问:
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$
For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

$\sin x \approx x$

$\cos x \approx 1-\frac{x^2}{2}$

$\tan x \approx x +\frac{x^3}{3}$

I am looking for a justification of these approximations without the use of series expansions.

By purely geometric considerations, it is easy to see that for small values of $x$, we have

$$ \sin x < x < \tan x.$$

Division by $\cos x$ and an application of the squeeze lemma yield

$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$

and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.

Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?

I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.

答:
Look at the trigonometric circle and approximate the chord $s$ with the angle $\theta$.

Here $x=\cos \theta$ and $y=\sin \theta$.

From the big triangle, you have $x^2+y^2=1$ which is obvious, and from the small circle, you have $s^2 = y^2 + (1-x)^2$. But with approximation $s \approx \theta$ the above is

$$\theta^2 \approx y^2 + (1-x)^2 \tag{1}$$

Now sub $y=\sqrt{1-x^2}$ to get the well-known approximation

$$ \theta^2 \approx 1-x^2 + (1-x)^2 = 2 (1-x) $$

or

$$ x = \cos \theta \approx 1 - \tfrac{1}{2} \theta^2 \tag{2}$$

Now use (2) in (1) to get $\theta^2 \approx y^2 + \tfrac{1}{4} \theta^4$, but since we are keeping this a second order approximation $\theta^4 \approx 0$ and

$$ y = \sin \theta \approx \theta \tag{3} $$

Finally, $\tan \theta = \tfrac{y}{x} \approx \tfrac{\theta}{1-\tfrac{1}{2} \theta^2}$. But use the property $\tfrac{1}{1-z} = 1 + \frac{1}{1-z} z \approx 1 + z$ with $z=\tfrac{1}{2} \theta^2$ to get the final approximation.

$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{1}{2} \theta^2 \right) \tag{4a} $$

But note that the last one is based on two approximations.

Another approach with $\tan$ would be to approximate once $\tan \theta = \tfrac{y}{x} = \theta \frac{\sqrt{1-\tfrac{1}{2}z}}{1-z}$ with $z=\tfrac{1}{2}\theta^2$ and the approximation $ \tan \theta \approx \theta \left( 1 + \tfrac{3}{4} z \right)$

$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{3}{8} \theta^2 \right) \tag{4b} $$

For example with $\theta=0.6$ you have $\tan(0.6) = 0.68413681$, $0.6 \left( 1 + \tfrac{1}{2} 0.6^2 \right)=0.708$ an 3.5% error, and $0.6 \left( 1 + \tfrac{3}{8} 0.6^2 \right)=0.681$ an 0.45% error.

So (4b) is at least an order of magnitude better at approximating $\tan \theta$ than (4a). A graphical comparison of the error $f(x)/\tan x-1$ is shown below:


问:
Maximize $\sum_{i=0}^{\infty}p^i\log(x_i)$ where $x_i\ge 0, \sum x_i\le 1.$
I came across this problem while trying to learn some economics. I have a $p\in (0, 1)$ and I have to maximize the sum $$U((x_i)):=\sum_{i=0}^{\infty}p^i\log(x_i),$$
where $x_i\ge 0$ and $\sum x_i\le 1.$

Since the maximizer, if it exists, for the above problem and $(1-p)\sum_{i=0}^{\infty}p^i\log(x_i)$ would be the same. I use Jensen's inequality to conclude that $U(x_i)\le 0.$ Therefore, a natural thing to try is to come up with a suitable $(x_i)$ such that the above sum is indeed zero. My guess is that the maximizer does not exist. My argument is as follows.

It is easy to see that $x_i>0$ for all $i$ and $x_i\ge x_{i+1}$ (otherwise just interchanging $x_i$ and $x_{i+1}$ yields a better estimate). Using these two facts, I argue that $$log(x_0)+p\log(x_1)+\sum_{i=2}^{\infty}p^i\log(x_i)\le \log(x_0+\epsilon)+p\log(x_1-\epsilon)+\sum_{i=2}^{\infty}p^i\log(x_i),$$
for sufficiently small $\epsilon.$ In other words, given any vector $x_i$ with $x_i>0$ for all $i,$ we can modify it to obtain a vector with better $U.$ Is this argument correct?

答:
Let's treat this as a straight constrained maximization problem.  Most econ problems would probably have some return to $x_i$ you don't eat today, but this seems to mean there is no loss or gain to storage.

Form the Lagrangean,
$$L=\sum_0^\infty p^i log(x_i)-\lambda(\sum_0^\infty x_i -1).$$
$$\frac{\partial L}{\partial x_i}=p^i\frac{1}{x_i} -\lambda = 0.$$
$$\frac{p^i}{x_i}=\lambda.$$
$$x_i=\frac{p^i}{\lambda}.$$
Now use the constraint
$$\sum\frac{p^i}{\lambda}=1$$
or
$$\lambda = \sum p^i$$
or
$$x_i = (p^i)(1-p)$$.
Intuitively, you want to smooth your consumption, that is, don't pig out today if it means you starve tomorrow, but today is worth more than  tomorrow, so spread things out in a declining manner.

ContentLicense: CC BY-SA 4.0


问:
Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$.
Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.

Plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
and assuming the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.

I've not been able to find the value for $d$, putting the three points in the equation of sphere I got $3$ equations with $4$ variables.

答:
Given the points

$$
\cases{
A=(a,0,0)\\
B=(0,b,0)\\
C=(0,0,c)\\
p = (x,y,z)
}
$$

the sphere center $p_0$ is determined by the intersection of plane $\Pi$ with the line $L$ such that

$$
\cases{
\Pi\to (p-A)\cdot\vec n=0\\
\vec n = (A-B)\times(B-C)\\
L\to \left((p-\frac{A+B}{2})\cdot(A-B)=0\right)\cap \left((p-\frac{A+C}{2})\cdot(A-C)=0\right)\\
L\to p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)
}
$$

then

$$
p_0 = \frac{1}{2(a^2b^2+a^2c^2+b^2 c^2)}\left(a^3 \left(b^2+c^2\right),b^3 \left(a^2+c^2\right),c^3\left(a^2+b^2\right)\right)
$$

and

$$
r = \|p_0-A\| = \frac{1}{2} \sqrt{\frac{\left(a^2+b^2\right) \left(a^2+c^2\right) \left(b^2+c^2\right)}{a^2b^2+a^2c^2+b^2 c^2}}
$$

Attached a plot for $a=1,b=2,c=3$. In pink the plane $(p-A)\cdot \vec n=0$. In green the plane $(p-\frac{A+B}{2})\cdot(A-B)=0$ and in yellow the plane $(p-\frac{A+C}{2})\cdot(A-C)=0$. In red the line $p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)$

ContentLicense: CC BY-SA 3.0


问:
Proving one function is greater than another
How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?

My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?

Body:
Try this: $${ x }^{ 2 }+2x+1-4xq\ge 0\\ { x }^{ 2 }-2x\left( 2q-1 \right) +1+{ \left( 2q-1 \right)  }^{ 2 }-{ \left( 2q-1 \right)  }^{ 2 }\ge 0\\ { \left( x-2q+1 \right)  }^{ 2 }\ge { \left( 2q-1 \right)  }^{ 2 }-1$$

You can use this for any $q$


问:
How do I calculate an integral like $\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$?
$$\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$$

This is Gauss' law from physics but my question is more maths related.

Say I have $\vec{E}^{\,}=3.5\times 10^3N/C\times e_x$ and $A = 0,35\times0,70m^2$. The plane that has that area, where the electric field $E$ is applied, is parallel to $YZ$. So it has a coordinate in $e_x$. Right?

What does the circle in the integral mean in practice?

What is the value of $dA$? Why is it a vector?

How do I solve this integral?

答:
It looks like you're overcomplicating things massively for this simple example.

In this case, the answer is just multiply the two numbers together. No integral solving, nothing sophisticated. This is because you just have a constant electric field passing perpendicular through an area. I am not sure how your course defines electric flux, but for a uniform field passing through a flat surface, you multiply the area by the component of your field that is perpendicular to your surface. In your question, field and surface are perpendicular, so you don't even need to worry about any potential $\cos\theta$ factors.

For your other questions, these are all things from vector calculus, which I'm guessing you'll learn about sooner or later. The above result is a very special case of the general Gauss law with vector integrals etc. For the purpose of this question, you don't need to worry about all that. I can expand more on the mathematics/vector calculus if you wish.

OK so the following is a super rapid overview of vector integrals.

You no doubt know how differentiation/integration for functions $f:\mathbb R\to\mathbb R$ works. The idea of vector calculus is to generalise things to functions $\mathbb R^3\to\mathbb R$, $\mathbb R\to\mathbb R^3$, $\mathbb R^3\to\mathbb R^3$ etc. There's nothing special about $\mathbb R^3$, it just so happens that the world we live in has three spatial dimensions, so physics people care an awful lot about $\mathbb R^3$.

Let's consider a concrete example. Suppose we have some material contained in a volume $V$, with surface $S$. We are interested in finding how much heat flows out of this block. We can find this by computing the heat flow out of the surface $S$.

To do this, we somehow need to integrate the heat flow over the surface. We write $\mathop{}\!\mathrm{d}S$ for the differential corresponding to a tiny chunk of this surface (similar to how $\mathop{}\!\mathrm{d}x$ represents a small change in $x$ in single variable calculus). If our surface happened to be the $yz$-plane, then we would have $\mathop{}\!\mathrm{d}S=\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z$; I hope it's intuitively clear why.

Suppose $\mathbf H(\mathbf v)$ represents the heat flow at some point $\mathbf v$ in space. Note that $\mathbf H$ takes a vector for the position as input and spits out a vector for the heat flow at that position (per unit area and time) as output. The heat flow out of a chunk of surface $\mathop{}\!\mathrm{d}S$ is the area times the component of $\mathbf H$ perpendicular to $\mathop{}\!\mathrm{d}S$. The reason we only take the perpendicular component is that the component of $\mathbf H$ that flows parallel to the surface doesn't flow in or out of the material, so doesn't contribute to the quantity we care about.

If $\mathbf n$ represents the unit length vector normal to $S$ (so $\mathbf n$ may vary depending on where on $S$ you are standing), then the component of $\mathbf H$ perpendicular to the surface is just $\mathbf H\cdot\mathbf n$, so the heat flow through that little chunk is $\mathbf H\cdot\mathbf n\mathop{}\!\mathrm{d}S$.

To find the total heat flow, we need to sum the contributions from all chunks of our surface, i.e. we integrate over the surface:
$$\int\mathbf H\cdot\mathbf n\mathop{}\!\mathrm{d}S.$$

Since $\mathbf n$ is intrinsic to the surface $S$, we often write as an abbreviation $\mathop{}\!\mathrm{d}\mathbf S=\mathbf n\mathop{}\!\mathrm{d}S$. This is the vector area element. With this notation, the total heat flow is
$$\int\mathbf H\cdot\mathop{}\!\mathrm{d}\mathbf S.$$
There's nothing special about heat of course: I just picked it because it's easy to visualise heat flowing out of stuff. The same expression above, applied to the electric field $\mathbf E$, just tells you how much of the electric field is flowing through the surface. We call this quantity the (electric) flux through the surface.

Let's apply all this to the question at hand. Your electric field is uniform $\mathbf E=3500\mathbf e_x$. Let's say the surface is a rectangle with vertices $(0,0,0)$, $(0,0,0.35)$, $(0,0.7,0.35)$, $(0,0.7,0)$. The normal vector is clearly just $\mathbf e_x$, anywhere on the surface. So $\mathop{}\!\mathrm{d}\mathbf S=\mathbf e_x\mathop{}\!\mathrm{d}S=\mathbf e_x\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z$, and applying the above,
\begin{align}
\int\mathbf E\cdot\mathop{}\!\mathrm{d}\mathbf S&=\int_0^{0.35}\int_0^{0.7}3500\mathbf e_x\cdot\mathbf e_x\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z\\
&=3500\cdot\int_0^{0.35}\int_0^{0.7}1\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z\\
&=3500\cdot\int_0^{0.35}0.7\mathop{}\!\mathrm{d}z\\
&=3500\times0.35\times0.7 \\
&=857.5.
\end{align}
So this integral is the long way of saying "just multiply everything". Of course, when the field $\mathbf E$ or the surface $S$ are more complicated, the integrals won't be nearly so trivial. But the basic method is the same:

Take some parametrisation of your surface, say $(u,v)$. In principal, you can just use Cartesians for everything, but it's often easier to use other coordinate systems, e.g. polars.

Use the parametrisation to find an expression for the normal to your surface.

Dot the normal with the vector field, and integrate over the values of $u$ and $v$.


