问:
Deducing a Function from a Series of Points
This is for a computer science thing, but I'd like to understand it from the raw mathematics.

Let's say we have a randomly selected function. It could be linear, logarithmic, exponential, or a sine wave, possibly with random efficient (and yes, we know it will be one of these 4 ahead of time). So you might have $y=x^2$, or $y=3x^2$. (These are not in stone, just examples; if possible I'd like a general solution for any collection of known functions.)

Now, given a series of points along that function, I'd like to guess which of the functions, in our collection, it is.

With 1 point, nothing can be predicted.
With 2 points, nothing can, also, be predicted. (I think)
But with 3+ points, we'll start to get a clearer and clearer picture.

What type of theorem, process, or even just Google search terms, should I look into find methods to solve this problem? (I'm not even sure how to tag this question-- I'm just starting out).

答: The space of all functions is much too large.  Most functions are wildly chaotic and you can't learn anything from even a large number of points.  The space of functions of interest to people depends sensitively on the population of people you are talking about.  You can always fit an $n^{th}$ degree polynomial through $n+1$ points.  Lagrange interpolation will tell you how.  If you know the form of the function but it has a number of parameters, function minimization will find the parameters.  Numerical analysis books will discuss these questions.


问:
Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$
I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.

ContentLicense: CC BY-SA 4.0

Body: by Huygen's inequality $$\frac{(1+x^2)(1+y^2)(1+z^2)}{xyz}\ge \frac{({1+\sqrt[3]{x^2y^2z^2})}^3}{xyz}={\left(\frac{1}{\sqrt[3]{xyz}}+\sqrt[3]{xyz} \right)}^3\ge 2^3$$


问:
What is $\int_0^1 f'(x)e^{f(x)}dx$?
Don't be fooled by this innocuous question and answer $e^{f(1)}-e^{f(0)}$! Allow me to trouble you with details.

Suppose $f$ is as smooth as you want, on $\mathbb (0,1)$, so $f'(x)$ exists and is smooth on $\mathbb (0,1)$. But define $f$ to be $-\infty$ outside $[0,1]$ so that $e^{f(x)}=0$ outside $[0,1]$. $f(0),f(1)$ are real numbers though.

I want to know, what is the value of $\int_\mathbb R f'(x)e^{f(x)}dx$? Should it still be $e^{f(1)}-e^{f(0)}$? Or should it be $0$? Think of $e^f$ as basically the truncation of a perfectly nice function to $[0,1]$.

答: If $f$ is only defined at $(0,1)$, then you have an improper integral. So $$\begin{align}\int_0^1 f'(x){\rm e}^{f(x)}\,{\rm d}x &= \int_0^c f'(x){\rm e}^{f(x)}\,{\rm d}x + \int_c^1 f'(x){\rm e}^{f(x)}\,{\rm d}x \\ &= \lim_{a \to 0^{+}} \int_a^c f'(x){\rm e}^{f(x)}\,{\rm d}x + \lim_{b \to 1^-} \int_c^b f'(x){\rm e}^{f(x)}\,{\rm d}x \\ &= \lim_{a \to 0^+} ({\rm e}^{f(c)} - {\rm e}^{f(a)}) + \lim_{b \to 1^-} ({\rm e}^{f(b)} - {\rm e}^{f(c)}) \\ &= {\rm e}^{f(c)} - \left(\lim_{a \to 0^+} {\rm e}^{f(a)}\right) + \left(\lim_{b \to 1^{-}}{\rm e}^{f(b)}\right) - {\rm e}^{f(c)} \\ &= \left(\lim_{b \to 1^{-}}{\rm e}^{f(b)}\right)  - \left(\lim_{a \to 0^+} {\rm e}^{f(a)}\right), \end{align}$$where the auxiliary $c$ we used was your favorite number in $(0,1)$. Whether the limits above exist or not (which seems to be your concern), "that's between them and god".

Title: What is $\int_0^1 f'(x)e^{f(x)}dx$?


问:
Solving for $\theta$ in a nested sine function
I currently have the function:

\begin{equation}
\cos(1.3\pi\cos(\theta)) - \cos(1.3\pi) = 0
\end{equation}

but am unsure if it is possible to solve this analytically. I have looked over possible trig identities and have not found one that may be useful.

答:
Using whole numbers, you want to solve
$$\cos \left(\frac{13}{10} \pi  \cos (t)\right)=\cos \left(\frac{13\pi  }{10}\right)=\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}$$ Taking the inverse
$$t=\cos ^{-1}\left(\frac{10 }{13 \pi }\cos ^{-1}\left(-\frac{1}{2} \sqrt{\frac{1}{2}
   \left(5-\sqrt{5}\right)}\right)\right)$$ modulo ... something.


问:
Show that for each composite integer n, there exists a prime $p$ dividing $n$ (with $p\leq n^{1/2}$)

Let $n$ be a composite integer. Show that there exists a prime $p$ dividing $n$, with $p\leq n^{1/2}$.

$n$ is a composite integer so $n= ab$ where $a$ and $b$ are larger than $1$, let $p$ be a prime, suppose $p^2 \leq a^2 \leq ab = n$, because $p \ | \ a$ and $a \ | \ n$ then $p \ | \ n$. Hence, $p\leq n^{1/2}$.

Am I right about this?

答: If $n$ is composite then $n=ab$ either $a<n^{\frac{1}{2}}$ or $b<n^{\frac{1}{2}}$ since otherwise $n>n$. How  does this relate to primes?


问:
The Subspace $M=\{f\in C[0,1]:f(0)=0\}$
Let $C[0,1]$ with the supremum norm.  It's easy to see that $M=\{f\in C[0,1]:f(0)=0\}$ is a closed subspace and so $C[0,1]/M$ is a Banach space. But I'm having trouble in finding a Banach space isometric to $C[0,1]/M$.</p>

答:
For a given $a\in[0,1]$ consider linear bounded functional
$$
\operatorname{ev}_a:C[0,1]\to\mathbb{C}:f\mapsto f(a)
$$
Obviously $M=\operatorname{Ker}(\operatorname{ev}_0)$, i.e. $M$ is a kernel of bounded linear functional so $C([0,1])/M\cong \mathbb{C}$.</p>

ContentLicense: CC BY-SA 3.0

Title: The Subspace $M=\{f\in C[0,1]:f(0)=0\}$


答:
One way to solve it is as follows:

$$\cos{(1.3\pi\cos{\theta})} = \cos{1.3\pi}$$

Using the identity I gave above,

$$1.3\pi\cos{\theta} = 2n\pi\pm\ 1.3\pi.....n\in Z$$

Dividing both sides by $1.3\pi$, we get,

$$\cos{\theta} = \frac{20n}{13}\pm1$$

Solve for inequalities
$$-1<\frac{20n}{13}+1<1$$

and
$$-1<\frac{20n}{13}-1<1$$

since $\cos{\theta} \in [-1,1]$. Then take inverse cosine for the solutions you get.

Title: Solving for $\theta$ in a nested sine function

Body:
I currently have the function:

\begin{equation}
\cos(1.3\pi\cos(\theta)) - \cos(1.3\pi) = 0
\end{equation}

but am unsure if it is possible to solve this analytically. I have looked over possible trig identities and have not found one that may be useful.


问:
$\sum_{k=2}^\infty \frac{1}{1+(-1)^k \sqrt k}$ converges or diverges?
$\frac{1}{1+\sqrt k (-1)^k}= \frac{1-(-1)^k \sqrt k}{1-k}$. Not quite sure how to move on from here.

答:
Apply Leibniz test to show that $-\sum\frac {(-1)^{k} \sqrt k} {k-1}$ is convergent. Show that $\sum \frac 1 {1-k}$ is divergent. It then follows that the sum of these two series is divergent.

[$\frac {\sqrt x} {x-1}$ is a decreasing function on $(0,\infty)$ as you can see by differentiation].


问:
Find solution of this ODE
I try to solve this ODE
$$
y^2\left(1+\left(\dfrac{dy}{dx}\right)^2\right)=C
$$
where $C$ is positive constant.

Then I have $y=\pm\sqrt{C}$.
Is it right $y=\pm\sqrt{C}$ is only the solution to that ODE?

答:
$y^2\left(1+\left(\dfrac{dy}{dx}\right)^2\right)=C$

$\dfrac{dy}{dx} = \pm \sqrt {\frac{C}{y^2} - 1} \,$ (for $|y| \lt C$)

$\dfrac{ydy}{\sqrt{C-y^2}} = \pm dx$

$x = C' \pm {\sqrt{C-y^2}} \implies (x - C')^2 + y^2 = C$

Title: Find solution of this ODE


问:
Show that in figure bisector of angle intersect at perpendicular
The following was asked by a high school student which I could not answer. Please help

In the figure below, show that the bisector of $\angle AEB$ and $\angle AFD$ intersect at perpendicular

答:
Here is a better figure:

Here $ABF, DCF, ADE, BCE$ are collinear, and $PE, PF$ bisects $\angle AEB, \angle AFD$ respectively.

We have:

\begin{align}180^\circ &= \angle BAE + \angle AEB + \angle EBA \\&= \angle BAE + \angle AEB + \angle AFC + \angle BCF\\&=\angle BCF + 2\angle PEB + 2\angle PFC + \angle BCF\end{align}

This shows that $\angle BCF + \angle PEB + \angle PFC = 90^\circ$.

Now consider the quadrilateral $PFCE$ and you will finish the proof:

\begin{align}360^\circ &= \angle PFC + \text{reflex}\angle FCE + \angle CEP + \angle EPF\\
&=\angle PFC + (180^\circ + \angle BCF) + \angle BEP + \angle EPF\\
&=180^\circ + 90^\circ + \angle EPF
\end{align}

giving $\angle EPF = 90^\circ$.


答:
Make the case more general looking at
$$t_n=\left(a+\frac{1}{n}\right)^{n+b}\implies \log(t_n)=(n+b)\log\left(a+\frac{1}{n}\right)$$
Now, using Taylor
$$\log(t_n)=(n+b)\left(\log (a)+\frac{1}{a n}-\frac{1}{2 a^2 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
$$\log(t_n)=n \log (a)+\left(b \log (a)+\frac{1}{a}\right)+O\left(\frac{1}{n}\right)$$
$$t_n=e^{\log(t_n)}=e^{\frac{1}{a}} a^b\, n^a+O\left(\frac{1}{n}\right)$$ So, as soon as $a < 1$, your conclusion.

Title: $\lim_{n \to +\infty}(0{,}999 + \frac{1}{n})^{n+3} = $. Did I do it right?

Body:
I need to find the limit of this expression:

$$\lim_{n \to +\infty} (0{,}999 + \frac{1}{n})^{n+3} = e^{(n+3)\ln(0{,}999+\frac{1}{n})} = e^{(n+3)[\ln(\frac{1}{n})+\ln(1+0{,}999n)]}$$

Therefore:

$\lim_{n \to +\infty} 0{,}999n = 0$

$\lim_{n \to +\infty} \ln(1+0{,}999n) = 0$

And I get:

$$e^{(n+3)[\ln(\frac{1}{n})+ln(1+0{,}999n)]} \implies e^{(n+3)\ln(\frac{1}{n})} = e^{\ln(\frac{1}{n^{n+3}})}$$

$\ln(\frac{1}{n^{n+3}}) \implies -\infty $

So:

$$e^{\ln(\frac{1}{n^{n+3}})} \implies e^{-\infty} \implies 0$$

$$\lim_{n \to +\infty}(0{,}999 + \frac{1}{n})^{n+3} = 0$$

Did I do it right?


问:
Expectation Values
Suppose that $\{X_n\}_{n\ge1}$ take values $-\dfrac{1}{2}$ and $\dfrac{1}{2}$ with probability $a$ and $1-a$ respectively and $0$ otherwise. Suppose further that they are independent and discrete and random.

Then, how do I calculate the limit of $E\left[\dfrac{X_1+X_2+...+X_n}{3^n}\right]$ as $n$ tends to infinity.

If I had just variables and probability, I would use $$E[X]=\sum_{n=1}^\infty x_ip_i$$. But I am not sure about distribution of these values or whatsoever.
I would appreciate any help.

答:
I'd say

$\displaystyle\left|\sum_{k=0}^n X_i\right|<n/2$, hence $\dfrac{E[\sum_{k=0}^n X_i]}{3^n}\to 0$ quite fast.

Title: Expectation Values


问:
Does $A$ spans $B$ mean the same thing as $\text{span}(A) = B$?
I think that $A$ spans $B$ means that any vector in $B$ is a linear combination of the vectors in $A$, and $\text{span}(A) = B$ means that the set of all linear combinations of vectors in $A$ is the identical to vector space $B$. So these two definitions would be different, right?
For example, if
$$A = \begin{bmatrix}1 \\1 \end{bmatrix},\begin{bmatrix}0 \\1 \end{bmatrix}
$$
and
$$
B = \begin{bmatrix}a \\0 \end{bmatrix}, a\in \mathbb R
$$
then I'm fairly sure that $A$ spans $B$, but span(A) does not equal $B$. since vector space $\text{span}(A)$ has vectors that $B$ does not.
Does $A$ spans $B$ mean the same thing as $\text{span}(A) = B$?

