问:
Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always?
My question is summarised as follows:

Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$
or can we only deduce values of limits using differentiation in the forward direction, ie we can't deduce that the value of $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ is the same as $\lim_{x\to a}\frac{f(x)}{g(x)} $?

Logically I cannot see why this shouldn't be true (ie I can't see why we can't also integrate the numerator and denominator). HOWEVER, the following result seems to throw some doubt on this:

Consider the identity
$$f(x)^2\frac{d}{dx}\frac{g(x)}{f(x)}+g(x)^2\frac{d}{dx}\frac{f(x)}{g(x)}\equiv0$$
$$\implies\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv-\frac{f(x)^2}{g(x)^2}$$
On applying L'Hopital's rule we obtain
$$\lim_{x\to a}\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}\implies\lim_{x\to a}\frac{\frac{f(x)}{g(x)}}{\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
$$\implies\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
which is obviously false.

Any help explaining this apparent proof would be appreciated.

答:
Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$

I can't see why we can't also integrate the numerator and denominator

As other answers have already noted, these limits are not always equal (even when they exist).

Let’s try to find the simplest possible explanation here. We start by noting that differentiation produces only one answer, but integration (in the sense of antidifferentiation) produces infinitely many answers, expressed using a “constant of integration”. Changing this constant does not affect the first limit in the question, but we expect that it does affect the second.

Let $f_1$ and $f_2$ be functions where $f_2(x) = f_1(x) + c$ for some constant $c \neq 0$. Then, assuming that the relevant limits exist and are finite,

\begin{align*}
\lim_{x \to a} \frac{f_2(x)}{g(x)} &= \lim_{x \to a} \frac{f_1(x) + c}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + \lim_{x \to a} \frac{c}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + c \cdot \lim_{x \to a} \frac{1}{g(x)} \\
&= \lim_{x \to a} \frac{f_1(x)}{g(x)} + c \cdot \frac{1}{\lim_{x \to a} g(x)} \\
&\neq \lim_{x \to a} \frac{f_1(x)}{g(x)}
\end{align*}

So we were right: the second limit changes (in at least some cases). Therefore, the limits are not always equal.


问:
Application of Cauchy-Schwarz on integrals
The application of the Cauchy-Schwarz inequality is of the following form:
\begin{align}
\bigg|\int_{\mathbb{R}_+}G_\epsilon(x,y)f(y)dy\bigg|^2\leq \int_{\mathbb{R}_+}G_\epsilon(x,y)dy\cdot\int_{\mathbb{R}_+}G_\epsilon(x,y)f(y)^2dy,
\end{align}
where $G_\epsilon(x,y)=p_\epsilon(x-y)-p_\epsilon(x+y)$ with $p_\epsilon(z)=\frac{1}{\sqrt{2\pi\epsilon}}\exp\{-\frac{z^2}{2\epsilon}\}$.

This is not the original form of Cauchy-Schwarz, obviously. I just don't get how to arrive at the above inequation.

答: Assuming that $x \in \mathbb R^{+}$ observe that $G_{\epsilon} \geq 0$. Define $\mu (E)=\frac {\int_E G_{\epsilon} (x,y) dy} {\int_{\mathbb R^{+}} G_{\epsilon} (x,y)dy} $. then $\mu $ is a probablity measure and hence $|\int f(y)d\mu(y)| \leq \int |f(y)|^{2} d\mu (y)$.

ContentLicense: CC BY-SA 4.0


问:
A man throws 4 dice. What is the probability that same number shows up in at least two faces?
I saw a question  in internet , it seems very trivial.I tried to solve it in two different way but i obtained two different result.

I know that my second solution is correct but why is the former wrong ? What am i missing ?

Question: A man throws $4$ dice. What is the probability that same number shows up in at least two faces?

First Solution:

The probability of getting ONLY $2 $ dice showing the same face : $\frac{C(4,2).6.5.4}{6.6.6.6}=\frac{120}{216}$

The probability of getting ONLY $3 $ dice showing the same face : $\frac{C(4,3).6.5}{6.6.6.6}=\frac{20}{216}$

The probability of getting ONLY $4 $ dice showing the same face : $\frac{C(4,4).6}{6.6.6.6}=\frac{1}{216}$

$\therefore \frac{141}{216}=0.652$

Second Solution: All situations - all dice appears different faces

$1- \frac{6.5.4.3}{6.6.6.6}=0.722$

What am i missing ?

答:
It's worth noting another approach to this solution that imagines each die being rolled individually and looks at the odds of the complimentary event that there are no matching dice.

The first rolled die can be anything, doesn't matter. Imagine its a 1 if it helps.

The second rolled die has a $\frac{1}{6}$ chance of matching the first (which we're not interested in), and a $\frac{5}{6}$ chance of differing from the first.

Given that the first two dice were different, the third rolled die has a $\frac{1}{3}$ chance of matching one of the first two, and a $\frac{2}{3}$ chance of differing from the first two.

Finally, given that we have no matches by the fourth die, there's a $\frac{1}{2}$ chance that the fourth die matches, otherwise, there are no matches. Given that this is the only way to not get a match, we can compute the probability of not having a match as $\frac{5}{6} \cdot \frac{2}{3} \cdot \frac{1}{2}$. This simplifies to $\frac{5}{18}$, and the compliment $\frac{13}{18} \simeq 0.7222$ agrees with the other approaches.


问:
Proving linear independence of intersecting planes.
If two planes intersect at a line passing through the origin and we are given two sets of vectors $\{\textbf{d}, \textbf{v}_i\}$ where $\textbf{d}$ is the direction vector of the line and the sets are the bases for the two planes, how do I show that the sum of the two planes forms $\mathbb{R}^3$?

答:
The only way $\text{span}\{v_i\}$ is 2-dimensional is for the basis vectors of one plane to be both linearly dependent to the basis of the first plane -- which makes the two plane the same plane.

Your question is thus trivially true.

P.S. Next time, please show what you have done. Especially so since the question seems to be asking for granularity instead of general ideas.


问:
Guessing pattern of Picard's iteration of ODE
Given the following initial value problem $x'(t) = x(t) + e^t$ with $x(0) = 0$. I've calculated its picard's iteration, but having trouble guessing the pattern: $$\begin{cases}
      x_1 = e^t - 1,  \\[3pt]
      x_2 = 2e^t - t - 2,  \\[3pt]
      x_3 = 3e^t - 2t - 3 - \frac{t^2}{2},  \\[3pt]
      x_4 = 4e^t - 3t - 4 - t^2 - \frac{t^3}{6},  \\[3pt]
      x_5 = 5e^t - 4t - 5 - \frac{3t^2}{4} - \frac{t^3}{3} -  \frac{t^4}{24}, \\[3pt]
      x_6 = 6e^t - 5t - 6 - 2t^2 - \frac{t^3}{2} -  \frac{t^4}{12} - \frac{t^5}{120},...
   \end{cases} $$

So maybe we have something like $$x_n(t) = n(e^t - 1) - (n-1)t - \sum_{k = 2}^{n-1} \frac{t^k}{??}$$
Since the solution is actually $x(t) = t e^t$, I also don't see how the $n$ in the first two elements should disappear as $n \to \infty$, because otherwise, it diverges.

Any hint/help would be appreciated. Thanks.

答:
If you expand with Maclaurin $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+O\left(t^6\right)$$
and plug this in the, let's say, $6$th step, you get
$$6 \left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}\right)-\frac{t^5}{120}-\frac{t^4}{12}-\frac{t^3}{2}-2 t^2-5 t-6+O\left(t^6\right)=\\=t+t^2+\frac{t^3}{2}+\frac{t^4}{6}+\frac{t^5}{24}+O\left(t^6\right)$$
the last one is the expansion of $te^t$


问:
Why $p^i/p^{i+1}$ is a vector space over $O_k/p$
Let $O_K$ be a number ring of number field $K$ and $p$ be a prime ideal of $O_K$. Consider the chain
$$p\subseteq p^2\subseteq ...\subseteq p^v$$
If we choose $a\in p^i\setminus p^{i+1}$, and $b=(a)+p^{i+1}$, so, $p^{i+1}\subset b\subseteq p^i$ so, $b=p^i$.

So, $p^i/p^{i+1}$ is generated by a single element, $a\;mod\;p^{i+1}$. But why it generate $p^i/p^{i+1}$ over $O_K/p$ is what I am not getting.

This is a specific proof form Newurich, page-35, prop-6.1.

答:
For $R$ a commutative ring and $I$ an ideal then  $I^n/I^{n+1}$ is a $R/I$-module: it is clearly a $R$-module and for $b\in I$ we have $b . I^n/I^{n+1}=0$.

Then for $K$ a number field and $P$ a maximal ideal of $O_K$ the point is that $P$ becomes principal $=(\pi)$ in $O_K/P^{n+1}$, so does $P^n=(\pi^n)$,
whence $b\in O_K\to \pi^n b \in P^n/P^{n+1}$ is surjective and $b\to \pi^n b$ is a $O_K/P$-module ($O_K/P$-vector space) isomorphism $O_K/P\to P^n/P^{n+1}$.

Compare with $R=\Bbb{Q}[x,y]$, the maximal ideal $m=(x,y)$ is not invertible, it is not principal in $R/m^{2}$ and $m/m^2 = \Bbb{Q}x +\Bbb{Q}y$ is a $2$-dimensional $\Bbb{Q}=R/m$-vector space.


答: Each plane is a 2-dimensional vector subspace of $\mathbb{R}^3$. There exists a vector $v$ in the second plane $P_2$, which doesn't belong to the first plane $P_1$, since we know they intersect only at a line. Thus, the subspace $P_1 + \operatorname{span} v$ is 3-dimensional. Since the whole vector vector space $\mathbb{R}^3$ is also 3-dimensional, it must be the case that $P_1 + \operatorname{span} v = \mathbb{R}^3$. ∎

Title: Proving linear independence of intersecting planes.

Body: If two planes intersect at a line passing through the origin and we are given two sets of vectors $\{\textbf{d}, \textbf{v}_i\}$ where $\textbf{d}$ is the direction vector of the line and the sets are the bases for the two planes, how do I show that the sum of the two planes forms $\mathbb{R}^3$?


问:
Which characterization of continuity in a point is used here?
I am currently studying a proof, where we have a function $f: A\to Y$ where $A\subseteq X$ is a topological space, and $Y$ is a metric space.

We want to show that $f$ is continuous in a point $x$, and the proof goes along to show that for a specific open neighborhood of $f(x)$ the preimage contains an open neighborhood of $x$.

For $\varepsilon >0$ we show that $f^{-1}(B_\varepsilon(f(x)))$ contains an open neighborhood of $x$.
Which characterization of continuity of a point is then used here?

Do we not have to show that for EVERY open neighborhood $V$ of $f(x)$, the preimage contains an open neighborhood $U$ of $x$.

A function $f:X\to Y$ of topological spaces is continuous in $x$ if the preimage of neighborhoods of $f(x)$ or neighborhoods of $x$.

But in the given proof it looks like that for a general type of open neighborhoods this is shown.
So to general in my opinion.

Or is this just an equivalent characterization that I am missing here?

Thanks in advance.