问:
Asymptotic behaviour of an integral
Let $I(r)$ be

$$\frac{1}{4\pi^2 r}\int_{m}^{\infty} d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$

How can I show that

$$I(r) \sim  e^{-mr} $$ for $r\to\infty$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{{1 \over 4\pi^{2}r}\int_{m}^{\infty}\dd\rho\, {\rho\expo{-\rho r} \over \root{\rho^{2} - m^{2}}}}
\\[5mm] \stackrel{\rho\ =\ m\cosh\pars{\theta}}{=}\,\,\,\,&\
{m \over 4\pi^{2}r}\int_{0}^{\infty}\cosh\pars{\theta}
\expo{-mr\cosh\pars{\theta}}\,\,\,\dd\theta
\\[5mm] \stackrel{{\rm as}\ r\ \to\ \infty}{\sim}\,\,\,&\
{m \over 4\pi^{2}r}\int_{0}^{\infty}
\exp\pars{-mr\bracks{1 + {\theta^{2} \over 2}}}\,\,\,\dd\theta
\\[5mm] = &\
{m \over 4\pi^{2}r}\expo{-mr}\root{2 \over mr}
{\root{\pi} \over 2}
\\[5mm] = &\
\bbx{{\root{2} \over 8}\,{m^{2} \over \pi^{3/2}}\,
\color{red}{\expo{-mr} \over \pars{mr}^{3/2}}} \\ &
\end{align}

Title: Asymptotic behaviour of an integral


问:
Möbius transformation are the same iff exists a constant

Let $T_{1} = \frac{az+b}{cz+d}$, $T_{2}=\frac{a'z+b'}{c'z+d'}$ be two Möbius transformation. Prove that $T_{1}(z)=T_{2}(z)$, $\forall z\in\mathbb{C}_{\infty}$ if and only if exsists $\lambda\neq 0$ such that $a=\lambda a'$, $b=\lambda b'$, $c=\lambda c'$, $d=\lambda d'$.

$(\Leftarrow)$ It's clear that $z_{0} = -\frac{b}{a}$ it's such that $T_{0}(z_{0})=0$, then $T_{1}(z_{0})=0$ and the same argument works to see that if $z_{1,2}$ are such that if  $T_{1}(z_{1})=\infty$, then $T_{2}(z_{1})=\infty$ and, if $T_{1}(z_{2})=1$, then $T_{2}(z_{2}) =1$. A Möbius transformation has at most two fixed point so $T_{1}=T_{2}$.

I have some problems to prove the other implication. Using the same argument, I have that $-\frac{b}{a} = -\frac{b'}{a'}$ but I don't know how to prove the existence of $\lambda$. Can someone help me? Thanks before!

答:
Idea: if $T_1 = T_2$:

$$ac'z^2 + (ad' + bc')z + bd' = (az + b)(c'z + d') = (a'z + b')(cz + d) = a'cz^2 + (a'd + b'c)z + b'd.$$
Equal polynomials have the same coefficients. Start form $bd' = b'd$ and define $\lambda = b/b' = d/d'$ (what happens if $b' = 0$ or $d' = 0$?).


问:
Existence of a $\mathbb{Z} _{15}$ in a group $G$ where $|G|=3 \cdot 5 \cdot 17$

Existence of a $\mathbb{Z} _{15}$ in a group $G$ where $|G|=3 \cdot 5 \cdot 17$.

This is my proof:

we know (from a previous point of the exercise) that $\mathbb{Z}_{17}$ is a characteristic subgroup of $G$. For Cauchy, we have that $\exists \mathbb{Z}_3,\ \mathbb{Z}_5 <G$. If we demonstrate that $\mathbb{Z}_3$ or $\mathbb{Z}_5$ is normal in $G$, then we have that $\mathbb{Z}_3\mathbb{Z}_5$ (or $\mathbb{Z}_5\mathbb{Z}_3$) is a subgroup of $G$ of order 15, therefore is isomorphic to $\mathbb{Z}_{15}$.

For Sylow, we have that $n_5 \equiv 1 \pmod{5}$ and $n_5 \mid 51$ where $n_5$ is the number of $5$-Sylow in $G$, so $n_5=1,51$.

If $n_5=1$ then the only $5$-Sylow in $G$ is normal.

If $n_5=51$ then there are $51 \cdot 4$ elements of order $5$ in $G$, and of the remaining $51$ elements there are $16$ of order $17$ (from the only $\mathbb{Z}_{17}$) and one of order $1$, that is the identity. So we have $34$ elements left.

At this point, I thought about two different conclusions, but I am not really sure which one is the correct one:

All the $34$ elements have to be of order $3$, therefore there are $17$ $3$-Sylow, but this is impossible because $17 \not\equiv 1 \pmod{3}$. So we can conclude that $n_5=1$.

We have $34$ elements, so we can have at most $17$ $3$-Sylow (in other words, $n_3\le 17$), and this means that $n_3$ can only be $1$. So if $n_5 \neq 1$ then $n_3=1$.

答:
$\textbf{Hint}:$ let's take a look at the number of groups in $Syl_{17}(G)$ and $Syl_3(G)$.

We know that $n_{17}\equiv1\text{ (mod 17)}\implies n_{17}\in \{1\not3,\not5,\not{15}\}\implies \exists!P_{17}\trianglelefteq G$.

While $n_3\equiv 1\text{ (mod 3)}\implies n_3\in\{1,\not5,\not17,85\}.$

If we now consider the action of conjugation of $G$ on its Sylow's groups we have that $$|Syl_{17}(G)|=|G:N_G(P_{17})|\implies|N_G(P_{17})|=|G|.$$
Furthermore $$\dfrac{|N_G(P_{17})|}{|C_G(P_{17})|}\bigg||\operatorname{Aut}(P_{17})|=\varphi(17)=16\implies\dfrac{3\cdot 5\cdot 17}{|C_G(P_{17})|}\big |2^4\iff|C_G(P_{17})|=|G|.$$
$P_{17}$ is abelian, so $P_{17}\le N_G(P_3)\implies17\big||N_G(P_{3})|$. We also have $\{1,85\}\ni n_3=|G:N_G(P_3)|$, so $n_3=1$. You can do something similar for $P_5$.


问:
$\sigma$-weak closure of $B(H) \otimes B(K)$.
Let $H$ and $K$ be Hilbert spaces. I'm  trying to prove that  $B(H) \otimes B(K)$ (= minimal tensor product of $C^*$-algebras) is $\sigma$-weakly dense (= weak$^*$-dense) in $B(H \otimes K)$.

My progress so far:

I managed to show:

$B(H) \otimes B(K)$ is strongly dense in $B(H \otimes K)$

$B(H) \otimes B(K)$ contains the compact operators on $H \otimes K$, hence it suffices to show that the compact operators are $\sigma$-weakly dense in $B(H \otimes K)$.

Any ideas how I can proceed?

答:
Here is a more or less alternative argument.

By looking at what the rank-one operators look like, it is easy to see that $\mathcal K(H\otimes K)=\mathcal K(H)\otimes \mathcal K(K)$, this last product as C$^*$-algebras (no ambiguity due to nuclearity).

Then $\mathcal K(H)\otimes \mathcal K(K)$ is $\sigma$-weakly dense in $B(H\otimes K)$ because the locally compact space $\mathcal K(H\otimes K)$ is weak$^*$-dense in its double dual $B(H\otimes K)$.


问:
Free abelian with basis definition
Let $F$ be an abelian group and $B$ a subset of $F$. $F$ is free abelian with basis $B$ if and only if the cyclic subgroup $\langle b\rangle $ is infinite cyclic for each $b\in B$ and $F$ is the direct sum of $\langle b\rangle $.

My question is, $\langle b\rangle $ is already infinite, so why does the author need to say that it is "infinite cyclic", especially given that the author states that it is a cyclic subgroup. Why doesn't the author just say the "cyclic subgroup $\langle b\rangle$ is infinite"?

Is there something I'm missing?

答:
I’m not sure what you mean by “is already infinite”...

Let’s look at the definition, and what we know before we get to that sentence:

Let $F$ be an abelian group and $B$ a subset of it.

So, we know $F$ is abelian, we know $B$ is a subset. That’s it. We don’t know yet if $F$ is finite or infinite, torsion, torsion-free, or mixed, or anything else. Just that $F$ is abelian.

$F$ is free abelian with basis $B$ if and only if the cyclic subgroup $\langle b\rangle$ is infinite cyclic for each $b\in B$, and $F$ is the direct sum of [the] $\langle b\rangle$.

We know $\langle b\rangle$ is cyclic; we do not know it is infinite. Perhaps that’s what you meant? That we know it’s cyclic so why do we need ot say it it is infinite and cyclic? I suspect that this is just a question of the “name” given to these groups. It is not uncommon to refer to/define the subgroup generated by a non-torsion element as “infinite cyclic”, so you should think of that not as two adjectives describing the group (“it is both infinite and it is cyclic”), but rather as a single compound adjective describing it (“it is infinitecyclic”).

Title: Free abelian with basis definition


问:
Fundamental group of a sphere in a cylinder
I'm currently studying the fundamental group and the applications of Seifert van Kampen's theorem. I'm doing several exercises but I really have difficulties in understanding how to choose the relations of the presentation of the fundamental groups. In this exercise the relation was (I think) trivial but I'd really like to learn a "standard" way to choose the relations on the presentation of the fg.

I have to calculate the fundamental group of of $X:=S^2\cup (S^1\times [-1,1]\cup D^2\times \{\pm 1\})$.

First I observed that $X$ is path connected and I chose two open sets in $X$ such that:
$$U=X\cap \Big\{z<\dfrac{1}{2}\Big\},V=X\cap \Big\{z>-\dfrac{1}{2}\Big\}.$$
The union $U\cup V$ is the set $X$, and the intersection is $$U\cap V=X\cap \Big\{z<\dfrac{1}{2}\Big\}\cap\Big\{z>-\dfrac{1}{2}\Big\}=X\cap\Big\{-\dfrac{1}{2}<z<\dfrac{1}{2}\Big\}.$$
Let $C$ be the set $S^1\times [-1,1]\cup D^2\times \{\pm 1\}$, then we have that $U\cong V$ and $U\cap C\underset{h}{\sim}D^2\times\{-1\}$, which is its deformation retract. Then I said that $U\underset{h}{\sim}S^2\vee S^1$. My (intuitive) idea is that we can consider a function $\rho$ that maps the points of the inferior surface of the cylinder to the surface of the inferior semi-sphere, while this last surface is mapped to the semisphere with $z>0$.

Then, using the fact that a segment is contractible, I can extend the "south" point $S$ to the segment $\overline{SN}$ on the whole sphere $S^2$, where $N$ is the "north" pole.

As loop for the fundamental group of $U$, I chose a generic path $\alpha$ that contains $\overline{SN}$ on the surface of the sphere obtained by the trasformation I described before.

So, $\Pi_1(U)\cong\Pi (S^2\vee S^1)=\langle \alpha \rangle\cong \mathbb Z$ and similarly $\Pi_1(V)=\langle \beta \rangle \cong \mathbb Z$.

Retracting the intersection $U\cap V$, we have that $U\cap V\underset{h}{\sim} S^1\implies \Pi_1(U\cap V)=\langle \gamma \rangle \cong \mathbb Z$.

If $i_{1}*$ is the inclusion $U\cap V\hookrightarrow U$ and $i_2*$ is the inclusion $U\cap V\hookrightarrow V$, then $i_1*\gamma=i_2*\gamma=1$ (constant path).

The relation on the intersection is trivial and, since $U\cap V$ is path connected, I concluded that $$\Pi_1(U\cap V)=\langle\alpha,\beta|\emptyset\rangle=\langle \alpha,\beta\rangle\cong\mathbb Z*\mathbb Z.$$

答:
$X$ is the union of the unit sphere (presumably) and the capped unit cylinder (presumably).  (I say "presumably" because $S^2$ can be any sphere and $S^1$ can be any circle.)  It looks like

The intersections of the sphere and capped cylinder are highlighted in green.  There are three of them -- two points at the centers of the caps, $(0,0,\pm 1)$, and one circle, $S^1 \times \{0\}$.

$U$ is all of that up to $z < 1/2$, the lower half of $X$ with two collars attached to $S^1 \times \{0\}$ on the exterior of the volume enclosed by the lower half of $X$.  Picture:

The collars deformation retract onto the circle intersection, leaving the lower half of $X$.  It should be easy to see that this is a sphere with two of its points identified.

Another name for  this shape is horn torus.  Normally, a torus has two generators, but in a horn torus, the longitudinal cycle (homotopic to the green circle) is nullhomotopic, specifically homotopic to the identified points in the sphere.  The meridional cycle survives (shown in red).

Then $U \cap V$ is the circle intersection with four collars attached.

The collars deformation retract onto the circle.  So $\pi_1(U \cap V) \cong \Bbb{Z}$ and $S^1 \times \{0\}$ (with an orientation and one of its points chosen to be the basepoint) generates $\pi_1(U \cap V)$.

$\pi_1(U) \cong \pi_1(V) \cong \Bbb{Z}$, where a generator is the red path (with an orientation and, in order to be convenient for all three fundamental groups, the basepoint being the intersection of the red path and green circle).  Siefert-van Kampen allows us to realize the one generator in $U \cap V$ is nulhomotopic in each of $U$ and $V$ (although it would have been sufficient for it to be nulhomotopic in either since any homotopic version of it in $U$ can be moved to $U \cap V$ and from there to $V$ and vice versa).  So of the three generators, for $\pi_1(U)$, $\pi_1(V)$, and $\pi_1(U \cap V)$, only the two meridional generators survive to the fundamental group of $X$.