答:
There may be some variation in usage.

For example, https://en.wikipedia.org/wiki/Linear_span says that a set of vectors $S$ spans a subspace $W$ if $W$ is the [unique] subspace spanned by $S$.
This would imply that "$A$ spans $B$" means the same thing as "$\mathrm{span}(A)=B$."

On the other hand, in
Minimum number of vectors that span a linear subspace
an exercise in a linear algebra course asks for the minimum number of linearly independent vectors that still span $V$, where $V$ is a subspace of some unnamed vector space.
If "$A$ spans $B$" means the same thing as "$\mathrm{span}(A)=B$," then the only sets of linearly independent vectors that span $V$ will contain exactly the same number of vectors as $V$ has dimensions, that is, there is no such thing as a "non-minimal" number of linearly independent vectors that span $V.$
That does not seem to be the intent of the exercise: rather, it seems to be assumed that you need to say "minimal" in order to exclude sets of vectors that span $V$ and also span a larger subspace.

So I think the answer to the question is context-dependent. The context is how the phrase "$A$ spans $B$" was introduced into the particular mathematical work where you see it.

The usage where the statements have the same meaning (or at least are indistinguishable in effect) seems to be far more common however.
(An example of "indistinguishable in effect" is when you ask if a subset $S$ of vectors in a space $V$ spans $V$. Since the set is a subset of $V$ it cannot span any larger vector space, so either $\mathrm{span}(S)=V$ or $S$ does not span $V.$)


问:
How do you isolate $\theta$ from $\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$?
This equation derives from physics force equations, and I have verified to make sure that the equation works. No actual physics pertains to this question. The part I need help on only requires math skills. Here is the equation:

$$\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$$

where $k$, $m_1$, and $m_2$ are variables, specifically the kinetic friction coefficient, mass one, and mass two respectively (irrelevant).

I have tried squaring both sides to then use double angle rules on the $2\sin\cos$ that yields, but all of my attempts have failed to simplify the equation further.

Could someone help me derive an equation for theta?

答:
Let $sin{\theta} = x$.

Then your equation becomes

$$x-k\sqrt{1-x^{2}} = \frac{m1}{m2}$$

Separate the linear terms and square both sides.
$$(x-\frac{m1}{m2})^{2} = k^{2}(1-x^{2})$$

It is now a simple quadratic equation which you can solve. But remember to choose only those solutions that lie within [-1,1] or can lie within [-1,1] for some values of $k,m1, m2$.

Now when you have obtained $\sin{\theta}$, use $\sin^{-1}$ to obtain $\theta$


答:
You might want to post the physics question which led you to this equation. It is likely there was a mistake in the derivation. Nonetheless,

$$\sin \theta - k \cos \theta = m_1/m_2 = \gamma$$
$$\sin \theta - k\sqrt{1-\sin^2 \theta} = \gamma$$
$$\sin \theta - \gamma = k \sqrt{1-\sin^2 \theta}$$
$$(\sin \theta -\gamma)^2 = k^2(1-\sin^2 \theta)$$
$$\sin^2 \theta - 2\sin \theta \gamma + \gamma^2 = k^2-k^2 \sin^2 \theta$$
$$(k^2+1) \sin^2 \theta  +(-2 \gamma) \sin \theta + \gamma^2 -k^2=0$$
$$\sin \theta = \frac{2\gamma \pm \sqrt{4\gamma^2 - 4(k^2+1)(\gamma^2 -k^2)}}{2(k^2+1)}$$

Title: How do you isolate $\theta$ from $\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$?

Body:
This equation derives from physics force equations, and I have verified to make sure that the equation works. No actual physics pertains to this question. The part I need help on only requires math skills. Here is the equation:

$$\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$$

where $k$, $m_1$, and $m_2$ are variables, specifically the kinetic friction coefficient, mass one, and mass two respectively (irrelevant).

I have tried squaring both sides to then use double angle rules on the $2\sin\cos$ that yields, but all of my attempts have failed to simplify the equation further.

Could someone help me derive an equation for theta?


Title: $\sum_{k=2}^\infty \frac{1}{1+(-1)^k \sqrt k}$ converges or diverges?

Body: $\frac{1}{1+\sqrt k (-1)^k}= \frac{1-(-1)^k \sqrt k}{1-k}$. Not quite sure how to move on from here.

Body:
Rewrite

\begin{align}\sum_{k=2}^{2N+1}\frac{1}{1+(-1)^k \sqrt k}&= \sum_{k=1}^N\left(\frac{1}{1+(-1)^{2k} \sqrt {2k}}+\frac{1}{1+(-1)^{2k+1} \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1}{1+\sqrt {2k}}+\frac{1}{1- \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1- \sqrt {2k+1}+1+\sqrt {2k}}{(1+\sqrt {2k})(1- \sqrt {2k+1})}\right)
\\&<\sum_{k=1}^N\left(\frac{2}{(1+\sqrt {2k+1})(1- \sqrt {2k+1})}\right)
\\&=\sum_{k=1}^N\left(\frac{2}{-2k}\right)\to-\infty
\end{align}


问:
Showing that the inflection points of $x\sin x$ lie on a certain curve

Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$

I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?

答:
An inflection point is where $f''$ changes sign,
so we need to find where $f''(x)=0$.

$f(x)=x\sin(x),
f'(x)=\sin(x)+x\cos(x),
f''(x)=\cos(x)+\cos(x)-x\sin(x)
=2\cos(x)-x\sin(x)
$.

So we want to show
that this satisfies
$y^2(x^2+4)=4x^2$.

If $f''(x)=0$ then
$2\cos(x)=x\sin(x)
$
or $x=2/\tan(x)$.

Since $f(x)=x\sin(x)$,
we want
$x^2\sin^2(x)(4/\tan^2(x)+4)=4x^2
$
or
$\sin^2(x)(1/\tan^2(x)+1)=1
$
or
$1=\sin^2(x)(\cos^2(x)/\sin^2(x)+1)
$
or
$1=\cos^2(x)+\sin^2(x)
$
which is true.


Title: Showing that the inflection points of $x\sin x$ lie on a certain curve

Body:
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$

I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?

Body:
$f(x)$ has an inflection point only if $f''(x)=0.$
$$f'(x) = \sin x + x \cos x$$
$$f''(x) = \cos x + \cos x -x \sin x=2\cos x - f(x)$$
$$f''(x) = 2\sqrt{1-\sin^2 (x)} - f(x)$$
$$f''(x) = 2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$0=2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$y=2\sqrt{1-y^2/x^2}$$
$$y^2 = 4-4y^2/x^2$$
$$...$$


问:
Compute ND Triangulation from Pairwise Distance Matrix
Is there a way to compute a triangulation of points (e.g. triangles that approximate the surface of a 2-sphere) from a pairwise distance matrix in n-dimensions?

So far, what I've found is from Building a graph from pairwise distances:

I'm aware that you can take the distance matrix and obtain (or attempt to obtain) a layout in however many dimensions you like by taking the eigenvectors (by say the power iteration) of the all pairs shortest path matrix computed by the Floyd-Warshall algorithm from that distance matrix.

答:
These two articles should cover this well enough.

(1) Boissonnat, J. D.; Dyer, R.; Ghosh, A.; Oudot, S. Y. Only Distances Are Required to Reconstruct Submanifolds. Computational Geometry: Theory and Applications 2017, 66, 32–67. https://doi.org/10.1016/j.comgeo.2017.08.001.

(2) Connor, R.; Vadicamo, L.; Rabitti, F. High-Dimensional Simplexes for Supermetric Search. Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) 2017, 10609 LNCS, 96–109. https://doi.org/10.1007/978-3-319-68474-1_7.


答:
It can be shown that any expression of the form $a\cos(\theta)+b\sin(\theta)$ ($a$ and $b$ are constants) can be written in the form $m\cos(\theta+c)$, where $m$ and $c$ are constants. Specifically, by taking $m=\text{sgn}(a)\sqrt{a^2+b^2}$ and $c=-\tan^{-1}\left(\frac{b}{a}\right)$, where $\text{sgn}(a)$ denotes the sign of $a$, we can write

$$a\cos(\theta)+b\sin(\theta)=\text{sgn}(a)\sqrt{a^2+b^2}\cos\left(\theta-\tan^{-1}\frac{b}{a}\right)$$

Substituting $a=-k$ and $b=1$ gives

\begin{align*}
\sin(\theta)-k\cos(\theta) &= \text{sgn}(-k)\sqrt{(-k)^2+1^2}\cos\left(\theta-\tan^{-1}\frac{1}{-k}\right)\\
&=-\text{sgn}(k)\sqrt{1+k^2}\cos\left(\theta+\tan^{-1}\frac{1}{k}\right)
\end{align*}

so if $k>0$, your equation boils down to

$$-\sqrt{1+k^2}\cos\left(\theta+\cot^{-1}k\right)=\frac{m_1}{m_2}$$

Dividing both sides by $-\sqrt{1+k^2}$ gives $\cos\left(\theta+\cot^{-1}k\right)=-\frac{m_1}{m_2\sqrt{1+k^2}}$, provided that $\left|-\frac{m_1}{m_2\sqrt{1+k^2}}\right|\leq 1$.

It can be shown that for any real number $y$ with $|y|\leq 1$, the only solutions to the equation $\cos(x)=y$ are of the form $x=2\pi m\pm\arccos(y)$ for integer $m$, so

$$\theta+\cot^{-1}(k)=2\pi m\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$

The desired result immediately follows.

$$\theta=2\pi m-\cot^{-1}(k)\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$


问:
Does $\sum_i e_i \otimes e_i$ in $V\otimes V$ have a name?
If $V$ is a vector space with orthogonal basis $(e_i)$,  consider the element
$\sum_i e_i \otimes e_i  \in V\otimes V$.
It can be shown that this element is independent of the choice of orthogonal basis.

Does this element have a name?

答:
You could call it the Casimir element. If $V$ is finite-dimensional, then under the identification $V \otimes V \cong V \otimes V^{\ast} \cong \text{End}(V)$ it corresponds to the identity

$$V \otimes V^{\ast} \ni \sum e_i \otimes e_i^{\ast} \mapsto \text{id}_V \in \text{End}(V)$$

and can also be called the unit.


问:
$X_n$ is a sequence and uniformly distributed on $(-n,n)$. Does it converge in $L^p$?
If $X_n$ is a sequence and uniformly distributed on $(-n,n)$. Does it converge in $L^p$ and if it does then what would be the limit?

MY attempt consisted of finding the characteristic function which leads us to

the characteristic function  $$ \varphi_{X_n}(t) = \frac{\sin(tn)}{tn}$$ and so:

$$\varphi_{X_n}(t)=\begin{cases}\frac{\sin(tn)}{tn} &t \ne0\\ 1&t = 0\end{cases}.$$

and then:
$$\lim_{n\to\infty} \varphi(t)=\begin{cases}0 &t \ne0\\ 1&t = 0\end{cases}.$$

Now, I can see that when $t = 0$ for  $t \in \mathbb{R}  $, the limiting characteristic function $\varphi(t)$ is discontinuous. So the limit exists except for when $t=0$.
Hence we see that there is no convergence in distribution. From here can we conclude that it does not converge in $L^p$?

Is there any other way to prove this.
Thanks

答: $$E|X_n|^{p}=\frac 1 {2n}\int_{-n}^{n} |t|^{p}dt$$ $$ \geq \frac 1 {2n}\int_{n/2}^{n} |t|^{p}dt$$ $$ \geq \frac  1 {2n} {\frac n 2} |\frac  n 2|^{p} \to \infty$$ as $n \to \infty.$ Hence $(X_n)$ is not even bounded in $L^{p}$ for any $p >0$.


答:
Let:

$$k=\tan \alpha \ \iff \ \alpha:=\tan^{-1}k.$$

The equation

$$\sin \theta - k \cos \theta = \underbrace{m_1/m_2}_{\gamma}$$

becomes, by multiplication by $\cos \alpha$:

$$\cos \alpha \sin \theta - \sin \alpha \cos \theta =\gamma \cos \alpha$$

$$\sin(\theta - \alpha)=\gamma \cos \alpha \tag{1}$$

If $\gamma \cos \alpha \in [-1,1]$, we can set

$$\sin \delta:= \gamma \cos \alpha \ \iff \ \delta=\sin^{-1}(\gamma \cos \alpha)$$

Equation (1) becomes:

$$\sin(\theta - \alpha)=\sin \delta \tag{2}$$

$$\iff \begin{cases}\theta - \alpha&=&\delta+k 2 \pi\\
\theta - \alpha&=&\pi-\delta+k' 2 \pi\end{cases}$$

whence two expressions for $\theta$.


答:
$f(x)=x\sin x \implies f''(x)=2\cos x-x\sin x$ let $x=h$ be the point of inflexion then $f''(h)=0 \implies h=2\cot h ~~~(1)$, The point $(h,k)$ falls on the curve hence $k=f(h)= h \sin h~~~(2)$ from (1) and (2) the point $(h,k)$ will satisfy ans algebraic equation when we eliminiate the periodic function from (1) and (2).
From (2) we have $\cos h=\frac{h}{\sqrt{4+h^2}}~~~(3)$ squaring ans adding (1) and (2) we get $$\frac{h^2}{4+h^2}+\frac{k^2}{h^2}=1 \implies k^2(h^2+4)=4h^2$$
So the required curve ids $y^2(x^2+4)=4x^2$


问:
How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$?
Numerically, it looks that the limit is

$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,

but I have not been able to demonstrate it analytically. Does anyone have a idea on how to deal with this limit?

