问:
Unbounded open set whose characteristic function is integrable in extended sense
Put simply, I would like to know if there is an unbounded open set (of $\mathbb{R}^n$) whose characteristic function is integrable in extended sense. I get the suspicion that something like the area between the graph of $f(x)=\frac{1}{x^2}$ for $1<x<\infty$ in $\mathbb{R}^2$ would do the trick, but I'm stuck extending this to higher dimensions.</p>

答:
Expanding on Dennis's answer:

Let $n \geq 2$ and let $f: \mathbb{R}^{n-1} \to [0,\infty)$ be a continuous integrable function whose support is all of $\mathbb{R}^{n-1}$. Its subgraph $U = \{(x,y) \in \mathbb{R}^{n-1}\times \mathbb{R}\,:\,0 \leq y \leq f(x)\}$ is an example of the kind of set you're looking for. Indeed, by Tonelli's theorem,

$$\mu(U) = \int_{\mathbb{R}^{n}} \chi_{U}\, d\mu^{n} = \int_{\mathbb{R}^{n-1}} \int_{\mathbb{R}} \chi_{U}\, d\mu^{1} \, d\mu^{n-1}  = \int_{\mathbb{R}^{n-1}} f \, d\mu^{n-1} \lt \infty$$

since Lebesgue measure $\mu^{n}$ on $\mathbb{R}^{n}$ is the product measure $\mu^{n-1} \times \mu^{1}$.

To make this completely explicit, take e.g. $f(x) = e^{-\|x\|^2}$ whose integral is $\pi^{(n-1)/2}$ (after Poisson).

A way of constructing an example that works for all $n \geq 1$: let $x_{k} = (k,0,\ldots,0)$ for $k =1,2,\ldots$. Choose an open ball or cube $U_{k}$ of volume $1/{2^k}$ centered around $x_{k}$. Then the union $U = \bigcup U_{k}$ will have volume $1$ in $\mathbb{R}^{n}$ since the sets $U_{k}$ have pairwise empty intersection.


问:
Prerequisites for Quantum cryptography
Sir i am from mathematics background. I am aware of pure mathematics subjects. My university is offering quantum cryptography in Jan semister. I am an absolute beginner. Can you please list down the prerequisite courses for quantum cryptography?. I have recently joined graduate studies. I want to pursue research in a new area. So this word of "quantum" struck me. When i googled i found another course called quantum computing. Apart from linear algebra, probability(till what extent?), what are other prerequisites? how quantum cryptography differs from normal cryptography course?( i have not done cryptography but i am aware of basic number theory concepts). How quantum cryptography differs from quantum computing? Do i need to have quantum theory background of physics? Is there any connection between this quantum cryptography and blockchain or machine learning please englighten me in layman language. Please tell me possible research areas in this subjects and its interconnection with graph theory, combinatorics, machine learning etc if any Please suggest me some books/resources to get acquantance with the subject?</p>

答:
Well, in view of quantum computing, you need functional calculus (incl. tensor products) and complex numbers. This is sufficient at the algorithmic level.

It will depend whether the course will include quantum models for the implementation of the cryptographic protocols in question. Then you would need some quantum mechanics as well.

ContentLicense: CC BY-SA 4.0

Title: Prerequisites for Quantum cryptography


问:
Surface of a cylinder is a differentiable manifold
I'm struggling to understand the concept of a differentiable manifold. Lets look at an example. Let $S\subset\Bbb{R^3},S=\{(x_1,x_2,x_3)\vert \enspace x_1^2+x_2^2=1\}.$ This is a cylindrical surface with a radius of $1$. Let $g:U_1=(0,2\pi)\times\Bbb{R}\to \Bbb{R^3},$ and $h:U_2=(-\pi,\pi)\times \Bbb{R} \to \Bbb{R^3}$, such that $$g(\phi,z) = (\cos\phi, \sin\phi, z)$$

and $$h(\phi,z) = (\cos\phi,\sin\phi, z).$$

Now $S=g(U_1) \cup h(U_2).$ Since $g,h$ are continuously differentiable homeomorphisms, $S$ is a differentiable manifold. Both $g$ and $h$ look like cylindrical coordinate transforms, but why do we need two of them here? What would happen if we exclude $h$?

答:
You're right that the formulas for $g$ and $h$ are identical, but we really need both. For example, if we only used $g$, then because the domain is $(0,2\pi)\times \Bbb{R}$, this parametrization doesn't cover the entire cylinder $S$; i.e $S\setminus g(U_1)$ is non-empty (geometrically, it is a vertical line). The concept of a manifold requires us to show that around EVERY point of $S$, we can find an appropriate parametrization (very roughly speaking, if you take any "small" region of $S$, then it must "look" very nice).

But like I said, if we use $g$ alone, we fail to describe what happens at all those points on the line. This is why we need $h$ to guarantee that even at those points which we originally missed, everything is still nicely behaved.

Of course, when we use $h$, we also fail to cover the whole of $S$, we again miss out a vertical line. The thing is this is a different vertical line; so although both $g$ and $h$ miss out stuff, together their images cover $S$.

Otherwise, think of the graph of $f(x)=|x|$. This is "nice and smooth" everywhere except at the origin, where it is pointy and "badly behaved".


问:
Where does the following formula come from
Where does the following formula come from:
$\sum_{i=1}^n i \cdot  2^i $
$= ( − 1)2^{n+1}  + 2$

Body:
$$S=x+2x^2+3x^3+4x^4+....+nx^n~~~~(1)$$
Is AGP, multiply by x and write
$$xS=~~~~~~~~~~~~~~~~~~~~~~~~~~x^2+2x^3+3x^4+.....(n-1)x^n+nx^{n+1}~~~(2)$$
Suntract (2) from (1), to get
$$(1-x)S=x+(x^2+x^3+x^4+...+x^n)-nx^{n+1}$$
Use the sum of GP here
$$x+x^2+x^3+...x^n=x\frac{x^n-1}{x-1}$$
Then $$S=\frac{x^{n+1}-x}{-(x-1)^2}-\frac{n x^{n+1}}{1-x}$$
Pur here $x=2$, to get $$S=2^{n+1}(n-1)+2$$


问:
Show that $f$ is constant on $\mathbb{C}$

Suppose $f$ is entire. Let $a\in C $ and $f$ does not take any
values in the open disc $D_r(a)$. Show that the function $f$ is
constant.

Please tell me if my solution is correct:

$f $ is entire $\rightarrow h(z) = f(z) - a$ is entire and non zero $\rightarrow g(z) = \frac{1}{f(z) - a}$ is entire.

since $f$ does not take any values in the open disc $D_r(a) \rightarrow |f(z) - a| \geq r \rightarrow |\frac{1}{f(z)-a}| \leq \frac1r \rightarrow |g(z)| \leq \frac1r$

So, $g$ is entire and bounded. So, $g$ is constant (non zero) and hence $f$ is constant.

答:
That proof is completely correct.

The same approach is used to proof the Casorati–Weierstrass theorem: If an analytic function omits values from some disk in the neighbourhood of an isolated singularity then that singularity is not essential (i.e. removable or a pole).

Conversely, you could use the Casorati–Weierstrass theorem to prove your statement: If $f$ is entire and omits values from an open disk then it cannot have an essential singularity at infinity. It follows that $f$ is a polynomial, and therefore constant.


问:
Dividing by zero
I'm studying this function and I have this limit.

I think the domain is $x\neq0$ so it can't be calculated the limit for $x \rightarrow 0^-$.

$\lim\limits_{x \rightarrow 0^+} \dfrac{ 1-2lnx }{ x^2 }$

Now I'm stuck because $1-2lnx$ tends to $-1$ and $x^2$ tends to $0^+$ so I tought the result would be $-\infty$ but Wolfram says it's generic infinite $\infty$ and in the graph it shows $+\infty$.

答:
In fact the domain of $f(x) = \frac{1-2 \ln x}{x^2}$is not $\mathbb R \setminus \{0\}$ but $\mathbb R_+$ as $\ln x$ is only defined for $x \gt 0$.

Now as both $\lim\limits_{x \to 0^+} 1-2 \ln x = \lim\limits_{x \to 0^+} \frac{1}{x^2} = \infty$, you get $\lim\limits_{x \to 0^+} f(x) = \infty$.

Title: Dividing by zero


问:
Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge
$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure

答:
By Stirling $$\frac{{{n^n}}}{{n!}} \approx \frac{{{n^n}}}{{{n^n}{e^{ - n}}\sqrt {2\pi n} }} = \frac{{{e^n}}}{{\sqrt {2\pi n} }}$$ so the series cannot converge.

Alternatively, $${a_n} = \frac{{{n^n}}}{{n!}} \to \frac{{{a_{n + 1}}}}{{{a_n}}} = {\left( {1 + \frac{1}{n}} \right)^n} \to e > 1$$ so $a_n\to\infty$.

ContentLicense: CC BY-SA 3.0


问:
Map irregular quadrangle to square
Say I map an arbitrary quadrangle ABCD to a square A'B'C'D', A will be mapped to A'(0,0), B mapped to B'(1,0), C mapped to C'(1,1), D mapped to D'(0,1), P mapped to P'(u,v). How do I get P'(u,v) in the square given the coordinates of A B C D P ?

答:
I don't know how well this will work, but you could use the following simultaneous equations, which are a kind of bilinear interpolation:

$$P_x = (1-u)(1-v) A_x + u(1-v) B_x + uv C_x + (1-u)v D_x\\
P_y = (1-u)(1-v) A_y + u(1-v) B_y + uv C_y + (1-u)v D_y$$

Title: Map irregular quadrangle to square


问:
Software for drawing geometry diagrams
What software do you use to accurately draw geometry diagrams?

答:
Python Matplotlib

https://matplotlib.org/

Although Matplotlib's focus is on data plotting, it has become so featureful that you can generally produce good 2D illustrations with it.

Being written in Python is also a huge plus over domain specific languages like gnuplot.

It must be said that since it's main focus is not illustration, sometimes you have to Google a bit for the solution, but one often exists, or at least a reasonable workaround. And when it doesn't, it is often not hard to code it yourself and submit a patch.

Consider for example this simple educational graph demo I've made:

Source code

Also note that nothing in that demo is mandatory: you could easily rip off the axes or title for example.

It must be said that Matplotlib is not perfect for 3D however, as it does not support an OpenGL backend. As a result, the 3D interface is particularly clumsy and rather incomplete.

But you can get away with sometimes if your requirements are not too strict. For example, here's my attempt at a Bloch sphere illustration that I made for this other answer:

Source code.

In that example, we can see how I was unable to place the little circles above the plot as I would have wanted, showing how 3D is not perfect. But there is even a draft pull request for it.

One advantage of Matplotlib is that it has its own LaTeX subset parser, and thus dispenses a full LaTeX installation for mathematics as mentioned at https://matplotlib.org/3.3.3/tutorials/text/mathtext.html:

Note that you do not need to have TeX installed, since Matplotlib ships its own TeX expression parser, layout engine, and fonts. The layout engine is a fairly direct adaptation of the layout algorithms in Donald Knuth's TeX, so the quality is quite good (matplotlib also provides a usetex option for those who do want to call out to TeX to generate their text

Tested on matplotlib==3.2.2.

3blue1brown's Manim

https://github.com/ManimCommunity/manim/

This is based on the engine that 3blue1brown uses for his amazingly looking videos, which include complex moving graphs and formulas.

I'll let a sample video speak for itself: https://www.youtube.com/watch?v=r6sGWTCMz2k&t=1s

And there's an in-tree demo:

```<code>git clone https://github.com/ManimCommunity/manim/
cd manim
git checkout 1b12f076be49a677add97346eed6900e3cf18873
cd example_scenes
manim basic.py OpeningManimExample
xdg-open ./media/videos/basic/480p15/OpeningManimExample.mp4
</code>```

Inkscape demos

Inkscape had been previously mentioned at: Software for drawing geometry diagrams but here are some demos that felt relevant.

https://graphicdesign.stackexchange.com/questions/141237/how-to-color-2-circles-intersection-in-different-color-in-draw-io shows how it supports doing intersection of figures (you have to potentially copy objects to merge them):

Intuition behind normal subgroups shows a more free-form block diagram I've put up myself:

Inkscape currently doesn't support constraints however unfortunately: https://gitlab.com/inkscape/inbox/-/issues/1465 e.g. "same widths, parallel, etc.", but you can generally get decent results by just snapping to a grid in the editor. The "Align and Distribute" tool also helps, acting as "one time constraints", but there's no way to make real permanent constraints right now.

There isn't LaTeX math support in an extremely convenient out-of-box way it seems, but there are some possible methods available:

https://tex.stackexchange.com/questions/61274/is-there-any-way-to-type-latex-code-directly-into-the-text-boxes-inkscape

https://graphicdesign.stackexchange.com/questions/88027/where-to-find-latex-formula-in-inkscape

FreeCAD SVG export

This is usually a bit overkill, but it is really cool.

FreeCAD, as a CAD software, actually supports explicit constraints with a constraint solver.

This is a bit along the lines of what Software for drawing geometry diagrams mentioned kseg can do, except that kseg seems a bit abandoned, e.g. last release on SourceForge is from 2011.

FreeCAD is not focused on doing nice looking SVGs, but a possible workflow would be to export a perfect geometry to SVG, and then import the SVG into Inkscape to finish more visual details. This might be a reasonable workaround until the blessed day when Inkscape will implement constraints.

To export to SVG, you have to create a sketch. FreeCAD is a bit intimidating since it does so much more than 2D figures, so just watch this video: https://www.youtube.com/watch?v=sxnij3CkkdU

With that, I obtain a perfectly specified figure, you can see some constraints showing on the FreeCAD GUI, e.g. equality of edges of that triangle:

And then, I export to SVG via:

Select the sketch on the right menu

File

Export

Files of type: Flattned SVG

And then after conversion to PNG for uploading here:

```<code>inkscape -b 'FFF' -z -w 512 eye.svg -e eye.png
</code>```

I obtain:

TODO: why is it so messed up?

