问:
Alternative Proof of Equivalent definitions of continuity
The proof that we often see for the equivalence of $\epsilon$-$\delta$ definition of a limit and the definition of a limit in terms of sequences often use proof by contradiction. As an exercise I wanted to try a direct proof.

I'll only prove one direction as this is the part I had a question on. We consider function $f: \mathbb{R} \rightarrow \mathbb{R}$ and assume that it is continuous at $(x,y)$ in the sequential sense.

Choose any $\epsilon > 0$. Further, choose any sequence $\{x_n\}$ where $\lim x_n = x$ and $x_n \neq x$ for all $n$. We know that there must exist some $n_1$ where $n > n_1$ implies $|f(x_n) - y| < \epsilon$. Also, there must be some $n_2$ for which $n > n_2$ implies $|x_n - x| < \epsilon$. Now, we let

$$\delta = \sup \{|x_n - x| : n > \max(n_1, n_2)\}.$$

Clearly, $\delta$ is finite and larger than zero. Further if we have $x'$ such that $|x' - x| < \delta$, then there must exist some $x_{n'}$ for which $|x' - x| < |x_{n'} - x|$.

I want to conclude that $|f(x') - y| < \epsilon$, but I'm not sure how to proceed from here.

答:
Suppose that $f$ is given by

$$f(x)=\begin{cases}
0,&\text{if }x=\frac1n\text{ for some }n\in\Bbb Z^+\\
2x,&\text{otherwise;}
\end{cases}$$

this is continuous at $x=0$. We might happen to choose the sequence $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$, so that $x_n=\frac1n$ for each $n\in\Bbb Z^+$. Now take $\epsilon=\frac12$. We can take $n_1=1$ and $n_2=2$, so that $\delta=\frac13$. Let $x'=\frac3{10}$; then $|x'|<\delta$, and $x'<\frac13=x_3$, so we can take $n'=3$. But then

$$|f(x')-f(0)|=\frac35>\frac12=\epsilon\,.$$

Thus, on the basis of what you’ve done so far you cannot hope to prove in general that $|f(x')-y|<\epsilon$. I am not at all sure that a direct proof is possible, because somehow you really do need to use the fact that $f$ behaves at $x$ for all sequences converging to $x$.


问:
Generators increasing the word metric
Let $G$ be an infinite group with a finite generating set $T$ which is symmetric ($T=T^{-1}$) and let $|\cdot|$ be the corresponding word metric, i.e. $|g|$ is the minimal number of (not necessarily distinct) elements from $T$ one needs to multiply to get $g$.

Is it true that for every $g\in G$ there is a $t\in T$ such that $|gt|>|g|$?

答: Consider $G=\mathbb{Z}$ with $T=\{\pm2,\pm3\}$. Then $|1|=2$, and $|1+t|\leq 2$ for all $t\in T$.

ContentLicense: CC BY-SA 4.0

Title: Generators increasing the word metric


问:
"Interpretation" versus "judgement" versus "assertion"?
I believe "assertion" and "judgment" mean the same thing (at least according to Per Martin Lof). I understand that they are meta statements (above that of say formal propositions). How, if at all, do these meanings differ from that of "interpretation", which is also on top of a proposition?</p>

答:
An interpretation is a situation in which a statement can become true or false.

The claim that the statement is true in that situation is a judgement.

E.g., "$P$ means 'is even' and $a$ means '1'" is an interpretation.

"Under the interpretation where $P$ means 'is even' and $a$ means '1', the statement $P(a)$ is false" is a judgement or assertion.

There are plenty kinds of assertions other than truth under an interpretation, e.g. "$X$ is provable", "$X$ logically entails $Y$", $X$ is a formula", etc.


问:
I'm having a hard time finding the $c_1$ and $c_2$ coordinates of a system after finding the Eigenvalues and Eigenvectors.
My task is to solve a system given the initial value.

Solve System : $\frac{{d\vec{x}}}{dt} = \left | \matrix{-4 & 3 \\ 2 & -3 } \right |$ with init value $\vec{x_0} = \left | \matrix{1 \\ 0}\right | $

What I have done so far:

So I have a matrix :

$\frac{{d\vec{x}}}{dt} = $$\left | \matrix{-4 & 3 \\ 2 & -3 } \right | $

I found the Eigenvalues, $\lambda_1 = -6$ , $\lambda_2 = -1$
I found the Eigenvectors by solving the system of the Eigenvalues

Which are $\vec{v1} = span(\left | \matrix{3 \\ -2} \right |)$

and $\vec{v2} = span(\left | \matrix{1 \\ 1} \right |)$

So now, I have to find a c(t) = $S^{-1}\vec{x(t)}$

What I did, then was applied my eigenvectors to a new matrix
$S =\left | \matrix{3 && 1 \\-2 && 1} \right |$

and get the $S^{-1}=\det(\left | \matrix{3 && 1 \\-2 && 1} \right | )$

which is $\frac{1}{5}$

The point where I am confused:

So, according to the theorem in my book, to find the scalars $c_1 , c_2,...,c_n$ according to the basis
$\vec{v_1}, \space \vec{v_2}, \space ,...\vec{v_n}$ I have to write the matrix form as.

$S \lambda^{t}_t S^{-1}\vec{x_0} = S =\left | \matrix{3 && 1 \\-2 && 1} \right |\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right| \frac{1}{5}\left | \matrix{3 && 1 \\-2 && 1} \right |\left|\matrix{1 \\ 0}\right| $

with $\lambda^{t}_t$ as a stand in for

$\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right|$

So now, I would do the math and end up with :

$\left| \matrix{\frac{9}{5}e^{-6t} & \frac{-2}{5}e^{-t} \\ \frac{-6}{5}e^{-6t} & \frac{-2}{5}e^{-t}}\right|$

But I look in the back of the book, aaaand the coordinates $c_1$ && $c_2$ with respect to $\vec{v1}$ &&  $ \vec{v2}$ are actually $c_1 = \frac{-1}{5}$ and $c_2 = \frac{2}{5}$

I'm clearly doing something wrong here, but I do not know what exactly.

答:
$\frac {dx}{dt} = Ax\\
\frac {dx}{dt} = e^{At}x_0\\
A = PDP^{-1}\\
e^{At} = P e^{Dt}P^{-1}\\
\frac{dx}{dt} = P e^{Dt}P^{-1}x_0\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&-2\\1&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\&e^{-t}\end{bmatrix}\begin{bmatrix} 1& -1\\2&3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&1\\-2&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\&e^{-t}\end{bmatrix}\begin{bmatrix} 1\\2\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&1\\-2&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\2e^{-t}\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3e^{-6t}+2e^{-t}\\-2e^{-6t}+2e^{-t}\end{bmatrix}\\
\frac{dx}{dt} = \begin{bmatrix} \frac {3}{5}e^{-6t}+\frac {2}{5}e^{-t}\\-\frac{2}{5}e^{-6t}+\frac {2}{5}e^{-t}\end{bmatrix}$
$


问:
Integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n \in \mathbb{Z}$
Find integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n$ is an integer.

So far, I have just listed some squares which are larger than $2005.$ The first few are $2025,2116, 2209,2304,$ etc. I can just plug in values of $n$ which match up the square minus $2005.$ However, this seems way too bashy. Is there any good way to do this problem? Thanks in advance.

答:
HINT

I would start by noticing that

\begin{align*}
\sqrt{n} + \sqrt{n+2005} = \frac{2005}{\sqrt{n + 2005} - \sqrt{n}}
\end{align*}

Now you can solve the equation $\sqrt{n + 2005} - \sqrt{n} = k$, where $k$ divides $2005$.

Can you take it from here?


问:
How to write in $2^x=5$ in logarithmic form?
How do I write:
$$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.

答: you can see all the identity used in other answer in this answer of mine in Write the expressoin in terms of $\log x$ and $\log y \log(\frac{x^3}{10y})$</p>

ContentLicense: CC BY-SA 3.0


问:
How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$
The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$

My attempt

I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.

答:
Let $t=x-\frac\pi6$ to get

\begin{align}
& I= \int \frac{\cos x+ \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{d}x
=\int \frac{\frac{\sqrt3}2(2+\cos t)-\frac12\sin t }{(1+ 2 \cos t)^2}{d}t
\end{align}
Note that $\left( \frac{\sin t}{1+2\cos t}\right)’ = \frac{2+\cos t}{(1+2\cos t)^2}$ and $\left( \frac1{1+2\cos t}\right)’ = \frac{2\sin t}{(1+2\cos t)^2}$. Then
$$I = \frac{\sqrt3}2 \frac{\sin t}{1+2\cos t}- \frac14\frac{1}{1+2\cos t}  = \frac{2\sqrt3\sin t-1}{4(2\cos t+1)}+C
$$


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

答:
You may like this. Suppose $x\ge0$. Note
$$ 1\le(1+t)^\frac14 \le 1+\frac1{2}t$$
which is derived from
$$ 1\le 1+t\le(1+\frac12t)^4. $$
So
$$ x\le\int_0^x(1+t)^\frac14dt \le \int_0^x(1+\frac12t)dt=x+\frac14x^2. $$
Since
$$(1+x)^{5/4}=1+\frac{5}{4}\int_0^x(1+t)^{\frac14}dt $$
you hav
$$ 1+\frac54x\le(1+x)^\frac54 \le 1+\frac54x+\frac5{16}x^2.$$


答:
The denominator of $1+e\cos\theta$ occurs in inverse $r^2$ force problems and when $e\gt1$ as here, corresponds to a hyperbolic orbit, like Rutherford scattering or ʻOumuamua. We can zap this denominator depending on whether the orbit is elliptical, parabolic, or hyperbolic. In the latter case we may let
$$\sin y=\frac{\sqrt{e^2-1}\sinh\theta}{e\cosh\theta-1}=\frac{\sqrt3\sinh\theta}{2\cosh\theta-1}$$
$$\cos y=\frac{e-\cosh\theta}{e\cosh\theta-1}=\frac{2-\cosh\theta}{2\cosh\theta-1}$$
The reader may check that $\cos^2y+\sin^2y=1$. Taking differentials we have
$$\cos y\,dy=\frac{e-\cosh\theta}{e\cosh\theta-1}dy=\frac{\sqrt{e^2-1}(e-\cosh\theta)}{(e\cosh\theta-1)^2}$$
So
$$dy=\frac{\sqrt{e^2-1}}{e\cosh\theta-1}d\theta=\frac{\sqrt3}{2\cosh\theta-1}d\theta$$
And then
$$1+e\cos y=1+2\cos y=\frac{e^2-1}{e\cosh\theta-1}=\frac3{2\cosh\theta-1}$$
$$\cosh\theta=\frac{\cos y+e}{1+e\cos y}=\frac{\cos y+2}{1+2\cos y}$$
$$\sinh\theta=\frac{\sqrt{e^2-1}\sin y}{1+e\cos y}=\frac{\sqrt3\sin y}{1+2\cos y}$$
So we can just clean up the denominator a little by letting $x=y+\frac{\pi}6$ so that $\sin\left(x+\frac{\pi}3\right)=\sin\left(y+\frac{\pi}2\right)=\cos y$ and $\cos x=\cos\left(y+\frac{\pi}6\right)=\frac{\sqrt3}2\cos y-\frac12\sin y$ so we have
$$\begin{align}I&=\int\frac{\cos x+\sqrt3}{\left(1+2\sin\left(x+\frac{\pi}3\right)\right)^2}dy=\int\frac{\frac{\sqrt3}2\cos y-\frac12\sin y+\sqrt3}{(1+2\cos y)^2}dy\\
&=\int\frac{(2\cosh\theta-1)^2}9\left[\frac{\sqrt3}2\frac{(2-\cosh\theta)}{(2\cosh\theta-1)}-\right.\\
&\quad\left.\frac{\sqrt3}2\frac{\sinh\theta}{(2\cosh\theta-1)}+\sqrt3\frac{(2\cosh\theta-1)}{(2\cosh\theta-1)}\right]\frac{\sqrt3\,d\theta}{(2\cosh\theta-1)}\\
&=\int\left(\frac12\cosh\theta-\frac16\sinh\theta\right)d\theta=\frac12\sinh\theta-\frac16\cosh\theta+C_1\\
&=\frac{\frac{\sqrt3}2\sin y-\frac16\cos y-\frac13}{1+2\cos y}+C_1=\frac{\frac{\sqrt3}2\sin y-\frac14}{1+2\cos y}+C\\
&=\frac{\frac34\sin x-\frac{\sqrt3}4\cos x-\frac14}{1+2\cos\left(x-\frac{\pi}6\right)}+C=\frac{\frac34\sin x-\frac{\sqrt3}4\cos x-\frac14}{1+2\sin\left(x+\frac{\pi}3\right)}+C\end{align}$$

Title: How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$

Body:
The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$

My attempt

I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.


答:
If $a$ and $b$ are non-negative integers, then $\sqrt{a}+\sqrt{b}$ is an integer if and only if $a$ and $b$ are perfect squares.

So if $\sqrt{n}+\sqrt{n+2005}$ is a perfect square, $n = k^2$ and $n+2005 = m^2$ for some non-negative integers $k,m$.