答:
Here is how you advance, compute the Taylor series of $\frac{1}{\sqrt{x-y}}$ at the point $x=1$

$$ \frac{1}{\sqrt{x-y}} = {\frac {1}{\sqrt {1-y}}}-\frac{1}{2}\, \left( 1-y \right) ^{-3/2} \left( x-1
 \right) +O \left(  \left( x-1 \right) ^{2} \right) $$

$$ \implies \frac{1}{\sqrt{x-y}} \sim {\frac {1}{\sqrt {1-y}}}. $$

So, when $x$ close to $1$, our integral behaves as

$$\int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {x-y}}dy \sim \int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {1-y}}dy$$

$$=\int _{0}^{1}\!{\frac{ \left( -u+1 \right)^{3/2}}{\sqrt {-u+2}u}}{du}\sim \int _{0}^{1}\!{\frac {1}{\sqrt {2}u}}{du}.$$

The last integral does not converge. Note that, we used the change of variables $1-y=u$ and

$$ \frac{ ( -u+1 )^{3/2}}{\sqrt{-u+2}u} \sim \frac{1}{\sqrt{2}\,u} \quad \mathrm{when}\quad u\sim 0.$$


问:
How do I determine from a picture of a vector field if it's a possible formula for the vector field and conservative or not?

I have an image here of a vector field $F(x,y)$ and am tasked to do the following things:

True or false: : A possible
formula for $F(x, y)$ is $F(x, y) = <−y, x>$

Is $F$ (the vector field in the picture) conservative?

So the first one I really have no clue how to tell if that's a possible formula. The second one I think maybe it's conservative because it's circular so the curl = 0. But how do I actually know?

How do I solve these problems with just a picture?

答:
I will answer the second part of the question about the plots (first part seems settled by now).

Thought experiment No. $1$ or circulation:

Imagine yourself to be piloting a plane flying in a closed loop within the vector field, discard gravity, and ask yourself whether the total energy spent when traveling against the direction of the wind will be possibly cancelled out by the free ride you will catch when flying with the wind. In other words, is it possible that the circulation (the line integral in a closed loop) is zero? We can guess whether it is at all possible or if it is impossible by visual inspection - if it is impossible, the field is not conservative.

So draw a box in a somewhat deliberate fashion and observe whether circulating counterclockwise the field is consistently either flowing against or with you. Say that the vector field in your example is $\langle y,-x\rangle,$ and that you draw a square centered below the horizontal:

The field is fighting your progress throughout the path, resulting in a negative circulation of $-190,$ represented as an orange stripe along the path. This random closed path is enough to rule out that the field is conservative.

Let's look at a counterexample of a conservative field, $\langle x, y \rangle:$

It is not possible to tell whether the circulation is zero by visual inspection, but it is certainly impossible to rule out, since the field alternates between being aligned with (in green) and flowing against (in orange) the motion along the rectangular path.

Another way to look at it is to detect the presence of curl, which is zero in path-independent vector fields. The presence of negative curl can be easily noticed on the plot:

This can be tricky in different ways (macroscopic circulation without microscopic and simple connectedness).

Thought experiment No. 2 or flux:

Conservative fields are the gradient of a scalar function (so-called "potential" function), and gradients are examples of co-vector fields - they indicate the directions of steepest ascent, and are often represented as isolines or equipotential lines (uniting points with the same value in the potential function). These can be dotted with vectors to obtain directional derivatives. Equivalently, a common visualization is with stacks of parallel lines at every point.

Here the experiment is running from the bottom of valleys to the top of mountains along the steepest paths: following the direction of the gradient field, which portrayed in vectorial form, is orthogonal to the isolines. If the vector field is indeed the gradient of some other function, it may be possible to visualize on a plot of the vector field the isolines orthogonal to the vectors, not crossing at any points and packed more closely together as the magnitude of the vectors increases.

The mental image is of many alpinists climbing away (perpendicular) to the isobars from the valley to the mountains, along the steepest paths. The idea is the two-dimensional flux integral through an isoline.

In the case of the conservative field $\langle x, y\rangle,$ the flux across a circle centered at the origin is seen as a consistent pink stripe of uniform width:

And the representation with stacks (co-vectors) does convey the idea of isolines with each coordinate-specific stack being denser in proportion to the length of the vector in that position. The flux across the isoline in the plot is in pink, and consistent in sign along the isoline:

In contradiction, the example in the original post expressed as stacks would not lend itself to joining in an isoline equally dense covectors:

An example / counterexample dual that parallels the graph in the original question is the magnetic field created by a straight electrical wire, which spirals following the right-hand rule, and is non-conservative, since by Maxwell's equation $\nabla \times \mathbf{B} =  \mathbf{J}_\mathbf f +  \frac{\partial\mathbf{E}}{\partial t}$ (Ampère's circuital law) dictates that the field is only curl-free in the absence of free-currents or time-varying electric fields (see here or here) (left plot with arrows and stacks) in contradistinction to the diverging electric field created by a positive charge (right plot):


问:
Residue theorem of rational function with real roots in the denominator
I am studying residue calculations and until now had learned that one should avoid using it for integral of rational functions P/Q when Q has real roots. However, I found an exercise concerning this integral :
$$\int_{-\infty}^{\infty}\frac{1}{(x+1)(x^2-x+1)}dx $$
which obviously has a real roots for $x=-1$... The answer I saw was to include one half the residue for pole $Z=-1$ when going complex. I never heard of that : could anyone point to any source and explain why it is so ?
Thanks and have a good day !

Marc

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{r \equiv }$:
\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{-\infty}^{\infty}
{\dd x \over \pars{x + 1}\pars{x^{2} - x + 1}}}
\\[5mm] = &\
\overbrace{2\pi\ic\lim_{\large x\ \to\ \expo{\ic\pi/3}}\,\,\,
{x - \expo{\ic\pi/3} \over \pars{x + 1}\pars{x^{2} - x + 1}}}
^{\ds{{2 \over 3}\,\pi\,\ic\expo{-2\pi\ic/3}}}
\\[2mm] - &\
\underbrace{\lim_{\large \epsilon \to 0^{+}}\,\,\int_{\pi}^{0}
{\epsilon\expo{\ic\theta}\ic\,\,\dd\theta \over
\epsilon\expo{\ic\theta}
\bracks{\pars{-1}^{2} - \pars{-1} + 1}}}_{\ds{-\,{\pi \ic \over 3}}}
\\[5mm] = &\
\bbx{{\root{3} \over 3}\,\pi} \approx 1.8138 \\ &
\end{align}


问:
Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $
I have been trying to derive

$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$

with $G$ being the the Catalan constant.

I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.

Body:
$$I=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx$$

$$=\underbrace{\int_{0}^{1}\frac{log(1-x)log(x)}{1+x^2}dx}_{I_1}+\underbrace{\int_{0}^{1}\frac{log(1-x)log(1+x)}{1+x^2}dx}_{I_2}$$

Let's start evaluating $I_2$ using Leibniz Rule:
$$I_2={\int_0^1}\frac{log(1-x)log(1+x)}{1+x^2}dx=-{\int_0^1}{\int_0^1}{\int_0^1}\frac{x^2}{(1+x^2)(1-yx)(1+xz)}dydzdx$$

$$={\int_0^1}{\int_0^1}{\int_0^1}dxdydz\\\left(\frac{(1+yz)+x(y-z)}{(1+y^2)(1+z^2)(1+x^2)}-\frac{y}{(1+y^2)(y+z)(1-yz)}-\frac{z}{(1+z^2)(y+z)(1+zx)}\right)$$

$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\underbrace{\frac{log(1-y)}{(1+y^2)(y+z)}}_{y\rightarrow z\\ z\rightarrow y}+\frac{log(1+z)}{(1+z^2)(y+z)}\right)$$

$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\frac{log\left(\frac{1-z}{1+z}\right)}{(1+z^2)(y+z)}\right)$$

$$=\int_0^1\left(\frac{\frac{\pi^2}{16}+\frac{log^2(2)}{2}}{1+z^2}+\underbrace{\frac{log\left(\frac{1-z}{1+z}\right)log\left(\frac{1+z}{z}\right)}{1+z^2}}_{z\rightarrow \frac{1-x}{1+x}}\right)dz$$

$$=\frac{\pi}{4}\left(\frac{\pi^2}{16}+\frac{log^2(2)}{4}\right)+log(2)\int_0^1\frac{log(x)}{1+x^2}dx-\underbrace{\int_{0}^{1}\frac{log(x)log(1-x)}{1+x^2}dx}_{I_1}$$

$$=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)-I_1$$

Hence:
$$I=I_1+I_2=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx
=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)\ \blacksquare$$


问:
Proving complex numbers identity using, $\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$
Show that $\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$ can be expressed in the form $\alpha$ or $\alpha^2$, Hence prove $$(a+b\alpha+c\alpha^2)(a+b\alpha^2+c\alpha)=\frac12[(a-b)^2+(b-c)^2+(c-a)^2]$$

My try:

I wrote the expression as,$$y=\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$$
$$y^2+y+1=0$$
$$y=\frac{-1+\sqrt{3}i}{2}\text{ or } y=\frac{-1-\sqrt{3}i}{2}$$
This is in the form of $\alpha$ and $\alpha^2.$I have trouble solving the second part. Can someone please explain it to me? Thank you in advance.

答:
The LHS is $a^2+b^2 \alpha^3+c^2 \alpha^3+ab(\alpha^2+\alpha)+ac(\alpha^2+\alpha)+bc(\alpha^4+\alpha)$.

Use the fact that $\alpha^4=\alpha$,$\alpha^2+\alpha=-1$,$\alpha^3=1$.

So The LHS becomes $$a^2+b^2+c^2-ab-ac-bc\\=\frac 1 2\left[a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac\right]\\=\frac 1 2\left[(a-b)^2+(b-c)^2+(a-c)^2\right]$$


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答:
Under development but already working very nicely is Sympy, a symbolic math package for Python.

The usage is similar to other math software, there is even a live version available. A session might look like the following:

```<code>>>> from sympy import *
>>> x = symbols('x')
>>> f = cos(x)
>>> f.series()
1 - x**2/2 + x**4/24 + O(x**6)
>>> f.series(n=12)
1 - x**2/2 + x**4/24 - x**6/720 + x**8/40320 - x**10/3628800 + O(x**12)
>>> latex(f.series(n=12))
</code>```

$1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \frac{x^{6}}{720} + \frac{x^{8}}{40320} - \frac{x^{10}}{3628800} + \mathcal{O}\left(x^{12}\right)$

```<code>>>> diff(f, x)
−sin(x)
</code>```

An overview of the possible functions is given in the online documentation. It is free software and has quite readable source code.

Title: Free online mathematical software

Body:
Under development but already working very nicely is Sympy, a symbolic math package for Python.

The usage is similar to other math software, there is even a live version available. A session might look like the following:

>>> from sympy import *
>>> x = symbols('x')
>>> f = cos(x)
>>> f.series()
1 - x**2/2 + x**4/24 + O(x**6)
>>> f.series(n=12)
1 - x**2/2 + x**4/24 - x**6/720 + x**8/40320 - x**10/3628800 + O(x**12)
>>> latex(f.series(n=12))

$1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \frac{x^{6}}{720} + \frac{x^{8}}{40320} - \frac{x^{10}}{3628800} + \mathcal{O}\left(x^{12}\right)$

>>> diff(f, x)
−sin(x)

An overview of the possible functions is given in the online documentation. It is free software and has quite readable source code.


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答: The way I see it intuitively as a venn diagram. Two spheres represent the arbitrary amount moved around, I. This case a tea spoon amount. So when they overlap, you ask which area is greatest of the two spheres which aren't overlapping. But you see any area taken from one must be taken from the other and the area is the same.</p>


问:
How to determine distance between two points in Poincare?
I study Poincare model as hyperbolic model. But honestly, I don't understand with this topic. I try to find some examples but nothing. Here is my problem.

Find the equation of the hyperbolic line (euclid circle) through $P(1/3,0)$ and $Q(0,1/3)$.

Distance between $P$ and $Q$.

In my textbook, distance in Poincare is $\left\lvert\ln\dfrac{(PQ')(QP')}{(PP')(QQ')}\right\rvert$. But how to find $P'$ and $Q'$?

答:
First, I use geogebra to solve it.

Draw a circle $C$ with $x^2+y^2=1$. Suppose $R$ and $S$ are inverse points $P,Q$ of circle $C$. By using inversion formula $|OP||PR|=1$ so $R=(3,0)$ and $S=(0,3)$. Now, suppose circle $D$ through $P,Q,R$. If $\ell$ is perpendicular bisector of $P,Q$ then $\ell$ pass the centre of circle $D$ in $(1.67,1.67)$. So the equation of the circle $D$ is $(x-1.67)^2+(y-1.67)^2=4.56$.