问:
If ${ x }^{ 4 }+{ y }^{ 2 }=1$ then $x$ and $y$ can be both rational numbers?
Can you give two numbers $x,y\in\mathbb{Q}$ such that  ${ x }^{ 4 }+{ y }^{ 2 }=1$?

I don't know if exists or not. I derived this equation questioning that if $\sin { \alpha  } ={ x }^{ 2 }$ for $x\in \mathbb{Q}$ then for which $\alpha$, $\cos{ \alpha  }=y$ and $y\in \mathbb{Q}$?

Edit: $x,y\neq0$

答: Denote $x=\dfrac{b}{a},y=\dfrac{d}{c},a,b,c,d>0,$ then $(\dfrac{b}{a})^4-(\dfrac{d}{c})^2=1,$ hence $a^4=c^2,c=a^2,b^4-d^2=a^4,$ this equation has no positive integer solution, see Solving $x^4-y^4=z^2$</p>


问:
calculating norm of a linear functional
I was solving a problem regarding Hanh-Banach extension of a linear functional and I encountered the following problem as a subpart of the original one -

Consider the normed linear space ($ \mathbb C^2$, $ $|| ||$_\infty$). $f$ is a linear functional on the space given by, $f(x,y)=ax+by$, for some constant a,b. Calculate norm of the functional $f$.

Now, it can be easily shown that $\frac {|f(x,y)|} {||(x,y)||_\infty} \leq |a|+|b|$ $\Rightarrow$ $||f|| \leq |a|+|b|$ . I know that the norm of the functional is $|a|+|b|$. However I am not being able to establish the equality $||f|| = |a|+|b|$. Someone kindly help me in proving the opposite inequaity. I have seen many hints that were available online and still have not been able to figure it out. So please be a little elaborate.

Thank you for your time.

答:
Note that the map
$$(\Bbb{C}^2,\|\cdot\|_1) \to (\Bbb{C}^2, \|\cdot\|_\infty)^*$$
where $(a,b) \in \Bbb{C}^2$ maps to the functional $f(x,y) = ax+by$ is an isometric isomorphism. In particular the operator norm of $f$ is equal to the $\|\cdot\|_1$-norm of the pair which induces the functional:
$$\|f\| = \|(a,b)\|_1 = |a|+|b|.$$

Title: calculating norm of a linear functional


问:
Probability with uncommon event.
In a lottery, there are $25$ choosable numbers and you have to choose $15$, but you have to hit all the $15$ numbers, the probability for this is $1$ in $$\binom{25}{15}=3,268,760$$

Let's assume that the Lottery owner gives you one number as a spoiler, the number $1$.

I need to know the probability of choosing $15$ numbers of $25$ hitting all the $15$, but already know one of them.

Thank you!

答: As you already know one of the numbers, you can remove that of your Lottery. Thus your new "game" is to hit the correct 14 numbers of 24 numbers, which is $1$ in $$\binom{24}{14}=1,961,256.$$ So the probability is $$\frac{1}{1961256}.$$

Title: Probability with uncommon event.


答:
The circle in the integral specifies that integration is being performed over a closed surface (i.e. one without a boundary). The surface of a sphere would be considered a closed surface, but the surface of a plane would not.

Gauss's law states that the net electric flux through any arbitrary closed surface enclosing charge is proportional to the total charge.

We obtain the electric flux through an infinitesimal element of a surface by taking the dot product $\overrightarrow{E} \cdot d\overrightarrow{A}$. The vector d$\overrightarrow{A}$ refers to the unit normal vector of the infinitesimal surface element. The net electric flux is obtained by integration over the entire surface.

If we were to consider your surface to be closed (which a plane is not) and that the electric field was constant and always normal to that surface, then $\oint \overrightarrow{E} \cdot d\overrightarrow{A}= |\overrightarrow{E}|A$.

Title: How do I calculate an integral like $\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$?

Body:
$$\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$$

This is Gauss' law from physics but my question is more maths related.

Say I have $\vec{E}^{\,}=3.5\times 10^3N/C\times e_x$ and $A = 0,35\times0,70m^2$. The plane that has that area, where the electric field $E$ is applied, is parallel to $YZ$. So it has a coordinate in $e_x$. Right?

What does the circle in the integral mean in practice?

What is the value of $dA$? Why is it a vector?

How do I solve this integral?


问:
Elevator stops in building
Bill is working in the 38th floor of an 100-floors building. This building has a strange elevator which only has 2 buttons: the green one which takes you to the next floor every time you press it (and if you are at the last floor, it takes you to the 1st) and a red button which takes you to a random floor every time you press it. Once you enter the elevator, the first stop is selected at random, without pressing any button. Bill wants to eventually arrive at his office at the 38th floor. How is he going to do it, with the least number of steps?

I am thinking of a very simplistic approach: Since the first stop is decided at random, depending on the number of the floor where the elevator stops, Bill will decide which button to use: If he is near the 38th and before it, he will keep pressing the green button until he reaches the 38th. If it stops in a floor $>38$, he will press the red button until the elevator stops at some floor $<38$, but I assume the problem involves the expected number of stops etc probabilities, which I am not very familiar with!
Any help is really appreciated!

答:
Well, let me first extract from your question an important idea for solving this - you have identified, essentially, a set of reasonable strategies (which I've parameterized in some unknown integer $0 \leq n \leq 37$):

If Bill is at or below floor $n$, he should press the red button. If he is above floor $n$ and below floor $38$, he should press the green button. If he is above floor $38$, he should press the red button.

The question we're faced with is finding the best $n$ - for instance, one extreme case would be $n=0$, where he always presses the green button* when it'll eventually lead to the desired floor. On the other extreme, if $n=37$, Bill just presses the red button until he gets to his location. One might wish to justify why the strategy must be of this form - but it's not too hard to see that, since the green button is entirely non-random, it'd be silly to start pressing it and then to later decide, having learned nothing new, to press the red button.

Now, we can do our analysis. Let's imagine $n$ is fixed and evaluate the strategy. What we need to do is to define a series of variables $T_k$ which tells us how many more buttons we expect to press if we find ourselves on the $k^{th}$ floor.

Clearly, $T_{38}=0$. If $k$ is a floor on which we'd choose the green button, we have simply:
$$T_k = 1 + T_{k+1}$$
since we need to press the green button, then do whatever we need to do to finish from the $(k+1)^{th}$ floor. So, we'd get $T_{37}=1$ and $T_{36}=2$ provided that $n$ is no more than $35$ - or, more generally
$$T_k = (38 - k)$$
if $n < k \leq 38$.

If $k$ is a floor on which we'd choose the red button - so, any floor where $k\leq n$ or $38<k$ - then
$$T_k = 1 + \frac{1}{100}\left(T_1+T_2+T_3+\ldots+T_{100}\right)$$
since we use one button press for sure and then end up, with probability $\frac{1}{100}$ each, at some other floor - and then should expect however many button presses it requires to get from that floor on. Then, the probabilistic part of the question is over - the rest is algebra. Note that our goal can be stated to minimize $T_{39}$ (or any floor above $38$), since doing so just takes one more button press than it takes to get to $38$ from a random floor.

Our equations can be simplified by realizing that, if we're going to press the red button, it doesn't really matter which floor we're on right now - that is, all of these values of $T_k$ are equal - let's call them all $$R=T_1=T_2=\ldots = T_{n-1} = T_n = T_{39} = T_{40} = \ldots = T_{100}.$$
There are $62 + n$ floors on which this random strategy holds, and we can thus simplify that
$$R = 1 + \frac{62+n}{100}\cdot R + \frac{1}{100}\left(T_{n+1}+T_{n+2}+\ldots + T_{38}\right)$$
where we just pull out every term equal to $R$ from the original sum.

Then a bit of algebra gives
$$R = \frac{100}{38 - n}\cdot \left(1 + \frac{1}{100}\left(T_{n+1}+T_{n+2}+\ldots + T_{38}\right)\right)$$
We can also recognize that the inner sum is known as just the sum of the first $38-n$ natural numbers, which is $\frac{(38-n)(37-n)}2$. Substituting this in and cancelling what needs to be cancelled gives:
$$R = \frac{100}{38-n} + \frac{37-n}2$$
Note that we could get this result faster by realizing that we'll spend some amount of time moving randomly then have to go one step at a time. Formally, you could phrase this in random variables by writing that the time taken is $T$ which is $T_R + T_G$ where $T_R$ is the number of presses of the red button and $T_G$ is the number of presses of the green button. $T_R$ follows a geometric distribution with $p=\frac{38-n}{100}$. $T_G$ follows a uniform distribution on $\{0,1,2,\ldots,37-n\}$. The means of these distributions are $\frac{100}{38-n}$ and $\frac{37-n}2$ respectively. However, if you're uncertain of how to approach these questions, it's often better to work from first principles and not involve fancy constructions using random variables and linearity of expectation, as this fast path does.

However we arrive here, we finish by minimizing that expression for $R$ over $n$. We don't need to be clever here - there's not so many values and if you try all of them, you find the best value is $n=24$ and that $R=13+\frac{9}{14}$, meaning that if you press the green button on floors $25$ to $38$, you will, on average, require $13+\frac{9}{14}$ presses to get from any floor on which we'd press the red button to floor $38$. In terms of the original question, since the first random movement is free (and forced), this means Bill will need $12+\frac{9}{14}$ button presses on average.

(*I'm assuming the floors are $1,2,3,\ldots,100$ not $0,1,2,\ldots,99$ - you'll need to check your nearest skyscrapper to see if your country uses zero based indexing of floors)

Title: Elevator stops in building


问:
Linear independence of matrices with different rank
We have 4 5x7 matrices $A_1, A_2, A_3, A_4$ of rank 1, 2, 3, 4  that are in row echelon form. Is it true that the set of these matrices is necessarily linearly independent?

I would think that the echelon form condition ensures that their nonzero rows are ordered in such a way that by doing linear combinations of these matrices you'd at best get 4 matrices of rank 1 with 1 nonzero row and zeroes everywhere else. E.g. say you can't zero out the 4th row of matrix $A_4$ using the other three. Which would suggest they are LI. Is that correct reasoning? Thanks for any help.

答:
Essentially, yes. The form of each matrix is something like
\begin{align*}
A_1 &= \left( \begin{matrix}
1 & ? & ? & ? & ? & ? & ? \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0& 0 & 0 & 0
\end{matrix} \right) \\
A_2 &= \left( \begin{matrix}
1 & ? & ? & ? & ? & ? & ? \\
0 & 1 & ? & ? & ? & ? & ? \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0& 0 & 0 & 0
\end{matrix} \right) \\
A_3 &= \left( \begin{matrix}
1 & ? & ? & ? & ? & ? & ? \\
0 & 1 & ? & ? & ? & ? & ? \\
0 & 0 & 1 & ? & ? & ? & ? \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0& 0 & 0 & 0
\end{matrix} \right) \\
A_4 &= \left( \begin{matrix}
1 & ? & ? & ? & ? & ? & ? \\
0 & 1 & ? & ? & ? & ? & ? \\
0 & 0 & 1 & ? & ? & ? & ? \\
0 & 0 & 0 & 1 & ? & ? & ? \\
0 & 0 & 0 & 0& 0 & 0 & 0
\end{matrix} \right) \\
\end{align*}

where $?$ can be any scalar from your field. (I could use variables, but that would be a pain. Note that the $?$ may be distinct from each other and do not represent any particular quantity.) Take $\alpha_i$ to be four scalars. Then

\begin{align*}
&\text{the 4rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\
&=
\left(\begin{matrix}
0 & 0 & 0 & \alpha_4 & \alpha_4 ? & \alpha_4 ? & \alpha_4 ? \\
\end{matrix} \right)
\end{align*}

Obviously, this is a zero vector iff $\alpha_4 = 0$. Then consider

\begin{align*}
&\text{the 3rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\
&=
\left(\begin{matrix}
0 & 0 & \alpha_3 + \alpha_4 & \alpha_3 ? + \alpha_4 ?& \alpha_3 ? + \alpha_4 ? & \alpha_3 ? + \alpha_4 ? & \alpha_3 ? + \alpha_4 ? \\
\end{matrix} \right)
\end{align*}

Since $\alpha_4 = 0$ though,

\begin{align*}
&\text{the 3rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\
&=
\left(\begin{matrix}
0 & 0 & \alpha_3  & \alpha_3 ? & \alpha_3 ?  & \alpha_3 ?  & \alpha_3 ?  \\
\end{matrix} \right)
\end{align*}

so, once again, this is only a zero vector iff $\alpha_3 = 0$. You can continue this to show that

$$\sum_{i=1}^4 \alpha_i A_i = \mathcal O \implies \alpha_i = 0 \; \forall i \in \{1,2,3,4\}$$

where $\mathcal O$ is the $5 \times 7$ all-zeroes matrix, the identity of the vector space $M_{5 \times 7}(F)$. Thus, the set of $A_i$ is linearly independent.