Substituting $n = k^2$ into $n+2005 = m^2$ yields $k^2+2005 = m^2$, i.e. $2005 = m^2-k^2 = (m+k)(m-k)$.

Hence, $m+k$ and $m-k$ must be complimentary factors of $2005$. Since the prime factorization of $2005$ is $2005 = 5 \cdot 401$ and $m+k \ge m-k$, there are only two possibilities.

Case 1: $m+k = 2005$ and $m-k = 1$. Solve this to get $(m,k) = (1003,1002)$, i.e. $n = 1002^2$.

Case 2: $m+k = 401$ and $m-k = 5$. Solve this to get $(m,k) = (203,198)$, i.e. $n = 198^2$.

Title: Integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n \in \mathbb{Z}$

Body:
Find integer values of $\sqrt{n}+\sqrt{n+2005}$ where $n$ is an integer.

So far, I have just listed some squares which are larger than $2005.$ The first few are $2025,2116, 2209,2304,$ etc. I can just plug in values of $n$ which match up the square minus $2005.$ However, this seems way too bashy. Is there any good way to do this problem? Thanks in advance.


答:
There's a simple example that is:

$$\color{green}{10}^\color{red}{3}=\color{blue}{1000}\tag{1}$$

$$\log_{\color{green}{10}} \color{blue}{1000}=\color{red}{3}\tag{2}$$

Notice that the question changed of point:

For $(1)$ you have: what's the result of $10$ multiplied by itself $3$ times?

For $(2)$ you have: what number should be the exponent of $10$ so that $10^{x}=1000$?

Although Isomorphic's answer is way better because it introduces properties of logarithms, this answer will help you to reflect on the nature of logarithms. I've found this particularly useful on high school.

Title: How to write in $2^x=5$ in logarithmic form?

Body:
How do I write:
$$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.


问:
Prove using Greens Theorem that the line integral is independent of curve C
Let f :
$R^2$ → R be a twice continuously differentiable scalar field which is harmonic,
that is, $∇^2$
f = 0. Use Green’s Theorem to prove that the line integral
$$\int_c {df\over dy}dx-{df\over dx}dy$$

is independent of curve C in the (xy)-plane.

$$so far $$
I have worked out that
∇ x f=0 for the graph being irrotational
I'm not sure this is relevant to be honest
I know f is conservative if it is the gradient of a scalar field as well and I think conservative and independent are the same in this sense I'm not sure how to take the gradient I've tried to apply greens theorem making ${df\over dy}$ =     and $-{df\over dx}$=Q but the answers just look plain wrong to me can anyone help

my application gave me $2 \int \int_R {d^2f\over dxdy}$dxdy  if that is right I don't know how it proves independence.

答:
To apply Green's Theorem, we have to have a closed curve that encloses a region. $C$ itself might not be closed. The phrase "independent of curve $C$" means that as long as two curves $C_1$ and $C_2$ start and end at the same point, then they will give the same line integral. On a piece of paper, draw a random curve $C_1$ and then draw another curve $C_2$ starting and ending at the same point. Notice that this now encloses a region. Call it $R$.

Then Green's Theorem says that along the curve $C=C_1-C_2$, which is the path along $C_1$ and then back to the beginning along $C_2$, we have $$\int_C Pdx+Qdy=\iint_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$$

So we actually get $$\iint_R\left(-\frac{\partial^2 f}{\partial x^2}-\frac{\partial^2 f}{\partial y^2}\right)dA=-\iint_R\left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\right)dA$$

Now we need to use the fact that $f$ is harmonic. What does this tell us?


问:
Kronecker's famous quote about integer numbers
I am working with the excellent book of Stephen Abbott's Understanding Analysis.
The author (p. 3) quotes Kronecker:

The natural numbers are the work of God. All the rest is the work of the mankind.

As far as I know the exact quote attributed to Kronecker reads:

God made the integers; all else is the work of man.

[Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk

In German ganzen Zahlen stands for integer numbers;  natürlichen Zahlen means natural number]

Is it this just a simple misquote of the author or am I missing something here? Thanks in advance.

答: Found a free preview. In context, the author does not think the reader yet knows what the word integers means. The quote is paraphrased, very much on purpose.


问:
If $V$ is completely reducible and $End_{A}(V)$ is a division ring then $V$ is irreducible
Since V is completely reducible we can write V as a direct sum of irreducible sub modules or a direct sum of 2 complement sub modules.
Can anyone give a hint on which way to write V and how can I make use of $End_{A}(V)$ as a division ring.

I think  writing $V = W \oplus U$ is better. Since W is any submodule.
I think I should choose a map from V to W and show that W is equal to {0} or V

答:
Suppose that $V$ is not irreducible, i.e. $V=V_1\oplus V_2$.

Then $(Id_{V_1}\oplus0)\circ (0\oplus Id_{V_2})=0$. We deduce that $End_A(V)$ is not a division ring.


问:
solving $\sin\left(x+\frac{3\pi }{5}\right)=2\sin\left(\frac{\pi }{5}-\frac{x}{2}\right)$
Trying to solve this:
$$\sin \left( x+\frac{3\pi}{\:5} \right) =2\sin \left( \frac{\pi}{5}-\frac{x}{2} \right) $$

I tried to divide $\sin\left(\frac{\pi }{5}-\frac{x}{2}\right)$ and expanded to get

$$\frac{\sin \left( x \right) \cos \left( \frac{3\pi}{5} \right) +\cos \left( x \right) \sin \left( \frac{3\pi \:}{5} \right)}{\sin \left( \frac{\pi}{5} \right) \cos \left( \frac{x}{2} \right) -\cos \left( \frac{\pi}{5} \right) \sin \left( \frac{x}{2} \right)}=2$$

I don't know how to proceed. Appreciate any help!

答:
$\sin\left(x + \frac{3\pi}{5}\right) = \sin\left(2(\frac x2 + \frac{3\pi}{10})\right) = 2\sin \left(\frac x2 + \frac{3\pi}{10}\right)\cos \left(\frac x2 + \frac{3\pi}{10}\right)$

Using $\sin(\frac{\pi}{2} -y) = \cos y $ and $\cos(\frac{\pi}{2} -y) = \sin y $ we can write,

$\sin\left(x + \frac{3\pi}{5}\right) = 2\sin \left(\frac{\pi}{5}-\frac x2 \right)\cos \left(\frac{\pi}{5}-\frac x2 \right)$.

So you're left with,

$2\sin \left(\frac{\pi}{5}-\frac x2 \right)\cos \left(\frac{\pi}{5}-\frac x2 \right) = 2\sin \left(\frac{\pi}{5}-\frac x2 \right)$

$\sin \left(\frac{\pi}{5}-\frac x2 \right) = 0 $ or $\cos \left(\frac{\pi}{5}-\frac x2 \right)=1$

Could you solve now?


问:
Continuous Function + open set
A mapping $T$ of a metric space $X$ into a metric space $Y$ is continuous iff the inverse image of any open subset $Y$ is open subset of $X$.

Proof:
(a)Suppose that $T$ is continuous. Let $S \subset Y$ be open and $S_0$ the inverse image of $S$. If $S_0 = \emptyset$, it is open. Let $S_0 \neq \emptyset$. For any $x_0\in S_0$ let $y_0=Tx_0$. Since $S$ is open, it contains an $\epsilon$-neighborhood $N$ of $y_0$. Since $T$ is continuous, $x_0$ has a $\delta$-neighborhood $N_0$ which is mapped into $N$. Since $N\subset S$, we have $N_0\subset S_0$ ...

Please You will be able explain me in detailed this part "Since $N\subset S$, we have $N_0\subset S_0$", I don't understand "we have $N_0\subset S_0$".

答: If $x \in N_0$, then $T(x) \in N$ (as "$N_0$ is mapped into $N$") so $T(x)$ is in $S$ (as we assume that $N \subset S$). So by definition, $x \in S_0$, as $S_0$ is defined the be the inverse image of $S$ ($S_0 = T^{-1}[S]$), so exactly the set of all $p$ with $T(p) \in S$, and the $x$ we start with is such a point.</p>

Title: Continuous Function + open set


答:
HINT

I would start by noticing that

\begin{align*}
\sin\left(x + \frac{3\pi}{5}\right) = \sin\left(\pi - \frac{3\pi}{5} - x\right) = \sin\left(\frac{2\pi}{5} - x\right) = 2\sin\left(\frac{\pi}{5} - \frac{x}{2}\right)\cos\left(\frac{\pi}{5} - \frac{x}{2}\right)
\end{align*}

Consequently, the proposed equation is equivalent to
\begin{align*}
2\sin\left(\frac{\pi}{5} - \frac{x}{2}\right)\cos\left(\frac{\pi}{5} - \frac{x}{2}\right) = 2\sin\left(\frac{\pi}{5} - \frac{x}{2}\right)
\end{align*}

Can you take it from here?

Title: solving $\sin\left(x+\frac{3\pi }{5}\right)=2\sin\left(\frac{\pi }{5}-\frac{x}{2}\right)$

Body:
Trying to solve this:
$$\sin \left( x+\frac{3\pi}{\:5} \right) =2\sin \left( \frac{\pi}{5}-\frac{x}{2} \right) $$

I tried to divide $\sin\left(\frac{\pi }{5}-\frac{x}{2}\right)$ and expanded to get

$$\frac{\sin \left( x \right) \cos \left( \frac{3\pi}{5} \right) +\cos \left( x \right) \sin \left( \frac{3\pi \:}{5} \right)}{\sin \left( \frac{\pi}{5} \right) \cos \left( \frac{x}{2} \right) -\cos \left( \frac{\pi}{5} \right) \sin \left( \frac{x}{2} \right)}=2$$

I don't know how to proceed. Appreciate any help!


问:
Question about how to interpret arbitrary elements
I have a question about how to interpret arbitrary elements.

When we say:

"let $x$ be an arbitrary integer"

"let $x$ be any integer"

"let $x$ be an integer"

at the beginning of proofs, are we saying:

"let the symbol $x$ represent one of the integers, although it is not clear which specific integer the symbol $x$ represents".

If not, please let me know how I should think about the above statements.

Thank you for your time.

Edit:

I had a follow up question to the one I asked earlier. What do mathematicians mean by “arbitrary element” or “any element”. I can understand when mathematicians say “let $x$ be $4$” as this means “the symbol $x$ represents $4$”. Could someone please elaborate on why the specific terms “arbitrary” and “any” are used?

答:
Your statement is somewhat correct, but maybe not that useful.

Consider a statement like the following, which we'll call Statement 1:

Let $x$ be an integer. The number $4x+3$ is odd.

The power of this statement is its universality: it applies to every integer $x$. If I wanted, I could say Statement 2:

If $x=-26$, then $4x+3$ is odd.

This is a less general statement that only applies to a single integer $x$. I could also say Statement 3:

If $x=55$, then $4x+3$ is odd.

The power of a construction like Statement 1 is that it implies Statements 2 and 3: once we've proven it, we can use it for any integer we'd like. Statement 1 is not intended to be used for a specific value of $x$ that we decided not to name -- it can be used for any integer $x$ we want.

A side-note is that we can also apply such a statement to other "generic" integers. Consider Statement 4:

If $t$ is a positive integer, then $2^{t-1}$ is an integer.

We can combine the two to say that, if $t$ is a positive integer, then $2^{t-1}$ is an integer, and so if we set $x=2^{t-1}$ in Statement 1, we get that $4\cdot 2^{t-1}+3$ is an integer. So, Statement 4 and Statement 1 can be combined to give another general statement:

If $t$ is a positive integer, then $2^{t+1}+3$ is odd.

If we only cared about Statement 1 for a single integer $x$, doing something like the above wouldn't really be possible.


问:
Solving Dudeney's "A Question of Cubes": sum of consecutive cubes is a square
The following is a puzzle of Dudeney:

Professor Rackbrane pointed out one morning that the cubes of successive numbers, starting from $1,$ would sum to a square number... He stated that if you are forbidden to use the $1,$ the lowest answer is the cubes of $23,24,25,$ which together equal $204^2.$  He proposed to seek the next lowest number using more than three consecutibe cubes and as many more as you like excluding $1.$

There is an answer provided but a proof is not included:

 The cubes of $14,15,$ up to $25$ inclusive (twelve in all) add up to... the square of $312.$ The next lowest answer is the five cubes of $25,26,27,28,$ and $29,$ which together equal $315^2.$

My query is more general: Classify all finite sets of consecutive positive integers, the sum of whose cubes is a square. Any idea how we can do this? If no one manages to answer this question, I will accept an answer that shows how Dudeney came to the minimal solutions.

Here are my thoughts on the matter so far:
Clearly the sum of first $n$ positive cubes works. The sum of the cubes of $m+1,m+2,\ldots,n$ for positive $n$ and non-negative $m$ is $$(1^3+2^3+\cdots+n^3)-(1^3+2^3+\cdots+m^3) = \left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2.$$
If this is equal to a square, it is equivalent to seeking all Pythagorean triples such that at least two of the elements of the triple are triangular numbers. At that point, I tried to use the classification of all (primitive) Pythagorean triples, but that went nowhere.

