3},
           {6, 8}, {4, 9}, {1, 4},
           {0, 0}}];

Graphics[
     {FaceForm[None], EdgeForm[Blue], RR},
     Axes -> Automatic]

Here are candidate points that span the region:

pts = Flatten[Table[{i, j}, {i, 0, 10}, {j, 0, 10}], 1];

Select the points inside the region (including boundary):

insidepts= Select[pts, RegionMember[RR]]

{{0, 0}, {1, 1}, {1, 2}, {1, 3}, {1, 4}, {2, 1}, {2, 2}, {2, 3}, {2,
4}, {2, 5}, {3, 1}, {3, 2}, {3, 3}, {3, 4}, {3, 5}, {3, 6}, {3,
7}, {4, 1}, {4, 2}, {4, 3}, {4, 4}, {4, 5}, {4, 6}, {4, 7}, {4,
8}, {4, 9}, {5, 1}, {5, 2}, {5, 3}, {5, 4}, {5, 5}, {5, 6}, {5,
7}, {5, 8}, {6, 2}, {6, 3}, {6, 4}, {6, 5}, {6, 6}, {6, 7}, {6,
8}, {7, 3}}

Show them:

Graphics[{Red, PointSize[0.03], Point[insidepts]}]

Note that there is no restriction on the region being convex, or not having holes...  you just have to be careful defining such regions.

You seem concerned about timing, but this is blazingly fast on a Mac laptop..... 0.0032 seconds.  For the entire code on $10^6$ points:  5.52 seconds.


问:
Splitting the dollar Nash equilibrium
I'm working on a game theory problem I can't seem to figure out.

Players 1 and 2 are bargaining over how to split $\$10$. Each player names an amount $s_i$, between 0 and 10 for herself. These numbers do not have to be in whole dollar units. The choices are made simultaneously. Each player's payoff is equal to her own money payoff. In all the cases below, if $s_1+s_2\leq 10$, then the players get the amounts they named (and the remainder is destroyed).

(a) In the first case, if $s_1+s_2 >10$, then both players get zero and the money is destroyed.

(b) In the second case, if $s_1+s_2 >10$ and the amounts named are different, then the person who names the smaller amount gets that amount and the other person gets the remaining money. If $s_1+s_2 >10$ and $s_1=s_2$, then both players get $\$5$.

(c) In the third case, the games (a) and (b) must be played such that only integer dollar amounts can be named by both players.

Determine the pure strategy Nash Equilibria for all games (a) - (c).

I'm pretty sure I've figured out (a), noting that if player 2 chooses a strategy such that $0 \leq s_2 < 10$, the best response of player 1 is $BR_1(s_2) = 10- s_2$. If $s_2 = 10$, then $BR_1(s_2) = [0,10]$ since the payoff is $0$ regardless. The same holds for player 2. The Nash Equilibria is the intersection of the BR lines, and occur at $(10,10)$ and the line $s_1+s_2=10$.

For (b), my thought is that if player 2 chooses $s_2 \leq 5$, then $BR_1(s_2) = 10-s_2$ as in (a). However, if $s_2 > 5$, I feel that $BR_1(s_2) = s_2 -\epsilon$ for some very small $\epsilon >0$. This way, the total amount will be over $\$10$, but player 1 will have the smaller amount and thus get his money. However, I'm not sure if this is right or how to find the Nash Equilibrium in this case. Any help would be greatly appreciated.

答:
So I'm working through Ben Polack's "Game Theory" course (free from Open Yale Courses, by the way, and highly recommended).  He has an identical problem in problem set 2, number 4.

(Disclaimer: I only have a BS in pure math, so I could be missing something here but I'll take a crack at this.)

As I understand Nash Equilibrium (NE), it's when each player's response is a best response to every other player's response.  More formally, for a strategy profile $(s_1^*, s_2^*,...,s_N^*)$, it is a NE if and only if for each player $i \in \{ 1, 2 , ... , N \}$ her choice $s_i^*$ is a best response to $s_{-i}^*$.  (Sidenote about notation: Polack uses $s_{-i}^*$ to denotes all strategies besides the strategy of player $i$.  I'm not sure how common this notation is, so I thought I'd point it out.)

Since we're only dealing with two players, we only need player 1's strategies that are best responses to player 2's best responses.

Unfortunately, I have a slight nit-pick on the OP's 4(a) answer, so I'll address that first.

4(a): It is true that the line $s_1 + s_2 = 10$ can give NE's.  But I think it's more precise to say that they're strategy profiles of the form $(s_1, 10-s_1)$ or equivalently $(10-s_2, s_2)$ for $s_1,s_2 \in [0,10)$.  It's also true that if $s_2 = 10$ then $BR_1(s_2)=x$ where $x \in [0,10]$.  By symmetry, if $s_1 = 10$ the $BR_2(s_1)=y$ where $y \in [0,10]$. So, like the OP said, there's another NE for the strategy profile $(10, 10)$.