答: Suppose that you have indeed proved that, for every $\varepsilon>0$, $f^{-1}\bigl(B_\varepsilon\bigl(f(x)\bigr)\bigr)$ is a neighborhood of $x$. Now, let $V$ be any neighborhood of $f(x)$. Then, there is some $\varepsilon>0$ such that $B_\varepsilon\bigl(f(x)\bigr)\subset V$ and therefore $f^{-1}\bigl(B_\varepsilon\bigl(f(x)\bigr)\bigr)\subset f^{-1}(V)$. Since $f^{-1}\bigl(B_\varepsilon\bigl(f(x)\bigr)\bigr)$ is a neighborhood of $x$, then so is $f^{-1}(V)$.


问:
Show that $n | \sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$
I must show that $n$ divides $\sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$.

When $n$ is prime, Fermat's theorem gives us the answer easily because then $\sum_{k=1}^n m^{\textrm{gcd}(k, n)} = m^n + (n-1)m \equiv 0  [n]$.

In the general case, how to show this? I would like to use mostly integers/arithmetics arguments, not too high level theories.

答:
Consider the set $$\mathcal{F} = \left( \mathbb{Z}/m\mathbb{Z}\right)^{\mathbb{Z}/n\mathbb{Z}}$$

of all functions $\varphi : \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$ (not group morphisms, just functions). Then you can define a group action of $\mathbb{Z}/n\mathbb{Z}$ over $\mathcal{F}$ by
$$\begin{array}{}
 & \mathbb{Z}/n\mathbb{Z} \times\mathcal{F} & \longrightarrow & \mathcal{F} \\
    & \quad \quad( k,\varphi)  & \longmapsto & (x \mapsto \varphi(x+k)) \end{array}$$

Let's count the number of fixed points of an element $k \in \mathbb{Z}/n\mathbb{Z} $ for this action. A function $\varphi$ is fixed by $k$ if $\varphi$ is constant on the orbits of the natural (additive) action of $k$ on $\mathbb{Z}/n\mathbb{Z}$ : by a classical fact, this action has exactly $\gcd(k,n)$ orbits, so you have exactly $m^{\gcd(k,n)}$ functions from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$ that are constant on these orbits, i.e. $k$ has exactly $m^{\gcd(k,n)}$ fixed points for the action of $\mathbb{Z}/n\mathbb{Z}$ on $\mathcal{F}$.

You can then apply Burnside's formula to get that
$$\mathrm{Card}(\mathcal{F}/(\mathbb{Z}/n\mathbb{Z})) = \frac{1}{n} \sum_{k=1}^n m^{\gcd(k,n)}$$

so, since the LHS is an integer, $n$ must divide $\sum_{k=1}^n m^{\gcd(k,n)}$.


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

答:
If you  believe in the Monodromy theorem and are comfortable with the Riemann sphere $\overline{\mathbb C}$, the following short argument suffices:
$$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = z\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$$
Here $\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$ admits analytic continuation in $\overline{\mathbb C}\setminus [1,3]$, which is a simply-connected domain. Therefore, it has a single-valued branch there.

If you are not comfortable with the Riemann sphere, but still  believe in the Monodromy theorem, then move the segment to infinity, for example by letting $w=1/(z-1)$. This transforms $\overline{\mathbb C}\setminus [1,3]$ into $ {\mathbb C}\setminus [1/2,\infty)$. We get
$$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = \Big(  w^{-1}(w^{-1}-1) (w^{-1}-2) \Big)^{1/3}
=w^{-1} ( (1-w)(1-2w) )^{1/3}
$$
Here $( (1-w)(1-2w) )^{1/3}$  admits analytic continuation  in $ {\mathbb C}\setminus [1/2,\infty)$, which is a simply-connected domain. Therefore, it has a single-valued branch there.

ContentLicense: CC BY-SA 3.0


问:
If $f$ is a bounded convex function on $(a,b)$, then $f$ is continuous on $(a,b)$
In the proof of this statement, we need to prove that $f$ is continuous at an arbitrary $x_0\in(a,b)$. My professor has "shifted the interval $(a,b)$" so that we can set $x_0=0$. We then fix $c>0$ such that both $c, -c\in(a,b)$.

From the definition of convexity, we write $$\theta c=\theta c+(1-\theta)0$$ so we have $$f(\theta c)\le\theta f(c)+(1-\theta)f(0).~~~~(1)$$

My professor then claims that if we take $\limsup$ as $\theta\to0$, we get $$\limsup_{x\to0^+}f(x)\le f(0).~~~~(2)$$

Similarly, from writing $$0=\theta(-c)+(1-\theta)\epsilon,$$ where $$\epsilon=\theta\frac{c}{1-\theta}.$$
Then from convexity we have $$f(0)\le \theta f(-c)+(1-\theta)f(\epsilon).~~~~(3)$$
Taking $\liminf$ as $\theta\to0$ we get $$f(0)\le \liminf_{x\to0^+}f(x).~~~~(4)$$

I understand how to use $(2)$ and $(4)$ to conclude that $f(0)=\lim_{x\to0^+}f(x)$ and similarly for $f(0)=\lim_{x\to0^-}f(x)$, but I don't understand how we go from $(1)$ to $(2)$ and from $(3)$ to $(4)$. What does $\limsup$ do for us in $(2)$ that $\lim$ doesn't, because to me, the right hand side of $(2)$ looks precisely like $$\lim_{\theta\to0^+}(\theta f(c)+(1-\theta)f(0)).$$
So can someone explain what it means to be taking $\limsup$ of $(1)$ and $\liminf$ of and $(3)$?

答:
If $\epsilon >0$ the  $\theta f(c)+(1-\theta)f(0) <f(0)+\epsilon$ provided $|\theta| <\frac {\epsilon} {|f(c)+|f(0)|}$. Hence, for any such $\theta$ we get $ f(\theta c)\leq f(0)+\epsilon$. Replacing $\theta$ by $\frac {\theta } c$ this says $f(\theta) <f(0)+\epsilon$ whenever $|\theta| < \frac {\epsilon |c|} {|f(c)+|f(0)|}$. Can use definition  of $\lim \sup$ now to conclude that $\lim \sup_{\theta \to 0} f(\theta) \leq f(0)$?

[(3) implies (4) is proved in a similar way].


问:
Action of $\operatorname{Aut}(A) \times \operatorname{Aut}(G)$ on $\operatorname{Hom}(G,\operatorname{Aut}(A))$
Let $G$ and $A$ be two finite abelian groups. Let $T:= \operatorname{Hom}(G,\operatorname{Aut}(A))$ be the set of all injective group homomorphisms from $G$ to $\operatorname{Aut}(A)$.

Take $g := (\sigma,\rho)\in \operatorname{Aut}(A) \times \operatorname{Aut}(G)$ and let $$T^g := \{\alpha \in T :\alpha \circ \rho = \gamma_{\sigma} \circ \alpha \}.$$
where $\gamma_{\sigma}$ denotes the conjugation by $\sigma$ in $Aut(A)$.
Are there some conditions on $G$ and $A$, or examples, under which we can compute easily the cardinality of $T^g$?

Thank you very much.

答:
Just some very partial work, that I couldn't accomodate into a comment. So:

\begin{alignat}{1}
T^{(\sigma,\rho)} &:=\{\alpha\in T\mid \alpha(\rho(g))=\sigma^{-1}\alpha(g)\sigma, \space\forall g\in G\} \\
&=\{\alpha\in T\mid \sigma\alpha(\rho(g))=\alpha(g)\sigma, \space\forall g\in G\} \\
\tag 1
\end{alignat}

As special cases:

$(\sigma,\rho)=(\iota_A,\iota_G)$

\begin{alignat}{1}
T^{(\iota_A,\iota_G)} &=\{\alpha\in T\mid \alpha(g)=\alpha(g), \space\forall g\in G\} \\
&= T
\tag 2
\end{alignat}

$(\sigma,\rho)=(\iota_A,\rho\ne\iota_G)$

\begin{alignat}{1}
T^{(\iota_A,\rho)} &=\{\alpha\in T\mid \alpha(\rho(g))=\alpha(g), \space\forall g\in G\} \\
\tag 3
\end{alignat}

$(\sigma,\rho)=(\sigma\ne\iota_A,\iota_G)$

\begin{alignat}{1}
T^{(\sigma,\iota_G)} &=\{\alpha\in T\mid \sigma\alpha(g)=\alpha(g)\sigma, \space\forall g\in G\} \\
&=\{\alpha\in T\mid \alpha(g)\in C_{\operatorname{Aut}(A)}(\sigma), \space\forall g\in G\} \\
&=\{\alpha\in T\mid \alpha(G)\le C_{\operatorname{Aut}(A)}(\sigma)\} \\
\tag 4
\end{alignat}


问:
How do I find this root?
How can the third root of $\sqrt{x} - 2x + x^a$ be expressed in terms of a when $a>1.5$? (Obviously there is always one at 1 and 0)

答:
Beside the trivial $x=0$ and $x=1$, the third solution of the equation
$$\sqrt{x} - 2x + x^a=0$$ is close to $\frac 14$ as soon as $a >3$.

So, we can have approximations using Taylor series. For example
$$0=4^{-a}+\left(4^{1-a} a-1\right) \left(x-\frac{1}{4}\right)-2^{-2 a} \left(-8 a^2+8
   a+4^{a}\right)
   \left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)$$ which is a quadratic in $\left(x-\frac{1}{4}\right)$.

Using only the first term (this would be equivalent to the first iterate of Newton method
$$x_1=\frac{1}{4}+\frac{1}{4^a-4 a}$$ is quite decent. For $a=3$, it would give $x=\frac 7{26}=0.269231$ while the "exact" solution is $0.269143$.

But we can continue with series reversion and have
$$x_2=\frac 14+\frac{1}{4^a-4 a}+\frac{8 (a-1) a-4^a}{\left(4^a-4 a\right)^3}+\cdots$$

For $a=3$, this gives $x=\frac{2365}{8788}=0.269117$

Another one which is not bad
$$x_3=\frac 14+\frac{4^a-4 a}{8 a \left(a-4^a+1\right)+4^a+4^{2a}}$$

For the case where $a$ is close to $1.5^+$, the third solution is close to $1$ but, using the same method, it can be approximated as
$$x=1-\frac{4(2a-3)}{4 (a-1) a-1}-\frac{8 (2 a-3)^3 \left(4 a^2-6 a-1\right)}{3 (4 (a-1) a-1)^3}-$$
$$\frac{4 (2 a-3)^3(320 a^6-1728 a^5+3440 a^4-2784 a^3+452 a^2+300 a+27 )}{9 (4 (a-1) a-1)^5}$$

For example, for $a=2$, the above formula gives $x=\frac{6639}{16807}=0.395014$ while the exact solution is $x=\frac{3-\sqrt{5}}{2} =0.381966$.

Title: How do I find this root?


问:
Proving interiorality and exteriorality of curves using the Jordan curve lemma

How do I formally go about proving (i), (ii) and (iii)?
Can I get a formal proof for all three parts(I am new to topology). This is what I have for part(i) - γ1 divides C into interior, exterior and boundary of γ1.Since the image of γ2 is disjoint from γ1, γ2 should lie in the interior of γ1 or its exterior. Is this adequate?

For part (ii) I have sketch of proof - its not formal and it is derived from pictorial intuition. Again I would appreciate a formal proof. This is what I have: γ2∩ext(γ2)=∅. C is divided into int, ext and boundary of γ1, therefore since γ1 and γ2 have disjoint boundaries, int(γ1⊆int(γ2) and ext(γ1)⊃ext(γ2)

For part(iii) I do not have even geometric intuition

Can I please get some help?