Body:
Euclid's formula $\quad A=x^2-y^2\quad B=2xy\quad C=x^2+y^2\quad$ is the one most commonly used for generating Pythagorean triples. Your expression

$$\left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2$$
is the same as that for generating the A-component above. To find triples for a given side-A, we can solve the A-function for y and test x-values to see which, if any yield integers.

\begin{equation}
A=x^2-y^2\implies y=\sqrt{x^2-A}\qquad\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le x \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $y\in\mathbb{N}$ and the upper limit ensures $x>y$.
$$A=81\implies \lfloor\sqrt{81+1}\rfloor=9\le x \le \frac{81+1}{2} =41\\\quad\land\quad x\in\{15,41\}\implies y \in\{12,40\} $$
$$f(15,12)=(81,360,369)\qquad \qquad f(41,40)=(81,3280,3281) $$

You can substitute any square number into the formula but the difficulty is finding x and y where both are the sum of integers $\big(\frac{n(n+1)}{2}\big)$ as shown inside your square brackets. E.g.
\begin{equation}
f(5,4)=(9,40,41)\quad f(5,3)=(16,30,34)\\
 f(13,12)=(25,312,313)\quad f(10,8)=(36,160,164)\\
f(25,24)=(49,1200,1201)\quad \mathbf{f(10,6)=(64,120,136)}\\
f(17,15)=(64,510,514)\quad f(15,12)=(81,360,369)\\
f(41,40)=(81,3280,3281)\quad f(26,24)=(100,1248,1252)\\
 f(61,60)=(121,7320,7321)\quad f(13,5)=(144,130,194)\\
 f(15,9)=(144,270,306)\quad f(20,16)=(144,640,656)\\
 \mathbf{f(45,36)=(729,3240,3321)}\quad f(123,120)=(729,29520,29529)\\
  f(365,364)=(729,265720,265721)\quad f(221,85)=(41616,37570,56066)\\
  f(255,153)=(41616,78030,88434)\quad \mathbf{f(325,253)=(41616,164450,169634)}\\
\end{equation}

Here, among these examples are
$$f(10,6)=f\bigg(\frac{4(5)}{2},\frac{3(4)}{2}\bigg)
\implies A=(10^2-6^2)=64=4^3=8^2
$$
and
$$f(45,36)=f\bigg(\frac{9(10)}{2},\frac{8(9)}{2}\bigg)
\implies A=(45^2-36^2)=729=9^3=27^2
$$
and
$$f(325,253)=f\bigg(\frac{25(26)}{2},\frac{22(23)}{2}\bigg)
\implies A=(325^2-254^2)=23^3+24^3+25^3=204^2
$$

Note that in these examples, the 2nd number of the y-product and the first number of x-product suggest, perhaps by coincidence, the range of cubes to be summed. E.g.
$$4(5),3(4)\rightarrow 4^3\quad 9(10),8(9)\rightarrow 9^3\quad 25(26),22(23)\rightarrow 23^3+24^3+25^3$$

A search is needed if all squares are used but the square of any of the numbers in sequence A126200 will yield x,y values that meet the criteria above.


问:
Determine the generating function for the sequence $a_n = 2^n n^2$
This is an exercise in my course notes.

Exercise 3.1 : Find the generating function for the sequence $a_{n}=2^{n}n^{2}$. Hint: Look at the solutions to the generating functions $\displaystyle\frac{1}{(1 - 2x)^{k}}$ and the fact that $n^{2}$ can be written in terms of $\displaystyle\binom{n}{2}$, $n$, and $1$.

I think I have made some progress towards the solution but I haven't gotten all the way. I will present my work below.

I think that the "term formula" for a generating function $\frac{1}{(1 - 2x)^k}$ is $a_n = \binom{n - 1 + k}{n}2^n$.

I figure I'm going to have to pick a value of $k$ to solve this problem. If I let $k = 2$, then the binomial coefficient becomes $\binom{n + 1}{n} = n + 1$. If I let $k = 3$ then the binomial coefficient becomes $\binom{n + 2}{n} = \frac{(n + 2)(n + 1)}{2}$. I'm leaning towards $k = 3$ at the moment but I'm not sure.

Regarding the hint about $\binom{n}{2}, n$ and $1$, I looked at the fact that $\binom{n}{2} = \frac{n^2 - n}{2}$, and I then added $\frac{n^2 + n}{2}$ in order to make it equal $n^2$. Thus I determined that $n^2 = \binom{n}{2} + \binom{n + 1}{2}$.

I thought there might be some binomial coefficient identity that I could take advantage of here, but I didn't see anything relevant at Wikipedia. I think I need to somehow manipulate $\binom{n - 1 + k}{n}$ and $\binom{n}{2} + \binom{n + 1}{2}$ into the same thing, and/or maybe multiply $\frac{1}{(1 - 2x)^k}$ by some constant. I'm not sure what to do, though.

I appreciate any help.

Body: Hint: $\sum n^2 z^n =  \dfrac{z (z + 1)}{(1-z)^3}$ and take $z=2x$.


答:
Your observation is right.

Hint: Take a look at
$$
\frac{2}{\left( 1-2x \right) ^3}-\frac{3}{\left( 1-2x \right) ^2}+\frac{1}{1-2x}
$$

Compare the coefficients.

Title: Determine the generating function for the sequence $a_n = 2^n n^2$

Body:
This is an exercise in my course notes.

Exercise 3.1 : Find the generating function for the sequence $a_{n}=2^{n}n^{2}$. Hint: Look at the solutions to the generating functions $\displaystyle\frac{1}{(1 - 2x)^{k}}$ and the fact that $n^{2}$ can be written in terms of $\displaystyle\binom{n}{2}$, $n$, and $1$.

I think I have made some progress towards the solution but I haven't gotten all the way. I will present my work below.

I think that the "term formula" for a generating function $\frac{1}{(1 - 2x)^k}$ is $a_n = \binom{n - 1 + k}{n}2^n$.

I figure I'm going to have to pick a value of $k$ to solve this problem. If I let $k = 2$, then the binomial coefficient becomes $\binom{n + 1}{n} = n + 1$. If I let $k = 3$ then the binomial coefficient becomes $\binom{n + 2}{n} = \frac{(n + 2)(n + 1)}{2}$. I'm leaning towards $k = 3$ at the moment but I'm not sure.

Regarding the hint about $\binom{n}{2}, n$ and $1$, I looked at the fact that $\binom{n}{2} = \frac{n^2 - n}{2}$, and I then added $\frac{n^2 + n}{2}$ in order to make it equal $n^2$. Thus I determined that $n^2 = \binom{n}{2} + \binom{n + 1}{2}$.

I thought there might be some binomial coefficient identity that I could take advantage of here, but I didn't see anything relevant at Wikipedia. I think I need to somehow manipulate $\binom{n - 1 + k}{n}$ and $\binom{n}{2} + \binom{n + 1}{2}$ into the same thing, and/or maybe multiply $\frac{1}{(1 - 2x)^k}$ by some constant. I'm not sure what to do, though.

I appreciate any help.


问:
What is the complexity of $f(n)=f(n-1)+f(n/2)+1$
$$f(1) = 1 : f(n)=f(n-1)+f(\lceil n/2\rceil)+1$$
What is the computational complexity of this recursive function? I have not been able to find anyone explaining a function like it in my searching, and I'm completely stuck as to what to do with it.

It seems like it should be exponential, but if you try to substitute that you get
$$2^{n-1}+2^{n/2}+1<2^n$$
But it also must be greater than polynomial, since
$$(n/2)^k+(n-1)^k=n^k/2^k+n^k-{k \choose 2}n^{k-1}+{k \choose 3}n^{k-2}-{k \choose 4}n^{k-3}...>n^k$$
Edit: I am NOT asking for you to redefine the function so that it can be calculated more efficiently, I already know that can be done in linear time. I am asking what the complexity of the function as written is. You can't take the recursive function and redefine it in a non-recursive way because that changes its complexity.

If I had told you the function was $f(x)=2f(x-1)$ I'm sure you'd have happily jumped at the chance to tell me it's exponential, you wouldn't say it's constant because it can be calculated using a single bitshift operation. Some examples.

答:
First thoughts: a worst case scenario is that in order to calculate $f(n)$ you need to calculate $f()$ of all integers from $1$ to $n$. That is $n$ calculations, and each of those calculations is $O(1)$ , because if you consider that you are building the sequence from bottom up then for each $f(n)$ line, you have previously calculated $f(n-1)$ and $f(\lceil \frac{n}{2} \rceil)$ . In this sense, the complexity of this recursive function should be $O(n)$ .

Second thoughts: you might even be able to find that the worst case is not as bad as we reviewed in the above paragraph (although I doubt that, because for each $n$ we need $f(n-1)$ ).

Edit: Based on the new details added in an edit to the text of the problem, I believe the problem is not asking for computational complexity as meant when using the O-notation. Rather, it is asking for an approximate but explicit form of the given recursive function.


问:
Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?
My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I was wondering if it's fair to say that every ordinal has cardinality no greater than $\aleph_\mathbb{0}$?

Alternatively, I guess it's possible that that the sequence of infinite ordinals listed above, does not include some of the ordinals with a cardinality greater than the naturals.. but I wasn't sure.

答: A hint to start you off. Take all the countable ordinals, ordered in size. Ask: are they well-ordered? if so, ask: What is the order-type of that sequence of objects? Is it an ordinal? Must it be bigger than any countable ordinal?


问:
How to square a velocity vector?
I've just looked at Bernoulli's equation $\frac 12 \mathbf V^2- \Phi+\frac{p}{\rho}=\rm constant$. Since $\mathbf V$ is a velocity vector how is it squared? Is each component squared separately?

答: $\mathbf V^2$ simply means $\mathbf V \cdot \mathbf V $, where $\cdot $ is the Dot product. It implies the squared  norm $\|\mathbf V\|^2$.

Title: How to square a velocity vector?


问:
group algebra and its center
Let F be a field and G group so the group algebra is defined as:

$F[G] = {\sum_{g\in G} c_{g} g: c_{g} \in F, g\in G}$

take the element $a= \sum_{g \in G} g$, clearly $a\in G$. Can I say that a is in the center of $F[G]$, i.e a comutes with every element in $F[G]$

答:
Yes. $\sum_{g \in G}g$ is a central element of $F[G]$ assuming that $G$ is finite and this sum is well-defined.

Let $\sum_{g \in G} a_g g \in F[G]$, then $(\sum_{h \in G} h)(\sum_{g \in G} a_g g)=\sum_{g \in G}a_g (\sum_{h \in G}hg)$. Now the map $G \to G, h \mapsto hg$ is a bijection, as is the map $G \to G, h \mapsto gh$, so we have $\sum_{h \in G}hg=\sum_{h \in G} gh$.
Thus $\sum_{g \in G}a_g(\sum_{h \in G} hg)=\sum_{g \in G}a_g(\sum_{h \in G} gh)=(\sum_{g \in G}a_g g)(\sum_{h \in G} h)$. This shows that $\sum_{h \in G}h$ is central.

Title: group algebra and its center


问:
If $f^{-1}$(a) $\in$ $\chi$ for all a $\in$ $\mathbb{R}$ then f is a measurable function
$f$ : $X$ $\to$ $\mathbb{R}$ is a function on a measurable space ($X$, $\chi$).

I would like to show that this statement is $false$:

If $f^{-1}$(a) $\in$ $\chi$ for all a $\in$ $\mathbb{R}$ then f is a measurable function

I know that if $f$ is a measurable function then $f^{−1}$(a) ∈ $\chi$ for all a $\in$ $\mathbb{R}$

but I am having trouble showing the converse is false.

I can see that $f^{-1}$(B) $\in$ $\chi$, B being a Borel subset, but just a $\in$ $\mathbb{R}$?

any help would be appreciated thanks.

答:
The sigma algebra generated by $\{a| a\in \mathbb{R}\}$ is the countable-cocountable sigma algebra on the real line, i.e. $\{A\in \mathcal{P}(\mathbb{R})|$either $A$ is countable or $A^c$ is countable$\}$, which is a sub-sigma algebra of the Borel sigma algebra.

In this case, a simple counterexample is any interval, say $[0,1]$, since both $[0,1]$ and $[0,1]^c$ are uncountable. So, even if $f^{-1}(a)\in\chi, \forall a\in \mathbb{R}$, there is no guarantee that $f^{-1}([0,1])\in\chi$.


问:
Pullback in the top dimension
I am solving exercises from Loring Tu.

Show that if $L : V \rightarrow V$ is a linear operator on a vector space V of dimension n, then the pullback $L^{\wedge} : A_n(V) \rightarrow A_n(V)$ is multiplication by determinant of L.

Attempt:

This is a linear operator between the same spaces of the dimension 1. It follows that it must be multiplication by a constant. I don't understand why is it the determinant ?

答:
Here are some options:

Pick a basis $v_1, \ldots, v_n$ for $V$.