Intersection of two circles are $P'=(0.94,-0.34)$ and $Q'=(-0.34,0.94)$. Calculate $|PQ'|, |QP'|, |PP'|, |QQ'|$ by using distance formula in euclidean. Then

$PQ=\left\lvert\ln\dfrac{(PQ')(QP')}{(PP')(QQ')}\right\rvert=\left\lvert\ln\dfrac{(1.16)(1.16)}{(0.7)(0.7)}\right\rvert\approx 1.01019$.


问:
for $S^{-1}A$-module $M$, $S^{-1}A\otimes_A M\to M$
Sorry for my bad English.

Let $A$ be commutative ring, and $S\subset A$ be multiplicatively closed set.

And let $M$ be $S^{-1}A$-module.

Now we have counit of adjunction of "change of rings"

$M\otimes_A S^{-1}A\to M$

I think this is isomorphism of $S^{-1}A$-module, is it true?

答:
Now counit $\varphi: M\otimes_A S^{-1}A\to M$ is $\varphi(x\otimes \frac{a}{s})=\frac{ax}{s}.$

Let $\psi:M\to M\otimes_A S^{-1}A$ be $\psi(x):=x\otimes 1$.

Now $\psi$ is morphism of $S^{-1}A$-module.
Indeed, for any $a\in A, s\in S, x\in N,$

$s \psi(\frac{a}{s}x)=s(\frac{a}{s}x\otimes 1)=\frac{a}{s}x\otimes s=ax\otimes 1=x\otimes a=a(x\otimes 1)=a\psi(x)=s(\frac{a}{s}\psi(x)). $

multiply $s$, we get $\psi(\frac{a}{s}x)=\frac{a}{s}\psi(x)$, so $\psi$ is morphism of $S^{-1}A$-module.

Morover, $\varphi(\psi(x))=\varphi(x\otimes 1)=y. $

$\psi(\varphi(x\otimes\frac{a}{s}))=\psi(\frac{ax}{s})=\frac{ay}{s}\otimes 1=\frac{y}{s}\otimes a=\frac{1}{s}\cdot s(\frac{y}{s}\otimes a)=\frac{1}{s}(y\otimes a)=y\otimes \frac{a}{s}.$

From the above, $\varphi$ and $\psi$ is inverse morphism each other.


问:
Evaluate the line integral of the given vector field
The formula I'm using is $\int_C \mathbf F \bullet d\mathbf r$

The given vector field is
$$F(x,y) = \frac{x}{(x^2 + y^2)^{3/2}}\vec i +\frac{y}{(x^2 + y^2)^{3/2}} \vec j$$
and the curve C given parametrically by
$$r(t)=e^{2t}\cos(3t)\vec i + e^{2t}\sin(3t)\vec j,\space 0\leq t\leq 2\pi$$
$$r^{'}(t)= -3e^{2t}\sin(3t)+2e^{2t}\cos(3t)\vec i + 2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\vec j$$

The integral will then be
$$\int_0^{2\pi}\left<\frac{e^{2t}\cos(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}+ \frac{e^{2t}\sin(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}\right> \bullet \left<-3e^{2t}\sin(3t)+2e^{2t}\cos(3t), \space2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\right>dt$$
After working that whole thing out I got
$$\int_0^{2π} \frac{2}{e^{2t}}dt=1-\frac{1}{e^{4\pi}}$$
Am I correct?

Body:
$F(x,y)$ is a scalar valued function.

We don't use an inner product, we use the magnitude of $dr$

$\int F(x(t),y(t))\|r'(t)\|\ dt$

$r'(t) = e^{2t} \langle 2\cos 3t - 3\sin 3t, 2\sin 3t + 3\cos 3t\rangle$

$\|r'(t)\| = \sqrt{2^2 + 3^2} e^{2t}$

$\int_0^{2\pi} \frac {e^{2t}(\sin t + \cos t)}{e^{3t}} e^{2t}\sqrt 13 \ dt\\
\sqrt {13} \int_0^{2\pi} e^{t}(\sin t + \cos t) \ dt\\
\sqrt {13} (\frac {e^t(\cos 3t + 3\sin 3t)}{10} + \frac {e^t(\sin 3t - 3\cos 3t)}{10})|_0^{2\pi}\\
\frac {\sqrt {13}\ e^t}{10}(4\sin t - 2\cos t)|_0^{2\pi}\\
\frac {\sqrt {13} (1-e^{2\pi})}{5}\\
$


问:
If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?

Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$

$(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$

$(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$

$(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated.

Thanks for your time.

答:
Note the Sum of two squares theorem states

An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no term $p^k$, where prime $p\equiv 3 \pmod{4}$ and $k$ is odd.

For your $(2)$, with $p$, since it's distinct from $q$, then the power of $p$ in $a$ would be $1$ plus $2$ times the power of $p$ in $d$, i.e., the exponent of $p$ is odd. Thus, since $p \equiv 3 \pmod{4}$, then the theorem quoted above says the value cannot be the sum of two squares.

Note the Wikipedia's article's proof link to the Internet Archive says "This item is no longer available". I did some searching, but couldn't find any other link to it. However, there's basically the equivalent theorem stated in Sum of Two Squares near the bottom of page $4$:

Theorem $6$. A positive integer n is a sum of two squares iff $\operatorname{ord}_p(n)$ is even for all primes $p \equiv 3 \pmod{4}$.

It then follows with a remark about an equivalent statement (which is also equivalent to Wikipedia's theorem statement I quoted initially):

Remark: An equivalent statement of the theorem, which we will use in the proof,
is: $n$ is a sum of two squares iff it factors as $n = ab^2$, where $a$ has no prime factor $p \equiv 3 \pmod{4}$.

This statement is possibly somewhat confusing as it's basically the equivalent of saying $a$ is square-free. Anyway, the linked paper goes on to state & prove a lemma it then uses to prove its theorem $6$.


问:
Cups in boxes- Probability of choosing a box with no cup broken
Suppose there are 4 boxes of cups, each box having 3 cups and also given that 5 cups are broken. What is the probability of choosing a box with no cup broken in that box ?

I tried using conditional probability i.e. if A is the event of choosing a box with no cup broken then

P(A)=P(A|5 cups distributed in 2 boxes)+P(A|5 cups distributed in 3 boxes)+ P(A|5 cups distributed in 4 boxes)

I am not sure what mistake I am doing here. Is there any easier method to do this ?

答:
Tricky problem.  Believe it or not, WLOG, you choose the 1st box.

Number of ways that 5 broken cups can be distributed into 12 slots is

$$D = \binom{12}{5}.$$

Number of ways that 5 broken cups can be distributed into the 9 slots of the other 3 boxes is

$$N = \binom{9}{5}.$$

Final answer

$$\frac{N}{D}.$$

Addendum

First of all, with the original problem as stated, the analysis of $\frac{\binom{9}{5}}{\binom{12}{5}}$ will apply regardless of which box you choose.  Once you choose a box, the question is whether any of the 3 slots in your box will receive a broken cup.

As for your hypothetical:

Suppose there are 4 boxes of cups with 3 cups in 3 boxes and 2 cups in one box such that 5 cups are broken.

With this hypothetical, I would take the following approach.

If I choose a box with 3 cups, my chance of no broken cup, for reasons similar to the start of my answer, will be

$$\frac{\binom{8}{5}}{\binom{11}{5}}.$$

If I happen to choose the box with 2 cups, my chance of no broken cup will be

$$\frac{\binom{9}{5}}{\binom{11}{5}}.$$

My chance of choosing the box with 2 cups is $\frac{1}{4}.$

Therefore, my answer to the hypothetical would be

$$\left[\frac{1}{4} \times \frac{\binom{9}{5}}{\binom{11}{5}}\right] ~+~
\left[\frac{3}{4} \times \frac{\binom{8}{5}}{\binom{11}{5}}\right].$$


问:
Solve the given system of quadratic equations

$$x^2 + xy + xz = 2$$
$$y^2 + yz + xy = 3$$
$$z^2 + zx + yz = 4$$

I tried solving it but i just ended up with
$$(x-y)(x+y+z) = -1$$
$$(y-z)(x+y+z) = -1$$
$$(x-z)(x+y+z) = -2$$

I'm not sure what to do. Any hints?

答:
Your equations are equivalent to $$x(x+y+z)=2,\quad y(x+y+z)=3,\quad z(x+y+z)=4.$$ Set $$x+y+z=t,$$ then $t^2=2+3+4=9$ and we have $t=\pm 3.$ Hence $$x=\pm\dfrac23,\qquad y=\pm 1,\qquad z=\pm\dfrac43.$$

Edit: Btw, here we hove only two solutions for the system.


问:
Cotangent in integral form
Could you help me to establish the following relation, without going through the Eulerian development
$\displaystyle \pi \operatorname{cotan}(\pi z) = \frac{1}{z} + \sum_{n=1}^{+\infty} \left( \frac{1}{z+n} + \frac{1}{z-n} \right)$ :
$$\pi \operatorname{cotan}(\pi z) = \int_0^{+\infty} \frac{e^{(1-z)t} - e^{tz}}{e^t-1} dt. \quad 0<\Re(z)<1$$

答:
Hoping that you do enjoy hypergeometric functions
$$I=\int\frac{e^{(1-z)t} - e^{tz}}{e^t-1} \,dt$$
$$ I=\frac {e^{-t z} \, _2F_1\left(1,-z;1-z;e^t\right)+e^{t z} \,
   _2F_1\left(1,z;z+1;e^t\right)-e^{-t z} } z$$

$$J=\int_0^\infty\frac{e^{(1-z)t} - e^{tz}}{e^t-1} \,dt=\psi (-z)-\psi (z)-\frac{1}{z}=\pi  \cot (\pi  z)$$

Title: Cotangent in integral form


问:
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ .
I am trying to find $x+y$ given that

$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$

It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side

$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$

and see if a term $x+y$ comes out, but it looks hopeless at this point.

答:
Hint: Calculate $(-t+\sqrt{t^2+3})(t+\sqrt{t^2+3})$.

Details: The product in the hint is $3$.  So for every $y$, by putting $x=-y$ we get a solution of the equation
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3}=3)=3\tag{$1$}$$
with $x+y=0$. We now show these are the only solutions,

One of the terms in $(1)$ is $\le \sqrt{3}$, say the first one. Then the second one is $\ge \sqrt{3}$.  But for positive $t$, the function $t+\sqrt{t^2+3}$ is increasing, so there is a unique value of $y$ corresponding to $x$.


问:
Using the l'hopital's rule for a $1^∞$ case
I need to find the limit of:

$$\lim_{x \to 0} (1-2x)^{1/x}$$

I can see that this is a $1^∞$ case, so I will let $y =$ the equation and use the natural log on both sides:

$$\ln y = \lim_{x \to 0} \frac{1}{x}\ln(1-2x) $$

Now it is a $∞\times0$ case, so I can do:

$$\ln y = \lim_{x \to 0}\frac{\frac{1}{x}}{\frac{1}{\ln(1-2x)}}$$

and now it is a $\frac{∞}{∞}$ case. I think my mistake happens while I use the l'hopital's rule:

$$\ln y = \lim_{x \to 0}\frac{-x^{-2}}{-[\ln(1-2x)]^{-2}\times\frac{1}{1-2x}\times-2x}$$

$$\ln y = \lim_{x \to 0}\frac{-[\ln(1-2x)]^2}{\frac{2x^3}{1-2x}}$$

It is now a 0/0 case, and I have tried using the l'hopital's rule a few more times but to no such luck. Where did I make a mistake, and is there a better way to solve this?

答:
Note that

$$ \frac{1}{x}\ln(1-2x)= \frac{\ln(1-2x)}{x},$$

which is the indeterminate $\frac{0}{0}$ in the limit as $x \rightarrow 0.$ This is what you should apply L'Hopital's rule to. Why? Because differentiation will make the expression simpler -- it will get rid of the logarithm. You never want to apply L'Hopital's rule if you have to differentiate $1/\ln(f(x))$, as differentiation will not get rid of the logarithm in such a case -- it won't make the expression simpler. Any time you end up with a more complicated expression after L'Hopital's rule, you've made a mistake.


问:
help reducing by modulo
I'm an engineer, not a math guy.  Please use small words if possible.  ;-)

I am going through this neat paper on a method of partial fraction decomposition by repeated synthetic division.  On page 157 it is covering case with repeated irreducible quadratics in denominator - and it says this:

"Multiply both sides by the denominator...then
We reduce the right hand side modulo $x^2+bx+c $ by sending it to the field $\mathbb R[x]/(x^2+bx+c) =
$ {$px+q|p,q\in \mathbb{R},x^2+bx+c=0$}. Modulo $x^2+bx+c$....

I have no clue how to "reduce the r.h.s. modulo by sending it to the field".   Can someone give me example of how to do such?  Preferably with real equations i can ponder like an engineer.

