问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
If you'd love to, you may take it as a comment. I am just putting up a a try, so sorry if I were wrong.

$\because 0 < x < 2^k , x = 2^k - m$

$2^{2k} - x^2 \mid 2^{2k} - 1 \implies \dfrac {2^{2k} - 1}{2^{2k} - x^2} \in \mathbb{N}$ for now. You may place $\mathbb{Z}$ insteasd of $\mathbb{N}$ as well.

If $m$ is even(i.e., $m = 2n$):

$2^{2k} - x^2 = 2^{2k} - (2^k - 2n)^2 = 2^{2k} - 2^{2k} + 4 \times 2^k n - 4n^2 = 4n(2^k - n)$

$2^{2k} - 1$ is obviously odd, so its factors are odd and none of its factors are divisible by $4$, so we can easily reject the case that $m$ is even.

If $m$ is odd (i.e., $m = 2n - 1$):

$2^{2k} - x^2 = 2^{2k} - (2^k - (2n - 1))^2 = 2^{2k} - 2^{2k} + 2 \times 2^k (2n - 1) - (2n - 1)^2 = (2n - 1)(2^{k+1} - (2n - 1)) \longrightarrow(1)$

Let's put back the $m$ :

$(1) = m(2^{k + 1} - m)$

Let's take $u = 2^k, \implies (1) = m(2u - m)$ and the fraction becomes $\dfrac {u^2 - 1}{m(2u - m)} = \dfrac{(u + 1)(u - 1)}{m(2u - m)}$.

Obviously, for $(2u - m) \nmid (u \pm 1)$,$(2u - m) > (u \pm 1)$ ($\because x = 2^k - m = u - m, m \in [0,2^k)$ ; $m$ can never be $0$ so as to avoid the denominator from becoming $0$. Plus, the opposite [$(2u - m) < (u \pm 1)$] needn't always be true as $m = u + 1$ is a solution in that case)$\forall m : m \in [0, 2^k]$ even if $m \mid (u \pm 1)$ and hence, our objective here must be to find an $m$ that reverses or breaks the inequality. Since the denominator must be small enough (or even equal to the factors of the expression in the numerator) we can conclude from the inequality that $m(2u - m)\nmid (u^2 - 1) \space \forall m : m < u \pm 1$. Also, if $m > u \pm 1$, we get to see that $x \leq 0$ and that goes against our constraints that $0 < x < 2^k$.  This simplifies things a bit and thus tells us that if we put $m = u \pm 1$, you'll get an integer quotient and out of the values we have for $m$, we can only take $m = u - 1$ since $0 < x < 2^k$ and $x = 2^k - m$. Thus we end up with $x = 1$ as the only solution.

This is not like a common approach of factoring $2^{2k} - 1$ into $2^k + 1$ and $2^k - 1$, and the denominator into $2^k + x$ and $2^k - x$ and matching corresponding parts (or using the fact that $(x -y)\mid (x^n - y^n)$. Here, we assume that we want more $x$'s and see if we can find them.

Note : I haven't mentioned about the variables I had to use here for substitution, so I'd like to mention it. The least significant variable is $n$, and I lay more of an emphasis is on $m$, since its value can directly affect that of $x$. The only thing : $m \in \mathbb{Z}$ while $n \in \mathbb{R}$.

I admit that this is not a perfect solution, but a weak indication to the fact that it is hard to find whether there exists other $x$ for which the divisibility holds for all $k$.

Edit :  A Better Way to Deal With the Indivisibility Condition

The condition I am referring to is the condition for which $(2u - m) \nmid (u \pm 1)$ if $m \mid (u \pm 1)$, as mentioned in the second case (i.e., $m$ is odd).

Since $(2u - m) \nmid (u \pm 1)$, we can consider two cases : $(2u - m) > (u \pm 1)$ and $(2u - m) < (u \pm 1)$ (I didn't take $(2u - m ) = u \pm 1$ since that gives only a straightforward answer $m = u \pm 1$).

To simplify these conditions, let's suppose that $m \mid (u + 1)$. Here, if $(2u - m)\nmid (u - 1)$ , let's take the two cases :

$(2u - m) > u - 1 \implies (u + x) > (u - 1) [\because m = u - x] \implies x > -1 $ - perfectly in line with our constraints.

$(2u - m)< u - 1 \implies (u + x) < (u - 1) \implies x < -1 $, which is against our constraints.

From 1. , we can see that the fraction can be simplified to $\dfrac{y(u + 1)}{(u + x)}$ (since $m \mid (u - 1)$, let $y = \dfrac{u - 1}m$; but $\because m = u - x,\dfrac{u - 1}{u - x}$ is only possible if $x = 1$ $\implies y = 1$) $ = \dfrac{u + 1}{u + x}$ which is possible only if $x = 1$.

Now , to prove why $x = 1$ is the only feasible value here :

Let $\exists k : k \in \mathbb{Z}, \space u + 1 \equiv 0 (\mod{u + k})$ and let's set $x = k$

If $k < 0$ (according to the fact that the factor of a number is lesser than or equal to itself) , $$u - k = u - (-|k|) = u + |k| > u - 1 \space \forall k$$ and thus the factor in the denominator gets bigger than the factor in the numerator.
If it is $u - 1 \equiv 0 (\mod{u + k})$ ($x = k$ as usual), $$u - k = u - (-|k|) = u + |k| \geq u + 1 \space \text{for} \space |k| \geq 1$$
Now since $u + |k| > u + 1$ for $|k| > 1$, the only value we can choose for $|k|$ is $1$.

The same thing can be applied by assuming $k > 0$:

If $k > 0$ and $u + 1 \equiv 0 (\mod{u + k})$, $$u - k \leq u - 1 \space \forall k \geq 1$$
If $k > 0$ and $u - 1 \equiv 0 (\mod{u + k})$, $$u - k < u + 1 \forall k$$

An experimental approach : suppose that $x$ has a value greater than $1$ for which $(u - x)\mid(u - 1)$. Since our $x > 1$, $(u + 1) < (u + x)$ thus making division impossible. Also, if $(u - x) \mid (u + 1)$, $(u + x) > (u - 1)$ for all $x > 1$ , again making division impossible.

From this point, we can say that only $x = 1$ is the only possible solution.


问:
Anti-Cauchy Sequences
We say a real valued sequence $(a_{n})_{n\geq0}$ is anti-Cauchy if $\forall\varepsilon>0$, there exists some $N\in\mathbb{N}$ such that :
$$
|a_{n}-a_{m}|>\varepsilon\quad\forall m,n>N,\;m\neq n
$$
For natural numbers K, we define the $K^{th}$ difference sequence $D_{i}(k)$ of a sequence $a_{n}$ and the $K^{th}$ sum sequence $S_{i}(k)$ of a sequence $a_{n}$ as follow : (note that $i\geq0$)
$$
\begin{align}
D_{i}(k)=&
\begin{cases}
D_{i}(0)=a_{i}\\
D_{i}(k+1)=D_{i+1}(k)-D_{i}(k)&\text{for $k\geq0$}
\end{cases}
\\
\\
\qquad
S_{i}(k)=&
\begin{cases}
S_{i}(0)=a_{i}\\
S_{i}(k+1)=\displaystyle\sum_{r=0}^{i}S_{r}(k)&\text{for $k\geq0$}
\end{cases}
\end{align}
$$
$1)$ Does there exist for each natural number $K$, a sequence such that its $k^{th}$ difference sequence is actually anti-Cauchy if and only if $k\leq K$ ?

$2)$ Does there exists a real valued sequence, not identically $0$ such that its $K^{th}$ sum sequence is never anti-Cauchy for any natural $K$?

UPDATE :

I would like to thank mathworker21 for his partial answer, I would also thank for T.H. Shehadi for his guess for part $(1)$ yet I am unsure if it is a correct answer or not I am still looking for a credible answer for $(1)$.

PROPOSITION :

T. H. Shehadi has offered an alternative definition for the $K^{th}$ difference sequence and $K^{th}$ summation sequence that would describe difference sum an in terms of partial sum I hope it could help in solving $(1)$  :

$(1)$ : Let Λ map sequences to sequences so that $\Lambda(a_{0})=a_{0}$ and $\Lambda(a_{n+1})=a_{n+1}−a_{n}$. We say that $\Lambda^{k}(a)$, i.e. $\Lambda$ composed with itself $k$ times, is the $k^{th}$ difference sequence of $a$.

$(2)$ : Let $\Omega$ map sequences to sequences such that $\Omega(a_{n})=a_{0}+a_{1}+\cdots +a_{n}$. We say that $\Omega^{k}(a)$ i.e. $\Omega$ composed with itself $k$ times, is the $k^{th}$ summation sequences of $a$ where $\Omega^{0}(a)=a$

答:
The answer to (2) is yes.

Consider the pattern $x_1,x_2,\dots = 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,\dots$.

Our real-valued sequence will start off with $a_1 = 1$. With $a_1,\dots,a_{n-1}$ chosen, we choose $a_n$ so that $S_n(x_k) = S_{n-1}(x_k)$, which is possible, since $S_{n-1}(x_k)$ just depends on $a_1,\dots,a_{n-1}$ and since $S_n(x_k)$ depends linearly on $a_n$.

We claim the $K^{\text{th}}$ sum of $(a_n)_n$ is anti-cauchy for all $K \ge 1$. Indeed, fix $K \ge 1$, and just note that, by construction, we have $S_n(K) = S_{n-1}(K)$ for infinitely many $n$.

ContentLicense: CC BY-SA 4.0

Title: Anti-Cauchy Sequences


问:
question on determining values of an angle where the tangent is a certain value
How would I solve this question?

Determine the values of $\theta \in [0,2\pi]$ for which $\tan\theta = -\sqrt{\sec^2{\theta} - 1}$ is true.

答:
Hint 1: It will not be true if $\tan \theta \gt 0$

Hint 2: $\sec \theta = 1/ \cos \theta$

Hint 3: $\sin^2 \theta + \cos^2 \theta =1$

Hint 4: $\tan \theta = \sin \theta / \cos \theta$

ContentLicense: CC BY-SA 3.0


问:
Number of real roots of $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$
Prove the following, without induction. Is it possible?

The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.

  And if $n$ is odd, it has only one real root.

I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$".
But failed. What is the way to search the related topics?

答:
Look up "roots of truncated exponential series" in Google.

In particular look at http://demonstrations.wolfram.com/SzegoeCurve/.


问:
Is there a single term for the *pair* (domain, codomain) of function $f$, or generally the (source, target) of morphism $f$?
This is just about finding concise terminology. So if $f:A \to B$, is there a single generic name for the pair (A,B)?

What about for the pair (domain, image)?

答: The only proposal suggested in the comments was "type signature" or type.</p>


问:
Suppose $A$ is an n by n matrix whose eigenvalues have a magnitude larger than 1. $P$ is a covariance matrix. Is it true that $APA' \geq P$?
Here, $A$ is an $n$ by $n$ matrix, whose eigenvalues all have a magnitude larger than 1. $P$ is a covariance matrix of dimension $n$. I guess it should be true that $APA' - P$ is positive semi-definite because $A$ is 'amplifying' the covariance $P$.

I guess to shows its correctness we may leverage on matrix decomposition such as $APA' - P = AQTQ'A' -QTQ'$. But I am not sure how to proceed especially how to incorporate the fact that the eigenvalues of $A$ all have magnititude larger than $1$.

答:
This is true for all positive semidefinite $P$ if and only if $A=cI$ for some scalar $c$ with $|c|\ge1$.

Suppose the inequality is true for all PSD matrices $P$. By passing $P$ to the limit, the inequality is true for all rank-one PSD matrices $P=vv'$. Hence $(Av)(Av)'\succeq vv'$ and $Av$ is parallel to $v$. Since $v$ is arbitrary, $A$ must be a scalar matrix. Conversely, suppose $A=cI$ for some scalar $c$ of modulus $\ge1$. Then $APA'-P=(|c|^2-1)P\succeq0$ for all $P\succeq0$.


问:
Change of variables to polar coordinates with constant $r$ evaluate definite integral.
Question:

Use the change of variables $x = \cos\theta$ and $y=\sin\theta$ to find
$$\int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2 + y^2} \,dy\,dx$$

Attempt 1:
Assuming coordinates are supposed to be $\,dr\,d\theta$.

As $r=1$ (constant), the Jacobian determinant turns the integral to $$\iint0\,dr\,d\theta$$
Is this the way to go?

Attempt 2:
Differentiating x and y with respect to $\theta$,
$$dx=-\sin\theta d\theta$$
$$dy=\cos\theta d\theta$$

But when trying to evaluate limits for $x$ we get
$$2=\cos\theta$$
But this can't be possible??

Also does an integral increment of $d\theta^2$ make sense?

答:
I think you'd better again read your textbook with more effort, you apparently misled the concept of changing variables.

The integration domain is
$$
\bigg\{{(x,\,y) | 0\leq x\leq 2,\;0\leq y\leq \sqrt{2x-x^2}}\bigg\},
$$
which is the upper semi-disc enclosed by the circle $(x-1)^2 + y^2 = 1$.

By letting
$$
\begin{cases}
x = r\cos\theta \\
y=r\sin\theta,
\end{cases}
$$
since $0\leq x\leq 2$ and $y\geq 0$, it follows that $0\leq\theta\leq\frac{\pi}{2}$, also, since $y\leq\sqrt{2x-x^2}$, it follows that $r\leq 2\cos\theta$.

Therefore, by changing the variables, the original integral becomes
$$
\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\cos\theta}r^2\,dr\,d\theta.
$$

Note: In your way of changing variables, you should note that $r$ is not a constant. You are not integrating over a circle but a semi-disc.


