问:
Two dice and one dice rolling game

Let's say we are paying a game. I have two standard dice and you have one. We rolls the dices, if the sum of the pips on two die roll is greater than twice the pips on your single roll dice, I get the difference and vice a versa. For ex, I roll a (3,6) and you roll 4 then I get (3+6-2*4 = 1) and vice a versa. What is the expected value of the payout of this game.

I calculated the expected value of the sum of two dice rolls to be 7 using the following guidelines:

Rolling a 2: one chance (1&1)

Rolling a 3: two chances (2&1)(1&2)

Rolling a 4: three chances (3&1)(1&3)(2&2)

Rolling a 5: four chances (4&1)(1&4)(3&2)(2&3)

Rolling a 6: five chances (5&1)(1&5)(4&2)(2&4)(2&2)

Rolling a 7: six chances (6&1)(1&6)(5&2)(2&5)(4&3)(3&4)

Rolling an 8: five chances (6&2)(2&5)(5&3)(3&5)(4&4)

Rolling a 9: four chances (6&3)(3&6)(5&4)(4&5)

Rolling a 10: three chances (6&4)(4&6)(5&5)

Rolling an 11: two chances (6&5)(5&6)

Rolling a 12: one chance (6,6)

And the expected value of twice the pips on a single roll as 7 (2*3.5).
Then took the difference (zero in this case) to be the expected value of the earnings in this game. My first question is this correct?

Now let's say we raise the stakes a little bit, With same structure as above, we decide the winner would get double the difference(so in the above (3,6) and 4 rolls, I would get 2$). Here's the catch though, I roll the first of my two dice and based on that decide whether to raise the stakes to double the difference or to continue the game as mentioned in the first part.
i. What would my strategy be in this case?
I thought if first dice shows 1,2,3 then I would continue the game as is because in that case my sum to be higher than your average(7) would have lower probability than when I roll a 4, 5, 6. Is this the right way to approach it?
ii. What would be the expected winnings in this case?
I couldn't solve this but here's my approach:

$\frac{1}{6}*0+\frac{1}{6}*0+\frac{1}{6}*0+\frac{3}{6}*{expected earnings in the high stakes game}$

Wouldn't the expected earnings in the high stakes game also be zero though? Because I double my earnings but the opponent does it too.

答:
Your analysis in problem 1 is correct.

In problem 2, your analysis of the optimal strategy (don't double unless roll is greater than 3) is also correct.

However, in problem 2, assuming that you use the optimal strategy, you have a positive expectation.  First, from an intuitive perspective, this must be the case because in problem 1, when you roll a 3 or less on your first die, you have a negative expectation that is perfectly balanced by your positive expectation when you roll a 4 or more on your first die.

As to the actual math, all I have to do is prove that you have a positive expectation assuming that you roll a 4 or more and assuming that the payout is as in problem 1.

When you roll a $4$, your chance is $(1/6)$ of having the two dice total 5,6,..., 10.

When you roll a $5$, your chance is $(1/6)$ of having the two dice
total 6,7,..., 11.

When you roll a $6$, your chance is $(1/6)$ of having the two dice total 7,8,...,12.

Therefore, under the assumption that you roll $4$ or higher, the probabilities are

$5 : (1/18)~~$ Win (3 + 1), Lose (1 + 3 + 5 + 7) : Net = -12.

$6 : (2/18)~~$ Win (4 + 2 + 0), Lose (2 + 4 + 6) : Net = -6.

$7$ : $(1/6)~~$ Win (5 + 3 + 1), Lose (1 + 3 + 5) : Net = 0.

$8$ : $(1/6)~~$ Win (6 + 4 + 2 + 0), Lose (2 + 4) : Net = 6.

$9$ : $(1/6)~~$ Win (7 + 5 + 3 + 1), Lose (1 + 3) : Net = 12.

$10$ : $(1/6)~~$ Win (8 + 6 + 4 + 2 + 0), Lose (2) : Net = 18.

$11 : (2/18)~~$ Win (9 + 7 + 5 + 3 + 1), Lose (1) : Net = 24.

$12 : (1/18)~~$ Win (10 + 8 + 6 + 4 + 2 + 0) : Net = 30.

The above chart makes the math obvious.

Title: Two dice and one dice rolling game


问:
Positive semidefiniteness of a block matrix of positive semidefinite matrices
Given any symmetric matrix $\mathbf{M} = \begin{pmatrix}
\mathbf{A} & \mathbf{B}\\
 \mathbf{B}^\mathrm{T}& \mathbf{C}
\end{pmatrix}$, the following conditions are equivalent:

(1) $\mathbf{M}\succeq 0$ ($\mathbf{M}$ is positive semidefinite)

(2) $\mathbf{A}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;\mathbf{C} - \mathbf{B}^T\mathbf{A}^{\dagger }\mathbf{B} \succeq 0$.

(3) $\mathbf{C}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, $\;\;\mathbf{A} - \mathbf{B}\mathbf{C}^{\dagger }\mathbf{B}^T \succeq 0$.

Then, if we are given that $\mathbf{A}\succeq 0$, $\mathbf{B}\succeq 0$ and $\mathbf{C}\succeq 0$, is there any way to verify (2) or (3) and therefore conclude that $\mathbf{M}$ will be positive semidefinite as well?

答:
Hint By definition a matrix $M\in\mathbb{R}^{n\times n}$ is positive semidefinite if $v^TMv\geq 0$ for all $v\in\mathbb{R}^n$. Use the fact
$
\det M=  \det\Big(\mathbf{C} - \mathbf{B}^T\mathbf{A}^{-1}\mathbf{B} \Big)
$

and  proof by inductin in $n\in\mathbb{N}$ for $n\geq 1$.

ContentLicense: CC BY-SA 3.0


问:
Prove or disprove that if X is homeomorphic to Y in $\mathbb{R}^n$ then $\mathbb{R}^n\setminus X$ is homeomorphic to $\mathbb{R}^n \setminus Y$
Intuitively if I imagine homeomorphisms to be continuously transforming from one set to the other, it feels like it is true. However, I'm struggling to find an explicit homeomorphism f.</p>

答:
Knots.  They're all homeomorphic to a circle.  But they are commonly studied by putting tubular neighborhoods around them and studying their complements in $S^3\cong\Bbb R^3\cup\{\infty\}$ (the so-called "knot complement").

Thus all that remains is to convince ourselves that tubular neighborhoods of "different" knots are homeomorphic.  They are all homeomorphic to a torus.

There are also higher dimensional knots.

ContentLicense: CC BY-SA 4.0


问:
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$
The question is:

Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$

I did the question without using the Hint, but I don't know how to do it using the hint.

Quick working out of what I've done:

\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}

\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}

\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}

$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$

But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !

答:
It isn't necessary to use the hint. What you have done is correct and valid! Nonetheless, here's the intended method using hint.

Let $\theta=2\tan^{-1}2$ and $\alpha=\pi-\cos^{-1}\frac 3 5$. Then,
$$\tan\theta=\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}$$
$$\tan\theta=-\frac{4}{3}\quad (*)$$
and using the hint,
$$\tan\alpha=-\tan(\cos^{-1}\frac{3}{5})$$
Using $\cos^{-1}\frac{A}{\sqrt{A^2+B^2}}=\tan^{-1}\frac{B}{A}$, we get
$$\tan\alpha=-\tan(\tan^{-1}\frac{4}{3})$$
$$\tan\alpha=-\frac 4 3\quad (**)$$
From $(*)$ and $(**)$, we conclude
$$\theta=\alpha$$
Or $\text{LHS}=\text{RHS}$ as desired.

Hope this is clear :)


答: $$\tan \alpha =2 \implies \tan 2\alpha = \frac{2\cdot 2}{1-2^2}=-\frac{3}{4} \implies tan(\pi-2\alpha)=\frac{3}{4} \\\implies \cos (\pi-2\alpha)=\frac 35. \blacksquare$$

Title: Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$

Body:
The question is:

Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$

I did the question without using the Hint, but I don't know how to do it using the hint.

Quick working out of what I've done:

\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}

\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}

\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}

$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$

But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !


问:
Translating the apex of $-x^2$
I am attempting to write a function with the following properties:

The highest possible value is 100, and the further you get from a certain value, the lower the result the function will return...
I created a simple bell-like curve by subtracting x^2 from an apex of 100:

$$f(x)=100-x^2  $$

I wanted the curve to span from 0 to 100, so I found it necesary to make the apex occur at x=50:

$$f(x)=100  - \frac{1}{25} \cdot (x-50)^2  $$

The function produces exactly what I'm looking for, but only for one value, 50.  I would like to shift the apex horizontally while maintaining the x intercepts of 0 and 100.  That is, I would like a curve similar to this one, except the apex can be any number greater than 0 and less than 100.  Can this be accomplished simply?

答: One choice is to squash your parabola on one side of the peak and expand it on the other.  Say you want the peak at $m$.  Let $$g(x)=\begin {cases} 50\frac xm & x \le m \\100-50\cdot \frac {100-x}{100-m}& x \ge m \end {cases}$$ and your function is then $$f(g(x))=100-\frac 1{25}(g(x)-50)^2$$</p>

Title: Translating the apex of $-x^2$


问:
Going to people's house combinatorics
Many students in Mr. Noeth’s BC Calculus class missed their first test, and to avoid taking a makeup, have decided to never leave their houses again. As a result, Mr. Noeth decides that he will have to visit their houses to deliver the makeup tests. Conveniently, the 17 absent students in his class live in consecutive houses on the same street. Mr. Noeth chooses at least three of every four people in consecutive houses to take a makeup. How many ways can Mr. Noeth select students to take makeups?

I realized that every group of 4 consecutive students intersects with 3 of the next 4. This means that the next set can have all students already picked or one spot left.After that i have no clue what to do

答: If he chooses the 17th student, then any acceptable set of choices of the first $16$ students remains acceptable. If he doesn't choose the 17th student, then he must choose the 16th, 15th, and 14th, but any acceptable choice of the first $13$ remains acceptable. So if $a_n$ is the number of ways to solve the $n$-student problem, then $a_{17}=a_{16}+a_{13}$. More generally, $a_n=a_{n-1}+a_{n-4}$. Now if you can just work out $a_1,a_2,a_3,a_4$, you can use the recurrence to calculate $a_{17}$.

Title: Going to people's house combinatorics


问:
Maximal order of magnitude of Prime Omega Function
Let $\omega(n)=\sum_{p|n}1$ be the prime omega function.

Prove that for any $\epsilon > 0$,

There exists $N>0$ such that for all $n > N$,

$$\omega (n)<\frac{(1+\epsilon)\log{n}}{\log{\log{n}}}$$

There exists infinitely many $n$ such that

$$\omega (n)>\frac{(1-\epsilon)\log{n}}{\log{\log{n}}}$$

I don't even know how to begin the question.

I only know that $\displaystyle \left| \frac{\sum_{i=1}^n\omega(i)}{n}-\log{\log{n}}\right|<C$ for some constant $C$ and $n\geq 3$.

For (1), I don't know how to come up with $N$ with $\epsilon$;

For (2), I suppose somehow we can construct a formula for those $n$'s, but I don't know how to deal with the $\log{\log{n}}$.

Thank you for any help.