Approach 1:  The vector $d=v_1^*\wedge \ldots \wedge v_n^*$ is a basis for $A_n(V)$, and $L^{\wedge}(d)=L(v_1)^*\wedge \ldots \wedge L(v_n^*)$. Now if $L$ has matrix $M$ with entries $m_i^j$ in this basis, i.e. $L(v_i)=\sum_j m_i^j v_j$ then $L^{\wedge}(d)=(\sum_{j_1} m_1^{j_1} v_{j_1}) \wedge \ldots \wedge (\sum_{j_n} m_n v^{j_n})$, which you can rewrite as sum over permutations by property of wedge product, recovering precisely the sum over permutations formula for $det M$ - times $v_1^*\wedge \ldots \wedge v_n^*$. Thus the rescaling factor is $\det M$ (aka $\det L$).

Approach 2: Let $^{\wedge}$ be the map that sends any linear map $L:V\to V$ to the rescaling factor of $L^{\wedge}$. Since we have fixed a basis for $V$, we get a map sending matrices to numbers. Then we check that this map: 1) is linear in each column of $M$, 2) is totally antisymmetric in permuting the columns of $M$ and 3) sends the identity matrix to $1$. Then a) this determines such a map from matrices to numbers uniquely, and  b) $M\to \det M$ has all these properties.
Thus $^{\wedge}$ sends $L$ to the determinant of $M$ (aka the determinant of $L$).

Title: Pullback in the top dimension


问:
Markov Chain - Snakes and Ladders
A simple game of snakes and ladders is played on a board of nine squares. At each turn a player tosses a fair coin and advances one or two places according to whether the coin lands heads or tails. If you land at the foot of a ladder you climb to the top, but if you land at the head of a snake you slide down the tail. how many turns on averages does it take to complete the game? What is the probability that a player who reaches the middle square will complete the game without slipping back to square 1?

This is a very famous Markov chain question. The original problem and the its image is here. it's the problem 1.3.3 in Markov Chains.

We can represent the chain with the following table
$$\begin{array}{ccc} \text{Start Square} & \text{If Head go to} & \text{If Tail Go to}  \\
1&2&5\\
2&5&4 \\
3&4&5\\
4&5&1\\
5&1&7\\
6&7&4\\
7&4&9\\
8&9&9\\
9&9&9
\end{array}$$

答:
There are some general methods, but at that stage in the book I think you're supposed to work it out manually.

Let $p_i$ be the probability of reaching square nine before square one given that I'm on square $i$.

I need to come up with an equation involving $p_5$ that I can solve.
From square five I can reach only squares four five and seven before I reach square one or nine.

Looking at the table we may write
$$\begin{array}{rcl}
p_4 &=& \tfrac 12 p_5 \\
p_5 &=& \tfrac 12 p_7 \\
p_7 &=& \tfrac 12 p_4 + \tfrac 12
\end{array}$$

It remains to solve this system of equations, which you shouldn't have much trouble with.

Title: Markov Chain - Snakes and Ladders


问:
Showing derivative operator is self-adjoint
Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain
$$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$
I’m showing that it’s self-adjoint. It’s not hard to show that it’s symmetric, and suffice to show that $D(A^*) \subseteq D(A)$, which I don’t know how to deal with.
Any help will be appreciated.

答:
On the Hermite Function basis, $A$ is a tri-diagonal matrix
$$const \times
\begin{bmatrix}
0 & -i\sqrt{1} & 0 & 0 & 0 & \cdots \\
i\sqrt{1} & 0 & -i\sqrt{2} & 0 & 0 & \cdots \\
0 & i\sqrt{2} & 0 & -i\sqrt{3} & 0 & \cdots \\
0 & 0 & i\sqrt{3} & 0 & -i\sqrt{4} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$
(Wikipedia).
Since the off-diagonal elements are  $\mathscr O(\sqrt n)$,
by Carleman's test $A$ is essentially self-adjoint on the span of the Hermite Functions. Since the Hermite Functions are complete, $A$ is essentially self-adjoint on the larger domain.


问:
For a function $f(x) = x^2-(n-a)x+a$, where $a$ and $n$ are integers $1 \le a < n \le 49$. No of values of $n$ such thats $f(x)$ has irrational roots
For a function $f(x) = x^2-(n-a)x+a$, where $a$ and $n$ are integers $1 \le a < n \le 49$. For how many values of $n$ does $f(x)$ have irrational roots for all possible values of $a$?

I first found that if $r$ and $s$ are the roots, then they must be integers as $f(x)$ is monic. Then, I found that $r+s = a-n$, and $rs=n$, so $rs-r-s+1=(r-1)(s-1)=n+1$. I don't know what to do next from here on. Any suggestions would be welcome

答:
If it has rational roots $r_1,r_2$ it will factor as $(x-r_1)(x-r_2)$ and so $r_1+r_2 = n-a$ and $r_1r_2 = a$.

In other words $r_1+r_2+r_1r_2=n$.  At this point we can do brute force on what $r_1, r_2$ can be.  (You noted they must both be integers.

Wolog $r_1 \le r_2$ so if $r_1 = 1$ we can have: $n = 1+r_2 + 1*r_2 = 1+2r_2$. That can be any odd number.  If $r_1 = 2$ we can have $n=2+r_2=2*r_2 = 2+3r_2$ i.e $n = 8,11,14,....,47$. If $r_1=3$ we can have $n=3+4r_2$ and so on up to $n = 6+7*r_2=48$.

We can use exclusion/inclusion to count the number of terms omitted.


问:
Show $\int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds, x > 0$
I am trying to show that  $\displaystyle \int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds$ where $x > 0$

I have tried relating it to the erf function, it has to do with standard normal random variables.

Additionally I have tried re-writing the RHS but nothing comes of it.

答:
I think it is simpler than you think:

$$e^{-\tfrac{s^2}{2}} = e^{-\tfrac{s^2}{4}-\tfrac{s^2}{4}}\leq e^{-\tfrac{x^2}{4}} e^{-\tfrac{s^2}{4}}$$
since $s\geq x > 0\implies -x^2\leq -s^2.$


问:
In a mixed company of Poles, Italians, Greeks, Turks and Germans.... How many Germans are there?
In a mixed company of Poles, Italians, Greeks, Turks and Germans. The Poles are one less than 1/3 of the Germans; and 3 less than half the Italians; the Germans and Turks outnumber the Greeks and Italians by 3; the Greeks and the Germans form one less than half the company and the Greeks and the Italians form 7/16 of the company. How many Germans are there?

I have come up with the following equations but I'm not sure how to proceed.

P = (Gm / 3) - 1

P = (I / 2) -3

Gm + T = Gk + I + 3

Gk + Gm = (Total / 2) -1

Gk + I = (7/16) Total

P + I + Gk + T + Gm = Total

答:
In Wolfram language, you can solve the equations like this:

In[1]:=Solve[{P == (Gm / 3) - 1,P == (It / 2) -3,Gm + T == Gk + It + 3,Gk + Gm == (Total / 2) -1,Gk + It == (7/16) Total,P + It + Gk + T + Gm == Total },{P,It,Gk,T,Gm, Total}]

And here is the output:

Out[1]={{P->7,It->20,Gk->15,T->14,Gm->24,Total->80}}

Which means the company contains 7 Poles, 20 Italians, 15 Greeks, 14 Turks and 24 Germans. There are 80 people in the company.

('I' represents the imaginary unit in Wolfram language, so we should use 'It' for Italians in the equation. )


问:
Poisson Distribution - sum of RVs
Question:

$X$ balls are thrown to $n$ bins (each ball has an equal chance to get to each bin). Let $X_1,\dots, X_n$ be the amount of balls in each cell.
a. Show that if $X \sim \text{Poisson}(\lambda)$ then $\displaystyle X_i \sim \text{Poisson} \left(\frac \lambda n\right)$ and that the $X_i$'s are independent.

My thought:
I know how to prove it the other way around, but not from the sum to the parts of it...

Here's a chance to reverse the steps (I'm not sure how I can justify the last move, and how I should figure indepndence:
$P(X=k)=e^{-\lambda} \frac {\lambda^k}{k!}=
e^{-(\frac \lambda n)n} \frac {((\frac \lambda n)n)^k}{k!}=
\frac {e^{-(\frac \lambda n)n}}{k!} \sum_{m=0}^k \frac {k!}{m!(k-m!)}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=
 {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac 1{m!(k-m)!}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=$
$ {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac {(\frac \lambda n)^m}{m!}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$ \sum_{m=0}^k {e^{-(\frac \lambda n)}} \frac {(\frac \lambda n)^m}{m!} {e^{-(\frac \lambda n(n-1))}}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$\sum_{m=0}^k P(X_1=m)P(\sum_{i=2}^nX_i=k-m)$

答:
First we calculate the probability of : $P(X_i=k | X=m)=\binom mk (\frac 1n)^k(1-\frac 1n)^{m-k}$ . This is true because we pick k out of m balls to put in the $i$'th bin, so k balls fall into that bin with probability $(\frac 1n)$ and the rest of the balls are are not in it with probability $(1-\frac 1n)^{m-k}$.

Now using the law of total probability:
$ P(X_i=k)=
\sum _{m=k}^\infty P(X_i=k | X=m) P(X=m)=
\sum _{m=k}^\infty \binom mk (\frac 1n)^k(1-\frac 1n)^{m-k} e^{-\lambda} \frac {\lambda ^m}{m!}=
\sum _{m=k}^\infty \frac 1{k!(m-k)!} (\frac 1n)^k(1-\frac 1n)^{m-k} e^{-\lambda} {\lambda ^m}=$
$\frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k\sum _{m=k}^\infty \frac 1{(m-k)!} (1-\frac 1n)^{m-k} \lambda ^{m-k}=$
$\frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k\sum _{l=0}^\infty \frac {(\lambda -\frac {\lambda }n)^l} {l!} =$
$ \frac {e^{-\lambda}}{k!} (\frac {\lambda}n)^k e^{(\lambda -\frac {\lambda }n)} =
 \frac {e^{\frac {\lambda }n}}{k!} (\frac {\lambda}n)^k \Rightarrow X_i \sim Poisson(\frac np) $

Now I still need to find out how to show independence...

Title: Poisson Distribution - sum of RVs


问:
Multivariable Calculus - second derivate using chain rule
$f(x,y,z) = g(2x+y-z,x+2y-z)$ which $g(u,v)$

$f_{y}=g_{u}(u,v)* u_{y} + g_{v}(u,v)*v_{y}$

$f_{y}=g_{u}(u,v) + 2g_{v}(u,v)$

We want $f_{yz}$ and I have no idea how to find the 2nd derivate using the chain rule.

for reference the answer is,
$f_{yz} = -g_{uu} - g_{uv} - 2g_{vu} - 2g_{vv} $

答:
Remember that $g_u$ and $g_v$ follow the "tree diagram" of $g$. So,

$$f_{yz} = \frac{\partial}{\partial z} \, (g_u) + 2 \, \frac{\partial}{\partial z} \, (g_v)$$

$$f_{yz} = g_{uu} \, u_z + g_{uv} \, v_z + 2 \, g_{vu} \, u_z + 2 \, g_{vv} \, v_z$$

$$f_{yz} = - g_{uu} - g_{uv} - 2 \, g_{vu} - 2 \, g_{vv} $$


问:
Is $f(S) = (\mathbb{N} \setminus A)$ surjective or injective?
Based on Cantor's Theorem for any set $A$ there is no surjective function $A\to\mathcal P(A)$. How about $\mathcal P(A)\to\mathcal P(A)$? For instance, if $\mathcal P(\mathbb{N})\to \mathcal P({\mathbb{N}})$, where $\forall S \in P(\mathbb{N}): f(S) = (\mathbb{N} \setminus S$) then is f(S) surjective or injective? I think:

It is surjective since for every element of co-domain $\mathcal P(A)$, there is an element from domain $\mathcal P(A)$ to be mapped to.

It is injective since for any two different inputs from domain, the corresponding outputs are not the same.

But I don't know how to prove each of them, if my claims are correct.

Note: "$\setminus$" denotes difference i.e. $A \setminus B = \{x | (x \in A) \wedge (x \not\in B)\}$

答:
Yep -- this is definitely both injective and surjective. A simpler example of an injective and surjective function on $\mathcal P(A)$ is $f(S)=S$.

A cool way to prove this is using that $f(f(S))=S$, which you can prove by showing that both sides contain exactly the same elements. From this,

For every $S$, there exists $T$ for which $f(T)=S$ (take $T=f(S)$), so $f$ is surjective.

If $f(S_1)=f(S_2)$, then $S_1=f(f(S_1))=f(f(S_2))=S_2$, so $f$ is injective.


答:
Since $x^2\leq s^2$, then $-x^2\geq -s^2$, we get $\exp(-x^2/4)\geq \exp(-s^2/4)$.

$$
\int_x^{\infty}{\exp}\left( \frac{-s^2}{2} \right) ds\\=\int_x^{\infty}{\exp \left( \frac{-s^2}{4} \right) \exp \left( \frac{-s^2}{4} \right)}ds\le \int_x^{\infty}{\exp \left( \frac{-x^2}{4} \right) \exp \left( \frac{-s^2}{4} \right)}ds \\=\exp \left( \frac{-x^2}{4} \right) \int_x^{\infty}{\exp}\left( -\frac{s^2}{4} \right) ds
$$

And we are done.

Title: Show $\int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds, x > 0$

Body:
I am trying to show that  $\displaystyle \int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds$ where $x > 0$

I have tried relating it to the erf function, it has to do with standard normal random variables.