Tested on FreeCAD 0.18.4, Ubuntu 20.04.

MathMod

https://github.com/parisolab/mathmod

Not a huge community as of 2020, but it does seem to be able to produce some really nice 3D parametric/isocurve graphs with OpenGL and customizable shading.

You can use a custom scripting language to automate it (shame not python), and the GUI has widgets to vary parameters, including animations in time.

I got it to run on Ubuntu 20.10 by downloading the 10.1 prebuilt from the official download page on SourceForge (unfortunately). Then just unzip and run with:

```<code>./RunMathMod.sh
</code>```

Here's a screenshot from the 10.1 release:

ContentLicense: CC BY-SA 2.5

Title: Software for drawing geometry diagrams

Body:
Python Matplotlib

https://matplotlib.org/

Although Matplotlib's focus is on data plotting, it has become so featureful that you can generally produce good 2D illustrations with it.

Being written in Python is also a huge plus over domain specific languages like gnuplot.

It must be said that since it's main focus is not illustration, sometimes you have to Google a bit for the solution, but one often exists, or at least a reasonable workaround. And when it doesn't, it is often not hard to code it yourself and submit a patch.

Consider for example this simple educational graph demo I've made:

Source code

Also note that nothing in that demo is mandatory: you could easily rip off the axes or title for example.

It must be said that Matplotlib is not perfect for 3D however, as it does not support an OpenGL backend. As a result, the 3D interface is particularly clumsy and rather incomplete.

But you can get away with sometimes if your requirements are not too strict. For example, here's my attempt at a Bloch sphere illustration that I made for this other answer:

Source code.

In that example, we can see how I was unable to place the little circles above the plot as I would have wanted, showing how 3D is not perfect. But there is even a draft pull request for it.

One advantage of Matplotlib is that it has its own LaTeX subset parser, and thus dispenses a full LaTeX installation for mathematics as mentioned at https://matplotlib.org/3.3.3/tutorials/text/mathtext.html:

Note that you do not need to have TeX installed, since Matplotlib ships its own TeX expression parser, layout engine, and fonts. The layout engine is a fairly direct adaptation of the layout algorithms in Donald Knuth's TeX, so the quality is quite good (matplotlib also provides a usetex option for those who do want to call out to TeX to generate their text

Tested on matplotlib==3.2.2.

3blue1brown's Manim

https://github.com/ManimCommunity/manim/

This is based on the engine that 3blue1brown uses for his amazingly looking videos, which include complex moving graphs and formulas.

I'll let a sample video speak for itself: https://www.youtube.com/watch?v=r6sGWTCMz2k&t=1s

And there's an in-tree demo:

git clone https://github.com/ManimCommunity/manim/
cd manim
git checkout 1b12f076be49a677add97346eed6900e3cf18873
cd example_scenes
manim basic.py OpeningManimExample
xdg-open ./media/videos/basic/480p15/OpeningManimExample.mp4

Inkscape demos

Inkscape had been previously mentioned at: Software for drawing geometry diagrams but here are some demos that felt relevant.

https://graphicdesign.stackexchange.com/questions/141237/how-to-color-2-circles-intersection-in-different-color-in-draw-io shows how it supports doing intersection of figures (you have to potentially copy objects to merge them):

Intuition behind normal subgroups shows a more free-form block diagram I've put up myself:

Inkscape currently doesn't support constraints however unfortunately: https://gitlab.com/inkscape/inbox/-/issues/1465 e.g. "same widths, parallel, etc.", but you can generally get decent results by just snapping to a grid in the editor. The "Align and Distribute" tool also helps, acting as "one time constraints", but there's no way to make real permanent constraints right now.

There isn't LaTeX math support in an extremely convenient out-of-box way it seems, but there are some possible methods available:

https://tex.stackexchange.com/questions/61274/is-there-any-way-to-type-latex-code-directly-into-the-text-boxes-inkscape

https://graphicdesign.stackexchange.com/questions/88027/where-to-find-latex-formula-in-inkscape

FreeCAD SVG export

This is usually a bit overkill, but it is really cool.

FreeCAD, as a CAD software, actually supports explicit constraints with a constraint solver.

This is a bit along the lines of what Software for drawing geometry diagrams mentioned kseg can do, except that kseg seems a bit abandoned, e.g. last release on SourceForge is from 2011.

FreeCAD is not focused on doing nice looking SVGs, but a possible workflow would be to export a perfect geometry to SVG, and then import the SVG into Inkscape to finish more visual details. This might be a reasonable workaround until the blessed day when Inkscape will implement constraints.

To export to SVG, you have to create a sketch. FreeCAD is a bit intimidating since it does so much more than 2D figures, so just watch this video: https://www.youtube.com/watch?v=sxnij3CkkdU

With that, I obtain a perfectly specified figure, you can see some constraints showing on the FreeCAD GUI, e.g. equality of edges of that triangle:

And then, I export to SVG via:

Select the sketch on the right menu

File

Export

Files of type: Flattned SVG

And then after conversion to PNG for uploading here:

inkscape -b 'FFF' -z -w 512 eye.svg -e eye.png

I obtain:

TODO: why is it so messed up?

Tested on FreeCAD 0.18.4, Ubuntu 20.04.

MathMod

https://github.com/parisolab/mathmod

Not a huge community as of 2020, but it does seem to be able to produce some really nice 3D parametric/isocurve graphs with OpenGL and customizable shading.

You can use a custom scripting language to automate it (shame not python), and the GUI has widgets to vary parameters, including animations in time.

I got it to run on Ubuntu 20.10 by downloading the 10.1 prebuilt from the official download page on SourceForge (unfortunately). Then just unzip and run with:

./RunMathMod.sh

Here's a screenshot from the 10.1 release:


问:
Prove that if $~\sum a_n=A~$ , $~\sum b_n=B~$ , and $~\sum c_n=C$
Let $\{a_n\}$, $\{b_n\}$ be sequences. Define $\displaystyle c_n=\sum_{k=1}^n a_kb_{n+1-k}$.

Prove that if $~\sum a_n=A~$ , $~\sum b_n=B~$ , and $~\sum c_n=C~$ (so they are all convergent series) then $C=AB$. (Note that we do not need $\sum a_n$ to be absolutely convergent).

Hello everyone. I am stuck on how to start this problem. I don't want the answer, just a hint on how to get started.

答:
I'm sorry that I misunderstood the question previously. What you are looking for is probably this , which says :

Let $\sum a_{n}~$ ,  $\sum b_{n}$  are conditionally convergent complex series,   $\sum c_{n}$ is the Cauchy product of $\sum a_n$, $\sum b_n$ such that $\sum c_n$ converges. Then, $$\sum c_{n} = \left(\sum a_{n}\right)\left(\sum b_{n}\right)$$

For a complete proof, please refer to same link as above.

EDIT : Updated the links. Sorry for inconvenience.


问:
Partial fraction decomposition with a nonrepeated irreducible quadratic factor
I'm trying to do a partial fraction decomposition on the following rational eqn with a nonrepeated irreducible quadratic factor:

$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}$$

I've broken it down into an identiy: $-28x^2 -92 = A((x-4)(x^2-1))+B(x^2+1)+(Cx+D)(x-4)^2$

and solved for B by setting x to 4 ($B-12$), distributed the -12 and moved the resulting quadratic to the left side, leaving me with:

$$12x^2-28x-80 =  A((x-4)(x^2-1)) + (Cx+D)(x-4)^2$$

Now I want to solve for A by setting x to something that will make the coefficient on $Cx+D$ zero, but I'm stumped - setting it to 4 also gets rid of my A. What do I do? Or have I messed up somewhere along the way?

答:
To find out an equation of A and c compare coefficient of $x^3$.put x=0 and compare coefficient of $x^2$.after doing these all operation you will find equation and some values.If you stuck then leave a comment.I'll do it.
 you have this eqn
$$-28x^2-92=A(x-4)(x^2+1)+B(x^2+1)+(Cx+D)(x-4)^2$$
then putting $x=4$ you will get $B=\dfrac{-540}{17}$

compare coefficient of $x^3$ you will get : $0=A+C$

compare coefficient of $x^2$ you will get : $-28=-4A+B-8C+D$

putting  $x=0$ you will get: $-92=-4A+B+16D$

you have value of B so these eqn will become litte more  easy and I hope you will take it to final from now on.


问:
Is my method of solving equation correct?
The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$

Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!

答:
When you know $a+b$, you can express $a^n+b^n$ in terms of powers of $c=ab$. A few cases

$a^2+b^2=(a+b)^2-2ab$

$a^3+b^3=(a+b)^3-3ab(a+b)$

What about $a^5+b^5$? The relations are obtained from the expansion of $(a+b)^5$:
\begin{align}
(a+b)^5&=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)\\
&=a^5+b^5+5ab((a+b)^3-3ab(a+b))+10a^2b^2(a+b)\\
&=a^5+b^5+5ab(a+b)^3-5a^2b^2(a+b)
\end{align}
In your case you get
$$
a^5+b^5=32-40c+10c^2
$$
which yields
$$
32=32-40c+10c^2
$$
hence $c=0$ or $c=4$.

The first case yields $16-\sqrt{x}=0$; the second case yields
$$
256-x=4^5
$$

Alternative approach. Since $a=\sqrt[5]{16+\sqrt{x}}\ne0$, we can set $t=b/a$ and from $(a+b)^5=a^5+b^5$, we get
$$
f(t)=(t+1)^5-t^5-1=0
$$
We have $f'(t)=5(t+1)^4-5t^4$, which vanishes only at $t=-1/2$, and $f(-1/2)<0$. Thus $f$ only vanishes twice, but one root is negative. Since $0$ is a root, it is the only nonnegative one. Thus $b=0$.


问:
For any point in $\Bbb R^2$ $f$ has a local inverse. Can I say $f$ is invertible on its range?

Define $f:\Bbb R^2\to \Bbb R^2$ by $$\begin{align}f(x,y)=(e^{2x+y},4x^2+4xy+y^2+6x+4y)\ . \end{align}$$
Define $U:=f(\Bbb R^2)$.
Prove that the inverse function of $f$ (say $f^{-1}:U\to \Bbb R^2$) exists.

I want to apply the inverse function theorem. I know that $f$ is $C^1$ and the Jacorbian determinant $J_f(x,y)=2e^{2x+y}>0$ for all $(x,y)\in \Bbb R^2$. But I have a question. If I know that for all $(x,y)\in \Bbb R^2$, I can find a local inverse, does this imply $f$ has a inverse from its range $U$ onto $\Bbb R^2$? Or I need to find some ways to prove that $f$ is injective from $\Bbb R^2$ to $U$?

答:
Local invertibility without injectivity cannot give you global invertibility.
But here injectivity is easy to prove, as well as an explicit formula for the inverse. If
$$(e^{2x+y}, 4x^2+4xy+y^2+6x+4y) = (a, b)$$
then
$$y = \ln(a) - 2x$$
and you can check that
$$b = 4x^2+4xy+y^2+6x+4y = - 2x + 4 \ln(a) + \ln(a)^2.$$
You can easily conclude with this.


问:
Help with calculating the product of cosets of a polynomial ring.
Description of Ring and Ideal

First I'm sorry this is my first post and I don't how to entirely format math equations yet, but I'm trying to figure it out.

But I am having a difficult time understanding this specific case of a quotient ring. I know that generally speaking the product of elements of a quotient ring are

${(a + I)(b + I) = (ab + I)}$

where I is an ideal in the ring.

So for my specific question in the quotient ring R[x]/((x^2)-4) the problem states that each coset has the unique form ${a + bx + (x^2-4)}$ given that a and b are real numbers. The problem then asks to express the product of two of these cosets

${[a + bx + (x^2-4)] * [c + dx + (x^2-4)]}$

in the same form, that is, to find the product in the form ${e + fx + (x^2-4)}$ and write out what e and f are in terms of a,b,c, and d.

What I have so far is this:

I'm essentially taking ${(a + I)(b + I) = (ab + I)}$ to be

${(a + bx + (x^2-4))(c + dx + (x^2-4))}$

and then

${(a + bx)(c + dx) = ac + x(ad + cb) + bdx^2}$

But from my understanding of how the quotient ring works is that since the principal ideal is ${(x^2-4)}$ then that leads to an additional condition that we have ${x^2-4 = 0}$ which implies ${x^2 = 4}$. So then my equation above becomes