答:
Using the definition of limit, the problem asks you to prove that

$$
\lim_{n\to\infty}\omega(n)\left/\log n\over\log\log n\right.=1
$$

Now, let's define the following functions:

$$
\pi(x)=\sum_{p\le x}1\tag{Prime-counting function}
$$
$$
\vartheta(x)=\sum_{p\le x}\log p\tag{Chebyshev's theta function}
$$

As a result, $\exp\vartheta(x)$ gives the product of all prime numbers within $x$, meaning

$$
\pi(x)=\omega[\exp\vartheta(x)]\tag1
$$

By the prime number theorem, we have

$$
\pi(x)\sim{x\over\log x}\quad\text{or}\quad\lim_{x\to\infty}{\pi(x)\log x\over x}=1\tag2
$$

Now, via Riemann-Stieltjes integration, we obtain the following formula:

$$
\begin{aligned}
{\vartheta(x)\over x}
&=\frac1x\sum_{p\le x}\log p=\frac1x\int_{2-\varepsilon}^x\log t\mathrm d\pi(t) \\
&={\pi(x)\log x\over x}-\frac1x\int_2^x{\pi(t)\over t}\mathrm dt \\
&={\pi(x)\log x\over x}+\mathcal O\left(\frac1x\int_2^x{\mathrm dt\over\log t}\right)
\end{aligned}
$$

By taking $x\to\infty$ and employing (2), we deduce the following relation:

$$
\vartheta(x)\sim x\quad\text{or}\quad\lim_{x\to\infty}{\vartheta(x)\over x}=1\tag3
$$

This means $\log[\exp\vartheta(x)]\sim x$, allowing us to make the following substitution:

$$
\omega[\exp\vartheta(x)]\sim{\log[\exp\vartheta(x)]\over\log x}\tag4
$$

All we need here is to perform substitution on the denominator. Taking logarithm on (3) gives

$$
\lim_{x\to\infty}[\log\vartheta(x)-\log x]=0
$$

As a result,

$$
\lim_{x\to\infty}{\log\vartheta(x)\over\log x}=\lim_{x\to\infty}{\log x+\log\vartheta(x)-\log x\over\log x}=1
$$

and plugging this into (4) gives

$$
\omega[\exp\vartheta(x)]\sim{\log[\exp\vartheta(x)]\over\log\log[\exp\vartheta(x)]}
$$

Finally, substitute $\exp[\vartheta(x)]$ with $n$ produces

$$
\omega(n)\sim{\log n\over\log\log n}\quad\text{or}\lim_{n\to\infty}\omega(n)\left/\log n\over\log\log n\right.=1
$$

$
\Huge\square
$


问:
Show that the group ordering in the direct product is doesn't matter to the group structure.
Show that the group ordering in the direct product is doesn't matter to the group structure.

For instance, for $A,B,C \ne \emptyset$, $A×B×C$ and $A×C×B$ is same (That is, the group structure of $A×B×C$ is same with $A×C×B$).

Any idea?

答:
First, $A,B,C$ are not only nonempty, they must be groups. This is because empty set can never be a group.

Second, in group theory, we always say that two groups have same structure if they are isomorphic.

Hint:

To show that $A\times B\times C$ is isomorphic to $A\times C\times B$, we need to construct an isomorphism between them. The obvious one should be $\phi:A\times B\times C \rightarrow A\times C\times B$ defined by $$(a,b,c)\mapsto (a,c,b)$$
where $(a,b,c)\in A\times C\times B.$

In general, let $A_1,\dots,A_n$ be groups and $\sigma\in S_n$, then $$A_1\times\dots\times A_n \cong A_{\sigma(1)}\times\dots\times A_{\sigma(n)}$$
and this can be proven in a similar way.


Body:
Note
\begin{align}
& \tan(2\tan^{-1}2)-\tan(\pi-\cos^{-1}\frac35)\\
=&\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}+\tan(\cos^{-1}\frac35)
=\frac{2\cdot2}{1-2^2}+ \frac43=0
\end{align}
which leads to
$$2\tan^{-1}2=\pi-\cos^{-1}\frac35$$


问:
Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$?
As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?

答:
For me it's just Vieta's formulas (or comparing coefficients as if $x^2+5x$ were a single variable):

$(x^2 + 5x + \color{red}4)(x^2 + 5x + \color{red}6) \color{red}{-48} = (x^2 + 5x + \color{red}{12})(x^2 + 5x \color{red}{-2})$?

$4+6=12+(-2)$, and $4\cdot 6 - 48 = 12 \cdot (-2)$, so yes.


问:
$\xi$ be a primitive cube root of unity
Could anyone tell me first of all the below problem is wrong or okay?I am not able to figure out how to solve

$\xi$ be a primitive cube root of unity define $A=\begin{pmatrix}\xi^{-1}&0\\0&\xi\end{pmatrix}$, for a vector $v=(v_1,v_2)\in\mathbb{R}^2$ define $$|v|_A=\sqrt{|vAv^T|}$$ if $w=(1,1)$ what is the value of $|w|_A$?

答:
An idea with a probably different, simpler approach:

First, pay attention to the fact that $\,|\xi|=1\iff \xi^{-1}=\bar\xi\;$ , so

$$(1\;1)\begin{pmatrix}\bar\xi&0\\0&\xi\end{pmatrix}\binom11=\bar\xi+\xi=2\,\text{Re}(\xi)=2\cos\frac{2\pi}3=-1\implies|w|_A=1$$

Title: $\xi$ be a primitive cube root of unity


问:
Exchanging definite integral and infinite sum in context of Fourier series
I am writing a mathematics paper on the convergence of the Fourier series for periodic functions, and in the first section (where I define the Fourier series), I also derive the standard coefficients $a_n, b_n$. The standard way of doing this is to multiply both sides of the function by $\cos{kx}$ or $\sin{kx}$ depending on which coefficient you are calculating, and integrating from $-\pi$ to $\pi$, as follows:

$$\int_{-\pi}^\pi f(x)\cos(kx) \, dx = \int_{-\pi}^\pi a_0 \cos{kx} \, dx + \int_{-\pi}^\pi \left(\sum_{n=1}^\infty a_n\cos{nx}\cos{kx} + b_n\cos{nx}\sin{kx} \right)\,dx.$$

However, from this point, the standard derivation of the coefficients involves interchanging the integral and the sum, but to me this is not inherently obvious. For example, if $a_n = 1$ for all $n$, the summation does not converge and we would be integrating something infinite, which doesn't make sense. I've looked at a few posts that talk about interchanging the order of the sum and integral, and most of them don't really seem to apply.

Monotone convergence and Vitali convergence don't seem to apply since we can't make statements about pointwise convergence of $a_n\cos{kx}$ or $b_n\sin{kx}$ without being circular. Fubini's theorem tells us that if $\int\sum_n |f_n(x)| \,dx $ is finite, then we can interchange the summation and integral. However, we can't guarantee that such an integral is finite unless $\int_{-\pi}^\pi f(x) \cos(x)\, dx$ is finite--which led me to assume that the finiteness of the sum is a sufficient condition for interchanging the integral and sum, but my professor said this was insufficient.

Another thing I am thinking about is that pointwise convergence does not preserve integrability, but uniform convergence does. Does this mean that we can only derive the coefficients $a_n, b_n$ when $f$ is uniformly convergent? In that case, why would anyone care about pointwise convergence with the Dirichlet kernel proof?

I guess put simply, my question is, what can we stipulate about $f$ so that the interchange above is valid?

答:
When we talk about infinite series, one always has to specify what type of convergence is intended, for example: pointwise convergence/ a.e pointwise convergence/ uniform convergence/ convergence in $L^2$ norm etc. Once you specify this then we can start discussing further. Now, the way I've usually seen things, pointwise convergence of Fourier series is a very subtle topic, and depending on how much regularity you assume on your function $f$, you get all sorts of (weird and crazy but cool) results.

I don't really feel like working with $\sin$ and $\cos$; so let $\phi_n(x)= e^{inx}$ (perhaps with appropriate normalization factors, which I can never remember, and don't feel like calculating now). Given $n\in \Bbb{N}$, let
\begin{align}
s_{n,f}= \sum_{l=-n}^n \hat{f}(l)\phi_l = \sum_{l=-n}^n \left(\int_{-\pi}^{\pi} f(t)\overline{\phi_l(t)}\, dt\right) \phi_l
\end{align}
This is the $n^{th}$ partial sum. Now, here's a couple of results:

If $f$ is of class $C^p$, where $p\geq 1$ is an integer, then $\lVert s_{n,f}-f \rVert_{\infty} = O\left(n^{\frac{1}{2}-p}\right)$ as $n\to \infty$. In particular, this means $s_{n,f}\to f$ uniformly on $[-\pi,\pi]$. As a result of this, you can always interchange the series with the integral $\int_{-\pi}^{\pi}$.

Usually, you may not want to assume such strong results. Another common thing we do is to suppose $f\in L^2([-\pi,\pi])$. In this case, $s_{n,f}$ need not converge to $f$ pointwise. However, what is true is that $s_{n,f}\to f$ in the $L^2$ norm; i.e $\lVert s_{n,f} - f \rVert_{L^2([-\pi,\pi])} \to 0$ as $n\to \infty$. So, in this case, if we write something like $f=\sum_{n=-\infty}^{\infty}\hat{f}(n)\phi_n$, then the limit which defines the series must be interpreted as an $L^2$ limit. In this case, for any $g\in L^2([-\pi,\pi])$, we can calculate the integral $\int_{-\pi}^{\pi}fg$ by bringing the series out. You can justify this as follows:
\begin{align}
\int_{-\pi}^{\pi}fg &= \left\langle f, \overline{g} \right\rangle =\left\langle \lim_{n\to \infty}s_{n,f}, \overline{g} \right\rangle = \lim_{n\to \infty}\left\langle s_{n,f}, \overline{g} \right\rangle = \sum_{n=-\infty}^{\infty}\hat{f}(n)\int_{\pi}^{\pi}\phi_n(t)g(t)\, dt,
\end{align}
where I used the fact that the inner product $\langle \cdot, \cdot\rangle$ on $L^2$ is a bilinear bounded function (because of Cauchy Schwarz inequality), or equivalently, it is continuous, so we can pull the limit outside.

Btw, note that if you take the sequence of coefficients to be all $1$, then you do not have convergence in $L^2$ even so that's just a very bad situation. Again, you may want to impose other types of regularity assumptions, in which case you get corresponding results. I obviously haven't written an exhaustive list, but hopefully this points you in the right direction.


问:
If $x$ is rational and $\sin(x)$ is irrational, then is $\cos(x)$ irrational?
I am wondering if the following implication is true. I know it is not true if $x$ is irrational, as clearly $x =\pi/3$ is a counterexample.

$$\text{$x$ is rational and $\sin(x)$ is irrational} \implies \cos(x) \text{ is irrational}$$

I know irrationals are not closed under multiplication or addition. However, this seems true.

If there is an obvious counterexample I am missing, I apologize.

答:
A form of the Lindemann-Weierstrass theorem states that if $a_1,...,a_n$ are algebraic numbers and $\alpha_1,...,\alpha_n$ are distinct algebraic numbers, then the only solution to :
$$
a_1e^{\alpha_1} + ... + a_n e^{\alpha_n} = 0
$$

is $a_i = 0$ for all $i$.

Suppose that for some $x$ non-zero rational (in fact, algebraic), we have $\sin x$ is rational. In particular, $\sin x = \frac{e^{ix} - e^{ix}}{2i}$, so we get $e^{ix} - e^{-ix} = 2iq$ for some rational $q$. Rearranging gives us :
$$
(1)e^{ix} + (-1)e^{-ix} + (-2iq)e^0 = 0
$$

If $x$ is non-zero , then the above contradicts the LW theorem because $ix,-ix$ and $0$ are distinct algebraic numbers. This shows that $\sin x$ is in fact irrational. But we can say more : see if you can prove that $\sin x$ is transcendental using this fact.

From here, a similar trick can be used to show that if $x$ is non-zero algebraic then $\cos x$ is irrational. Thus your implication is clear since both sides of it are always true unless $x=0$ in which case both sides are false, so the implication holds always.


问:
Given the characteristic (and minimal) polynomial of $T:V\to V$, how many distinct Jordan forms are possible?
I was solving some routine problems about determining the possible Jordan forms of a linear operator, given the characteristic and minimal polynomials, and an interesting thought came to my mind! All the combinatorics enthusiasts out there should take a look.

Is there a way to comment on the number of Jordan forms, given the characteristic polynomial of $T:V\to V$?

Let's say $$p_T(t) = \prod_{i=1}^k(t-\lambda_i)^{n_i}$$

is the characteristic polynomial of $T:V\to V$. Is there a closed-form solution to describe the number of Jordan Forms corresponding to this polynomial? Two Jordan forms are considered the same iff they consist of the same Jordan blocks (any permutation).

What if I am also given the minimal polynomial of $T$, namely $m_T(t)?$
$$m_T(t) = \prod_{i=1}^k(t-\lambda_i)^{m_i}$$
where $1\leq m_i\leq n_i$ for all $i=1,2,...,k$

The answer definitely reduces since we have imposed more constraints but by how much? What is the number, exactly?