问:
Self Adjoint operator $\Rightarrow$ Idempotent Operator?
If $P\in\mathcal{L}(H,H)$, with $H$ a Hilbert space, such that $P = P^*$, Is possible to show that $P^2 = P$?

If that is possible, then $P$ is a projection operator, right?

Thanks in advance.

答:
An operator $P$ satisfying $P = P^{\ast}$ is called self-adjoint. There are plenty of self-adjoint operators that do not satisfy $P^2 = P$; for example, $P = 2I$, as proximal mentioned.

An operator $P$ satisfying $P = P^2$ is called a projection. Not all projections are self-adjoint, such as $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$.

An operator that satisfies both is called an orthogonal projection.


问:
Entropy is subadditive
Let $(X,\mathcal{B},\mu)$ be a probability space, abd let $\alpha$ and $\beta$ be countable-measurable paratition.

Let $H_\mu(\alpha)$ and $H_\mu(\beta)$ be the (Shanon) entropy of $\alpha$ and $\beta$.

Let $\alpha\bigvee\beta:=\{A\cap B\mid A\in\alpha, B\in \beta\}$

Is it true that $H_\mu(\alpha\bigvee\beta)\leq H_\mu(\alpha)+H_\mu(\beta)?$

My tries:

$$H_\mu(\alpha\bigvee\beta)=-\sum_{A\cap B\in\alpha\bigvee\beta}\mu(A\cap B)log(\mu(A\cap B))=$$

$$(Absolute-convergence)$$

$$=-\sum_{B\in\beta}\sum_{A\in\alpha}\mu(A\cap B)log(\mu(A\cap B))$$

But I have no idea how to continue from here.

I would be happy for hint or advice.

Thanks in advance.

答:
Let $\gamma:=\alpha\vee\beta$. First, note that
\begin{align}
I(\alpha,\beta)=\sum_{(A,B)\in\gamma}\mu(A\cap B)\log\!\left(\frac{\mu(A\cap B)}{\mu(A)\mu(B)}\right)=H(\alpha)+H(\beta)-H(\gamma).
\end{align}
Thus, it suffices to show that $I(\alpha,\beta)\ge 0$. Using Jensen's inequality,
$$
\sum_{(A,B)\in\gamma}\mu(A\cap B)\log\!\left(\frac{\mu(A)\mu(B)}{\mu(A\cap B)}\right)\le \log\!\left(\sum_{(A,B)\in\gamma}\mu(A)\mu(B)\right)=\log(1)=0.
$$

Title: Entropy is subadditive


问:
How to create a modulo operator that changes?
Suppose I have the rule

$$k \text{ mod } 2$$

Then $$0,1,2,3,4,5,6,...$$ produces $$0,1,0,1,0,1,0,...$$

I want to create a function that would allow me to do a sequence like 2-2-3

So I would have

$$0,1,2,3,4,5,6,...$$

and this would yield

$$0,1,0,1,0,1,2,...$$

What's the most elegant way to do this?

I intend to do this with arbitrary integers, so instead of 2-2-3, I might have 2-4-7-8, etc. So a solution stated generally would be best.

答:
Trivial answer for your first example: take a polynomial $P$ s.t.:
$$P(0) = 0, P(1) = 1,..., P(6) = 2.$$
The function
$$n\mapsto P(n\mod 7)$$
is a "changed modulo operator" for you?


问:
Constrained $L^2([0,1],\mathbb{R})$ space
The space $L^2([0,1],\mathbb{R})$ admits an orthonormal basis:

$1, \sqrt{2} cos(2\pi n x), \sqrt{2} sin(2\pi n x), \quad n= 1,2,...$

What happens to $L^2([0,1],\mathbb{R})$ if we add a constrain, say $u(x)=0 , x \in [0,\epsilon]$ for every function $u \in L^2([0,1],\mathbb{R})$, with $\epsilon$ a small positive number?

Clearly the above trigonometric basis, can not be a basis for the new constrained space. So what is a new basis for this space ? Is the a rigorous way to determine the new basis of the constrained space ?

Also, it is clear that the dimensionality of the space is still infinite. But the space has changed, so is there any notion that can capture this change ?

Thanks a lot

答:
Disclaimer: This is in response to your comment and not an answer to the edited but rather the original question. In fact I am not yet sure about the answer to that. Nevertheless, I hope this is helpful to you.

Recall the construction of the Lebesgue space: $$ L^2([0, 1]) = \mathcal{L}^2([0,1]) / N $$where $\mathcal{L}^2([0,1])$ is the space of all square-integrable functions $f: [0,1] \to \mathbb{R}$ and $N=\{f=0 \, \text{a. e.}\}$. This means that elements in $L^2$ are equivalence classes of functions and thus, two functions are called equal if they differ only on a set of measure zero. In other words, assigning a value at a certain point is not something that's "interesting" or meaningful. You can do this if you really want to but it doesn't change much; say you want the set of all $[u] \in L^2([0,1])$ with $u(0)=0$ (where we identify $u$ as an $\mathcal{L}^2$-function and $[u]$ as the corresponding equivalence class). Then clearly this is a subset of $L^2([0,1])$. On the other hand we can realise this by choosing a representative for every $[u] \in L^2([0,1])$ such that $u(0)=0$, giving us the other subset relation. Hence they are equal and you can choose the same basis (by choosing appropriate equivalence class representatives that you need).


问:
Relations on Surjective Functions
Let $f:A\to B$ be a surjective function. I am asked to find a relation $R$ such that

$\bar{f}: A/R \to B$

$[a]\mapsto \bar{f}([a]) = f(a)$

My thoughts on this problem are that, for $\bar{f}$ to be well defined, it cannot depend on the representative i.e. if $[a]=[b]$ then $\bar{f}([a]) = \bar{f}([b])$. But I don't know how to keep going.

I have defined $R$ to be $aRb \iff f(a) = f(b)$. I have also proved that $R$ is an equivalence relation, but I don't know how to prove the $\bar{f}$, with $R$ defined like this, is bijective. Does anyone know how?

答:
Hint:

$\bar f$, defined by $\bar f\bigl([a]\bigr)=f(a)$ is injective by construction, and it is not hard to check that it is surjective because $f$ is by  hypothesis.

Title: Relations on Surjective Functions


问:
Identification of a plane transformation
We have four points  $ A , B, A', B'$ in the Euclidean plane.
Let $O$ be the center of the direct similitude $s: AB \mapsto A'B'$.

Now, the points $A', B, B'$ being fixed, how can I find the expression in terms of complex numbers of the plane transformation $A \mapsto O$  ?

It should be an homography but I can't find its exact analytical form...
Many thanks for any suggestions.

答: In $\mathbb{C}$ a direct similitude is $z \mapsto h(z) = \alpha z + \beta$ with $\alpha \neq 0, 1$.Let $a, a', b, b'$ the complex affixes of $A, A', B, B'$. Let us suppose $A \neq B \iff a \neq b$. The center $O$ (affixe $o$ ) is such that $h(o)= o \iff o = \frac{\beta}{1- \alpha}$. Now $\alpha$ and $\beta$ are defined by $h(a)= a'$ and $h(b) = b'$ hence $o = \frac{a b' - a'b}{a-b - (a'-b')}$. The map $a \mapsto o$ is an homography.


问:
Finding matrices that satisfy equation
I am trying to find the set of square matrices that satisfy:
$$M=M^{2n-1}-\sum_{k=1}^{2n-2}{M^k}$$
$$n\in \mathbb{N}$$
I have tried simplifying and trying to treat the sum as a geometric series but this hasn't lead me anywhere. How could I go about solving this?.

答:
I am not sure about fully describing the set of matrices with this property. But we can use the Cayley-Hamilton Theorem to know that if $p(\lambda)$ is the characteristic polynomial of a matrix $A$, then $p(A)=0$. If you consider all matrices $M$ with characteristic polynomials $p$ which divide polynomial $q$ given by $$q(\lambda)= \lambda^{2n-1}-2\lambda - \sum_{k=2}^{2n-2}\lambda^k$$
Then $q(M)=0$ and we can rewrite the equation to infer that $q(M)=0$ implies that
$$M=M^{2n-1}-\sum_{k=1}^{2n-1}M^k$$
So if you are attempting to find matrices such that $q(M)$ is equal to $0$, you can form a polynomial $p$ by taking $n$ distinct roots (including multiplicities) of $q$. This will give you the eigenvalues of a matrix. Then you just need to construct a matrix with those eigenvalues. This will at least give you a subset and I am not sure if it will fully describe all matrices in the set.

Title: Finding matrices that satisfy equation


问:
Wolfram Alpha's Complex Integration involving Log in the antideriviative, gives a different result
I need to compute:

$$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{i}{9 e^{i 2 \theta} + 1}d\theta $$

The indefinite integral above is:

$$ F(\theta)= i \theta - \frac{1}{2} \log(1 + 9 e^{2 i \theta}) $$

According to Wolfram Alpha and my own computation.

The problem is that if I assign the values of the limit myself to it I get:

$$ F(\frac{\pi}{2}) - F(-\frac{\pi}{2}) = \\
i (\frac{\pi}{2} - (-\frac{\pi}{2}))-\frac{1}{2} \left (\log(1+9 e^{2 i\frac{\pi}{2}}) -\log(1+9e^{2 i \cdot (-\frac{\pi}{2})}\right ) = \\
i \pi -\frac{1}{2}\left ( \log(\frac{1-9}{1-9})\right ) = i\pi
$$

And wolfram's solution for the definite integral is $0$ (link).

Why? Is it related to the principal value of the natural logarithm?

答:
Definitely related to the natural logarithm. When $\theta=-\pi/2$ we may take the argument to be $9(-1)+1=8e^{-\pi i}$. As $\theta$ increases from $-\pi/2$ to $\pi/2$, $9e^{2i\theta}+1$ circles the point $z=1$ counterclockwise in the complex $z$-plane so that when $\theta=0$ it will be $10e^{i0}$ and by the time $\theta=\pi/2$ it will be $8e^{\pi i}$ so
$$F(\pi/2)-F(-\pi/2)=\left(i\frac{\pi}2-\frac12\left(\pi i+\ln8\right)\right)-\left(-i\frac{\pi}2-\frac12\left(-\pi i+\ln8\right)\right)=0$$
Since the magnitude is never $0$ the phase must change continuously.

EDIT: Another way to think about it to say that
$$\begin{align}F(\theta)&=i\theta-\frac12\ln\left(1+9e^{2i\theta}\right)\\
&=-\frac12\ln\left(e^{-2i\theta}\right)-\frac12\ln\left(1+9e^{2i\theta}\right)=-\frac12\ln\left(e^{-2i\theta}+9\right)\end{align}$$
And now as $\theta$ changes the path doesn't circle a branch point so it's not so critical to think about the phase of the argument of the natural logarithm.


问:
What is the difference between a predicate and function?
I need to to understand the difference between predicates and functions in the context of Clasual Form Logic in order to define the Herbrand universe.

If I have p(x) :- q(f(x)) would I be right in saying that p and q are predicates while f is a function because it is "nested"?  By this thinking then if I have p(x) :- q(x) both p and q are predicates and I have no functions?

If this is incorrect then how can I tell the difference between a predicate and function?

答:
A predicate is a box that takes an argument and returns a Boolean value. For example, "$x \mapsto x \text{ is even}$".

A function is a box that takes an argument and returns a value. For example, "$x \mapsto x^2$".

Edit (following Amy's suggestions): There is some domain over which all variables range. A function takes zero or more arguments of that domain and returns another argument from that domain. A predicate takes zero or more arguments of that domain and returns a Boolean value.


问:
Percentage Problem: Chance of being killed in a lifetime calculation
If the average lifetime of a person is $72.6$ and if the person has a $0.00016\%$ chance of being killed by doing a certain activity in a year.

How do you calculate the chance of him being killed doing that activity in a lifetime?

Is it $(72.6 * 0.00016)=0.011616\% ?$
(It might sound simple but If possible, please provide a clue/explanation on how to calculate such problems.)

答:
If the probability of a single person experiencing a certain condition (such as being killed, in your example) each year is $p$ then we can say that the probability of a single person not experiencing that condition in a year is $1-p$.

Then, the probability of a person never experiencing that condition throughout their $n$ years of life can be calculated as the product of probabilities of not experiencing that condition in every single one of those $n$ years. In your example, if a person is to not ever get killed in their life, that person should not be killed in their year 1, and not in year 2, and not in year 3, etc. So, the probability of not experiencing that certain condition in $n$ years is $(1-p)^n$ .

Now, the probability of experiencing that condition (in your example, getting shot) will be:
$$1 - (1-p)^n$$

In your example, $p=\frac{405}{250.4million}=0.0000016$ and $n=72.6$ , so:
$$1 - (1-p)^n = 0.000117$$
In other words, in your example a person has a nearly $0.01$ percent chance of experiencing that particular condition in their life.