4(b): joriki makes some really good points on this in the continuous case, so read his answer, too. I'll address this in the discrete case (specifically if we're dealing with cents).  It's not specified in the problem set if fractional cents are allowed, but I'll go with the assumption that there are not fractional cents.

There are two cases to think about:

(1) $s_1 + s_2 \leq 10$

(2) $s_1 + s_2 > 10$

Like joriki argues, any choices by player 1 where $s_1 < 5$ are strictly dominated by $s_1'=5$. Why?  Because in case (1), the payoff of player 1 is $u_1(5,s_2) = \$5$ since both players receive a payoff of whatever number they choose. In case (2), $s_2 > 5$ so player one gets a payoff of the lower number.  Thus, it's still true that $u_1(5,s_2)= \$5$.

But what about $s_1 > 5$ in the discrete case using cents?  Well, this is were things become a bit different.  You end up with NE's for the strategy profiles $(5,5)$, $(5,5.01)$, and $(5.01,5)$, and $(5.01,5.01)$.

It might be easier see this if we start with a case where there isn't a best response from both players, so let's consider when player 2 chooses $s_2=5.02$. So what's player 1's best response? To slightly undercut player 2's choice by a penny (that devious devil).  So $BR_1(5.02)=5.01$ which yields $u_1(5.01,5.02)= \$5.01$ and $u_2(5.01,5.02)=\$4.99$.

So what would be player 2's best response to $s_1=5.01$? A couple options, actually.

Player 2 could pick $s_2=5.01$ and they'd both get a payoff of $u_1=\$5$ and $u_2=\$5$ since they both get a payoff of \$5 when $s_1+s_2>10$ and $s_1=s_2$.

Another choice player 2 could make is $s_2=5.00$ (which would still maximize $u_2$ against $s_1=5.01$).  Why? Because $u_2=\$5$ since it's the smaller number and the player who chooses the smaller value gets that amount of money as their payoff.

Hence, player 2 has two best responses for $s_1=5.01$. The argument is the same for player one, so we have NE's $(5,5.01)$, $(5.01,5)$, and $(5.01,5.01)$.

Lastly, similar to joriki's argument, if $s_1=5$ then the way for player 2 to maximize their utility function (that is, their best response) is to pick 5 or higher.  The argument is symmetric for player 1, and the only overlapping best responses are when $s_1=5$ and $s_2=5$, therefore the strategy profile $(5,5)$ is also a NE.

Hope that's helpful!

Title: Splitting the dollar Nash equilibrium


问:
Example of Non-Measurable Bijection from $\mathbb R \to \mathbb R$
The title is fairly self explanatory. I am looking for an example of a function from $\mathbb R\to\mathbb R$ that is bijective, but not measurable with respect to Lebesgue measure, if such an example exists.

答: Take a none measurable subset $X\subset [0, 1]$ whose cardinal is the same as that of $[0, 1]$ - for example let $X$ be a non-measurable subset in $[0, \frac 12]$ union with $[\frac 12, 1]$. Let $f$ be a bijection from $X$ to $[0, 1]$. Now the cardinal of ${\mathbb R}-X$ equals to the cardinal of ${\mathbb R}-[0, 1]$ (both equals the cardinal of ${\mathbb R}$ since both contains an interval), so we can take $f: {\mathbb R}-X \to {\mathbb R}-[0, 1]$ to be a bijection. Fitting together we get the bijection $f:{\mathbb R}\to{\mathbb R}$ but $f^{-1}([0, 1])$ is not measurable.


问:
Sample mean distribution for Gamma random variables
I am curious as to what the sample mean distribution for a set of iid Gamma random variables is.

I know $\bar{X}$ has a Gamma distribution when $X_1, X_2 ... X_n$ are iid ~ exponential with mean $\theta$ and gamma distributed random variables $X_1 + X_2$ also follow a gamma distribution with parameters $(\alpha_1 + \alpha_2, \theta)$, for $X_1$ ~ $(\alpha_1, \theta)$ and $X_1$ ~ $(\alpha_1, \theta)$, but I can't find any information regarding the distribution of the sample mean for a set of gamma distributed random variables.

As a sidenote, when can we approximate $\bar{X}$ with a normal distribution? Ex: when can we say that the distribution of $\bar{X}$ for $X_1, X_2 ... X_n$ iid ~ exponential$(\theta)$ is approximately normal with mean $\theta$ and variance $\frac{\theta}{n}$, instead of saying it follows a gamma distribution?

Thank you for your answers in advance.

答:
We can in fact relax the requirement that the sample be identically distributed:  suppose for each $i = 1, 2, \ldots, n$, we have $$X_i \sim \operatorname{Gamma}(n_i, \theta),$$ where each $n_i$ is a positive integer and $\theta$ is a common scale parameter.  Then $$S = \sum_{i=1}^n X_i \sim \operatorname{Gamma}\left(\sum_{i=1}^n n_i, \theta\right).$$  This is because each $X_i$ is the sum of $n_i$ iid exponential variables with mean $\theta$; e.g., $$X_i = Y_{i1} + Y_{i2} + \cdots + Y_{in_i},$$ thus the total sum is $$S = \sum_{i=1}^n \sum_{j=1}^{n_i} Y_{ij}.$$
In fact, we do not even require that the $n_i$ are positive integers; they may be arbitrary positive reals, and the result still holds.  This can be shown by considering the moment-generating function of a gamma distribution.  From here, the sample mean $\bar X = S/n$ is also gamma distributed, with shape $\sum n_i$ and scale $\theta/n$.