I think the following ideas will be very important in determining the answer, though I wasn't able to quite figure out something concrete using them:

The sum of sizes of all Jordan blocks corresponding to $\lambda$ is equal to the multiplicity of $\lambda$ in $p_T(t)$.

The size of the largest Jordan block corresponding to $\lambda$ is equal to the multiplicity of $\lambda$ in $m_T(t)$.

Thank you, and I look forward to an interesting discussion!

答:
There is not really much more that can be said than what you have observed at the end.  A multiset of positive integers whose sum is $n$ is called a partition of $n$, and the number of such partitions is commonly written $p(n)$.  So, a Jordan normal form with characteristic polynomial $\prod_{i=1}^k(t-\lambda_i)^{n_i}$ just consists of a partition of $n_i$ for each $i$, so the number of them is $$\prod_{i=1}^kp(n_i).$$  However, there is no known closed form for $p(n)$ (and in the case $k=1$, your problem is equivalent to finding a closed form for $p(n)$).

Similarly, the number of partitions of $n$ into parts such that the largest part is $m$ can be written as $p_m(n)$, so if you additionally require the minimal polynomial to be $\prod_{i=1}^k(t-\lambda_i)^{m_i}$ then the number of such Jordan normal forms is $$\prod_{i=1}^kp_{m_i}(n_i).$$  Again, though, there is no known closed form for $p_m(n)$.


问:
Probability that $3$ hands of a clock in the same semi-circle?
I am wondering how to calculate the probability that $3$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $n/2^{(n-1)}$ with $n = 3$.

But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one.
If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $3/4$?

Anyone can help? Thanks.

答:
The hands in fact move deterministically; it is just the time when we view them that can be random. The movement repeats ever twelve hours, so assume we view the hands at some moment uniformly distributed between noon and midnight. Then the probability they will be within a semicircle is the number of hours they spend in a semicircle, divided by $12.$

Starting at noon, the hands are within one semicircle until the second hand is opposite the hour hand. The second hand makes $12\times 60 = 720$ revolutions in $12$ hours while the hour hand makes one revolution in the same time,
so the time until the second and hour hands first align after noon is $12/719$ hour,
which is $720/719$ minutes;
but the time from noon until the hands are exactly opposite is half that time,
$360/719$ minute.

For a short time after the first $360/719$ minute, the hands are not within a semicircle. They are in a semicircle again starting at the instant when the second hand is opposite the minute hand.
The second hand aligns with the minute hand for the first time after noon at
$60/59$ minutes after noon, so the first time they are opposite is at
$30/59$ minute after noon.

Let $t_h = 360/719$ and $t_m = 30/59,$
and let's take the convention that "at time $t$" means at $t$ minutes past noon.
So the first time interval when the hands are not in a semicircle
starts at $t_h$ and ends at $t_m.$ The length of this interval is $t_m - t_h$ minutes;
specifically, the time during which the hands are continuously not in a semicircle,
in minutes, is
$$
\frac{30}{59} - \frac{360}{719} = \frac{330}{42421}.
$$

The second such time interval starts at $3t_h$ and ends at $3t_m$ for
$3(t_m - t_h)$ minutes not in a semicircle,
the third starts at $5t_h$ and ends at $5t_m$, and so forth as long as the minute hand remains less than $180$ degrees clockwise from the hour hand.

This is a promising pattern: after $n$ intervals like this, we will have spent a total of
$n^2(t_m - t_h)$ minutes during which the hands do not fit in a semicircle.

Alas, the full solution is not that simple.
The intervals increase in length until they are almost half a minute long,
and then this simple pattern is broken.

The minute hand first becomes exactly opposite the hour hand at $6/11$ hour after noon,
that is, at $360/11$ minutes after noon.
Shortly before that, at $65t_h$ minutes, the second hand passes opposite to the hour hand, making the hands not within a semicircle;
but at $360/11$ minutes after noon the second hand is suddenly finds itself between the minute and hour hands so that the hands are all within a semicircle again,
even though $65t_m$ minutes have not yet passed.

After this abbreviated interval of not-in-a-semicircle time,
the next interval is almost half a minute long, and the next several intervals decrease in length. But once they have decreased enough, they start increasing again, and the pattern repeats but with different intervals each time as the relative position of the second hand is different each time the minute hand overtakes the hour hand or comes opposite to it.

At about this point in the analysis I decided I would rather have a computer count the time for me. The events where the hands start fitting in a semicircle or stop fitting in a semicircle all occur at exact integer multiples of
either $1/11,$ $1/59,$ or $1/719$ minute after noon.
Since those fractions have a common denominator of
$11\times 59\times 719 = 466631,$
every event of interest (where the hands start or stop being within a semicircle) occurs at some multiple of $1/466631$ minute after noon.
By taking $720\times 466631$ equal timesteps of that length we cover twelve hours.
At the start of each timestep we know from the relative positions of the three hands whether the hands will fit in a semicircle or not during that timestep. (It is one way for the entire duration of the timestep.)

I wrote the analysis in the python script shown below.
I had the script count the first six hours, take a checkpoint, and then do the last six hours. Everything after $6$ o'clock is a mirror image of what happened before, in reverse order, so the same sequence of lengths of intervals occurs in reverse order after $6$ o'clock. I had the script record the start, end, and duration of each interval to verify this. Note that since each interval starts about a minute after the previous one, there are almost $360$ intervals in six hours, but not quite $360$ due to the way the hands "catch up" with each other.

According to my script, it takes
$720\times 11\times 59\times 719 = 335974320$
timesteps to complete twelve hours, and during $83982960$ of those timesteps the three hands do not fit within a semicircle. The proportion of time not in a semicircle is therefore
$$
\frac{83982960}{335974320} = \frac{1977}{7909} \approx 0.2499683904412694.
$$
So the probability that the hands are not within a semicircle is almost, but not quite, $1/4$;
the probability that they are within a semicircle is therefore slightly over $3/4.$

To relate these calculations to G Cab's method, let $x$ be the number of units of $12/11$ hour since noon (that is, $x$ is $0$ at noon and $11$ at midnight). If $k = \lfloor x\rfloor$ (an integer),
then for each $k$ (that is, for each period of $12/11$ hour) there are two sequences of intervals during which the three hands are not within a semicircle.
During the first half of this period, while the angle from the hour hand to the minute hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle whenever
$$
m + \frac12 \leq \frac{15(2k)}{22} + \frac{719}{11} \tau < m + \frac12 + \tau
$$
for some integer $m,$ where $\tau = x - k,$ and during the second half of this period, while the angle from the minute hand to the hour hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle when
$$
n + \tau \leq \frac{15(2k + 1)}{22} + \frac{719}{11} \tau < n + \frac12
$$
for some integer $n,$ where $\tau = x - k - \frac12.$
In either case we also have the restriction that $0 \leq \tau < \frac12.$

In the typical case, the beginning or end of an interval occurs when one of the inequalities becomes an equation; the exceptions occur when the equation is true only for $\tau < 0$ or $\tau > \frac12$.
Solving these equations for $\tau,$ we find
$$\tau = \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag1$$
at the start and
$$\tau = \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag2$$
at the end of a typical interval in the first sequence, but
$$\tau = \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right)\tag3$$
at the start and
$$\tau = \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)\tag4$$
at the end of one in the second sequence.
The length of the first interval (in the same units as $\tau$)
is therefore
\begin{align}
\Delta\tau_1
&= \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)
   - \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right) \\
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
\end{align}
and the length of the second interval is
\begin{align}
\Delta\tau_2
&= \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)
   - \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right) \\
&= \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right)
\end{align}

The combined length of two intervals, one from each sequence, is
\begin{align}
\Delta\tau_1 + \Delta\tau_2
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
 + \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \\
&= \frac{11}{708(719)} \left(367 + 11(m - n)\right) \\
\end{align}

If we hold $m - n$ constant, $\Delta\tau_1 + \Delta\tau_2$
is independent of the value of $m$ as long as all of the $\tau$ values from which we computed $\Delta\tau_1$ and $\Delta\tau_2$ are between $0$ and $\frac12.$
For example, if we set $n = m$, we get a pairing of intervals in which each pair of intervals has total length $\frac{11(367)}{708(719)}$ except in some cases where an interval starts at $\tau=0$ or ends at $\tau=\frac12$ or where an interval has no corresponding interval in the other sequence.
Trial and error shows that if we set $n=m+1$ we minimize the number of intervals that are unmatched or that are in a pair where one interval ends at $\tau=0$ or $\tau=\frac12$ instead of where the equations above say it will
(there are $14$ such intervals);
each matched pair then has total length
$\frac{11(367-11)}{708(719)} = \frac{11(89)}{3(59)(719)}$
in the units of $\tau.$
This is equal to $\frac{356}{59(719)}$ hour,
which is $234960$ timesteps of the brute-force method.

According to my spreadsheet calculations, if $n = m+1$ there are $32$ matched pairs of intervals for each value of $k.$ since there are $11$ values of $k,$ the total time in intervals that belong to typical matched pairs is
$352 \times 234960$ timesteps. It then remains to account for the $14$ intervals that are not in typical matched pairs.
Those intervals can be identified for each $k$ by finding the maximum $m$ such that right-hand side of Equation $(1)$ is negative
and the minimum $n$ such that the right-hand side of Equation $(4)$
is greater than $\frac12.$
A relatively quick hand calculation should therefore be possible, though it would still be longer than I would like to write in detail in this answer.

Update: The calculations are not as long as I feared after all, as shown in G Cab's edited answer.