UPDATE 1:  Adding real example we can use.
Let,
$$\frac{N(s)}{D(s)}=\frac{10(s+1)}{(s+1)^3(s^2+5s+6)^3}=\frac{K_1}{s+1}+\frac{K_2}{(s+1)^2}+\frac{K_3}{(s+1)^3}+\frac{M_1s+C_1}{s^2+5s+6}+\frac{M_2s+C_2}{(s^2+5s+6)^2}+\frac{M_3s+C_3}{(s^2+5s+6)^3}$$

Multiplying both sides by D(s) I get,
$$N(s) = 10(s+3) = K_1(s+1)^2(s^2+5s+6)^3+K_2(s+1)(s^2+5s+6)^3+K_3(s^2+5s+6)^3+(M_1s+C_1)(s+1)^3(s^2+5s+6)^2+(M_2s+C_2)(s+1)^3(s^2+5s+6)+(M_3s+C_3)(s+1)^3$$

How do i proceed now?

Title: help reducing by modulo

Body:
$$\begin{align}
   10(s+3)
        &= K_1(s+1)^2(s^2+5s+6)^3 \\
        & +K_2(s+1)(s^2+5s+6)^3 \\
        &+ K_3(s^2+5s+6)^3 \\
        &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\
        &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\
        &+ (M_3s+C_3)(s+1)^3
\end{align}$$

Let $s = -1$ and you get $20 = 8K_3$. So $K_3 = \dfrac 52$

$$\begin{align}
   10(s+3)
        &= K_1(s+1)^2(s^2+5s+6)^3 \\
        & +K_2(s+1)(s^2+5s+6)^3 \\
        &-\dfrac 52(s^2+5s+6)^3 \\
        &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\
        &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\
        &+ (M_3s+C_3)(s+1)^3
\end{align}$$

$$\begin{align}
   -\dfrac 52 (s + 3) (s + 1) (s^4 + 11 s^3 + 46 s^2 + 88 s + 68)
        &= K_1(s+1)^2(s^2+5s+6)^3 \\
        & +K_2(s+1)(s^2+5s+6)^3 \\
        &+ (M_1s+C_1)(s+1)^3(s^2+5s+6)^2 \\
        &+ (M_2s+C_2)(s+1)^3(s^2+5s+6) \\
        &+ (M_3s+C_3)(s+1)^3
\end{align}$$

$$\begin{align}
   -\dfrac 52 (s + 3)(s^4 + 11 s^3 + 46 s^2 + 88 s + 68)
        &= K_1(s+1)(s^2+5s+6)^3 \\
        & +K_2(s^2+5s+6)^3 \\
        &+ (M_1s+C_1)(s+1)^2(s^2+5s+6)^2 \\
        &+ (M_2s+C_2)(s+1)^2(s^2+5s+6) \\
        &+ (M_3s+C_3)(s+1)^2
\end{align}$$

Let $s=-1$ again and you get...

Added 11/22/2020

Your original equation is wrong. It should look like

\begin{align}
   \frac{10}{(s+1)^2 (s+2)^3 (s+3)^3} &=\\
   &\frac{K_1}{(s+1)} + \frac{K_2}{(s+1)^2} + \\
   &\frac{L_1}{(s+2)} + \frac{L_2}{(s+2)^2} + \frac{L_3}{(s+2)^3} + \\
   &\frac{M_1}{(s+3)} + \frac{M_2}{(s+3)^2} + \frac{M_3}{(s+3)^3}
\end{align}

You then get the equation

\begin{align}
   10 &=
   K_1(s+1)(s+2)^2 (s+3)^3 \\
   &+ K_2(s+2)^3 (s+3)^3 + \\
   &+ L_1(s+1)^2 (s+2)^2 (s+3)^3 \\
   &+ L_2(s+1)^2 (s+2) (s+3)^3 \\
   &+ L_3(s+1)^2 (s+3)^3 \\
   &+ M_1 (s+1)^2 (s+2)^3 (s+3)^2 \\
   &+ M_2 (s+1)^2 (s+2)^3 (s+3) \\
   &+ M_3 (s+1)^2 (s+2)^3  \\
\end{align}

Now let $s=-1$. You get $10=8K_2 \implies K_2 = \dfrac 54$. Simplifying, we get

\begin{align}
   \dfrac 58 (s + 1) (9 s^5 + 124 s^4 + 685 s^3 + 1902 s^2 + 2668 s + 1528) &=
   K_1(s+2)^3 (s+3)^3 \\
   &+ L_1(s+1) (s+2)^2 (s+3)^3 \\
   &+ L_2(s+1) (s+2) (s+3)^3 \\
   &+ L_3(s+1) (s+3)^3 \\
   &+ M_1 (s+1) (s+2)^3 (s+3)^2 \\
   &+ M_2 (s+1) (s+2)^3 (s+3) \\
   &+ M_3 (s+1) (s+2)^3  \\
\end{align}

Again let $s=-1$. You get $-45 = 8K_1 \implies K_1 = -\dfrac{45}{8}$. Simplifying, we get

\begin{align}
   -\dfrac 54 (4 + s)(52 + 70 s + 39 s^2 + 10 s^3 + s^4)
   &= L_1 (s+2)^2 (s+3)^3 \\
   &+ L_2 (s+2) (s+3)^3 \\
   &+ L_3 (s+3)^3 \\
   &+ M_1 (s+2)^3 (s+3)^2 \\
   &+ M_2 (s+2)^3 (s+3) \\
   &+ M_3 (s+2)^3  \\
\end{align}


问:
$H$ and $K$ are two normal subgroups of group $G$, $H\cap K=\{1\}$, $G=HK$, how to prove $G$ is isomorphic to $H\times K$
If group $G$ has two normal subgroups $H$ and $K$ satisfying $H\cap K=\{1\}$ and $G=HK$, how to prove that $G$ is isomorphic to $H\times K$?

Since the factoring of any $g\in G$ as product $hk$ is unique under given conditions, I tried to prove that the map $\phi:G\ni g\mapsto (h,k)\in H\times K$ is the isomorphism. I can show that $\phi$ is a bijection. Next, I need to show it preserves group operation. Assuming $\forall g_1,g_2\in G$, $g_1=h_1k_1$ and $g_2=h_2k_2$ (the factoring is unique), we have $g_1g_2=h_1k_1h_2k_2$. But $\phi(g_1)=(h_1,k_1),\phi(g_2)=(h_2,k_2)$, so in product group $H\times K$, the operation $\phi(g_1)\phi(g_2)=(h_1h_2,k_1k_2)$. Its pre-image of $\phi$ in $G$ should be the  product $h_1h_2k_1k_2$. So we must establish $h_1k_1h_2k_2=h_1h_2k_1k_2$, which is equivalent to $k_1h_2=h_2k_1$. But I cannot prove this commutative relation based on the given conditions. The normality of $H$ implies $k_1^{-1}h_2k_1\in H$, but it does not necessarily equal $h_2$. This is where I got stuck. Is the above commutative relation true and how do I prove it? Or, was I wrong at the very beginning about what the isomorphism is? Could you please give me some help? Thank you.

答: Hint. For $h ∈ H$, $k ∈ K$, you have $hk = kh \iff hkh^{-1}k^{-1} = 1$.


问:
Surprisingly Countable or Uncountable Sets
By Cantor, we know that there are different cardinalities of infinite sets. $\mathbb{N}$ is countable, $\mathbb{Z}$ is countable and - rather surprisingly - $\mathbb{Q}$ is countable, too. And even more surprisingly, $\mathbb{R}$ is uncountable.

I sometimes show Cantor's arguments in high school (where I teach) and they and their consequences are - indeed and expectedly - very surprising to learn about.

What are some other infinite sets which are surprisingly countable or uncountable?

答:
The set of natural numbers is countable as you have noted, also the set of all orderd pair's of numbers are countable
(Intuitively this is because the size of this set is very close to the one of all rationals). Since a countable set A "generates" the countable set (A,A) (Thats not quite correct notation) we can use this argument to show that ((A,A),A)==(A,A,A) is alse countable, so any set of pairs or triples or millions of numbers is countable.</p>


Title: For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ .

Body:
I am trying to find $x+y$ given that

$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$

It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side

$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$

and see if a term $x+y$ comes out, but it looks hopeless at this point.

Body:
$$
\begin{align}
x+\sqrt{x^2+3}
&=\frac3{y+\sqrt{y^2+3}}\\
&=\frac3{y+\sqrt{y^2+3}}\frac{y-\sqrt{y^2+3}}{y-\sqrt{y^2+3}}\\
&=\frac3{-3}\left(y-\sqrt{y^2+3}\right)\\
&=-y+\sqrt{y^2+3}\tag{1}
\end{align}
$$
Similarly
$$
y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\tag{2}
$$
Add $(1)$ and $(2)$ and cancel the radicals.


问:
How do I find the smallest integer of $n$ such that $\sigma(n) = 24$?
How do I find the smallest integer such that $\sigma(n) = 24$?

My attempt :
I listed out all the number one by one until I have $n = 14$.
But can I get the smallest integer number of $n$ by using the theorem instead of by listing it out one by one.

Can anyone help me? Will appreciate it alot...

答:
I suppose you could "filter" some "reasonable guesses" in the following sense:

if $n = p$ is prime then $24 = 1 + p$, so $p = 23$ (and this is of course the unique prime with this property).

What other possibilities are there? Well, you can look at semiprimes less than $24$. Notice that if $n = pq$ with $p, q$ primes (at first sight not necessarily distinct!) then

$$\sum_{d \mid n} d = 1 + p + q + pq = (p+1)(q+1).$$

Write $24 = (p+1)(q+1)$ and see if you can find primes $p, q$ for which this is true. You'll notice that $p \neq q$, as otherwise $23 = p(p+1)$. Now look at the divisors of $24$ and you'll have your answer.


问:
proving a binary relation on $\mathbb{Z} × \mathbb{Z}$ is an equivalence relation.
Define a binary relation ∼ on $\mathbb{Z}×\mathbb{Z}$ by $(a, b) ∼ (c, d)$ if and only if $ad = bc$.

Prove that ∼ is an equivalence relation.

I'm trying to prove symmetric of this relation. Do I just show that $ab=bc$ and $bc=ab$ are equivalent by commutative property? how can I prove it?

答:
We have to show that $(a_1,a_2)\sim(b_1,b_2)\implies (b_1,b_2)\sim(a_1,a_2)$.

Now $$(a_1,a_2)\sim(b_1,b_2)\\\implies a_1b_2=a_2b_1\\\implies a_2b_1=a_1b_2\;\;\text{,since '=' is symmetric}\\\implies b_1a_2=b_2a_1\;\;\text{,since multiplication is commutative for reals}\\\implies (b_1,b_2)\sim(a_1,a_2)\,\,\square$$


问:
Show that in figure bisector of angle intersect at perpendicular
The following was asked by a high school student which I could not answer. Please help

In the figure below, show that the bisector of $\angle AEB$ and $\angle AFD$ intersect at perpendicular

答:
We claim that $\triangle FHP \cong \triangle FGP$. So $\angle FPG = \angle FPH$. So $\angle FPE = 90^0$.

Now how are they congruent?

$\angle FHP = \angle FAE + \angle AEH$

Also, $\angle FGP = \angle DCE + \angle CEP = \angle FAE + \angle AEH$

So, $\angle FHP = \angle FGP$ and we know $\angle PFH = \angle PFG$.


问:
Determine the number of these subsets for which the sum of the three elements is a multiple of 3. [Read Full Q. Below]

Consider the collection of all three-element subsets drawn from the set ${1,2,3,4,..., 299, 300}$. > Determine the number of these subsets for which the sum of the three elements is a multiple of 3.

This is a question from the Chapter Combinatorics but from what I understood of the question, I tried to find a pattern between subsets from which I found that the sum of all consecutive 3 positive integers is a multiple of 3, meaning if repetition isn't allowed, the three-element subsets would be $100$

I can't find a solution if repetition is allowed so please verify my answer if repetition isn't allowed and answer the question thinking it is allowed.

EDIT/UPDATE: IF repetition is allowed and we take the fact that the sum of all 3 consecutive integers is a multiple of 3 then each consecutive set {1, 2, 3},
{4, 5, 6}, will have 3 more sets for example the first will also be viable for
{2, 3, 4}, {3, 4, 5}; also, a set with the same number repeating 3 times would also be a multiple of 3 so the final answer would be $3 * 100 + 300 = 600$
Would this be correct?

答:
For your $S=\{1,2,3,\ldots, 300\}$ write
$S=A\cup B\cup C$ with
$$A=\{3,6,9,\ldots, 300\}=S\cap 3\Bbb Z  $$
$$B=\{1,4,7,\ldots, 298\}=S\cap(3\Bbb Z+1)  $$
$$C=\{2,5,8,\ldots, 299\}=S\cap(3\Bbb Z+2)  $$

There are $300\choose 3$ 3-subsets of $S$.
Subtract the $100\choose 3$ sets that are $\subset A$.
Subtract the $100\choose 3$ sets that are $\subset B$.
Subtract the $100\choose 3$ sets that are $\subset C$.
Subtract the $100^3$ sets that correspond to triples in $A\times B\times C$.


问:
So Ive been trying to solve the infamous 'question 6'
so I've set myself this question as a challenge and because of that do not want to read any solutions on the internet, so it would be appreciated if no one spoiled the answer, please just tell me if I'm incorrect

the question goes:
$$\frac{a^2+b^2}{ab+1}$$
prove that the expression above is equal to the square of an integer if $a$ and $b$ are positive integers.

Now, I've been able to prove that when $b = a^3$, the expression will equal $a^2$, but am I mistaken to believe that the expression has many different sets of numbers that will satisfy the question?