问:
Need to know what this theorem is asking
Theorem
Let (Y,$\tau_1$) be a a Subspace of (X,$\tau$)
If (X $\tau$) satisfies Second Countable axiom
Show  (Y,$\tau_1$) does too

My question is this : is this theorem asking us
to prove the hereditary of the Second Countable axiom?

答:
It is indeed asking to show that the property “$X$ is second countable “ is hereditary with respect to all subspaces.

Not that it helps with the proof.


问:
Finding the area of the 4th triangle, given the areas of the other 3, and all the 4 form a rectangle
In one of my tutorial classes, when I was studdying in 9th class (I am in 10th now), our tutor gave us a problem saying it’s a difficult one, and to him, it was incomplete.

This is that problem:

In a rectangle ABCD, point B is joined to the points P and Q on the sides AD and DC respectively. And P and Q are joined to form three distinct triangle; namely PDQ or x, QCB or y, BAP or z and BPQ or f.

Given the area of x = 3m^2, y = 4m^2 and z = 5m^2, find the area of f.

Note: all the information regarding this problem has been provided. The dimensions of any of the line segments are not known. This also means that the exact positions of  P and Q  on their respective line segments are also not known.

Because of the lack of information (See Note), my tutor declared it as an incomplete problem (he is just a high-school Science and Mathematics teacher, please don’t expect him to think beyond the limits of a typical high-school student, that is, in case he is wrong).

After a few hours of brainstorming, I came up with what I called an ingenious method. But then I felt that my method actually was not accurate.

So, can anybody tell me if this problem is complete and solvable, if yes, is my method correct (see below), if no, what is the best/better method to solve it.

MY METHOD:

As all the triangles with their areas given are on the edge of the rectangle, we know that they all are right triangles. So we can say that one of the arms of the right angle is the altitude and the other is the base.

Now, we can use the formula (1/2) * base * altitude to calculate their areas. Lets take a as altitude and b as base.

(Here’s the first part of the trick) We now try all possible _a_s and _b_s for the three triangles, where a and b are natural numbers only. (This is also where the major weakness of my method is. More numbers, i.e., any real numbers should have been tried, but that is infeasible). Note that all possible values are those values that form a triangle of that specified area.

For triangle x, we have the following dimensions:
  -a = 1, b = 6
  -a = 2, b = 3
  -a = 3, b = 2
  -a = 6, b = 1 #

For triangle y, we have the following dimensions:
  -a = 1, b = 8
  -a = 2, b = 4
  -a = 4, b = 2#
  -a = 8, b = 1

For triangle z, we have the following dimensions:
  -a = 1, b = 10
  -a = 2, b = 5
  -a = 5, b = 2
  -a = 10, b = 1#

(Here’s the other part of the trick) Then, we select those specific dimensions of triangles, which when placed together form a rectangle.

The condition we require:
  When the triangles are placed together to form a quadrilateral the opposite sides of it must be equal.

If this quadrilateral’s opposite sides are equal, we know that it is a parallelogram. And as all of the triangles that form the edges are right angled, we have at least one of the angles of the parallelogram right angled_, so we know that this parallelogram is a rectangle.

The correct set of dimensions (that we will be using here) are high-lighted in bold, while the once suffixed with a ‘#’ can also be used.

So, in triangle x we take PD as altitude and DQ as base.
  In triangle y we take BC as altitude and CQ as base.
  In triangle z we take PA as altitude and AB as base.

Henceforth, we now have:
  (PA + PD) = BC
  (DQ + CQ) = AB
  Therefore, the condition is matched.

WE NOW HAVE THE CORRECT DIMENSIONS OF THE THREE TRIANGLES

Now we can either find the area of f by finding the area of the rectangle and subtracting it by the areas of the three triangles, or by finding the three triangle’s hypotenuse using the Pythagorean formula and then apply the _Herron’s formula _on the hypotenuses. I prefer the former.

>After calculating, we have, _f = 8m^2_

Thank you =)

答:
Let $P = (0, 2), Q = (3, 0)$ and $B = (a,b)$ to force $x = 3$. Using the shoelace formula, the area of $\Delta PQB$ is $\frac{1}{2}(3b+2a-6)$, which is equal to the area of the rectangle minus the three smaller triangles, or $ab - (3+4+5)$.

Thus $3b+2a-6 = 2ab - 24$, or $2ab - 2a - 3b - 18 = 0$. This factors as $(2a-3)(b-1) = 21$, and as $B$ must have a larger $x$-coordinate than $Q$, and a larger $y$-coordinate than $P$, the only option is the pair $(7, 3)$ which gives $a = 5, b = 4$. Thus the area is $\frac{1}{2}(3b+2a-6) = \boxed{8}$.