Additionally I have tried re-writing the RHS but nothing comes of it.


问:
How to prove that two equations in complex numbers have no solution in common?
So my complex numbers are:

$$r_1^n = 1+i$$
$$r_2^m = 2-i$$

$$r_1, r_2 \in \mathbb{C} \quad;\quad n,m \in \mathbb{N}$$

I wrote down the general $z^n=r^n(\cos(n\theta)+i\sin(n\theta))$ formula for these, but I don't know what to do after this step to prove these two equations don't have any common solution.

答:
Assume that there exists some $z$ for which $z^n=1+i$ and $z^m=2-i$ for some integers $m$ and $n$. You're right to think about converting between rectangular and polar, but in this case it's polar form that's useful:
$$1+i=\sqrt{2}e^{i\theta_1},\ \ 2-i=\sqrt{5}e^{i\theta_2}.$$
(I haven't bothered to find $\theta_1$ or $\theta_2$, since they happen not to be relevant). So, by comparing magnitudes,
$$|z|=2^{\frac{1}{2n}}=5^{\frac{1}{2m}}.$$
Can you show that this can never happen?


答: $r_1=(1+i)^{1/n}$ and and $r_2=(2-i)^{1/m}$  implies $|r_1|=\sqrt{2}^{1/n}$ and $|r_2|=\sqrt{5}^{1/m}.$. Therefore $|r_1|\neq |r_2|$ for any values of $m,n$ which intern implies that  $r_1\neq r_2$ for any values of $m,n.$

Title: How to prove that two equations in complex numbers have no solution in common?

Body:
So my complex numbers are:

$$r_1^n = 1+i$$
$$r_2^m = 2-i$$

$$r_1, r_2 \in \mathbb{C} \quad;\quad n,m \in \mathbb{N}$$

I wrote down the general $z^n=r^n(\cos(n\theta)+i\sin(n\theta))$ formula for these, but I don't know what to do after this step to prove these two equations don't have any common solution.


问:
A sequence that converges pointwise but not uniformly.
I am reading through Roydens Real Analysis text and I have not studied, at least in depth, sequences of functions. Thus, I have taken a slight aside from reading the text in order to look at sequences of functions. I am curious if the following is an example of a sequence of functions on $[0,1]$ that converges pointwise but not uniformly. Moreover, if the follow up argument provides a proof of this claim.

Define the sequence of functions $\{f_n\}$ as:

$$f_n(x):=n\chi_{(0,1/n]}.$$

Or written differently:

$$
f_n:=\begin{cases}
n & \text{ if } 0< x \leq 1/n \\
0 & \text{ otherwise } \\
\end{cases}
$$

Naturally, we have the following for a fixed $x$:

$$\lim\limits_{n\to\infty}f_n = \lim\limits_{n\to\infty}n\chi_{(0,1/n]} = \lim\limits_{n\to\infty}
\begin{cases}
n & \text{ if } 0 < x \leq 1/n \\
0 & \text{ otherwise } \\
\end{cases} =
\begin{cases}
\infty & \text{ if } 0 < x \leq 0 \\
0 & \text{ otherwise } \\
\end{cases} =
0.$$

Therefore, we can conclude that $\{f_n\}\overset{point.}{\to} 0$. However, we claim that
$\{f_n\}\overset{unif.}{\not\to} 0$. To see this, define the following:

$$\Delta_n:=\sup\limits_{x\in[0,1]}\left|f_n(x)-f(x)\right| = \sup\limits_{x\in[0,1]}\left|n\chi_{[0,1/n]} - 0\right| = \sup\limits_{x\in[0,1]}\left|n\chi_{[0,1/n]}\right| =
\sup\limits_{x\in[0,1]}n\chi_{[0,1/n]} =
n.$$

Since

$$\lim\limits_{n\to\infty}\Delta_n = \lim\limits_{n\to\infty}n\neq 0,$$

we can conclude that $\{f_n\}\overset{unif.}{\not\to} 0$.

答:
This is essentially correct. In some places you use $\chi_{[0,1/n]}$ instead of $\chi_{(0,1/n]}$ -- for these functions, the sequence doesn't converge at $x=0$.

You can make your functions converge pointwise on all of $[0,1]$ bounded by removing the factor of $n$: if $f_n=\chi_{(0,1/n]}$, then $f_n(x)$ converges to $0$ for all $x$.

You can also make the functions continuous if you wish by replacing the drop from $n$ to $0$ (or from $1$ to $0$ as in the previous paragraph) a very short line segment, and moving your interval to somewhere in the middle of $[0,1]$, so that you can do this for both endpoints.


问:
Proper constants for $\alpha, \beta$
Here is the problem:

For what values of $\alpha$ and $\beta$, the function

$$\mu(x,y)=x^{\alpha}y^{\beta}$$

is an integrating factor for the OE $$ydx+x(1-3x^2y^2)dy=0.$$ I am working on it just knowing the definition. :(

答:
Let $$M(x,y)dx+N(x,y)dy=0$$

If exists $F$ such that $$dF=M(x,y)dx+N(x,y)dy$$

then

$$ F(x,y)=C$$

for some $C \in \mathbb{R}$ defines an implicit solution of the ODE

When does $F$ exists? Exists if

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$

and $$M(x,y)=\frac{\partial F}{\partial x}, N(x,y)=\frac{\partial F}{\partial y} $$

Multiply your equation by $\mu(x,y)=x^{\alpha}y^{\beta}$

$$ x^{\alpha}y^{\beta+1}dx+x^{\alpha+1}y^{\beta}(1-3x^2y^2)dy=0$$

Identify $M(x,y)$ and $N(x,y)$ and apply the above theory. I found that $\alpha=\beta=-3$

Title: Proper constants for $\alpha, \beta$


问:
Can you find an expected value for a R.V. that is not non-negative when R.V. has a cdf but no pdf?
Can you find an expected value for a random variable that is not non-negative when random variable has a cdf but no pdf?

For instance, say we have that the cdf of X is the following

$$F(x)=\begin{cases} 0 & \text{if }x\in(-\infty,-5)\\
\frac{3x+20}{68}&x \in [-5,3)\\
1&x\in[3,\infty)
\end{cases}$$

X is not a non-negative random variable in this case. We also can't find a pdf of X since X is not absolutely continuous; it has two transition points at x=-5 and x=3.

Do we have any insight into the average value of a random variable that is not non-negative given its CDF?

I was thinking that we wouldn't be able to find the expected value since the random variable is not non-negative and there is no pdf.

答:
I'm sure Greg was assuming you would use the following formula for nonnegative random variables $Z$:
$$ E[Z] = \int_0^{\infty} P[Z>z]dz$$
In general you can use $X=X^+-X^-$ and
$$ E[X] = E[X^+] - E[X^-]$$
which exists (possibly being $\infty$ or $-\infty$) if and only if the right-hand-side avoids the undefined case $\infty + -\infty$.

If we allow the use of impulse functions, we can often express a PDF for a random variable $X$ in the following form:
$$ \boxed{f_X(x) = \sum_{i \in A} w_i \delta(x-x_i) + g(x)\quad \forall x \in \mathbb{R}}$$
where $A$ is a finite or countably infinite set; $\{x_i\}_{i \in A}$ are the points where $X$ has a point mass; $\{w_i\}_{i \in A}$ are the corresponding point masses (so $P[X=x_i]=w_i$); $\delta(x-x_i)$ is a unit impulse function centered at point $x_i$; $g(x) \geq 0$ for all $x \in \mathbb{R}$. In this case we have
\begin{align}
1 &=  \sum_{i\in A} w_i + \int_{-\infty}^{\infty} g(x)dx\\
E[X] &= \sum_{i \in A} w_ix_i + \int_{-\infty}^{\infty} xg(x)dx\\
E[X^2] &=  \sum_{i \in A} w_ix_i^2 + \int_{-\infty}^{\infty} x^2g(x)dx
\end{align}
If you allow the use of impulse functions, then almost all random variables you will ever see (i.e., random variables that apply to practical problems) have PDFs of the form boxed above.  There are some academic counter-examples of random variables that do not have PDFs, even if we allow the use of impulse functions.


答:
Suppose $n, m \in \mathbb{N}$ are such that

$$
z^n = 1 + i \\
z^m = 2 - i
$$

for some complex number $z = r e^{i\theta}$. Note that $n > 0$ and $m > 0$. Taking the absolute value of both equations we get

$$
r^n = \sqrt{2} \\
r^m = \sqrt{5}
$$

and so

$$
r^{nm} = \sqrt{2^m} \\
r^{nm} = \sqrt{5^n}.
$$

But this implies that

$$
2^m = 5^n
$$

which is impossible since the left hand side is even and the right hand side is odd.


问:
Artin's Algebra: 'Fibres of the Absolute Value Map'
I'm self-studying Algebra, 2nd Ed., by Michael Artin. In section 2.7 he shows this picture:

What are these regions?

Clarification:
The inverse image of an element $t$ of $T$ is the subset of $S$ consisting of all elements of $s$ such that $f(s) = t$. It is denoted symbolically as:

$$f^{-1}(t) = \{s \in S | f(s) = t\}$$

These inverse images are defined to be the fibres of the map.

答:
The fiber of a map $f\colon A\to B$ at a point $b\in B$ is the set consisting of all $a\in A$ for which $f(a)=b$. When one speaks of the fibers of $f$, this means "all the different sets you can get as fibers of various points $b\in B$." This is a useful concept since the fibers partition $A$: every $a\in A$ is in the fiber of $f$ at $b$ for exactly one $b\in B$.

In this particular case, the fiber of the absolute value function at some $b\in\mathbb R_{>0}$ is the circle in the complex plane centered at $0$ with radius $b$ (I've stipulated $b>0$ since for $b<0$ the fibers are empty). The given diagram shows some of these fibers for some (roughly) arbitrarily chosen values of $b$, which are simply a bunch of concentric circles on the complex plane.


问:
Proof that if $n^2/r^2 = p$, with $p$ prime, then $n$ is divisible by $p$
Given positive integers $n, r$, with $p$ a prime, and $n^2/r^2 = p$, prove that $n$ is divisible by $p$.  Use proof by contradiction (I am sure there are easier ways to prove this, but this is what the book requests, since the chapter is exercises on proof by contradiction).

I started off by assuming that $n$ is not divisible by $p$, therefore:

$n = ap + b$ for a non-negative integer $a$ and a positive integer $b$.

$n^2 = pr^2$, so $(ap+b)^2 = pr^2$

I expanded this out to get a quadratic polynomial in $b$:

$b^2 + 2apb + a^2p^2-r^2p = 0$

Using the quadratic formula, I found:

$b = -ap \pm r\sqrt{p}$

Since $a$, $r$, and $p$ are integers, and $p$ is prime, then both of these values of $b$ would therefore be irrational, violating the condition that $b$ must be a positive integer.

I'm not sure if this proof works or not.  It doesn't feel very elegant (compared to some of the examples in the book).  Could you please critique it?

答:
Your proof is really very overcomplicated. It's also missing something: note that you aren't assuming that $b < p$ anywhere. What goes wrong if I choose $b = p = 2, a = 1, n = 4$?

For a much easier solution: Multiply by $r^2$ to obtain $n^2 = pr^2$. Apply the definition of "prime".


答: You proof does not work because you did not use the fact that $n$ is not divisible by $p$ (you could have $b=0$ for instance). Also, when defining $b$, I think it would be more convenient to say that $n=ap+b$ is the euclidean division of $n$ by $p$ (this also implies that $0<b<p$). What is wrong in you proof is that the initial equation is false. Indeed you found a contradiction without using the fact  that $n$ is not divisible by $p$ so the initial hypothesis is false, that is $n^2=pr^2$. You can see that by writing $2v_p(n)=1+2v_p(r)$ which is not possible because the LHS is even and the RHS is odd.

Title: Proof that if $n^2/r^2 = p$, with $p$ prime, then $n$ is divisible by $p$

Body:
Given positive integers $n, r$, with $p$ a prime, and $n^2/r^2 = p$, prove that $n$ is divisible by $p$.  Use proof by contradiction (I am sure there are easier ways to prove this, but this is what the book requests, since the chapter is exercises on proof by contradiction).

I started off by assuming that $n$ is not divisible by $p$, therefore:

$n = ap + b$ for a non-negative integer $a$ and a positive integer $b$.

$n^2 = pr^2$, so $(ap+b)^2 = pr^2$

I expanded this out to get a quadratic polynomial in $b$:

$b^2 + 2apb + a^2p^2-r^2p = 0$

Using the quadratic formula, I found:

$b = -ap \pm r\sqrt{p}$

Since $a$, $r$, and $p$ are integers, and $p$ is prime, then both of these values of $b$ would therefore be irrational, violating the condition that $b$ must be a positive integer.

I'm not sure if this proof works or not.  It doesn't feel very elegant (compared to some of the examples in the book).  Could you please critique it?


答:
Hint: there are errors in your derivation.

$$rs=a, r+s=n-a \implies (r+1)(s+1) = n+1 $$

So $n+1$ must be a composite number.

Can you take it from here?