答:
We will need the following facts: 0) image of a Jordan curve $\gamma$ is compact, hence closed, and $int(\gamma)$ and $ext(\gamma)$ are open; 1) for open sets, connected is the same as path connected; 2) for a Jourdan curve $\gamma$ any two points joined by a path $\beta$ disjoint from $\gamma$ are either both in $int(\gamma)$ or are both in $ext(\gamma)$ (the intersection of $\beta$ with either of these is clopen in $\beta$, so is either empty or the  whole thing).

Thus, by 1), $ext(\gamma)$ is precisely the set of points in the plane connected by a path "to infinity" (i.e. there exists a large ball such that for any point $P$ in a complement of this ball, $ext(\gamma)$ is precisely the set of points  $p$ such that there exists a path from  $p$ to $P$ avoiding $\gamma$).

Now first part of the exercise is obvious: any two points of $\gamma_2$ are connected by a path avoiding $\gamma_1$, namely a piece of $\gamma_2$, so by property 2) lie either both in $int(\gamma_1)$ or both in $ext(\gamma_1)$.

For the second part of the exercise, assume that $p$ is in the exterior of $\gamma_1$. Then it can be connected to infinity by a path $\beta$ in exterior of $\gamma_1$ (i.e. to a point $P$ sufficiently far to be in exterior of both $\gamma_1$ and $\gamma_2$). Since $\gamma_2$ avoids the exterior of $\gamma_1$, $\beta$ is either fully in interior or fully in exterior of $\gamma_2$; but it has endpoint $P$ in the exterior, so it is fully in the exterior. Thus $p$ is also in the  exterior of $\gamma_2$, and $ext(\gamma_1)\subseteq ext(\gamma_2)$, as wanted. Now for the other part, suppose $q$ is in the interior of $\gamma_2$. We want to show its in the interior of $\gamma_1$. Well, $q$ has a ball around it all in the interior of $\gamma_2$, thus disjoint from the exterior of $\gamma_2$, and  by what we have just proved this ball is disjoint from the exterior of $\gamma_1$. Since $\gamma_1$ is the boundary of its exterior, this is only possible for points on the interior of $\gamma_1$. So $int(\gamma_2)\subseteq\int(\gamma_1)$, as wanted.

For the third part, we first prove that for any Jordan curve $\gamma$ and any point $p$ in $int(\gamma)$ there exists a path $\beta$ from $p$ to a point $b \in \gamma$ which avoids $ext(\gamma)$. Indeed, pick $q\in ext(\gamma)$ and connect $p$ to $q$ by a path (say, a straight segment $
\beta(t)=tp+(1-t)q$), and take $b$ to be the first time this path hits $\gamma$ (i.e. $b=\beta(T)$ where $T=\inf \{t| \beta_t \notin int(\gamma)\}$). Then the part of this path up to that point is as we want ($b=\beta(T)$ is in closure of $int(\gamma)$ and in closure of $ext(\gamma)\cap \gamma$, so is in $\gamma$, and all $t<T$ have $\beta(t)\in int(\gamma)$).
Now the exercise is easy: given $p\in int(\gamma_1)$ connect it to $b\in \gamma_1$ by a path in $int(\gamma_1)$, and thus avoiding $\gamma_2$. Now $b\in \gamma_1 \subset ext(\gamma_2)$, so by property 2, $p$ is also in $ext(\gamma_2)$.


问:
Existence and uniqueness of homoclinic orbit

Consider the following 2D autonomous system of ODEs:
$$
\left\{
\begin{array}{ll}
\dot{x} = x^2 + 2y - x \\
\dot{y} = 3xy/2 - 3x^2 - y + 2x
\end{array}
\right.
$$

How can we prove the existence and uniqueness of a homoclinic orbit (i.e. a solution $X$ of the system
for which $\lim_{t \to \infty} X(t) = \lim_{t \to -\infty} X(t) = x_0,$ where $x_0$ is an equilibrium point of the system) for the system?

It is not hard to determine the equilibrium points: $(0,0), (-3, -6)$ and $(2/3, 1/9)$. The latter two are asymptotically stable equilibrium points, so there cannot be a homoclinic orbit for those points.

However, $(0,0)$ is a saddle equilibrium, so perhaps there is a homoclinic orbit here.
We can also see that the function $H(x,y) = - y^2 + x^2(1-x)$ is constant on the solutions of the system.
The set $H(x,y) = 0$ seems to be comprised of three solutions of the system, one of which is indeed a homoclinic orbit, so this proves existence.

However, how do we prove that this is unique? I don't really know how to approach this part.

答:
Actually, proving uniqueness of the homoclinic orbit is much easier than expected, but it requires some heavy machinery.

The Hartman-Grobman theorem says that, near a hyperbolic equilibrium point, any system of ODEs is topologically conjugate to its linearization.

In the linear case, for a saddle equilibrium point, there are $2$ trajectories going out of the equilibrium and two trajectories coming in. So, this is the behaviour for a general system as well.

Using you function $H$, you already determined your trajectories, so the above discussion tells you that there is only one homoclinic orbit.


问:
a problem on nonconstant holomorphic function has a zero or no in the closed unit disk
Let $f:D \to \mathbb{C}$ be a non-constant holomorphic function ($D$ is the closed unit disk) such that $|f(z)|=1$ for all $z$ satisfying $|z|=1$ . Then prove that there exist $z_0 \in D$ such that $f(z_0)=0$

My thought:-

By Maximum Modulus Theorem $|f(z)|$ has Maximum value on the curve which is $1$.

By Minimum Modulus Theorem if $f(z)\ne0$ for all $z\in D$, then it has its minimum value on the boundary which is $1$.

Then $|f(z)|=1$ for all $z\in D$.

Hence $|f(z)|$ is constant which is a contradiction.

Is my thinking correct?

答: By Maximum Modulus Theorem $|f(z)|$ has Maximum value on the curve which is $1$ . by Minimum Modulus Theorem if $f(z)\ne0$ for all $z\in D$ Then it has its minimum value on the boundary which is $1$.Then $|f(z)|=1$ for all $z\in D$.Then $|f(z)|$ is constant. Hence $f(z)$ becomes constant which gives a contradiction. </p>


问:
discrete math binary relations - why is set transitive
The question is:

Let $R$, $S$ be partial orders defined on $A$. Is $R \cap S$ a partial order too? Justify.

The school answer given states that $R$ and $S$ are transitive since if $(x,y), (y,z) \in R \cap S$, then $(x,y), (y,z) \in R$  and $(x,y), (y,z) \in S$. Hence, $(x,z) \in R ∩ S$.

In my answer, I gave an example on transitivity seen in $R$ and $S$ respectively, but transitivity not observed in $R ∩ S$. Why is it wrong?

答:
The only pairs $(a,b)$ and $(b,c)$ that would be useful to refute transitivity of $R\cap S$ would be those where $(a,b),(b,c)\in R\cap S$, but $(a,c)\not\in R\cap S$.

In your example, $(2,4)\in R$ but $(2,4)\not\in S$ so $(2,4)\not\in R\cap S$. Similarly, $(4,5)\in S$ but $(4,5)\not\in R$, therefore $(4,5)\not\in R\cap S$. This means those two pairs are not useful in determining (non)-transitivity. And, actually $R\cap S$ is transitive because $R\cap S=\{(3,3),(4,4)\}$.

I am also completely at loss about why you claim $R$ and $S$ are transitive in the first place. $(2,4), (4,3)\in R$ but $(2,3)\not\in R$, and, similarly, $(3,4),(4,5)\in S$ but $(3,5)\not\in S$.


问:
Prove that the normals at A, B, C incribed in $xy=c^2$ will meet in a point if $cot2\alpha + cot2\beta + cot 2\gamma = 0$
The sides of a triangle ABC, inscribed in a hyperbola $xy = c^2$, makes angles $\alpha, \beta, \gamma $ with an asymptote.

Prove that the normals at A, B, C will meet in a point if $cot2\alpha + cot2\beta + cot 2\gamma = 0$

My approach: Assuming $A \left(ct_1,\frac{c}{t_1}\right), B \left(ct_2,\frac{c}{t_2}\right) and C \left(ct_3,\frac{c}{t_3}\right)$, we can work out $t_1t_2=-cot\alpha$ and so on

Equation of a normal to rectangular hyperbola is $t^3x-ty=ct^4-c$ which is fourth degree equation which doesn't help the cause. Using determinants for concurrency doesn't yield any fruitful results.

答:
I worked out the answer myself and since I couldn't get the solution elsewhere online, I have decided to post it

We have already worked out $t_1t_2=-cot\alpha, t_2t_3=-cot\beta,$ and $t_3t_1=-cot\gamma.$

Also, equation of normal to hyperbola is $ct^4-xt^3+ty-c=0$

From the equation we get $S_4=t_1t_2t_3t_4=-1.$ We can write $t_4=\frac{-1}{t_1t_2t_3}$

Also, $S_2=t_1t_2+t_2t_3+t_3t_1+t_4(t_1+t_2+t_3)=0$

We can write, $S_2=t_1t_2+t_2t_3+t_3t_1+\frac{-1}{t_1t_2t_3}(t_1+t_2+t_3)=0$ which is $\sum \left( t_1t_2-\frac{1}{t_1t_2} \right)=0$

$\sum \left( cot\alpha-tan\alpha \right)=0 \Rightarrow \sum cot2\alpha=0$


问:
continuous functions inequality
Let f:R$\rightarrow$R be a continuous function such that rfor any two two real numbers x and y,$$|f(x)-f(y)|\leq7|x-y|^{201}$$.Then
which one is true $$(A)f(101)=f(202)+8$$.$$(B)f(101)=f(201)+1$$.$$(C)f(101)=f(200)+2$$.(D)None of these.please explain.

答:
If $f(x)=0$ for all $x$, then $|f(x)-f(y)|\leq |x-y|^{201}$ for all $x,y$.

(D) is the correct option.

Title: continuous functions inequality


问:
Change of basis Matrix in a subspace
I've been investigating this for quite a long time (also looking at other posts), and I still cannot figure it out. Let's try it.

I do have a subspace $F$ belonging to $\mathbb{R}^{4}$, which is having a base $A=\{\vec{a_{1}}, \vec{a_{2}}, \vec{a_{3}}\}$, each of them having 4 components. There is another base $B$ from the same subspace, and they're asking how the change of basis matrix from B to A would look like. I understand that this matrix should be a 3x3 because:

\begin{equation*}
\vec{a_{1}}=w_{11}\vec{b_{1}}+w_{21}\vec{b_{2}}+w_{31}\vec{b_{3}}\\
\vec{a_{2}}=w_{12}\vec{b_{1}}+w_{22}\vec{b_{2}}+w_{32}\vec{b_{3}}\\
\vec{a_{3}}=w_{13}\vec{b_{1}}+w_{23}\vec{b_{2}}+w_{33}\vec{b_{3}}
\end{equation*}

Then I get the nice change of basis matrix from B to A formed by the different $w_{ii}$, so then I see this $M_{B\rightarrow A}$ is a 3x3 matrix, OK. Then they're giving me specific values for the base A, and also a matrix $M_{B\rightarrow A}$, and the question is, "given the matrix $M_{B\rightarrow A}$ and the specific values for A, provide the coordinates of the basis A in basis B".

My rationale then is to proceed like this: If I have $M_{B\rightarrow A}$ but I want $M_{A\rightarrow B}$ i do $M_{B\rightarrow A}^{-1}$ and then $B=M_{B\rightarrow A}^{-1}\cdot A$. But of course this is not working as $M_{B\rightarrow A}^{-1}$ is 3x3 whilst $A$ is 4x3. And here is where I'm stuck. I know that probably there is a fundamental misunderstanding on the last step. I've done quite a lot of this kind of exercises but not with subspaces but with "complete" spaces.