问:
What do we call the angular arcs between two edges of triangles?
I've been trying to find a geometry library for java which is as high level as describing angles between adjacent sides of triangles given 3 sides. So, what do we call such kind of arcs. In many geometry books we see arcs describing the angle between edges of triangle and squares for right angles. What are the formal terms for them. Because, I failed to search google successfully due to the insufficiency of vocabulary lately.
I need the terminology to search for java library.
I don't know if my question fits here but any body used such libraries in java here, perhaps some of you might have involved in math projects.
Picture depicts what I wanted...

Any help would be appreciated...

答:
Referring to a software, Geometer's Sketchpad, it may be called angle marker.

However, I'm not sure if there's any formal terminology for this.


问:
Exercise 1.5.7 (b) in Abbott’s Understanding Analysis
In the second edition, this exercise is as follows:

Find a 1-1 mapping from $S$, the open unit square $\{(x,y) : 0<x,y<1\}$, to the open interval $(0,1)$. Use the fact that each real number has a decimal expansion.

The answer I’ve been seeing the most, and is also the one I thought of, was to consider the decimal expansions of $x$ and $y$, $0.x_1x_2x_3...$ and $0.y_1y_2y_3...$, respectively. Map $(x,y)$ to the number in $(0,1)$ of the form $0.x_1y_1x_2y_2...$

Since numbers of the form $0.a_1a_2...a_n$ are equal to $0.a_1a_2...a_{n-1}(a_n-1)99999...$, I noticed that this creates the following issue:

Suppose I have $(.2,.7)$ which is equal to the pair $(.1999...,.6999...)$. Upon applying the previous mapping to both, they map to $.27$ and $.16999...$, respectively. Since $.16999...$ is the same as $.17$, this means the ordered pairs $(.1,.7)$ and $(.2,.7)$ map to the same element of $(0,1)$. So it isn’t 1-1? Is this problem avoided by just referring to $x$ and $y$‘s finite decimal expansion if it exists?

答: You are correct. He gives this as the solution in the solution manual, but it isn't right. Even his own example to "keep in mind" does not pass. You can fix this by choosing the map $0.x_1 y_1 0 x_2 y_2 0 x_3 y_3 0 ...$ such that you will never get an infinite sequence of 9's and therefore no room to play.


问:
How to solve an initial value problem?
For this question, I'm having trouble solving this equation. I'm not sure how to proceed further with the substitution. Here is what I have so far. Can anyone please help me out?

For this question, I'm having trouble solving this initial-value problem. Here is what I have so far. Can anyone please help me out?

$$
\frac{du}{dt} = \frac{{u}^{3/2} + \arctan(t)+2u}{1+t^2}, u(0)=4\\
\frac{du}{dt} = u^{3/2} (\frac{\arctan(t) + 2u^{-1/2}}{1+t^2})
$$
Let $x = u^{-1/2}\,dx = \frac{-1}{2}u^{-3/2} \,du$

答:
The differential equation being
$$u'=\frac{u^{3/2}+2 u+\tan ^{-1}(t)}{1+t^2}$$ consider first the homogeneous equation
$$u'=\frac{u^{3/2}+2 u}{1+t^2}$$ which is separable and leads to
$$\log \left(\frac{\sqrt{u}}{2+\sqrt{u}}\right)=\tan ^{-1}(t)+C$$ There are potentially two (nasty) solutions for $u$.

Now, I suppose that you need to define the integration factor.

Title: How to solve an initial value problem?


问:
Jacobson Radical of integer matrices
Let $R = M_2(\mathbb Z)$

Find $J(R)$

The reason I'm asking is that I know by Artin Wedderburn that $M_n(D)$ is semisimple for any division ring D and hence $J(D)=0$. But here $\mathbb Z$ is of course not a division ring.

答:
For definiteness, let's consider left $R$-modules. First choose a prime $p\in\mathbb{Z}$, and consider the set $A_p=\left(\mathbb{Z}\big/p\mathbb{Z}\right)^2$ as a space of $2\times1$ column vectors. A matrix $(m_{ij})_{ij}\in R$ acts on an element $(\overline{a},\overline{c})\in A_p$ by sending it to $(\overline{m_{11}a}+\overline{m_{12}c},
\overline{m_{21}a}+\overline{m_{22}c})$.

Exercise: Show that $A_p$ is simple. $\square$

Thus $J(R)\subseteq\operatorname{ann}A_p$. What form does this latter ideal take? Well, if $(m_{ij})_{ij}\in\operatorname{ann}A_p$, then in particular it sends $(\overline{1},\overline{0})$ and $(\overline{0},\overline{1})$ to $(\overline{0},\overline{0})$. Thus we must have $\overline{m_{ij}}=\overline{0}$ for each $i$ and $j$, and hence $m_{ij}\in p\mathbb{Z}$ for each $i$ and $j$. Conversely, if an element of $R$ is of that form (ie has all entries in $p\mathbb{Z}$), then it annihilates $A_p$. (Why?) So the annihilator of $A_p$ is precisely the left ideal $M_2(p\mathbb{Z})$. Now, we know that $$J(\mathbb{Z})=\bigcap_{p\text{ prime}}p\mathbb{Z}=\{0\}.$$ What can we thus conclude about $J(R)$, which is a subset of $\bigcap_{p\text{ prime}}\operatorname{ann} A_p$?