In that case, is this a proper solution or is there another one that can somehow describe all sets at once?

I haven't looked at any solutions on the internet as I don't want to ruin the challenge for myself.

答:
(Possibly too big) hint: Fix an integer $k$, for which there is a solution to $\frac{a^2+b^2}{ab+1}=k$. Let $a$ be unknown and see that there are two solutions with the same $b$ and $k$.

This forms a single link in a chain of solutions. Study this chain, and you have solved the problem.


问:
Infinite sample space in probability
My text book on probability says "sample spaces with an infinite number of elements are quite common. As an example, consider throwing a dart on a square target and viewing the point of impact as the outcome."

Assuming 'point of impact' denotes the point in the square target where the dart hits, does this infiniteness pertain to infinite number of points in a square? Can anyone give any other example of

Infinite number of elements in sample space, for a single well defined experiment (not like throwing a dice infinite number of times.)

"When dealing with probabilistic models
involving an uncountably infinite sample space, there are certain unusual subsets for
which one cannot associate meaningful probabilities." Any example for this case?

答:
Yes, the example given is making reference to the fact that there are an uncountable number of points inside a square on the real number plane.

A classic example of a sample space of infinite size is any continuous measurement, my favourite being human height. Theoretically, if your single experiment is measuring the height of the next person you meet, you could expect to see any measurement (of which there are an uncountable amount) between (let's be conservative here) 50 and 500 cm.

And I think the second point is hinting at the fact that not every subset of the real numbers is measurable. (See Vitali Set - Wikipedia for an example.) This is quite advanced stuff, but basically the foundation of probability is measure theory, which is the mathematical tool for measuring the size of a subset. Unfortunately, for the kind of measurement we usually use on subsets of the real numbers, (in short, for an interval $(a,b)$ its size is $b-a$) one can construct a subset of the reals like the Vitali set for which there is no sensible notion of its size.

Title: Infinite sample space in probability


问:
Decay rate of tensor product of series
Suppose I have a series $a_n = n^{-1}$. Now I define a double-indexed series $b_{m,n}:=a_m\cdot a_n$. And I reindex $b_{m,n}$ to a signle-indexed series $c_n$ so that $c_n$ is decreasing w.r.t. $n$. For example, the first several elements of $c_n$ is
\begin{equation}
c_1 = 1;\ c_2=c_3=1/2;
\end{equation}
Can I tell how $c_n$ depends on $n$ asymptotically? Is it $c_n\sim n^{-1}$? $c_n\sim n^{-2}$? or $c_n\sim exp(-n)$?

答:
For convenience of notation I am going to reindex the sequence so that $c_0 = 1, c_1 = c_2 = 1/2$, etc.

First notice that the sequence $(c_n)$ consists only of numbers of the form $1/m$, $m \in \mathbb{N}$, and that $1/m$ appears in the sequence exactly $d(m)$ times, where $d(m)$ is the number of divisors of $m$.  Therefore we can describe the general term of the sequence $(c_n)$ as
$$
c_n = \frac{1}{k(n)},
$$
where $k(n)$ is defined to be the largest $k$ such that $\sum_{m=1}^{k} d(m) \leq n$.

Now we can use the well-known elementary estimate $\sum_{m=1}^{k} d(m) = k\log k + O(k)$ (see for example here) to establish the asymptotic $k(n) \sim \frac{n}{\log n}$.  First we note that
$$
\sum_{m=1}^{n/\log n} d(m) = \frac{n}{\log n} \log \left( \frac{n}{\log n} \right) + O\left(\frac{n}{\log n}\right) = n - n \frac{\log\log n}{\log n} + O\left(\frac{n}{\log n}\right) < n
$$
for $n$ sufficiently large, thus $k(n) \geq \frac{n}{\log n}$.  For the reverse inequality, suppose that $k(n) > (1+\epsilon) \frac{n}{\log n}$ for infinitely many $n$.  Then we would have
\begin{align*}
n &\geq \sum_{m=1}^{(1+\epsilon)n/\log n} d(m) = (1+\epsilon) \frac{n}{\log n} \log \left( (1+\epsilon) \frac{n}{\log n} \right) + O\left(\frac{n}{\log n}\right) \\
&=  (1+\epsilon)n - (1+\epsilon) n \frac{\log \log n}{\log n} + O\left(\frac{n}{\log n}\right) \\
&= (1+\epsilon)n + o(n),
\end{align*}
a contradiction once $n$ is sufficiently large.

Thus we have established $k(n) \sim \frac{n}{\log n}$ and therefore the desired asymptotic is
$$
c_n \sim \frac{\log n}{n}.
$$

Title: Decay rate of tensor product of series


问:
If $1\in\bigcup_{i=1}^\infty A_i$ while $A_i=\{\sum_{j=1}^i\frac{9}{10^j}\}$?
Hi I found the general definition of union of sets in here.

So I would like to ask if $1\in\bigcup_{i=1}^\infty A_i$ while $A_i=\{\sum_{j=1}^i\frac{9}{10^j}\}$

From the definition, there must exists a set that make $1\in A_i$ so that $1\in\bigcup_{i=1}^\infty A_i$ sustains.

And for set it does not have the definition of distance. So does it simply implies that $1\not\in\bigcup_{i=1}^\infty A_i$ or there would be a definition of $i=\infty$ ??

答:
$\bigcup_{i=1}^\infty A_i$ is a notation for $\bigcup\limits_{i\in\Bbb N\setminus\{0\}}A_i$, i.e. for the set $X$ such that, for all $x$, $x\in X$ if and only if there is some natural number $i\ge1$ such that $x\in A_i$.

Therefore your $\bigcup_{i=1}^\infty A_i$ does not contain $1$.


问:
Common eigenvector of two linear transformation matrices
I have two linear transformation matrices

\begin{pmatrix}
3 & 2 \\
-2 & 1
\end{pmatrix}

and

\begin{pmatrix}
1-a & -a \\
a & 1
\end{pmatrix}

How to find out what the value of $a$ should be for them to have a common eigenvector?

答:
If two matrices $ \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}
$ have common  eigen vector the we will have :(suppose $(x_1,x_2)$ be that eigen vector):

$\begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_1 I \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_1\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

and

$\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_2 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

then we will have :

$3x_1+2x_2=\lambda_1 x_1=\frac{\lambda_1}{\lambda_2}\lambda_2 x_1=\frac{\lambda_1}{\lambda_2}((1-a)x_1-ax_2)$ $\to$

$(1-a)=\frac{\lambda_2}{\lambda_1}3$

and

$-a=\frac{\lambda_2}{\lambda_1}2$

so $a=-3\frac{\lambda_2}{\lambda_1}+1$ and $a=-2\frac{\lambda_2}{\lambda_1}$
so $-3\frac{\lambda_2}{\lambda_1}+1=-2\frac{\lambda_2}{\lambda_1}$ $\to $ $\frac{\lambda_2}{\lambda_1}=1$ and so $a=-2$

since a=-2 sarisfy in $-2x_1+x_2=\frac{\lambda_1}{\lambda_2}(ax_1+x_2)$ too they this can accept as unic solution.


问:
Hypergeometric probability of having less than half in a sample of X elements
I am solving a problem similar to this:

In a sample of 100 balls there are 33 red balls and 67 blue balls. What is the probability of picking randomly $X$ balls and have less than half red balls

I was just introduced to hypergeometric distirbutions and my solution was something like this:

(Imagining that $X$ = 13 and $K$ = number of red balls)

$P( K \leq 6) = \displaystyle\sum_{n = 0}^{6}\frac{{33 \choose n}{67 \choose 13-n}}{{100 \choose 13}}  \approx 91,63\%$

My problem (if my solution has no problem) is that as i am increasing the sample size, i notice a weird pattern in my probabilities of finding less than half of red balls (only the sample size increases, the number of balls for each color is the same)

This is the results i'm getting

$P(X = 14, K \leq 6) \approx 87,43\%$

$P(X = 15, K \leq 7) \approx 93,29\%$

$P(X = 16, K \leq 7) \approx 89,93\%$

Why everytime i increase my sample size but my $K$ remains the same the probability of having less than half of red balls lowers? Shouldn't it increase because i have a higher number of blue balls?

答:
There is nothing wrong with your answer, but your observation that the probability should increase as the sample size increases is incorrect.

Remember, we are computing cumulative probabilities. To understand what is happening,
let us analyse the first two cases, computing probabilities directly.

With sample size $13, P(0\; red) = \frac{67}{100}\frac{66}{99}...\frac{55}{88}$

wheras for sample size $14$, we need one more multiplier, $\frac{54}{87}=\frac{18}{29}$

On the other hand, for $P(6\;red)$ the multiplier for sample size $14$ taking into account changes in permutations will be $\frac{60}{87}\frac{14}{8} = \frac{35}{29}$


问:
Is entropy of prime numbers smaller?
Seems that entropy (in information theory) can be expressed as a measure of how unpredictable is each bit of information.

I have done a little experiment: I've measured entropy of the binary representation of numbers, and averaged the results for primes and composite numbers.

To calculate entropy, I use Shannon entropy, which can be expressed as $-\sum_{i}P(x_i)·log_2 (P(x_i))$.  This relates the appearance of each symbol (0 or 1) with how much it was expected to appear. For example, 5 is 101 in binary, so 1 is expected 2 times and 0 is expected once. From there, $P(0)=1/3$ and $P(1)=2/3$, and the entropy of 5 is 0.918296 bit/symbol.