Here is the script for the brute-force method:

```<code># Program to figure out what proportion of the time the three hands
# of a clock are not within a semicircle.

# Assume there are 59*719*720 equally-spaced positions around the clock face.
# Note: 59*719 == 42421; 42421*11 = 466631.

# Each timestep of this program is 1/42421 of a minute.
# In one timestep, the second hand moves 720 positions, the minute hand
# moves 12 positions, and the hour hand moves 1 position.
# In 42421 timesteps, one minute has passed and the second hand returns
# to the starting position,

numPositions = 11*59*719*720

# Returns the new position after having moved the given amount forward
# from the given position. That is, the return value is the sum of
# the two input values, modulo numPositions.
# It is assumed the input values are in the range 0 to numPositions - 1.
def newPosition(fromPosition, movement):
    toPosition = fromPosition + movement
    if toPosition >= numPositions:
        toPosition -= numPositions
    return toPosition

# Returns the number of steps clockwise to get from position1 to position2.
# positions measured in number of steps from the start.
def stepsClockwise(position1, position2):
    distance = position2 - position1
    if distance < 0:
        distance += numPositions
    return distance

# Returns 1 if the hands are at the start of a timestep in which the
# smallest angle encompassing all three hands will be greater than 180 degrees,
# 0 if the angle will be less than 180 degrees.
def isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
    # A 180 degree angle
    halfCircle = numPositions // 2
    # A position exactly opposite the seconds hand
    antiSecondsPosition = newPosition(secondsPosition, halfCircle)
    minutesOffset = stepsClockwise(hoursPosition, minutesPosition)
    if minutesOffset < halfCircle:
        # The minutes hand is < 180 degrees clockwise from the hour hand
        antiOffset = stepsClockwise(hoursPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < minutesOffset):
            # The anti-seconds hand will be between the hours and minutes
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the hour hand but less than 180 degrees
            # before the minute hand.
            return 1
        else:
            return 0
    else:
        # The hour hand is <= 180 degrees clockwise from the minute hand.
        # During the next timestep the short arc between the hands will be
        # clockwise from the minute hand to the hour hand.
        hoursOffset = stepsClockwise(minutesPosition, hoursPosition)
        antiOffset = stepsClockwise(minutesPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < hoursOffset):
            # The anti-seconds hand will be between the minutes and hours
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the minute hand but less than 180 degrees
            # before the hour hand.
            return 1
        else:
            return 0

# Runs the given number of timesteps from the given starting position
# and returns the number of times the hands spanned more than 180 degrees.
def countLargeAngles(hoursPosition,
                     minutesPosition,
                     secondsPosition,
                     startingTimestep,
                     numTimesteps):
    numLargeAngles = 0
    wasLargeAngle = 0
    largeAngleStarted = 0
    numIntervals = 0
    for i in range(numTimesteps):
        if isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
            numLargeAngles += 1
            if wasLargeAngle == 0:
                largeAngleStarted = startingTimestep + i
                wasLargeAngle = 1
        else:
            if wasLargeAngle == 1:
                largeAngleStopped = startingTimestep + i
                numIntervals += 1
                print(">180 degrees interval ", numIntervals,
                      " from ", largeAngleStarted,
                      " to ", largeAngleStopped, " for ",
                      (largeAngleStopped - largeAngleStarted), " timesteps")
                wasLargeAngle = 0
        hoursPosition   = newPosition(hoursPosition, 1)
        minutesPosition = newPosition(minutesPosition, 12)
        secondsPosition = newPosition(secondsPosition, 720)
    print("After ", numTimesteps, " timesteps")
    print("hour hand at ", hoursPosition)
    print("minute hand at ", minutesPosition)
    print("second hand at ", secondsPosition)
    print(numLargeAngles, " positions with > 180 degrees")
    return numLargeAngles, hoursPosition, minutesPosition, secondsPosition

# Run 360*466631 timesteps to simulate 6 hours,
# then another 360*466631 timesteps to simulate the next 6 hours.
halfTotalTimesteps = numPositions // 2
hoursPosition = 0
minutesPosition = 0
secondsPosition = 0

(numLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     0,
                                     halfTotalTimesteps)
(additionalLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     halfTotalTimesteps,
                                     halfTotalTimesteps)
numLargeAngles += additionalLargeAngles

print("Total number of timesteps was ", numPositions)
print("Total number of > 180 degrees was ", numLargeAngles)
</code>```

Body:
The hands in fact move deterministically; it is just the time when we view them that can be random. The movement repeats ever twelve hours, so assume we view the hands at some moment uniformly distributed between noon and midnight. Then the probability they will be within a semicircle is the number of hours they spend in a semicircle, divided by $12.$

Starting at noon, the hands are within one semicircle until the second hand is opposite the hour hand. The second hand makes $12\times 60 = 720$ revolutions in $12$ hours while the hour hand makes one revolution in the same time,
so the time until the second and hour hands first align after noon is $12/719$ hour,
which is $720/719$ minutes;
but the time from noon until the hands are exactly opposite is half that time,
$360/719$ minute.

For a short time after the first $360/719$ minute, the hands are not within a semicircle. They are in a semicircle again starting at the instant when the second hand is opposite the minute hand.
The second hand aligns with the minute hand for the first time after noon at
$60/59$ minutes after noon, so the first time they are opposite is at
$30/59$ minute after noon.

Let $t_h = 360/719$ and $t_m = 30/59,$
and let's take the convention that "at time $t$" means at $t$ minutes past noon.
So the first time interval when the hands are not in a semicircle
starts at $t_h$ and ends at $t_m.$ The length of this interval is $t_m - t_h$ minutes;
specifically, the time during which the hands are continuously not in a semicircle,
in minutes, is
$$
\frac{30}{59} - \frac{360}{719} = \frac{330}{42421}.
$$

The second such time interval starts at $3t_h$ and ends at $3t_m$ for
$3(t_m - t_h)$ minutes not in a semicircle,
the third starts at $5t_h$ and ends at $5t_m$, and so forth as long as the minute hand remains less than $180$ degrees clockwise from the hour hand.

This is a promising pattern: after $n$ intervals like this, we will have spent a total of
$n^2(t_m - t_h)$ minutes during which the hands do not fit in a semicircle.

Alas, the full solution is not that simple.
The intervals increase in length until they are almost half a minute long,
and then this simple pattern is broken.

The minute hand first becomes exactly opposite the hour hand at $6/11$ hour after noon,
that is, at $360/11$ minutes after noon.
Shortly before that, at $65t_h$ minutes, the second hand passes opposite to the hour hand, making the hands not within a semicircle;
but at $360/11$ minutes after noon the second hand is suddenly finds itself between the minute and hour hands so that the hands are all within a semicircle again,
even though $65t_m$ minutes have not yet passed.

After this abbreviated interval of not-in-a-semicircle time,
the next interval is almost half a minute long, and the next several intervals decrease in length. But once they have decreased enough, they start increasing again, and the pattern repeats but with different intervals each time as the relative position of the second hand is different each time the minute hand overtakes the hour hand or comes opposite to it.

At about this point in the analysis I decided I would rather have a computer count the time for me. The events where the hands start fitting in a semicircle or stop fitting in a semicircle all occur at exact integer multiples of
either $1/11,$ $1/59,$ or $1/719$ minute after noon.
Since those fractions have a common denominator of
$11\times 59\times 719 = 466631,$
every event of interest (where the hands start or stop being within a semicircle) occurs at some multiple of $1/466631$ minute after noon.
By taking $720\times 466631$ equal timesteps of that length we cover twelve hours.
At the start of each timestep we know from the relative positions of the three hands whether the hands will fit in a semicircle or not during that timestep. (It is one way for the entire duration of the timestep.)

I wrote the analysis in the python script shown below.
I had the script count the first six hours, take a checkpoint, and then do the last six hours. Everything after $6$ o'clock is a mirror image of what happened before, in reverse order, so the same sequence of lengths of intervals occurs in reverse order after $6$ o'clock. I had the script record the start, end, and duration of each interval to verify this. Note that since each interval starts about a minute after the previous one, there are almost $360$ intervals in six hours, but not quite $360$ due to the way the hands "catch up" with each other.

According to my script, it takes
$720\times 11\times 59\times 719 = 335974320$
timesteps to complete twelve hours, and during $83982960$ of those timesteps the three hands do not fit within a semicircle. The proportion of time not in a semicircle is therefore
$$
\frac{83982960}{335974320} = \frac{1977}{7909} \approx 0.2499683904412694.
$$
So the probability that the hands are not within a semicircle is almost, but not quite, $1/4$;
the probability that they are within a semicircle is therefore slightly over $3/4.$

To relate these calculations to G Cab's method, let $x$ be the number of units of $12/11$ hour since noon (that is, $x$ is $0$ at noon and $11$ at midnight). If $k = \lfloor x\rfloor$ (an integer),
then for each $k$ (that is, for each period of $12/11$ hour) there are two sequences of intervals during which the three hands are not within a semicircle.
During the first half of this period, while the angle from the hour hand to the minute hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle whenever
$$
m + \frac12 \leq \frac{15(2k)}{22} + \frac{719}{11} \tau < m + \frac12 + \tau
$$
for some integer $m,$ where $\tau = x - k,$ and during the second half of this period, while the angle from the minute hand to the hour hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle when
$$
n + \tau \leq \frac{15(2k + 1)}{22} + \frac{719}{11} \tau < n + \frac12
$$
for some integer $n,$ where $\tau = x - k - \frac12.$
In either case we also have the restriction that $0 \leq \tau < \frac12.$

In the typical case, the beginning or end of an interval occurs when one of the inequalities becomes an equation; the exceptions occur when the equation is true only for $\tau < 0$ or $\tau > \frac12$.
Solving these equations for $\tau,$ we find
$$\tau = \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag1$$
at the start and
$$\tau = \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag2$$
at the end of a typical interval in the first sequence, but
$$\tau = \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right)\tag3$$
at the start and
$$\tau = \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)\tag4$$
at the end of one in the second sequence.
The length of the first interval (in the same units as $\tau$)
is therefore
\begin{align}
\Delta\tau_1
&= \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)
   - \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right) \\
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
\end{align}