问:
Implicit differentiation $(\sin x +y)/(\cos y +x)=1$
I am currently stuck on this problem

Find $\frac{dy}{dx}$ when
$$\frac{\sin x+y}{\cos y+x}=1$$

I've been using the quotient rule to try and differentiate this, but my online homework rejected my answer of:

$$\frac{-x\cos x - \cos x\cos y +y+\sin x}{\cos y +x-\sin y \sin x+y\sin y}$$

Can anyone tell me where I went wrong, or if there is another way to do this besides the quotient rule?

答:
Actually, Quotient Rule makes it hell complicated. The simple and intended way would be writing the given equation as
$$\sin x+y=\cos y +x$$
And then differentiate both sides w.r.t. $x$ to get
$$\cos x+\frac{dy}{dx}=-\sin y\frac{dy}{dx}+1$$
$$\implies (1+\sin y)\frac{dy}{dx}=1-\cos x$$
$$\implies \frac{dy}{dx}=\frac{1-\cos x}{1+\sin y}$$
which should be accepted as the answer.

Another thing, if you did it the correct way, then after some damn long simplifications, you will arrive at the same result using Quotient Rule.

Hope this helps :)


问:
Proving that a subset $\mathscr{H}$ is dense in $\mathbb{R}$

Consider the subset $\mathscr{H} =\displaystyle\left\{\frac{a}{10^{b}} \ |\: a,b\in\mathbb{Z}\: \right\}$. Then is $\mathscr{H}$ dense in $\mathbb{R}$.

To prove that $\mathscr{H}$ is dense I have to show that given $x,y \in\mathbb{R}$  there is a $m,n\in\mathbb{Z}$ such that $x < \frac{m}{10^{n}} <y$.

Thats it I am hopelessly stuck.

答: Hint: For instance, to get 0.563, choose $a=5*10^2 + 6*10^1 + 3*10^0 = 563$ and $b=3$. Generalizing this method should give you all decimal numbers with a finite number of decimals.</p>


答:
Please check the limit of integration to understand the region you are integrating over.

$0 \leq y \leq \sqrt{2x-x^2}$

$y = \sqrt{2x-x^2} \implies y^2 = 2x - x^2 \implies (x-1)^2+y^2 = 1$

So the region is a circle with center at $(1,0)$ and radius $1 \, (0 \leq x \leq 2)$ .

In polar coordinates, this circle is represented as $2 \cos \theta$.

So the limits of integration will be $0 \leq r \leq 2 \cos \theta$ and $0 \leq \theta \leq \pi/2$ (as $y \geq 0)$.

In polar coordinates, $x = r \cos \theta, y = r \sin \theta$. Substitute in your integrand.

Can you take it from here?

Title: Change of variables to polar coordinates with constant $r$ evaluate definite integral.

Body:
Question:

Use the change of variables $x = \cos\theta$ and $y=\sin\theta$ to find
$$\int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2 + y^2} \,dy\,dx$$

Attempt 1:
Assuming coordinates are supposed to be $\,dr\,d\theta$.

As $r=1$ (constant), the Jacobian determinant turns the integral to $$\iint0\,dr\,d\theta$$
Is this the way to go?

Attempt 2:
Differentiating x and y with respect to $\theta$,
$$dx=-\sin\theta d\theta$$
$$dy=\cos\theta d\theta$$

But when trying to evaluate limits for $x$ we get
$$2=\cos\theta$$
But this can't be possible??

Also does an integral increment of $d\theta^2$ make sense?


问:
Second partial derivatives at (0, 0) for the function
For the function $$f(x,y)=\begin{cases}
\frac{x^{4}y^{2}}{x^{4}+y^{2}},x^{2}+y^{2}\neq0\\
0,x^{2}+y^{2}=0
\end{cases}$$

I need to find $f_{x}^{\prime}(0,0)$ and $f_{y}^{\prime}(0,0)$

I don't know the answer but I was told it is not $0$ and $0$. However, I don't understand why. First, $f(0, 0) = 0$ so it's sort if constant and if you differentiate a constant, it is $0$. Second, even if I apply the regular technique, there will always be a fraction with only $x$'s and $y$'s in both the numerator and denominator and but in this case I don't know what to do because it is $0/0$.

How to find $f_{x}^{\prime}(0,0)$ and $f_{y}^{\prime}(0,0)$?

答:
Hint

Use the following definition twice

$$
f_x(x,y)=\lim_{h\to 0}{f(x+h,y)-f(x,y)\over h}
$$

$$
f_y(x,y)=\lim_{h\to 0}{f(x,y+h)-f(x,y)\over h}
$$


问:
Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$
Month ago I encounter a nice result numerically checked by Mathematica
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2
$$
where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.

At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like
$$
\int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi}
$$
just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.

So I question it here for some further discussion, and thanks in advance for any suggestion.

答:
Here is a solution based on @G Cab's remarkable observation.

Reduction. Let $I$ denote the integral, and apply the substitution $(\theta,\varphi)\mapsto(2\theta,2\varphi)$ to write

$$ I = 4 \int_{0}^{\pi/4} \int_{0}^{\pi/4} \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \, \mathrm{d}\theta\mathrm{d}\varphi. $$

If we define the $1$-form $\omega$ by

$$ \omega = \left( \log\frac{\cot\theta}{\cot\varphi} \right) \biggl( \frac{\mathrm{d}( (\theta-\varphi)^2 )}{2\sin^2(\theta-\varphi)} - \frac{\mathrm{d}( (\theta+\varphi)^2 )}{2\sin^2(\theta+\varphi)} \biggr), $$

then $\omega$ is smooth in the region $(0, \pi/2)^2$ via continuation along the diagonal $\varphi = \theta$. Moreover, @G Cab's computation for the differential reduces to1)

$$ \mathrm{d}\omega
= 4 \left( \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \right) \, \mathrm{d}\theta \wedge \mathrm{d}\varphi. $$

Now let $\epsilon > 0$ be small and consider the triangular region

$$ \mathcal{T}_{\epsilon} = \bigl\{ (\theta, \varphi) : \epsilon \leq \varphi \leq \theta \leq \tfrac{\pi}{4} \bigr\}. $$

Its boundary $\partial\mathcal{T}_{\epsilon}$ consists of three oriented line segments

Then by the symmetry and the Stokes' theorem,

\begin{align*}
I
&= 2 \lim_{\epsilon \to 0^+} \int_{\mathcal{T}_{\epsilon}} \mathrm{d}\omega
= 2 \lim_{\epsilon \to 0^+} \int_{\partial \mathcal{T}_{\epsilon}} \omega
= 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}),
\end{align*}

where $\rightarrow$, $\uparrow$, and $\swarrow$ denote the oriented bottom/height/hypothenus of $\partial\mathcal{T}_{\epsilon}$, respectively, and

$$ I_{\bullet} =  \lim_{\epsilon \to 0^+} \int_{\bullet} \omega $$

for each $\bullet \in \{ \rightarrow, \uparrow, \swarrow \}$.

Integral Computations. We will evaluate each $I_{\bullet}$:

1. It is not hard to check that $\omega = 0$ along $\varphi = \theta$, and so, we get $I_{\swarrow} = 0$.

2. For $I_{\uparrow}$, we note that

\begin{align*}
I_{\uparrow}
&= \int_{\varphi = 0}^{\varphi = \pi/4} \omega|_{\theta=\pi/4} \\
&= \int_{\varphi = 0}^{\varphi = \pi/4} \left( - \log\cot\varphi \right) \biggl( \frac{\mathrm{d}( (\pi/4-\varphi)^2 )}{2\sin^2(\pi/4-\varphi)} - \frac{\mathrm{d}( (\pi/4+\varphi)^2 )}{2\sin^2(\pi/4+\varphi)} \biggr) \\
&= 2 \int_{0}^{\pi/4} \left( \log\cot\varphi \right) \biggl( \frac{\pi/4-\varphi}{1-\sin(2\varphi)} + \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \biggr) \, \mathrm{d}\varphi \\
&= 2 \int_{-\pi/4}^{\pi/4} \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \log \left| \cot\varphi \right|  \, \mathrm{d}\varphi \\
&= \frac{1}{2} \int_{0}^{\pi} \frac{x}{1-\cos x} \log \left| \cot \left(\frac{x}{2} - \frac{\pi}{4}\right) \right|  \, \mathrm{d}x,
\end{align*}

where the last step is the result of the substitution $\varphi=\tfrac{x}{2}-\tfrac{\pi}{4}$. Then by noting that

$$ \log \left| \cot \left(\frac{\pi}{4} - \frac{x}{2} \right) \right|
= \operatorname{Re}\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)
= i (\arctan(e^{-ix}) - \arctan(e^{ix})), $$

we get

\begin{align*}
I_{\uparrow}
&= \frac{1}{2}\int_{0}^{\pi} \frac{ix}{1-\cos x} (\arctan(e^{-ix}) - \arctan(e^{ix}))  \, \mathrm{d}x \\
&= \frac{1}{2}\int_{-\pi}^{\pi} \frac{ix}{1-\cos x} \left(\frac{\pi}{4} - \arctan(e^{ix})\right)  \, \mathrm{d}x \\
&= i \int_{\mathcal{C}} \frac{(\log z )(\frac{\pi}{4} - \arctan z)}{(z-1)^2}  \, \mathrm{d}z,
\end{align*}

where the substitution $z = e^{ix}$ is applied to the last step and $\mathcal{C}$ is the path parametrized by $e^{ix}$ for $-\pi < x < \pi$. Now deforming the contour $\mathcal{C}$ so as to enclose the principal branch cut for $\log$,

\begin{align*}
I_{\uparrow}
&= 2\pi \int_{0}^{1} \frac{\frac{\pi}{4} + \arctan t}{(t+1)^2}  \, \mathrm{d}t
= 2\pi \int_{0}^{1} \frac{1}{(t^2+1)(t+1)} \, \mathrm{d}t
= \frac{\pi^2}{4} + \frac{\pi}{2} \log 2.
\end{align*}

3. For the integral along the bottom,

\begin{align*}
\int_{\rightarrow} \omega
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \omega|_{\varphi=\epsilon} \\
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\mathrm{d}( (\theta-\epsilon)^2 )}{2\sin^2(\theta-\epsilon)} - \frac{\mathrm{d}( (\theta+\epsilon)^2 )}{2\sin^2(\theta+\epsilon)} \biggr) \\
&= \int_{\epsilon}^{\pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\theta-\epsilon}{\sin^2(\theta-\epsilon)} - \frac{\theta+\epsilon}{2\sin^2(\theta+\epsilon)} \biggr) \,  \mathrm{d}\theta.
\end{align*}

Substituting $\theta = \epsilon x$,

\begin{align*}
\int_{\rightarrow} \omega
&= \int_{1}^{\pi/4\epsilon} \log\left( \frac{\cot(\epsilon x)}{\cot\epsilon} \right) \biggl( \frac{\epsilon^2 (x-1)}{\sin^2(\epsilon(x-1))} - \frac{\epsilon^2 (x+1)}{2\sin^2(\epsilon(x+1))} \biggr) \, \mathrm{d}x.
\end{align*}

It is not hard (although not entirely trivial) to show that the above integral converges to

\begin{align*}
I_{\rightarrow}
= \lim_{\epsilon \to 0^+} \int_{\rightarrow} \omega
= \int_{1}^{\infty} (-\log x) \biggl( \frac{1}{x-1} - \frac{1}{x+1} \biggr) \, \mathrm{d}x
= -\frac{\pi^2}{4}.
\end{align*}

Conclusion. Combining all these observations altogether, we have

$$ \bbox[10px,#ffd]{ I = 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}) = \pi \log 2 } $$

as desired.

${}^{1)}$ We include the sketch of computation for $\mathrm{d}\omega$, which is essentially @G Cab's computation in backward direction. Substituting $u=\theta-\phi$ and $v=\theta+\phi$,

$$ \omega = \left( \log \left( \frac{\sin v - \sin u}{\sin v + \sin u} \right) \right)\biggl( \frac{u \, \mathrm{d}u}{\sin^2 u} - \frac{v \, \mathrm{d}v}{\sin^2 v} \biggr) . $$

Taking $\mathrm{d}$,

\begin{align*}
\mathrm{d}\omega
&= \mathrm{d}\left( \frac{u}{\sin^2 u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}u \\
&\qquad - \mathrm{d}\left( \frac{v}{\sin^2 v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}v \\
&= \frac{u}{\sin^2 u} \left( \frac{\partial}{\partial v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}v  \wedge \mathrm{d}u\\
&\qquad - \frac{v}{\sin^2 v} \left( \frac{\partial}{\partial u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}u \wedge \mathrm{d}v \\
&= 4 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) \, \mathrm{d}u \wedge \mathrm{d}v.
\end{align*}

By noting that $\mathrm{d}u \wedge \mathrm{d}v = 2 \, \mathrm{d}\theta \wedge \mathrm{d} \varphi$, the question boils down to showing

$$ 2 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) = \frac{(v+u)\cot(v+u) - (v-u)\cot(v-u)}{\cos(v+u) - \cos(v-u)}. $$

This can be verified by expanding the right-hand side.


问:
An amazing property of the Catenary
I discovered that if we want an arc of catenary  in the interval $[a,b]$ we solve
$$\int_a^b \sqrt{\cosh '(x)^2+1} \, dx=\int_a^b \cosh x \, dx$$
which means that the "result" of the length is equal to the result of the area in the same interval, though in different units.

So I asked myself if there are any other curve with the same property.