Title: Jacobson Radical of integer matrices


问:
A simple riddle related to addition of odd numbers
I'm not sure if this type of question can be asked here, but if it can then here goes:

Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.

PS : The inception of this question is a result of a random discussion that I was having during the break hour :)

答:
A direct approach:

Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like
$${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$
(the integers $a,b,\ldots,i$ may or may not be the same).
Grouping things together, this is equal to
$$2(a+b+c+d+e+f+g+h+i+4)+1.$$
Thus, the result is odd.

An simpler approach would be to prove these three simple facts:
$$\begin{align*}
\mathsf{odd}+\mathsf{odd}&=\mathsf{even}\\
\mathsf{odd}+\mathsf{even}&=\mathsf{odd}\\
\mathsf{even}+\mathsf{even}&=\mathsf{even}
\end{align*}$$
Thus, starting from
$$\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}$$
and grouping into pairs,
$$\mathsf{odd}+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})$$
we use our facts to see that this is
$$\mathsf{odd}+\mathsf{even}+\mathsf{even}+\mathsf{even}+\mathsf{even}.$$
Grouping again,
$$\mathsf{odd}+(\mathsf{even}+\mathsf{even})+(\mathsf{even}+\mathsf{even})$$
becomes
$$\mathsf{odd}+\mathsf{even}+\mathsf{even}$$
becomes
$$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}=
\mathsf{odd}$$


问:
Conditional Probability vs Dependent Events
I'm conceptually/intuitively confused about when we would use the joint probability (multiplication rule) versus a conditional probability, specifically when the dependent event has no chance of occurring without the other event.

Here's an example that illustrates my confusion:

Suppose we have a weird city, where there's a 20 percent chance of rain (so a 80% chance of any other weather). Now suppose that it can only rain frogs when it rains, and there's a 10% chance of frogs.

Why can we not use bayes rule to model:

$$Pr(frogs | rain) = \frac{Pr(rain | frogs)\cdot Pr(frogs)}{Pr(rain)} = \frac{1 \cdot 0.1}{0.2}$$

I know the correct way of doing it would be the multiplication rule, but why can we not use a conditional probability? We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains, and if there are raining frogs, we must have rain so $Pr(rain | frogs)=1$. And we also have $Pr(rain)=0.2$, so it seems like we have everything we need for conditional probability.

The correct answer would of course be $0.2\cdot0.1=0.02$.

What am I missing aside from the nonsensical results?

Edited for sensical numbers, but assume that the probability of raining frogs is just the proportion of days on which it rains.

答:
"We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains"

This is false. $P(frogs|rain)=.1$ but $P(frogs)\ne.1$!

So it is incorrect to plug it into Bayes' Theorem as you did.

"The correct answer would of course be 0.2⋅0.1=0.02."

You found P(frog), not P(frog|rain). Indeed,

$\begin{split}P(frog)&=P(frog|no rain)P(no rain)+P(frog|rain)P(rain)\\
&=0+.1*.2\\
&=.02\end{split}$

Conclusion

You switched P(frogs|rain) and P(frogs) around. One is the conditional probability of frogs given that it rains and one is the overall probability that it rains frogs.


问:
Find the $f^{(n)}(0)$ of $\arctan(x)$.
I wish to solve this by using the method as prescribed here:

https://diendantoanhoc.net/topic/192848-lim-sum-frac2n-12n/?p=742824

$$\arctan(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nx^{2n+1}}{2n+1}$$

The general formula of Maclaurin series is

$$\displaystyle\sum_{n=0}^{+\infty} \dfrac{f^{(n)}(0)}{n!}x^n$$

Since both series are equal, their coefficients ought to equal,

So I have

$$\dfrac{f^{(n)}(0)}{n!}=\dfrac{(-1)^nx^{2n+1}}{2n+1}$$

$$f^{(n)}(0)=\left[\dfrac{(-1)^nx^{2n+1}}{2n+1}\right]n!$$

Something is wrong, the n! shouldn't be there.

Reference from:

https://socratic.org/questions/how-do-you-find-f-6-0-where-f-x-arctanx-x

答:
In fact,
$$\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1} x^{2n+1} = x-\frac13x^3+\frac15x^5-\frac17x^7+\cdots = \sum_{m=0}^\infty a_mx^m$$ where $$a_m = \begin{cases} 0 & \textrm{if $m$ is even,} \\[2mm] \dfrac1m & \textrm{if $m \equiv 1 \!\pmod{\!4}$,} \\ -\dfrac1m & \textrm{if $m \equiv 3 \!\pmod{\!4}.$}\end{cases}$$

Title: Find the $f^{(n)}(0)$ of $\arctan(x)$.