and the length of the second interval is
\begin{align}
\Delta\tau_2
&= \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)
   - \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right) \\
&= \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right)
\end{align}

The combined length of two intervals, one from each sequence, is
\begin{align}
\Delta\tau_1 + \Delta\tau_2
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
 + \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \\
&= \frac{11}{708(719)} \left(367 + 11(m - n)\right) \\
\end{align}

If we hold $m - n$ constant, $\Delta\tau_1 + \Delta\tau_2$
is independent of the value of $m$ as long as all of the $\tau$ values from which we computed $\Delta\tau_1$ and $\Delta\tau_2$ are between $0$ and $\frac12.$
For example, if we set $n = m$, we get a pairing of intervals in which each pair of intervals has total length $\frac{11(367)}{708(719)}$ except in some cases where an interval starts at $\tau=0$ or ends at $\tau=\frac12$ or where an interval has no corresponding interval in the other sequence.
Trial and error shows that if we set $n=m+1$ we minimize the number of intervals that are unmatched or that are in a pair where one interval ends at $\tau=0$ or $\tau=\frac12$ instead of where the equations above say it will
(there are $14$ such intervals);
each matched pair then has total length
$\frac{11(367-11)}{708(719)} = \frac{11(89)}{3(59)(719)}$
in the units of $\tau.$
This is equal to $\frac{356}{59(719)}$ hour,
which is $234960$ timesteps of the brute-force method.

According to my spreadsheet calculations, if $n = m+1$ there are $32$ matched pairs of intervals for each value of $k.$ since there are $11$ values of $k,$ the total time in intervals that belong to typical matched pairs is
$352 \times 234960$ timesteps. It then remains to account for the $14$ intervals that are not in typical matched pairs.
Those intervals can be identified for each $k$ by finding the maximum $m$ such that right-hand side of Equation $(1)$ is negative
and the minimum $n$ such that the right-hand side of Equation $(4)$
is greater than $\frac12.$
A relatively quick hand calculation should therefore be possible, though it would still be longer than I would like to write in detail in this answer.

Update: The calculations are not as long as I feared after all, as shown in G Cab's edited answer.

Here is the script for the brute-force method:

# Program to figure out what proportion of the time the three hands
# of a clock are not within a semicircle.

# Assume there are 59*719*720 equally-spaced positions around the clock face.
# Note: 59*719 == 42421; 42421*11 = 466631.

# Each timestep of this program is 1/42421 of a minute.
# In one timestep, the second hand moves 720 positions, the minute hand
# moves 12 positions, and the hour hand moves 1 position.
# In 42421 timesteps, one minute has passed and the second hand returns
# to the starting position,

numPositions = 11*59*719*720

# Returns the new position after having moved the given amount forward
# from the given position. That is, the return value is the sum of
# the two input values, modulo numPositions.
# It is assumed the input values are in the range 0 to numPositions - 1.
def newPosition(fromPosition, movement):
    toPosition = fromPosition + movement
    if toPosition >= numPositions:
        toPosition -= numPositions
    return toPosition

# Returns the number of steps clockwise to get from position1 to position2.
# positions measured in number of steps from the start.
def stepsClockwise(position1, position2):
    distance = position2 - position1
    if distance < 0:
        distance += numPositions
    return distance

# Returns 1 if the hands are at the start of a timestep in which the
# smallest angle encompassing all three hands will be greater than 180 degrees,
# 0 if the angle will be less than 180 degrees.
def isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
    # A 180 degree angle
    halfCircle = numPositions // 2
    # A position exactly opposite the seconds hand
    antiSecondsPosition = newPosition(secondsPosition, halfCircle)
    minutesOffset = stepsClockwise(hoursPosition, minutesPosition)
    if minutesOffset < halfCircle:
        # The minutes hand is < 180 degrees clockwise from the hour hand
        antiOffset = stepsClockwise(hoursPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < minutesOffset):
            # The anti-seconds hand will be between the hours and minutes
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the hour hand but less than 180 degrees
            # before the minute hand.
            return 1
        else:
            return 0
    else:
        # The hour hand is <= 180 degrees clockwise from the minute hand.
        # During the next timestep the short arc between the hands will be
        # clockwise from the minute hand to the hour hand.
        hoursOffset = stepsClockwise(minutesPosition, hoursPosition)
        antiOffset = stepsClockwise(minutesPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < hoursOffset):
            # The anti-seconds hand will be between the minutes and hours
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the minute hand but less than 180 degrees
            # before the hour hand.
            return 1
        else:
            return 0

# Runs the given number of timesteps from the given starting position
# and returns the number of times the hands spanned more than 180 degrees.
def countLargeAngles(hoursPosition,
                     minutesPosition,
                     secondsPosition,
                     startingTimestep,
                     numTimesteps):
    numLargeAngles = 0
    wasLargeAngle = 0
    largeAngleStarted = 0
    numIntervals = 0
    for i in range(numTimesteps):
        if isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
            numLargeAngles += 1
            if wasLargeAngle == 0:
                largeAngleStarted = startingTimestep + i
                wasLargeAngle = 1
        else:
            if wasLargeAngle == 1:
                largeAngleStopped = startingTimestep + i
                numIntervals += 1
                print(">180 degrees interval ", numIntervals,
                      " from ", largeAngleStarted,
                      " to ", largeAngleStopped, " for ",
                      (largeAngleStopped - largeAngleStarted), " timesteps")
                wasLargeAngle = 0
        hoursPosition   = newPosition(hoursPosition, 1)
        minutesPosition = newPosition(minutesPosition, 12)
        secondsPosition = newPosition(secondsPosition, 720)
    print("After ", numTimesteps, " timesteps")
    print("hour hand at ", hoursPosition)
    print("minute hand at ", minutesPosition)
    print("second hand at ", secondsPosition)
    print(numLargeAngles, " positions with > 180 degrees")
    return numLargeAngles, hoursPosition, minutesPosition, secondsPosition

# Run 360*466631 timesteps to simulate 6 hours,
# then another 360*466631 timesteps to simulate the next 6 hours.
halfTotalTimesteps = numPositions // 2
hoursPosition = 0
minutesPosition = 0
secondsPosition = 0

(numLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     0,
                                     halfTotalTimesteps)
(additionalLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     halfTotalTimesteps,
                                     halfTotalTimesteps)
numLargeAngles += additionalLargeAngles

print("Total number of timesteps was ", numPositions)
print("Total number of > 180 degrees was ", numLargeAngles)


问:
Ceiling of exponential distribution
Let $X \sim \text{Exp}(\lambda)$ and $T = \lceil{X}\rceil$.
What is the distribution of $T$?

My approach:
$$
\begin{align*}
  P(T \leq t) &= P(\lceil{X}\rceil \leq t) \\
  &= P(X \leq t-1) + P(t-1 < X \leq t) \\
  &= \left(1 - e^{-\lambda(t-1)}\right) + \left(1-e^{-\lambda t}\right) - \left(1 - e^{-\lambda(t-1)}\right)
\end{align*}
$$
But this obviously can't be true. How can I transform this inequality involving a ceil into an ordinary one I can apply my PDF on?

It should be a discrete random variable.

答:
Your mistake lies in the second step, and becomes apparent if you choose a specific numeric value for $t$; e.g., $t = 2.5$:  $$\Pr[T \le 2.5] = \Pr[\lceil X \rceil \le 2.5] \ne \Pr[X \le 1.5] + \Pr[1.5 \le X \le 2.5].$$  The last step above does not follow because for instance, if $X = 2.4 \le 2.5$, then $T = \lceil X \rceil = \lceil 2.4 \rceil = 3 > 2.5$.  So the event $T \le 2.5$ is not satisfied, but the event $X \le 2.5$ is satisfied.  What went wrong?  Well, by working with the CDF, you're not restricting $t$ to be an integer.

Instead of doing it this way, it is much more preferable to work with the probability mass function of $T$.  Specifically, let $t \in \{1, 2, 3, \ldots \}$ be a positive integer. Then $$\Pr[T = t] = \Pr[\lceil X \rceil = t] = \Pr[t-1 < X \le t] = (1 - e^{-\lambda t}) - (1 - e^{-\lambda(t-1)}) = e^{-\lambda (t-1)}(1- e^{-\lambda}).$$  Now if we let $p = e^{-\lambda}$, then we obtain $$\Pr[T = t] = p^{t-1} (1-p), \quad t \in \{1, 2, 3, \ldots\}.$$  What distribution does this look like?

Title: Ceiling of exponential distribution


问:
Finding all normal subgroups of a group
On my homework today, we had to find all the normal subgroups of $D_{n}$, the dihedral group of order 2n. I solved the problem by looking at how the conjugacy classes change based on whether n is even or odd and then constructed the normal subgroups as unions of the conjugacy classes.

I have 2 questions:

(i) Is there a better way to approach the problem than looking at the conjugacy classes?

(ii) Could someone explain why we want to find all the normal subgroups of a particular group?   How does this provide us additional insight into the structure of the group we are studying? (Right now, this exercise feels more like a 'computation' to me than a way of understanding $D_{n}$)

Thanks :)

答:
(i) Conjugacy classes are certainly useful, since a normal subgroup must be a union of conjugacy classes. There are other ideas that can come into play: once you know a proper normal subgroup $N$, by taking the quotient you can obtain information about $G$ from information you may find in $G/N$, which will hopefully be easier than working directly in $G$ because $G/N$ will be smaller than $G$.

(ii) In fact, that is one important reason for finding all normal subgroups: being able to look at the quotient, and thus obtain information about $G$ by looking at groups that are smaller than $G$. Another is that the normal subgroups are closely associated with all possible images of $G$ under homomorphisms. Both of these facts are part of the Isomorphism Theorems.

Right now you are probably engaged in computation and getting familiar with techniques for working with groups in general, and $D_n$ in particular; the importance of normal subgroups will likely emerge as you start using groups. For example, normal subgroups are extremely important when you are considering Galois groups (which help you study fields and field extensions).

ContentLicense: CC BY-SA 2.5

Title: Finding all normal subgroups of a group


问:
Proof of sylow theorem (1)
Let $G$ be a group of order $n\cdot p^m$. Let $K$ be the set of all subsets of $G$ of size  $p^m$, and let $G_j$ be the group stabilizing $j$ in $K$. Why is $\rm{ord}(G_j) \leq |j|$?</p>

答: Fix any $a \in j$ consider the map $\varphi:G_j \to j$ defined by $\varphi(g)=ga$. This map is an injection since if $g_1a=g_2a$, then you can multiply by $a^{-1}$ from the right and get $g_1=g_2$. Hence $|G_j| \leq |j|$.</p>

Title: Proof of sylow theorem (1)


问:
Using Archimedean Property to Prove the following
So I've worked through a few of the properties of Archimedas.  That is, I understand that for every real number $x$, there exists a natural number $n$ such that $n>x$

And I've also been able to show that, as a consequence of this fact, that for every positive real number $x$ there exists a natural number $n$ such that $1/n<x$

I'm now trying to (and fairly unsuccessfully) show that, as a consequence of the above that for each positive $x<5$ there exists a natural number $n$ such that $5-1/n>x$