I set
$$y=\sqrt{y'^2+1}\to y^2=y'^2+1; y(0)=1$$
then
$$y'=\sqrt{y^2-1}\to dx=\frac{dy}{\sqrt{y^2-1}}\to x=\cosh^{-1} \,y$$
hence, arbitrary constant is zero,
$y=\cosh x$

But I am not sure how to deal with the other solution $y'=-\sqrt{y^2-1}$ even if Mathematica   gives the same result $y=\cosh x$

I'd like someone checking  this proof, you know: I am not a pro, I am just an (almost) retired high school teacher :)

Update 9/1/2020. Now I am officially retired :)

答:
From

$$\frac{y'}{\sqrt{y^2-1}}=\pm1$$ you draw

$$\text{arcosh}(y)=c\pm x$$

and

$$y=\cosh(c\pm x).$$

With the initial condition $y(0)=1$,

$$y=\cosh(\pm x)$$ which is $$y=\cosh(x).$$

Title: An amazing property of the Catenary


问:
Area under curve equals product of arc length and its projection
Find equation of a curve whose  area under it equals product of arc length  $L$ and its projection $(b-a)$ on the x-axis.

$$ A = L (b-a)$$

I was trying to establish a corollary to the Amazing Catenary property

答:
Shift the function such that $ a \to 0$

$$ \int_0^x f(t) dt =x  \int_0^x \sqrt{ 1 + (f'(t))^2} dt $$

d.w.r.t.x

$$ f(x) =  \sqrt{ 1 + (f'(x))^2} + \frac{xf''(x)}{\sqrt{ 1 + f'(x)^2} }$$

Sub: $ f(x) \to y$

$$  y = \sqrt{1 + (y')^2} + \frac{ x y''}{ \sqrt{ 1 + y'^2} }$$

Now all you have to do is solve this DE.


问:
If $E_{1}$ and $E_{2}$ satisfy the condition of caratheodory, then the intersection also satisfies
The condition of caratheodory tells us the following:

A set $E\subset { R }^{ n }$ satisfies the condition of caratheodory if for any set $S\subset { R }^{ n }$ , ${ m }_{ e }\left( S\cap E \right) +{ m }_{ e }\left( S-E \right) ={ m }_{ e }\left( S \right) $.

I must prove that ${ m }_{ e }\left( S\cap \left( { E }_{ 1 }\cap E_{ 2 } \right)  \right) +{ m }_{ e }\left( S-\left( { E }_{ 1 }\cap { E }_{ 2 } \right)  \right) ={ m }_{ e }\left( S \right)$

But since $E_{1}$ and $E_{2}$ satisfy the condition we have the following

${ m }_{ e }\left( S\cap { E }_{ 1 } \right) +{ m }_{ e }\left( S-{ E }_{ 1 } \right) ={ m }_{ e }\left( S \right) $

${ m }_{ e }\left( S\cap { E }_{ 2 } \right) +{ m }_{ e }\left( S-{ E }_{ 2 } \right) ={ m }_{ e }\left( S \right) $

I could use the following equivalence but I can't get to the result

$S-{ \left( { E }_{ 1 }{ \cap E }_{ 2 } \right) = }\left( (S\cap { E }_{ 1 })-{ E }_{ 2 } \right) \cup \left( \left( S-{ E }_{ 1 } \right) \cap { E }_{ 2 } \right) \cup \left( \left( S-{ E }_{ 1 } \right) -{ E }_{ 2 } \right) $

答:
The sets that satisfy the Caratheodory condition form an algebra (in fact a $\sigma$-algebra but that isn't necessary here). To see this, let $\mathcal{M}$ be the collection of all such sets. Since $S \setminus E = S \cap E^c$, the Caratheodory condition is symmetric wrt $E$ and $E^c$ so $\mathcal{M}$ is closed under complements. Now take any $E,F \in \mathcal{M}$ and any $S \subset \mathbb{R}^n.$ Then

\begin{align*}
m(S)&=m(S \cap E)+m(S \cap E^c)\\
&=\left(m(S \cap E \cap F)+m(S \cap E \cap F^c) \right) + \left(m(S \cap E^c \cap F) + m(S \cap E^c \cap F^c))\right)
\end{align*}

where the second equality holds because $S \cap E$ and $S \cap E^c$ are subsets of $\mathbb{R}^n$ and $F \in \mathcal{M}.$ Now observe that

$$E \cup F = (E \cap F) \cup (E \cap F^c) \cup (E^c \cap F),$$

and hence

$$S \cap (E \cup F) = (S \cap E \cap F) \cup (S \cap E \cap F^c) \cup (S \cap E^c \cap F).$$

So by subadditivity

$$m(S \cap (E \cup F)) \leq m(S \cap E \cap F) + m(S \cap E \cap F^c) +m(S \cap E^c \cap F).$$

And therefore,

\begin{align*}
m(S)&=m(S \cap E \cap F)+m(S \cap E \cap F^c) + m(S \cap E^c \cap F) + m(S \cap E^c \cap F^c)\\
& \geq m(S \cap (E \cup F))+m(S \cap E^c \cap F^c)\\
&= m(S \cap (E \cup F))+m(S \cap (E \cup F)^c).
\end{align*}

This shows $E \cup F \in \mathcal{M}$ since $m(S) \leq m(S \cap (E \cup F))+m(S \cap (E \cup F)^c)$ by subadditivity (note that $A=(A \cap B)\cup(A \cap B^c)$ for any sets $A,B$). So $\mathcal{M}$ is closed under unions. Now by DeMorgan's law,

$$E \cap F = (E^c \cup F^c)^c \in \mathcal{M}.$$


问:
Expected number of days
If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another B?

What I think is right: $E(days)=0.7(1)+0.12(2)+0.072(3)+...$

Is this the right way? If it is, how would I sum this? If not, please let me know.

答:
Whenever we have a bad day, we count that day and stop otherwise with probability $p$ we continue to the next day. As today is a bad day, for tomorrow $p = 0.3$ of having a good day. There on if we have a good day, probability of another good day is $0.6$.

$E(X) = 1 + 0.3 \times (1 + 0.6 \times (1 + 0.6 \times(1 + ...) + ..._)$

$E(X) = 1 + 0.3 \times \frac{1}{1-0.6} = 1.75$ (using sum of infinite geometric series $ = \frac{a}{1-r} \, ,a = 1, r = 0.6)$.

If the day when the next bad day happens should not be counted in, subtract $1$.

Title: Expected number of days


问:
Why can I write two unknowns $a,b$ as $K\cos(p)$ and $K\sin(p)$?
If I have two real numbers $a$ and $b$, why is it possible to express those numbers as $a=K\cos(\phi)$ and $b=K\sin(\phi)$? I came across this problem in Riley, Hobson and Bence (2006:14-15). Here is the question and answer in the book.

Question:

Solve for $\theta$ the equation:
$$a\sin\theta + b\cos\theta = k, $$
where $a$, $b$ and $k$ are real quantities.

Answer:

To solve this equation we make use of result (1.18) [the compound angle formulae] by setting $a=K\cos\phi$ and $b=K\sin\phi$ for suitable values of $K$ and $\phi$.

My question is: how do you know that, for arbitrary real numbers $a$ and $b$, there exists $K$ and $\phi$ such that $a=K\cos(\phi)$ and $b=K\sin(\phi)$? Additionally: does this mean that I can express two unknowns in terms of other trig functions, e.g. $a=R\sec(\alpha)$ and $b=R\cot(\alpha)$? Why or why not?

Reference:

Riley, K. F., Hobson, M. P., and Bence, S. J. (2006), Mathematical Methods for Physics and Engineering (3e). Cambridge: CUP.

答: Divide both sides by $\sqrt{a^2+b^2}$. $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ can be considered as sine and cosine of a some angle.


问:
Large n approximation of double factorial
A text I am reading claims that

$$(-g)^{n} \frac{(4 n-1) ! !}{n !} \stackrel{n \gg 1}{\sim}\left(-\frac{g n}{e}\right)^{n}$$

I know the stirling approximation $n!\stackrel{n \gg 1}{\sim} n^n e^{-n}$. But how do I go about approximating the double factorial in the numerator?

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\pars{-g}^{n}\,{\pars{4n - 1}!! \over n!}} =
{\pars{-g}^{n} \over n!}\,\prod_{k = 0}^{2n - 1}\pars{2k + 1} =
{\pars{-g}^{n} \over n!}\,2^{2n}\prod_{k = 0}^{2n - 1}\pars{k + {1 \over 2}}
\\[5mm] = &\
{\pars{-g}^{n} \over n!}\,2^{2n}\,\pars{1 \over 2}^{\overline{2n}} =
{\pars{-g}^{n} \over n!}\,2^{2n}\,{\Gamma\pars{1/2 + 2n} \over \Gamma\pars{1/2}} =
{2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\,{\pars{2n - 1/2}! \over n!}
\\[5mm] &\
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\,
{\root{2\pi}\pars{2n - 1/2}^{2n}\expo{-2n + 1/2} \over \root{2\pi}n^{n + 1/2}\,\,
\expo{-n}}
\\[5mm] = &\
{2^{2n}\,\pars{-g}^{n} \over \root{\pi}}\,
{2^{2n}n^{2n}\,\, \bracks{1 - \pars{1/2}/\pars{2n}}^{\,2n}\expo{1/2} \over
n^{n + 1/2}}\,\expo{-n}
\\[5mm] &\ \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{2^{4n}\,\pars{-g}^{n} \over \root{\pi}}\,
n^{n - 1/2}\,\expo{-n} =
\bbx{{1 \over \root{\pi}}\pars{-16gn \over \expo{}}^{n}\,{1 \over n^{1/2}}}
\\ &
\end{align}


问:
When doing finding missing sides in trigonometry do I have to divde or multiply?
so I have a test in like an hour and I'm confused.

Let's suppose that the hypotenuse is $8$ $\mathrm{cm}$ and our angle is $40^\circ$ and the opposite side is missing.

So $\sin B = \frac{b}{8\textrm{ cm}}$ is $5$ $\mathrm{cm}$ when I multiply.

But when I divide on my calculator I get this $12$ $\mathrm{cm}$.

I was told I had to multiply but $b$ is bigger than the hypotenuse...

答:
Your equation $\sin40^\circ = \frac{b}{8}$ is correct.

So multiply both sides by 8.

$b = \sin(40^\circ)\times 8 = 5.142cm$

EDIT:

Suppose the opposite side was given as 8, and you're told to find the hypoteneuse. That's a different problem. In that case

$\sin(40^\circ) = \frac{8}{h}$

Then multiply both sides by h, and divide both sides by $\sin(40^\circ)$ to get

$h = \frac{8}{\sin(40^\circ)} = 12.45cm$


问:
How to calculate $\sum_{i=2}^n {\frac 1{\log_2 i}}$
How to calculate
$$\sum_{i=2}^n {\frac 1{\log_2 i}}$$</p>

答:
If you meant $$\sum_{i=2}^n \dfrac1{\log_2(i)},$$ then there is no closed form; though you can compute its asymptotic using Euler-Maclaurin formula:
$$\sum_{k=2}^{n} \dfrac1{\log_2(k)} \sim \log 2\times\int_2^n \dfrac{dx}{\log(x)} \sim \dfrac{n}{\log_2(n)}$$

However, if you meant $$\sum_{i=2}^n \dfrac1{\log_2(n)}$$ note that the term inside the summation is independent of $i$ and hence equals (exactly) $$\dfrac{n-1}{\log_2(n)}$$


答:
Is this the right way? : E(days)=0.7(1)+0.12(2)+0.072(3)+...

Yes, it is!

In other words, the sequence, expressed in days, is the following

$1 \xrightarrow{\text{with probability}} p=0.7$

$2\xrightarrow{\text{with probability}} p=0.3\times0.4$

$3\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.4$

$4\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.4$

$5\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.6\times0.4$

...

$n \xrightarrow{\text{with probability}} p=0.3\times0.4\times\underbrace{0.6\times0.6\times\dots\times0.6}_{\text{(n-2) times}}$

Thus the expectation is the following

$$ \bbox[5px,border:2px solid red]
{
\mathbb{E}[X]=0.7+0.3\mathbb{E}[Y]=0.7+0.3\Bigg[1+\frac{1}{0.4}\Bigg]=1.75
\qquad (1)
}
$$

Where $Y$ is a geometric distribution with success parameter $0.4$ and starting from $k=2$

In any case, you can solve the series in an analytical way:

Setting $0.6=p$ your series becomes

$$\mathbb{E}[X]=0.7+\frac{0.3\times0.4}{0.6}\underbrace{\sum_{x=2}^{\infty}x p^{x-1}}_{=S}$$

$$S=\sum_{x=2}^{\infty}\frac{d}{dp}p^x=\frac{d}{dp}\sum_{x=2}^{\infty}p^x=\frac{d}{dp}\frac{p^2}{1-p}=\frac{p(2-p)}{(1-p)^2}\Bigg]_{p=0.6}=5.25$$

Thus

$$\mathbb{E}[X]=0.7+\frac{0.3\times0.4}{0.6}\times 5.25=1.75$$

As alreay obtained in (1).

Title: Expected number of days

Body:
If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another B?

What I think is right: $E(days)=0.7(1)+0.12(2)+0.072(3)+...$

Is this the right way? If it is, how would I sum this? If not, please let me know.


问:
Why six Trigonometric Ratios
Why should we use six trigonometric ratios when all the three sides of a right angled triangle (opposite, adjacent and hypotenuse) are connected with the three rations sine, cosine and tangent?
what is the significance of inverses of these ratios(cosec, sec and cot)?</p>

答:
Short answer: there are three sides (adjacent $A$, hypotenuse $H$ and opposite $O$) we can choose in three unordered pairs or six ordered pairs, giving ratios such as $\sec x=\frac{H}{A}$ etc.