Here are the results, showing range and average entropy for primes and composites:

```<code>1-9  Avg:  Primes: 0.4796 bit/symbol (4 items)  Composites: 0.7296 bit/symbol (5 items)
1-99  Avg:  Primes: 0.7795 bit/symbol (25 items)  Composites: 0.8763 bit/symbol (74 items)
1-999  Avg:  Primes: 0.8801 bit/symbol (168 items)  Composites: 0.9210 bit/symbol (831 items)
1-9999  Avg:  Primes: 0.9276 bit/symbol (1229 items)  Composites: 0.9403 bit/symbol (8770 items)
1-99999  Avg:  Primes: 0.9481 bit/symbol (9592 items)  Composites: 0.9541 bit/symbol (90407 items)
1-999999  Avg:  Primes: 0.9580 bit/symbol (78498 items)  Composites: 0.9625 bit/symbol (921501 items)
</code>```

Seems that primes have less entropy than composite numbers. I cannot go any further with my computer but, is there any reason why entropy in prime numbers should follow a particular rule?

(Source available at: https://gist.github.com/jjmontesl/2dcf59267b1a5f3827a4)

答:
This can be easily explained as this:

Shannon entropy:  The distance of an n dimensional rectangular solid object from an n dimensional hypercube of the same circumference.

If you take a close look at the formula of entropy: $-\sum_{i}P(x_i)*log_2(P(x_i))$, you can see that it has a maximum when the probabilities of all $x_i$ are exactly even, in this case it is for $x_1=x_2=1/2$.

Shannon entropy is here nothing more than the area of a rectangle with some rearranged sites of a circumference 4, the nearer to a perfect square the bigger will be it's entropy since a square always has the maximal area.

For the general case, Shannon entropy is the volume of the n-dimensional hyperrectangle of a circumference $3*2^{n-1}$, the further the arrangement of the sites is apart of a hypercube, the less is the volume and the entropy as such.

Now your question is easy to explain, since binary numbers like $111000, 110100,110010,101100,101010,100110$ can never be a prime number, since they are divisible by 2, while at the same time, the remaining combinations which are not divisible by two: $110001,101001,100101,100011$, can still be composite.

Title: Is entropy of prime numbers smaller?


问:
To solve an arctan sum $\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)}$
I was suggested (without any background setting) to prove a numerically checked result, which is dramatically refined
$$
\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)} = \frac{\pi}{12}
$$
where $a_n=(1+\sqrt3)^{4^n}$, with $4^n$ on shoulder.

Usually we deal with arctan series using telescoping or gamma function technique, but this $a_n$ holds a geometrical growth and I have no idea how to do the reduction. Thanks in advance for any suggestion.

答:
Let $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. Since $a_{n+1} = a_n^4$, we have
$b_{n+1} = \frac{\sqrt{3}}{1 + 2a_{n+1}} = \frac{\sqrt{3}}{1 + 2a_n^4}$.

With help of a CAS, one can verify

$$\frac{\frac{\sqrt{3}}{1+2x} - \frac{\sqrt{3}}{1+2x^4}}{1 + \frac{\sqrt{3}}{1+2x}\frac{\sqrt{3}}{1+2x^4}}
= \frac{\sqrt{3}x(x-1)}{(x+1)(2x^2-3x+2)}
$$
Substitute $x$ by $a_n$, $x^4$ by $a_{n+1}$ and sum over $n$. The sum
at hand becomes

$$\begin{align}\sum_{n=0}^\infty \tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}
&= \sum_{n=0}^\infty\left(\tan^{-1}b_n - \tan^{-1}b_{n+1}\right)
= \tan^{-1}b_0\\ &= \tan^{-1}\frac{\sqrt{3}}{3+2\sqrt{3}}
= \tan^{-1}\frac{\frac12}{1+\frac{\sqrt{3}}{2}}
= \tan^{-1}\frac{\sin\frac{\pi}{6}}{1+\cos\frac{\pi}{6}}\\
&= \frac12\cdot \frac{\pi}{6} = \frac{\pi}{12}
\end{align}
$$

Update

In case anyone wonder how the magic $b_n$ is chosen. The basic idea is to rewrite the summands into the form $\tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}$ so that one can use  addition formula of tangent to turn the sum into a telescoping one. Since the summands have the form $\tan^{-1}\sqrt{3}(\cdots)$, one expect $b_n$ equals to $\sqrt{3}$ multiply something simple in $a_n$.

We know the series sum to $\tan^{-1}b_0$. Since $\tan\frac{\pi}{12} = \frac{\sqrt{3}}{3+2\sqrt{3}}$ and $a_0 = 1+\sqrt{3}$. The simplest candidate of $b_n$ comes to my mind is $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. It turns out this works.


问:
Finding one solution of second order DE using another solution using Wronskian
I want to check whether following method works in general for finding a solution of the $2^{nd}$ order DE when one solution is given.

$$y''+P(x)y'+Q(x)y=0$$

Let $y_1$ be a solution of the DE and assume that $y_2=v(x)\cdot y_1$ is a solution where $v(x)$ is a function of $x$ and not any constant in $\mathbb{R}$.
Hence $y_1$ and $y_2$ will be linearly independent thus making the Wronskian non-zero. Thus,
$$\begin{vmatrix}
y_1 & y_2 \\ y_1' & y_2'
\end{vmatrix}=\begin{vmatrix}
y_1 & v\cdot y_1 \\ y_1' & v\cdot y_1' +v'\cdot y_1
\end{vmatrix}\neq 0\\
\therefore\ v'y_1^2=c_1,\qquad c_1\in \mathbb{R}\\
\therefore\ dv=c_1(y_1^{-2})dx\\
\therefore\ v=c_1\int \frac{dx}{y_1^2}\\
\therefore\ y_2=y_1c_1\int \frac{dx}{y_1^2}\\
\therefore\ y_2=y_1c_1\cdot (I +c_2)$$

So, the only question that remains is finding $c_1\ \&\ c_2$ which can be found by putting some constraints, right?

答:
The Wronskian of this ODE
$$y''+Py'+Qy=0~~~~(1)$$
is given as
$$y_1y_2'-y_1'y_2=\exp[-\int P dx]$$
Let $y_2(x)=v(x)y_1(x)$, then it follows that
$$v'y_1^2=\exp[-\int P dx] \implies v=\int \frac{\exp[-\int {P dx}]}{y_1^2} dx~~~(2)$$
Hence if $y_1$ is one solution of (1), then $y_2(x) = v(x) y_1$ is its other solution. The total (general) solution is given by $y(x)=C_1y_1(x)+C_2 v(x) y_1(x)$, where $C_1, C_2$ are calculated by the additional given particular conditions in a question.

Note that OP's expression for $v(x)$ is incorrect, it should be as given in (2).


答:
Here are some thoughts on the matter: I believe there is no general algorithm except trial and error.

By "formula" you probably mean $\sigma(n) = \prod_{p|n, p \text{ prime}}(1+p+p^2+\dots+p^k)$. This could help in some cases, in particular if $\sigma (n)$ has very few divisors:

Suppose we are to find the smallest $n$ such that $\sigma(n) = 124 = 2^2\times 31$.

The factors of $124$ and $1,2,4,31,62,124$, so the sum of prime powers must equal to one of them.

$2$ is never attained, so we do not need to consider $62$. By brute force:

\begin{align}
4 &=1+3\\31&=1+2+4+8+16\\31&=1+5+25
\end{align}

and $124$ is never attained. This shows that $n = 3\times 16$ or $n = 3 \times 25$, so the minimum $n$ is $48$.

However this does not help with $\sigma(n)=24$, since $24$ has too many divisors.

There are other properties of the sigma function that can narrow your search:

$\sigma(p) = 1+p$ for prime $p$

$\sigma(n)$ is odd iff $n$ is a square or twice a square

A naive bound: $\sigma (n) \le n+T_{n/2}=n+\frac {(n/2)(n/2+1)}2=\frac18(n^2+10n)$

In this case, the first criteria eliminates $2,3,5,7,11,13$, and the second elminiates $4,8,9$, and so $6,10,12,14$ remains. The bound can only eliminate the cases $n \le 9$.

Also check out https://oeis.org/A051444: Smallest k such that sigma(k) = n, or 0 if there is no such k

Title: How do I find the smallest integer of $n$ such that $\sigma(n) = 24$?

Body:
How do I find the smallest integer such that $\sigma(n) = 24$?

My attempt :
I listed out all the number one by one until I have $n = 14$.
But can I get the smallest integer number of $n$ by using the theorem instead of by listing it out one by one.

Can anyone help me? Will appreciate it alot...


问:
If $I_n=\int_{-\pi}^{\pi} \frac{\sin nx}{(1+\pi^x)(\sin x)}dx$, then prove the following

$I_n=I_{n+2}$

$ \sum_{m=1}^{10} I_{2m+1} =10 \pi$

$\sum _{m=1}^{10} I_2m =0$

My simplification of the given integral is
$$I_n =\int_0^{\pi} \frac{\sin nx}{\sin x} dx$$

Where $n$ is an integer

I believe the expression needs further simplification before the above statements can be proven, but I am unable to achieve that. How should I proceed?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\left.\vphantom{\Large A}I_{n}\,
\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq 1}}\ & \equiv
\bbox[5px,#ffd]{\int_{0}^{\pi}{\sin\pars{nx} \over \sin\pars{x}}\dd x} =
\int_{-\pi/2}^{\pi/2}{\sin\pars{nx + n\pi/2} \over \cos\pars{x}}\dd x
\\[5mm] & =
\int_{0}^{\pi/2}{\sin\pars{nx + n\pi/2} + \sin\pars{-nx + n\pi/2} \over \cos\pars{x}}\dd x
\\[5mm] & =
2\sin\pars{n\pi \over 2}\
\underbrace{\int_{0}^{\pi/2}{\cos\pars{nx} \over \cos\pars{x}}\dd x}
_{\ds{\cal J_{n}}}
\\ & \underline{\mbox{Note that}\ \color{red}{I_{n} = 0}\ \mbox{whenever}\ n\
\mbox{is}\ \color{red}{even}}
\\[2mm] & \mbox{and}\
\left.2\sin\pars{n\pi \over 2}\,\right\vert_{\,n\ \color{red}{odd}} = 2\pars{-1}^{\pars{n - 1}/2}.
\label{1}\tag{1}
\end{align}

With $\ds{n \equiv 2k + 1}$ where $\ds{k \in \mathbb{N}_{\,\geq\ 0}\,\,}$:
\begin{align}
\left.\vphantom{\Large A}{\cal J}_{n}
\,\right\vert_{\ds{\,n\ odd}} & =
{\cal J}_{2k + 1}\
\equiv
\bbox[5px,#ffd]{\int_{0}^{\pi/2}
{\cos\pars{\bracks{2k + 1}x} \over
\cos\pars{x}}\,\dd x}
\\[5mm] & =
\Re\int_{0}^{\pi/2}{\expo{\ic\pars{2k + 1}x} - \expo{\ic\pars{2k + 1}\pi/2} \over \cos\pars{x}}
\,\,\dd x
\\[5mm] & =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\
{z^{2k + 1} - \pars{-1}^{k}\,\ic \over
\pars{z + 1/z}/2}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
\left.-2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\
{\pars{-1}^{k}\,\ic - z^{2k + 1} \over
1 + z^{2}}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
2\ \underbrace{\Im\int_{1}^{0}\
{\pars{-1}^{k}\,\ic - \ic^{2k + 1}\,\,y^{2k + 1} \over
1 - y^{2}}\,\ic\,\dd y}
_{\ds{= \color{red}{0}\ \mbox{because the integrand}\ \in\ \mathbb{R}}}
\\[2mm] &\
+ 2\,\Im\int_{0}^{1}
{\pars{-1}^{k}\,\ic - x^{2k + 1} \over
1 + x^{2}}\,\dd x
\\[5mm] = &\
2\pars{-1}^{k}\int_{0}^{1}
{\dd x \over 1 + x^{2}}
\bbx{{\pi \over 2}\pars{-1}^{k}} \\ &
\end{align}

Finally, see (\ref{1}),
$$
\mbox{}\\
\left.\vphantom{\Large A}I_{n}\,
\right\vert_{\,n\ \in\ \color{red}{\mathbb{Z}}}\ =
\bbx{\left\{\begin{array}{lcl}
\ds{-I_{-n}} & \mbox{if} & \ds{n \leq 0}
\\[2mm]
\ds{0} & \mbox{if} & \ds{n > 0\ \mbox{is}\ even}
\\[2mm]
\ds{\pi} & \mbox{if} & \ds{n > 0\ \mbox{is}\ odd}
\end{array}\right.} \\
$$

The three statements at the beginning of the OP question are $\underline{correct}$ but the first one when $\ds{n = -1}$.


问:
Group actions and natural isomorphisms
Let $G$ be a group-as-category, and let $S$ be the image of $G$ by the $Hom(G,-)$ functor. The $Hom$ functor defines a bijection $\chi: G \to S$ between elements of $S$ and morphisms of $G$, and thus $S$ is equipped with a canonical group action $\psi$ given by $\psi(g,s) = \chi(g\cdot\chi^{-1}(s))$.

However, one can define many other group actions, given an element $h$ of G, by $\psi'(g,s) = \chi(h \cdot g \cdot h^{-1} \cdot \chi^{-1}(s))$. It seems to me that these group actions are related to the natural isomorphisms of the $Hom(G,-)$ functor. If $h=(\chi^{-1}(s))^{-1}$, we can even get the right group action on $s$.

My question is: what framework should I use if I'd like to use all these group actions on $S$ and not just the canonical one ?

答:
I'm not sure if it is what you are looking for. However, considering the category $\mathcal C$ with a unique object $\ast$ and the monoid $\hom(\ast,\ast)$ being the group $G$, the category $\mathbf{Sets}^\mathcal{C}$ of functors
$$ \mathcal C \to \mathbf{Sets} $$
is the category of $G$-sets and $G$-sets morphisms.

Then, the functor $\hom(\ast, -) : \mathcal C \to \mathbf{Sets}$ is the $G$-set $G$ together with the action $G\times G \to G,\, (g,x) \mapsto gx$. Yoneda's lemma states that for each $G$-set $S$, there is a $G$-sets isomorphism
$$ \hom_{G-\mathrm{sets}} (G, S) \cong S $$
natural in $G$. Moreover, this isomorphism is given by
$$ \varphi \mapsto \varphi(e) \qquad (e\text{ unit of }G).$$

Title: Group actions and natural isomorphisms


问:
$\int_0^\infty \frac{\sin x}{x}dx$ using Feynman's technique
While watching this video, he says that

$$\lim_{b\to\infty}\int_0^\infty \frac{\sin x}{x}e^{-bx}dx\to 0$$

since the integrand tends to $0$. However I think that it must be an indeterminate form, since $\infty\times0$ may not always be $0$

Am I right in saying this? If yes, how else can we evaluate the integration constant?

Reference picture:

答:
As Professor Vector comments, since $\left|\frac{\sin(x)}x\right|\le1$,
$$
\begin{align}
\left|\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x\right|
&\le\int_0^\infty e^{-bx}\,\mathrm{d}x\tag1\\
&=\frac1b\tag2
\end{align}
$$
Thus, the integral vanishes as $b\to\infty$.

We can then evaluate the integral using Feynman's technique:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}b}\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x
&=-\int_0^\infty\sin(x)\,e^{-bx}\,\mathrm{d}x\tag3\\
&=-1+b\int_0^\infty\cos(x)\,e^{-bx}\,\mathrm{d}x\tag4\\
&=-1+b^2\int_0^\infty\sin(x)\,e^{-bx}\,\mathrm{d}x\tag5\\
&=-\frac1{1+b^2}\tag6
\end{align}
$$
Explanation:

$(3)$: differentiate under the integral (Feynman's technique)

$(4)$: integrate by parts

$(5)$: integrate by parts

$(6)$: add $\frac{b^2}{1+b^2}$ times $(1)$ to $\frac1{1+b^2}$ times $(3)$

Using $(2)$, solving $(6)$ gives
$$
\begin{align}
\int_0^\infty\frac{\sin(x)}x\,e^{-bx}\,\mathrm{d}x
&=\frac\pi2-\tan^{-1}(b)\tag{7}\\
&=\tan^{-1}\left(\frac1b\right)\tag{8}
\end{align}
$$


问:
Fourier transform of integral operator
I would like to know which is the fourier transform of integral operator:

$$Tf(x)=\int_{-\infty}^{+\infty}\quad f(x)dx\rightarrow \hat{T}\hat{f}(k)$$