问:
limit of sequence of functions define inductively using integral
Let $f_0 \in L^1([0,1])$. Define $f_n(x) = (\int\limits_0^x f_{n-1}(y) dy)^{1/2}$. Assume that $f_0 \geq 0$ $a.e$ and $f_1(x) \leq f_0(x)$ $a.e$. I have proved the following:

$f_n(x)$ converges pointwise to a function $f(x)$ that is finite everywhere on $[0,1]$.

$f(x) = (\int\limits_0^x f(y)dy)^{1/2}$.

The next question asks me to prove that if $f(x) > 0$, then $f$ is differentiable at $x$ and compute $f'(x)$. I already proved that $f'(x) = 0$. However, I failed to see why this does not hold for all $x$. Is there a counter-example?

Finally, I need to find an explicit formula for $f(x)$ where $f(x) \neq 0$. Using the previous part, I found that in fact $f$ is identically $0$, which seems wrong. I hope someone can provide a counter example.

答: $f^{2}(x)=\int_0^{x} f(t)dt$ gives $2f(x)f'(x)=f(x)$ so $f'(x)=\frac 1  2$ when $f(x) \neq 0$. Hence, if $f(x) >0$ for all $x$, then  $f(x)=\frac x 2 +c$ where $c$ is a constant.  Also $f(0)=0$ so $c=0$.


问:
Modify the covariance matrix of a vector to obtain a new vector with a given variance after matrix multiplication
I have a covariance matrix A that is used to generate a vector X. The vector X is then multiplied with a transformation matrix T to give a new vector Y. The vector Y now needs to have a pre-defined variance, sigma. How can I edit the matrix A such that after the transformation, Y has a variance sigma?</p>

答: If $X$ is a random vector in $\mathbb{R}^n$ with covariance matrix $A$, and if $T$ is a $m \times n$ matrix, then $Y=TX$ is a random vector with covariance matrix $TAT^\top$.


问:
How do I calculate an integral like $\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$?
$$\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$$

This is Gauss' law from physics but my question is more maths related.

Say I have $\vec{E}^{\,}=3.5\times 10^3N/C\times e_x$ and $A = 0,35\times0,70m^2$. The plane that has that area, where the electric field $E$ is applied, is parallel to $YZ$. So it has a coordinate in $e_x$. Right?

What does the circle in the integral mean in practice?

What is the value of $dA$? Why is it a vector?

How do I solve this integral?

答:
Gauss' law involves a surface integral. The answers to your questions come from understanding how this surface integral works.

First off, in an integral of the form
$$
\int_S \vec{F} \cdot \vec{dA}
$$
we have $2$ elements at play:

$\vec{F}$ is a vector field, which is just a function that gives you a vector as an output.

$\vec{dA}$ is the area vector. Explicitly, it represents the normal vector to the surface multiplied by the area differential $dS$. In practice you constantly have to use $\vec{dA} =\hat{n} \ dS$ to solve the integrals.

These 2 quantities are being multiplied using the dot product, which by definition will transform our integrand from vectors being multiplied into a scalar function.

Here's a diagram to better understand these components:

where $S$ is the surface we're integrating over and $dS$, $\hat{n}$ and $\vec{F}$ have the same interpretation as above. Now to answer the questions:

What does the circle in the integral mean in practice?

The circle just means that your surface $S$ of integration is closed. This just means that Gauss' law doesn't apply to any blanket-like surface you want, it HAS to completely enclose the charge (which is the physical reason for this). This is why you'll find that you usually take spheres or cylinders as the surface $S$ in these problems. In practice, it's no different as to solving any other surface integral, it only affects the limits of integration.

What is the value of $\vec{dA}$? Why is it a vector?

By definition, it's the area vector we described previously.

How do I solve this integral?

I'll use your data to try and show the general approach. Our vector field $\vec{E}$ is given by
$$
\vec{E} = (3500,0,0)
$$
Now, the way I understand your question as written is that our surface of integration $S$ in this case is a segment of a plane parallel to the $y-z$ plane. This means that the normal vector to this plane is the unitary vector in the $x$ direction, or in other words:
$$
\hat{n} = (1,0,0)
$$
Note here that in this case our surface is not closed, since a plane does not enclose some volume like a sphere, for example. This is why it wouldn't be correct to use $\oint$ in this case, we should just use $\int_S$. Having said this, the integral we want becomes
\begin{align*}
\int_S \vec{E} \cdot \vec{dA} &\overset{\color{blue}{2}}{=} \int_S\left( \vec{E} \cdot \hat{n}\right) \ dS \\
&= \int_S\left( (3500,0,0) \cdot(1,0,0)\right) \ dS \\
&= \int_S 3500\ dS \\
& = 3500 \int_S \ dS
\end{align*}
but by definition of surface integral, if we're integrating the area differential over the whole area we're interested in, we just get the area we're interested in as a result. In other words:
$$
\int_S \ dS = 0.35 \times 0.70 = 0.245
$$
which is the area of our plane segment. And so
\begin{align*}
\int_S \vec{E} \cdot \vec{dA} & = 3500 \int_S \ dS\\
& = 3500 (0.245)\\
& = 857.5
\end{align*}
I've dropped the units in the proceadure, but this last result would have units of $\frac{N m^2}{C}$.

As a final note, this kind of vector-surface integral is also known as the flux. It's called this way because it essentially tells you "how much field" passes through the surface we're integrating over. You can see this by the use of the dot product between $\vec{F} \cdot \vec{dA}= \vec{F} \cdot \hat{n} \ dS$ in the diagram at the beginning since when $\vec{F}$ and the normal vector to the surface are parallel the dot product will give a bigger number, whereas if they don't point in the same direction we get a smaller dot product.

Lastly, in a general approach to solve the integral we usually can't easily calculate the area or factor out the integrand. In these cases you would proceed to parametrize the surface (in cartesian coordinates, for example) and then use that $dS = dy \ dx$ to solve the integrals explicitly.


问:
If $f$ is a contraction then it is continuous.
The proof might seem intuitive if just has one or more jump points which have a distance $d$ from each other.

But I am struggling, with the following problem:

If f is not continuous, such that for all intervals: $$[a,b], \text{with } a < b < c < d: f \text{ is not continuous in } [b,c]$$
where $a,d$ are constant and $b,c$ freely selectable.

Can it then be a contraction?

Edit:

$$\exists a,d: \forall b,c: (a < b < c  <d \implies f \text{ is not continuous in } [b,c])$$

I would prefer to use the definition of continuity using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition.

答:
Welcome to MSE!

Hint:

If $f$ is a contraction, then $d(fx,fy) < d(x,y)$ for every $x,y$.

To show $f$ is continuous, we want to show we can make $d(fx,fy) < \epsilon$ small by controlling $d(x,y)$... Of course, this is exactly the flavor of control that contractibility buys us.

Formally, say $\epsilon > 0$. We want to find a $\delta$ so that whenever $d(x,y) < \delta$, we're guaranteed $d(fx,fy) < \epsilon$...
By contractibility, we're guaranteed $d(fx,fy) < d(x,y) < \delta$.

Do you see where to go from here? What's a good choice of $\delta$?

I hope this helps ^_^


问:
Showing that these algebras are isomorphic
This paper (page 157) diagonalized circulant matrix $S$ like this where $\psi$ is an eigenvalue and $\Omega$ is composed of the eigenvectors as columns:

$$
\Omega^{-1}S\Omega =
\begin{bmatrix}
\psi_0 & 0 & \dots & 0 \\
0 & \psi_1 & \dots & 0 \\
\vdots &  & \ddots & \vdots \\
0 & \dots & 0 & \psi_{n - 1}
\end{bmatrix}
$$

It was then stated that this is a bijective and linear transformation between 3 algebras:

The algebra of columns of $\mathbb{C}^n$, identified with the maps from $\mathbb{Z}_n$ to $\mathbb{C}$

The algebra of circulating matrices

The algebra of diagonal matrices

It is said that the product in the first algebra is the convolution product. It is also stated that the last isomorphism stems from the fact that all circulant matrices are polynomials in

$$
J = \begin{bmatrix}
0 & 0 & \dots & 1 \\
1 & 0 & 0 \ \dots & 0 \\
0 & 1 & \dots & 0 \\
\vdots & & \ddots & \vdots \\
0 & & \dots & 0
\end{bmatrix}
$$
since $J^n = I_n$, its eigenvalues will be the $n$th roots of unity that appear in the DFT formula, and the eigenvectors of $J$ remain eigenvectors for every polynomial in $J$, e.g. for every circulating matrix $S_v = \sum_{k = 0}^{n - 1}v_kJ^k$.

Firstly, I don't exactly understand what the paper means to be a bijective and linear transformation here and how we should even go on about proving how algebras are isomorphic. I also don't understand what it means for a matrix to be a polynomial in another matrix. In particular, the fact that circulant matrices are polynomials in the permutation matrix $J$. In linear algebra, the only polynomials I've come across are characteristic polynomials of matrices.

I probably need to learn more topics to understand exactly what is going on here. But I would like to make sense of this and get a grasp of it if possible.

答:
The first algebra is the vector space $\operatorname{Map}(\mathbb{Z}_n, \mathbb{C})$ of functions $f \colon \mathbb{Z}_n \to \mathbb{C}$. The set $\mathbb{Z}_n$ is a group under addition, and so we can equip $\operatorname{Map}(\mathbb{Z}_n, \mathbb{C})$ with a convolution product $\star$, defined by either of these two familiar formulas:
$$ (f \star g)(z) = \sum_{\substack{x, y \in \mathbb{Z}_n \\ x + y = z}} f(x) f(y) = \sum_{x \in \mathbb{Z}_n} f(x) f(z - x).$$
The convolution product is commutative and associative, and the indicator function $\delta_0 \colon \mathbb{Z}_n \to \mathbb{C}$ defined by $\delta_0(0) = 1$ and $\delta_0(n) = 0$ for $n \neq 0$ is the unit for the product. On indicator functions in general, we have for all $x, y \in \mathbb{Z}_n$ that $\delta_x \star \delta_y = \delta_{x + y}$.

The second algebra is the algebra of $n \times n$ circulant matrices, which we will call $\operatorname{Circ}_n$. By polynomials in $J$, they mean simply that any $n \times n$ circulant matrix $C$ may be written as $C = c_0 + c_1 J + c_2 J^2 + \cdots + c_{n - 1} J^{n-1}$, in other words the vectors $I, J, J^2, \ldots, J^{n-1}$ form a basis for the vector space of circulant matrices. Hence the circulant matrices form a vector space, but more is true: multiplying two circulant matrices produces another circulant matrix. If $C = c_0 + c_1J + \cdots + c_{n-1}J^{n-1}$ and $D = d_0 + d_1 J + \cdots + d_{n-1} J^{n-1}$, then $CD$ is again a polynomial in $J$, with at most $n-1$ terms because $J^n = I$. Let $CD = E = e_0 + e_1 J + \cdots + e_{n-1}J^{n-1}$, then just from the product of polynomials and $J^n = I$ we get
$$ e_z = \sum_{\substack{x, y \in \mathbb{Z}_n \\ x + y = z}} c_x d_y,$$
which looks like the convolution product.

In order to show that the first and second algebras are isomorphic, we must give a bijective linear map $\varphi \colon \operatorname{Map}(\mathbb{Z}_n, \mathbb{C}) \to \operatorname{Circ}_n$ such that $\varphi(f \star g) = \varphi(f) \varphi(g)$ for all $f, g \in \operatorname{Map}(\mathbb{Z}_n, \mathbb{C})$. This shows that the vector spaces and products are really the same, and all that has changed is our labelling. After the discussion above, the natural map to try is
$$ \varphi(f) = \sum_{x = 0}^{n-1} f(x) J^x.$$
We now try to show that $\varphi(f) \varphi(g) = \varphi(f \star g)$:
$$ \begin{aligned}
\varphi(f) \varphi(g)
&= \left(\sum_{x \in \mathbb{Z}_n} f(x) J^x \right) \left(\sum_{y \in \mathbb{Z}_n} f(y) J^y \right) \\
&= \sum_{x, y \in \mathbb{Z}_n} f(x)f(y) J^{x + y} \\
&= \sum_{z \in \mathbb{Z}_n} \left(\sum_{\substack{x, y \in \mathbb{Z}_n \\ x + y = z}} f(x) f(y)\right) J^z \\
&= \sum_{z \in \mathbb{Z}_n} (f \star g)(z) J^z \\
&= \varphi(f \star g).
\end{aligned} $$
Since $\varphi$ is a bijection preserving the product, the algebras are isomorphic.

Can you take it from here to try to show that the third algebra is isomorphic to either of these two? The third algebra is basically just noticing that all circulant matrices have the same eigenvectors, but different eigenvalues, so rather than writing $C = c_0 + c_1 J + \cdots + c_{n-1} J^{n-1}$, instead you can pick some ordering $v_0, \ldots, v_{n-1}$ of the circulant eigenvectors and write down the list of eigenvalues of $C$.