Long answer: You could argue only one trigonometric function is needed, $t:=\tan\frac{x}{2}$ so$$\sin x=\frac{2t}{1+t^2},\,\cos x=\frac{1-t^2}{1+t^2},\,\tan x=\frac{2t}{1-t^2}.$$Or you could argue it's convenient to name not only the above functions and their reciprocals, but also other functions such as $\operatorname{versin}x$ etc. (That link shows geometric significances of all these functions, albeit from the perspective of a circle in which we embed a right-angled triangle of hypotenuse $1$.) History has settled on a compromise where a few functions are widely memorised while others aren't (e.g. a mathematician might not know off the top of their head that $\operatorname{exsec}x:=\sec x-1=\frac{2t^2}{1-t^2}$). But history turned out that way because of how often each function is useful.

Title: Why six Trigonometric Ratios


问:
Simultaneous algebra question
Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$

Find $$x^3+3y^3$$

Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$

I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$

答:
Being lazy, I changed a little the notations
$$ax+by=v \tag 1$$ $$ax^2+by^2=w \tag 2$$

From $(1)$, you have $y=\frac{v-a x}{b}$. Plug in $(2)$ to face
$$\left(v^2-b w\right)-2 a vx+a\left(a+ b\right)x^2=0$$
$$x_\pm=\frac{av\pm\sqrt{a b \left(w (a+b)-v^2\right)}}{a (a+b)}\qquad \qquad y_\pm=\frac{v-a x_\pm}{b}$$

Now, just a bit of patience for the simplifications and
$$a x_+^3+b y_+^3=\frac{a^2 \left(bv+\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3+b^2 \left(a
   v-\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3}{a^2 b^2 (a+b)^3}$$

Title: Simultaneous algebra question


问:
A basic doubt on the sojourn time of a CTMC
By using the memoryless property, I find the CDF of sojourn time of a CTMC as follows :

$$F_X(t) = 1-e^{-F_X'(0)t}$$

I am slightly confused about the intuitive meaning of the term $F_X'(0)$. How this term represents the rate of leaving that state.

答:
These are the basics of the exponential distributions, probably explained on the WP page.

If $T$ is exponential with parameter $a$, then $F_T(t)=1-\mathrm e^{-at}$ for every $t\geqslant0$ hence $a=F_T'(0^+)$ and $F_T(t)=1-\mathrm e^{-F_T'(0^+)t}$.

Furthermore, for every $t\geqslant0$, $P[t\leqslant T\leqslant t+s\mid T\geqslant t]=1-\mathrm e^{-as}\sim as$ when $s\to0$ hence, indeed, $a$ represents the rate of $T$ happening at $t$. Exponential distributions are characterized by the fact that this rate does not depend on $t\geqslant0$.


问:
Understanding Screw Axes in Crystallographic Space Groups
I'm having trouble understanding how a 'screw axis' actually acts as on $\mathbb{E}^3$ as an element of a space group. Everything I talk about here will be in 3 dimensions. nLab defines a space (or crystallographic) group as a subgroup $G$ of the Euclidean group $\mathrm{E}(3)$ for which there exists a lattice $L$ in $\mathbb{E}^3$ such that $G\subseteq\mathrm{Aut}(L)$ and such that $G$ contains all translations by elements of $L$.

Let's consider a very simple example of a space group which contains a screw axis—$P222_1$ (note that I am using Wikipedia as my source for lists of space groups). If I understand Hermann–Mauguin notation correctly (which I probably don't), we choose 3 positive numbers $a,b$ and $c$, and then the group $P222_1$ with respect to this basis is generated by: all translations by elements of the lattice $L=\mathbb{Z}\langle (a,0,0),(0,b,0),(0,0,c) \rangle$; rotation by $\pi$ about the $x$-axis; rotation by $\pi$ about the $y$-axis; and the combination of rotation by $\pi$ about the $z$-axis and translation by $(0,0,c/2)$.

Now, if this is correct, then it is quite clear to me that the orbit of the origin under $P222_1$ is not $L$, but the lattice $L'=\mathbb{Z}\langle (a,0,0),(0,b,0),(0,0,c/2) \rangle$; and $P222_1$ contains all translations by elements of this lattice and preserves this lattice. Thus it is exactly the same as the group $P222$ with respect to the basis $a,b$ and $c/2$, requiring no screw axes to generate it. So $P222_1$ is a completely redundant symbol.

What have I done wrong here? It's driving me crazy.

答:
Hermann–Mauguin notation is sneaky. It tells you the nature of a crystal's symmetry when viewed along certain axes, but not the precise realisations of those symmetries. It does not tell you the positions of rotation and screw axes in the unit cell, for example.

In your case, the symbols $\mathrm P222_1$ do not tell you that the "rotation by $\pi$ about the $x$-axis" should instead be a rotation by $\pi$ about the line $(0,0,\color{red}{1/4})+(1,0,0)\lambda$ (which is parallel to the $x$-axis).

With this replacement, the generating set you have described is $\{(x,-y,-z+c/2),(-x,y,-z),(-x,-y,z+c/2)\}$ and correctly describes $\mathrm P222_1$. But now note that the first generator is the composition of the other two modulo the lattice translations. Hence a minimal generating set for $\mathrm P222_1$ is $\{(-x,y,-z),(-x,-y,z+c/2)\}$, showing that there are $4$ fundamental domains in its unit cell.


问:
Number of ways of arranging stained glass panes on a $2\times 4$ grid(upto rotations and flips).
Suppose you have eight squares of stained glass, all of different colors, and you would like to make a rectangular stained glass window in the shape of a $2 × 4$ grid.

How many different ways can you do this, taking symmetry into account? (Note that any pattern may be rotated 180 degrees, flipped vertically, or flipped horizontally. You should count all the possible resulting patterns as the same window)

I computed $\frac{8!}{8}$ since for each window, there are $2$ windows that are identical when rotating one of the windows by $180$ degrees, $2$ windows identical when flipping one of the windows vertical, and $2$ windows identical when flipping horizontally. Thus, $8$ identical windows for each $8!$ number of initial windows. Is there anything wrong with this solution?

答:
You can view an arrangement as an element of a set $S$. The total number of elements in $S$ is $8!$ (without any restrictions). Consider any element, for example;
$$
\begin{bmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8
\end{bmatrix}
$$
(each number corresponds to a color). Now, we can get $3$ additional elements of $S$ just by a $180^\circ$ rotation, a vertical flip or a horizontal flip, giving you
$$
\begin{bmatrix}
8 & 7 & 6 & 5\\
4 & 3 & 2 & 1
\end{bmatrix},
\begin{bmatrix}
4 & 3 & 2 & 1\\
8 & 7 & 6 & 5
\end{bmatrix},
\begin{bmatrix}
5 & 6 & 7 & 8\\
1 & 2 & 3 & 4
\end{bmatrix}
$$

Notice that applying these operations to any one of $4$ elements results in an element from one of these only. Hence, you can partition $S$ into such independent groups of $4$. Therefore the total number of arrangements identical upto rotations and flips is $\dfrac{8!}{4}$.

Problem in your solution:

You can now see why the solution $\dfrac{8!}{8}$ is not correct. When you first divide by $2$, let's say to remove $180^\circ$ rotations, half of the elements are rightly  removed. The next time you divide by $2$, the horizontal flips are removed. To remove the vertical flips, you've divided by $2$ once more, but notice that a $180^\circ$ rotation followed by horizontal flip is essentially a vertical flip, so by just dividing by $4$ you've removed even the vertically flipped versions of every element in the process, hence dividing by $2$ once more leads to a miscount.


问:
Discrete math, antisymmetric proof question
I am so confused on this one question on my practice work. it's: Show that if R is an antisymmetric relation on S then any relation R' ⊆ R is also antisymmetric. I think how you solve it is if (x,y) are in R, they must be in R' since R' is a subset of R, thus (x,y) are in R'. To be antisymmetric x has to be related to y and y has to be related to x such that x = y. And since R is antisymmetric and x,y are in R and R' is a subset then x,y in R' follow the same condition of R making them antisymmetric. That's one way I think it's solved but not 100% sure.</p>

答:
Mistake: If $(x,y) \in R$, then they must be in $R'$ since $R'$ is a subset of $R$.

This reasoning is not correct. If you see a black cat, you certainly see cat. If you see a cat, it need not be black.

If $(x,y) \in R'$, then $(x,y) \in R$ since $R' \subset R$, since $R$ is antisymmetirc, $(y,x)\notin R$ and hence we conclude that $(y,x)\notin R'$. That is $R'$ is antisymmetric.


问:
$\sum^\infty_{n=0} \frac{e^n}{n!}$
$$\sum^\infty_{n=0} \frac{e^n}{n!}$$
I am pretty sure already it converges, I'm trying to find to what.
$$\sum^\infty_{n=0} \frac{e^n}{n!}=\sum^\infty_{n=1}\prod^n_{j=1}\frac{e}{j}$$
$$\sum^\infty_{n=0}\frac{\sum^\infty_{j=0}\frac{n^j}{j!}}{n!}=\sum^\infty_{n=0}\sum^\infty_{j=0}\frac{n^j}{n!j!}=\sum^\infty_{j=0}\frac{1}{j!}\sum^\infty_{n=0}\frac{n^j}{n!}$$
from here, i have no idea.

答:
We will use the Taylor series fromula:
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
This formula converges to the value of the function if and only if the function is analytic in an open disk centered at 0.

Now, let $f(x)=e^x$. We know that
$$\frac{d^n}{dx^n}e^x=e^x\quad \forall n\ge0$$
so that
$$\frac{d^n}{dx^n}e^x\bigg\rvert_{0}=e^0=1$$
Substituting this in the original Taylor series formula yields
\begin{align}
e^x&=\sum_{n=0}^{\infty}\frac{\frac{d^n}{dx^n}e^x\big\rvert_{0}}{n!}x^n\\
&=\sum_{n=0}^{\infty}\frac{x^n}{n!}
\end{align}
Substituting $x=e$ gives
$$\sum_{n=0}^{\infty}\frac{e^n}{n!}=e^e\approx15.142622$$

Title: $\sum^\infty_{n=0} \frac{e^n}{n!}$


问:
Commutativity of Contractions
I'm working through Pavel Grinfeld's book 'Introduction to Tensor Analysis and the Calculus of Moving Surfaces' and I am having trouble understanding the example given in section 4.9.2 - Commutativity of Contractions:

For an illustration, consider an object $T_{kl}^{ij}$ with four indicies and the double contraction $T_{ij}^{ij}$. The expression $T_{ij}^{ij}$ can be interpreted in three ways. First, we can imagine that contraction on i takes place first, yielding the object $U_l^j = T_{il}^{ij}$. Fully expanding, $U_l^j$ is the sum

$U_l^j = T_{1l}^{1j} + T_{2l}^{2j} + T_{3l}^{3j}$

What I don't understand is the l that now appears in $U_l^j$ instead of j because surely now there can be no contraction on j because there are different indices on the top and bottom? Why is there the need to introduce the index l?

答:
I think you are somehow confused by the fact that $j$ and $l$ are seemingly different: they could be but what if they're equal then you can contract them.

We are given a tensor of type $(2,2)$, $T^{ij}_{kl}$, element of some vector space $V \otimes V\otimes V^*\otimes V^*$, with 2 contravariant elements and 2 covariant elements.

The point of this short paragraph in the book is to illustrate that we can first contract $i$ and $k$ (that is setting $i=k$), getting a mixed tensor of type $(1,1)$ and then contract $j$ and $l$ (that is setting $j=l$) or first contract $j$ and $l$ (again setting $j=l$) and then after that $i$ and $k$ (setting $i=k$), that is: contractions commute.

Suppose first that $i=k$ and contract these indices. One then obtain a $(1,1)$ mixed tensor with one covariant element and one contravariant element and it can be written as $$U^j_l= T^{1j}_{1l}+T^{2j}_{2l}+T^{3j}_{3l}.$$

Now next to operate the second contraction, let $j=l$, and smilarly one gets the following:

$$T^{11}_{11}+T^{12}_{12}+T^{13}_{13}+T^{21}_{21}+T^{22}_{22}+T^{23}_{23}+T^{31}_{31}+T^{32}_{32}+T^{33}_{33}.$$

You can do the contractions either way first. Alternatively you can do the contractions simultneously: just write down all possible combinations for $T^{ij}_{ij}$ after having set $i=k$ and $j=l$.

Title: Commutativity of Contractions


问:
A particular disease affects 0.1% of the population.

A particular disease affects 0.1% of the population.

The test is 95% accurate.

You test positive.

What is the chance that you have the disease?

How I approached the problem:

Since I already tested positive, and the question clearly states that the test is 95% accurate, there should be a 95% probability that I have the disease. But, a YouTube video says that the chance I have the disease is 2%. How is my reasoning wrong?

答:
Take a population of $1.000.000$.

This means that $1.000$ people have the disease and $999.000$ don't have the disease.
With the test we get groups of false and true positives and false and true negatives. We aim to find the portion that are true positives compared to the false and true positives.

From the $1000$ people that have the disease, $950$ people will get a true positive and $50$ people will get a false negative (they have the disease, but because the test is $95$% accurate, they get a negative result).
From the $999.000$ people that don't have the disease, $949.050$ will get a true negative and $49.950$ people will get a false positive.

Now the probability you actually have the disease while being positive is simply:

$\frac{950}{49.950+950}$ = $\frac{950}{50.900}$=0.018664...which is roughly $2$%.