Title: For a function $f(x) = x^2-(n-a)x+a$, where $a$ and $n$ are integers $1 \le a < n \le 49$. No of values of $n$ such thats $f(x)$ has irrational roots

Body:
For a function $f(x) = x^2-(n-a)x+a$, where $a$ and $n$ are integers $1 \le a < n \le 49$. For how many values of $n$ does $f(x)$ have irrational roots for all possible values of $a$?

I first found that if $r$ and $s$ are the roots, then they must be integers as $f(x)$ is monic. Then, I found that $r+s = a-n$, and $rs=n$, so $rs-r-s+1=(r-1)(s-1)=n+1$. I don't know what to do next from here on. Any suggestions would be welcome


问:
Sequence $(a_n)$ s.t $\sum\sqrt{a_na_{n+1}}<\infty$ but $\sum a_n=\infty$
I am looking for a positive sequence  $(a_n)_{n=1}^{\infty}$ such that  $\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}<\infty$ but  $\sum_{n=1}^{\infty} a_n>\infty$.

Thank you very much.

答: The simplest example I can think of is $\{1,0,1,0,...\}$. If you want your elements to be strictly positive, use some fast-converging sequence such as $n^{-4}$ in place of the zeroes.</p>


问:
Find basis and dimension in $\mathbb{R}^6$
Given a set $A=(x_1,x_2,x_3,x_4,x_5,x_6)$ in $\mathbb{R}^6$, that follows

$$5x_1+x_2+x_3+x_4+x_5+5x_6=0$$

How to find a basis and dimension to its subspace?

By now, I only proved that the set is linear dependent, but I have no idea about how to continue it.

Body:
HINT

Take a vector $x\in\mathbb{R}^{6}$ and apply the restriction related to the set $A$. After so, you'll get

\begin{align*}
x & = (x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})\\\\
& = (x_{1},-5x_{1} - x_{3} - x_{4} - x_{5} - 5x_{6},x_{3},x_{4},x_{5},x_{6})\\\\
& = x_{1}(1,-5,0,0,0,0) + x_{3}(0,-1,1,0,0,0) + x_{4}(0,-1,0,1,0,0) + x_{5}(0,-1,0,0,1,0)\\\\
& + x_{6}(0,-5,0,0,0,1)
\end{align*}

Based on such result, we conclude that
\begin{align*}
A = \text{span}\{(1,-5,0,0,0,0),(0,-1,1,0,0,0),(0,-1,0,1,0,0),(0,-1,0,0,1,0),(0,-5,0,0,0,1)\}
\end{align*}

It also can be proven that such set is LI. Consequently, it is also a basis. Finally, we deduce that $\dim A = 5$.

Hopefully this helps!


问:
Show that some group must have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$
I have two different questions, but they are related.

The first question is, Let $G$ be a finite abelian group. show that if $G$ contains (atleast) $2^n-1$ distinct elements of order 2, then there must be a subgroup of $G$ isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.

The second question is, Let $H$ be a finite abelian group.  show that if $H$ contains $\mathfrak{p}_1,...,\mathfrak{p}_n$ which are n distinct elements of order 2, such that for any $I\subsetneq\{1,2,\cdots,n\}$ we have $\prod_{i\in I}\mathfrak{p}_i\neq 1$ Show that $G$ contains a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.

Some background and context: This may be a fairly easy question to do by induction, but I haven't done group theory in a little bit, and I was never that good at it in the first place. The point of this question is that I was trying to show that a certain quadratic number field had a (ring of integers with a) class number divisible by $2^n$.

edit
It was pointed out that the second question follows fairly easily from the first one. I believe there should be a relatively easy way to show the first question by induction, but I am struggling with that.

I think the base case is trivial, if a group contains 1 element of order 2, then that element will generate a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

So now we assume any (finite abelian) group with (atleast) $2^n-1$ elements of order 2, will have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$. And let $G$ be a finite abelian group with $2^{n+1}-1$ elements of order 2. We want to show that $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ is isomorphic to a subset of $G$.

I think we can also show pretty easily that, since we have more than $2*(2^n-1)$ elements of order 2, we can find 2 distinct injective homomorphisms $\varphi_1$ and $\varphi_2$ from $(\mathbb{Z}/2\mathbb{Z})^n$ into G, such that their images are disjoint.

So can we just take $\varphi_3$ from $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ to G, and then define it somehow based on the last 2 functions?

答:
Hint: For your first question: show there is an injective group homomorphism $$\varphi: (\mathbb{Z}/2\mathbb{Z})^n\to G.$$ For the second question: show that the group has $2^n-1$ elements of order $2.$

Added: In order to setup a such group homomorphism, lets look at the  subgroup in $G$ generated by these order two elements. If they were multiplicatively closed then this subgroup is exactly of order $2^n,$ and otherwise bigger than that. In both cases there must be at least $n$ generators for it. So we can simply pick some non-trivial distinct $n$ generators and look at (possibly smaller) subgroup generated by them. If you write additively, say, $$H=\langle e_i: 2e_i=0, 1\le i\le n\rangle.$$ A general element $h\in H$ is of the form $$h=\sum_{1\le i \le n}\epsilon_ie_i$$ where $\epsilon_i$ is either zero or one, hence $|H|=2^n$. Now we can simply define the desire group homomorphism by

$$\varphi((\underbrace{0, 0,\cdots,1,\cdots, 0}_{i\text{-th coordinate is one}}))=e_i$$ and extending linearly. Clearly this is a homomorphism and by construction its image is $H.$

Hence a bijection onto the image.


问:
How to understand "The collection of all sets satisfying a given property may not be a set"
I'm a beginner in set-theory, and I got to read:

"The collection of all sets satisfying a given property may not together form a set"

I would like to ask, could this statement be translated as follows (in general):

"The collection of all sets satisfying a given property may not together form a set that also satisfies the given property"

Or is the added qualification not needed ? --- meaning that the collection may form "something" that does not follow from the Axioms of Set Theory ?

答:
Its actually stronger than that. The collection of all sets that satisfy a certain property may form what is known as a proper class. A proper class is a collection of sets which is to big to be a set itself. Which means ZF can prove that it does not exist. (ZF can prove that no set exists such that this set contains all sets which satisfy the aforementioned property)

For example, if we apply this to the property, x=x, then the collection of all sets which satisfy this property, is just the proper class containing all sets. ZFC can prove that a set containing all sets cannot exist, because sets are not allowed to contain themselves in ZFC, due to the axiom of foundations.

This is slightly informal, but every time you have the collection of all sets satisfying $\varphi(x)$, is not a set itself, this will happen because you can show that the "number" of sets which satisfy that property is the same size as the total "number" of sets.
(calling it a number is very very much problematic, but I think its a useful way to think about it)


答:
The added modification makes the statement weaker. Let $P$ denote the given property. The original statement asserts that the collection of all sets satisfying $P$ is not a set at all. Whereas your modification asserts that the collection of all sets satisfying $P$ either fails to be a set (as in the original statement) or is a set, albeit one failing to satisfy $P$.

Note that the original version implies your modified version. However, by Russel's paradox the original version is true. Therefore, the possibility that you have added - that the collection of sets satisfying $P$ is a set failing to satisfy $P$ - is unnecessary.

Remark: In set theory, a collection that is not a set is called a proper class. Therefore, another way to rephrase the original statement is

"The collection of all sets satisfying a given property may be a proper class."

Title: How to understand "The collection of all sets satisfying a given property may not be a set"

Body:
I'm a beginner in set-theory, and I got to read:

"The collection of all sets satisfying a given property may not together form a set"

I would like to ask, could this statement be translated as follows (in general):

"The collection of all sets satisfying a given property may not together form a set that also satisfies the given property"

Or is the added qualification not needed ? --- meaning that the collection may form "something" that does not follow from the Axioms of Set Theory ?


问:
NP combination puzzle (Klotski)
I've written a C++ program to solve sliding puzzles games such as UnblockMe and Car Parking. I'm quite happy about it, since it solves various schemes in less than a second. Recently I fed the game with Klotski schema (https://en.wikipedia.org/wiki/Klotski) that requires no less than 81 moves to solve it. My code respects the following algorithm:

```<code>Enqueue the current board
while Q not empty:
    Dequeue a board and examine it if not already examined
    can the big block escape?
        "solution found"
    it cant?
        for each possible board that can arise out of this one
            add board to END of Q
</code>```

With my surprise things go well for about 45-50 moves, then the process is extremly slow. I know the game is NP-complex, but I have the suspect my code could be wrong. Considering all pieces different, how many permutation are possible? My code output the following in about one second:

```<code>step 1 new: 6 duplicated: 0
step 2 new: 14 duplicated: 14
step 3 new: 25 duplicated: 46
step 4 new: 38 duplicated: 112
step 5 new: 63 duplicated: 213
step 6 new: 107 duplicated: 378
step 7 new: 160 duplicated: 659
step 8 new: 239 duplicated: 1094
step 9 new: 332 duplicated: 1750
step 10 new: 491 duplicated: 2665
step 11 new: 668 duplicated: 4009
step 12 new: 804 duplicated: 5897
step 13 new: 853 duplicated: 8226
step 14 new: 869 duplicated: 10534
step 15 new: 869 duplicated: 12813
step 16 new: 918 duplicated: 14921
step 17 new: 955 duplicated: 17058
step 18 new: 1003 duplicated: 19174
step 19 new: 1097 duplicated: 21343
step 20 new: 1215 duplicated: 23674
step 21 new: 1267 duplicated: 26391
step 22 new: 1272 duplicated: 29299
step 23 new: 1493 duplicated: 32276
step 24 new: 1561 duplicated: 35949
step 25 new: 1624 duplicated: 39871
step 26 new: 1595 duplicated: 43830
step 27 new: 1693 duplicated: 47669
step 28 new: 1940 duplicated: 51725
step 29 new: 2280 duplicated: 56435
step 30 new: 2816 duplicated: 62037
step 31 new: 3166 duplicated: 69297
step 32 new: 3420 duplicated: 77335
step 33 new: 4166 duplicated: 85621
step 34 new: 5490 duplicated: 95771
step 35 new: 7398 duplicated: 109271
step 36 new: 9630 duplicated: 127499
step 37 new: 12538 duplicated: 151293
step 38 new: 16240 duplicated: 182497
step 39 new: 20068 duplicated: 223457
</code>```

Last line i.e. step 39 new: 20068 duplicated: 223457 tells me that with 39 moves player can reach 223457 + 20068 possible disposition of pieces (but 223457 positions are already analized, so availabe also with less moves).

Are these numbers legit? Or I am missing something?

答:
Doubtful you're still working on your program, but I only just came across this because I'm presently writing a klotski solver program too and was also wondering if my numbers were, as you say, legit.

The magnitudes of the reachable positions shown in your output are close to my own program's, but the very first line of your output is curious. If I understand correctly, "step 1 new: 6 duplicated: 0" means that your algorithm found 6 new arrangements of the pieces by moving just one piece from the starting arrangement. It's easy to see that only the four smallest pieces can be moved in the very beginning. If each of these pieces is only permitted to move one space at a time, then only 4 new arrangements are possible. If additionally each of them is allowed to move two spaces in a row (such as moving the piece in the lower left corner two spaces to the right) while still counting as one "move", then I count a total of 8 possible new arrangements for the opening move. Indeed my program lists 8 reachable positions in the beginning. But your output says there are exactly 6 opening moves. Why 6?

Title: NP combination puzzle (Klotski)


问:
How do I find the tangent to a parameterized curve at a point $t$?
So I have a question that goes like this:

The parametric equations of a curve are
$$x = \frac{1}{\cos^3t}, y = \tan^3t$$
where $0 \le t \lt \frac{1}{2}\pi$.

(i) Show that $\frac{dy}{dx} = \sin t $

(ii) Hence show that the equation of the tangent to the curve at the point with parameter t is $y = x\sin t - \tan t$.

(i) is easy enough, but I have no idea where to even start with (ii), considering I don't know what the value of $t$ at this point is.

答: Use $\frac{y-y(t)}{x-x(t)}=\frac{dy}{dx}\Big|_t$. You don't need to know the value of $t$, we can calculate the tangent for a general $t$.


问:
Schwarz derivative and Möbius transformation.
In https://en.wikipedia.org/wiki/Schwarzian_derivative#Properties ,
some properties about Schwarz derivative
\begin{align}
(Sf)(z)&=
\left(\dfrac{f''(z)}{f'(z)}\right)'
-\dfrac{1}{2} \left( \dfrac{f''(z)}{f'(z)} \right)^2 \\
&=
\dfrac{f'''(z)}{f'(z)}-\dfrac{3}{2}\left(\dfrac{f''(z)}{f'(z)}\right)^2
\cdots \ast
\end{align}
are described.

And it is said that "The Schwarzian derivative of any Möbius transformation
$g(z)=\dfrac{az+b}{cz+d}$ is zero. Conversely,
the Möbius transformations are the only functions with this property."

I could understand the fact that the Schwarzian derivative of any Mobius transformation is zero.
But I cannot understand the fact that the Möbius trasformations are the only function with this property.