答:
You have your definitions all mixed up. You have already stated that $P(\text{frogs}|\text{rain})=0.1$, so you don't need Bayes Theorem to prove it.

What you need is the law of total probability, which states that:

$$P(\text{frogs})=P(\text{frogs}|\text{rain})P(\text{rain}) + P(\text{frogs}|\text{no rain})P(\text{no rain})$$

which gives

$$P(\text{frogs})=(0.1\times0.2) + (0.0\times0.8)=0.02$$

Title: Conditional Probability vs Dependent Events

Body:
I'm conceptually/intuitively confused about when we would use the joint probability (multiplication rule) versus a conditional probability, specifically when the dependent event has no chance of occurring without the other event.

Here's an example that illustrates my confusion:

Suppose we have a weird city, where there's a 20 percent chance of rain (so a 80% chance of any other weather). Now suppose that it can only rain frogs when it rains, and there's a 10% chance of frogs.

Why can we not use bayes rule to model:

$$Pr(frogs | rain) = \frac{Pr(rain | frogs)\cdot Pr(frogs)}{Pr(rain)} = \frac{1 \cdot 0.1}{0.2}$$

I know the correct way of doing it would be the multiplication rule, but why can we not use a conditional probability? We know that the probability of raining frogs is 0.1 because it cannot occur in any other scenario other than when it rains, and if there are raining frogs, we must have rain so $Pr(rain | frogs)=1$. And we also have $Pr(rain)=0.2$, so it seems like we have everything we need for conditional probability.

The correct answer would of course be $0.2\cdot0.1=0.02$.

What am I missing aside from the nonsensical results?

Edited for sensical numbers, but assume that the probability of raining frogs is just the proportion of days on which it rains.


问:
Prove that the sequence is a Cauchy Sequence
The Sequence ${a_n}$ satisfies the following condition ::

$|a_{n+1} - a_n|$ $\leq$ $\alpha$$|a_n-a_{n-1}|$,   $0<\alpha <1 $ for all natural numbers $n\geq2$

I approached the question like this ::

In a field of real numbers, A sequence is convergent iff it's cauchy. So I tried to prove that the sequence is convergent using monotone convergence theorem. Using Induction, I first proved that the sequence is monotonous. And again using induction, trying to prove that it is bounded, but I am stuck here.

Apart form this, I think applying Cauchy's general principle of convergence to prove this would be a more neater technique.
Cauchy's General principle of Convergence states that the necessary and sufficient condition for convergence of a sequence ${S_n}$ is that for each $\epsilon$, there exists an $m$ such that

$|S_{n+p}-S_n| < \epsilon$, $\forall n>m$ and   $p\geq1$

Please let me know how this can be approached by either of the ways. Thanks.

答:
The argument is telescoping in nature as evident by what is given. See below a sketch. Also, recall that $\alpha^n \to 0$ when $0<\alpha < 1$

Suppose $m > n $, then

$$ |a_m - a_n| = | a_m - a_{m-1} + a_{m-1} - a_{m-2} + a_{m-2} - ... + a_{n+1} - a_n| \leq |a_m - a_{m-1}| + |a_{m-1} - a_{m-2}| + ... + |a_{n+1} -a_n| \leq \alpha^m |a_3-a_2| + \alpha^{m-1} |a_3-a_2| + ... + \alpha^n |a_3 - a_2| = \sum_{k=n}^m \alpha^k |a_3-a_2| $$


答:
Here’s a tip for the special case of a triangle ( e.g., as obtained from a triangulation of the original polygon:
Given vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ find a matrix $A\in SL_2(\Bbb Z)$ that maps the $(x_i,y_i)-(x_1,y_1)$ to $(0,0),(a,0),(0,b)$. Here the enumeration is very straightforward and can be mapped back via $A^{-1}$

Title: How to get collection of points inside polygon without holes?

Body: I have 2D polygon (integer vertices) without holes. I would like to have a collection of all the points (integer coordinates) inside polygon. Can I do it faster that checking all points within polygon's boundary and applying https://math.stackexchange.com/a/3441442/725387 for each of them? Can I apply any precomputation on polygon?</p>


Title: A simple riddle related to addition of odd numbers

Body:
I'm not sure if this type of question can be asked here, but if it can then here goes:

Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.

PS : The inception of this question is a result of a random discussion that I was having during the break hour :)

Body:
$$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$

$$41=2\sum_{k=1}^9n_k$$

There is no integer $\sum_{k=1}^9n_k$ that satisfies this.


问:
Factoring $a^4+b^4+(a-b)^4$
I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.

I rewrite it as:

$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.

Title: Factoring $a^4+b^4+(a-b)^4$

Body: Hint: $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$


问:
Rectangle in polar coordinates

Suppose that we have $D=[-a,a]\times [-b,b]\subseteq \mathbb{R}^{2}$. How can I transform that region into a new region described by polar coordinates?