问:
Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.
I want to prove the following statement:

Two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear vector space $X$ are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

(Note that going from equivalent to bounded is straightforward, so I don't need help with that.)

As per my book, two norms are equivalent if there exists $\alpha,\beta > 0$ such that
$$
\alpha \| x \|_1 \leq \| x \|_2 \leq \beta \| x \|_1 \hspace{1cm} \forall x \in X
$$

Attempt at solution

Proof by contradiction, i.e. for all $\alpha,\beta > 0$, we have the following inequalites
$$
\| x_0 \|_2 < \alpha\| x_0 \|_1
$$
or
$$
\beta\| x_0 \|_1 < \| x_0 \|_2
$$
for some $x_0$.

I want to use the boundedness assumption here, like, the LHS is bounded by the RHS, thus the RHS is bounded too. But the problem is that this assertion is for $x_0$ specifically, not the entire domain $X$.

Any ideas?

答:
Suppose bounded sets for the two norms are the same. Let $A=\{x: \|x\|_1 \leq 1\}$. This is bounded in the first norm. So it must be bounded in then second norm. Hence ther exists $\beta \in (0,\infty)$ such that $\|x\|_2 \leq \beta$ whenever $\|x\|_1 \leq 1$. Now take any non-zero vector $y$. Let $x=\frac y {\|y\|_1}$. Then $\|x\| \leq 1$ so $\|x\|_2 \leq \beta$. This gives $\|y\|_2 \leq \beta \|y|\|_1$. A similar argument gives $\|y\|_1 \leq c \|y|\|_2$ for some $c>0$ and this can be writen as $\alpha \|y\|_1 \leq \beta \|y|\|_2$ where $\alpha =\frac 1  c$. Hence the norms are equivalent.

Conversely if the norms are equivalent it is evident that boundedness of one the norms for a set of vectors implies boundedness of the set for the other norm.


问:
$R=K[[x]]$, show that for every $0\neq f\in R,f|x^n $ and $x^n|f$ where $n\in\mathbb{N}$
$K$ is a field

I have to show the existence of a unit $u$, i.e. an element where the constant part of the formal power series is not $0$ such that $f=ux^n$, i.e $x^n=fu$

From the excercise sheet I cannot tell if this should hold for every $n\in\mathbb{N}$ or just one $n\in\mathbb{N}$. Is it possible to sho it for every $n\in\mathbb{N}$ ?

How can I start?

答:
Assume it would hold for different $n$. So there exist $u,v$ units in $K[[x]]$ such that
$$ u x^n = f = v x^m $$
with $m,n\in \mathbb{N}$ and $n\neq m$. Wlog assume $n>m$, then we have
$$ 0 = vx^m - u x^n = vx^m(1- uv^{-1} x^{n-m)}. $$
As $K[[x]]$ is an integral domain and $ux^n \neq 0$ we get $1-uv^{-1} x^{n-m}=0$, thus $x^{n-m}$ is a unit, which is a contradiction.
So, indeed, for every $f$ there is a unique $n$ with the desired property.

To show the general statement it might be helpful to first trying to solve the case
$$ f= \sum_{n\geq 1} x^n. $$
From here you should get the intuition how to proceed.


答: Consider the unit ball in the $\|\cdot\|_1$ norm. It is bounded in the $\|\cdot\|_2$ norm, which means that for any $x$ with $\|x\|_1\le 1$ you have $\|x\|_2\le \beta$ for some $\beta>0$. But since both norms are homogeneous, the same is true for any $x$ in your space. In the same way you prove that for any $x$ with $\|x\|_2\le 1$ you have $\|x\|_1\le \gamma$ for some $\gamma>0$.

Title: Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

Body:
I want to prove the following statement:

Two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear vector space $X$ are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

(Note that going from equivalent to bounded is straightforward, so I don't need help with that.)

As per my book, two norms are equivalent if there exists $\alpha,\beta > 0$ such that
$$
\alpha \| x \|_1 \leq \| x \|_2 \leq \beta \| x \|_1 \hspace{1cm} \forall x \in X
$$

Attempt at solution

Proof by contradiction, i.e. for all $\alpha,\beta > 0$, we have the following inequalites
$$
\| x_0 \|_2 < \alpha\| x_0 \|_1
$$
or
$$
\beta\| x_0 \|_1 < \| x_0 \|_2
$$
for some $x_0$.

I want to use the boundedness assumption here, like, the LHS is bounded by the RHS, thus the RHS is bounded too. But the problem is that this assertion is for $x_0$ specifically, not the entire domain $X$.

Any ideas?


问:
Solve a recurrence relation in two variables
Suppose the function $f : \mathbb{N}\to\mathbb{N}$ is such that $f(1)=1$ and
$$f(2m) + f(2n) = 2(f(m+n) + f(m-n))$$
for all $m>n>0$. I would like to show that this implies $f(n) = n^2$ for all $n\in\mathbb{N}$.

It is easy to show that $f(n)=n^2$ is a solution, just by substituting, but I'm having trouble proving uniqueness. I know about mathematical induction but in this case, the fact that there are two variables $m$ and $n$ is giving me trouble. Any help appreciated.

Added: the part that's giving me trouble is that I cannot assume that $f(0)$ exists. If I could assume that, then $f(0)=0$ by putting $m=n=0$, and then by putting $m=n$ I could show that $f(2m) = 4f(m)$, from which it's not hard to continue.

I guess my question is: am I allowed to extend $f$ by defining $f(0)=0$, and use that in my reasoning? It seems like cheating somehow.

答:
Ok, I try to explain completely my idea.

First, taking $n=1$, $n=2$, $m\geq 3$, eliminating $f(2m)$, we obtain that

$$f(m+2)-f(m+1)-f(m-1)+f(m-2)=(f(4)-f(2))/2 $$ and we put $d=(f(4)-f(2))/2$. This is a linear recurrence equation; we solve it completely, we do not care about particular values of $f(m)$ for the moment. The general solution is of the form (particular solution+general solution of the homogeneous equation).

We try to find a particular solution of the form $cm^2$, and we find $cm^2$, with $c=d/6$.

Now the homogeneous equation ;  the polynomial characteristic of the recurrence is $(x-1)^2(x^2+x+1)$, with a double root $1$, and simple roots $j$, $j^2$ (notation for the cubic roots of unity). So the general solution of the homogeneous recurrence equation is of the form $u+vm+wj^{m}+qj^{2m}$ for some constants $u,v,w,q$. (note that for example, $j^m+j^{2m}$ is in $\mathbb{R}$, and a solution of the homogeneous recurrence equation ; we cannot, at this point, eliminate these terms).

So, what we have proven is that a solution of the original equation is of the form

$$f(m)=cm^2+u+vm+wj^{m}+qj^{2m}$$
Now we replace in the original equation:

$$4cm^2+u+2mv+wj^{2m}+qj^{4m}+4cn^2+u+2nv+wj^{2n}+qj^{4n}$$
is equal to
$$2[c(m+n)^2+u+v(m+n)+wj^{m+n}+qj^{2(m+n)}+c(m-n)^2+u+v(m-n)+wj^{m-n}+qj^{2(m-n)}]$$

Now the expression of a recurrence sequence (here we fix $n$, and $m$ is the variable) as $\sum\alpha \beta^m$ is unique. So we look at the coefficients of $1,m,m^2, j^m, j^{2m}$. The coefficient of $m$ on the first side is $2v$; on the second side, $4v$. Hence $v=0$. The coefficient of $j^{m}$ on the first side is $q$, on the second side $2w(j^n+ j^{-n})$, hence $q=2w(j^n+ j^{-n})$; the coefficients of $j^{2m}$ give also $w=2q(j^{2n}+j^{-2n})$; choosing $n=3$ gives $q=w=0$. The constant coefficient is (using $v=w=q=0$) $2u+4cn^2$ in the first expression,and $4u+4cn^2$ in the second, hence we get $u=0$, and we have proven that a solution of the original equation must be of the form $c m^2$ for some constant $c$. It is easy to finish.


问:
Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$
I am still not convinced by the post that the function$$\sqrt[3]{(z-1)(z-2)(z-3)}$$ can be defined so it is analytic on $\mathbb{C}\backslash [1,3]$. We define for each $z\in \mathbb{C}\backslash (-\infty,3]$ the function $$f(z)=\int_4^z \frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}\,dz +\ln 6$$ and it claims that function $$\exp\left(\frac{1}{3}f(z)\right)$$ is continuous on $(-\infty,1)$. This is the part I don't understand. How do we prove that it is continuous on $(-\infty,1)$. I computed the integral from 4 to $-4$ along the upper semicircle of $|z|=4$, and the integral from 4 to $-4$ along the lower semicircle of $|z|=4$ and they are not equal. So I don't see how the function can be continuous at $-4$, for example.

答:
Given that
$$\frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}=\frac1{z-1}+\frac1{z-2}+\frac1{z-3}$$ you see that all the three residues of the integrand are equal to one, so their sum is $3$. The "upper" and "lower" integrals thus differ by $6\pi i$.

After multiplication by $1/3$ they differ by $2\pi i$, and after $\exp$ they agree. IOW $f(z)$ won't be continuous, but $\exp(f(z)/3)$ still is.


问:
expressing class of inversions in einstein summation convention
I'm working on a personal project where I'm studying the set of inversions in $\Bbb R^n$
that preserve the unit sphere centered at the origin; these transformations can be defined based on a center of inversion $c$, like so:

$$I_{c}(x) = (c\cdot c-1){{x-c}\over{(x-c)\cdot(x-c)}}+c$$

I'm curious about how this would be expressed in Einstein summation convention, but I'm not too familiar with it and my attempts have failed.

Body: It would be$$[I_c(x)]_i=\frac{c_jc_j-1}{(x_k-c_k)(x_k-c_k)}(x_i-c_i)+c_i.$$


问:
A confusion regarding Rayo's number and Busy Beaver function.
I am slightly confused about the definition of the Rayo function and Rayo's number, and how it relates to the Busy Beaver function. I know that ZFC can't pin down the precise value of even $BB(7918)$. However, ZFC can define the Busy Beaver function itself, in less than, say, a billion symbols. So, we can form the number $BB(7918)$, and even $BB(BB(7918))$, using way less than a googol symbols of ZFC. But now here is where I am confused. Since ZFC can't determine the value of $BB(7918)$, is it even legitimate to give a lower bound for Rayo's number as $BB(7918)$? Do we only use descriptions of numbers that ZFC can prove define a number uniquely? I would be grateful if someone clarified this matter for me, and told me the exact definition of the Rayo function and whether it is a well-defined function at all.

答:
Short version: Rayo's function doesn't relate at all to the Busy Beaver function. (Or rather there is an obvious relation - Rayo is way way way way WAY bigger than the Beaver - but the Busy Beaver just isn't relevant to Rayo.) Rayo is fundamentally about truth rather than provability, and so the relevant "logical obstacle" is Tarski's undefinability theorem rather than the incomputability of the halting problem.

In more detail:

Rayo's function $R$ has nothing to do with provability, in $\mathsf{ZFC}$ or otherwise. Simply stated, $R(n)$ is $1$ + the supremum of the natural numbers which are definable in set theory by formulas of length $<1$. Here, we say that a natural number $a$ is defined by a formula $\varphi$ iff $a$ is the unique object satisfying $\varphi$ in the sense of $V$, that is, the unique thing such that $V\models\varphi(a)$. The question of whether $\varphi(a)$ is provable in some theory or other doesn't arise. So $BB(7918)$ is indeed a lower bound for Rayo's number $R(10^{100})$.

However, this does mean that we run into Tarski's theorem: Rayo's function cannot itself be defined within $V$ (and in particular it doesn't even make sense to ask what $\mathsf{ZFC}$ does or does not prove about it). Rather, to talk about $R$ we need to work in a context rich enough to define "truth in $V$" - such as some appropriate class theory. So it does make sense to ask, for example, whether $\mathsf{NBG}$ proves that $R(10^{100})>BB(7918)$ (which it indeed does, quite easily).

It may help to first consider the "arithmetic Rayo function" - this is the map $R_{arith}$ sending $n$ to $1$ + the supremum of the natural numbers which are definable by formulas of length $<n$ in the structure $\mathfrak{N}=(\mathbb{N};+,\times)$. The function $R_{arith}$ is (again, per Tarski) not definable in $\mathfrak{N}$ itself; however, it is easily definable in set theory, and so it makes sense to ask what $\mathsf{ZFC}$ proves about $R_{arith}$.

It may also help to contrast $R$ with the "provability analogue" of the Busy Beaver function: let $P_{\mathsf{ZFC}}$ be the function sending $n$ to $1$ + the supremum of the natural numbers $k$ such that there is a formula $\varphi$ in set theory of length $<n$ such that $\mathsf{ZFC}$ proves that $\varphi$ holds of $k$ and only $k$. The function $P_{\mathsf{ZFC}}$ may superficially seem like Rayo's function, but it's really just a Busy Beaver variant (and in particular is easily definable in set theory).

Finally, there is a "local version" of the Rayo function: for $n\in\omega$ and $M$ (for simplicity) a transitive model of $\mathsf{KP}$, let $R_{loc}(n,M)$ be $1$ + the supremum of the natural numbers $k$ such that there is some formula $\varphi$ with $\{x\in M: M\models\varphi(x)\}=\{k\}$. The function $R_{loc}$ is definable in set theory again, since we're only ever asking about truth in set-sized structures, but for each appropriate $M$ the function $n\mapsto R_{loc}(n,M)$ is not definable in $M$.