${ac + x(ad + cb) + bdx^2 = ac + x(ad + cb) + 4bd}$

and so in the form the question is asking me it would then be that for the product of the two cosets in terms of a,b,c, and d we have that

$${e + fx + (x^2-4)}$$
$${e = 4bd + ac}$$
$${f = ad + cb}$$

Is this the correct way of going about multiplying two elements of this quotient ring? Or am I entirely missing the process here?

答:
You can't do $x^2=4$, only “sort of”. What you do know is that
$$
x^2-4\in(x^2-4)=I
$$

and therefore $x^2+I=4+I$.

So when you multiply
$$
(a+bx+I)(c+dx+I)=ac+(ad+bc)x+bdx^2+I
$$
you can use $bdx^2+I=4bd+I$ and you end up with
$$
(a+bx+I)(c+dx+I)=(ac+4bd)+(ad+bc)x+I
$$


答:
Spoiler alert: this is the solution.

Let's take $ a = \sqrt[5]{16+\sqrt x} $ and $ b = \sqrt[5]{16-\sqrt x} $ . Note that because $ \sqrt x \ge 0 $, $ a $ is always positive (technically greater than or equal to $ \sqrt[5]{16} $), but $ b $ can be positive, zero, or negative. We have:
$$ a + b = 2 $$
$$ (a + b)^5 = 32 $$
We also have:
$$ a^5 = 16 + \sqrt x $$
$$ b^5 = 16 - \sqrt x $$
$$ a^5 + b^5 = 32 $$
Aha! Let's see where this takes us:
$$ (a + b)^5 = a^5 + b^5 $$
$$ a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 = a^5 + b^5 $$
$$ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 = 0 $$
For this sum to be zero, either all of its terms should be zero, or some of them should be positive and some others negative. The former case happens if $ b $ is zero (we noted earlier that $ a $ is always positive, so it can't be zero). The latter case can only happen if $ b $ is negative, because again as we noted earlier, $ a $ is always positive and cannot make negative terms. Let's keep that in mind and continue.
$$ 5ab(a^3 + 2a^2b + 2ab^2 + b^3) = 0 $$
The expression in the parentheses looks a bit like $ (a+b)^3 $, but not exactly! Let's add $ a^2b + ab^2 - a^2b - ab^2 $ to it:
$$ 5ab[(a+b)^3 - a^2b - ab^2] = 0 $$
$$ 5ab[(a+b)^3 - ab(a+b)] = 0 $$
$$ 5ab(a+b)[(a+b)^2-ab] = 0 $$
And recalling that $ a+b = 2 $:
$$ 10ab[4-ab]=0 $$
Now, for this equality to hold, we must have either $ ab = 4 $ or $ ab = 0 $.

But $ ab = 4 $ is not possible, because if it were so, considering that $ a $ is positive, $ b $ should necessarily be positive as well (otherwise $ ab $ would not be positive). And if both $ a $ and $ b $ were positive, then the sum $ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 $ would necessarily be positive, whereas it should be zero.

So we are left with the conclusion that $ ab = 0 $. As we mentioned before, $ a $ is positive and not zero. So:
$$ b = 0 $$
$$ \sqrt[5]{16-\sqrt x} = 0 $$
$$ 16 - \sqrt x = 0 $$
$$ x = 16^2 = 256 $$

Title: Is my method of solving equation correct?

Body:
The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$

Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!


问:
Does the following simultaneous differential equations has an analytical solution?
I'm wondering whether the following simultaneous differential equations has an analytical solution?

$$\begin{cases} \cfrac{dx}{dt}=\cfrac{2v}{x^2+v^2} \\ \cfrac{dv}{dt}=\cfrac{-2x}{x^2+v^2} \end{cases}$$

答:
Use symbol $y$ in place of $v$ for quicker recognition.

$$ \cfrac{dx}{dt}=\cfrac{2y}{x^2+y^2} ;\cfrac{dy}{dt}=\cfrac{-2x}{x^2+y^2} ;$$

Divide to eliminate parameter $t$

$$\dfrac{dy}{dx}=\dfrac{-x}{y} , \text { integrate } \to x^2+y^2=c_1^2$$

$$ x = c_1 \cos (t + c_2)  ; y = c_1 \sin (t + c_2);$$

Alternately recognize second order DEs

$$ \ddot x + x=0 \;;\ddot y + y=0\;;$$


问:
How can I prove or disprove this for conditional probability?
P(E|S) >= P(E) - P(S)

I assumed it to be true, and started off like this, but am stuck.

P(E) - P(S)

= P(E) - (P(E ∩ S) / P(S))

Not sure how to proceed from here, or if I am even correct in assuming the given statement is true. Please help.

答:
Take two disjoint events $E,S$ with $P(E) >P(S)$. Then $P(E|S)=0 <P(E)-P(S)$. So the inequality is false.

Specific example:  on $(0,1)$ with Lebesgue measure take $E=(0,\frac 1 2)$ and $F=(\frac 3  4,1)$.


问:
question about solution of $1,(x-5),(x-5)^2,...,(x-5)^m$ as a basis for $\mathcal{P}_m(\mathbf{R})$
axler 3.F.8  here is another answer for but I have problem for understanding highlighted part:

first why $\varphi_j(0)$??
second isn't any problem for $j=0$?

答:
We have an equality of polynomials
$$
a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m = 0
$$
Applying $\varphi_j$ to both sides, we keep the equality and get
$$
\varphi_j(a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m) = \varphi_j(0)
$$
The main difficulty here is actually recognizing that the left-hand side is equal to $a_j$, not that the left-hand side and right-hand side are equal.

As for what happens when $j = 0$, there is no issue. Differentiating $0$ times is the same as just not doing anything, so $p^{(0)}$ is just $p$. And $0!$ is $1$. So $\varphi_0(p)$ turns out to just be $p(5)$. Which for our particular (left-hand side) polynomial turns out to be just $a_0$.


答:
The $j$th derivative of $$a_0+a_1(x-5)+\ldots +a_m(x-5)^m$$ is $$j!\cdot a_j+b_{j+1}(x-5)+b_m(x-5)^{(m-j)}$$ (for some constants $b_{j+1},\ldots ,b_m$ that we don't need to know, but that one can easily calculate), so when you evaluate it a $x=5$, you get $j!\cdot a_j$, and then $\varphi_j(p)=a_j$. On the other hand, exactly as in the question you have cited, here we are supposing that $p(x)=a_0+a_1(x-5)+\ldots +a_m(x-5)^m$ is the null vector, so $\varphi_j(p)=\varphi_j(0)=0$.

When $j=0$, conventionally one have to take $p^{(0)}$ as $p$, see for example here (find "zeroth derivative" in the first paragraph) or here (paragraph 1.2, "Taylor series").

Note that actually there is a typo, the coefficients $a_0, \ldots ,a_m$ are in $\mathbb{R}$ not in $\mathbb{F}$, but actually you can consider also polynomials over any infinite field (viewed as an extension of $\mathbb{Q}$).

Title: question about solution of $1,(x-5),(x-5)^2,...,(x-5)^m$ as a basis for $\mathcal{P}_m(\mathbf{R})$

Body:
axler 3.F.8  here is another answer for but I have problem for understanding highlighted part:

first why $\varphi_j(0)$??
second isn't any problem for $j=0$?


问:
Conformal Map from Vertical Strip to Unit Disc
I haven't found a similar question on here, though I suspect the question may be rather well-covered.

I want to find a conformal map from the vertical strip $\{z:-1<Re(z)<1\}$ onto the unit disc.

Under the exponential map the region is taken to the annulus with radii $e$ and $e^{-1}$, but I'm not sure how useful this will be.

Can anyone advise on what the conformal map may be?

Thanks.

答:
First apply $z \mapsto iz$ to map the strip onto the corresponding horizontal strip $-1 < \operatorname{Im} z < 1$.

Next step, apply $z \mapsto \exp(\frac{\pi z}2)$. This will give you the right half-plane $\operatorname{Re} z > 0$.

To finish off, the Möbius transformation $z \mapsto \dfrac{z-1}{z+1}$ takes the half-plane onto the unit disc.


问:
Find the derived subgroup of $A_4$

Find the derived subgroup of $A_4$.

Since it is $A_4$, for a permutation $\sigma$ to be in $A_4$, $\sigma$ must have a cycle structure of $2$ cycles. Therefore, $\sigma=(ab)(cd)$.

The commutator of such elements, would be obviously another permutation of $2$-cycle (of length $2$).

Does it mean that $A'_4=A_4$?

答:
I don't have enough reputations to comment Alan Wang's solution.
But it should be checked that each factor of a 2 disjoint transpositions is actually single (i.e. odd) permutation which is not in $A_4$.
ex.) (1 2)(3 4) $\in A_4$, but (1 2) $\notin A_4$ nor (3 4) $\notin A_4$.

If (a b c)S = (a b)(c d) for some S$\in A_4$ then
S = (b c d) such that, (a b c) (b c d) = (a b)(c d) where
(d a b) (b c d) (d b a) = (a c b).
Thus, V < $A'_4$.
And, V > $A'_4$ since $A'_4/V\simeq Z_3$ is abelian, we have $A'_4 = V$.

Title: Find the derived subgroup of $A_4$


问:
derivative for the following matrix
I have a question about the following derivative. Let us have $X\in\mathbb{R}^{m\times n}, z\in\mathbb{R}^{n}$ and I would like to find the derivative $$\frac{\partial (z^{T}X^{T}Xz)}{\partial z}. $$ Any idea? It gives me a hard time. Thank you.

答:
Let $f: \mathbb{R}^n\to \mathbb{R}$ the map  $f(z)= z^TX^TXz $ for $X\in\mathbb{R}^{m\times n}$ and $z\in \mathbb{R}^{n\times 1}$. We have
$$
Df(z)\cdot v= v^TX^TXz+z^TX^TXv
$$
In fact,
$$
f(z+v)=(x+v)^TX^TX(x+v)= z^TX^TXz+v^TX^TXz+z^TX^TXv+v^TX^TXv.
$$
Then
$$
\lim_{v\to 0}\frac{f(z+v)-f(z)- v^TX^TXz+z^TX^TXv}{\|v\|}
=
\lim_{v\to 0}\frac{v^TX^TXv}{\|v\|}
=0
$$
The derivatie is a linear map
$$
\mathbb{R}^n\ni v\longmapsto Df(z)\cdot v=v^TX^TXz+z^TX^TXv\in \mathbb{R}^n
$$

Title: derivative for the following matrix


问:
Solve differential equation using 'variation of parameters'
Given the differential equation $L \ u = f$, with $$L \ u = a_2(t) \frac{d^2u}{dt^2}+a_1(t)\frac{du}{dt}+a_0(t)u$$
with $a_i(t)$ sufficiently smooth and $a_2(t) \neq 0$ for every $t$.
Suppose that $u_0$ is a solution to the homogenenous equation $L \ u=0$, with $u_0(t)\neq0$ for every $t$.
 How can I then use the variation of parameters method by Lagrange to find the solution $u =u_0v$ ? And how can I find the general solution if the equation isn't homogeneous?

Update: I've substituted $u=u_0v$ into the equation and found a new $2^{nd}$ order differential equation of the form
$$f = \frac{d^2v}{dt^2}(a_2(t)u_0)+\frac{dv}{dt}(2a_2(t)u_0'+a_1(t)u_0)+va_0(t)u_0$$ which I'm supposed to solve for $w$ using $w=v'$. But I don't seem to be able to proceed as I'm not sure how to solve the system
$$\begin{equation}
\begin{split}
w =&\ v'\\
w'a_2(t)u_0+w(2a_2(t)u_0'+a_1(t)u_0)+v(a_0(t)u_0)=&\ f\ \\
\end{split}
\end{equation}$$

答:
With

$$
L(u) = u''+\frac{a_1}{a_2} u'+\frac{a_0}{a_2} u=u''+b_1 u'+b_0 u
$$

given a non null $u_0(t)$ such that $L(u_0) = 0$ being a second order $L$, we can obtain another independent solution $L(u_1) = 0$ as follows: if $c_0$ is a generic constant then $L(c_0u_0) = 0$ so choose $u_1 = c_0(t)u_0$ then we have

$$
L(c_0(t)u_0) = c_0(t)L(u_0) + u_0c''_0+(b_1u_0+2u'_0)c'_0=u_0c''_0+(b_1u_0+2u'_0)c'_0=0
$$

making $\phi = c'_0$ we have a first order DE as $\frac{\phi'}{\phi} + \frac{u_0}{(b_1u_0+2u'_0)}=0$ so obtaining $\phi\to u_1$ is direct.

Having $u_0, u_1$ for the homogeneous case, let us approach the case $L(u) = \frac{f}{a_2(t)}$. The method of variation of constants now will be applied to

$$
L(c_0(t)u_0+c_1(t)u_1)=\frac{f}{a_2(t)}
$$

so developing

$$
L(c_0(t)u_0+c_1(t)u_1)=c_0L(u_0)+c_1L(u_1) + u_0c''_0+(b_1u_0+2u'_0)c'_0+c''_1u_1+(b_1u_1+2u'_1)c'_1-\frac{f}{a_2(t)}=0
$$

Here as $c_0, c_1$ are independent we can choose them as

$$
\cases{
u_0c''_0+(b_1u_0+2u'_0)c'_0-\frac{f}{a_2(t)}=0\\
c''_1u_1+(b_1u_1+2u'_1)c'_1=0
}
$$

finally determining $c_1(t), c_2(t)$ for the non-homogeneous case.


问:
Markov's like inequalities
This sounds trivial but unfortunately I kind of feel lost in proving these:

which of the followings inequalities holds:

\begin{align}
&P(X\ge a)\leq\frac{E(X)}{a}\\
&P(X\ge a)<\frac{E(X)}{a}\\
&P(X> a)\leq\frac{E(X)}{a}\\
&P(X> a)<\frac{E(X)}{a}
\end{align}

For a non-negative constant $a$ and a random variable $X$.

I know the first one is Markov's inequality so it does hold, but what about the others?

答:
All of them are false. If $0<a<EX$ the $\frac {EX} a >1$ and probability of an event cannot exceed $1$.

Answer for the reviser version of the question:

You also need $X$ to be non-negative. Otherwise the first one is also false.

Take $X=a$ w.p. $1$ to see that the second one  is false.

$P(X>a) \leq P(X \geq  a) \leq \frac {EX } a$.

To see the the fourth inequality can fail take $X=\pm \frac a 2$ with probability $\frac 1 2 $ each.

Title: Markov's like inequalities


答:
By Stirling formula
$$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$
we have