If we start by making a graph, we can see that the graph will be a rectangular region that can be partitioned by the diagonals of the rectangle into 4 isosceles triangles. So the region in polar can be written as 4 regions in polar coordinates and one of them is of the form $$D_{1}=\{(r,\theta): \theta\in [-\arctan(b/a),+\arctan(b/a)]; r\in [0,a/\cos(\theta)] =D_{3}$$ $$D_{2}=\{(r,\theta): \theta\in [-\arctan(a/b),+\arctan(a/b)]; r\in [0,b/\cos(\theta)]=D_{4}$$

Is it correct? Would the other regions have a similar scheme or should I change the approach?

答:
We always express the range of $\theta$ with respect to positive $x$ axis. In $D_2$, note that the angle $\beta=2\arctan(a/b)$ is distributed symmetrically about the positive $y$ axis, i.e. $\theta=\pi/2$ with respect to positive $x$ axis. So the range of $\theta$, would be $[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]$ with respect to $x$ axis.

For the range of $r$, note that $y=b\implies r\sin\theta=b$ giving $r\in[0,b/\sin\theta]$.

$$D_2=\{(r,\theta):r\in[0,b/\sin\theta],\theta\in[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]\}$$

Further note that due to the symmetricity of the region, a simple tweak in the equation for $D_2$ would get you $D_2\cup D_4$ if negative values of $r$ are allowed.$$D_2\cup D_4=\{(r,\theta):-b/\sin\theta\le r\le b/\sin\theta, \theta\in[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]\}$$and similarly you can tweak $D_1$ to include $D_3$.

Title: Rectangle in polar coordinates


问:
Prove $\lim\limits_{n\to\infty}(\frac{a}{n})=0$
I'm very new to the epsilon definition of a limit and I was hoping I could get some feedback to see if I have the right idea.

For any $a\in\mathbb{R}$, $$\lim\limits_{n\to\infty}\left(\frac{a}{n}\right)=0$$

Proof:

Let $\epsilon >0$. Choose $N>\frac{|a|}{\epsilon}$. Then for all $n\geq N$, since $n\geq N>\frac{|a|}{\epsilon}$, we have

$$n>\frac{|a|}{\epsilon}\;\;\;\Rightarrow\;\;\;\epsilon >\frac{|a|}{n}$$

Therefore,

$$\left|\frac{a}{n}-0\right|=\frac{|a|}{n}<\epsilon$$

$\blacksquare$

答:
Your answer is $100\%$ correct. One can generalize your theorem for a sequence $\displaystyle\sigma_{n}=\frac{\alpha}{n^{r}}$ with $\alpha\in\mathbb{R}$ and $r\in\mathbb{R}^{+}$ as follow :

Let $\varepsilon>0$, then by the archimedean property, $\exists N\in\mathbb{N}$ such that $\displaystyle\frac{\alpha}{N}<\sqrt[r]{\varepsilon}$. Now we show that starting this index $N$, elements of $\sigma_{n}$ are in the neighborhood of $0$. For $n\geq N$, we have that :
$$
\left|\frac{\alpha}{n^{r}}\right|=\frac{|\alpha|}{n^{r}}\leq\frac{|\alpha|}{N^{r}}<\varepsilon
$$
which means that $|\sigma_{n}|<\varepsilon\implies\displaystyle\lim_{n\to\infty}\sigma_{n}=0$


问:
Solve one nonlinear ODE
Can we find an explicit solution to the following nonlinear ODE
$2yy^{\prime\prime}+(y^\prime)^2+1=0$ ?

答:
For simplicity, let us consider $y(0) = 1$ and $y'(0) =1$. The general idea is the same.

Observe we have that

\begin{align}
0=y'(2yy''+(y')^2+1) = 2yy'y''+(y')^3+y' = (y(y')^2+y)'
\end{align}
then it follows that
\begin{align}
y((y')^2+1) = 2 \ \ \implies \ \ y'= \sqrt{\frac{2-y}{y}}.
\end{align}

Then you solve this by separating the variables.

Title: Solve one nonlinear ODE


问:
Divisibility of combinatorial number with Lucas' Theorem
For how many non-negative integers n from 0 to 300 (inclusive), the combinatorial number ${300\choose n}$ is not divisible by 5?

I know how to use Lucas' Theorem to solve easy questions like finding remainder of ${100\choose 18}$ divided by 13, but can't figure out this one.

答:
By Lucas's theorem, if we write $n=125n_3+25n_2+5n_1+n_0$, then $\binom{300}n$ is divisible by $5$ if and only if $\binom{2}{n_3}\cdot\binom{2}{n_2}\cdot\binom0{n_1}\cdot\binom0{n_0}$ is. Note that this can only be divisible by $5$ if the product is actually $0$.

In other words, $\binom{300}n$ is not divisible by $5$ exactly when $n_2,n_3\le2$ and $n_1=n_0=0$. This gives us $3\cdot3\cdot1\cdot1=9$ options.

In particular, we have the following list of $n$ for which $\binom{300}n$ isn't divisible by $5$:
$$0,25,50,125,150,175,250,275,~\textrm{and}~300.$$

In general, if we replace $300$ with $m$ and $5$ with $p$, then the answer is exactly $\prod(m_i+1)$, where $m_i$ are the unique nonnegative integers with $m=\sum m_ip^i$.