I've started out by considering a positve $x<5$.  Then by the results above we can find a natural number $n$ such that $1/n<x$

Do I need to consider cases at this point?  That is consider when $0<x<4$ and then when $4<x<5$?  I'm a little stuck with how to properly apply Archimedes to reach my proposed result.  Help?

答: Making Hagen's comment above into an answer (until and unless he posts it as an answer and informs me of that), put $y:=5-x$. Note that $y>0$ if (and only if) $x<5$. Using your corollary to the Archimedean property, if $x<5$, we may find a natural number $n$ such that $$\frac1n<y.$$ Recalling the definition of $y$ and adding $x$ to both sides of the above inequality gives us the desired result.</p>


问:
Show that if $a, b$ are relatively prime integers, each of which divides $n$, then $ab$ divides $n$

Let $n$ be an integer. Show that if $a, b$ are relatively prime integers,
each of which divides $n$, then $ab$ divides $n$

Solution:

If $a, b$ are relatively prime, then $\gcd(a,b) = 1$, so there are $s, t \in \Bbb Z$ such that $as + bt = 1.$ $\color{red}{(1)}$

If $a$ and $b$ divide $n$, then there are $u,v\in\Bbb Z$ such that $n = au$  and $n = bv$. Multiplying $\color{red}{(1)}$ by $n$
shows $n = nas + nbt = (bv)as + (au)bt = (vs)ab + (ut)ab = (vs + ut)ab.$

Hence, $ab$ divides $n$.

I don't understand why from $(vs + ut)ab$ deduce $ab$ divides $n$ ?

答:
It looks like $c$ was typed instead of $n$ in some places.  Try this instead:

If $a, b$ are relatively prime, then $\gcd(a,b) = 1$, so there are $s, t \in \Bbb Z$ such that $as + bt = 1.$ $\color{red}{(1)}$

If $a$ and $b$ divide $n$, then there are $s, t \in \Bbb Z$ such that $n = au$  and $n = bv$; multiplying $\color{red}{(1)}$ by $n$
shows that $n = nas + nbt = (bv)as + (au)bt = (vs)ab + (ut)ab = (vs + ut)ab.$

Hence, $ab$ divides $n$.


问:
Compactness of diagonal operator $ T_\alpha(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,...)$
Let $\alpha=(\alpha_1, \alpha_2, \alpha_3...)$ a sequence of complex numbers  that converges to zero.

I need to show the compactness of the diagonal operator $T_\alpha:l^2\to l^2$ defined by $$ T_\alpha(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,...)$$

For this:

I define the sequence the operator of finite range $T_n$ so that for $n\in \mathbb{N}$

$$T_n(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,..., \alpha_n x_n,0,0,0,... )$$

I need to show that $T_n\to T_\alpha$ and then $T_\alpha$ is compact.

I have that $$\|(T_\alpha-T_n)(x_1, x_2,x_3...)\|^2=\sum_{j=n+1}^\infty|\alpha _j x_j|^2$$

But I dont know how to conclude that $\|T_\alpha-T_n\|$.

I think that I must limit the sum by an expression in terms of $n$ using that the sequence $\alpha=(\alpha_1, \alpha_2, \alpha_3...)$ converges to zero. But I do not know how to do it. I appreciate your suggestions.

答:
Fix $\varepsilon>0$. Since $a_n\to0$, there exists $n_0$ such that $|a_n|<\varepsilon^2$ whenever $n>n_0$. So, for $n>n_0$,
$$
\|(T-T_n)x\|^2=\sum_{j=n+1}^\infty|a_jx_j|^2\leq \varepsilon^2\,\sum_{j=n+1}^\infty|x_j|^2
\leq\varepsilon^2\,\|x\|^2.
$$
This shows that $\|T-T_n\|<\varepsilon$ whenver $n>n_0$, so $\lim T_n=T$.


问:
Find the inverse for arbitrary k
I need to find a, b, c, d, e, f, g, h (all of which are not zero)

such that for all k is in Real number, show A is invertible or this can't happen

$$A = \left(\begin{array}{ccc}
a&b&c\\d&k&e\\f&g&h
\end{array}\;\begin{array}{c}\end{array}\right)$$

My answer was this can't happen, but I don't know what the answer is.
And I couldn't provide a complete proof to support my answer.

I assume that a, ~, h are all 1. Then use det and element row operation I got 0 determinant for all k.

And the general determinant for this with Sarrus's rule is

det(A) = k(ah-cf) + g(cd - ae) + b(ef - dh)

What should I do next? This is invertible or non-invertible?

Thanks,

答:
It is possible to choose $a,b,c,d,e,f,g,h$ such that the determinant is non-zero for all $k$, i.e., the matrix is invertible for all $k$.

Note that as you have $$\det(A) = k(ah-cf) + g( cd-ae) + b(ef-dh)$$
First lets cut the dependence on $k$ in the determinant, i.e., set $ah-cf = 0$, i.e., $ah=cf$. Hence, now the determinant is (eliminating $a$)
$$\det(A) = g\left(cd-\dfrac{cfe}h\right) + b(ef-dh) = - \dfrac{(dh-ef)(bh-cg)}h$$
Now choose $b,c,d,e,f,g,h$ such that the determinant is non-zero.

Title: Find the inverse for arbitrary k


Title: Compactness of diagonal operator $ T_\alpha(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,...)$

Body:
Let $\alpha=(\alpha_1, \alpha_2, \alpha_3...)$ a sequence of complex numbers  that converges to zero.

I need to show the compactness of the diagonal operator $T_\alpha:l^2\to l^2$ defined by $$ T_\alpha(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,...)$$

For this:

I define the sequence the operator of finite range $T_n$ so that for $n\in \mathbb{N}$

$$T_n(x_1, x_2,x_3...)=(\alpha_1x_1, \alpha_2x_2,\alpha_3 x_3,..., \alpha_n x_n,0,0,0,... )$$

I need to show that $T_n\to T_\alpha$ and then $T_\alpha$ is compact.

I have that $$\|(T_\alpha-T_n)(x_1, x_2,x_3...)\|^2=\sum_{j=n+1}^\infty|\alpha _j x_j|^2$$

But I dont know how to conclude that $\|T_\alpha-T_n\|$.

I think that I must limit the sum by an expression in terms of $n$ using that the sequence $\alpha=(\alpha_1, \alpha_2, \alpha_3...)$ converges to zero. But I do not know how to do it. I appreciate your suggestions.

Body:
$$\|(T_\alpha-T_n)(x)\|^2=\sum_{j=n+1}^\infty|\alpha _j x_j|^2 \leq \left(\sup_{m>n} |a_m|^2\right)\sum_{j=n+1}^\infty|x_j|^2 \leq \left(\sup_{m>n} |a_m|^2\right) \|x\|^2$$

Therefore, $$\|T_\alpha-T_n\| \leq \sup_{m>n} |a_m|$$


问:
Checking if a quotient group is cyclic
Is there an easy way to check if the group $(\mathbb{Z}_9 \times \mathbb{Z}_9)/\langle(3,3)\rangle$ is cyclic? I know I could determine this by writing out each of the elements and checking if one of them is a generator. But there are 27 elements (if my math is correct), which would make that approach very tedious. Is there an easier way?

答: It isn't.  In the quotient, the order of any element's image under the canonical projection $\pi$ divides its order in $\Bbb Z_9\times\Bbb Z_9$.  Thus it divides $9$.

Title: Checking if a quotient group is cyclic


问:
To show $\int_{[-2,2]} \frac{x^{2n}}{1+x^{2n}} d \mu \to 2$ using convergence theorem
Let $\mu$ be the Lebesgue measure. Then use convergence theorem to evaluate :

$(1)$ To show $\int_{[0,\pi/2]} (\sin x)^n d \mu \to 0$,

$(2)$ To show $\int_{[-2,2]} \frac{x^{2n}}{1+x^{2n}} d \mu \to 2$.

Answer:

$(1)$  For $0 \leq x < \pi/2$, we have $ 0 \leq \sin x <1$, and so $(\sin x)^n \to 0$ while for $x=\pi/2$, we have $\sin (\pi/2)=1$.

Thus $\int_{[0,\pi/2]} (\sin x)^n d \mu=\int_{[0,1)} (\sin x)^n d \mu+ \int_{\{\frac{\pi}{2}\}} (\sin x)^n d \mu=0+\mu (\{\frac{\pi}{2}\})=0+0=0$.

How to prove the 2nd one ?

$\int_{[-2,2]} \frac{x^{2n}}{1+x^{2n}} d \mu=2 \int_{[0,2]} \frac{x^{2n}}{1+x^{2n}} d \mu=2 \leq \int_{[0,2]} \frac{x^{2n}}{1+x^{2n}} d \mu=2 \mu([0,2])=4 $  but we have to show $2$.

Help me

答:
Your mistake is the last part  of that line : $\int_{0}^2 \frac{x^{2n}}{1+x^{2n}} d \mu \not \to \mu([0,2])$, because you have to be careful about the pointwise limit.

In question $2$, DCT applies since the (absolute value of the) given function is uniformly bounded by $1$. Hence, we need to look at the limit as $n \to \infty$ of $\frac{x^{2n}}{1+x^{2n}}$. Note that if $|x|<1$ then this limit is $0$ as $x^{2n} \to 0$, and if $|x| > 1$ this limit is $1$ as $x^{2n} \to \infty$. Of course, $|x|=1$ is a set of measure zero, hence ignored.

Therefore, the pointwise limit of the functions $\frac{x^{2n}}{1+x^{2n}}$ is(upto the set of measure zero) :
$$
f(x) =\begin{cases}
0 & |x| <1\\
1 & |x| > 1
\end{cases}
$$
therefore the answer is the measure of $[-2,2] \cap \{|x| >1\} = [1,2] \cap [-1,-2] = 2$. You got this question wrong by exactly half because you thought the whole function converges to $1$ so the answer is the entire measure of $[-2,2]$ which is twice the correct answer.


问:
The integrals $\frac{1}{2\pi} \int_{-\infty}^\infty\frac{ e^{-i z u}}{z \pm i c }dz $
I would like to evaluate the two integrals
$$I_\pm(u) = \frac{1}{2\pi}\int_{-\infty}^\infty\frac{ e^{-i z u}}{z \pm i c }dz  $$
where $c$ is a positive real constant. I expect these each to be a real exponential times some Heaviside function, but I'm having trouble showing this. Can anyone offer guidance? I figure this is trivial for someone with more contour integration experience.

So far, concentrating on the integral $I_-(u)$, I attempted to use the semicircle contour depicted below:

Evaluating the residue at $z=ic$, this provides

$$I_-(u)  = i e^{uc} - \lim_{R\rightarrow \infty} \frac{1}{2\pi}\int_0^\pi d\theta \frac{iRe^{i\theta} \exp(-i R e^{i\theta} u )}{Re^{i\theta} - ic}$$

I suppose this limit depends on the sign of $u$ and therefore introduces a heaviside function? Shooting in the dark here...

答: I don't understand why people don't realize rectangular contours are more natural than semicircles.. It is mostly obvious that for $u <0$, $\lim_{R\to \infty} \int_{\partial [-R,R] + i[0,R]} \frac{e^{-izu}}{z-ic}dz = \int_{-\infty}^\infty \frac{e^{-izu}}{z-ic}dz$. Apply the residue theorem to the LHS. For $u>0$ the integrand in the LHS has exponential growth so we look instead at $\lim_{R\to \infty} \int_{\partial [-R,R] - i[0,R]} \frac{e^{-izu}}{z-ic}dz = -\int_{-\infty}^\infty \frac{e^{-izu}}{z-ic}dz$.  The LHS is $0$ by Cauchy's integral theorem.


问:
Prove the condition for system of linear equations to be solvable
It is clear that if the system of linear equations

$$
\left\{
\begin{array}{c}
x_1-x_3=c_1 \\
x_2-x_1=c_2 \\
x_3-x_2=c_3
\end{array}
\right.
$$

is solvable, then we have $c_1+c_2+c_3=0$.

How could we prove for the backward direction? That is, how could we prove that if $c_1+c_2+c_3=0$, then the above system of linear equations must be solvable without any corner cases?

答:
The augmented form of our system is
$$
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
$$
Row-reducing gives
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
-1 & 1 & 0 & c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]
\xrightarrow{R_2+R_1\to R_2}\left[\begin{array}{rrr|r}
1 & 0 & -1 & c_{1} \\
0 & 1 & -1 & c_{1} + c_{2} \\
0 & -1 & 1 & c_{3}
\end{array}\right]  \\
\xrightarrow{R_3+R_2\to R_3}\left[\begin{array}{rrr|r}
\fbox{1} & 0 & -1 & c_{1} \\
0 & \fbox{1} & -1 & c_{1} + c_{2} \\
0 & 0 & 0 & \fbox{$c_{1} + c_{2} + c_{3}$}
\end{array}\right]
\end{align*}
The first two boxed positions are guaranteed to be pivots in the reduced form of our system. The third boxed position is a pivot position if and only if $c_1+c_2+c_3\neq 0$. This third position is in the augmented column, so our system is solvable if and only if $c_1+c_2+c_3=0$.


问:
Is $h(n)$ independent of $n$?
Let $n$ be a positive integer.
Let $f(n) = \lfloor\frac{8n+13}{25}\rfloor$ and $g(n) = \lfloor\frac{n-17}{25}\rfloor$.

Now consider $$h(n) = f(n) - \lfloor\frac{n - 12 - g(n)}{3}\rfloor$$
Then I calculated values of $h(n)$ for various values of $n$ and I'm getting $h(n)=4 $ everytime. Infact I ran a C code also and verified it till $n = 10^6$. But I am not able to prove this equality. How can I prove this?

答:
Let $n=25k+r$ for some non-negative integers $k,r$ with $r<25$. We then have

$$\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{24k+r-12-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor = 8k-4+\left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor.$$

We also have

$$\left\lfloor\frac{8n+13}{25}\right\rfloor=8k+\left\lfloor\frac{8r+13}{25}\right\rfloor.$$

So it suffices to show that for any non-negative integer $r<25$,

$$\left\lfloor\frac{8r+13}{25}\right\rfloor - \left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor=0.$$

Try showing that this is true.

Title: Is $h(n)$ independent of $n$?


问:
Linear algebra(Linear transform and matrix)
with condition

$T:V \rightarrow V$

$\tilde{A}$={A1, A2, A3, A4, A5}

$\tilde{B}$={A1, A1+A2, A1+A2+A3, A1+A2+A3+A4, A1+A2+A3+A4+A5}

$T(A_i)=A_{i+1}, i = 1, ..., 4$

$T(A5) = O$

(1) Get $(T)$$ \tilde{A} \over \tilde{A}$

(2) Get $(T)$$ \tilde{B} \over \tilde{B}$