$$\left|(-1)^{n}\frac{n^n}{n!}\right|\sim\frac{e^n}{\sqrt{2\pi n}}\not\to0$$
hence your series is divergent.</p>

Title: Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge

Body:
$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure


问:
Can subtraction be commutative
In primary and secondary school we all learnt that subtraction is not commutative, as only addition and multiplication is.

However, could subtraction be commutative for all real numbers that are subtracted from itself.

E.g 5-5=0
And 5-5=0

There are an infinite number of these pairs, so could subtraction be commutative when:
x=y
As in x-y?

Answers will be appreciated

答:
I think if you keep the sign with its number then subtraction could be commutative. For example $5-2=3$ and $-2+5=3$. If you keep the negative with the $2$ and the positive with the $5$ then it can be commutative.

Edited: subtraction itself is not commutative eg 5-2=3 and 2-5 =-3. But if you keep the number with its sign, then the order of adding or subtracting does not matter, eg +5-2=3 and -2+5=3.

Your example ($X-Y=0$ and $Y-X=0$ when $X=Y$) does not show  the commutative property. It simply shows that any number subtracted from itself is zero.

Title: Can subtraction be commutative


问:
Prove $\mathbb{Z[i]}/(1+4i) \cong \mathbb{Z_{17}}$
Prove that $\dfrac{\mathbb{Z[i]}}{(1+4i)} \cong \mathbb{Z_{17}}$.

My attempt:

$\phi:\mathbb{Z} \rightarrow \mathbb{Z_{17}}$ be the natural (onto) ring homomorphism, sending $a \mapsto \bar{a}$. Hence by universal property of polynomial rings there exists a (onto) ring homomorphism $\phi':\mathbb{Z[X]} \rightarrow \mathbb{Z_{17}}$ such that $\phi'(x)=\bar{4}$  and $\phi'(n)=\phi(n)$ for all $n \in \mathbb{Z}$.

Clearly, $(1+x^2) \subset \operatorname{Ker}\phi' \implies$ we can obtain a (onto) ring homomorphism $\phi'':\mathbb{Z}[X]/(1+x^2) \rightarrow \mathbb{Z_{17}}$ where $\bar{x} \mapsto \bar{4}$

Since $\mathbb{Z}[X]/(1+x^2) \cong \mathbb{Z[i]}$ with $i \mapsto \bar{x}$, one can conclude there is an onto ring homomorphism $\Phi:\mathbb{Z[i]} \rightarrow \mathbb{Z_{17}}$ where $i \mapsto \bar{4}$.

$1+4i \mapsto 0$ one has $(1+4i) \subset \operatorname{Ker}\Phi$.

How can I conclude the reverse inclusion?

答:
With these problems, the third isomorphism theorem is usually your friend (divide out by $17$ and $x-4$ before you divide out by $x^2 +1$). Here is a proof of what you want to show without doing any explicit quotienting, although it works just as well if you rephrase it to be about the map $\Bbb Z[x]/(17, x-4, x^2+1)\to \Bbb Z_{17}$.

The kernel of $\phi'$ clearly contains $(17, x^2+1, x-4)$, as you have shown. We want to show that this is the entire kernel. Note that $x^2+1\in (17, x-4)$, because $(x-4)(x+4) + 17 = x^2 + 1$. So that's a superfluous generator. Now we just have $(17, x-4)$, and things are much simpler.

Take an element $f$ in the kernel of $\phi'$. We want to study the coset $f + (17, x-4)$. It has a representative among the elements $\{0, 1, 2, \ldots, 16\}$ (we can use the $x-4$ generator to get rid of any $x$'s, and then the $17$ generator to bring the remaining constant term into this range). And since any element in this coset is in the kernel of $\phi'$, the chosen representative must be in the kernel. But among these, only $0$ is in the kernel of $\phi'$. So $$f + (17, x-4) = 0 + (17, x-4)$$and we are done.


问:
Proof of divisibility of 3 sequential cubic numbers by induction
The task is to prove that the sum of three subsequent cubic numbers is divisible by 9.
My approach looks like this:

$n^3+(n+1)^3+(n+2)^3 = 9M$

Proof that the equation is true for $n=1$

$1^3+2^3+3^3 = 1+8+27 = 36;$

$ 36/9 = 4$

Assume that the equation is true for $n<=k$

$k^3+(k+1)^3+(k+2)^3 = 3k^3+9k^2+15k+9$

Show that the equation is true for $n = k+1$

$(k+1)^3+(k+2)^3+(k+3)^3=3k^3+18k^3+42k+36$

And this is the point where I am stuck. With a simpler formula I would simply replace the $k$ with $9M $ but how can you do it in this case?

Body:
For all $n \in \{1,2,3,\dots\}$, $n^3 + (n+1)^3 + (n+2)^3$ is a multiple of $9$.

It's true for $n=1$.

$\qquad 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$, which is a multiple of $9$.

If its true for $n=k$, then it's true for $n=k+1$.

Suppose $k^3 + (k+1)^3 + (k+2)^3 = 9x$. Then

\begin{align}
   (k+1)^3 + (k+2)^3 + (k+3)^3
   &= [k^3 + (k+1)^3 + (k+2)^3] + (k+3)^3 - k^3 \\
   &= 9x +  (k+3)^3 - k^3 \\
   &= 9x + (k^3 + 9k^2 + 27k + 27) - k^3 \\
   &= 9x + 9k^2 + 27k + 27 \\
   &= 9(x + k^2 + 3k + 3)
\end{align}


问:
Determine the number of three-digit numbers that are divisible by 3
Determine the number of three-digit numbers that are divisible by 3 and that you can form with 3 different numbers among 1, 2, 3, 5, 6, 8

I found 5 numbers or 5 different ways with a sum that is divisible by 3. And each of them has 6 combinations because 3* 2* 1=6 so the total number should be 5*3=30 three-digit numbers that are divisible by 3. But I'm not sure if it's right or if there's a better solution.

答:
$$\begin{align}2,5,8&\equiv -1\pmod 3\\3,6&\equiv 0\pmod 3\\1&\equiv 1\pmod 3\end{align}$$
There are $3\times 2\times 1=6$ ways to select one number from each group and $1$ way to select three numbers from a single group, for a total of $7$ combinations.

There are $3!=6$ permutations of each combination, giving a total of $7\times 6=42$ three-digit numbers.


问:
Approximation by a polynomial degree 3 centered at a=2?
Approximate $\:\mathbf{f(x) = 1/\sqrt{x^3 + 1}}\:$ by a polynomial of degree 3 centered at a = 2.

What exactly is this question asking when it says "approximate"? What is the end result supposed to be I'm so lost

I found the derivatives(up to the 3rd) and I substituted x with 2, but now what?

I'd appreciate any help!

Body:
The taylor polynomial of $f$ of dgree $3$ at $2$ is given by

$$f(2)+\frac{f'(2)}{1!}(x-2)+\frac{f''(2)}{2!}(x-2)^2+\frac{f'''(2)}{3!}(x-2)^3.$$


问:
Every element of a local ring is a sum of two units iff the cardinality of the ring is less than twice the cardinality of units
Is it true that every element of a local ring with unity can be written as a sum of two units iff the cardinality of the ring is strictly less than twice the cardinality of units of the ring?

It is known that the unique maximal ideal of the local ring consists entirely of non-unit elements. Using this fact how can we arrive at a resolution of the above question. I think it is true. Does it follow simply from the fact that any such representation of a non-unit is unique? Any hints? Thanks beforehand.

答:
This does not hold in $\mathbb{F}_2$, since $1 \neq 1+1$. If $k$ is a field besides $\mathbb{F}_2$, pick some $b \in k \setminus \{0,1\}$. Then $(x-b) + b$ expresses $x$ as a sum of two units for all $x \neq b$, and $b = (b-1)+1$ is also a sum of two units.

Next, let $R$ be a commutative local ring with maximal ideal $m$. First, suppose $R/m$ is isomorphic to $\mathbb{F}_2$. Then we cannot write $1$ as a sum of two units of $R$, or else we would be able to write $1$ as a sum of two units in $R/m$. Thus, $R$ does not have the desired property. On the other hand, suppose $R/m$ is not isomorphic to $\mathbb{F}_2$. Then, let $x \in R$ be arbitrary, and write $\overline{x}$ (the image of $x$ in the residue field) as a sum of two units in $R/m$. This means that there exist units $a,b \in R$ such that $x-(a+b) \in m$. Let $y = x-(a+b)$. Then $x = (y + a) + b$ expresses $x$ as a sum of two units.

We conclude that a local ring $R$ with maximal ideal $m$ has the property that each element is a sum of two units if and only if $R/m$ is not isomorphic to $\mathbb{F}_2$.


问:
Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge
$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure

答: If a series $\sum_{n = 0}^\infty a_n$ converges then $\displaystyle \lim_{n\to \infty}a_n = 0$ is necessary condition. Calculate $\lim_{n\to \infty}a_n$. Stirling approximation might be helpful to calculate the limit.</p>


问:
Trivial sequence: $32,21,14,\dots$
My little brother (third grade) asked me for help with this math problem on his homework, which was:

Find the next number in the sequence $32,21,14,\dots$

I was not able to see a trivial solution, so I ran a linear regression (for an equation for $a_n$) which turned up an equation with an $r^2$ of .$98$, which (considering I had three points) was not suitable, so I ran a quadratic regression and got the polynomial $$2n^2-17n+47$$ which yielded an $r^2$ of 1, and so was a perfect fit. Therefore, I said the answer was $11$.

My question is, considering that the question assumes no knowledge of algebra, or exponents, is there a simple (recursive, maybe?) formula for the next number in the sequence?

答:
There are many possible answers. One possible pattern is subtraction of primes leading to a finite end, yet an infinite beginning.
...
81-19=62
62-17=45 45 <-- preceding number of sub-sequence
45-13=32 32 <-- start of given sub-sequence
32-11=21 21
21- 7=14 14 <-- end of given sub-sequence
14- 5= 9  9 <-- next number of sub-sequence
9- 3= 6
6- 2= 4 <- end of sequence</p>

Title: Trivial sequence: $32,21,14,\dots$


问:
Showing a solution to this differential equation is never zero
Say I want to find a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for some $k \neq 0$, $f' = kf$. Every time I have seen a book solve this, the expression is rewritten  as $$\frac{1}{f}f' = k$$ and then solved using u-substitution. To be technical though, must one first know that $f$ is never zero? If so, how could I go about showing that $f$ is never zero? I know that $f$ must be infinitely differentiable and if for some $a \in \mathbb{R}$, $f(a) = 0$, then for each $n \in \mathbb{N}_0$, $f^{(n)}(a) = 0$. I think I remember from complex analysis, if $f$ is holomorphic with this property, then $f$ is zero everywhere, but I'm not sure about the usual case of $\mathbb{R} \rightarrow \mathbb{R}$.

答:
More generally, for an ODE of the form $f'=g(f)$, any zero $f^*$ of the function $g$ gives a constant solution $f(x)=f^*$. And if $g$ is nice enough so that one has uniqueness of solutions (say that $g$ satisfies a Lipschitz condition, so that the Picard–Lindelöf theorem applies), then any other solution can never be equal to $f^*$ at any point $x=x_0$, since that would mean that you would have (at least) two solutions to the initial value problem $f'=g(f)$, $f(x_0)=f^*$.

So when using separation of variables, you should always first note “by hand” the constant solutions $f(x)=f^*$, and only then proceed to find all the other solutions (which are never equal to any $f^*$, so that it's safe to divide by $g(f)$) from $\int \frac{df}{g(f)}= \int dx$.


问:
Does the following simultaneous differential equations has an analytical solution?
I'm wondering whether the following simultaneous differential equations has an analytical solution?

$$\begin{cases} \cfrac{dx}{dt}=\cfrac{2v}{x^2+v^2} \\ \cfrac{dv}{dt}=\cfrac{-2x}{x^2+v^2} \end{cases}$$

Body:
\begin{align}
  z &= v+xi \\
  |z|^2 &= v^2+x^2 \\
  \frac{d|z|^2}{dt}  &= 2v \frac{dv}{dt}+2x\frac{dx}{dt} \\
  &= 0 \\
  |z| &= k \\
  \frac{dz}{dt} &= \frac{2(-x+vi)}{v^2+x^2} \\
  &= \frac{2iz}{|z|^2} \\
  &= \frac{2iz}{k^2} \\
  \frac{dz}{z} &= \frac{2i}{k^2} dt \\
  \ln z &= \frac{2it}{k^2}+c \\
  z &= A\exp \left( \frac{2it}{k^2} \right)
  \tag{$c=\ln A$} \\
  &= k\exp \left[ i\left( \frac{2t}{k^2}+\phi \right) \right]
  \tag{$A=ke^{i\phi}$} \\
  v(t) &= k\cos \left( \frac{2t}{k^2}+\phi \right) \\
  x(t) &= k\sin \left( \frac{2t}{k^2}+\phi \right) \\
\end{align}


问:
Inverse of projection map in product topology which defines sub-basis.