Any little hint here will be appreciated.

Thanks!

答:
Let's see it this way:

We will call the vectors in $B$ as $b_1, b_2, b_3$.

Coordinates in $B$ means you have a linear decomposition of any vector $v = v_{b_1}b_1+v_{b_2}b_2+v_{b_3}b_3$.

To get coordinates in $A$, you can start with the above and decompose $b_i$ in $A$.

Suppose $b_i = \sum_j M_i^j a_j$. You can see $v$ expressed in $A$ is $\sum_i v_{b_i} \sum_j M_i^j a_j$. Now please write the sums out yourself and look closely how the coefficients $v_{b_i}$ and $M^j_i$ interact. It may help to put arrow signs on the vectors to emphasize abstract vectors. (Coefficients = coordinates. Vectors are abstract.)

The coefficients actually coincide with the matrix-vector multiplication of $[M_i^j][v_{b_i}]$ where I used $[M_i^j]$ to denote the matrix formed by $M_i^j$ etc.

That is why the change of coordinate matrix is formed by combining the coordinates of $\{b_i\}$ in basis $A$ which are 3-dimensional column vectors.

To clarify how dimension works, a vector in the subspace start with 3 coordinates in basis $B$. Hence, the coordinates form a 3-d vector. NOT 4-d!

Title: Change of basis Matrix in a subspace


问:
How do you graph $x + y + z = 1$ without using graphing devices?
How can I graph $x + y + z = 1$ without using graphing devices?

I equal $z = 0$ to find the graph on the xy plane. So I got a line, $y = 1-x$
But when I equal 0 for either the $x$ or the $y,$ I get $z = 1-y$ or $z = 1-x$ , and those are two different lines from different angles. Different graphing websites were telling me different answers...

Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeros and etc.

答: I hope you don't mind if I post here a silence solution. I think it is telling you everything, however, other nice answers give you the theoretical points. :-)


问:
What is the largest possible value of $E[X] - (E[X^{-1}])^{-1}$?

Suppose $X$ is a random variable on $[A; B]$ ($0 < A < B$). What is the largest possible value of $E[X] - (E[X^{-1}])^{-1}$?

On one side it is rather obvious, that $E[X] - (E[X^{-1}])^{-1} \leq E[X] \leq B$. But this bound is clearly not tight.

On the other side, the largest value for which I managed to build example for is $(\sqrt{B} - \sqrt{A})^2$. It is the random value $X$ with following distribution:

$$P(X = A) = \frac{\sqrt{A}}{\sqrt{A} + \sqrt{B}}$$

$$P(X = B) = \frac{\sqrt{B}}{\sqrt{A} + \sqrt{B}}$$

I suspect that this lower bound is tight, but do not know how to prove it.

答:
Here is the demonstration for the tight bound $( \sqrt{A}-\sqrt{B})^2$

Demonstration

Given our bound on $X$, we have the following inequality:
$$ \mathbb{E}( (X-A)( \frac{1}{B}-\dfrac{1}{X})) \ge 0 $$
Which gives us $ A+B -\mathbb{E}(X) \ge AB\mathbb{E}( \frac{1}{X})$, so the left hand side of our expression is :
$$ LHS \le \mathbb{E}(X) - \dfrac{AB}{ A+B-\mathbb{E}(X) }= A+B- \left[\left(A+B- \mathbb{E}(X)\right)+ \dfrac{AB}{ A+B-\mathbb{E}(X) } \right] \le A+B-2\sqrt{AB}  $$

$\square$


问:
$\sigma (xy) \subseteq \sigma (x)\sigma (y)$ in Banach algebra.
Let $A$ be a Banach algebra(with unit $e$), $x,y \in A$ and $xy=yx$. Prove that $\sigma(xy) \subseteq \sigma (x) \sigma (y)$, also $\sigma (x+y)\subseteq \sigma(x)+\sigma(y)$. Where $\sigma$ means the spectrum.

Here $\sigma(x) \sigma(y)=\{ \lambda_1 \lambda_2 \in \mathcal{C}: \lambda_1 \in \sigma(x), \lambda_2 \in \sigma(y)\} $, and similarly for $\sigma(x)+\sigma(y)$.

For my attempt, let $B$ be the closed subalgebra generate by $e$, $x$ and $y$. Then $B$ is commutative. Denote $\Delta_{B}$ by the maximal ideal space of $B$. Then we have $\sigma_B(x)=\{ \phi(x): \phi \in \Delta_{B} \}$, and similarly for $y$,$x+y$, $xy$. Use $\phi$ is multiple linear functional,it’s clear $\sigma_{B}(x+y)\subseteq \sigma_{B}(x)+\sigma_{B}(y)$ and $\sigma_{B}(xy)\subseteq \sigma_{B}(x)\sigma_{B}(y)$. But $\sigma_{B}(x)$ maybe larger than $\sigma(x)$. And I don’t know how to continue. I know that the spectrum radical is keep invariant for subalgebra, so $r(x+y)\leq r(x)+r(y)$ and $r(xy)\leq r(x)r(y)$, this is a standard result on Rudin’s Functional Analysis.

This is an exercise from another book. But I heavily doubt the claim of this exercise is not true...

Any help or hint? Thanks!

答:
Instead of subalgebra generated by $e, x, y$, consider some maximal commutative subalgebra $C$ that contains $x, y$ (it exists because union of ascending chain of such subalgebras is also commutative subalgebra).

We can show that $a \in C \leftrightarrow \forall b \in C: ab = ba$ - this implies $C$ is closed and if $a \in C$ is inversible then $a^{-1} \in C$.

So if $a \in C$, then $a - \lambda e$ is inversible in $C$ iff it's inversible in $A$, and your proof works.


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

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

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

答:
Catenary satisfies the given problem

Consider $f(x)=\cosh x+k$. We have
$$A=\int_a^b (k+\cosh x) \, dx=k (b-a)+\sinh b-\sinh a$$
and
$$(b-a)L=(b-a) \int_a^b \sqrt{1+\sinh ^2 x} \, dx=(b-a) \int_a^b \cosh x \, dx=(b-a) (\sinh b-\sinh a)$$
therefore for $k= \frac{(b-a-1) (\sinh b-\sinh a)}{b-a}$ we have $A=(b-a)L$.

we get the result $$f(x)=\cosh x+\frac{(b-a-1) (\sinh b-\sinh a)}{b-a}$$


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

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

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

Thats it I am hopelessly stuck.

答:
From how I've learned to do density proofs, the easist way to go about this is by expanding the interval $(x,y)$ (note I will assume from here on - w.l.o.g. ofc - that $x<y$) and forcing some number in between (one of the form $\frac{m}{10^n}$ in this case).

So let's start by expanding the interval so that we get an interval with a length greater than 1 (it's pretty easy to fit a number into an interval of length greater than 1, of any form). So let's say we find $r \in \mathbb{R}$ so that
$$
ry - rx > 1
$$
Now from this we can see that
$$
\lfloor rx \rfloor + 1 > rx
$$
and also from that first inequality we get the following
$$
ry - 1 > rx \ge \lfloor rx \rfloor \implies ry > \lfloor rx \rfloor + 1
$$
so we have found some integer $q = \lfloor rx \rfloor$ s.t.
$$
rx < q < ry
$$
Our next step is to find an exact formula for $r$, however we must be specific - we will be dividing this above inequality by $r$ so we will want $r$ to be of the form $10^m : m \in \mathbb{Z}$. I first want to note that
$$
10^m > m \, \forall \, m \in \mathbb{Z}
$$
you can use induction on that if you insist (noting that for any negative m the inequality is trivially true). Next we go back to our original inquality
$$
ry - rx > 1 \iff r > \frac{1}{y-x}
$$
So we can easily pick $r = \left\lfloor \frac{1}{y-x} \right\rfloor$ however we require $R$ to be of the form $10^m$. We now take into account that $10^m > m$ so it must be true that $$
10^{\left\lfloor \frac{1}{y - x} \right\rfloor} > \left\lfloor \frac{1}{y - x} \right\rfloor > \frac{1}{y-x}
$$
So if we choose $r = 10^{\left\lfloor \frac{1}{y - x} \right\rfloor}$ then our inequality becomes as follows
$$
10^{\left\lfloor \frac{1}{y - x} \right\rfloor} y > \left\lfloor 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x \right\rfloor + 1 > 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x
$$
Now dividing through by $10^{\left\lfloor \frac{1}{y - x} \right\rfloor}$ we end up with the following
$$
y > \frac{\left\lfloor 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x \right\rfloor + 1}{10^{\left\lfloor \frac{1}{y - x} \right\rfloor}} > x
$$
and so we have found a number of the form $\frac{a}{10^b}$ between any two given real numbers so it follows that $\mathscr{H}$ is dense in $\mathbb{R}$.


问:
Probability - Continuous Joint Probability
I am taking a probability class and I am not sure how to solve this problem:

Suppose $Z=X+Y$, find $f_Z$ if
$$
 f_X(x)=f_Y(x)=
\begin{cases}
x/2, \text{ when }0<x<2 \\ 0, \text{ otherwise}
\end{cases}
$$

I know the solution is $f_Z=x^3/24$ when 0<x<2 and $f_Z=x-x^3/24-4/3$ when 2<x<4

I understand that f(x,y)=xy/4 when 0<x,y<2

I am looking for a simplified answer to get to the bottom line of the concept. Is the Jacobian matrix necessary for this? The limits of integration have always been challenging when doing these problems. Thanks!

答:
First, you have to assume independence between X and Y.

Then, the Jacobian method is not the only one possible but it is a good choice (IMHO)

Setting

$$
\begin{cases}
u=x+y, \\
v=x,
\end{cases}\rightarrow \begin{cases}
x=v, \\
y=u-v,
\end{cases}$$

The jacobian is 1 and the joint density of $(U;V)$ is the following

$$f_{UV}(u,v)=\frac{v(u-v)}{4}$$

defined in the following parallelogram

Do you think you are able to integrate $f(u,v)$ over $V$-support to obtain the marginal $f_U(u)$?


问:
Sum of $\sum_{n=1}^{+\infty}\frac{1}{(n-1)! +n!}$
I have to compute the nature of this series, and if it converges, the sum of it. I just can't find the proper way to get to the partial form, from where I can simplify the parts. I've tried writing $(n+1)!+n!$ as $(n-1)!(1+n)$, but that didn't work either.

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

答:
Almost there. Observe that

$$\frac 1{(n-1)!+n!} = \frac 1{(n-1)!(n+1)} = \frac n{(n+1)!} = \frac {n+1}{(n+1)!} - \frac 1{(n+1)!} = \frac1{n!} - \frac1{(n+1)!}$$

Do you see how it telescopes?


问:
Examples of non-polynomial analytic homeomorphisms
For any odd positive integer $k$, the map $f_k(x):=x^k$ is both an analytic function and a homeomorphism on $\mathbb{R}$, since $f_k^{-1}(x) = \sqrt[k]{x}$ is bijective an analytic.  In fact, this shows that it is a biholomorphism.

Are there examples of biholomorphisms from $\mathbb{R}$ to itself, more generally examples of analytic homeomorphisms (not there is no requirement for the inverse to be more than continuous here), which are not polynomials?

答:
Yes. Take $\sinh$, for instance.

On the other hand, $x\mapsto\sqrt[k]x$ isn't even differentiable at $0$. So, it's not analytic.