Now you may ask: why have I defined $R$ and $P_{arith}$ in terms of suprema above, rather than maxima? Well, thinking set-theoretically it becomes quite natural to consider a "super-Rayo function" $S$ which applies to arbitrary ordinals: for an ordinal $\alpha$ let $S(\alpha)$ be the supremum of the ordinals $\beta$ such that there is some $\mathcal{L}_{\infty,\infty}$-formula of length $<\alpha$ which defines $\beta$.

Note that $S(n)$ is generally much greater than $R(n)$ even when $n$ is finite - e.g. $\omega$ is pretty easily definable in the language of set theory by a finite-length formula.

I don't know offhand of a particular application of $S$, but to me it actually seems a bit more interesting than $R$. For example, consider the "ultra-Rayo" function $U$, defined as $S$ but using $\mathcal{L}_{\infty,\omega}$ instead of $\mathcal{L}_{\infty,\infty}$. This gives a slower-growing function than $S$ which is still pretty fast-growing, and at a glance it seems like there might be some nontrivial comparisons to be drawn. And these are merely two of the many extensions of first-order logic which let us talk about arbitrarily large ordinals. So I'm happy to use "supremum" rather than "maximum" even in the finite case, in order to smooth the way for these variants.

Wait, what is the "length" of an infinitary formula? Well, we'll define it recursively, similarly to quantifier rank or similar complexity notions. For example, we should probably set $$len("\bigvee_{\alpha<\theta}\varphi_\alpha")=(\sum_{\alpha<\theta}len(\varphi_\alpha))+1$$ or something similar. The point is that it's not hard to extend the notion of length to infinitary formulas, even if it's not immediately clear which extension is the "right" one, and as soon as we do we get the ultra- and super-Rayo functions emerging as natural objects of study.


问:
Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........?
Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?

I am not well versed with binomial theorem hence I am not able to proceed. I am also interested in the different ways this problem can be solved.

Any help would be appreciated.

Body:
From binomial expansion it follows that
$$[(2+\sqrt{3})^n+(2-\sqrt{3})^n]$$ $$=[2^n+{n \choose 2} 2^{n-2} 3+{n \choose 4}2^{n-4} 3^2+...]=2K, K \in N$$
And $$[(2+\sqrt{3})^n-(2-\sqrt{3})^n]$$ $$=2\sqrt{3}[2^{n-1}+{n \choose 1} 2^{n-1}+{n \choose 3}2^{n-3} 3+...]=2\sqrt{3}L, L \in N$$
$$S_1=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}=2^n[(2+\sqrt{3})^n+(2-\sqrt{3})^n]=2^{n+1} K.$$
$$S_2=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}=S_1+2^n \sqrt{3}[(2+\sqrt{3})^{n}-(2-\sqrt{3})^n]=3.2^{n+1}L$$
Hence, both $S_1$ and $S_2$ are divisible by $2^{n+1}$.

By binomial expansio


问:
Proving that a subset $\mathscr{H}$ is dense in $\mathbb{R}$

Consider the subset $\mathscr{H} =\displaystyle\left\{\frac{a}{10^{b}} \ |\: a,b\in\mathbb{Z}\: \right\}$. Then is $\mathscr{H}$ dense in $\mathbb{R}$.

To prove that $\mathscr{H}$ is dense I have to show that given $x,y \in\mathbb{R}$  there is a $m,n\in\mathbb{Z}$ such that $x < \frac{m}{10^{n}} <y$.

Thats it I am hopelessly stuck.

答:
We write out an argument in fairly formal style.  We show there is an integer $m$, and a positive integer $n$, such that $x\lt \frac{m}{10^n}\lt y$.

Suppose that $x\lt y$, and let $d=y-x$. There is a positive integer $n$ such that $\frac{1}{10^n}\lt d$.

Consider the set $S$ of integers $k$ (positive, negative, or $0$) such that $\frac{k}{10^n}\lt y$. The set $S$ is bounded above, so there is a largest integer $m$ in $S$.

We show that $x\lt \frac{m}{10^n}\lt y$. It is obvious from the choice of $m$ that $\frac{m}{10^n}\lt y$. We show that $x\lt \frac{m}{10^n}$.

Suppose to the contrary that $\frac{m}{10^n}\le x$. Then
$$\frac{m+1}{10^n}\le  x+\frac{1}{10^n} \lt x+d\lt y,$$
contradicting the choice of $m$ as the largest integer such that $\frac{m}{10^n}\lt y$.


问:
How to show that a bounded subset of a metric space is contained within a closed sphere using the definition of a bounded set?
Source: Introduction to Topology and Modern Analysis by GF Simmons(Pg-69, Q.No.3)

How to show that a bounded non-empty subset of a  metric space is contained within a closed sphere using the definition of a bounded set as given below?

A set is bounded if its diameter is finite (a real number less than infinity).

Also see that,
given a metric space $(X,d)$, the diameter of nonempty subset $ M \subseteq X$ is defined as follows:
$$\mathrm{diam}(M)=\displaystyle\sup_{x, y \in M} d(x, y)$$

答: Just pick a point in the given subset, $P\in A$ and consider the ball centered at $P$ with radius $R$ equal to the diameter of $A$. Then clearly for any $Q\in A$, $dist (Q,P)\le R$ by definition of diameter.


问:
How to find $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$

Let $c \neq 1$ be a postive real number. Find the following limit

$$\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n.$$

I know that $\lim_{n \to \infty} \Big( 1 + \frac{c}{n} \Big)^{bn} = e^{bc}$. But it cannot be used here, since $\ln(n)$ is not a constant.  I also know how to find the $\lim_{n \to \infty} \Big( 1 + \frac{1}{n^2} \Big)^{n}$, which is maybe closer to our limit.

I tried to use some methods (e.g., L'Hopital's rule) that appear in the proofs of the above limits to solve also $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$ but without success. The graphs of $\Big (1 - \frac{c \ln(n)}{n} \Big)^n$ indicate that the limit is $\infty$ if $c < 1$ and the limit is 0 otherwise.

I appreciate any suggestions on how to solve this limit. Also, any readings on this topic (I did not find any such example anywhere) are appreciated.

答:
$\frac {\ln (1-x)} x \to -1$ as $x \to 0$. To find the limit of $n \ln (1-\frac {c \ln n} n)$ it is therefore enough to find the limit of $-n \frac {c \ln n} n$ which is clearly $-\infty$ for any $c>0$ and $\infty$ for $c <0$. Now take exponential.

Answer: The limit is $0$ if $c>0$, $\infty$ if $c <0$ and $1$ if $c=0$.


问:
When is this norm defined?
This question was part of my linear algebra assignment and I was unable to solv eit. SO, I am asking you for help.

Consider the vector space V of real polynomials of degree less than or equal to n. Fix distinct real numbers $a_0 , …, a_k$ . For $p \in V$ define $max${${|p(a_j)| : 0\leq j \leq k}$} . It defines a  norm on V if:

A only if k<n

B only if $k\geq n$

C if $k+1 \leq n $

D if $k\geq n+1$

I tried thinking on which result / reasoning I should use but I am clueless on when the norm will not be defined .

Can you please help me through it?

答:
Let $n=2$. If $k=1$, let $\|.\|$ be the corresponding defined norm.

If $a_0=1, a_1=-1$, then we have $\|x^2-1\|=0$, even thought $x^2-1 \ne 0$.

In fact, if $n=2$ and $k=1$, then we have $\|(x-a_0)(x-a_1)\|=0$.

In general, if $k+1 \le n$, then we have $\|\prod_{i=0}^k (x-a_i) \|=0$ and $\prod_{i=0}^k (x-a_i)$ is of degree $k+1$ and non-zero.

Hence we need $k \ge n$ due to the positive definite requirement.

Now to verify sub-additivity,

\begin{align}
\|f + g\| &= \max_{0 \le i \le k} \{|(f+g)(a_i)| \} \le \max_{0 \le i \le k} \{|f(a_i)| + |g(a_i)| \}  \le \|f\| +  \|g\|
\end{align}

Verification of absolutely homogeneous is direct.

As an exercise, you might like to answer the question when is it a seminorm.

Title: When is this norm defined?


问:
Does this proof of the Schröder–Bernstein theorem work?
$ f : A \mapsto B $ and $ g : B \mapsto A $ with both functions being injective. Prove that there exists a bijective function $ h $ on $A$ onto $B$.

Suppose $ \exists A_1 $ such that  $ A_1  \subseteq A $ and $ g[ B - f [ A_1 ] ] = A - A_1 $ , in which case we can define a $ h $ such that $ h : A \mapsto B $ where $ x \in A_1 \implies h(x) = f(x) $ and $ x \in A - A_1 \implies h(x) = g^{-1} (x) $ with $h$ being bijective.

Constructing the h is dependent an making an  $ A_1 $ which satisfies $ g[ B - f [ A_1 ] ] = A - A_1 $ . Let $ A_1 = \bigcap \{ A_0 | A - g [B] \subseteq A_0 $ and $ g [ f[ A_0 ] ] \subseteq A_0 \} $.

By defining $ A_1 $ in this way, is $h$ a bijective function? Does this $A_1$ satisfy $ g[ B - f [ A_1 ] ] = A - A_1 $ ?

Here are some thoughts I have on the definition of $ A_1 $ proposed.

Wanting the $ A_1 $ to satisfy the property $ g [ f[ A_1 ] ] \subseteq A_1 $ seems obvious as a direct consequence of injectivity from the statement $ g[ B - f [ A_1 ] ] = A - A_1 $.

The property $ A - g [B] \subseteq A_1 $ probably stems from the fact that if every  element in $A$ that is not in the range of  $g$ is not inside of $ A_1 $ then we could end up having a blindspot in constructing $ h $ to be bijective since we're using $h(x)= f(x)$ for $ x\in A_1 $

I'm slightly confused about how to prove things when the big intersection $ \bigcap $ sign involved.

答:
To find such a set $A_1$, we do not need to find the intersection of all sets $\{ A_0 \mid A-g[B] \subset A_0 \, \, \& \, \, g[f[A_0]] \subset A_0 \}$; in fact, each of such sets can be considered as a good candidate for the set $A_1$. In the discussion below, we construct one of such sets and see that it satisfies the requirement of the proof.

Let us review the idea of the proof of the theorem. Let $A_1$ be a subset of $A$, $f[A_1]=B_1 \subset B$, $B_2=B-B_1$, and $g[B_2]=A_2 \subset A$. If we have some $A_1$ such that $A_1 \cup A_2 =A$, then we have the following bijective mapping between the sets $A$ and $B$:$$h(x)=\begin{cases}f(x) & \text{if }x \in A_1; \\ g^{-1}(x) & \text{if } x \in A_2 \end{cases}.$$Thus, our original problem is reduced to finding the set $A_1$. So, let us find it.

According to the description of the set $A_1$, it must contain the elements of the set $A$ which are not in the range of the function $g$, that is, $A-g[B] \subset A_1$. So one may conclude that the set $C_0=A-g[B]$ is a good candidate for the set $A_1$. But, the set $C_0$ does not satisfy our required condition because$$B_1=f[C_0], \quad B_1 \cap B_2=\varnothing \\ \Rightarrow \quad g[f[C_0]] \cap g[B_2]= \varnothing \quad \Rightarrow \quad C_0 \cup A_2 = A - g[f[C_0]]$$(Please note that in the above argument we have used the fact that $g$ is one-to-one and $g[B_2]=A_2$).

So, we need to add the elements of the set $C_1=g[f[C_0]]$ to the set $C_0$. So one may conclude that the set $C_0 \cup C_1$ is a good candidate for the set $A_1$. But, the set $C_0 \cup C_1$ does not satisfy our required condition because$$B_1=f[C_0 \cup C_1], \quad B_1 \cap B_2=\varnothing , \quad C_1 \cap g[f[C_1]]= \varnothing \\ \Rightarrow \quad g[f[C_0 \cup C_1]] \cap g[B_2]= \varnothing \\ \Rightarrow \quad (C_0 \cup C_1) \cup A_2 = A - g[f[C_1]]$$(Please note that in the above argument we have used the fact that $g$ and $f$ are one-to-one and $g[B_2]=A_2$).

$$\vdots \qquad \vdots \qquad \vdots$$

So, we need to add the elements of the set $C_n=g[f[C_{n-1}]]$ to the set $\bigcup_{i=0}^{n-1}C_i$. So one may conclude that the set $\bigcup_{i=0}^n C_i$ is a good candidate for the set $A_1$. But, the set $\bigcup_{i=0}^n C_i$ does not satisfy our required condition because$$B_1=f \left [\bigcup_{i=0}^n C_i \right ], \quad B_1 \cap B_2=\varnothing , \quad C_n \cap g[f[C_n]]= \varnothing \\ \Rightarrow \quad g \left [f \left [\bigcup_{i=0}^n C_i \right ] \right ] \cap g[B_2]= \varnothing \\ \Rightarrow \quad \left (\bigcup_{i=0}^n C_i \right ) \cup A_2 = A - g[f[C_n]]$$(Please note that in the above argument we have used the fact that $g$ and $f$ are one-to-one and $g[B_2]=A_2$).

$$\vdots \qquad \vdots \qquad \vdots$$

Hence, we can conclude that the set $\bigcup_{n=0}^{\infty } C_n$ satisfies our required condition because$$g \left [f \left [ \bigcup_{n=0}^{\infty } C_n \right ] \right ] \subset \bigcup_{n=0}^{\infty } C_n \\ \Rightarrow \quad \left ( \bigcup_{n=0}^{\infty }C_n \right ) \cup g \left [B- f \left [ \bigcup_{n=0}^{\infty } C_n \right ] \right ]=A.$$Thus,$$A_1=\bigcup_{n=0}^{\infty }C_n, \quad \text{ where } \, C_0=A-g[B], \quad C_n=g[f[C_{n-1}]].$$

Addendum

Please note that the idea of the foregoing proof is nothing but the one we see in the König's proof of the Cantor-Schroeder-Bernstein Theorem. To see that fact, let us first review the König's proof.