I have $(X,\tau_X)$ and $(Y,\tau_Y)$ - two topological spaces. Consider $U\subseteq X$ and $V\subseteq Y$ and two projection maps $\pi_X: X\times Y \to X$ and $\pi_Y: X \times Y \to Y$. Then it is well known that  $$\mathcal B = \{\pi_X^{-1}(U): U \in \tau_X\} \cup \{\pi_Y^{-1}(V):  V \in \tau_Y\}$$ is a sub-basis for the product topology on $X\times Y$. I am wondering about that inverse mapping because I can take $U_1 \times V_1$ and $U_1 \times V_2$ then use $ \pi_X(X \times V) $ and I have $ U_1 $ which is open. Then I am using $ \pi_X^{-1}(U) $ an I have ambiguity in result as I can have two results $ U_1 \times V_1 $ and $ U_1 \times V_2 $, then inverse is not map. Where my logic fails, point me please?

答:
By definition of the inverse mapping for $U \in \tau_X$,

$$\pi_X^{-1}(U) = \{(x,y) \in X \times Y \mid \pi_X(x,y) \in U\} = U \times Y.$$

So I don't really see where the ambiguity is. To be more specific, you don't have the equality $\pi_X^{-1}(U) = U \times V$ for $V \subsetneq Y$.


问:
$f:S^n\to X$ is nulhomotopic if and only if there is an extension $\tilde{f}:D^{n+1}\to X$ of $f$.

$f:S^n\to X$ is nulhomotopic if and only if there is an extension $\tilde{f}:D^{n+1}\to X$ of $f$. ($X$ is a topological space)

I know how to prove the forward direction, but I don't know how to prove the reverse direction directly. Could you give any hint? (I'm a beginner of Algebraic topology so I don't know the higher concept of Algebraic topology. I just wanted to extend some theorem in Munkres' topology textbook.)

答: If the map $f$ factor through $D^n$ then it is nullhomotopic as $D^n$ is contractible, hence every map out of $D^n$ is null homotopic. For the other direction you have a homotopy $h:S^{n} \times I \rightarrow X$ where $h(s,1) = *$. Therefore we get a factorization throught the qoutient $(S^n \times I)/(S^n \times {1} )$ which is the cone of $S^n$ hence the disc.


问:
What is the dependent formula of the vectors
Are vectors $(1,0,3,4)$, $(1,0,1,1)$, $(4,1,2,3)$ and $(6, 1,8,11)$ linearly
dependent? If yes, what is  the dependence formula.

I found out the determinant is $0$ isn't is supposed to be independent and what is dependance formula.
What is the dependance formula in this case?

Body:
We have, $$(6,1,8,11)=2(1,0,3,4)+0(1,0,1,1)+(4,1,2,3).$$

So if the vectors are dependent, then we must can write one of them as $(6,1,8,11)$ as linear combination of others:

$$c_1(1,0,3,4)+c_2(1,0,1,1)+c_3(4,1,2,3)=(6,1,8,11)$$

By solving this system, we will have $$c_1=2,c_2=0,c_3=1.$$


问:
What is the most efficient way to tile a regular shape inside an irregular shape?
Suppose you have a roughly circular irregular 2D shape, and want to tile as many regular squares (or rectangles) inside this shape as possible.

What would be the most efficient way to go about this?

Some constraints include:

The regular shapes can extend outside the irregular shape, though would prefer not to.

The regular shapes cannot overlap.

You do not need to use 100% of the irregular shape's area, though less wasted space is preferred.

It is preferable to maintain whole regular shapes inside the irregular shape resulting in some non-tiled space rather than having lots of regular shapes hanging outside the irregular shape.

Any reading materials would be appreciated too!

答:
If the squares or rectangles are all the same size and orientation, then the problem is the same as rasterizing the input shape. The solution depends on how the shape is given. Abstractly, it relies on a predicate that can decide whether a square is inside or outside the shape.

If the squares or rectangles can have different sizes but same aspect and orientation, then a good solution is to compute a quadtree approximation for the shape. If needed, neighbor squares of the same type can be merged into larger squares. Again, the predicate above is key. For shapes given implicitly, interval methods are simple to implement; see here for instance.

(image found here)


问:
A morphism of sheaves is surjective if it is surjective at the level of stalk for every point.
I would like to show that if $\phi:\mathscr{F} \rightarrow \mathscr{G}$ be a morphism of sheaves (on a topological space $X$) such that $\phi_P:\mathscr{F}_{P} \rightarrow \mathscr{G}_{P}$ is surjective for all $P \in X$ then $\phi$ is also surjective.

My effort: I am using the fact $\text{Im}(\phi)_P=\text{Im}(\phi_P)$ for all $P \in X$. Then we have $\text{Im}(\phi)_P=\mathscr{G}_P$ implying that $(\mathscr{G}/\text{Im}(\phi))_P=\mathscr{G}_P/\text{Im}(\phi)_P=0$. Now since $\mathscr{G}/\text{Im}(\phi)$ is a sheaf we've $\mathscr{G}/\text{Im}(\phi)=0$. From here how do I conclude that $\text{Im}(\phi)=\mathscr{G}$.

I need some help. Thanks.

答:
By $im(\phi) = \mathscr G$ I will assume you mean that the canonical map $im(\phi) \longrightarrow \mathscr G$ is an isomorphism. This argument holds in any abelian category.

As you have proven, $\mathscr G/im(\phi) = 0$. This is precisely the cokernel of $\phi$, and the image of a map in an abelian category is defined to be the kernel of its cokernel. We have the following universal property:

In symbols, $\pi \circ \psi = 0$ implies the existence and uniqueness of the dashed map making the diagram commute.

Now, as we said, $im(\phi) = ker(cok(\phi))=ker(\pi)$. So by "the canonical map $im(\phi) \longrightarrow \mathscr G$", I mean $\iota$. We will proceed to show that it is an isomorphism.

Since $\mathscr G/im(\phi)=0$, we necessarily have $\pi = 0$. We will therefore show that $id: \mathscr G \longrightarrow \mathscr G$ satisfies the universal property of $ker(\pi)$. Indeed, let $\psi: \mathscr H \longrightarrow \mathscr G$ satisfy $\pi \circ \psi = 0 \circ \psi$. Thus, we have the commutative diagram:

and certainly, only $\psi$ can fill in for the dashed map and allow the diagram to commute. Hence, this satisfies the same universal property as $ker(\pi)$. Hence, they are isomorphic. This arises by applying the universal property in both directions. In fact, this yields the following commutative diagram.

Hence, we have shown that the canonical map $im(\phi)=ker(\pi)\longrightarrow \mathscr G$ is an isomorphism. This is the closest to equality we can get in an abelian category.


问:
Number theory in sequence $x_{n+1}=x_n^3-2x_n^2+2$
$x_1=5, x_{n+1}=x_n^3-2x_n^2+2$
Prove that, there is no prime $p=4k+3(k>1)$ and $p\mid x_n^2-3x_n+3$

I think I can have $p\mid t^2+1$ and then I have QED.

But $p\mid x_n^2-3x_n+3$ means that $p\mid (2x_n-3)^2+3=t^2+3$

What should I do next?

答:
I found this to be a quite interesting (and challenging to solve) question. It's asking about the primes $p$ where

$$p \mid x_n^2 - 3x_n + 3 \tag{1}\label{eq1A}$$

As you showed, multiplying by $4$ gives

$$p \mid 4x_n^2 - 12x_n + 12 = (2x_n - 3)^2 + 3 \tag{2}\label{eq2A}$$

Since $p \neq 3$ (note $x_n \equiv 5 \pmod{72}$ for all $n \ge 1$, with this giving $x_n^2 - 3x_n + 3 \equiv 13 \pmod{72}$), this shows $-3$ is a quadratic residue modulo $p$.

Next, multiplying \eqref{eq1A} by $x_n - 3$ gives

$$\begin{equation}\begin{aligned}
(x_n - 3)(x_n^2 - 3x_n + 3) & = x_n^3 - 3x_n^2 + 3x_n - 3x_n^2 + 9x_n - 9 \\
& = x_n^3 - 6x_n^2 + 12x_n - 8 - 1 \\
& = (x_n - 2)^3 - 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

This means

$$(x_n - 2)^3 \equiv 1 \pmod{p} \tag{4}\label{eq4A}$$

If $n \gt 1$, substituting $x_n = x_{n-1}^3 - 2x_{n-1}^2 + 2$ gives

$$\begin{equation}\begin{aligned}
(x_{n-1}^3 - 2x_{n-1}^2)^3 & \equiv (x_{n-1}^2(x_{n-1} - 2))^3 \\
& \equiv x_{n-1}^6(x_{n-1} - 2)^3 \\
& \equiv 1 \pmod{p}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Repeating the substitutions for each $x_i - 2 = x_{i-1}^2(x_{i-1} - 2)$ for $i$ from $n - 1$ down to $2$ gives a product of $0$ or more $x_{i-1}^6$ values multiplied by $(x_1 - 2)^3 = 3^3$, i.e.,

$$\left(\prod_{i=1}^{n-1}x_i\right)^6(3^3) \equiv 1 \pmod{p} \tag{6}\label{eq6A}$$

Multiplying both sides by $3$ gives

$$\left(\left(\prod_{i=1}^{n-1}x_i\right)^3 3^2\right)^2 \equiv 3 \pmod{p} \tag{7}\label{eq7A}$$

This shows $3$ is also a quadratic residue. Thus, $3^{-1}(-3) \equiv -1 \pmod{p}$ is a quadratic residue, so there's an integer $x$ with $x^2 \equiv -1 \pmod{p} \implies x^4 \equiv 1 \pmod{p}$. Therefore, $4$ is the multiplicative order of $x$ modulo $p$, so $4 \mid p -1 \implies p \equiv 1 \pmod{4}$. This proves that $p$ cannot be of the form $4k + 3$.


问:
Evaluating a trigonometric integral with square root
I want to evaluate the integral

$ \int_0^{2\pi} \sqrt{1+a\cdot \cos(x)} ~dx$  where $|a| \leq 1$.

I have already tried to split the integral using the periodicity of $\cos$ into

$ \int_0^{\pi} \left( \sqrt{1+a\cdot \sin(x)} + \sqrt{1-a\cdot \sin(x)} \right) ~dx$,
but that does not seem to make things easier.

Do you have any suggestion on how to proceed?

答:
Unless $a = 1$ exactly, your integral is what turns out to be an Elliptic integral. No analytical solution can be found. Yet it is a well know integral. Indeed in literature we have that

$$\int \sqrt{1 + a\cos(x)}\ \text{d}x = 2 \sqrt{a+1} E\left(\frac{x}{2}|\frac{2 a}{a+1}\right)$$

As long as $a > 0$.

The notation $E\left( \cdot | \cdot \right) $ denotes the Elliptic integral of the second kind.

More here: https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind

In your case, it is well known that

$$\int_0^{2\pi} \sqrt{1 + a \cos(x)}\ \text{d}x = 4 \sqrt{a+1} E\left(\frac{2 a}{a+1}\right)$$


答:
The ratio of the absolute value of consecutive summands is
$$
\frac{(n+1)^{n+1}/(n+1)!}{n^n/n!} = \bigg(1+\frac1n\bigg)^n,
$$
the limit of which is $e$ as $n\to\infty$. Since $e>1$, the Ratio Test tells us that the series diverges.

It is true that the summand does not tend to $0$ (which also implies divergence of the series); but establishing that the summand does not tend to $0$ is more complicated than this Ratio Test argument, I believe.


问:
Result due to Cohn, unique division ring whose unit group is a given group?
This is a result of Cohn I would like to understand.

Let $G$ be a group, $g\in G$, and $\varphi$ a map of the subset $H=\{x\in G:x\neq 1\}$ into itself which has the properties

$\varphi(yxy^{-1})=y(\varphi x)y^{-1}$ for $x\in H$, $y\in G$.

$\varphi^2(x)=x$

$\varphi(x^{-1})=g(\varphi x)x^{-1}$

$\varphi(xy^{-1})=(\varphi(\varphi(x)\varphi(y^{-1})))\varphi(y^{-1})$ for $x,y\in H, x\neq y$.

Then there exists a unique division ring $D$ such that $D^*=G$ for the set of nonzero elements $D^*$ of $D$, and in $G$, $\varphi x=1-x$, $x\in H$, $g=-1$.

I did notice that the first three properties agree with $\varphi x=1-x$ for $x\in H$. However, for the fourth property to be satisfied, would require
$$
\begin{align*}
1-xy^{-1} &= (1-(1-x)(1-y^{-1}))(1-y^{-1})\\
&= x+y^{-1}-2xy^{-1}-y^{-2}+xy^{-2}
\end{align*}
$$
which seems like an odd relation to have.

I thought about this, but really had no idea where to begin. Is there a reference for this result, or even a sketch I could try to work through? Thank you.

答:
To make this exercise from Jacobson's Basic Algebra I correct, replace the last condition on the function $\varphi$ with the following one:

$y\varphi(y^{-1}\varphi(x))=x\varphi(x^{-1}\varphi(y))$ for $x,y\in H, x\neq \varphi(y)$ (and so $y \neq \varphi(x)$).

This is saying, in effect, that $y-(1-x) = x-(1-y)$, which, when combined with the other assumptions, is enough to imply associativity of addition. I'm guessing that this--or something equivalent to it--is what Cohn stated and Jacobson intended.


答:
"Approximate" means: make use of some mathematical tool that you have mastered, or studied, that will give you a good approximation of a certain function. The most known method is Taylor series. So what you have to do is make use of Taylor Series. You know that

$$f(x) \approx \sum_{k = 0}^{+\infty} \dfrac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

Where $(k)$ stands for the derivative of $f$ of order $k$, evaluated in the given center $x_0$.

In your case, you find the three derivatives of $f$ as Fred wrote, centered at $x_0 = 2$ (your $a$ is my $x_0$).

Title: Approximation by a polynomial degree 3 centered at a=2?

Body:
Approximate $\:\mathbf{f(x) = 1/\sqrt{x^3 + 1}}\:$ by a polynomial of degree 3 centered at a = 2.