问:
Difference of funcions converging in measure to the same function
Suppose $f_n \to h$ and $g_n \to h$ in measure. Is it true that $|f_n - g_n| \to 0$ in measure? I tried using the triangle inequality to get to this result, but it implies only that:
$$\{ x \in X: |f_n(x) - g_n(x)| \geq \alpha\} \supseteq \{x \in X: |f_n(x) - h(x)|\geq \alpha/2 \}\cap \{x \in X: |g_n(x) - h(x)|\geq \alpha/2\}$$
I don't know if this is true, I guess it must be - can anyone help me?

Body:
$$\{ x \in X: |f_n(x) - g_n(x)| \geq \alpha\}$$ $$ \subseteq \{x \in X: |f_n(x) - h(x)|\geq \alpha/2 \}\cup \{x \in X: |g_n(x) - h(x)|\geq \alpha/2\}$$ and hence

$$\mu\{ x \in X: |f_n(x) - g_n(x)| \geq \alpha\}$$ $$ \leq \mu(\{x \in X: |f_n(x) - h(x)|> \alpha/2 \})+\mu( \{x \in X: |g_n(x) - h(x)|\geq \alpha/2\}) \to 0.$$


问:
$\lim\limits_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k},$
What is
$$\lim\limits_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k},$$
where $ k\in \mathbb{N}$?

答:
When we deal with limits involving the $n^{\text{th}}$ root of an expression, a useful result is this one: https://math.stackexchange.com/a/3159844/629594.

With this in mind, we may write that
$\displaystyle\lim_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k}=\lim_{n\to \infty}\frac{1^k+2^k+...+(n+1)^k}{1^k+2^k+...+n^k}\stackrel{\text{Stolz-Cesaro}}{=}\lim_{n\to \infty} \frac{(n+2)^k}{(n+1)^k}=1$.

The other answer is nicer, because it just uses the squeeze theorem, but I wanted to show that this can roughly be done by applying Stolz-Cesaro two times.


问:
Differential linear equation first order?
I have the equation $(1+y^2)\,dx+(xy+1)\,dy=0$.
It is a linear differential equation of the first order.
Now the problem is that this doesnt have the regular form of this type of equations which is $$\frac{dy}{dx}+p(x)y=q(x) \ .$$
I know I can divide both sides by $dx$ and then I'll have $(1+y^2) +
(xy+1)y'=0.$ Still, it should be $(xy+1)+y'$ not $(xy+1)y'$.</p>

答: If you set $M(x,y)=(1+y^2),~~N(x,y)=xy+1$, you see that $$M_y=2y,~~N_x=y$$ and $$\frac{M_y-N_x}{-M}=\frac{-y}{1+y^2}$$ is a function respect to $y$. Now think of an integrating factor $\mu(y)$ and then make your OE exact.</p>


问:
Subsets of $\Omega$ which are not events
If $\Omega = [0,1]$ is my sample space and we have the usual measure on it,

what are some pathological cases (subsets of $\Omega$) which cannot be treated

as events?

I am looking simply for a few examples to better understand which sub-sets can be treated as events and which not.

Obviously if A is e.g. a sub-interval of $\Omega$ or a countable union of sub-intervals, it is an event, we have no problem with it. Right? So which subsets we do have a problem with and thus we cannot treated as events?

答:
An usual example of non-Borel subset is the following : consider the equivalence relation $\sim$ on $[0,1]$ defined by $x \sim y \Longleftrightarrow x-y \in \mathbb{Q}$.

Then consider a subset $A$ of $[0,1]$ constituted of exactly one element from each equivalence class for the relation $\sim$. To show that $A$ is not Borel, do the following :

Consider $(r_n)_{n \in \mathbb{N}}$ an enumeration of the elements of $\mathbb{Q} \cap (-1,1)$. For each $n \in \mathbb{N}$, let $A_n = A + r_n$.

Show that for $i \neq j$, $A_i \cap A_j = \emptyset$, and that $$[0,1) \subset \bigcup_{n \in \mathbb{N}} A_n \subset (-1,2] \quad \quad (*)$$

Assuming that $A$ is Borel, show that $A_n$ is also Borel and deduce from $(*)$ that
$$1 \leq \sum_{n \in \mathbb{N}} \lambda(A_n) \leq 3$$

Finally see that  $\lambda(A)=\lambda(A_n)$ for each $n$, to get a contradiction from the last inequalities.


问:
How to find the angle situated in the intersection between two bisector segments?
The problem is as follows:

Find the value of the angle labeled $x$ in the figure from below.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&70^{\circ}\\
2.&85^{\circ}\\
3.&120^{\circ}\\
4.&95^{\circ}\\
5.&130^{\circ}\\
\end{array}$

I've attempted to solve this problem but the only thing which I could recall is that the sum of the interior angles in a quadrilateral is $360^{\circ}$ but nothing else than that. What sort of identity can be used to solve this problem?.

Please include a drawing in your answer as I think this might require some sort of construction. But more importantly, can this be solved relying only in euclidean geometry?. Please try to give an explanation step-by-step as it becomes difficult for me to catch up with figures.

答:
There is no need for extra constructions. Recall that the sum of interior angles of a polygon is given by
$$(n-2)\times180^\circ$$

where $n$ is its number of sides. This applies for convex and concave polygons.

First consider the angle sum of the triangle. We have:

$$80^\circ + (180^\circ - 2\theta) + (180^\circ - 2\beta) = 180^\circ$$

$$\theta + \beta = 130^\circ$$

Now consider the angle sum of the concave hexagon. We have:
$$x+90^\circ + \theta + (360^\circ - 80^\circ) + \beta + 90^\circ = (6-2)\times 180^\circ = 720^\circ$$

$$x+\theta + \beta = 260^\circ$$

$$x = 260^\circ - (\theta + \beta) = 260^\circ - 130^\circ =130^\circ$$

The rest of your recent questions can be solved by very similar methods.


问:
Show there exists a set $E \subset \mathbb{R}^2$ such that the cross sections are open subsets but $E \notin \mathcal{B}_2$.

Show there exists a set $E \subset \mathbb{R}^2$ such that the cross sections $[E]_a$ and $[E]^a$ are  open subsets for every $a \in \mathbb{R}$ but $E \notin \mathcal{B}_2$.

This is a question from Axler's book on pg 144, exercise 2 and $\mathcal{B}_2$ is the $\sigma$-algebra of Borel subsets of $\mathbb{R}^n$.

I know that in general, <s>if a set $E \in \mathbb{R}^n$ is open, the cross sections aren't generally open</s> that if the cross sections are open, then it isn't necessarily true that the set itself is open.

The condition that $E$ can't be in the Borel $\sigma$-algebra is really bothering me. I also thought maybe trying to use the Cantor function generalized to $\mathbb{R}^n$, but didn't see immediately how to keep cross sections as open subsets.

答:
Let $A$ be a non-Borel subset of $\mathbb R$.  Then
$$
E = \mathbb R^2 \setminus \{(x,x): x \in A\}
$$
is your set.  It is not Borel.  But all the cross-sections are either $\mathbb R$ or $\mathbb R$ minus a single point.


问:
Non exact differential equation - integration factor problem
Can you help me with this one?

I have checked it and it is non-exact:

$$d/dy(-ycos(ln(y/x))=-cos(ln(y/x))+sin(ln(y/x))$$
$$d/dx(x)=1$$
So, I should find integrating factor with solving this:

But I can't figure out how to solve this.

答:
$$x y'-y \cos \left(\log \left(\frac{y}{x}\right)\right)=0$$

Let $y=x e^z$ to get
$$x z'-\cos (z)+1=0$$ Now, switch variables
$$\frac x{x'}-\cos(z)+1=0\implies \frac {x'}x=\frac 1 {\cos(z)-1}$$ which looks workable.


问:
How to find the angle of a right triangle which is situated in between to bisector segments?
The problem is as follows:

Using the figure from below find $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&40^{\circ}\\
2.&20^{\circ}\\
3.&10^{\circ}\\
4.&36^{\circ}\\
5.&18^{\circ}\\
\end{array}$

I've attempted to solve this problem but I came stuck with how to relate the angle in the right triangle, as it is suspiciously half of what it is in the intersection between two bisector segments.

But I don't know what to do from there. Can someone help me here?. Please include a drawing in your answer as trying to understand solely from reading algebraic expressions is difficult for me. The thing is I think this might require a construction or something like that. Can this be solved relying only in euclidean geometry?. Please explain this in a step-by-step approach as I'm not sure what sort of identity can be used to solve this.

Body:
In quadrilateral $ABCD, $

$360^0 = 180^0 - 2 \alpha + 180^0 - 2 \theta + 90^0 + x + 90^0$

So, $2 \theta + 2 \alpha -x = 180^0$ ...(i)

In $ \triangle ADE$,

$\theta + \alpha + 2x = 180^0$ ...(ii)

From $(ii) \times 2 - (i), 5x = 180^0$,

So $x = 36^0$


问:
How to find the angle in between two triangles when one of them is the double of the other?
The problem is as follows:

Using the figure from below find $x$

The problem is as follows:

$\begin{array}{ll}
1.&60^{\circ}\\
2.&40^{\circ}\\
3.&60^{\circ}\\
4.&70^{\circ}\\
5.&30^{\circ}\\
\end{array}$

This peculiar problem is confusing to me, what sort of identity can be used to solve this?. The only thing which I could spot there was:

$\beta+2x+\theta=180$

But that's it. I don't know what other relation can be established from the given information. Can someone help me here?. Please include a drawing in your answer as it becomes difficult for me to read from only algebraic expressions. Please be the most descriptive as possible and it would help me a lot to have a step-by-step solution as I am struggling with this figure.

答:
Note that we have $2x+\beta + \theta = \pi$. Notice that $\angle BDF+\angle CEG = \pi - \angle DAE = \pi - x$. Notice that the sum of the angles of $\triangle BDF$ and $\triangle CEG$ is $2\pi$: $2(\beta+\theta)+\gamma+\phi+\pi-x=2\pi$. Note that the sum of the angles of $\triangle AFG$ is $\pi$, so $2\pi-2(\gamma + \phi)+x=\pi$ Using these 3 equations we derived:
$$ $2(\beta+\theta)+\gamma+\phi+\pi-x=2\pi$$
$$2x+\beta + \theta = \pi$$
$$2\pi-2(\gamma + \phi)+x=\pi$$
We can find $x=\frac{\pi}{3}$. (Substitute $y=\beta+\theta,z=\gamma+\phi$, you have three equations and three variables)


问:
Probably $x_1 \leq x_3 \leq x_2$
From set  ${1,2...n}$ we choose three random different numbers, $x_1, x_2,x_3$. If $x_1 \le x_2$, find probability $x_1 \le x_3 \le x_2$.
I set A is event  $x_1 \leq x_3 \leq x_2$, and B $x_1 \leq x_2$, and find P(AB)=$\frac{1}{6}$, P(B)=$\frac{1}{2}$. Is it okay?