König's Proof

Let $a \in A$, then $a \in A-g[B]$ or $a \in g[B]$. In the first case $a$ has no inverse image under $g$ while in the second case it has an inverse image $g^{-1}(x)$. In the second case, either $g^{-1}(a) \in B- f[A]$, in which case $g^{-1}(a)$ has no inverse image under $f^{-1}$, or $g^{-1}(a) \in f[A]$. Again, in the latter cases there exists a unique inverse image $f^{-1}(g^{-1}(a))$, ... . We call the elements $g^{-1}(a)$, $f^{-1}(g^{-1}(a))$, ... the ancestors of $a$. By similar analogy, such a description also holds for an arbitrary element $b \in B$.

Now, we can partition $A$ into three subsets:

$A_e$: the set of all elements of $A$ having an even number of ancestors,

$A_o$: the set of all elements of $A$ having an odd number of ancestors,

$A_i$: the set of all elements of $A$ having an infinite number of ancestors.

Similarly, we can partition $B$ into such subsets. It is clear that if $x\in A_e$ then $f(x) \in B_o$, if $x\in A_o$ then $g^{-1}(x) \in B_e$, and if $x\in A_i$ then $f(x) \in B_i$. So, we can easily conclude that the restricted functions $f:A_e \to B_o$, $g^{-1}: A_o \to B_e$, and $f:A_i \to B_i$ are bijective.

Here is the idea of the proof. Since $f$ and $g$ are injective functions into $B$ and $A$, respectively, one may want to map any element of $A$ into $B$ by $f$ and cover the remaining elements of $B$, i.e., $B-f[A]$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)=\begin{cases}f(x) & \text{if } x \in A-g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[B-f[A]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).

But, the obtained mapping is not onto $B$ because its range lacks the subset $f[g[B-f[A]]]$ (that is, the values of $f$ of the $1$-ancestor elements of $A$). So, one may want to cover such a subset of $B$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)= \begin{cases}f(x) & \text{if }x \in A-(g[B-f[A]] \cup g[f[g[B-f[A]]]]); \\ g^{-1}(x) & \text{if }x \in g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[f[g[B-f[A]]]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]] \cup g[f[g[B-f[A]]]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).

But, the obtained mapping is not onto $B$ because its range lacks the subset $f[g[f[g[B-f[A]]]]]$ (that is, the values of $f$ of the $3$-ancestor elements of $A$).

$$\vdots \qquad \vdots \qquad \vdots$$

But, the obtained mapping is not onto $B$ because its range lacks the subset $(f \circ g)^n[B-f[A]]$ (that is, the values of $f$ of the $2n-1$-ancestor elements of $A$). So, one may want to cover such a subset of $B$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)= \begin{cases}f(x) & \text{if }x \in A-(g[B-f[A]] \cup g[f[g[B-f[A]]]] \cup \cdots \cup (g \circ f)^n[g[B-f[A]] ); \\ g^{-1}(x) & \text{if }x \in g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[f[g[B-f[A]]]]; \\ \vdots & \vdots \\ g^{-1}(x) & \text{if } x \in (g \circ f)^n[g[B-f[A]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]] \cup g[f[g[B-f[A]]]] \cup \cdots \cup (g \circ f)^n[g[B-f[A]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).

But, the obtained mapping is not onto $B$ because its range lacks the subset $(f \circ g)^{n+1}[B-f[A]]$ (that is, the values of $f$ of the $2n+1$-ancestor elements of $A$).

$$\vdots \qquad \vdots \qquad \vdots$$

This pattern motivates us to define mapping $h$ as follows.$$h(x)=\begin{cases}f(x) & \text{if } x \in A_e \cup A_i; \\ g^{-1}(x) & \text{if } x \in A_o \end{cases}$$(Please note that $\bigcup_{n=0}^{\infty }(g \circ f)^n[g[B-f[A]]=A_o$ and $A-\bigcup_{n=0}^{\infty }(g \circ f)^n[g[B-f[A]]=A_e \cup A_i$).

Since $h$ maps the set $A=A_e \cup A_i \cup A_o$ onto the set $B=B_o \cup B_i \cup B_e$ (as already explained) and the sets $A_e \cup A_i$ and $A_o$ are disjoint, we conclude that the mapping $h$ is bijective. $\square$

Now, comparing the mentioned proofs with each other, we can easily see that the $C_n$'s constructed in our proof are nothing but $A_{2n}$'s in the König's proof, that is, the sets of elements of $A$ having $2n$ ancestors. So, it is clear that$$\bigcup_{n=0}^{\infty }C_n=A_e \cup A_i.$$

You can also find other approaches to prove the Cantor-Schroeder-Bernstein theorem in this post.


答:
Let $a_n = (1+\sqrt{3})^n + (1 - \sqrt{3})^n$ for $n \ge 0$.

Since $1 \pm \sqrt{3}$ are roots of $(\lambda - 1)^2 - 3 = \lambda^2 - 2\lambda - 2$, $a_n$ satisfies a recurrence relation:

$$a_{n+2} = 2(a_{n+1} + a_n)$$

Define two auxiliary sequences $b_n, c_n$ by
$\begin{cases}
b_n &= 2^{-(n+1)} a_{2n}\\
c_n &= 2^{-(n+1)} a_{2n+1}
\end{cases}$, we have

$$\begin{align}
b_{n} &= 2^{-(n+1)}a_{2n} = 2^{-n}(a_{2n-1} + a_{2n-2}) = c_{n-1} + b_{n-1}\\
c_{n} &= 2^{-(n+1)}a_{2n+1} = 2^{-n}(a_{2n} + a_{2n-1}) = 2b_n + c_{n-1} = 2b_{n-1} + 3c_{n-1}
\end{align}$$
or in matrix form:

$$\begin{bmatrix}b_n \\ c_n\end{bmatrix} =
\begin{bmatrix} 1 & 1\\ 2 & 3\end{bmatrix} \begin{bmatrix}b_{n-1} \\ c_{n-1}\end{bmatrix}$$
Since $a_0 = a_1 = 2 \implies b_0 = c_0 = 1$, this leads to following representation of $(a_{2n}, a_{2n+1})$:

$$\begin{bmatrix}a_{2n} \\ a_{2n+1}\end{bmatrix}
= 2^{n+1}\begin{bmatrix}b_n \\ c_n\end{bmatrix}
= 2^{n+1}\begin{bmatrix} 1 & 1\\ 2 & 3\end{bmatrix}^n \begin{bmatrix}1 \\ 1\end{bmatrix}$$

From above expression, it is easy to see $(b_n, c_n)$ are integers. This implies $2^{n+1}$ divides both $a_{2n}$ and $a_{2n+1}$.
In fact, since $c_0 = 1$ is odd and
$$c_{n} = 2b_{n-1} + 3c_{n-1} \equiv c_{n-1} \pmod 2$$
all $c_n$ are odd. From this, we can deduce $2^{n+1}$ is the highest factor of $2$ that divides both $a_{2n}$ and $a_{2n+1}$.

Title: Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........?

Body:
Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?

I am not well versed with binomial theorem hence I am not able to proceed. I am also interested in the different ways this problem can be solved.

Any help would be appreciated.


答:
If $x_n\to \infty$, and $c>0$, then
$$
\left(1-\frac{c}{x_n}\right)^{x_n}\to \mathrm{e}^{-c}
$$
In the OP case
$$
\left(1-\frac{c\ln n}{n}\right)^{n}=
\left(1-\frac{c}{\frac{n}{\ln n}}\right)^{\frac{n}{\ln n}\cdot \ln n}=
\left(\left(1-\frac{c}{x_n}\right)^{x_n}\right)^{\ln n}
$$
where $x_n=\frac{n}{\ln n}$. Hence
$$
\lim_{n\to\infty}\left(1-\frac{c\ln n}{n}\right)^{n}=\lim_{n\to\infty}\left(\mathrm{e}^{-c}\right)^{\ln n}=\lim_{n\to\infty}n^{-c}=0.
$$

Title: How to find $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$

Body:
Let $c \neq 1$ be a postive real number. Find the following limit

$$\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n.$$

I know that $\lim_{n \to \infty} \Big( 1 + \frac{c}{n} \Big)^{bn} = e^{bc}$. But it cannot be used here, since $\ln(n)$ is not a constant.  I also know how to find the $\lim_{n \to \infty} \Big( 1 + \frac{1}{n^2} \Big)^{n}$, which is maybe closer to our limit.

I tried to use some methods (e.g., L'Hopital's rule) that appear in the proofs of the above limits to solve also $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$ but without success. The graphs of $\Big (1 - \frac{c \ln(n)}{n} \Big)^n$ indicate that the limit is $\infty$ if $c < 1$ and the limit is 0 otherwise.

I appreciate any suggestions on how to solve this limit. Also, any readings on this topic (I did not find any such example anywhere) are appreciated.


问:
Number of non-singular $3\times3$ matrices with elements $1$ or $-1$
I am searching for number of non-singular $3×3$ matrices with elements $1$ or $-1$. I tried using basic combinatorics but I couldn't count in a systematic manner and therefore got confused. Would someone please help?

Thanks in advance!

答:
A matrix is non-singular when the row (column) vectors are linearly independent.

Total number of 3-vectors is only $2^3=8$, namely,

$$(1,1,1), (-1,-1,-1) $$
$$(1,1,-1), (-1,-1,1) $$
$$(1,-1,1), (-1,1,-1) $$
$$(-1,1,1), (1,-1,-1) $$

Notice I have arranged these in pairs, so one in each pair is negative of another.

To make a non-singular matrix, we can choose one from each pair and arrange them in any order. Their number is

$$ {4\choose 3} \cdot 2^3 \cdot 3! = 192$$

as confirmed by OEIS link mentioned in comment.

Edit :

Here one should show that no three of ${(1,1,1), (1,1,-1),(1,-1,1),(-1,1,1)}$ are linearly dependent i.e., none of these is linear combination of any other two. This can be argued geometrically - no plane through any of these three points passes through origin.


问:
Functions of a Markov Chains
can does anybody know if the following expectations are available in closed for...

Let $\{ X_t : t = 1, 2, 3 \dots \}$ be a random variable defined on a Markov chain with m -step transition matrix $P_m^{i,j}$. I'm trying hunting for a closed form expression
for the following expectations which are telescoping:

$m = 1 : E_t \Big [ \frac{1}{X_{t+1}} \Big ]$

$m = 2 : E_t \Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}}  \Big ]$

$m = 3 : E_t \Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}} \frac{1}{X_{t+3}}\Big ]$

A clue would be lovely - thanks.

Title: Functions of a Markov Chains

Body:
By definition,
$$
E\Big [ \frac{1}{X_{t+1}} \frac{1}{X_{t+2}}  \Big ]=\sum_{x,y,z}P[X_t=x]\frac{P_1^{x,y}P_1^{y,z}}{yz},
$$
where
$$
P[X_t=x]=\sum_{u}P[X_0=u]P_t^{u,x}.
$$


答:
This is really more of a comment than an answer, but I can't fit it in a comment box.  There are $2^3=8$ choices for the first row,  The second row can't be the same as the first row, nor the negative of the first row, leaving $6$ choices for the second row.  The third row can't be a scalar multiple of either of the first two, leaving $4$ choices, so we have at most $8\cdot6\cdot4=192$ matrices.  Of course, it's possible that the third row is a linear combination of the first two, even though it's not a scalar multiple of either of them.

I wrote a little python script to count the number of non-singular $3\times3$ matrices with $\pm1$ entries by brute force, and surprisingly, it cam up with $192$.  So, if we can somehow show that in this case, if no two rows are linearly dependent, then all three rows are linearly independent, we are done.  I've been thinking about this for a bit, but I don't see how to do it, and I'm going to bed now.