问:
equation of an involute gear
I'm trying to find the equation of an involute profile gear in polar coordinate system:

```
<img src="https://i.stack.imgur.com/5GngZ.png" width="400">
```

I know

$$ r \cos(\alpha - \theta) = R \tag{1}$$
$$l = \alpha R \tag{2}$$
$$r \sin(\alpha - \theta) = l \tag{3}$$

so far I can deduct the equations to

$$\theta = \alpha - \arctan(\alpha) \tag{4}$$

What I need to have

$$r = f(\theta) \tag{5}$$

but I can't find the inverse of equation $(4)$:

$$\alpha = g(\theta) \tag{6}$$

If I had $g(\theta)$ then

$$f(\theta) = \frac{R}{\cos(g(\theta) - \theta)} \tag{7}$$

I would appreciate it if you could help me know if there are analytical/closed-form solutions for $g(\theta)$ and/or $f(\theta)$. Thanks for your support in advance.

答:
Polar coordinates isn't a good choice for involute shapes. It is unnecessarily complex. I prefer to work with cartesian coordinates and declare the curve in a parametric form $(x(t),\,y(t))$. The reason a polar curve is difficult is that there is no analytical solution to $r(\theta)$, but rather a parametric form of $r(t)$ and $\theta(t)$ exist, similar to the cartesian case.

First, consider an aligned coordinate system (shown below) resting on the generating circle (blue dashes). The involute shape (red curve) starts at the origin of the coordinate system and moves radially out as well as to the right. At any point on the curve, the distance to the tangent point (purple line) to the generating circle equals the arc length from the origin to the tangent point.

Define a parameter $t$ for the curve and note the equation of the involute curve in terms of the coordinate system shown above

$$ \pmatrix{ x(t) \\ y(t)} = \pmatrix{ R_0 \sin t - (R_0 t) \cos t \\ -R_0 + R_0 \cos t + (R_0 t) \sin t } \tag{1}$$

and here is what this curve looks like

The next part is the rotate the gear tooth in order to make sure you get the correct gear tooth width and contact pressure angle $\alpha$ at the pitch diameter (or radius $R_p$)

You need to rotate the coordinate system to the left a certain amount to achieve this, and part of this rotation is to account for the angular separation between the generating point and the contact point indicated above as ${\rm inv}\alpha$. This is the same notation in the ISO standard for involute gears. Use $\alpha$ for the desired pressure angle in radians and calculate the following

$$ {\rm inv}\alpha = (\tan \alpha) - \alpha \tag{2}$$

Now if the gear has $n$ teeth then each tooth must have arc length width of $s = \frac{\pi}{n} R_p$ encompassing an angle $\varphi = \tfrac{\pi}{n}$ as seen below

The total rotation of the coordinate system to generate the proper gear flank is thus

$$ \text{rot.} = {\rm inv}\alpha + \varphi/2 \tag{3e}$$

Note that the above is valid for external geometry, and for internal geometry, if you do a similar sketch you can find the corresponding rotation to be

$$ \text{rot.} = -{\rm inv}\alpha + \varphi/2 \tag{3i}$$

Title: equation of an involute gear


问:
A square appears randomly within a square of ten time its area. What is probability that the smaller square contains the larger square's center?
The alignment relative to the larger square can be anywhere between 0° to 90° (0° or 90° being identical orientation to the larger square, 45° being all the way skewed), and the probability distribution of the alignment is based on percentage of total possibility given the constraint guaranteeing that it will fit within the larger square.

<s>The largest discrete probability, for an alignment of 0°=90°, should be simple enough to figure-out‡. The smallest discrete probability, for alignment of 45°, is a bit trickier but straightforward enough.</s>Edit:The off-kilter version would be more likely to contain the center, since it has less possible places to exist and a larger proportion of them are the center. Intuitively this should correspond directly (though not necessarily uniformly) to what I am totally clueless about, which is:

how to determine the probability  distribution for between 0° to 45°,

‡ 2. how this continuum corresponds to the probability of the original problem (e.g. {1/17?} for 0° up to {2/9?} for 45° back down to {1/17?} for 90°, along finite part of some sort of curve function presumably),

and 3. how then to apply these equations formulated into the original problem to find the final probability that a square true-randomly generated wholly within a larger square with sides √(10)-times longer would surround (or contain exactly on an edge or corner) the center of the larger square.

‡ I haven't calculated the maximum or minimum ‘discrete’ probabilities (let alone any inbetween) yet. Those guesstimate values are placeholders. I am especially interested in the processes required to solve this, and appreciate any insight or clues.

答:
Without Rotation

Suppose a square of side $s\le\frac12$ is placed randomly inside a unit square, without rotation. The probability that the $x$-coordinate of the center of the unit square is within the $x$-span of the smaller square is $\frac{s}{1-s}$. That is, when covering the center of the unit square, the left side of the smaller square can be $s$ to $0$ units to the left of the center of the unit square, while, without restriction, the left side of  the smaller square can go from $1$ to $s$ units to the left of the right side of the unit square. The same is true for the $y$-coordinate. So the probability that the center of the larger square is within the smaller square is
$$
\left(\frac{s}{1-s}\right)^2\tag1
$$
In your question, $s=\frac1{\sqrt{10}}$, which gives a probability of
$$
\bbox[5px,border:2px solid #C0A000]{\frac1{11-2\sqrt{10}}\approx0.213883399}\tag2
$$

With Rotation

Geometry and Probability

The smaller square rotated by $0\le\theta\le\frac\pi4$ will span a width and height of
$$
d=s(\cos(\theta)+\sin(\theta))\tag3
$$
Note that $(3)$ says that $d:\left[0,\frac\pi4\right]\mapsto\left[s,s\sqrt2\right]$ monotonically.

If the smaller square is to stay inside the unit square, its center is constrained to a non-rotated square of side $1-d$ concentric with the unit square (shown below as a red square).  If the center of the unit square is to be inside the smaller square, the center of the smaller square needs to be inside a square with the same size and rotation, but concentric with the unit square (shown below as a green square).

Thus, the locus of the center of the smaller square so that the smaller square is contained in the unit square, and so that the smaller square contains the center of the unit square is the intersection of the red and green squares (shown below in purple).

For a given size and rotation of the smaller square, given that the smaller square is contained in the unit square, the probability that the center of the unit square is inside the smaller square is the ratio of the area of the purple region divided by the area of the red square.

However, to get the proper weighting of the probabilities for the different angles, we need to weight the probabilities by the area of the red square since that is the probability that the rotated square is contained in the unit square. This means we compute the integral of the area of the purple region and divide that by the integral of the area of the red region.

Computation of the Areas

The area of the red square is
$$
\begin{align}
\text{area}_\text{red}
&=(1-d)^2\tag{4a}\\[6pt]
&=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{4b}\\[6pt]
&=1-2s(\sin(\theta)+\cos(\theta))+s^2(1+2\sin(\theta)\cos(\theta))\tag{4c}\\
\int_0^{\pi/4}\text{area}_\text{red}\,\mathrm{d}\theta
&=\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2\tag{4d}
\end{align}
$$
If $d\le\frac12$, then, because $d\lt1-d$, the green square is completely contained within the red square and so the area of the purple region is
$$
\begin{align}
\text{area}_\text{purple}
&=s^2\tag{5a}\\
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\theta s^2\tag{5b}\\
&=a_1(s,\theta)\tag{5c}
\end{align}
$$
We leave $\text{(5b)}$ indefinite since we may need to restrict $\theta$ due to the condition $d\le\frac12$.

For $d\gt\frac12$, we will refer to the following diagram:

Since the side length of the red square $1-d$ and the height of the rotated green square is $d$, the four corners of the green square that extend beyond the red square have altitude $\frac{d-(1-d)}2=d-\frac12$ and base $\frac{d-1/2}{\sin(\theta)\cos(\theta)}$. If the bases of these triangles are smaller than the sides of the red square, then $\frac{d-1/2}{\sin(\theta)\cos(\theta)}\lt1-d$, which is equivalent to $d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$.

Thus, if $\frac12\lt d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the area of the purple region is
$$
\begin{align}
\text{area}_\text{purple}
&=s^2-\frac{2(d-1/2)^2}{\sin(\theta)\cos(\theta)}\tag{6a}\\
&=s^2-\frac{2\left(s(\sin(\theta)+\cos(\theta))-\frac12\right)^2}{\sin(\theta)\cos(\theta)}\tag{6b}\\[3pt]
&=\scriptsize-\frac12\csc(\theta)\sec(\theta)+2(\csc(\theta)+\sec(\theta))s-(3+2\csc(\theta)\sec(\theta))s^2\tag{6c}\\[6pt]
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\scriptsize-\frac12\log(\tan(\theta))+2\log\left(\tan\left(\tfrac\theta2\right)\tan\left(\tfrac\pi4{+}\tfrac\theta2\right)\right)s-\left(3\theta+2\log(\tan(\theta))\right)s^2\tag{6d}\\[9pt]
&=a_2(s,\theta)\tag{6e}
\end{align}
$$
If $d\ge\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the red square is completely contained within the green square and so the area of the purple region is the same as the area of the red square:
$$
\begin{align}
\text{area}_\text{purple}
&=(1-d)^2\tag{7a}\\[3pt]
&=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{7b}\\[3pt]
&=1-2(\sin(\theta)+\cos(\theta))s+(1+2\sin(\theta)\cos(\theta))s^2\tag{7c}\\
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\theta-2(\sin(\theta)-\cos(\theta))s+\left(\theta+\sin^2(\theta)\right)s^2\tag{7d}\\
&=a_3(s,\theta)\tag{7e}
\end{align}
$$
Conditional Ranges

We need to apply each of $a_1$, $a_2$, or $a_3$ over particular parts of $\left[0,\frac\pi4\right]$. That is, we use $a_1$ over the green region of the chart below, $a_2$ over the yellow region, and $a_3$ over the red region.

To simplify the usage of the information in the chart above, define
$$
\theta_1(s)=\frac12\sin^{-1}\left(\frac{1-4s^2}{4s^2}\right)\tag8
$$
and
$$
\theta_2(s)=\frac12\sin^{-1}\left(\frac{1-4s^2+\sqrt{1-4s^2}}{2s^2}\right)\tag9
$$

If $s\le\frac1{2\sqrt2}$, we use $a_1$ for all $\theta\in\left[0,\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_1\!\left(s,\frac\pi4\right)-a_1\!\left(s,0\right)}{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}=\frac{\frac\pi4s^2}{\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2}\tag{10}
$$
If $\frac1{2\sqrt2}\lt s\le\frac{\sqrt2}3$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$ and $a_2$ over $\theta\in\left[\theta_1(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_2\!\left(s,\frac\pi4\right)-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)}
{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{11}
$$
If $\frac{\sqrt2}3\lt s\le\frac12$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$, $a_2$ over $\theta\in[\theta_1(s),\theta_2(s)]$, and $a_3$ over $\theta\in\left[\theta_2(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_3\!\left(s,\frac\pi4\right)-a_3(s,\theta_2(s))+a_2(s,\theta_2(s))-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)}
{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{12}
$$
In the question, $s=\frac1{\sqrt{10}}$, so we apply $(10)$ and get a probability of
$$
\bbox[5px,border:2px solid #C0A000]{\frac\pi{11\pi+2-8\sqrt{10}}\approx0.279022074}\tag{13}
$$
Here is a plot of the probability that a randomly rotated and positioned square of side $s$ inside the unit square contains the origin of the unit square:

Simulation

I ran a simulation placing the center of the smaller square, with side $s$, uniformly in a square, concentric with the unit square, and with side $1-s$, and rotating uniformly with $\theta\in\left[0,\frac\pi4\right]$.

All trials where the translated and rotated square did not fit completely in the unit square were rejected. The remaining trials were tallied counting those where the center of the unit square fell within the smaller square.

The results of $10,\!000,\!000$ trials with $s=\frac1{\sqrt{10}}$ were

non-rejected trials: $7,\!664,\!102$

successful trials: $2,\!136,\!729$

ratio: $0.278797046$

This is close to $0.279022074$.

Here is the Mathematica code for the simulation:

```<code>test[s_, x_, y_, \[Theta]_] :=
 Module[{c = {x - 1/2, y - 1/2} (1 - s),
   m = s {{Cos[\[Theta]], Sin[\[Theta]]}, {-Sin[\[Theta]], Cos[\[Theta]]}},
   u = 1/2 {{-1, 1}, {1, 1}, {1, -1}, {-1, -1}},
   cross = (#1.{{0, 1}, {-1, 0}}.#2) &, ul, ur, lr, ll},
   {ul, ur, lr, ll} = u.m + {c, c, c, c};
  If[Max[Abs[ul], Abs[ur], Abs[lr], Abs[ll]] > 1/2, {0, 0},
   If[Min[cross[ur, ul], cross[ul, ll], cross[ll, lr], cross[lr, ur]] < 0,
    {0, 1}, {1, 1}]]]
</code>```
```<code>Module[{s = 1/Sqrt[10], n = 10000000, data = {0, 0}},
 Do[data += test[s, RandomReal[1], RandomReal[1], RandomReal[Pi/4]], n];
 data]
</code>```

Body:
Without Rotation

Suppose a square of side $s\le\frac12$ is placed randomly inside a unit square, without rotation. The probability that the $x$-coordinate of the center of the unit square is within the $x$-span of the smaller square is $\frac{s}{1-s}$. That is, when covering the center of the unit square, the left side of the smaller square can be $s$ to $0$ units to the left of the center of the unit square, while, without restriction, the left side of  the smaller square can go from $1$ to $s$ units to the left of the right side of the unit square. The same is true for the $y$-coordinate. So the probability that the center of the larger square is within the smaller square is
$$
\left(\frac{s}{1-s}\right)^2\tag1
$$
In your question, $s=\frac1{\sqrt{10}}$, which gives a probability of
$$
\bbox[5px,border:2px solid #C0A000]{\frac1{11-2\sqrt{10}}\approx0.213883399}\tag2
$$

With Rotation

Geometry and Probability

The smaller square rotated by $0\le\theta\le\frac\pi4$ will span a width and height of
$$
d=s(\cos(\theta)+\sin(\theta))\tag3
$$
Note that $(3)$ says that $d:\left[0,\frac\pi4\right]\mapsto\left[s,s\sqrt2\right]$ monotonically.