I know that (is it right?):

$$\int_{-\infty}^{+\infty}\quad x^2f(x)dx\rightarrow -\frac{\partial^2}{\partial k^2}\hat{f}(k)|_{k=0}$$

Thank you

答:
You seem to be mistaken here. The Fourier transform is a  integral operator (when acting on certain function spaces). It takes a function and gives another function. So $\mathcal{F}:f\rightarrow \hat{f}$. What you have written here doesn't make sense because you are asking for the Fourier transform of a constant (which makes sense distributionally but not in the classical sense) since definite integrals are just constants. Additionally, all information is lost about $f$ in what you have written. What you want to say is that the Fourier transform maps differentiation to multiplication and vice versa which is true. To be more precise:

$$ \int_{\mathbb{R}} \frac{e^{-ikx}}{\sqrt{2\pi}} x^2f(x)dx = -\frac{d^2\hat{f}}{dk^2} $$

Title: Fourier transform of integral operator


问:
How do I find the smallest integer of $n$ such that $\sigma(n) = 24$?
How do I find the smallest integer such that $\sigma(n) = 24$?

My attempt :
I listed out all the number one by one until I have $n = 14$.
But can I get the smallest integer number of $n$ by using the theorem instead of by listing it out one by one.

Can anyone help me? Will appreciate it alot...

答:
Let's look for methods to find minimal $n$ with $\sigma(n)=m$ for larger $m$, where the one-by-one method is certainly not good enough or often not even feasible.

Recall that for prime powers $n=p^k$, we have $$\sigma(p^k)=1+p+p^2+\cdots +p^k=\frac{p^{k+1}-1}{p-1}=\frac{pn-1}{p-1}=n+\frac{n-1}{p-1}.$$
Also,  $\sigma$ is multiplicative.
So we might first look for prime $p$ that divide $m-1$ and check if $\frac{(p-1)m}p$ is a power of $p$. This will let us determine the set
$$A(m):=\{\,n\mid \sigma(n)=m\land n\text{ is a prime power}\,\}. $$
Do the same for all divisors $d$ of $m$, i.e., determine all $A(d)$ for $d\mid m$.
What we want is to find
$$ B(m):=\{\,n\mid \sigma(n)=m\,\}$$
(and ultimately its smallest element). To this end, note that
$$ B(m)=\bigcup_{d\mid m}\{\,xy\mid x\in B(d),y\in A(\tfrac md), \gcd(x,y)=1\,\},$$
which allows us to construct all $B(d)$ for $d\mid m$ recursively.


问:
What does $\mathbb Z[x]$ mean?
I’ve seen usage of the notation $\mathbb Z[x]$ before, where $x$ is some non-integer. For example in the Wikipedia article on the Gaussian integers it says that the set of Gaussian integers can be written as $\mathbb Z[i]$. But what does it actually mean?

I suspect that $\mathbb Z[x]$ refers to the set of numbers of the form $a+bx$, where $a,b\in\mathbb Z$, but I’m not sure.

答:
$\mathbb Z[x]$ is the ring of all polynomials over the variable $x$ with integer coefficients. So this includes $a+bx$ (for all $a,b \in \mathbb Z$), but also things like $x^2$, and the other higher order polynomials.

Whereas $\mathbb Z[i]$ is what we get when we evaluate the polynomials at $i$. That is, we take an integer polynomial $f(x)$ and plug in $i$ for $x$ to get $f(i)$, a complex number which we can always write in the form $a +bi$ for some integers $a,b$.

Title: What does $\mathbb Z[x]$ mean?


问:
For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ .
I am trying to find $x+y$ given that

$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$

It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side

$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$

and see if a term $x+y$ comes out, but it looks hopeless at this point.

答:
$$
p = x + \sqrt{x^2+3} \Rightarrow \sqrt{x^2+3} = p-x \\
p^2 = x^2 + x^2+3 + 2x \sqrt{x^2+3} = 2x^2+3 + 2x(p-x) = 2xp+3 \\
x = \frac {p^2-3}{2p}
$$
Analogously, if $y + \sqrt{y^2+3} = q$, then
$$
y = \frac {q^2-3}{2q}
$$
Initial expressions becomes $pq = 3$. Now, if calculate $x+y$ you'll get
$$
x+y = \frac {p^2-3}{2p} + \frac {q^2-3}{2q} = \frac 12 \left (\frac {p^2q-3q+q^2p-3p}{pq} \right ) = \frac {p+q}2 \left( pq-3\right) = 0
$$</p>


问:
Does completion of $(0,1)$ exists and does there is a metric where $[0,1]$ is not complete
This question was asked in a master's exam for which I am preparing and I was unable to find out which result I should use. So, I am posting it here.

Question: (i) does there exists a metric on which $(0,1)$ is complete? (More generally can you please tell if there always exits a metric such that the space which is fixed while finding the metric is complete?)

(ii) Does there exists a metric such that $[0,1]$ is not complete?

I studied analysis from Apostol but unable to think on how should I approach this particular question (i.e. which result I should) use. There were 2 more options in question but I worked them out.

答:
Suppose you have a metric space $(X, d)$, a set $Y$ (no topological features necessary), and a bijection $\varphi : Y \to X$. Then
$$d' : Y \times Y \to \Bbb{R} : (y_1, y_2) \mapsto d(\varphi(y_1), \varphi(y_2))$$
defines a metric such that $\varphi$ becomes an isometry between $(X, d)$ and $(Y, d')$. In particular, $(Y, d')$ is complete if and only if $(X, d)$ is complete.

In this case, we can put $(0, 1)$ in bijection with $[0, 1]$, both having cardinality of the continuum. If we inherit the metric from $[0, 1]$, then $(0, 1)$ becomes complete. If we inherit the metric from $(0, 1)$, then $[0, 1]$ becomes incomplete.

EDIT: You can also make $(0, 1)$ complete by putting it into bijection with $\Bbb{R}$ using a homeomorphism like
$$\varphi : (0, 1) \to \Bbb{R} : x \mapsto \tan\left(\pi\left(x - \frac{1}{2}\right)\right).$$
This shows that you can preserve the topology of $(0, 1)$, but yet change whether the space is complete.

On the other hand, we cannot form a homeomorphism between $[0, 1]$ and an incomplete space, as the continuous image of $[0, 1]$ is necessarily compact, and hence complete.


问:
Question about proof of $mG$ is subgroup of $G$
This is from Victor Shoup's book

Let $G$ be an abelian group, and let $m$ be an integer. Then
$mG := \{ma : a \in G\}$ is subgroup of $G$.

Proof. The set $mG$ is non-empty, since $0_G = m0_G \in mG$. For $ma, mb \in mG$, we have $ma + mb = m(a + b) \in mG$, and $−(ma) = m(−a) \in mG$.

My question about this is the above proof does not prove that $ma \in G$. Isn't that necessary for $mG$ to be a subgroup of $G$?

答:
As per OP's request, I am extending my comment into answer.

For $m \in \mathbb N$, we indeed write $ma := \underbrace{a+a+\dots+a}_{m \ a\text{'s}}$. However this is usually extended to $m\in\mathbb Z$:

For $m=0$, $ma:=0$, the identity of the group.

For $m<0$, $ma:=-((-m)a)$, where the first negative sign stands for the inverse.

In all these cases, we can see that $ma \in G$ by using the closure property of the group $G$ and induction on $m$. Hence $mG \subseteq G$.


问:
Prove that a planar and triangulated graph is $k$ connected if and only if it has no separating cycle of length at most $k−1$
Let G be a  planar graph with a fixed  planar drawing $Γ$. A separating cycle of G is a cycle $C$ such that the curve traced by $C$ in $Γ$  has at least  one vertex strictly inside and strictly outside.

Note: $uwv$ is a separating cycle .

I’d like to prove that a planar and triangulated graph is k connected  if and only if it has no separating cycle of length at most k−1.

The necessary condition is obvious.
I was stuck in a proof of sufficient condition.  I 'd like to know the reason why we will get $G$ is $k$ connected when the triangulated planar graph $G$ has no separating cycle of length at most $k−1$.

Some definitions and  theorems  that may be used:

A connected graph G is said to be
k-vertex-connected (or k-connected) if it has more than k vertices
and remains connected whenever fewer than k vertices are removed.

[Menger's theorem] An equivalent definition is that a graph with at least two
vertices is $k$-connected
if, for every pair of its vertices, it is possible to find k vertex-independent
paths connecting these vertices.

A planar graph $G$ is called triangulated if every face is bounded by a triangle.

答:
Suppose not, i.e. for some plane triangulation there is a separating set $S$ of size $k-1$ which does not contain a cycle.

Since $S$ is separating, we can divide the remaining vertices into two non-empty sets $T,U$ such that no edge connects $T$ to $U$. But since $S$ does not contain a cycle, the region of the plane remaining after removing $S$ and all edges spanned by $S$ is path-connected. In particular, there is a curve in the plane between a vertex in $T$ and a vertex in $U$.

Look at sequence of faces of the triangulation this curve passes through. Note that we can't move directly from a face with all vertices in $S\cup T$ to one with all vertices in $S\cup U$, since the common vertices to these faces would both be in $S$, but we never cross an edge between vertices of $S$. So the first face we encounter with a vertex in $U$ also contains some vertex in $T$. Since this face is a triangle, there is an edge between these vertices, contradiction.


问:
Why are circles of opposite directions not homotopic to each other?
I’m learning about homotopy and the fundamental groups, and I’m having a hard time visualizing the transformation. In particular, I can’t see when there is no homotopy between loops.

An example from my class is the following picture (the loops are all confined to $S^1$. I draw them that way just to make clear what paths are being travelled).

I can see how the first loop could be deformed into the second one by doing less and less of the “repeated” portion.

Why can’t we have something similar, for example as shown below, to deform a loop into another loop of the opposite direction? Is the condition of continuity violated somehow?

Apologize for the bad quality of the pics.

=======

Edit: I’m actually thinking of the fundamental group of the circle $S^1$. My understanding is that the group is comprised of the equivalence classes of loops with a base point $x_0$ (here I take it to be $(1,0)$).
In what sense is a loop goes a full circle from $(1,0)$ counterclockwise not the same as a loop goes a full circle from $(1,0)$ clockwise? I don’t see how the procedure in the second picture does not describe a homotopy between the two.

答:
It may take a while to identify the origin of the confusion. I can no longer help with that in the space of a comment, so an answer it is.

Here is the first problem that the extra image by the OP revealed. When going from picture 1 to picture 2, they seem to have in mind the homotopy shown in the animation below. Reload the page, if the animation has stopped. I have forgotten how to prepare a gif-file that runs continuously.

And now the problem is apparent. Instead of the picture in 1, one of the endpoints of this homotopy is the path around the circle that first traverses it counterclockwise, and then traces it back clockwise. This is homotopic to the identity element of $\pi_1(S^1,x_0)$, it is not the generator. After all, it is easy to see how to continue this homotopy and eventually shrink the path to the "constant" path at $x_0$.

In yet other words, we have a homotopy between this

and this (the path doing a 180 degree turn at the half point)

but the intended target path

in your figure 1 is nowhere to be seen.


问:
Big-O Estimate Problem ("Tightest Big-O Estimate")
For $f(n) = (2^n +n^n)(n^3+n\text{log}(n^n))$, what is the tightest big-O estimate?

My attempt is the following:

We can break this into two product statements: $(2^n +n^n) \text{ and }(n^3+n\text{log}(n^n))$

Focusing on the left most: $(2^n + n^n) \le (n^n + n^n) \text{ then } O(n^n)$

Now the right most: $n^3+n\text{log}(n^n) \rightarrow n^3 + n^2\text{log}(n) \le n^3 + n^3 \text{ then } O(n^3)$

Then all together we have: $O(n^{n+3})$

Is this the correct way to calculate this? Super new to this, and I have to admit, getting to the $O$ estimation is... strange, and I don't feel like my justification is correct.

答: The basic strategy is correct, but the way it's normally explain would be slightly different. Since $n\log(n^n)=n^2\log n\in o(n^3)$, $n^3+n\ln(n^n)\in O(n^3)$ (or which is stronger, $n^3+n\ln(n^n)\sim n^3$). Similarly, $2^n\in o(n^n)$ implies $2^n+n^n\sim n^n$. So $f\sim n^{n+3}$. This stronger than $f\in O(n^{n+3})$; it's also sufficient to prove no tighter big-$O$ characterization is possible.


问:
If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?

Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$

$(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$

$(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$

$(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated.

Thanks for your time.

答:
There already is an accepted answer, but I wanted to point out a more self-contained argument. A way to prove 2) is to use the following elementary lemma: if $p=3$ mod $4$ is a prime and $a,b$ are integers such that $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$, thus $p^2\mid a^2+b^2$.

This lemma implies that for any prime $p=3$ mod $4$ and any integers $a,b$, $v_p(a^2+b^2)$ is even, which shows 2).

Now, how to prove the lemma?

Assume we have $p\mid a^2+b^2$ and, say, $p$ does not divide $a$. Let $a'$ be its inverse mod $p$; take $b'=ba'$. Then $p\mid b'^2+1$. As $\frac{p-1}{2}$ is odd, $(b')^2+1\mid (b')^{2\times (p-1)/2}+1$, so $p\mid (b')^p+b'$. But by Fermat's little theorem, $p|(b')^p-b'$ so $p\mid 2b'$. As $p \neq 2$, $p\mid b'$, so (as $a'$ is coprime to $p$) $p\mid b$. Therefore $p\mid a^2$ so $p\mid a$, a contradiction.