Here's the script:

```<code>import numpy as np
from itertools import product

count =0
for p in product((-1,1), repeat=9):
    A = np.array(p).reshape((3,3))
    if np.linalg.det(A):
        count += 1
print(count)

</code>```

Title: Number of non-singular $3\times3$ matrices with elements $1$ or $-1$

Body:
I am searching for number of non-singular $3×3$ matrices with elements $1$ or $-1$. I tried using basic combinatorics but I couldn't count in a systematic manner and therefore got confused. Would someone please help?

Thanks in advance!

Body:
This is really more of a comment than an answer, but I can't fit it in a comment box.  There are $2^3=8$ choices for the first row,  The second row can't be the same as the first row, nor the negative of the first row, leaving $6$ choices for the second row.  The third row can't be a scalar multiple of either of the first two, leaving $4$ choices, so we have at most $8\cdot6\cdot4=192$ matrices.  Of course, it's possible that the third row is a linear combination of the first two, even though it's not a scalar multiple of either of them.

I wrote a little python script to count the number of non-singular $3\times3$ matrices with $\pm1$ entries by brute force, and surprisingly, it cam up with $192$.  So, if we can somehow show that in this case, if no two rows are linearly dependent, then all three rows are linearly independent, we are done.  I've been thinking about this for a bit, but I don't see how to do it, and I'm going to bed now.

Here's the script:

import numpy as np
from itertools import product

count =0
for p in product((-1,1), repeat=9):
    A = np.array(p).reshape((3,3))
    if np.linalg.det(A):
        count += 1
print(count)


问:
Prove the inverse map of an injective linear function with closed graph from $U$ to $V$ (Banach spaces) is continuous iff the image is closed in $V$.
Let $U$ and $V$ be Banach spaces, let $A⊆U$ be a subspace, and let $f:A⊆U→V$ be an injective linear map with closed graph. Show that the inverse map $f^{-1} : B \to U$ is continuous if and only if $B = f(A)$ is closed in $V$.

For the forward implication, I basically did what was outlined here Closed Graph of a function and the continuous inverse implies the image is closed.

For the other direction, I need to show that if $B$ is closed in $V$, then whenever $b_i \to b \in B$, we have that $f^{-1}(b_i) \to f^{-1}(b)$. I don't really know how to begin.. as continuity requires $b_i \to b \in B$ to begin with, I don't see how to use the fact that $B$ is closed.

答:
Let me denote the linear map by $T$ :)

Define the following norm on $A$ (graph norm of $T$):
$$
\|x\|_A:= \|x\|_X + \|Tx\|_Y.
$$
Then $(A,\|\cdot\|_A)$ is a Banach space. Moreover, $T: A\to T(A)$ is bijective,
$(A,\|\cdot\|_A)$ and $(T(A),\|\cdot \|_Y)$ are Banach spaces, so $T^{-1}$ is continuous and bounded. Then there is a constant $c>0$ such that
$$
\|y\|_Y \ge \|T^{-1}y\|_A = \|T^{-1}y\|_X + \|y\|_Y
$$
for all $y\in T(A)$. This implies that $T^{-1}$ continuous from $T(A)$ to $(A,\|\cdot\|_X)$.


问:
How to perform this Gaussian integral / sum of series?
I am trying to calculate the indefinite integral:

$\displaystyle\int dx \exp \left(-\dfrac{(a x + b)^2}{2}\right) {\rm erf} \left(\dfrac{x}{\sqrt{2}} \right), \quad x \in \mathbb{R}$

Context. This integral arises in the integration of a multivariate Normal distribution with finite integration extremes. Let be $I(x)$ the integral, so far I have been able to show that, if $b=0$:

$I(x) = \dfrac{1}{a}\sqrt{\dfrac{\pi}{2}} {\rm erf} \left( \dfrac{x}{\sqrt{2}} \right) {\rm erf}\left( \dfrac{a x}{\sqrt{2}} \right) + \\ \quad \dfrac{2}{a \sqrt{\pi}} \exp\left(-\dfrac{x^2}{2} \right) \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(2k+1)} \left(\dfrac{a}{\sqrt{2}}\right)^{2k+1} \displaystyle\sum_{j=0}^{k}\dfrac{2k!!}{2j!!}x^{2j}$

I have done this by expanding the exponential in the integrand in series, then calculating the first four of the resulting integrals, then figuring out the recurrence formulas in k and j. Part of the series can be summed to ${\rm erf}$, but the second term has proven hard to simplify. Even though the formula above looks like a step in the right direction (to me), I have no idea how to even reach this point if $b\neq0$.

答:
For $b=0$, the integral can be evaluated exactly in terms of Owen's T function (https://en.wikipedia.org/wiki/Owen%27s_T_function),
\begin{equation}
I = -\frac{\sqrt{\pi}}{\sqrt{2}a} \left[1+4 T\left(ax,\frac{1}{a}\right) \right].
\end{equation}
For $b\ne0$, no idea...


答: Since $$\lim_{n \to \infty}(1 - \frac{c \ln(n)}{n}) = 1 $$We can write the limit as $$L = \lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n = \lim_{n \to \infty} \exp(-\frac{c \ln(n)}{n})n = \lim_{n\to \infty} \exp(-c \ln(n)) = \lim_{n \to \infty} n^{-c}$$If $c \gt 0$ then $L = 0$. In the case $c\lt 0$ we have $L = \infty$ and $c = 0$ implies $L = 1$.


问:
Prove that Fibonacci sequence contains An infinite Sub-sequence That all its element are co-prime.
Prove that Fibonacci sequence contains An infinite Sub-sequence That all its element are co-prime.

Prove Two Sub-sequence different

Instruction: Fermat numbers.

$----------------------------$

attempt 1:
$F_n$ is strong divisibility sequence , $\gcd(F_m,F_n)$=$F_{\gcd(m,n)}$
we need to prove there an infinite sequence of coprime integers.

答: Hint: by the simple proof in this answer, the Fibonacci sequence $\,F_n\,$ is strong divisibility sequence, i.e. $\,\gcd(F_m,F_n) = F_{\gcd(m,n)},\,$ so the problem reduces to proving that there is an infinite sequence of coprime integers (indices), for which there are many obvious methods (some that don't require primes, e.g. Euclid's famous constructive algorithm that recursively generates an infinite sequence of (co)prime integers).


问:
Proving limit exist except for specific points
I stumbled upon this question that my prof asked during class

Let $f: \mathbb R \rightarrow \mathbb R$ with

$$f(x) = \begin{cases} x - 2 \cdot n, & 2 \cdot n \le x < 2 \cdot n + 1 \\ 2 \cdot n - x, & 2 \cdot n - 1 \le x < 2 \cdot n,\end{cases}$$

for every n $\in \mathbb Z$. Prove that $\lim_{x \to c} f(x)$ exist for all $c$ $\in \mathbb R$, except for even integer $c$

Now, I know that I need to prove that right-side limit and left-side limit are equals except for the even integer $c$. But I'm not sure how should I divide the cases I got here, for instance I've been thinking to divide by $c = 2a+1$ for odd and $c = 2a$ with $a \in \mathbb N$, but then I realized from the definition $c$ $\in \mathbb R$, so some cases might not included

any insight would really help, thanks beforehand.

答:
Your function is continuous everywhere.

First observe that if $x\in(\text{odd, even}) \text{or}x \in(\text{even, odd})$ then by definition graph of the function is a straight line. so it is continuous and limit exists.

Now suppose $c=\text{even}=2k$ for some k.

$$f(x)=\begin{cases}x-2k,x\in[2k,2k+1)\\2k-x,x\in[2k-1,2k)\end{cases}\implies f(2k)=0$$
The RHL is $f(2k^+)=\lim_{x\to 2k^+}x-2k=0$,similarly LHL is also zero. So the limit exists and the function is continuous at $x=\text{even numbers}$.

Similarly you can check for odd case.

Now suppose $c=\text{odd}=2k+1$ for some k.

$$f(x)=\begin{cases}2k+2-x,x\in[2k+1,2k+2)\\x-2k,x\in[2k,2k+1)\end{cases}\implies f(2k+1)=1$$
The RHL is $f(2k+1^+)=\lim_{x\to 2k+1^+}2k+2-x=1$,similarly LHL is also one. So the limit exists and the function is continuous at $x=\text{odd numbers}$.


问:
Find the equation of the plane, in $\bf{r}\cdot n = d$ form
I'll mark the vectors in bold.

$p_1 = \bf{i} - 2 \bf{j} + \bf{k}$

$p_2 = 2 \bf{i} + \bf{j} - \bf{k}$

$p_3 = \bf{i} + \bf{j} + \bf{k}$

Could someone please explain to me the way of finding the equation of a plane in $\bf{r}\cdot n = \bf{d}$ form for a plane containing $p_1, p_2, p_3$?

Thanks

答:
Presumably, what you are looking for is an equation in the form
$$ax + by + cz = d$$
with $a,b,c,d\in\mathbb{R}$, which can be expressed in the form
$$(x,y,z)\cdot\mathbf{n} = d$$
with $\mathbf{n}=(a,b,c)$ the normal vector of the plane.

To find the normal vector of the plane, construct two vectors that are in the plane, are not collinear, and take their cross product.

Since you have three points that are not collinear that are in the plane, note that $\mathbf{p1}-\mathbf{p2}$, $\mathbf{p1}-\mathbf{p3}$, and $\mathbf{p2}-\mathbf{p3}$ are all vectors that (after suitable translation) lie in the plane. Once you have the normal vector, finding $d$ is a matter of figuring out what value is required by plugging in at least one point in the plane.

Example. Suppose you want to find the plane that contains $\mathbf{p}_1=(1,1,1) = \mathbf{i}+\mathbf{j}+\mathbf{k}$, $\mathbf{p}_2=(-3,-2,0) = -3\mathbf{i}-2\mathbf{j}$, and $\mathbf{p}_3 = (0,3,1) = 3\mathbf{j}+\mathbf{k}$.

The plane contains vectors parallel to
$$\mathbf{p}_1-\mathbf{p}_2 = (1,1,1) - (-3,-2,0) = (4,3,1)$$
and to
$$\mathbf{p}_1 - \mathbf{p}_3= (1,1,1)-(0,3,1) = (1,-2,0).$$
So then the vector $(4,3,1)\times(1,-2,0)$ is perpendicular to plane:
$$(4,3,1)\times(1,-2,0) = \left|\begin{array}{rrr}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
4 & 3 & 1\\
1 & -2 & 0
\end{array}\right| = \Bigl( 0+2, -(0-1), -8-3\Bigr) = (2,1,-11).$$
Therefore, the plane that contains the three points will have equation of the form
$$2x + y - 11z = d$$
for some $d$. Plugging in $(1,1,1)$, which we know is in the plane, we get
$$d = 2(1) + 1 - 11(1) = -8$$
so the plane in question is the plane with equation
$$2x + y - 11z = -8.$$
Indeed, you can verify that $(1,1,1)$, $(-3,-2,0)$, and $(0,3,1)$ all satisfy the equation. To write in "$\mathbf{r}\cdot \mathbf{n}=\mathbf{d}$ form", we simply write
$$(x,y,z)\cdot(2,1,-11) = -8.$$


问:
Proving an equality
Let $f(n) = n^ {\log n}$. Let $p(n)$ and $q(n) \geq n$ be polynomials. I want to show that for
$n$ sufficiently large $f (n)$ satisfies
$$p(n) < f (n) < 2^{q(n)}$$

starting from the above inequality doesn't yield any satisfying result.

答:
Since $p(n)$ is a polynomial,
$p(n) < c n^d$ where $d$ is the degree of $p$
and $c$ is the sum of the absolute value of its coefficients.
So $p(n) < n^{d+1}$ for $n > c$.
Therefore,
if $\log n > d+1$,
$p(n) < f(n)$.

For the upper bound,
$f(n) = n^{\log n}
= e^{\log^2 n}
= 2^{(\log^2 n)/\log 2}
$,
so any polynomial $q$ such that
$q(n) > (\log^2 n)/\log 2$
will work.
Since $(\log^2 n)/n \to 0$ as $n \to \infty$,
any polynomial will work.

Title: Proving an equality


答:
Using the Maclaurin series of the logarithm and the exponential function, we have
\begin{align*}
\left( {1 - \frac{{c\log n}}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{{c\log n}}{n}} \right)} \right) = \exp \left( {n\left( { - \frac{{c\log n}}{n} + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{{n^2 }}} \right)} \right)} \right) \\ &= \exp \left( { - c\log n + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right)} \right) = n^{ - c} \left( {1 + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right)} \right).
\end{align*}
It is now easy to conclude.


问:
Proving the integral of the Dirac delta function is 1
Was wondering if my solution is mathematically accurate enough:

The question in the book yields:

Derive
  $$
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$
  From
  $$
f(x)=\int_{-\infty}^{\infty} f(x_i)\delta(x-x_i)\ dx_i
$$
  [Hint: let $f(x)=1$]

My method is:

$$
f(x)=\int_{-\infty}^\infty f(x_i)\delta(x-x_i)\ dx_i
$$

$$
f(x)=1
$$

so
  $$
1=\int_{-\infty}^{\infty} \delta(x-x_i)\ dx_i
$$

I think this is not correct. Does someone know is this is correct, or how to do it better?

Cheers

答:
I think you can look at it as a definition.

You can also look at it as the limit of any function that is symmetric around the y axis, which integral is 1, and you make it narrower and narrower, forcing its peak to go to infinity. This function can be a gaussian, like shown in the wikipedia article gif image, or a uniform function, as shown in this Khan Academy video.


问:
Why can I write two unknowns $a,b$ as $K\cos(p)$ and $K\sin(p)$?
If I have two real numbers $a$ and $b$, why is it possible to express those numbers as $a=K\cos(\phi)$ and $b=K\sin(\phi)$? I came across this problem in Riley, Hobson and Bence (2006:14-15). Here is the question and answer in the book.

Question:

Solve for $\theta$ the equation:
$$a\sin\theta + b\cos\theta = k, $$
where $a$, $b$ and $k$ are real quantities.

Answer:

To solve this equation we make use of result (1.18) [the compound angle formulae] by setting $a=K\cos\phi$ and $b=K\sin\phi$ for suitable values of $K$ and $\phi$.

My question is: how do you know that, for arbitrary real numbers $a$ and $b$, there exists $K$ and $\phi$ such that $a=K\cos(\phi)$ and $b=K\sin(\phi)$? Additionally: does this mean that I can express two unknowns in terms of other trig functions, e.g. $a=R\sec(\alpha)$ and $b=R\cot(\alpha)$? Why or why not?

Reference:

Riley, K. F., Hobson, M. P., and Bence, S. J. (2006), Mathematical Methods for Physics and Engineering (3e). Cambridge: CUP.

答:
Alternative approach.

$\tan(\phi)$ is a strictly increasing bijection from $(-\pi/2,\pi/2)$ to $\mathbb{R}.$

WLOG, $a \neq 0.$

Choose $\phi$ such that $\tan(\phi) = \frac{b}{a}.$

Then set $k = \frac{a}{\cos ~\phi}.$

Therefore,

$a = k \cos ~\phi~$ and

$b = a \tan \phi = k \sin \phi.$