If the smaller square is to stay inside the unit square, its center is constrained to a non-rotated square of side $1-d$ concentric with the unit square (shown below as a red square).  If the center of the unit square is to be inside the smaller square, the center of the smaller square needs to be inside a square with the same size and rotation, but concentric with the unit square (shown below as a green square).

Thus, the locus of the center of the smaller square so that the smaller square is contained in the unit square, and so that the smaller square contains the center of the unit square is the intersection of the red and green squares (shown below in purple).

For a given size and rotation of the smaller square, given that the smaller square is contained in the unit square, the probability that the center of the unit square is inside the smaller square is the ratio of the area of the purple region divided by the area of the red square.

However, to get the proper weighting of the probabilities for the different angles, we need to weight the probabilities by the area of the red square since that is the probability that the rotated square is contained in the unit square. This means we compute the integral of the area of the purple region and divide that by the integral of the area of the red region.

Computation of the Areas

The area of the red square is
$$
\begin{align}
\text{area}_\text{red}
&=(1-d)^2\tag{4a}\\[6pt]
&=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{4b}\\[6pt]
&=1-2s(\sin(\theta)+\cos(\theta))+s^2(1+2\sin(\theta)\cos(\theta))\tag{4c}\\
\int_0^{\pi/4}\text{area}_\text{red}\,\mathrm{d}\theta
&=\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2\tag{4d}
\end{align}
$$
If $d\le\frac12$, then, because $d\lt1-d$, the green square is completely contained within the red square and so the area of the purple region is
$$
\begin{align}
\text{area}_\text{purple}
&=s^2\tag{5a}\\
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\theta s^2\tag{5b}\\
&=a_1(s,\theta)\tag{5c}
\end{align}
$$
We leave $\text{(5b)}$ indefinite since we may need to restrict $\theta$ due to the condition $d\le\frac12$.

For $d\gt\frac12$, we will refer to the following diagram:

Since the side length of the red square $1-d$ and the height of the rotated green square is $d$, the four corners of the green square that extend beyond the red square have altitude $\frac{d-(1-d)}2=d-\frac12$ and base $\frac{d-1/2}{\sin(\theta)\cos(\theta)}$. If the bases of these triangles are smaller than the sides of the red square, then $\frac{d-1/2}{\sin(\theta)\cos(\theta)}\lt1-d$, which is equivalent to $d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$.

Thus, if $\frac12\lt d\lt\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the area of the purple region is
$$
\begin{align}
\text{area}_\text{purple}
&=s^2-\frac{2(d-1/2)^2}{\sin(\theta)\cos(\theta)}\tag{6a}\\
&=s^2-\frac{2\left(s(\sin(\theta)+\cos(\theta))-\frac12\right)^2}{\sin(\theta)\cos(\theta)}\tag{6b}\\[3pt]
&=\scriptsize-\frac12\csc(\theta)\sec(\theta)+2(\csc(\theta)+\sec(\theta))s-(3+2\csc(\theta)\sec(\theta))s^2\tag{6c}\\[6pt]
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\scriptsize-\frac12\log(\tan(\theta))+2\log\left(\tan\left(\tfrac\theta2\right)\tan\left(\tfrac\pi4{+}\tfrac\theta2\right)\right)s-\left(3\theta+2\log(\tan(\theta))\right)s^2\tag{6d}\\[9pt]
&=a_2(s,\theta)\tag{6e}
\end{align}
$$
If $d\ge\frac{1+\sin(2\theta)}{2+\sin(2\theta)}$, the red square is completely contained within the green square and so the area of the purple region is the same as the area of the red square:
$$
\begin{align}
\text{area}_\text{purple}
&=(1-d)^2\tag{7a}\\[3pt]
&=(1-s(\sin(\theta)+\cos(\theta)))^2\tag{7b}\\[3pt]
&=1-2(\sin(\theta)+\cos(\theta))s+(1+2\sin(\theta)\cos(\theta))s^2\tag{7c}\\
\int\text{area}_\text{purple}\,\mathrm{d}\theta
&=\theta-2(\sin(\theta)-\cos(\theta))s+\left(\theta+\sin^2(\theta)\right)s^2\tag{7d}\\
&=a_3(s,\theta)\tag{7e}
\end{align}
$$
Conditional Ranges

We need to apply each of $a_1$, $a_2$, or $a_3$ over particular parts of $\left[0,\frac\pi4\right]$. That is, we use $a_1$ over the green region of the chart below, $a_2$ over the yellow region, and $a_3$ over the red region.

To simplify the usage of the information in the chart above, define
$$
\theta_1(s)=\frac12\sin^{-1}\left(\frac{1-4s^2}{4s^2}\right)\tag8
$$
and
$$
\theta_2(s)=\frac12\sin^{-1}\left(\frac{1-4s^2+\sqrt{1-4s^2}}{2s^2}\right)\tag9
$$

If $s\le\frac1{2\sqrt2}$, we use $a_1$ for all $\theta\in\left[0,\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_1\!\left(s,\frac\pi4\right)-a_1\!\left(s,0\right)}{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}=\frac{\frac\pi4s^2}{\frac\pi4-2s+\left(\frac\pi4+\frac12\right)s^2}\tag{10}
$$
If $\frac1{2\sqrt2}\lt s\le\frac{\sqrt2}3$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$ and $a_2$ over $\theta\in\left[\theta_1(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_2\!\left(s,\frac\pi4\right)-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)}
{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{11}
$$
If $\frac{\sqrt2}3\lt s\le\frac12$, we use $a_1$ over $\theta\in[0,\theta_1(s)]$, $a_2$ over $\theta\in[\theta_1(s),\theta_2(s)]$, and $a_3$ over $\theta\in\left[\theta_2(s),\frac\pi4\right]$. That is, the probability of the randomly rotated and positioned smaller square containing the center of the unit square is
$$
\frac{a_3\!\left(s,\frac\pi4\right)-a_3(s,\theta_2(s))+a_2(s,\theta_2(s))-a_2(s,\theta_1(s))+a_1(s,\theta_1(s))-a_1(s,0)}
{a_3\!\left(s,\frac\pi4\right)-a_3\!\left(s,0\right)}\tag{12}
$$
In the question, $s=\frac1{\sqrt{10}}$, so we apply $(10)$ and get a probability of
$$
\bbox[5px,border:2px solid #C0A000]{\frac\pi{11\pi+2-8\sqrt{10}}\approx0.279022074}\tag{13}
$$
Here is a plot of the probability that a randomly rotated and positioned square of side $s$ inside the unit square contains the origin of the unit square:

Simulation

I ran a simulation placing the center of the smaller square, with side $s$, uniformly in a square, concentric with the unit square, and with side $1-s$, and rotating uniformly with $\theta\in\left[0,\frac\pi4\right]$.

All trials where the translated and rotated square did not fit completely in the unit square were rejected. The remaining trials were tallied counting those where the center of the unit square fell within the smaller square.

The results of $10,\!000,\!000$ trials with $s=\frac1{\sqrt{10}}$ were

non-rejected trials: $7,\!664,\!102$

successful trials: $2,\!136,\!729$

ratio: $0.278797046$

This is close to $0.279022074$.

Here is the Mathematica code for the simulation:

test[s_, x_, y_, \[Theta]_] :=
 Module[{c = {x - 1/2, y - 1/2} (1 - s),
   m = s {{Cos[\[Theta]], Sin[\[Theta]]}, {-Sin[\[Theta]], Cos[\[Theta]]}},
   u = 1/2 {{-1, 1}, {1, 1}, {1, -1}, {-1, -1}},
   cross = (#1.{{0, 1}, {-1, 0}}.#2) &, ul, ur, lr, ll},
   {ul, ur, lr, ll} = u.m + {c, c, c, c};
  If[Max[Abs[ul], Abs[ur], Abs[lr], Abs[ll]] > 1/2, {0, 0},
   If[Min[cross[ur, ul], cross[ul, ll], cross[ll, lr], cross[lr, ur]] < 0,
    {0, 1}, {1, 1}]]]

Module[{s = 1/Sqrt[10], n = 10000000, data = {0, 0}},
 Do[data += test[s, RandomReal[1], RandomReal[1], RandomReal[Pi/4]], n];
 data]


问:
Anomaly of frequency of last digits $1,3,7,9$ in consecutive primes
Consulting https://oeis.org/A032592 shows that there are $35$ quartets of consecutive primes ending in $1,3,7,9$, yet the probability of this is only $1$ in $256$ up to $23869$ which is prime number $2654$.

There should be just ten of them from $\frac{2654}{256}$, yet there are $3.5$ more than probability.

Will this anomaly continue for increasing primes?

One could also ask if the frequency of quartets containing in ANY order the last digits $1,3,7,9$ accords with the probability of $\frac{3}{32}$.

答:
You seem to be assuming that the last digit of two consecutive (but "randomly chosen") primes are independent. I see no reason to assume this, especially for as low primes as yours, where effects like twin primes (especially nice are the examples 11, 13 and 17, 19, and 101, 103 and 107, 109) are still relatively common, skewing the distribution toward even rather than uniformly random.

Another, or perhaps more general, effect is that as you reach 20 000, the average prime gap gets close to 10. Before that, having two consecutive primes with the same last digit is a bit rarer than pure random simply because the primes are too dense.


问:
Proof for the Ratio Test
I'd like to ask, if my below proof for the ratio test for the convergence of an infinite series is technically correct and rigorous.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Given a series $\sum_{n=1}^{\infty}a_n$ with $a_n \ne 0$, the Ratio Test states that if $(a_n)$ satisfies
\begin{align*}
    \lim \absval{\frac{a_{n+1}}{a_{n}}} = r < 1
\end{align*}
then the series converges absolutely.

(a) Let $r'$ satisfy $r < r' < 1$. Explain why there exists an $N$ such that $n \ge N$ implies that $\absval{a_{n+1}} \le \absval{a_n} r'$.

(b) Why does $\absval{a_N} \sum (r')^n$ converge?

(c) Now, show that $\sum \absval {a_n}$ converges, and conclude that $\sum a_n$ converges.

Proof.

(a) Since, $\lim (a_{n+1}/a_{n}) \to r$, given any $\epsilon > 0$, there exists $N \in \mathbf{N}$, such that :
\begin{align*}
    \frac{\absval{a_{n+1}}}{\absval{a_{n}}} < r + \epsilon
\end{align*}

If we set $r + \epsilon = r'$, we have the desired inequality,
\begin{align*}
    \absval{a_{n+1}} < r'\absval{a_n}
\end{align*}

for all $n \ge N$. Moreover, since the limit of the sequence of ratios is less than unity, the ratios themselves should also be less than unity (Using the fact that $a_n \le b_n \Longleftrightarrow \lim a_n \le \lim b_n$ ,the order limit theorem). Therefore, $\absval{a_{n+1}}/\absval{a_n} < 1$ for all $n \in \mathbf{N}$. Therefore, $r < r' < 1$.

(b) $\absval{a_N}\sum (r')^n$ is a geometric series. Since, the partial sums of this series, are monotone increasing, but bounded (by $a_{N}/(1-r')$), by the monotone convergence theorem, $\absval{a_N}\sum (r')^n$ is convergent.

(c) We have, $\absval{a_{n+1}} \le \absval{a_N}(r') \le \absval{a_N}(r')^n$ for all $n \ge N$. As $\absval{a_N}\sum (r')^n$ is convergent, by the Comparison test, $\sum \absval {a_n}$ is also convergent. By the Absolute value test, $\sum a_n$ is absolutely convergent.

答:
The ideas are right here. To make it clearer for (a) I would write: Let $r' \in (r,1)$, then there exists $\epsilon > 0$ such that $r + \epsilon = r'.$ Then continue from there as you have finding the $N$ you need for the $\epsilon$ given. In your answer it read as if you were only considering one value of $r'$ although of course this isn't what you mean. You also don't need to explain why $r < 1$ here since it's given in the question (although yes, that is right).