If $z \in \mathbb{R}$, I could understand this fact, owing to the $2$nd answer of this site, using "log". https://mathoverflow.net/questions/173182/schwarzian-derivative-of-a-diffeomorphism-is-zero-iff-linear-fractionals

However, "log" in $\mathbb{R}$ is not equal to "log" in $\mathbb{C}$.
So if $z \in \mathbb{C}$, I think I cannot use the approach using "log".

Summarizing the above, I wonder why Möbius transformations are the only solutions of $\ \ast$.

I want you give me some ideas.

答:
Per comment I will present a simple (though using a trick) proof of the fact that $S(f)=0$ implies $f$ Mobius. We work on a domain where $f'$ doesn't vanish so $S$ is analytic and we show that $f=g_1/g_2$ where $g_{1,2}''=0$ and that implies $f$ is Mobius on the given domain.

Consider a function $g$ for which $2f'g'+f''g=0$ on a small disc of center $w$ within the given domain - for example we can take $k(z)=\int_w^z \frac{f''(y)dy}{2f'(y)}$ which is analytic on a small disc centered at $w$ since $f'(z) \ne 0$ there; then take $g(z)=e^{-k(z)}$ and an easy check shows that $g'=-k'g=-f''g/(2f')$

But now differentiating the relation above we get:

$2f'g''+2f''g'+f'''g+f''g'=0$ and dividing by $f'$ and substituting $g'=-f''g/(2f')$ we get

$2g''+(f'''/f'-3(f'')^2/(2f'))g=0$ and obviously the $g$ coefficient is $S(f)=0$ so we indeed get $2g''=0$ hence $g$ is linear on the given small disc.

But now if $h=fg$ one has $h''=f''g+fg''+2f'g'=0$ since $g''=0$ and $2f'g'+f''g=0$, so $h$ is linear too and then $f=h/g$ is Mobius on the small disc, but by analytic continuation it is so on the given domain and of course we can extend it to the complex plane meromorphically with at most one pole!

Note that this proof shows how to solve $S(f)=p$ for any analytic $p$ as we get by the same trick and same computations that $f=h/g$ where now $2g''+pg=0$ and $2h''+ph=0$, so $f$ is a ratio of linearly independent solutions of a fairly simple differential equation (and conversely it is fairly straightforward to show that any such ratio gives an $f$ with $S(f)=p$)


问:
Intuitive reason why $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ when $a$ and $b$ are negative
As a rule of thumb I have learnt that $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ if $a$ and $b$ are negative.

I have not understood why this was a problem and took the word of my teacher.

Is this something so fundamental that it should be taken for granted or is there a reason why we can't

I'm sorry for this basic question but I would like to understand this concept a bit more .

答:
To disprove a statement, you need only provide one counterexample. We have

$$\sqrt{-1}\sqrt{-1} = -1 \neq 1 = \sqrt{(-1)(-1)}.$$


答:
Since you think (i) is easy enough, you should know what does the result in (i) means. It actually tells you the slope in (ii), that is to say, the slope of the tangent line to the curve is actually $\dfrac{dy}{dx}$, which equals to $\sin t$. Then for a line going through the point $(x(t),y(t))$ with slope $\sin t$, we can write the line equation as
$$
\frac{y-y(t)}{x-x(t)}=\sin t
$$
Thus
$$
y-\tan ^3t=\sin t\left( x-\frac{1}{\cos ^3t} \right)
\\
y=x\sin t+\tan ^3t-\frac{\sin t}{\cos ^3t} = x \sin t-\tan t
$$

As desired.

Notice that we are calculating for a general $t$, so we don't need to know what the value of $t$. We calculate for general points $(x(t),y(t))$.

Title: How do I find the tangent to a parameterized curve at a point $t$?

Body:
So I have a question that goes like this:

The parametric equations of a curve are
$$x = \frac{1}{\cos^3t}, y = \tan^3t$$
where $0 \le t \lt \frac{1}{2}\pi$.

(i) Show that $\frac{dy}{dx} = \sin t $

(ii) Hence show that the equation of the tangent to the curve at the point with parameter t is $y = x\sin t - \tan t$.

(i) is easy enough, but I have no idea where to even start with (ii), considering I don't know what the value of $t$ at this point is.


问:
A first countable pseudocompact extension space
Let $X$ be a first countable space which is embeded as a subsapce of a first countable pseudocompact extension space $Y=X\cup C$, where C is countable.

Question: If $X$ has countable extent, does $Y$ have countable extent?

Thanks for your help.

答: Your assumptions on $X$ and $Y$ seem irrelevant: if $Y = X \cup C$, with $C$ countable, and $A \subset Y$ is closed and discrete, and uncountable, then $A \cap X$ must be uncountable and it is still closed and discrete (in $X$). This cannot be, by assumption. So the extent of $Y$ is also countable. The same holds for spread as well. </p>


问:
Why ratio test for $\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \left(2n-1\right)}{2 \cdot 5 \cdot 8 \cdots \left(3n-1\right)}$
$$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \left(2n-1\right)}{2 \cdot 5 \cdot 8 \cdots \left(3n-1\right)}$$

Apparently the book says ratio test, but I'm unsure why that one is used. I originally thought it could be test for divergence since I'd have $n$'s on top and bottom having a limit of $\frac{2}{3}$ and that would mean the series diverges, but I was wrong.

I can't think of any similar series to justify using ratio test. I suppose I could use process of elimition since I won't use integral test, or root, and then since I can't think of anything to compare to then I wouldn't choose direct comparison or limit comparison. Alternating wouldn't work. And so I'm basically left with ratio test.

But is there a better reason to use ratio test in this case? (Preferably a reason that is more direct and "this is why ratio is used here..." (rather than "other tests won't work and so ratio is all you're left with"))

答:
You have the following summation:

$$\sum_{n=1}^\infty \prod_{k=1}^n \dfrac{2k-1}{3k-1}$$

That is not the same as the terms approaching $\dfrac{2}{3}$ as $n \to \infty$. Instead, that product is approaching 0 (since you approach an infinite product of numbers less than one, that product approaches zero).

$$0 \le \lim_{n \to \infty} \prod_{k=1}^n \dfrac{2k-1}{3k-1} \le \lim_{n \to \infty} \left(\dfrac{2}{3}\right)^n = 0$$

So, the infinite product definitely approaches zero. Since the $n+1$-st term contains all of the multipliers of the $n$-th term plus one new one, you can expect the ratio test to cancel everything out, which it does. That is the whole intuition behind using the ratio test. It is not that it is all that is left. It is more that you would expect it to be a simple solution.


问:
Understanding the Gram-Schmidt process
I would like to better understand the gram-schmidt process. The statement of the theorem in my textbook is the following:

The Gram-Schmidt sequence $[u_1, u_2,\ldots]$ has the property that $\{u_1, u_2,\ldots, u_n\}$ is an orthonormal base for the linear span of $\{x_1, x_2, \ldots, x_k\}$ for $k\geq 1$. The formula for $\{u_1, u_2,\ldots, u_n\}$ is:
  \begin{equation}
x_k   = \left|\left| x_k - \sum\limits_{i<k}\langle x_k, u_i\rangle u_i \right|\right|_2^{-1} \left(x_k - \sum\limits_{i<k}\langle x_k, u_i\rangle u_i\right)
\end{equation}

Note that I am primarily interested in how all of the vectors are orthogonal. The norm term in the above equation tells me that all the vectors will be unit vectors and hence we get an orthonormal set. Anyway, I see how this works algebraically; Let $v = x_k - \sum\limits_{i<k}\langle x_k, u_i\rangle u_i$. Now, take the dot product of $\langle v, u_j\rangle$ for some $j<k$:
\begin{equation}
  \langle v, u_j\rangle = \langle x_k, u_j\rangle - \sum\limits_{i<k}\langle x_k, u_i\rangle\langle u_i, u_j\rangle
\end{equation}
When we assume in the induction hypothesis that we have an orthonormal basis for $i<k$ then the sum is zero except when $i=j$. This leaves us with:
\begin{equation}
  \langle v, u_j\rangle = \langle x_k, u_j\rangle - \langle x_k, u_j\rangle = 0
\end{equation}

OK, I can logically follow algebra, but how can I see this geometrically? Can someone provide both 2D and 3D examples/plots? Since I am specifically interested in seeing how all the vectors meet at 90 degrees.

答:
Question has already been answered, but I just want to make an important point. Gram-Schmidt uses the fact that if you want to project a vector $v$ onto the span of two other vectors $a_1, a_2$, then you just add the projections of $v$ on each of these vectors independently. That is,

$$\text{proj}_{\text{span}(a_1, a_2)} v = \text{proj}_{a_1}v + \text{proj}_{a_2}v$$
BE CAREFUL!!! This formula only works of $a_1, a_2$ are already orthogonal to each other.

To see why this formula fails if they're not, consider $a_1 = (0, 1000, 0)$, $a_2 = (1, 1000, 0)$ and $v = (0, 1000, 50)$. Then
$$\text{proj}_{\text{span}(a_1, a_2)} v = (0, 1000, 0).$$

But
$$\text{proj}_{a_1}v = (0, 1000, 0)$$
and
$$\text{proj}_{a_2}v \approx (0, 1000, 0)$$
so adding these would give
$$\text{proj}_{\text{span}(a_1, a_2)} v \approx (0, 2000, 0).$$
which is wrong.

Title: Understanding the Gram-Schmidt process


问:
Why is $x^4+x^3+x^2+x+1$ irreducible over $\mathbb{Q}_3$?
I know that checking $f = x^4+x^3+x^2+x+1 \in \mathbb{Q}_3[x]$ over its residue field $\mathbb{F}_3$ is not enough since it has degree $4$. I also know that $f$ is supposed to be a polynomial where a primitive $5$-th root of unity vanishes but I would like to show the irreducibility more directly. Stuff like the Eisenstein criterion cannot be applied to.

Could you please help me with this problem?

答:
Here’s how you can do it, and I think it’s the best way.

First, because $\Bbb Z_3$, the $3$-adic integers, are a Unique Factorization Domain, irreducibility over $\Bbb Q_3$ is the same as irreducibility over $\Bbb Z_3$. (I’ve left off some details, but it’s the same as the story of $\Bbb Q$ versus $\Bbb Z$.)

Second, what you say about irreducibility over $\Bbb F_3$ not being enough to conclude irreducibility over $\Bbb Z_3$, that’s wrong. A factorization over $\Bbb Z_3$ will induce a factorization over $\Bbb F_3$.

Now I show why $x^4+x^3+x^2+x+1$, which you know is $(x^5-1)/(x-1)$, is $\Bbb F_3$-irreducible. Its roots are fifth roots of unity unequal to $1$. And it’s irreducible if and only if adjoining a fifth root of unity requires a degree-four extension of the base field $\Bbb F_3$. How do you decide this? See what the first field $\Bbb F_{3^m}$ is that contains fifth roots of unity.

How do you do that? Look at $(\Bbb F_{3^m})^\times$, the (cyclic) group of nonzero elements, $3^m-1$ in number, and see whether it has an element of period five, i.e. see whether $5|(3^m-1)$, i.e. see whether $3^m\equiv1\pmod5$. Oh of course, that’s $81=3^4$. (You could also have asked for the period of $3$ in $(\Bbb F_5)^\times$.) No matter what, there’s your proof.


答:
$T = \sqrt{a} \times \sqrt{b}$ where $a,b$ are each negative.

If the focus is on Real Analysis, then (for example) $\sqrt{a}$ is undefined, so the equation above is gibberish.

If the focus is on Complex Analysis, and $a < 0, i = \sqrt{-1}$, then $$\sqrt{a} = i\sqrt{|a|}, \tag1$$

so $$T = (i^2) \sqrt{|a|} \times \sqrt{|b|} = -1\sqrt{|ab|}.\tag2$$

It just so happens that when $a,b < 0$, then $(ab) = |ab|.$

In that event, within the Complex Analysis framework, you have

$$T = (-1)\sqrt{(ab)}.$$

Actually, the situation in Complex Analysis is somewhat more convoluted.  In Real Analysis, there is the generally accepted convention that (for example) $\sqrt{4}$ refers to $+2$ rather than $-2$.

In Complex Analysis, it is unclear (at least to me) how generally accepted the convention is that (for example) $\sqrt{-4} = 2i$, rather than $-2i$.

Title: Intuitive reason why $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ when $a$ and $b$ are negative

Body:
As a rule of thumb I have learnt that $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ if $a$ and $b$ are negative.

I have not understood why this was a problem and took the word of my teacher.

Is this something so fundamental that it should be taken for granted or is there a reason why we can't

I'm sorry for this basic question but I would like to understand this concept a bit more .


问:
cat and mouse maze - How long does the mouse survive?
The diagram below shows the ﬂoor plan of a house with six rooms: in room 1 is a mouse which will change rooms every minute, ﬁrst moving at t = 1 and choosing a door to an adjoining room at random. In room 6 is a sleeping but hungry cat which will instantly wake if the mouse should enter. How long on average can we expect the mouse to survive?

I attempted this problem by setting up and solving simultaneous equations in terms of 5 $e_i$'s, the average survival time from room $i$. However, when I simulated this in Sage, I got a different answer.