I have no idea how to get the answers. Can I use standard basis to get the answers?

答: Hint:  One way to get $T$ rel two bases is to apply it to the elements of the first and then write the results in terms of the second.   So, to get you started, the first column of the first one is $T(A_1)=A_2$, which in terms of $\tilde A$ is $(0,1,0,0,0)$.  Etc...


问:
How can one prove the worst case scenario of an algorithm purely mathematically?
How can someone prove purely mathematically that the worst case scenario (Big O) of the number of iteration of a recursive search algorithm is, let's say, 2 * log3(n)?

The only method that comes in my mind is by trying all possible combinations of a somewhat small list of elements (for example [1, 2, 3, 4, 5]), recording the time it takes, and find the worse time. Then, I would repeat it with a couple of different lists, then find, within the classical functions range (n^2, n^3, log(n), n*log(n), etc.) the most similar function according to the worse data points (worse time for each list).

Is it possible to prove any similar algorithm purely mathematically and not programmatically? What would be the steps for it?

答: Proving worst case complexity depends upon the particular algorithm, of course, but a general approach is to examine each branch of the algorithm and assume the ("worst case") data always happens to demand the longest (most complex) branch at each decision.  You needn't test all possible orders of data for such an analysis.


问:
Given the period of a trigonometric function, find the period of the transformed function
I would like someone to help me verify this exercise:

If the fundamental period of a periodic function $f$ is $4$, indicate the fundamental period of $3f(4x)+5$.

This is what I did:

$$Period=\frac{2\pi}{w_1}$$

$$4=\frac{2\pi}{w_1}$$

$$4w_1=2\pi$$

$$w_1=\frac{\pi}{2}$$

Then, let's say: $f(x)=sin(\frac{\pi}{2}x)$

$$3f(4x)+5=3sin(\frac{\pi}{2}(4x))+5=3sin(2\pi x)+5$$

Then, the new period is:

$$Period=\frac{2\pi}{w_2}=\frac{2\pi}{2\pi}=1$$

答:
I would have approached the problem differently.

The +5 term is irrelevant, because that is simply shifting the graph.

Similarly, the coefficient of 3 is irrelevant because that is merely stretching the graph.

Further, there is absolutely no reason to compare the function to trig functions, despite the fact that the original source of the function was a trig function.  Therefore, the factor of $\pi$ need not enter the discussion.

The two critical points are

period of $f$ is 4.

the new function [call it $g(x)$] relates to $f(4x)$.

Therefore, for example, since $f(4) = f(8)$, $g(1) = g(2).$

Therefore, it is immediate without math that the period of your function $3f(4x) + 5$ must be 1.


问:
the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$
I need to find:

$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \in \mathbb{N}$$

Looking at denominators, I see that [(...) represents any element between] : $$ 1 \le (\ldots) \le 1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$$

Then I take reverses and get:
$$ 1 \ge (\ldots) \ge \frac{1}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I put the other sequence on top of the former one (I see that the rightmost element is still the smallest one)
$$ 1 \ge (\ldots) \ge \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Then I take the sum of n elements on every end of inequality (to sum up n times the biggest element and n times the smallest element) and get:
$$ n \ge (\ldots) \ge \frac{n^2}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}$$

Ultimately I take into consideration $\frac{1}{\sqrt{2}}$ and get:
$$ \frac{n}{\sqrt{n}} \ge (\ldots) \ge \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})}$$

Now I can use the squeeze theorem and get:

$\lim_{n \to +\infty} \frac{n}{\sqrt{n}} = \infty$

$\lim_{n \to +\infty} \frac{n^2}{\sqrt{n}(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} = \lim_{n \to +\infty} \frac{\frac{n^2}{\sqrt{n}}}{(1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n})} \implies Stolz = \lim_{n \to +\infty} \frac{\frac{(n+1)^2}{\sqrt{n+1}}-\frac{(n)^2 }{\sqrt{n}}}{\sqrt{n+1}}$

And that is pretty disappointing - I think that the solution is wrong. Does anybody see an error in my way of thinking?

Unfortunately, I can not use integrals while doing that exercise.

答:
If you are familiar with generalized harmonic numbers.

The expression you are considering is
$$a_n=\frac{1}{\sqrt{n}}\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}$$ and, for large $p$, we have
$${H_p^{\left(-\frac{1}{2}\right)}}=\frac{2 p^{3/2}}{3}+\frac{\sqrt{p}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24 \sqrt{p}}+O\left(\frac{1}{
   p^{5/2}}\right)$$
$$\frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=\frac {3}{2\sqrt{p}}-\frac {9}{8p\sqrt{p}}+\cdots$$
$$\sum _{p=1}^n \frac{p}{H_p^{\left(-\frac{1}{2}\right)}}=3 \sqrt{n}+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
   \left(\frac{3}{2}\right)}{8}\right)+O\left(\frac{1}{n^{1/2}}\right)$$

$$a_n=3+\left(\frac{3 \zeta \left(\frac{1}{2}\right)}{2}-\frac{9 \zeta
   \left(\frac{3}{2}\right)}{8}\right)\frac 1 {\sqrt n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit and how it is approached.


问:
Find $f(x)$, given $f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2$
For a differentiable function $f(x)$, $$f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2.$$ Then find $f(x)$.

I am not able to think of any method which is working. I tried replacing $x$ and $y$, putting $x$ or $y = 0$, putting $y = x$, etc.
Can someone help me

Answer given at back is $f(x)=x+1$

答:
You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f ( x + y ) - 2 f ( x - y ) + f ( x ) - 2 f ( y ) = y - 2 \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = x + 1 $, even without assuming that $ f $ is differentiable. It's easy to check that the mentioned function is a solution. To prove the converse, first define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - x - 1 $. Then \eqref{0} can easily be rephrased in terms of $ g $ to get
$$ g ( x + y ) - 2 g ( x - y ) + g ( x ) - 2 g ( y ) = 0 \text . \tag 1 \label 1 $$
Letting $ x = y = 0 $ in \eqref{1} we get $ g ( 0 ) = 0 $. Now, letting $ x = 0 $ in \eqref{1} we get
$$ - 2 g ( - y ) = g ( y ) \text . \tag 2 \label 2 $$
Substituting $ - y $ for $ y $ in \eqref{2} we get $ g ( - y ) = - 2 g ( y ) $, which together with \eqref{2} itself shows that $ g ( y ) = 0 $, and thus $ f ( y ) = y + 1 $ for all $ y \in \mathbb R $.


问:
Is there always an equivalent metric which is not complete?
I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d)$. Let $d$ be complete. Does there exist an equivalent metric $d'$ in $X$ which is incomplete? When $d$ is compact, such $d'$ does not exist (Since compactness implies completeness). So answer me when $d$ is noncompact.</p>

答:
Here is a stronger theorem.

Theorem: Let $X$ be a metrizable space and let $(x_n)$ be a sequence in $X$ with no accumulation point.  Then $X$ embeds in a metrizable space $Y$ in which $(x_n)$ converges.

To answer your question from this theorem, note that if $X$ is not compact, it has a sequence $(x_n)$ with no accumulation point.  If you then take $Y$ as in the theorem and restrict its metric to $X$, you get a metric on $X$ for which it is not complete, since $(x_n)$ is Cauchy but does not converge.

Proof of Theorem: Let $d_0$ be a metric inducing the topology of $X$ and let $Y$ be $X$ with one new point $*$ added.  Let $d_1$ be a metric on the set $Z=\{x_n\}\cup\{*\}\subseteq Y$ such that $(x_n)$ converges to $*$ with respect to $d_1$. We define a metric $d$ on $Y$ as follows: for $a,b\in Y$, $d(a,b)$ is the infimum of all sums $$d_{i_0}(y_0,y_1)+d_{i_1}(y_1,y_2)+\dots+d_{i_n}(y_n,y_{n+1})$$ over all finite sequences of points $(y_0,\dots,y_{n+1})$ and sequences $(i_0,\dots,i_n)$ such that $y_0=a$, $y_{n+1}=b$, each $i_j$ is either $0$ or $1$, and if $i_j=0$ then $y_j,y_{j+1}\in X$ and if $i_j=1$ then $y_j,y_{j+1}\in Z$.  Intuitively, $d(a,b)$ is the "shortest distance" between $a$ and $b$ where you are allowed to jump between points using either the metric $d_0$ or the metric $d_1$.  It is easy to verify that $d$ is a pseudometric.  Also, $d\leq d_1$ when restricted to $Z$, so $(x_n)$ converges to $*$ with respect to $d$.  To verify that $d$ is in fact a metric (i.e., the distance between distinct points is positive) and that the restriction of $d$ to $X$ is equivalent to $d_0$, it suffices to prove the following claim:

Claim: For each $x\in X$, there exists $\epsilon>0$ such that $\{y\in X:d_0(x,y)<\epsilon'\}=\{y\in Y:d(x,y)<\epsilon'\}$ for all $\epsilon'\leq \epsilon$.

To prove this claim, note that since $(x_n)$ does not accumulate at $x$ in the topology of $X$, there exists $\epsilon_0>0$ such $d_0(x_n,x)>\epsilon_0$ for all $n$ such that $x_n\neq x$.  Also, if $x\in Z$, then $x$ is isolated in $Z$ with respect to $d_1$ (since $(x_n)$ converges to $*$), so there exists $\epsilon_1>0$ such that $d_1(x,z)>\epsilon_1$ for all $z\in Z\setminus\{x\}$.  Let $\epsilon=\min(\epsilon_0,\epsilon_1)$ (or just $\epsilon_0$ if $x\not\in Z$).

To show this value of $\epsilon$ works, suppose that $y\in Y$ and $d(x,y)<\epsilon'\leq\epsilon$; we must show that $y\in X$ and $d_0(x,y)<\epsilon'$ (note that the converse is trivial since $d\leq d_0$).  So, let $(y_0,\dots,y_{n+1})$ and $(i_0,\dots,i_n)$ be such that $y_0=x$, $y_{n+1}=y$, and $$d_{i_0}(y_0,y_1)+d_{i_1}(y_1,y_2)+\dots+d_{i_n}(y_n,y_{n+1})<\epsilon'.$$  We may assume that the $y_j$ are all distinct and that the $i_j$ alternate between $0$ and $1$ (we can collapse consecutive terms with the same $i_j$ value using the triangle inequality).  By our choice of $\epsilon_1$, there does not exist any $y_1\in Z$ such that $0<d_1(y_0,y_1)<\epsilon'$, so $i_0$ must be $0$.  If $n>0$, this means $i_1$ must be $1$, so $y_1\in Z$.  However, this is impossible, since $0<d_0(y_0,y_1)<\epsilon'$ and by our choice of $\epsilon_0$ there is no such $y_1$ that is in $Z$.  Thus we can only have $n=0$, so our sequence is just $(x,y)$, $y\in X$, and $d_0(x,y)<\epsilon'$.


问:
Quadrature formula
How can we find a quadrature formula $\int_{-1}^1 f(x) dx=c \displaystyle \sum_{i=0}^{2}f(x_i)$ that is exact for all quadratic polynomials?

Thanks for help.

答:
Write:
$$\int_{-1}^{1} f(x)\ dx =\int_{-1}^{1} ax^2+bx+c\ dx =\frac{2a}{3}+2c$$
Moreover you want to have three points all having the same weights. To make life simpler, let's also assume symmetry:

$$w\left(f(-x_0)+f(0)+f(x_0)\right) =\frac{2a}{3}+2c$$
$$w\left(2ax_0^2+3c\right)=\frac{2a}{3}+2c$$
It's clear that $w$ must be $2/3$, and so:
$$x_0^2 = 1/2 \to x_0=1/\sqrt{2}$$
All in all you have:
$$\int_{-1}^{1} f(x)\ dx \sim \frac{2}{3}\left(f(-1/\sqrt{2})+f(0)+f(1/\sqrt{2}) \right)$$

Edit: You can of course choose any three points and still get exact results, but this makes the calculation far less elegant.

Title: Quadrature formula


问:
Proof of Equivalence of $2$ seemingly Different Derivatives
In my post A very interesting question: intersection point of $x^y=y^x$ I demonstrated that for the graph $x^y=y^x$,
$$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$
HOWEVER

The following point has been brought to my attention by a friend: if we were to instead to calculate the derivative like this:
$$x^y=y^x\implies\frac{y}{\ln y}=\frac{x}{\ln x}\implies\frac{dy}{dx}\left(\frac{\ln y-1}{\ln^2y}\right)=\frac{\ln x-1}{\ln^2x}$$
$$\implies\frac{dy}{dx}=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}$$
So what we find is that if $x^y=y^x$ then
$$\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}$$

How do I prove this algebraically (ie NOT using calculus)? The only possible progress I have made is to observe that
$$\frac{\ln^2y}{\ln^2x}=\frac{y^2}{x^2}$$
on noting that
$$\frac{\ln^2y}{\ln^2x}=(\log_x y)^2=\frac{y^2}{x^2}$$
Thank you for your help. It is very much appreciated.

答:
For $x^y=y^x$, we have $y \ln x = x \ln y$ and $\dfrac {\ln^2 y}{\ln^2 x} = \dfrac {y^2}{x^2}$, as you have observed. Now:

\begin{align}&\quad\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}\\&= \frac{y^x \ln y -\frac yxx^y}{x^y \ln x -\frac xy y^x}\\&= \frac{y^x (\ln y -\frac yx)}{x^y (\ln x -\frac xy)}&\quad x^y=y^x\\&= \frac{\ln y -\frac yx}{\ln x -\frac xy}\\&= \frac{xy\ln y -y^2}{xy\ln x -x^2}\\&= \frac{y^2\ln x -y^2}{x^2\ln y -x^2}&\quad y \ln x = x \ln y\\&= \frac{y^2(\ln x -1)}{x^2(\ln y -1)}\\&=\frac{\ln^2y~(\ln x-1)}{\ln^2x~(\ln y-1)}&\quad \frac {\ln^2 y}{\ln^2 x} = \frac {y^2}{x^2}\end{align}


答:
Isn’t it quite easy? Start with the standard Laurent series
$$
\sin(z)=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\,,
$$
make your substitution $z\to2z^2$, and get
$$
\sin(2z^2)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}z^{4n+2}}{(2n+1)!}\,,
$$
in which your $a_n$-term is $(-1)^n2^{2n+1}z^{4n+2}/(2n+1)!$.
Now form $\mid a_{n+1}/a_n\mid$ to get
$$
\left|\frac {a_{n+1}}{a_n}\right|=\frac{2^{2n+3}z^{4n+6}\big/(2n+3)!}{2^{2n+1}z^{4n+2}\big/(2n+1)!}
=\frac{(2n+1)!}{(2n+3)!}\frac{2^{2n+3}z^{4n+6}}{2^{2n+1}z^{4n+2}}=\frac{4z^4}{(2n+2)(2n+3)}\,.
$$
That rightmost thing clearly goes to zero, no matter what $z$ is, so your radius is infinite, as I’m sure you expected.

Title: Find MacLaurin Series and radius of convergence

Body:
$$\sin{2z^2}$$
$$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$

Let's replace z with $2z^2$ in the MacLaurin series:

$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n
 \frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$

$$= z^2 - \frac{2^3}{3!}z^6 + \frac{2^5}{5!}z^{10} -+ ... = z^2 - \frac{4}{3}z^6 + \frac{4}{15}z^{10} -+...$$

But what about the radius of convergence? How do I find this?


问:
Inequality $(x^{-1}+\tan^{-1}x - \tanh^{-1}x )(\tan^{-1}x +\tanh^{-1}x )<2$
Can not figure out how to prove the inequality
$$(x^{-1}+\tan^{-1}x - \tanh^{-1}x )(\tan^{-1}x +\tanh^{-1}x )<2$$
over its domain.

Given the symmetry, it suffices to prove the inequality for $x\in (0,1)$. And, it is straightforward to verify that the limit of the LHS at $x=0$ is $2$. But the standard method to examine its first few derivatives does not appear to work, for it is not obvious to determine their signs due to the complex functional forms of the derivatives.

It is observed that the inequality around $x=0$ is very tight, which can be seen from the leading term $\frac{14}{15}x^4$ of its Taylor expansion.

答:
Let us prove that, for all $x$ in $(0, 1)$,
$$\frac{2}{\arctan x + \operatorname{arctanh} x} - \frac{1}{x} - \arctan x + \operatorname{arctanh} x \ge 0. \tag{1}$$
Denote LHS by $f(x)$. We have (the proof is given at the end)
$$f'(x) = -\frac{4}{(\arctan x + \operatorname{arctanh} x)^2(1-x^4)}  + \frac{1+x^4}{x^2(1-x^4)}\ge 0, \quad \forall x\in (0, 1). \tag{2}$$
Also, $\lim_{x\to 0} f(x) = 0$ (apply L'Hopital rule). Thus, we have $f(x) \ge 0$ for all $x$ in $(0, 1)$.

We are done.

$\phantom{2}$

Proof of (2): It suffices to prove that
$$\arctan x + \operatorname{arctanh} x - \frac{2x}{\sqrt{1+x^4}} \ge 0, \quad \forall x \in (0, 1).$$
Denote LHS by $g(x)$. We have, for all $x$ in $(0, 1)$,
$$g'(x) = \frac{2}{1-x^4} - \frac{2(1-x^4)}{(1+x^4)^{3/2}} \ge 0.$$
Also, $g(0) = 0$. Thus, $g(x) \ge 0$ for all $x$ in $(0, 1)$.


答: For a quadratic function, you can find an exact representation from any three samples $x_i$.  Thus, determine coefficients $a$, $b$, and $c$ of $y=a x^2+b x+c$ from these samples, and then integrate $y$ between $[-1,1]$.</p>

Title: Quadrature formula

Body:
How can we find a quadrature formula $\int_{-1}^1 f(x) dx=c \displaystyle \sum_{i=0}^{2}f(x_i)$ that is exact for all quadratic polynomials?

Thanks for help.


问:
$d(\bigcup_{i=1}A_i, \bigcup_{i=1}B_i)\leq \sum\limits_{i}d(A_i,B_i)$
Let $\mu:\mathcal {E} \rightarrow \mathbb{R}_{\geq0}$ be additive and regular.

Show that for countable families of subsets $A_i, B_i \subset \mathbb{R}^d, i \in \mathbb{N}:$

$$d(\bigcup_{i}A_i, \bigcup_{i}B_i)\leq  \sum\limits_{i}d(A_i,B_i)$$

答:
Hint:
$$d(A, B \cup C) \le \min\{d(A,B), d(A,C)\} \le  d(A,B) + d(A,C)$$


问:
Given that $BE$ is an angle bisector of $\angle ABC$ and $AD = 13.5$ cm, find $CD$ .

$\Delta ABC$ has $AB = 10$ cm, $BC = 40$ cm. Points $D$ and $E$ lie on side $AC$ and point $F$ on side $BC$ such that $EF \parallel AB$, $DF \parallel EB$. Given that $BE$ is an angle bisector of $\angle ABC$ and $AD = 13.5$ cm, find $CD$ .

What I Tried: Here is a picture :-

A first idea to immediately come in my mind was similarity, because one will find a lot of similar triangles here. For example $\Delta CFE \sim \Delta CBA$ , $\Delta CFD \sim \Delta CBE$ , $\Delta DFE \sim \Delta EBA$ and so on. I have also marked the angles based on the equality of them, noting that the green angle is equal to the sum of $1$ blue and $1$ brown angle. I also got $BF = FE$ .

I am not sure, however, how to deal with the sides. I could assume $BF = x$ and continue with similarity, or I could use $DE = y$ and continue, and there are a lot of possibilities and also time consuming. So is there a way to quickly deduce the value of $CD$ with a decent method of similarity? Thank You.

答:
$CD=x,DE=y,EA=z$
By similarity/thales theorem using$DF\parallel EB,EF\parallel AB$ $$\frac{x+y}{z}=\frac{CF}{FB}=\frac{x}{y}$$ also $$y+z=13.5$$ by angle bisector theorem $$\frac{z}{x+y}=\frac{10}{40}$$

Three variables three equations....


问:
How to solve equation $u^4 + bu + c=0 $
To solve for the roots of $x^2 + bx^\frac{1}{2} + c=0$,
I thought I might substitute
$u = x^\frac{1}{2}$ to get
$$u^4 + bu + c=0$$

But I don't really know what to do with that either. I plugged both into Wolfram Alpha, which provided some really long complicated solutions. However, I'm looking to implement this function in Julia and the translation to Julia would be ugly. I can determine a reasonable interval of interest, and can do a binary search for $0$ over that interval. So, I have that solution.

Is there a more elegant mathematical approach?

答:
The equation
$$u^4 + bu + c=0 $$
can be factorized as
$$\left( u^2+2s u +\frac{8s^3-b}{4s} \right)
 \left( u^2-2s u +\frac{8s^3+b}{4s} \right)=0
$$
where $s$ satisfies
$s^6-\frac c4s^2-\left(\frac b8\right)^2=0$, a cubic equation in $s^2$. The roots are then obtained via the familiar quadratic formula
$$u = -s \pm \sqrt{-s^2+\frac b{4s}},\>\>\>  s\pm \sqrt{-s^2-\frac b{4s}}
$$

Take the example $u^4+4u-1=0$ and solve for $s^2=\frac1{2}$.
Apply the formulas above to obtain the real roots

$$u=-\frac1{\sqrt2}\pm \sqrt{\sqrt2 -\frac12}$$

Title: How to solve equation $u^4 + bu + c=0 $


问:
List videos of interesting courses at the doctoral level.
Many mathematics departments has provided video lessons their courses (usually one semester) that are offered in their doctoral programs in mathematics. Most often these courses total average of 26 videos each containing a single class.  Below is a list of introdutory courses on video.

Introduction to Algebraic Topology

Introduction to Lie Algebras

Extremal and Probabilistic Combinatorics

I would appreciate if people could add on to this list.

Observation To a greater audience reach classes should preferably be in English language and introdutory curses.

Observation I will not be specific as to the area of investigation. Since I believe that the answers here will be useful for students who are not my area of interest. But I have a preference for introdutory courses of Ergodic Theory, Probability, Stochastic Processes, Combinatorics and Statistical Mechanics.

答: There is a very much related thread on MathOverflow with a lot of links.</p>


问:
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$
The question is:

Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$

I did the question without using the Hint, but I don't know how to do it using the hint.

Quick working out of what I've done:

\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}

\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}

\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}

$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$

But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !

答:
Let $\cos^{-1}\dfrac35=y,\cos y=?$

Using Principal values,  as $1>\dfrac35>0; 0<y<\dfrac\pi2$

$\tan y=\dfrac{+\sqrt{1-\left(\dfrac35\right)^2}}{\dfrac35}=?, y=\tan^{-1}\dfrac43$

$$\pi-2\tan^{-1}2=2\left(\dfrac\pi2-\tan^{-1}2\right)=2\cot^{-1}2$$

Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

and

Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$


答:
Hint:

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

$$2\tan^{-1}2=\pi+\tan^{-1}\dfrac{2\cdot2}{1-2^2}=\pi+\tan^{-1}\left(?\right)$$

Now using Principal values  $\tan^{-1}(-y)=-\tan^{-1}y$

Again, if $\tan^{-1}\dfrac43=u,0<u<\dfrac\pi2$  as $\dfrac43>0$

$\cos u=\dfrac1{+\sqrt{1+\tan^2y}}=?$

Can you take it home from here?


问:
Advocating base 12 number system
I had a calculus professor who suggested we should be using base 12 number system. What are the advantages of using such a system?   </p>

答: More factors: 1/2, 1/3, 1/4, 1/6 (and sometimes 1/12) are common fractions, which would turn out "even" (not infinite). The Babylonians where on to something with their base-60 system...</p>

Title: Advocating base 12 number system


问:
Subsets $Y$ of a partially ordered set $L$ need not have least upper bounds nor greatest lower bounds
I am currently studying the textbook Principles of Program Analysis by Flemming Nielson, Hanne R. Nielson, and Chris Hankin. Appendix A Partially Ordered Sets says the following:

Partially ordered set. A partial ordering is a relation $\sqsubseteq : L \times L \rightarrow \{ \text{true}, \text{false} \}$ that is reflexive (i.e. $\forall l : l \sqsubseteq l$), transitive (i.e. $\forall l_1, l_2, l_3 : l_1 \sqsubseteq l_2 \land l_2 \sqsubseteq l_3 \Rightarrow l_1 \sqsubseteq l_3$), and anti-symmetric (i.e. $\forall l_1, l_2 : l_1 \sqsubseteq l_2 \land l_2 \sqsubseteq l_1 \Rightarrow l_1 = l_2$). A partially ordered set $(L, \sqsubseteq)$ is a set $L$ equipped with a partial ordering $\sqsubseteq$ (sometimes written $\sqsubseteq_L$). We shall write $l_2 \sqsupseteq l_1$ for $l_1 \sqsubseteq l_2$ and $l_1 \sqsubset l_2$ for $l_1 \sqsubseteq l_2 \land l_1 \not= l_2$.

A subset $Y$ of $L$ has $l \in L$ as an upper bound if $\forall l^\prime \in Y : l^\prime \sqsubseteq l$ and as a lower bound if $\forall l^\prime \in Y : l^\prime \sqsupseteq l$. A least upper bound $l$ of $Y$ is an upper bound of $Y$ that satisfies $l \sqsubseteq l_0$ whenever $l_0$ is another upper bound of $Y$; similarly, a greatest lower bound $l$ of $Y$ is a lower bound of $Y$ that satisfies $l_0 \sqsubseteq l$ whenever $l_0$ is another lower bound of $Y$. Note that subsets $Y$ of a partially ordered set $L$ need not have least upper bounds nor greatest lower bounds but when they exist they are unique (since $\sqsubseteq$ is anti-symmetric) and they are denoted $\bigsqcup Y$ and $\sqcap Y$, respectively.

It is this part that I am unsure about:

Note that subsets $Y$ of a partially ordered set $L$ need not have least upper bounds nor greatest lower bounds but when they exist they are unique (since $\sqsubseteq$ is anti-symmetric) and they are denoted $\bigsqcup Y$ and $\sqcap Y$, respectively.

This isn't obvious to me. Would someone please explain/clarify this?

答:
To see that there are posets without least upper bounds nor greatest lower bounds consider $L = [-1,0) \cup (0,1) \cup (1,2]$ and $Y = (0,1)$. Let your partial order be the usual total order on $\mathbb{R}$. Then there is no least upper bound or greatest lower bound of $Y$ in $L$.

To prove this explicitly for the least upper bound case: suppose some $l \in L$ were a least upper bound of $Y$. Then we can select any $l_0 \in (1,l)$ and $l_0 < l$ is another upper bound for $Y$, but it's less than $l$. This is a contradiction, so no least upper bound can exist.

A similar argument works for the greatest lower bound.

For the uniqueness aspect, given existence, suppose both $l$ and $l^\prime$ were a least upper bound for $Y$ in $L$. Then this means in particular that they are both an upper bound. So $l \sqsubseteq l^\prime$ because $l$ is a least upper bound. But, $l^\prime \sqsubseteq l$ since $l'$ is also a least upper bound. By anti-symmetry this means we must have $l = l^\prime$.

A similar argument works for uniqueness of a greatest lower bound given its existence.