答:
The problem has two different solutions, depending on whether you take them with replacement or without it (your question is quite ambiguous in that sense: the word 'different' suggests that they are taken without replacement, and the sign '$\leq$' - that they are taken with replacement):

Solution for numbers taken without replacement:

$$P(x_1 < x_2) = \frac{1}{2}$$

$$P(x_1 < x_3 < x_2) = \frac{1}{6}$$

$$P(x_1 < x_3 < x_2| x_1 < x_2) = \frac{1}{3}$$

Solution for numbers taken with replacement:

$$P(x_1 \leq x_2) = P(x_1 = x_2) + P(x_1 < x_2) = P(x_1 = x_2) + \frac{1 - P(x_1 = x_2)}{2} = \frac{1}{n} + \frac{n-1}{2n} = \frac{n+1}{2n}$$

$$P(x_1 \leq x_3 \leq x_2) = P(x_1 < x_3 < x_2) + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1 - P(x_1 = x_3 < x_2) - P(x_1 < x_3 = x_2) + P(x_1 = x_3 = x_2) }{6} + P(x_1 = x_3 < x_2) + P(x_1 < x_3 = x_2) - P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5}{6}P(x_1 = x_3 < x_2) + \frac{5}{6}P(x_1 < x_3 = x_2) - \frac{5}{6}P(x_1 = x_3 = x_2) = \frac{1}{6} + \frac{5(n+1)}{12n^2} + \frac{5(n+1)}{12n^2} - \frac{5}{6n^2} = \frac{n+5}{6n}$$

$$P(x_1 \leq x_3 \leq x_2| x_1 \leq x_2) = \frac{n+5}{6n}\frac{2n}{n+1} = \frac{n+5}{3(n+1)}$$

Title: Probably $x_1 \leq x_3 \leq x_2$


问:
The interior of $\{f \leq 0\}$ is $\{ f < 0\}$
Let $X\subseteq \mathbb{R}^n$, $f: X \rightarrow \mathbb{R}$: what further conditions on $X$ and $f$ do we need to ensure that $\text{Int}\{x|f(x)\leq 0\}=\{x|f(x)<0\}$?

Context: I've encountered this kind of question when dealing with open bounded $X$ with $C^1$ boundary, in partial differential equations, but I'm sure that one could benefit from such a fact in optimization, for instance.

Clearly, continuity is not enough, since $f:[0,1]\rightarrow \mathbb{R} , f(x)=0$ is such that $[0,1]=\{f\leq 0\}$, but the interior of this set is not the empty set.

答:
If $f$ is continuous, it is easy to see that $$\tag
{1}\{x:f(x)<0\}\subseteq \text{Int}\{x:f(x)\le 0\}$$

This holds because the LHT is open and the RHT is, by definition, the biggest open subset of $\{x:f(x)\le 0\}$.

Let us first suppose $X=\mathbb{R}^n$:

One necessary and sufficient condition for equality, provided that $f$ is differentiable, is that $\nabla f\neq 0$ on $f^{-1}(0)$.

To prove it, let $x$ be a point inside $\text{Int}\{x:f(x)\le 0\}\backslash \{x:f(x)<0\}$. This means that $x$ is locally a maximum, and thus $\nabla f=0$

This clearly generalizes to open subsets of $\mathbb{R}^n$.
It is now easy to see that, for general $X$ and continuous $f$, a necessary and sufficient condition is that $f^{-1}(0)$ does not contain local maxima (although this is not a very useful condition)

Note that some kind of regularity hypotesis is necessary: as Whitney proved, given a closed subset $K$ of $\mathbb{R}^n$, there's always a $\mathcal{C}^\infty(\mathbb{R}^n)$ function with $f^{-1}(0)=K$. Given such an $f$, considering $f^2$ gives you a class of smooth examples with strict (or trivial) inclusion.


问:
Finding a function with a set of points
I have a function graph and I need to find the matching function. Here is the graph :

The graph

And here are three points from the graph :

(-2;0)

(0;-6)

(3;0)

Note: As you can see, this is not a linear function

Does anyone know how to achieve that?
I was thinking of interpolation, the problem is that I don't find the formula...

答:
Quick and simple answer:$$
y = x^2 -x -6
$$

Title: Finding a function with a set of points


答:
Solving a problem like this, at least by hand, usually involves figuring out the general family of functions that yours belongs to.  As the previous answer states, this is a quadratic.  However, being able to see that takes experience and there is in general no way to know.  Some hints that we have that indicate that it might be a quadratic:

It sounds like a pre-calculus course question, and they study polynomials extensively

The graph is obviously nonlinear, and if we are searching for polynomials we want to try the least complex one we can first.  The least complicated nonlinear polynomial is the quadratic.

You were given only three points.  A quadratic has the general formula $ax^2 + bx + c = y$.  There are three unknown parameters, so with three points the best you can do is identify those three parameters.

So, how do you do it? Ultimately, you want to turn this into a set of linear equations if you can.  Plug in each point to get three equations:

$$0 = a(-2)^2 + b(-2) + c$$
$$-6 = a(0)^2 + b(0) + c$$
$$0 = a(-3)^2 + b(3) + c$$

From these three equations, you have three unknowns, and you can solve for each of them.  This is a general technique that can sometimes work if your equation is nice (that is to say, if it nicely fits into this technique of producing linear equations): plug in the points, get the equations, and solve them simultaneously.  The REALLY hard part about this is determining what kind of function you're solving for.

I should mention that this problem has a nice property: you were given the two zeros of the parabola and a point on it, which might be the vertex (I'm not sure, I didn't look that closely), so this particular problem is much simpler to solve.  However, the method I show above works in general, no matter which points you're given.

Title: Finding a function with a set of points

Body:
I have a function graph and I need to find the matching function. Here is the graph :

The graph

And here are three points from the graph :

(-2;0)

(0;-6)

(3;0)

Note: As you can see, this is not a linear function

Does anyone know how to achieve that?
I was thinking of interpolation, the problem is that I don't find the formula...


问:
Finding the 3D coordinates of an unknown point from three known points

How can I find the coordinates $(x,y,z)$ in a $3d$ space when,

A) the unknown point is $(x,y,z)$;

B) there are three known points viz. $(a1,b1,c1)$, $(a2,b2,c2)$ and $(a3,b3,c3)$;

C) the distances from the three known points are $D1$, $D2$, $D3$, respectively.

答:
It would involve solving a system of 3 equations with 3 unknowns, where the solution $(x,y,z)$ represents the intersection point of the following three 3D spheres:
$$
(x-a_1)^2 + (y-b_1)^2 + (z-c_1)^2 = (D_1)^2\\
(x-a_2)^2 + (y-b_2)^2 + (z-c_2)^2 = (D_2)^2\\
(x-a_3)^2 + (y-b_3)^2 + (z-c_3)^2 = (D_3)^2
$$</p>


问:
how to complete my proof of $(A\cup B)\triangle(A\cup C) \subseteq A \cup (\ B \triangle C)$
\begin{align} (A\cup B)\triangle(A\cup C) \subseteq A \cup (\ B \triangle C) \end{align}

Here is my proof:

\begin{align} x \in (A\cup B)\triangle(A\cup C) \end{align}
def. $\triangle$
\begin{align} x \in (A\cup B)\setminus (A\cup C) \land  x \in (A\cup C)\setminus (A\cup B)  \end{align}
simp.
\begin{align} x \in (A\cup B)\setminus (A\cup C)\end{align}
def. $\setminus$
\begin{align} x \in (A\cup B) \land x \notin (A\cup C)\end{align}

and I'm stuck

答:
Take $x\in(A\cup B)\triangle(A\cup C)$. Then $x$ belongs to one of the sets $A\cup B$ and $A\cup C$, but not to both.

Suppose that $x\in A\cup B$. Then $x\in A$ or $x\in B$. But if $x\in A$, then $x\in A\cup C$, which is impossible, since we are assuming that $x$ doesn't belong to both sets $A\cup B$ and $A\cup C$. So, $x\in B$. And, since $x\notin A\cup C$, $x\notin C$. So, $x\in B\triangle C$. And so, since $A\cup(B\triangle C)\supseteq B\triangle C$, $x\in A\cup(B\triangle C)$.

The same argument shows that if $x\in A\cup C$, then $x\in A\cup(B\triangle C)$. Since this occurs for each $x\in(A\cup B)\triangle(A\cup C)$…


问:
Showing a function $G$ is surjective
This is one of the homework questions in an analysis course. The questions goes in the following way. Let Assume that for each $n \in \mathbb N$ the set $A_n$ is countable. In the lectures we have proven that there exists a surjective map $g:\mathbb N \mapsto A_n$.

Prove that the map
$$
G:\mathbb N \times \mathbb N \mapsto \bigcup_{n\in\mathbb N}A_n, G(n,m)=g_n(m)
$$
is surjective.

Now my prove goes in the following way:

First let us look at the codomain of $G$:
\begin{align*}
 \bigcup_{n\in \mathbb N} A_n = &\{A_1\cup A_2 \cup...\cup A_n\}\\
        = &\{\{a_{1,1},a_{2,1},\cdots,a_{|A_1|,1}\}
       \cup  \{a_{1,2},a_{2,2},\cdots,a_{|A_2|,2}\} \cup
         ...
         \{a_{1,n},a_{2,n},\cdots,a_{|A_n|,n}\}\}\\
        = &\{a_{1,1},a_{2,1},\cdots,a_{|A_1|,1},
          a_{1,2},a_{2,2},\cdots,a_{|A_2|,2},
         \cdots,
         a_{1,n},a_{2,n},\cdots,a_{|A_n|,n}\}\\
\end{align*}
We can now define $g_n(m)$ such that $g_n(m)=a_{m,n}$, meaning the $m$'th element of the $n$'th set. This means that $G(m,n)=a_{m,n}$. For any $a_{m,n}$ we know that $(m,n)$ maps onto that element and we know that $(m,n) \in\mathbb N$ because $m$ and $n$ are both integers, therefore $G$ is a surjective function.$\square$

However I have the feeling that this proof really isn't correct as I'm already defining $G$, however I'm not quite sure. Could someone tell me if something is going wrong and if yes, what?

答: This can be written in a more direct way : just take $y \in \bigcup_{n \in \mathbb{N}} A_n$. By definition, there exists $n_0\in \mathbb{N}$ such that $y \in A_{n_0}$. But because $g_{n_0} : \mathbb{N} \rightarrow A_{n_0}$ is surjective, there exists $m_0 \in \mathbb{N}$ such that $g_{n_0}(m_0)=y$. Now, you just have to check that $G(n_0, m_0)=y$ to conclude that $y$ is in the image of $G$, so $G$ is surjective.

Title: Showing a function $G$ is surjective


问:
The identity map in two metric spaces is a homeomorphism
We have given two metric spaces (M,$\tau_{d}$) and (M,$\tau_h$) whereby the metric $h$ is given as:

$h(x,y)=\frac{d(x,y)}{1+d(x,y)}$.

Now I have to show that the function

$id_{M}$ : $(M, \tau_{d}) \rightarrow  (M,\tau_h)$ which sends $x \rightarrow x$ is a homeomorphism.

I don't know if it would suffice to say that the identity function is a homeomorphism following the fact that it is a bijection and continuous and would we therefore be done or if there might be a more specific, elegant way.

Thanks in advance

Edit: I realised that my problem is the fact that I would only know how to show that the identity is contnuous in the case that we have the same metric spaces. But how can you show it if we have two metric spaces with two different induced topologies?

答: Those two metrics induce the same topology. In other words, those are equivalent metrics. You will find a proof of that assertion here. So, since the continuity of a map depends only upon the topologies that you are working with, $\operatorname{id}_M$ is continuous in both directions.


问:
$AX=C$: An Inconsistent Linear Equation

Question:

Let $A \in M_{n\times n}(F)$. Suppose that the system of linear equations $AX = B$
has more than one solution. Prove that there is a column $C \in F^n$ such that
the system of linear equations $AX = C$ is inconsistent.