Is the method below correct?

$$e_1=(e_4+1)/2+(e_2+1)/2$$$$e_3=e_2+1$$$$e_4=e_1+1$$$$e_2=(e_3+1)/3+(e_5+1)/3+(e_1+1)/3$$$$e_5=1/2+1/2(e_2+1)$$

Solving, $e_1=19$. Therefore the mouse survives 19 minutes on average.

In simulation, I got around 16 minutes average survival time.

```<code>def Mouse():
    counter=0
    room=1
    while room<6:
        if room==1:
            if randint(0,1)==0:
                room=4
            else:
                room=2
            counter=counter+1
        elif room==2:
            x=randint(0,2)
            if x==0:
                room=1
            if x==1:
                room=5
            if x==2:
                room=3
            counter=counter+1
        elif room==3:
            room=2
            counter=counter+1
        elif room==4:
            room=1
            counter+counter+1
        elif room==5:
            counter=counter+1
            if randint(0,1)==0:
                room=2
            else:
                room=6
    return(counter)
def MouseSim():
    total=0
    for i in range(10000):
        x=Mouse()
        total=total+x
    return N(total/10000)
MouseSim()
Out[19]:
15.9815000000000
</code>```

答:
Your calculations are correct, and the answer is $19$. In your "room=4" case, you have a line counter+counter+1 while it should say counter=counter+1 instead. If you change this, the code returns an answer around $19$, like it should.

You can be less likely to make these sorts of issues if your line incrementing the counter is placed outside all of the if statements, since you increment the counter regardless of the value of room.


问:
Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?
My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I was wondering if it's fair to say that every ordinal has cardinality no greater than $\aleph_\mathbb{0}$?

Alternatively, I guess it's possible that that the sequence of infinite ordinals listed above, does not include some of the ordinals with a cardinality greater than the naturals.. but I wasn't sure.

答:
You are not correct, of course, there are ordinals of unbounded cardinalities.

Note that it is sufficient to prove that there is a set of countable ordinals. If there is a set of countable ordinals, then Peter's hint completes the problem. To do that, note that you can find in $\mathcal P(\omega\times\omega)$ copies of every countable ordinal (as well-orderings of $\omega$). Conclude using replacement that countable ordinals make a set.

But there is another side to your question. Countable ordinals are closed under [definable] countable operations (and more if we assume the axiom of choice). This means that whenever we have a countable process (denoted in your post by $\ldots$) its limit is going to be countable as well. In order to get to $\omega_1$, the first uncountable ordinal we have to use a process which continues for uncountably many steps.

Therefore your suggestion that we increase by $1$, then increase by $\omega$, then multiply by $\omega$... all that is a description which is inherently countable, so you can't find a countable sequence whose limit is uncountable. In order to step up you need something stronger.

What two operations are stronger? Well, in $\sf ZFC$ there are two of them:

Cardinal successor, which means that we go from the ordinal $\alpha$ to the least $\beta$ such that $\alpha<\beta$ - and there is no bijection between $\alpha$ and $\beta$. In the case of $\omega$ we end up at the first uncountable ordinal, known as $\omega_1$.

Cardinal exponentiation, or its particular case - the power set operation. This assures us that we grow in cardinality, by Cantor's theorem. Assuming the axiom of choice, growing in cardinality means ordinals which are at least as large as the cardinal successor.

Note that I'm talking about $\alpha\mapsto\ ^\alpha\beta$ for some $\beta>1$, of course.

Also relevant:

How do we know an $ \aleph_1 $ exists at all?

Mechanical definition of ordinals


问:
How to calculate new box dimensions given a reduction in volume
I have a rectangular box with given dimensions $l, w, h$ with available volume $v = lwh$. If I can only pack up to 80% of $v$, how would I go about determining the new, smaller dimensions such that the new $v = lwh$ is 80% of original $v$ where the new dimensions should be reduced proportionally to retain the original box shape?

I'm trying to write some C# code to do some bin packing testing, but this algorithm doesn't consider a box volume threshold, it only packs according to dimensions. Thanks!

答: Suppose the new length $l'=kl$, new width $w'=kw$ and new height $h'=kh$ where $k$ is the ratio of new dimensions to old dimensions. We have $l'w'h'=k^3lwh=0.8lwh$, thus $k=0.8^{1/3}$.


问:
How to configure a sigma function so that it only counts every third number
I have a sigma function where the start value is k and the operation can only be counted when k is a multiple of 3. I know how to differentiate between even and odd numbers in a sigma function, but how do I differentiate between numbers that are multiples of 3, and numbers that are not?

答:
The answer given by Y. Tamer of modifying the index set is probably the most reasonable for most applications. However, if you don't want to modify the index set, you can use an indicator function:

Define $g:\mathbb Z\to\{0,1\}$ so that $g(n)=1$ if $n$ is a multiple of $3$ and $g(n)=0$ if $n$ is not a multiple of $3$. Then you can write
$$\sum_{\substack{i\in S\\3|i}}f(i)=\sum_{i\in S}f(i)g(i).$$


答:
You already know that $v = lwh$, so you want to determine new dimensions for which $v' = 0.8v$.  So you could reduce any one of the three dimensions, such as $w' = 0.8w$, to do that.  If you want to shrink the box but keep the proportions, just distribute the factor:

$$v' = 0.8v = 0.8lwh = \sqrt[3]{0.8}l\sqrt[3]{0.8}w\sqrt[3]{0.8}h$$
$$l' = \sqrt[3]{0.8}l$$
$$w' = \sqrt[3]{0.8}w$$
$$h' = \sqrt[3]{0.8}h$$

In other words, just shrink each dimension by $\sqrt[3]{80\%}$ or multiply each dimension by $\approx 0.9283$.

Title: How to calculate new box dimensions given a reduction in volume

Body:
I have a rectangular box with given dimensions $l, w, h$ with available volume $v = lwh$. If I can only pack up to 80% of $v$, how would I go about determining the new, smaller dimensions such that the new $v = lwh$ is 80% of original $v$ where the new dimensions should be reduced proportionally to retain the original box shape?

I'm trying to write some C# code to do some bin packing testing, but this algorithm doesn't consider a box volume threshold, it only packs according to dimensions. Thanks!


问:
Prove that $b_n$ is a Cauchy's sequence
Let $\{b_n\}$ is a sequence such that $b_1=1, b_{n+1} = b_n + (1+ 1/n)^{-n^2}$. Then prove that $~b_n~$ is a Cauchy's sequence

ok I wanna prove that it converges thus it meets Cauchy's criterion and not the opposite..

答: Hint: use the root test on the series $\sum_{k=1}^{\infty}(1+\frac{1}{n})^{-n^2}$

Title: Prove that $b_n$ is a Cauchy's sequence


答:
You can check that $$b_{n+1}=1+\sum_{k=1}^n \left(1+\frac 1 k\right)^{-k^2}$$
and since
$$\begin{split}\left(1+\frac 1 k\right)^{-n^2}&=\exp\left(-k^2\ln\left(1+\frac 1 k\right)\right)\\
&=\exp\left(-k^2\left(\frac 1 k+\mathcal O\left(\frac 1 {k^2}\right)\right)\right)\\
&=\exp\left(-k+\mathcal O\left(1\right)\right)\\
&=\mathcal O(e^{-k})
\end{split}$$
the series converges, and so does the $\{b_n\}$ sequence.

Title: Prove that $b_n$ is a Cauchy's sequence

Body:
Let $\{b_n\}$ is a sequence such that $b_1=1, b_{n+1} = b_n + (1+ 1/n)^{-n^2}$. Then prove that $~b_n~$ is a Cauchy's sequence

ok I wanna prove that it converges thus it meets Cauchy's criterion and not the opposite..


问:
What is the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$?
While working on a project, I came across the problem of finding the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$. Dirichlet's Unit Theorem stated that there should be $2$ multiplicatively independent elements among the unit group. However, aside from the generator $1+\sqrt{2}$ and $\frac{1+\sqrt{-3}}{2}$, which is not considered a generator since $\left(\frac{1+\sqrt{-3}}{2}\right)^6=1$. I cannot find any other generator. Can someone offer help or an explanation? Thanks.

答: Dirichlet's Unit Theorem states that the rank of the group of units $\mathcal O_K^\times$ is $r_1+r_2-1$ if $r_1$ is the number of real embeddings and $r_2$ is the number of (conjugate pairs of) complex embeddings of $K$. For $K=\mathbb Q(\sqrt2,\sqrt{-3})$, there are no real embeddings, since there are no real square roots of $-3$, and so $r_1=0$. As $r_1+2r_2=[K:\mathbb Q]=4$, we should have $r_2=2$, and so the unit group should be rank $1$ -- i.e. it should consist of the roots of unity in $K$ along with a single non-torsion generator (which you have found).


问:
Solving Dudeney's "A Question of Cubes": sum of consecutive cubes is a square
The following is a puzzle of Dudeney:

Professor Rackbrane pointed out one morning that the cubes of successive numbers, starting from $1,$ would sum to a square number... He stated that if you are forbidden to use the $1,$ the lowest answer is the cubes of $23,24,25,$ which together equal $204^2.$  He proposed to seek the next lowest number using more than three consecutibe cubes and as many more as you like excluding $1.$

There is an answer provided but a proof is not included:

 The cubes of $14,15,$ up to $25$ inclusive (twelve in all) add up to... the square of $312.$ The next lowest answer is the five cubes of $25,26,27,28,$ and $29,$ which together equal $315^2.$

My query is more general: Classify all finite sets of consecutive positive integers, the sum of whose cubes is a square. Any idea how we can do this? If no one manages to answer this question, I will accept an answer that shows how Dudeney came to the minimal solutions.

Here are my thoughts on the matter so far:
Clearly the sum of first $n$ positive cubes works. The sum of the cubes of $m+1,m+2,\ldots,n$ for positive $n$ and non-negative $m$ is $$(1^3+2^3+\cdots+n^3)-(1^3+2^3+\cdots+m^3) = \left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2.$$
If this is equal to a square, it is equivalent to seeking all Pythagorean triples such that at least two of the elements of the triple are triangular numbers. At that point, I tried to use the classification of all (primitive) Pythagorean triples, but that went nowhere.

答:
R. J. Stroeker, On the sum of consecutive cubes being a perfect square,  Compositio Mathematica 97: 295-307, 1995, is available at http://www.numdam.org/article/CM_1995__97_1-2_295_0.pdf The abstract reads,

In this paper estimates of linear forms in elliptic logarithms are applied to solve the problem
of determining, for given $n\ge2$, all sets of $n$ consecutive cubes adding up to a perfect square. Use is made of a lower bound of linear forms in elliptic logarithms recently obtained by Sinnou David. Complete sets of solutions are provided for all $n$ between $2$ and $50$, and for $n = 98$.

The author proves that nontrivial solutions exist for all odd values of $n$, which at least shows that there are infinitely many solutions.

The largest numbers that show up in the Stroeker paper are the $49$ cubes starting with $117576^3$, which add up to $282298800^2$.

There is more up-to-date information at https://oeis.org/A253679


问:
Calculating inverse of a function modulo m
Let f(p) = a*p + b (mod m)

Where a and m are relatively prime.
What is the inverse function of f?

This is confusing because generally we talk about inverse of a number not a function.
So what I understand is we have to find a function g() such that
f(p) *g() = 1 (mod m)

How to go about it?

答:
The same as any inverse fuction.

The functions so that $f(f^{-1}(p)) = p$.

Set $f^{-1}(p) = y$ (just so we can avoid typing

And $f(y) \equiv a*y + b \equiv p \pmod m$ and solve for $y$.

$ay + b \equiv p \pmod m$

$ay \equiv p-b \pmod m$

Now $a$ and $m$ are relatively prime so $a^{-1}$ exists.

$y \equiv a^{-1}(p -b)\pmod m$.

So $f^{-1}(p)  = a^{-1}(p-b)\pmod m$.

That's all there is to it.

For any $p \in \mathbb Z_m$ then if $K = a*y+b$ then $f^{-1}(K) = a^{-1}(K-b)= a^{-1}([ap+b] - b) \equiv a^{-1}(ap) \equiv p\pmod m$.


答:
According to Varberg, Purcell & Rigdon, the symbol $\sqrt{ \quad}$ means the positive root for a number, so, the property doesn't hold for negative numbers because there's no positive root. That's a possible reason why your teacher gave the property as a default.

On the other hand I would start proving that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ to get more knowledge of the property

Firstly I would say that, $\sqrt{a} = x $ and $\sqrt{b} = y$ and with some algebra we also have the following equalities $x^2 = a$ and $y^2 = b$ then to start the prove

$$
\begin{align*}
x^2    &= a \\
x^2y^2 & = ab \\
\sqrt{x^2y^2} & = \sqrt{ab} \\
\sqrt{x^2}\sqrt{y^2} & = \sqrt{ab} \\
xy & = \sqrt{ab} \\
\end{align*}
$$

On the other hand,

$$
\begin{align*}
x    &= \sqrt{a} \\
x \sqrt{b}  &= \sqrt{a}\sqrt{b} \\
xy  &= \sqrt{a}\sqrt{b} \\
\end{align*}
$$

So, we can conclude that

$$
\sqrt{a}\sqrt{b} = \sqrt{ab} \quad \dagger
$$

The proof shows that in order to the property to hold, squared numbers where used, so, those must be positive numbers.

Hope was useful