What exactly is this question asking when it says "approximate"? What is the end result supposed to be I'm so lost

I found the derivatives(up to the 3rd) and I substituted x with 2, but now what?

I'd appreciate any help!


问:
How to prove that a set is closed
Consider the set $C$ of vectors in $\mathbb{R}^{N^2}$ which have symmetric rank-one matrix representation in $\mathbb{R}^{N \times N}$. In other words,
$$
C = \{ u \in \mathbb{R}^{N^2} | \exists v \in \mathbb{R}^{N}, u = vec(vv^T) \}
$$
where $vec(.)$ stacks the columns of a matrix into a vector.

Is the set $C$ closed? how can I prove that?

Thanks in advance!

答:
Excluding the zero vector makes $C$ not closed, as you can have a sequence of vectors $v_n\in \Bbb  R^N$ with $v_n\to \vec 0$, which would make a sequence $u_n = \operatorname{vec}(v_nv_n^T)$ in $C$ that converges to the zero vector.

Including the zero vector in $C$, on the other hand, changes the story.

You can express symmetry as a set of equations (a bunch of entries that must be equal to one another, or more usefully to us, a bunch of differences between pairs of entries that must be $0$). You can express rank-at-most-1 as a set of equations (all the $2\times 2$ minors must be $0$).

Each difference and each minor mentioned above is a continuous function $\Bbb R^{N^2}\to \Bbb R$. The set of elements in $\Bbb R^{N^2}$ where any one of them is equal to $0$ is closed. Thus their intersection is closed. But this intersection is $C$. So $C$ is closed.

Title: How to prove that a set is closed


答:
First a remark: as $\mathbb R^{N^2}$ is isometric to the space of real matrices $M_N(\mathbb R)$, I will work in that later space.

Let $\{u_n\}$ be a sequence of $C$. By definition, it exists a sequence $\{v_n\}$ of vectors of $\mathbb R^N$ such that $u_n = v_n v_n^T$. If $\{u_n\}$ converges to $u \in M_N(\mathbb R)$, $\{u_n\}$ is bounded. As all norms are equivalent on a finite dimensional space, this implies that $\{v_n\}$ is also bounded. According to Bolzano–Weierstrass theorem, $\{v_n\}$ has a converging subsequence to let say $v \in \mathbb R^N$. As $v \mapsto vv^T$ is continuous, we get that $u = vv^T$ and finally that $C$ is indeed closed.

Title: How to prove that a set is closed

Body:
Consider the set $C$ of vectors in $\mathbb{R}^{N^2}$ which have symmetric rank-one matrix representation in $\mathbb{R}^{N \times N}$. In other words,
$$
C = \{ u \in \mathbb{R}^{N^2} | \exists v \in \mathbb{R}^{N}, u = vec(vv^T) \}
$$
where $vec(.)$ stacks the columns of a matrix into a vector.

Is the set $C$ closed? how can I prove that?

Thanks in advance!


答:
You are entering the world of elliptic integrals.

Start using $\cos(x)=1-2 \sin ^2\left(\frac{x}{2}\right)$ to make
$$I=\int \sqrt{1+a\, cos(x)}\,dx=\int\sqrt{(1+a)-2 a \sin ^2\left(\frac{x}{2}\right)}\,dx$$
$$I=\sqrt{1+a}\int\sqrt{1-\frac{2 a }{1+a}\sin ^2\left(\frac{x}{2}\right)}\,dx$$ Now, let $x=2y$
$$I=2\sqrt{1+a}\int\sqrt{1-\frac{2 a }{1+a}\sin ^2\left(y\right)}\,dy$$
Have a look here.

Title: Evaluating a trigonometric integral with square root

Body:
I want to evaluate the integral

$ \int_0^{2\pi} \sqrt{1+a\cdot \cos(x)} ~dx$  where $|a| \leq 1$.

I have already tried to split the integral using the periodicity of $\cos$ into

$ \int_0^{\pi} \left( \sqrt{1+a\cdot \sin(x)} + \sqrt{1-a\cdot \sin(x)} \right) ~dx$,
but that does not seem to make things easier.

Do you have any suggestion on how to proceed?


问:
Show that the sequence $(u_n(t))_{n\in \mathbb{N}}$ converges to $t^2$
For my homework assignment, I had to use Picard iteration to find a sequence $\{u_n(t)\}_{n \in \mathbb{N}}$ to estimate a solution for the Initial Value Problem $$u'(t) = \frac{2u(t)}{t}$$
The sequence I got (and proved to be correct) is:
$$u_n(t) = \sum_{i=0}^n \frac{2^i}{i!}\ln^i|t|$$
What is left for me to do is take the limit as $n \rightarrow \infty$. Since the ODE is seperabel, I know the $u_n(t) \rightarrow t^2$ as $n \rightarrow \infty$. However, I have failed to prove this.

I had ideas using the comparison and/or integral tests but then I have difficulties with finding the correct bounds to use the sandwhich principle. Can anybody give me a push in the right direction?

答: $u_n(t)$ is just the $n-$th partial sum of the series $\sum \frac {x^{i}} {i!}$ where $x =2\ln |t|$. Hence it converges to to $e^{2\ln |t|}=t^{2}$.


问:
Solve equation $x^2 \cdot |x| = 2 \cdot \bar{x}$ in complex numbers
Well, to solve it I firstly tried to use exponential form. However, it didn't give me expected result. Also, I tried using trigonometric form but I am a bit inexperienced in it, so I couldn't observe anything interesting with it either.

I am new to complex numbers, so I would love to hear some advices on how to think of solution in such equations and maybe some insights on how to solve this one.

答: Note that\begin{align}\require{cancel}x^2|x|=2\overline x&\implies\bigl|x^2|x|\bigr|=\left|2\overline x\right|\\&\iff|x|^3=2|x|.\end{align}Therefore, $|x|=0$ or $|x|=\sqrt2$. In the first case, we have $x=0$, which is indeed a solution of the equation. Otherwise, $x=\sqrt2e^{i\theta}$, for some $\theta\in\Bbb R$, and your equation becomes$$\cancel{2\sqrt2}e^{2i\theta}=\cancel{2\sqrt2}e^{-i\theta}.$$Can you take it from here?


问:
Let $f: \ \mathbb{R} \implies \mathbb{R} \ $ be defined by $f(x) = \frac{x}{2}$ is $f$ onto and one-to-one?
Let $f: \ \mathbb{R} \implies \mathbb{R} \ $ be defined by $f(x) = \frac{x}{2}$ is $f$

onto

one-to-one

Question 1, what is a better way to write this up?

$f$ is onto as there exists a value $x$ for every possible $f(x)$

Question 2, I know it is one-to-one but what is a nice way to write it up?

$f$ is one-to-one as there exists a unique $x$ for every possible $f(x)$

I understand that neither of my answers would suffice as proofs but what is a good way to write up the answers so that they would?

答:
If $y \in \mathbb R$, then $f(2y)=y.$  Conclusion ?

If $a,b \in \mathbb R$ and $f(a)=f(b)$, then $\frac{a}{2}=\frac{b}{2}$, hence $a=b.$ Conclusion ?


问:
Inequality $2(x^p+y^p)^{1/p} \le x+y$
Is it true that $2(x^p+y^p)^{1/p} \le x+y$ for all $x,y \in [0,+\infty)$ where $p \in [1,+\infty)$?
I would like to add something to this question, but I really have no idea how to deal with this inequality.
Any hint would be appreciated. Thank you!

EDIT ok it clearly is not true if for example $x=0$ and $y=1$, but then for which values of $x,y \in \mathbb{R}$ this inequality holds?

答:
You are (almost) looking at what is called a power mean or generalized mean. For these beasts, there exists the inequality   which holds for all real $p \ge 1$.
$$
\Big(\frac{x^p+y^p}{2}\Big)^{1/p} \color{red}{\ge} \frac{x+y}{2}
$$
in a well-defined sense: the inequality becomes an equality if and only if $x=y$, or (trivial) $p=1$.

Use this to write $$
\leftrightarrow  2(x^p+y^p)^{1/p}  \ge 2^{1/p} (x+y) > x+y
$$

So your LHS will never be smaller than the RHS, but larger.

If you want an upper bound, you can of course write $2(x^p+y^p)^{1/p}  \le 2 (x+y)$. This is equivalent to $ (1+ \frac{y}{x})^p > 1+ (\frac{y}{x})^p$ and an elegant proof for this can be found here: $ (1+a)^p - 1 = \int_0^a p(1+t)^{p-1} \, dt > \int_0^a pt^{p-1} \, dt = a^p$

Title: Inequality $2(x^p+y^p)^{1/p} \le x+y$


问:
What is a typical example of the tensor product of modules failing to be left exact?
I am looking for an example of an exact sequence of $R$-modules $$ 0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0 $$ and a $R$-module $N$, such that $$ 0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0 $$
fails to be exact.</p>

答:
The typical example is made up with $M=M'=\mathbb Z$, and $N= \mathbb Z/ n \mathbb Z$, with the injection $\mathbb Z \to \mathbb Z, x \mapsto n x$.
When tensorized by $Id_{\mathbb Z/ n \mathbb Z}$, this map becomes zero.</p>


问:
Simple proof of $\int_I \exp(-f(x))dx\geq \int_I \exp(-g(x))dx$ if $\int_I (f(x) - g(x))\exp(-g(x))dx = 0$
If for two functions $f(x)$ and $g(x)$ we have that:

$$\int_I (f(x) - g(x))\exp(-g(x))dx = 0 \tag{1}$$

where $I$ is an arbitrary interval on the real line, then

$$\int_I \exp(-f(x))dx\geq \int_I \exp(-g(x))dx \tag{2}$$

This follows in a straightforward way from the Bogoliubov inequality. It seems to be a reasonable powerful tool to get to sharp bounds for certain integrals or summations, but it doesn't seem to be used a lot for this purpose in mathematics. I was wondering if a more straightforward proof can be given for the special case of integrals and summations.

This can be used to obtain sharp lower bounds for integrals of the form $\int_I \exp(-f(x))dx$ by writing down a function $g(x)$ containing parameters, and then imposing the constraint (1) to eliminate one parameter and then maximizing $\int_I \exp(-g(x))dx$ w.r.t. the remaining parameters. A simple example is to take $f(x) = x^2$ and $g(x) = a + b x$ for $b>0$ and $I$ the positive real line, which yields the inequality $\pi \geq e$.

答:
Using $e^u \ge 1+u$ for $u \in \Bbb R$ we have
$$
 e^{-f(x)} - e^{-g(x)} = \bigl (e^{g(x)-f(x)} - 1 \bigr) e^{-g(x)} \ge \bigl(g(x)-f(x)\bigr) e^{-g(x)}
$$
so that
$$
 \int_I e^{-f(x)} \, dx - \int_I e^{-g(x)} \, dx \ge \int_I \bigl(g(x)-f(x)\bigr) e^{-g(x)} \, dx \, .
$$
If $(1)$ holds then the right-hand side is zero, and that implies $(2)$.

We can even conclude that strict inequality holds in $(2)$, unless $f(x) = g(x)$ for almost all $x \in I$.


答:
One way would we to substitute $x \in \mathbb{C}$ with $x = a + b i$, where $a,b \in \mathbb{R}$.
Put this in the equation:
$$
(a^2 + 2abi -b^2)(\sqrt{a^2+b^2}) = 2 (a - bi)
$$
Comparing reall and imaginary part leads to
$$
\begin{align}
a^2 \sqrt{a^2+b^2} - b^2 \sqrt{a^2+b^2}  &= 2 a \\
2 a b \sqrt{a^2+b^2} &= -2 b \\
\end{align}
$$

You can solve this and get to a couple of solutions:
$$
\begin{align}
a &= -\frac{1}{\sqrt{2}}, & b &= -\sqrt{\frac{3}{2}} \quad \lor \\
a &= -\frac{1}{\sqrt{2}}, & b &= \sqrt{\frac{3}{2}} \quad \lor \\
a &= 0, & b &= 0 \quad \lor \\
a &= \sqrt{2}, & b &= 0
\end{align}
$$

//EDIT: I forgot the "2" on the right side... fixed it

Title: Solve equation $x^2 \cdot |x| = 2 \cdot \bar{x}$ in complex numbers

Body:
Well, to solve it I firstly tried to use exponential form. However, it didn't give me expected result. Also, I tried using trigonometric form but I am a bit inexperienced in it, so I couldn't observe anything interesting with it either.

I am new to complex numbers, so I would love to hear some advices on how to think of solution in such equations and maybe some insights on how to solve this one.


问:
Differential equations basic problem
I know this is a basic Physics problems but somehow I can't solve it.
We have the differential equation: $2x''x^2 - 4 x^2x' - 2  x^3 = 0$

We have to conclude that the system:

$x' = y $

$y' = 2y + x$

..is equivalent to the differential equation. How can I do it?

Thanks in advance!

答:
If $x \neq 0$, then solving for $x^{\prime\prime}$ we get
$$
x^{\prime\prime} = 2x^{\prime} + x
$$
Now let $y = x^{\prime}$. Then
$$
y^{\prime} = x^{\prime\prime} = 2x^{\prime} + x = 2y + x
$$</p>

Title: Differential equations basic problem


问:
Lax-Friedrichs scheme stability necessary sufficient
We have the advection equation $q_t+aq_x = 0$ with constant velocity $a$ and the Lax-Friedrichs scheme

$Q_j^{n+1}=\frac{1}{2}(Q_{j-1}^n+Q_{j+1}^n)+\frac{ak}{2h}(Q_{j-1}^n-Q_{j+1}^n)$.

How to show that $|\frac{ak}{h}|\leq 1$ is necessary for stability and sufficient for stability in the discrete $L^1$ norm?

Edit:
The discrete $L^1$ norm is defined as

$\|Q^{n+1}\|_1 = h\sum\limits_j|Q_j^{n+1}|$.

So far I've noted the above as:

$\|Q^{n+1}\|_1=h\sum\limits_j\left|\frac{1}{2}(1-\frac{ak}{h})Q_{j-1}^n + \frac{1}{2}(1+\frac{ak}{h})Q_{j+1}^n\right|$

How should I take it from here?

答:
Necessity:

The stated condition is precisely the CFL condition.

Sufficiency:

It appears that you have a couple of sign errors in your final expression. It should read

\begin{align}
\| Q^{n+1} \|_1 &= h \sum_j \left| \frac{1}{2} \left( 1 + \frac{ak}{h} \right) Q^n_{j-1} + \frac{1}{2} \left( 1 - \frac{ak}{h} \right) Q^n_{j+1} \right| \\
%
&\leq \frac{h}{2} \left[ \sum_j \left| \left( 1 + \frac{ak}{h} \right) Q_{j-1}^n \right| + \sum_j \left| \left( 1 - \frac{ak}{h} \right) Q_{j+1}^n \right| \right]
\end{align}

Now, suppose that $\left| \frac{ak}{h} \right| \leq 1$. Then it holds that
$$
\left( 1 - \frac{ak}{h} \right) \geq 0 \quad \text{and} \quad \left( 1 + \frac{ak}{h} \right) \geq 0.
$$
Thus, these terms can be extracted from the absolute values to get

\begin{align}
\| Q^{n+1} \|_1 &\leq \frac{h}{2} \left[ \left( 1 + \frac{ak}{h} \right) \sum_j \left| Q_{j-1}^n \right| + \left( 1 - \frac{ak}{h} \right) \sum_j \left| Q_{j+1}^n \right| \right] \\
%
&= \frac{1}{2} \left[ \left( 1 + \frac{ak}{h} \right) \| Q^n \|_1 + \left( 1 - \frac{ak}{h} \right) \| Q^n \|_1 \right] \\
%
&= \| Q^n \|_1.
\end{align}

The method is therefore stable.


问:
Let $f:\mathbb{R}^{+}\to \mathbb{R}$ such that $f$ is differentiable and has the property $f(xy) =f(x) +f(y)$.Show that $f(x) =f′(1) \ln(x)$

Let $f:\mathbb{R}^{+}\to \mathbb{R}$ such that $f$ is differentiable and has the property $f(xy) =f(x) +f(y)$.Show that $f(x) =f′(1) \ln(x)$.

答:
Let $$(*) \quad f(xy)=f(x)+f(y)$$

for $x,y >0.$

If we differentiate $(*)$ with respect to $x$, we get

$$yf'(xy)=f'(x).$$

If we differentiate $(*)$ with respect to $y$, we get

$$xf'(xy)=f'(y).$$

These equations give

$$xf'(x)=yf'(y)$$

for all $x,y >0.$

With $y=1$ we derive

$$xf'(x)=f'(1),$$

hence $f'(x)= \frac{f'(1)}{x}.$ Therefore

$$f(x)= f'(1) \ln x+c.$$

Now show that $f(1)=0$, to see that $c=0.$


问:
Methods of solving an initial-value problem for a separable first-order ODE
In Schaum's outline of Differential Equations it is said that the initial-value problem of $A(x) d x+B(y) d y=0 $; $y\left(x_{0}\right)=y_{0}$ can be solved either $(1)$ integrating and applying the initial condition to evaluate $c$ or by using $(2)$:

$$
\int A(x) d x+\int B(y) d y=c; \quad y\left(x_{0}\right)=y_{0} \tag{1}
$$

$$
\int_{x_{0}}^{x} A(x) d x+\int_{y_{0}}^{y} B(y) d y=0 \tag{2}
$$

It is also said that this second method may not determine the solution of the initial-value problem uniquely; thal is, it may have many solutions, of which only one will satisfy the initial-value problem.

As I see it, both methods are equivalent: if $\widetilde A(x)$ and $\widetilde B(y)$ are the primitive functions of $A(x)$ and $B(y)$, then

$$(1) \rightarrow \widetilde A(x) + \widetilde B(y) =c \rightarrow c= \widetilde A(x_0) + \widetilde B(y_0)\rightarrow \\\widetilde A(x)- \widetilde A(x_0) + \widetilde B(y)- \widetilde B(y_0)=0$$

Which is the same result we get from $(2)$:

$$(2) \rightarrow \widetilde A(x)- \widetilde A(x_0) + \widetilde B(y)- \widetilde B(y_0)=0$$

Why does then the book say that the second method may may not determine the solution of the initial-value problem uniquely?

答:
This is true for both methods, $\tilde B$ need not be a monotonous function, there might be multiple points in its pre-images of the value $c-\tilde A(x)$.

Why that was only attached to the second variant can only be guessed, perhaps there  was a stylistic influence, a tendency towards a narrower narrative flow...


问:
Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in $(a,b)$ are topologically conjugate
I've read the proof of above theorem in L. Wen's book Differentiable Dynamical Systems (Thm.1.8 on p.11):

Theorem: Any two orientation-preserving homeomorphisms of $[a,b]$ without fixed point in$(a,b)$ are topologically conjugate.

Here is the proof :

Proof: Let $f$ and $g$ be two orientation preserving homeomorphisms of $[a,b]$ without fixed points in $(a,b)$. We assume $f(x)>x$ and $g(x)>x$ for any $x\in (a,b)$. For other cases the proofs are similar. Take any $p \in (a,b)$ and any homeomorphism $h_0:[p,f(p)] \to [p,g(p)]$ such that $h_0p=p$ and $h_0(f(p))=g(p)$. For each integer $n$ define :

\begin{align}
h_n:[f^n(p),f^{n+1}(p)] \to [g^{n}(p),g^{n+1}(p)], \, h_n = g^n \circ h_0 \circ f^{-n}
\end{align}

It is easy to see that these $h_n$ glue together to give a homeomorphism $ h: [a,b] \to [a,b]$ such that $hf=gh$.

Before I ask my questions: a homeomorphism  $f:[a,b] \to [a,b]$ is orientation preserving if it is strictly increasing and fixes the two end points of the interval.

Now my questions :

Why is $h_n$ a homeomorphism for $n \in \mathbb{Z}$? Is it because it's a bijection on a compact space $[a,b]$ ? How is it continuous?

How do $h_n$ really glue together to give a $h:[a,b] \to [a,b]$ such that $hf=gh$?

答:
For $\phi = f, g$ we set $\phi^0 = id$ and $\phi^{-m} = (\phi^{-1})^m$ for $m \in \mathbb N$. This defines $\phi^n$ for all $n \in \mathbb Z$. We have $\phi^n(p)  \in (a,b)$ for all $n$ and $\phi^{n+1}(p) = \phi(\phi^n(p)) > \phi^n(p)$ for all $n \in \mathbb Z$. We claim that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.

Clearly $\lim_{n \to +\infty} \phi^n(p) = b' \le b$. Note that all $\phi^n(p) < b'$. Assume $b' < b$. Then $b' \in (a,b)$ and $\phi(b') > b'$, i.e.  $\phi(b') - b' > 0$. By continuity there exists $\epsilon > 0$ such that $\phi(t) - t > 0$ for all $t \in (a,b)$ with $\lvert t - b' \rvert < \epsilon$. But $\lvert \phi^n(p) - b' \rvert < \epsilon$ for sufficiently large $n$, thus $\phi^n(p) > b'$ which is a contradiction. The case $n \to -\infty$ is treated similarly.

Each $h_n$ is a homeomorphism.

Proof. Since $\phi^n$ is a homeomorphism, the restriction $\overline{\phi^n} : [p,\phi(p)] \stackrel{\phi^n}{\to} [\phi^n(p),\phi^{n+1}(p)]$ is a homeomorphism (the notation $\overline{\phi^n}$ is used as a distinction to $\phi^n$). Note that $(\overline{\phi^n})^{-1}$ is the restriction of $\phi^{-n}$ to $[\phi^n(p),\phi^{n+1}(p)]$.

Thus $h_n = \overline{g^n} \circ h_0 \circ (\overline{f^n})^{-1}$ is a homeomorphism. This is what you wrote as $h_n = g^n \circ h_0 \circ f^{-n}$. In fact, for $t \in [f^n(p),f^{n+1}(p)]$ we have
$$h_n(t)  = g^n(h_0(f^{-n}(t))) \tag{1} .$$
By construction
$$h_n(f^n(p)) = g^n(p), h_n(f^{n+1}(p)) = g^{n+1}(p) \tag{2} .$$
For $t \in [f^n(p),f^{n+1}(p)]$ we get
$$h_{n+1}(f(t)) = g^{n+1}(h_0(f^{-(n+1)}(f(t)))) = g^{n+1}(h_0(f^{-n}(t))) = g(g^n(h_0(f^{-n}(t)))) \\ = g(h_n(t)) \tag{3}.$$

Define

$$h(t) = \begin{cases}  a & t = a \\ h_n(t) & t \in  [f^n(p),f^{n+1}(p)] \\ b & t = b\end{cases}$$

This is well-defined by $(2)$ because $[f^n(p),f^{n+1}(p)] \cap [f^{n+1}(p),f^{n+2}(p)] = \{f^{n+1}(p)\}$. It is also shows that $h$ is continuous on $\bigcup_{n \in \mathbb Z} [f^n(p),f^{n+1}(p)] =(a,b)$. Continuity in $a$ and $b$ follows from the fact that $\phi^n(p) \to b$ as $n \to +\infty$, $\phi^n(p) \to a$ as $n \to -\infty$.

By construction $h$ is a continuous bijection, thus a homeomorphism, such that $h \circ f = g \circ h$.


问:
Approximation by a polynomial degree 3 centered at a=2?
Approximate $\:\mathbf{f(x) = 1/\sqrt{x^3 + 1}}\:$ by a polynomial of degree 3 centered at a = 2.

What exactly is this question asking when it says "approximate"? What is the end result supposed to be I'm so lost

I found the derivatives(up to the 3rd) and I substituted x with 2, but now what?

I'd appreciate any help!

答:
To approximate a function usually means that you want to construct another function that 1) should be easier to compute and 2) takes values reasonably close to the original function. [The criterion 1) might be different, but this is rare.]

A method to obtain this is to start with a candidate expression (such as a polynomial) with adjustable parameters, and find the values of the parameters that minimize some measure of the average error. So you are trying a family of functions, and try to find the "best" of them.

Another method (probably the one meant to be used here) is to match the function closely in the neighborhood of a given $x$. This is done by ensuring that the value and derivatives (up to some order) of the function and its approximation are equal at the given $x$. This is the so-called Taylor approximation.

On the picture, the given function in blue, and Taylor approximations of increasing degree.


问:
Set of Homomorphisms as an $R-$ module
$\require{AMScd}$
I'm reading A first course of homogocial algebra by D.G. Northcott, and I don't quite get the Example 1 on page 25. Here's what it says:

Example 1

Let the module $A$ belongs to $\mathscr{C}_R$ (i.e the category of $R-$modules). Since $R$ is an $(R; R)-$bimodule. $\mbox{Hom}(R; A)$ is an $R-$module of the same type as $A$.

The bolded part is where I don't get it.

So, say I have $A$ as a left $R-$module, now I'll try to define the outer multiplication of $\mbox{Hom}(R; A)$, so to make it also a left $R-$module.

From this part on, I'll consider $f \in \mbox{Hom}(R; A)$; $r, \alpha \in R$.

Trial 1

I define $(rf)(\alpha) = f(r\alpha)$. I don't think this should work, as:

$\begin{CD}
(r_1f)(r_2\alpha) @= r_2(r_1f)(\alpha) @= r_2f(r_1\alpha) @= \color{red}{r_2r_1}f(\alpha) \\ @| \\ f(r_1r_2\alpha) \\ @| \\ \color{red}{r_1r_2}f(\alpha)
\end{CD}$

Trial 2

I define $(rf)(\alpha) = f(\alpha r)$. I don't think this should work, either:

$\begin{CD}
(r_1r_2f)(\alpha) @= (r_1(r_2f))(\alpha) @= r_1f(\alpha r_2) @= f(\alpha \color{red}{r_2 r_1}) \\ @| \\ f(\alpha \color{red}{r_1r_2})
\end{CD}$

I think I'm missing something really big here. I would be so glad if you guys can help me.

Thanks a lot in advance,

And have a good day,

答:
It might be visually helpful to track the module side in the notation. To do that, I'll write $A_R$ to indicate a right $R$ module $A$, or $_RA$ to indicate a left $R$ module.

It will also help to consider bimodules over two different rings, since it will help keep us from confusing the module operations.

Suppose $_SB_R$ is a bimodule.

$H=Hom(B_R,A_R)$ has a natural right $S$ module structure via $(f\cdot s)(x):=f(sx)$.

Thus if $r$ is another element of $R$, $(f\cdot s)(xr)=f(sxr)=f(sx)r=(f\cdot s)(x)r$. You can see that the module actions are taking place on opposite sides of the ring, and so they don't need to "cross over" each other.

If $_TA_R$ is a bimodule, then $H$ has a natural left $T$ module structure given by

$(t\cdot f)(x):=tf(x)$. Again, $(t\cdot f)(xr)=tf(xr)=tf(x)r=(t\cdot f)(x)r$. The same thing has happened here that the $R$ module action and $T$ module action do not have to cross each other.

Both can happen at once: If $_TA_R$ and $_SB_R$ are bimodules, then $_TH_S$ is a bimodule, using both operations above (the compatibility $(th)s=t(hs)$ has to be checked.)

The problems with your definitions are that you are crowding the side of modules in the Hom and the new module action on one side, where they need to "cross over" each other in order to come out right. They really should be taking place on opposite sides of each other, as with these two definitions.

Let's work out the "left hand version" of this, which it looks like you'll need, since you're considering $H=Hom(_RR,_RA)$. The natural left module structure would arise from the right module structure of $R_R$. That is, $(r\cdot f)(x):=f(xr)$ would be a left $R$ linear map: $(r\cdot f)(r_1x)=f(r_1xr)=x_1(f(xr))=x_1(r\cdot f)(x)$

Title: Set of Homomorphisms as an $R-$ module


问:
Special type of "Bipartite" graph
How do you call a graph that the bipartite property is satisfied for the one set of nodes and not the other?

Say you have two set of nodes $A$ and $B$ where there can be edges between nodes of set $A$ but there can be no edges between nodes of set $B$.Of course there can be edges between nodes of different sets.Is there a name for this type of graph?

答:
This is more a characteristic of $B$. A subset $B$ of the nodes of a graph that have no edges between each other is called an independent set.

To further elaborate:

Every independent subset fulfills the criterion of the question, and every $B$ that fulfills the criterion of the question is inversely an independent subset.

Note that there are actually no criteria on $A$ other that it's the remaining graph, after removing $B$.

Title: Special type of "Bipartite" graph


问:
changing argument of integrand of a periodic function inside an integral
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a $2\pi$-periodic function that is integrable on $[0,2\pi]$. Then for all $x \in \mathbb{R}$:
\begin{align}\int^{x+2\pi}_x f(y) \ dy =\int^{2\pi}_0 f(y) \ dy = \int^{2\pi}_0 f(y+x) \ dy. \end{align}
I get the first equation, because $F(x) = \int^{x+2\pi}_x f(y) \ dy$ is a constant function. However, can someone show me why the second equation is true ? I thought that $f(y) = f(y+x)$ is only true when $x = 2\pi$ , since it's a $2\pi$-periodic function ?

答:
We can set: $u=x+y$ and it follows $\frac{\textrm{d}u}{\textrm{d}y}=1  \rightarrow \textrm{d}u=\textrm{d}y$.

The new integration borders are [$x+0$ und $x+2\pi$].

After you put everything in your integral you get what you want:
\begin{align}\int^{x+2\pi}_x f(y) \ dy =\int^{2\pi}_0 f(y) \ dy = \int^{2\pi}_0 f(y+x) \ dy= \int^{x+2\pi}_x f(u) \ du
\end{align}


问:
Finding $abc$ for distinct reals satisfying $\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$
The problem is

Let $a$, $b$, $c$ be three distinct real numbers satisfying
$$\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ca}$$ Find $abc$.

I can solve this in the case of $a, b, c \ge 0$ by assuming $a =\max\{a, b, c\}$. In that case, $abc =0$. But I haven't solved it yet in the general case.

Please help me. Thanks.

答: As @fleablood observed, none of $a,b,c$ can be zero. Hence, notice that the condition is equivalent to $$\frac{1+ab}{a}=\frac{1+bc}{b}=\frac{1+ca}{c}\iff \frac1a+b=\frac1b+c=\frac1c+a$$ Yet this implies that $$\frac1a+b=\frac1b+c\iff b-c=\frac{a-b}{ab}$$ Similarly $$\frac1b+c=\frac1c+a\iff c-a=\frac{b-c}{bc}\quad\text{and}\quad  \frac1c+a=\frac1a+b\iff a-b=\frac{c-a}{ca}$$ Multiply the three resulting equations to obtain $$(b-c)(c-a)(a-b)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2}$$ Since $a,b,c$ are distinct real numbers, we have that $a-b\not= 0, b-c\not=0, c-a\not=0$ and thus $$a^2b^2c^2=1\iff abc=\pm1$$ Can you take it from here? Can both $1$ and $-1$ be attained?


问:
Formulate the Langrangian function of a non-linear optimization problem and solve it for $y\geq0$
Consider the (non-linear) optimization problem ($P$)

$$max \quad3x_1 + 4x_2$$

$$s.t. \quad x_1^2 + x_2^2 \leq 25$$

$$ \quad x_1,x_2 \geq 0$$

Formulate the Lagrangian function $\varTheta(y)$ and solve it for fixed $y>0$

This is the function that I obtained.
$\varTheta(y)=25y_1 + max\{(3-y_1x_1)x_1 + (4-y_1x_2)x_2; \quad x_1,x_2\geq 0\}$

However, Since the coefficents in the $max$ function still have $x_i$'s in them, I am not sure how to proceed. Could anyone please help me out?

答:
Since the objective function is increasing in the domain you just have to check the boundary an the problem reduces to
$$max_{x_1,x_2}\quad 3x_1+4x_2$$
$$s.t.\quad x_1^2+x_2^2=25$$
which has the Lagrangian
$$\varTheta(x_1,x_2,\lambda)=3x_1+4x_2-\lambda(x_1^2+x_2^2-25)$$
By applying first order conditions
$$\frac{\partial \varTheta}{\partial x_1}=3-2\lambda x_1=0$$
$$\frac{\partial \varTheta}{\partial x_2}=4-2\lambda x_2=0$$
$$\frac{\partial \varTheta}{\partial \lambda}=-x_1^2-x_2^2+25=0$$
which can be solved by $x_1=3$, $x_2=4$ and $\lambda=1/2$.</p>


问:
Determinant of symplectic matrix
A $2n \times 2n$ matrix $S$ is symplectic, if $SJ_{2n}S^T=J_{2n}$ where
\begin{equation}
J_{2n} = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}.
\end{equation}
My question is, how to prove that $\det S =1$, without using Pfaffian, as given in the wikipedia article.

答:
After the interruptions and warnings of some moderators, I rewrote my answer in more descent way

In Grove's book, Classical groups and geometric algebra, https://bookstore.ams.org/gsm-39 , He showed that symplectic transvections generate the symplectic group. This means that any symplectic matrix can be written as a product of symplectic transvections. I am explaining more below.

Let $(V,\{ .,. \})$ be a symplectic vector space. Let $a$ be a real number and $u \in V$. Then $T_{a,u}(v)=v+a\{v,u\}u$ is called symplectic transvection. Grove showed that any symplectomorphism can be written as a product of symplectic transvections. We already know that the set of symplectomorphisms is isormorphic to the set of symplectic matrices. So what we do for any symplectomorphism is valid for any symplectic matrix.

My aim is to show determinant of any symplectomorphism is 1. So it is equivalent to say that the determinant of product of symplectic transvections is 1.

Determinant of any matrix is product of its eigenvalues. Let us look at eigenvalues of any transvection. $T_{a,u}(v)=\lambda v \longrightarrow v+a\{v,u\}u = \lambda v \longrightarrow (1-\lambda)v=-a\{v,u\}u$.

We have $(1-\lambda)v=-a\{v,u\}u$. So they are linearly dependent if the coefficients are nonzero. But in this case $\{v,u\}$ vanishes because the form is skew symmetric linear. So $\lambda=1$. Assume they are linearly independent, then the coefficients must be zero, so again $\lambda$ is 1.

Therefore determinant of any symplectomorphism is $1$. Thus determinant of any symplectomorphism is $1$.

For details you can check my notes in the following link
https://ferhatkarabatman.github.io/positive_transvections.pdf

Title: Determinant of symplectic matrix


问:
Evaluate $\sum\limits_{n=1}^{+\infty} \frac{\left( \frac{3-\sqrt{5}}{2} \right)^{n}}{n^{3}}$
Evaluate

$$\sum\limits_{n=1}^{+ \infty} \frac{  \left( \frac{3-\sqrt{5}}{2} \right)^{n} }{n^{3}}$$

We can use the Fourier series to calculate this sum, because it converges.

Also, we know that $\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. What is going on about this number $\frac{3-\sqrt{5}}{2}$ ? we know something else ?

Thank you for your answer but we know something else without about trilogarithm function?

Body:
$$\sum_{n=1}^\infty \frac{a^n}{n^3}=\text{Li}_3(a)$$ Making $a=\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$, you just get a number
$$\text{Li}_3\left(\frac{1}{\varphi ^2}\right)=0.4026839629521090211599594481825111422197338\cdots$$


问:
sectionally/piecewise continuous functions
Assume
$f$
and
$g$
are two piecewise continuous functions on an interval
$(
a
,
b
)$
containing the point
$t_0$
. Assume further that
$f$
has a jump discontinuity at
$t_0$
while
$g$
is
continuous at
$t_0$
.How can i verify that the jump in the product
$fg$
at
$t_0$
is given by
“the jump in
$f$
at
$t_0$
”
×
$g
(
t_0
)$
?</p>

答: You can take the the two one-sided limits of $fg$, getting $\displaystyle \lim_{x \to t_0^+} f(x)g(x) - \displaystyle \lim_{x \to t_0^-} f(x)g(x)$.</p>


问:
What $i-1$ means in a summation formula?
What $i-1$ means when comes on the top of $∑$?
$$
\sum_{i=1}^{i-1}v_i
$$

(Encountered here)

If v is: $2, 4, 6$, then the sum will be: $2+4+6-2$?

Thanks.

答: There is no meaning. It is a typo.</p>

Body: There is no meaning. It is a typo.


问:
how to solve $\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$?
$$\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$$

I tried this this:
$$\int_{y}^{1} e^{x^2}dx=\sum_{0}^{\infty}\frac{1}{n!}(\frac{1}{2n+1}-\frac{y^{2n+1}}{2n+1})$$
I don't know what to do after or even if this way works...

答: You can do it as follows:\begin{align}\int_0^1\int_y^1e^{x^2}\,\mathrm dx\,\mathrm dy&=\int_0^1\int_0^xe^{x^2}\,\mathrm dy\,\mathrm dx\\&=\int_0^1xe^{x^2}\,\mathrm dx\\&=\left[\frac12e^{x^2}\right]_{x=0}^{x=1}\\&=\frac{e-1}2.\end{align}


问:
Showing a solution to this differential equation is never zero
Say I want to find a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for some $k \neq 0$, $f' = kf$. Every time I have seen a book solve this, the expression is rewritten  as $$\frac{1}{f}f' = k$$ and then solved using u-substitution. To be technical though, must one first know that $f$ is never zero? If so, how could I go about showing that $f$ is never zero? I know that $f$ must be infinitely differentiable and if for some $a \in \mathbb{R}$, $f(a) = 0$, then for each $n \in \mathbb{N}_0$, $f^{(n)}(a) = 0$. I think I remember from complex analysis, if $f$ is holomorphic with this property, then $f$ is zero everywhere, but I'm not sure about the usual case of $\mathbb{R} \rightarrow \mathbb{R}$.

答:
$$\frac{df}{dx}=kf(x)$$
Obviously a trivial solution is $$f(x)=0.$$
So we cannot say that $f$ is never equal to $0$.

Then we solve the ODE in the cases of $f\neq 0$. Note that is not the same as assuming that $f$ is never equal to $0$.

No need to repeat here the classical solving leading to
$$f(x)=C\:e^{kx}$$
with $\quad C\neq 0\quad$ in order to respect the case $f\neq 0$ considered.

This solution is completed with the trivial solution $f=0$ which is consistent with $C=0$. This leads to the general soltion :
$$f(x)=C\:e^{kx}\quad\text{any}\quad C.$$


问:
Question about a proof from Spivaks calculus book 4th edition
I am currently reading Spivaks Calculus book 4th edition, and am on the chapter with limits. Theorem one of this chapter says that as $x$ approaches $a$, you can not approach two different limits $l$ and $m$ unless they are equal. The proof of this theorem makes sense except for one thing. He ends up with a contradiction, but I have no idea what he contradicted. Would someone be able to explain it to me? I know I am a little vague, but their are pdf's of his book, so I figured it could be googled.</p>

答:
In the last line of calculations in the proof, look at the leftmost and rightmost expressions: you get

$$|l-m|<|l-m|$$

and this, of course, is a contradiction: no real number can be less than itself.


问:
Given a vector, get the coordinates of the same vector on a rectangle.
I have a circle and a known size vector starting from the center of that circle which comes out of a rectangle. My goal is to know the coordinates (x, y) of the vector being located on the rectangle and of the same size as the first vector.

Let's see an exemple:

exemple

The vector AB comes out of the rectangle, so I would like to know the coordinates of the vector AE of the same size as AB. If this is possible, I would like to get the closest vector (in this case AE but not AG)

答:
Say center of the circle is at the origin $O$.

Radius of the circle = $r$. Lengths of the sides of the rectangle are $c, d \, (c \geq d \geq r)$. $c$ is horizontal sides in your diagram and $d$ is vertical sides.

A position vector $A (a_1,a_2)$ is out of a given rectangle and $OA = \vec{a}$.

You can find points on the rectangle with the magnitude of vector $\vec{a}$, only if

$r \leq |\vec{a}| \leq \sqrt{c^2 + d^2} - r \sqrt2 \, \,$ ($c, d$ are sides of the rectangle).

Say we need to find a position vector $B (b_1, b_2)$ on the perimeter of the rectangle such that $\vec{b} = \vec{a} \, \, (\vec{OB} = \vec{b})$.

Any point on the sides of the rectangle are given by (depending on which side it is)

$(\pm x, r), (-(c-r), \pm y), (\pm x, -(d-r)), (r, \pm y)$.

Equate distance on each side of the rectangle to the magnitude of $|\vec{a}|$ to find the possible points.

For example, $x^2 + r^2 = |\vec{a}|^2$ should give you possible points on the upper horizontal side of the rectangle. Now how do you know this point works or not? Using the constraints for the points to be on the rectangle,

$-(c-r) \leq x \leq r$

$-(d-r) \leq y \leq r$

Once you know the points on the rectangle, you know all possible vector $\vec{b}$.

How do you know which one is closest? Do a dot product and find the angle between them. The lowest angle between them will give you the closest position vector $\vec{B}$ to $\vec{A}$.

$\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \, \cos \theta$