Attempt:

Lemma: Let $AX=C$ be a system of linear equations. Then the system is consistent if and only if $\text{dim Im}~(A)=\text{dim Im}~(A \mid C)$.

Let $Y$ and $Y'$ be solutions to the system $AX=B$. Then

$$ A(Y) = A(Y')=B,$$

or

$$ A(Y)-A(Y')=A(Y-Y')=0,$$

so $\dim \text{Ker}~(A)\geq 1$, which means $\dim \text{Im}~(A)<n$, or simply that

$$\text{span}~(\{A^{(1)},\dots,A^{(n)}\})\neq F^n.$$

In other words, there exists at least one $A^{(i)}$ such that

$$A^{(i)}=k_1A^{(1)}+\cdots +k_{i-1}A^{(i-1)}+k_{i+1}A^{(i+1)}+\cdots + k_nA^{(n)}\neq ce_i,$$

for some $e_i\in \{e_1,\dots,e_n\}$ the standard basis for $F^n$ with $c,k_i\in F$. Thus, by the lemma, we clearly see that such an $e_i$ is the desired $C$ for which $AX=C$ has no solutions, namely

$$\text{dim Im}~(A)\neq \text{dim Im}(A \mid e_i)=\text{dim Im}(A \mid C).$$

This is so because if $\text{dim Im}(A)=m$, then $\text{dim Im}~(A \mid e_i)=m+1$ for such an $e_i$ described above.

Acknowledgements: @jkn

答: The linear operator $A:V\to V$ is not injective. Therefore, since it acts on a finite-dimensional space $V$, it is not surjective. So there is a vector $C$, which is not contained in ${\rm Im} A$.</p>

Title: $AX=C$: An Inconsistent Linear Equation


问:
Why are these numbers coprime?
I have encountered the following argument in a number theory class:

Let $p$ be an odd prime and let $n,x$ and $y$ be integers with $\gcd(x,y) = 1$, $n > 0$ and $p \not \mid n$. Suppose that $p \mid x^2+ny^2$, it then follows that $\gcd(y,p) = 1$.

I do not understand why $\gcd(y,p) = 1$ should hold here. Could you give me a hint?

答: Since $p$ is prime, either it is coprime to $y$ or divides $y$. But it can't divide both $x$ and $y$, since they are coprime to each other, and it can't divide $y$ but not $x$, since then it would not divide $x^2+ny^2$.

Title: Why are these numbers coprime?


问:
What are the orders of a element $a + H$ of a quotient group $G/H$
I would like somebody to check my answers

$G = (\mathbb{Q, +}), H = (\mathbb{Z, +}), a = \frac{2}{3}$

I think the answer is $\infty$, because $a + H = \{\frac{2}{3} + h: h \in \mathbb{Z}\}$

$G = (\mathbb{Z_{12}, +_{12}}), H = \{0,3,6,9\}, a = 5$

$a +H  = \{5,8,11,2\}$, so the answer is 4.

答:
I think that you are misunderstanding the meaning of “order” here. It is the order of an element $g$ of a group $G$, which is the smallest natural $n$ such that $g^n=e_G$ (and it is $\infty$ if no such $n$ exists).

In your first question the order is $3$, because $3.(a+H)=H$, which is the identity element of $G/H$. And the answer is also $3$ in the other case, since$$5+H\ne H,\ 2.(5+H)=10+H\ne H\text{, and }3.(5+H)=H.$$


问:
Let $x_{n+1} = x_n + 1/(x_1 + x_2 +\ldots + x_n)$ with $x_1 = 1$. Show that $x_n\sim\sqrt{2\log(n)}$.
As the title states we have a sequence defined by

$$x_{n+1} = x_n + \dfrac{1}{x_1 + x_2 + \cdots + x_n}$$

with $x_1 = 1$.

The first few terms are: $1, 2, \frac{7}{3}, \frac{121}{48} \cdots$

Any ideas would be appreciated.

答:
I revisited this question and came up with a very different approach. It is so different, that I didn't think it could be incorporated into my previous answer; so, I have added a second answer.

Basic Bounds

Since $x_1=1$ and
$$
x_{n+1}-x_n=\left(\sum_{k=1}^nx_k\right)^{-1}\tag1
$$
$x_n$ is a positive, monotonically increasing, concave function of $n$. Therefore,
$$
\frac{n+1}2\,x_n\le\sum_{k=1}^nx_k\le nx_n\tag2
$$
The left-hand inequality is because $x_k\ge\frac knx_n$ ($x_1=1,x_2=2$, and $x_n$ is concave in $n$; concavity is maintained by extending $x_0=0$). The right-hand inequality is because $x_k\le x_n$.

We can modify the left hand inequality in $(2)$ by substituting $n\mapsto n+1$ and then subtracting $x_{n+1}$ from both sides to get
$$
\frac{n}2\,x_{n+1}\le\sum_{k=1}^nx_k\le nx_n\tag3
$$
Taking reciprocals of $(3)$ and applying $(1)$, we get
$$
\frac1{nx_n}\le x_{n+1}-x_n\le\frac2{nx_{n+1}}\tag4
$$
Noting that $2x_n\le x_n+x_{n+1}\le2x_{n+1}$, we can multiply $(4)$ by $x_n+x_{n+1}$ to get
$$
\frac2n\le x_{n+1}^2-x_n^2\le\frac4n\tag5
$$
Summing $(5)$ yields
$$
\sqrt{1+2H_n}\le x_{n+1}\le\sqrt{1+4H_n}\tag6
$$

Asymptotics

Apply $(4)$ and $(6)$ to get that $\frac{x_m}{x_n}\ge1+\epsilon$ requires $m-n\gt\frac{x_n\epsilon}{\frac2{nx_n}}\ge nH_n\epsilon$. Thus, if we set $m=n+nH_n\epsilon$, for $n\le k\le m$, we have $x_k\ge\frac{x_m}{1+\epsilon}$. Therefore,
$$
\begin{align}
\sum_{k=1}^mx_k
&\ge\frac{x_m}{1+\epsilon}\overbrace{\frac{mH_n\epsilon}{1+H_n\epsilon}}^{\substack{\text{number of}\\\text{terms}\\\ge\frac{x_m}{1+\epsilon}}}\tag7\\
&=mx_m\underbrace{\frac1{1+\epsilon}\frac{H_n\epsilon}{1+H_n\epsilon}}_{\substack{\text{can be made close to}\\\text{$1$ by choosing $\epsilon$ small}\\\text{and $n$ big}}}\tag8
\end{align}
$$
Due to $(8)$, for large $n$, $(3)$ becomes
$$
\sum_{k=1}^nx_k\sim nx_n\tag9
$$
which leads to $(6)$ becoming
$$
\begin{align}
x_{n+1}
&\sim\sqrt{1+2H_n}\tag{10}\\[3pt]
&\sim\sqrt{2\log(n)}\tag{11}
\end{align}
$$


问:
How to find the probability density function of a transformed standard normal RV, X= |Z + 1| for x=1?
Question: Let $Z$ be the standard normal random variable. Find the value of the probability density function  $f_{x}(x)$ of $X  =  |Z + 1|$ for x  =  1.

My approach:

The standard normal RV is Z~N(0,1) and its pdf is $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$

$$F_{x}(X) = P(X \le x) = P(|Z+1|\le x) = P(Z \le \pm x-1) = F_{z}(\pm x-1)$$

Then, $$f_{x}(x) = f_{z}(\pm x-1)\frac{d}{dx}(\pm x-1)$$
$$ f_{x}(x) = \frac{1}{\sqrt{2\pi}}e^{(\pm x-1)^2/2}\frac{d}{dx}(\pm x-1) $$

This is the part where I get stuck. When I substitute x=1 for both $-x-1$ and $x-1$ I get incorrect answers both times. Moreover, I am confused about finding the inverse of the absolute function in the beginning steps. I don't think there should be 2 cases (i.e. both $-x-1$ and $x-1$) but am unsure of how to proceed. Any help would be appreciated. Thanks!

答:
Suppose $Z$ is an arbitrary continuous random variable with PDF $f_z$. Let's express PDF of $X_c = |Z - c|$ for any constant $c \in \mathbb{R}$ through it:

Suppose $F_{X_c}$ and $F_Z$ CDFs of $X_c$ and $Z$ respectively.

Then $\forall t > 0$

$$F_{X_c}(t) = P(X_c < t) = P(|Z - c| < t) = P(Z \in [c-t; c+t]) = F_Z(c+t) - F_Z(c-t)$$

If we differentiate both sides of the equality by $t$ we get:

$$f_{X_c}(t) = f_Z(c+t) + f_Z(c-t)$$


问:
How to show this conclusion which concerns two distribution functions?
If $F$ and $G$ are two distribution functions, how to prove that $$\int_{0}^{1} \lvert F^{-1}(t)-G^{-1}(t)\rvert dt=\int_{-\infty}^{+\infty}\lvert F(x)-G(x)\rvert dx$$ with $$F^{-1}(p)=\inf\{x\in\mathbb{R},F(x)\geq p\}$$ I have no idea to deal with the integral of the absolute value of the difference of two distribution functions. The left side is in fact equal to $\mathbb{E}[\lvert F^{-1}(U)-G^{-1}(U)\rvert]$ where $U$ is in the uniform distribution on $[0,1]$.\
This is an important exercice in my course and I really need someone who can help me solve it.

答: What you need to do is to express $|F(x) - G(x)|$ as $\int 1_{(F(x),G(x))} (t)dt$ then you just make use of Fubini theorem


问:
Domain of $\frac{ln(y)}{\sqrt{x-y}}$?
i wanted to know why $x>y \land y>0 \ne x>y \lor y<0$

the left answer is the teacher, the right one is me

thanks for the answer

答:
The domain is given by the conditions: $y>0$ because of the $\log y$, $x-y>0$ because of the square root and the denominator different from $0$; so it must be $x>y>0$.
Your mistakes are:

you use $\lor$, that means "or" so it gives the meaning that one condition, other condition or both conditions are allowed, while you need that the conditions must be true all together, so you have to use $\land$ that means "and";

you say $y<0$ instead of $y>0$.

Title: Domain of $\frac{ln(y)}{\sqrt{x-y}}$?


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

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

This problem is also quoted here.

Here's my solution:

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

However, Arnol'd also says

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

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

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

答:
After step 1, there is always one spoonful of wine in your cup.

If, on step two:

You place one spoonful of wine back into your barrel, then there is 0 tea in your wine and 0 wine in your barrel.

You place one spoonful of tea back into in your barrel, then there is 1 tea in your wine and 1 tea in your barrel.

You place half a spoonful of tea and half of wine back into in your barrel, then there is 0.5 tea in your wine and 0.5 wine in your tea.

It seems that the quantity (not ratio) of tea in your wine and wine in your tea remains equal regardless.


问:
Series of integer part using fractionnal part fourrier series
I'm trying to calculate this sum:
$F(N,p) = \sum_{k=1}^\infty \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor $ with $ N \in \mathbb N^*$, $ p \in \mathbb P$ and $\beta >2$

Clearly, $ F(N,p)$ is finite.