For (b) just write that since $0 < r' < 1$ it's a geometric series. You don't need the other parts (even though it's correct).

The ideas in (c) look right but it's a little unclear. Let $N$ be from $(a)$ and note that

$$\sum_{n=1}^\infty |a_{N+n}| \leq \sum_{n=1}^\infty |a_N|r'^{n}.$$

Then continue as you have to explain why this converges. Note that we needed to shift our sum along by $N,$ so we aren't dealing with exactly the same series, so a sentence or two to explain how we then deduce that $\sum a_n$ indeed converges.

Title: Proof for the Ratio Test


问:
Proving one function is greater than another
How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?

My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?

答: Related to one of the above solutions, we know that there are positive values to $h(x)=(x+1)^2-4qx$ (why?) and so if there were also non-positive values, there would be roots to $h(x)=0$. But $h(x)=x^2+(2-4q)x+1$ so $h(x)=0$ has real roots if and only if $(2-4q)^2-4\geq 0$, which means $(1-2q)^2\geq 1$.  But if $q\in(0,1)$, $1-2q\in(-1,1)$ and thus $(1-2q)^2<1$.</p>


问:
Solution to recursive sequence involving floor function?
I want to find an explicit formula for the following recursive sequence:
$$ E_n =
\begin{cases}
E_0 & n=0 \\
\lfloor k E_{n-1} \rfloor & n \ge 1
\end{cases}, \ n \in \mathbb{N}_0 \ , \ k \in \left( 0 , 1 \right) $$
Without the floor function, the solution looks like:
$$ E_n = E_0 k^n, $$
but I'm not sure how to approach this with the floor function involved...

答:
The exact behavior of the sequence $\langle E_n:n\in\Bbb N\rangle$ seems very hard to pin down, but the long-term behavior is fairly simple.

Suppose first that $E_0\ge 0$. Then $$E_{n+1}=\lfloor kE_n\rfloor\le kE_n<E_n$$ for each $n\in\Bbb N$, so $0\le E_n\le k^nE_0$ for $n\in\Bbb N$, so $\lim\limits_{n\to\infty}E_n=0$, and since $E_n$ is an integer for each $n\in\Bbb Z^+$, there is an $n_0\in\Bbb N$ such that $E_n=0$ for all $n\ge n_0$.

If $E_0<0$, however, matters are somewhat different. For instance:

If $E_0=-1$, for instance, $E_n=-1$ for all $n\in\Bbb N$.

If $E_0=-2$, there are two cases. If $\frac12<k<1$, then $E_n=-2$ for all $n\in\Bbb N$, but if $0<k\le\frac12$, then $E_n=-1$ for all $n\in\Bbb Z^+$.

More generally, let $m\in\Bbb Z^+$. If $E_0=-m$ and $\frac{m-1}m<k<1$, then $E_n=-m$ for all $n\in\Bbb N$. If $0<k\le\frac{m-1}m$, there are a positive integer $m'<m$ and an $n_0\in\Bbb Z^+$ such that $E_n=-m'$ for all $n\ge n_0$. And with a little more thought it’s clear that if $r\in\Bbb R^+$, and $E_0=-r$, then there are a positive integer $m$ and an $n_0\in\Bbb Z^+$ such that $m\le\lceil r\rceil$ and $E_n=-m$ for all $n\ge n_0$.

In short, the sequence $\langle E_n:n\in\Bbb N\rangle$ is always eventually constant; its limit is $0$ if $E_n\ge 0$ and a negative integer greater than or equal to $\lfloor E_0\rfloor$ if $E_0<0$.


问:
$\left \{ u,v \right \}$ is linearly independent iff $\left \{ u+v,u-v \right \}$ is linearly independent
I know there are some questions about this exercise. Neverthless, I proved it by another way and I want you to check if my proof is correct, please.

Proof.

First part. $\left \{ u,v \right \}$ is linearly independent $\Rightarrow$ $\left \{ u+v,u-v \right \}$ is linearly independent

Let be $\lambda_1,\lambda_2 \in F$ and $u,v \in V$. We know that $\left \{ u,v \right \}$ is linearly independent, so

\begin{align*}
\lambda_1 u+\lambda_2 v=0
\end{align*}

implies that $\lambda_1=\lambda_2=0=-\lambda_1=-\lambda_2$. So it satisfies this

\begin{align}
\lambda_1u+\lambda_2v=\lambda_1 u+ \lambda_1 v = \lambda_1 (u+v) \ \ \ \ \ \ \ \ \ \ \ \ (1)
\end{align}

By other side, we have that it satisfies

\begin{align}
\lambda_1u+\lambda_2v=\lambda_2u+(-\lambda_2)v=\lambda_2(u-v) \ \ \ \ \ \ \ \ \ \ \ \ (2)
\end{align}

With (1) and (2), we have

\begin{align}
\lambda_1 (u+v)+\lambda_2(u-v)&=\lambda_1u+\lambda_1v+\lambda_2u-\lambda_2v\\ &=\underbrace{\lambda_1u-\lambda_2v}_{=0 \text{ since, $u,v$ are l.i.}}+\underbrace{\lambda_1v+\lambda_2u}_{=0 \text{ since $u,v$ are l.i.}}\\ &= 0+0\\ &= 0
\end{align}

Therefore,

\begin{align}
\therefore \left \{ u+v,u-v \right \} \text{ is linearly independent}
\end{align}

Second part. $\left \{ u+v,u-v \right \}$ is linearly independent $\Rightarrow$ $\left \{ u,v \right \}$ is linearly independent

Since $\left \{ u+v,u-v \right \}$ is l.i., we know that

\begin{align}
\lambda_1(u+v)+\lambda_2(u-v)=0 \Longrightarrow \lambda_1=\lambda_2=0
\end{align}

This is the same:
\begin{align}
\lambda_1u+\lambda_1v+\lambda_2u-\lambda_2v=0 \Longrightarrow \lambda_1=\lambda_2=0\\
\end{align}

Also, this is the same:
\begin{align}
(\lambda_1+\lambda_2)u+(\lambda_1-\lambda_2)v=0 \Longrightarrow \lambda_1=\lambda_2=0\\
\end{align}

With $\lambda_1+\lambda_2=\gamma_1 \in F$

and $\lambda_1-\lambda_2=\gamma_2 \in F$, we have:

\begin{align}
\gamma_1u+\gamma_2v=0 \Longrightarrow \lambda_1=\lambda_2=0 \Longrightarrow \gamma_1=\gamma_2=0\\
\end{align}

\begin{align}
\therefore \left \{ u,v \right \} \text{ is linearly independent}
\end{align}

By the first part and the second part, the proof is complete. Am I correct? I would be really very grateful if you help me to verify that it is correct.

答:
Neither proof looks correct to me - there's a bit of trouble here with variables sharing a name, allowing you to write a proof that looks ok, but is wrong. As far as I can tell, your first proof shows:

If $u$ and $v$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1 u + \lambda_2 v = 0$ then $\lambda_1(u+v)+\lambda_2(u-v) = 0$.

But what you're supposed to show is:

If $u$ and $v$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1(u+v) + \lambda_2(u-v) =0$, then $\lambda_1=0$ and $\lambda_2=0$.

These statements are really different - and I think this proof need to be entirely rewritten. Let me show the outline of how you do this; you should start your proof by just introducing the relevant variables:

Let $u$ and $v$ be linearly independent and $\lambda_1$ and $\lambda_2$ be such that $\lambda_1(u+v)+\lambda_2(u-v)=0$. We wish to show that $\lambda_1=0$ and $\lambda_2=0$.

This introduction mirrors the form of the statement you need to prove. The only thing you're allowed to do in proving this is to assume for any coefficients $\alpha_1$ and $\alpha_2$ that if $\alpha_1 u + \alpha_2 v= 0$ then $\alpha_1=\alpha_2=0$ - note that these $\alpha$'s do not have anything to do with the $\lambda$'s - they are bound variables rather than specific values.

You would probably make use of the fact that
$$(\lambda_1+\lambda_2)u + (\lambda_1-\lambda_2)v = \lambda_1(u+v)=\lambda_2(u-v)=0$$
and linear independence of $u$ and $v$ to note that $\lambda_1+\lambda_2=0$ and $\lambda_1-\lambda_2=0$. Finishing from there should be easy*.

Your second proof establishes:

If $(u+v)$ and $(u-v)$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $(\lambda_1+\lambda_2)u+(\lambda_1-\lambda_2)v=0$ then $\lambda_1+\lambda_2=0$ and $\lambda_1-\lambda_2=0$.

This is closer, but not quite right either. You need to instead show:

If $(u+v)$ and $(u-v)$ are linearly independent and $\lambda_1$ and $\lambda_2$ are such that $\lambda_1u+\lambda_2v=0$ then $\lambda_1=0$ and $\lambda_2=0$.

which is more general, since it loses the assumption that the pair $(\lambda_1, \lambda_2)$ can be written as $(x+y,x-y)$ - although, of course, it turns out that any pair can be written in that form*.

Mirroring the correct statement, your proof could begin:

Let $u$ and $v$ be vectors so that $(u+v)$ and $(u-v)$ are linearly independent. Suppose that $\lambda_1 u + \lambda_2 v = 0$. We wish to show that $\lambda_1=0$ and $\lambda_2=0$.

And, again, your strategy would need to be about rewriting $\lambda_1 u +\lambda_2 v$ as some linear combination of $(u+v)$ and $(u-v)$ and equating the coefficients of that sum to zero - and then doing more algebra to get to your conclusion.

Your second proof seems a bit closer to complete - basically, your $\gamma$'s take the place of what I'm using $\lambda$'s for - but your proof fails because you define the $\gamma$'s to be equal to something whereas the coefficients of $u$ and $v$ are the things that are given to you (i.e. you're proving a "for all" statement about them) - hence you may not exercise any influence over their values. The solution, in your notation, is just to write that $\lambda$'s in terms of the $\gamma$'s.

(*You may need to be careful about the base field, if you're doing this generally - things don't work out right in fields of characteristic $2$, where the statement is false. If this note is confusing, you don't need to worry about it)


问:
Proving $(AB)^T=B^TA^T$: Can you find the flaw in the reasoning?
Proving $(AB)^T=B^TA^T$ isn't difficult. An example of such a proof can be found here: http://www.math.ucdenver.edu/~esulliva/LinearAlgebra/ABT.pdf

However, if I take a slightly different approach, it doesn't work:

If  $B=(b_{ij})$, then $B^T=(b_{ji})$, and similarly, $A^T=(a_{ji})$.   Thus $B^TA^T=(b_{ji})(a_{ji}) = (\sum_k b_{jk}*a_{ki}).$ But the $(i,j) $ element of $(AB)^T$ is $(\sum_k b_{ki}*a_{jk})=(\sum_k a_{jk}*b_{ki})$, which is generally not equal to $(\sum_k b_{jk}*a_{ki})$.

What is the flaw in the reasoning? Any help is appreciated.

答: When you write $A^\top = (a_{ji})$, you have to remember that you're writing down the $ij$-entry. So, yes, the $ij$-entry of $B^\top A^\top$ is $\sum (B^\top)_{ik}(A^\top)_{kj} = \sum b_{ki}a_{jk}$. You just wrote symbols and didn't think about actual entries of actual matrices.


问:
Power series $S(x):=\sum\limits_{n=1}^{\infty}\frac{x^n}{{2n\choose n}}$ convergence on end points
Let's consider the real valued power series: $S_k(x):=\sum\limits_{n=1}^{k}\frac{x^n}{{2n\choose n}}$.

I found out that the radius of convergence is $4$ or in other words $S_k(x)$ converges absolutely for each $x\in(-4,4)$. However, I am not sure if I showed correctly what happens on the end points $x=-4$ and $x=4$, respectively.

With regard to $S_k(4)$ I tried to fiddle around with the binomial theorem:

$4^n=2^{2n}=(1+1)^{2n}=\sum\limits_{i=0}^{2n}{2n\choose i}\geq {2n\choose n}$, as ${2n\choose n}$ is always a summand of $\sum\limits_{i=0}^{2n}{2n\choose k}$. Hence, $\sum\limits_{n=1}^{k}\frac{4^n}{{2n\choose n}}\geq \sum\limits_{n=1}^{k}1=k\implies \lim\limits_{k\to\infty}\sum\limits_{n=1}^{k}\frac{4^n}{{2n\choose n}}=\infty$.

Plugging in $x=-4$, we notice that $S_k(-4)$ can't be a Cauchy sequence (which is equivalent to not being convergent in $\mathbb{R}$) because using the aboe result we get $|S_{2k+1}-S_{2k}|=\frac{4^{2k+1}}{{4k+2\choose 2k+1}}\geq1$ for all $k\in\mathbb{N}$.

Is this correct or is there a more elegant way?

答:
In general we have, that if $\sum_{i\geq 0} a_i$ converges, then $a_i$ converges to zero. Hence, if you can show that the summands are not a zero sequence, the series cannot converge. The estimate $\vert 4^n/ \binom{2n}{n} \vert \geq 1$ already does the trick.

Well, in a way that is the same as you did (you just proved it from scratch).