问:
Matrix exponential for non-commutative operator entries of matrix
I would like to find the matrix exponential $e^{iHt}$ of the Hermitian matrix $H$ where

$$
H=\begin{pmatrix}
\delta& \sqrt{2}a & 0\\
\sqrt{2}a^\dagger &0& \sqrt{2}a\\
0 &\sqrt{2}a^\dagger&-\delta
\end{pmatrix}
$$
$$[a,a^\dagger]=a a^\dagger-a^\dagger a=1$$
$$[a,\delta]=[a^\dagger,\delta]=0.$$

The dagger $\dagger$ denotes complex conjugate. My thought is to make use of the definition of matrix exponential $e^{X}=\sum_{k=0}^{\infty} \frac{1}{k !} X^{k}$ and calculate $H^2$ and $H^3$ to find some pattern to exploit. For example, in the special case where $\delta=0$, we can find a diagonal matrix $D$
$$
D=\begin{pmatrix}
4a^\dagger a+6 &0&0\\
0&4a^\dagger a+2&0\\
0&0&4a^\dagger a-2
\end{pmatrix}
$$
such that $H^{2n+1}=D^{n}H$ and $H^{2n+2}=D^{n}H^2$, with which we can obtain a neat result of $e^{i H t}$. However, it seems that no such diagonal matrix exists for $\delta \neq 0$ case. Are there any other ways to calculating matrix exponential like this with non-commuting entries? Any input is much appreciated!

答:
This is really just an extended comment. I'll introduce matrices
$$S_1
=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},\quad
S_2
=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix},\quad
S_3
=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$
which satisfy the familiar angular momentum algebra $[S_j,S_k]=i\epsilon_{jkl}S_l$. A quick calculation shows
$$S_1^2+S_2^2+S_3^2=2I_3,$$
corresponding to a spin of $s=1$. Continuing in this spirit, let $S_{\pm}:=S_1\pm i S_2$. This yields

$$[S_3,S_{\pm}]=\pm S_3,\qquad [S_+,S_-]=2S_3.$$
This renders the Hamiltonian as $H=a\,S_+ + a^\dagger \,S_-+\delta\,S_3$. One useful feature of this form is that $S_\pm^3=0$, which means that powers of $H$ have some possibility of simplifying.

One additional observation. The number operator $a^\dagger a I_3$ commutes with $S_3$, so we have a basis of states $|n,m_s\rangle=|n\rangle\otimes |m_s\rangle$ for nonnegative integer $n$ and $m_s=1,0,-1$. Neither of these commutes with the Hamiltonian. However, note that the action of $aS_+$ on a state $|n,m_s\rangle$ is to reduce $n$ by $1$ and raise $m_s$ by $1$ (if possible):

$$aS_+|n,m_s\rangle = a |n\rangle \otimes S_+|m_s\rangle \propto |n-1,m_s+1\rangle$$
So the action of $aS_+$  is to move one quanta from $n$ to $m_s$ and so does not change the value of the quantum number $n+m_s$. The same goes for $a^\dagger S_-$ (moves one quanta from $m_s$ to $n$) and $\delta S_3$ (leaves both numbers unchanged). This suggests introducing the combined number operator $N=a^\dagger a I +S_3$ where $I$ is the 3-by-3 identity matrix. A quick calculation verifies that this commutes with the Hamiltonian:

\begin{align}
[N,H]
&=[a^\dagger a I+S_3,a S_++a^\dagger S_-+\delta S_3]\\
&=[a^\dagger,a] aS_+ + a^\dagger [a,a^\dagger]S_- +a[S_3,S_+]+a^\dagger [S_3,S_-]\\
&=-aS_+ +a^\dagger S_-+a S_3-a^\dagger S_-\\&=0
\end{align}

Thus we can diagonalize $H$ and $N$ simultaneously.


问:
What does it mean to induce a map?
I'm reading about category theory and the term 'induce' is kind of mysterious to me. What does it mean for a particular map to induce another map? For instance, the following example was given:

$$
\require{AMScd}
\begin{CD}
X    @>>> X^I\\
@V{f}VV       @V{f_*}VV \\
Y @>>> Y^I
 \end{CD}$$

(I might be drawing it badly) But basically, there's a correspondence between the category of sets and the category of cartesian products of sets that associates a set $X$ and and it's cartesian product $X^I$. Then, if we have a map $f : X \to Y$ between two sets, this map will induce a map $f_*$ that acts on $X^I$ in a 'canonical' way, such that:

$$f_* ((x_i)_{i \in I}) := (f(x_i))_{i \in I}$$

it seems fine to just acknowledge this as a 'canonical' map as the map that is induced, but is there any way to actually formalise what it means for one map to induce another? What sets the map $f_*$ apart from the others?

答: In language form, we say an induced map arises from the composition map by restricting one of its components. Moreover, it is found that every continuous function gives an induced map.</p>

Title: What does it mean to induce a map?