We know that
$ \forall x \in \mathbb R , \lfloor x \rfloor = x - \{x\} $ and that for all $ p $ and $ N $, $\left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor \notin \mathbb Z $ (thanks to $\beta>2$)

We can use the fourrier series of $\{\}$: $ \forall x \notin \mathbb Z ,\{x\} = \frac12-\frac{1}{\pi}\sum_{l=1}^\infty \frac{sin\left(2\pi lx\right)}{l}$

Then
$$F(N,p) = \left(\frac{2N+1}{2}+\frac1\beta\right)\frac{1}{p-1} + G_\infty(N,p)$$
with
$$G_{a}(N,p) = \frac1\pi\sum_{k=1}^a\sum_{l=1}^\infty\frac{sin\left(2\pi l\left(\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}\right)\right)}{l}$$

And now, I don't know how to proceed rigorously. I tried somethings below but I don't know if it's correct

I try to use the fact that for every closed interval set without integer, the fourier series of fractional part converge uniformly to swap sums. In fact, it's easy to see that there can only be a finite number of terms in the sum of fractional part that are not equals to their value (values greater than 1...) and that $\lim_{k\to\infty}\left\{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}\right\}=\frac12$ It's what make me think that swapping was possible.
Then I used the series expansion of sin function to get something like this:

$$G_\infty(N,p)=\frac1\pi\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{k=1}^\infty\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi p^{-k}\left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}}{(2m+1)!}$$

And then

$$G_\infty(N,p)=\frac1\pi\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi l \left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}}{(2m+1)!(p^{2m+1}-1)}$$

What do you think?

EDIT 1 :

My assumptions were wrongs because there is no absolute convergence.
However, using this, and not summing over infinity, I can swap sums symbols.

First :

$$
\forall k> \left\lfloor\frac{ln(2N+1)}{ln(p)}\right\rfloor=k_0,\ \ \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor = 0
$$

So

$$
F(N,p) = \sum_{k=1}^{k_0} \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor = \left(\frac{2N+1}{2}+\frac1\beta\right)\frac{1-p^{-k0}}{p-1} + G_{k_0}(N,p)
$$

Using the proof here, and the fact that the Fourier series of fractional part is uniformly convergent for every close real interval without integer:

For clarity I pose $g$ as

$$
g_k(x)=\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi x \left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}\left(1-p^{-k(2m+1)}\right)}{(2m+1)!(p^{2m+1}-1)}
$$

Then:

$$
G_{k}(N,p)=\frac1\pi\sum_{l=1}^{\infty} \frac{(-1)^lg_k(l)}{l}
$$

Then I applied Ramanujan's master theorem :

$$
f_k(x)=\sum_{i=0}^\infty \frac{\psi_k(i)(-x)^i}{i!}
$$$$
\int_0^\infty x^{s-1}f_k(x)dx = \Gamma(s)\psi_k(-s)
$$

With some variable change, and posing :
$$
\psi_k(x)=\frac{\Gamma(x+1)\left(1-p^{-k(2x+1)}\right)}{\Gamma(2x+2)\left(p^{2x+1}-1\right)}
$$
$$
\alpha = 2\pi \left(\frac{2N+1}{2}+\frac1\beta\right)
$$

We get :

$$
\int_0^\infty x^{2(s-1)}g_k(x)dx=\frac{1}{2\alpha^{2s-1}}\Gamma(s)\psi_k(-s)
$$

And finally
$$
\lim_{s\to\frac12} \frac{1}{2\alpha^{2s-1}}\Gamma(s)\psi_k(-s) = \int_0^\infty \frac{g_k(x)}{x}dx = \frac{\pi k}{2}
$$

I suppose it's quite a good result but do you have any idea of how to evaluate $G_k(N,p)$ using this ?

答:
$$\lfloor \frac{x}{n}\rfloor = \lfloor \frac{\lfloor x\rfloor}{n}\rfloor, \qquad \lfloor 2x\rfloor =\lfloor x\rfloor+\lfloor x+\frac12\rfloor$$ gives for $\beta > 2$

$$\sum_{k\ge 1} \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor=
\sum_{k\ge 1}\left\lfloor{\frac{2N+1}{p^k}}\right\rfloor -\left\lfloor{\frac{N}{p^k}}\right\rfloor=
v_p ( \frac{(2N+1)!}{N!})$$


问:
Non exact differential equation - integration factor problem
Can you help me with this one?

I have checked it and it is non-exact:

$$d/dy(-ycos(ln(y/x))=-cos(ln(y/x))+sin(ln(y/x))$$
$$d/dx(x)=1$$
So, I should find integrating factor with solving this:

But I can't figure out how to solve this.

答:
$$xdy-y \cos \left(\log \left(\frac{y}{x}\right)\right)dx=0$$
The differential equation is homogeneous. Substitute $y=t x \implies y'=t'x+t$:
$$t'x+t-t \cos \left(\log t\right)=0$$
It's separable:
$$xt'=t(\cos \left(\log t\right)-1)$$
$$\int \dfrac {d \log t}{\cos \left(\log t\right)-1}=\int \dfrac {dx}{x}$$


问:
Maximum number of 'moves' required to correctly rack pool balls (for 8 ball)?
To correctly set up balls for 8-ball.
The balls must be arranged in a pattern like this:

```<code>    L
   H L
  L 8 H
 H L H L
L H L H H
</code>```

Where:

L are "the spots" numbered 1-7

8 is the 8-ball

H are "the stripes" numbered 9-15

There is a total of 12 valid set ups corresponding to 3 different 'symmetries?':

rotating 60 degrees either way like:

```<code>    L          H
   H H        L H
  L L L      H H L
 H H 8 H    L 8 L H
H L H L L  L H L H L
</code>```

inversion by swapping Hs and Ls like:

```<code>    H
   L H
  H 8 L
 L H L H
H L H L L
</code>```

flipping across axis going through center of 8 ball and its nearest corner like:

```<code>    L
   L H
  H 8 L
 L H L H
H H L H L
</code>```

A 'move' is swapping 2 balls.

What is then the maximum number of moves required to reach a correct set up from any set up?

答:
Well I turns out the answer is 4.

Since the maximum number of balls that can be in the wrong place is 8.

I brute forced solution by generating all 51480 permutations and compared each against the 12 solutions.

It would be trivial to find out distributions of distances to a solution, if anyone was interested.

Below is an example of a worst case scenario:

```<code>Nearest solution:
    L
   H L
  L 8 H
 H L H L
L H L H H

Permutation:
    H
   L H
  8 L L
 H L H L
H L L H H

Diff:
    H
   L H
  8 L L
 . . . .
H L . . .
</code>```

Below javacode (takes about 10 mins to run on a decent laptop):

```<code>import java.util.ArrayList;
import java.util.List;

public class Main {
    private static final int B = 0;
    private static final int L = 1;
    private static final int H = 2;
    public static void main(String[] args) {
        int[][] solutions = {
                {
                         L,
                        H,L,
                       L,B,H,
                      H,L,H,L,
                     L,H,L,H,H
                },
                {L, L, H, H, B, L, L, H, L, H, H, H, L, H, L},
                {H, L, H, H, B, L, L, H, L, H, H, L, H, L, L},
                {H, H, L, L, B, H, H, L, H, L, L, L, H, L, H},
                {L, H, H, L, L, L, H, H, B, H, H, L, H, L, L},
                {H, H, L, L, H, H, H, L, B, L, L, H, L, H, L},
                {H, L, L, H, H, H, L, L, B, L, L, H, L, H, H},
                {L, L, H, H, L, L, L, H, B, H, H, L, H, L, H},
                {H, L, H, H, H, L, L, B, L, H, L, H, L, H, L},
                {L, H, H, L, L, L, H, B, H, H, L, L, H, L, H},
                {L, H, L, L, L, H, H, B, H, L, H, L, H, L, H},
                {H, L, L, H, H, H, L, B, L, L, H, H, L, H, L}
        };

        final List<int[]> permutations = permutations(0, solutions[0]);
        int[][][] diffs = new int[permutations.size()][][];
        for (int i = 0; i < diffs.length; i++) {
            diffs[i] = diffs(permutations.get(i), solutions);
        }

        int[][] diffCounts = new int[diffs.length][];
        for (int i = 0; i < diffs.length; i++) {
            diffCounts[i] = diffCount(diffs[i]);
        }

        int[] minIndexes = new int[diffCounts.length];
        for (int i = 0; i < diffCounts.length; i++) {
            minIndexes[i] = indexOfMin(diffCounts[i]);
        }

        int[] mins = new int[minIndexes.length];
        for (int i = 0; i < mins.length; i++) {
            mins[i] = diffCounts[i][minIndexes[i]];
        }

        int indexOfMax = indexOfMax(mins);
        int max = diffCounts[indexOfMax][minIndexes[indexOfMax]];

        System.out.println("number of permutations:" + permutations.size());
        System.out.println("max of min diffs:" + max);
        System.out.println(permutationToString(solutions[minIndexes[indexOfMax]]));
        System.out.println(permutationToString(permutations.get(indexOfMax)));
        System.out.println(diffToString(diffs[indexOfMax][minIndexes[indexOfMax]], permutations.get(indexOfMax)));
    }

    private static int indexOfMax(int[] mins) {
        int maxIndex = 0;
        for (int i = 0; i < mins.length; i++) {
            if(mins[i] > mins[maxIndex]) {
                maxIndex = i;
            }
        }
        return maxIndex;
    }

    private static int[] diffCount(int[][] diffs) {
        int[] diffCount = new int[diffs.length];
        for (int j = 0; j < diffs.length; j++) {
            diffCount[j] = diffCount(diffs[j]);
        }
        return diffCount;
    }

    private static int[][] diffs(int[] test, int[][] solutions) {
        int[][] diffs = new int[solutions.length][];
        for (int i = 0; i < solutions.length; i++) {
            diffs[i] = sub(solutions[i], test);
        }
        return diffs;
    }

    private static int indexOfMin(int[] diffCounts) {
        int indexOfMin = 0;
        for (int j = 1; j < diffCounts.length; j++) {
            if(diffCounts[j] < diffCounts[indexOfMin]) {
                indexOfMin = j;
            }
        }
        return indexOfMin;
    }

    private static String toStringRaw(int[] test) {
        final StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < test.length; i++) {
            stringBuilder.append(i == 0 ? '{' : ',');
            stringBuilder.append(String.format("% d", test[i]));
        }
        stringBuilder.append('}');
        return stringBuilder.toString();
    }

    private static int diffCount(int[] ints) {
        int ret = 0;
        for (int i = 0; i < ints.length; i++) {
            if(ints[i] != 0) {
                ret++;
            }
        }
        return ret;
    }

    private static int[] sub(int[] ai0, int[] ai1) {
        int[] ret = new int[15];
        for (int i = 0; i < 15; i++) {
            ret[i] = ai0[i] - ai1[i];
        }
        return ret;
    }

    private static List<int[]> permutations(int offset, int[] ints) {
        final List<int[]> ret = new ArrayList<>();
        if(offset == ints.length - 1) {
            ret.add(ints);
        } else {
            List<int[]> p = permutations(offset + 1, ints);
            for (final int[] ints1 : p) {
                for (int i = offset; i < ints.length; i++) {
                    final int[] ints2 = swap(ints1, offset, i);
                    if(!contains(ret, ints2)) {
                        ret.add(ints2);
                    }
                }
            }
        }
        return ret;
    }

    private static boolean contains(List<int[]> list, int[] ints2) {
        boolean ret = false;
        for (int[] ints : list) {
            ret |= equals(ints, ints2);
        }
        return ret;
    }

    private static boolean equals(int[] ints, int[] ints2) {
        boolean ret;
        if(ints.length != ints2.length) {
            ret = false;
        } else {
            ret = true;
            for (int i = 0; (i <ints.length) && ret; i++) {
                ret &= ints[i] == ints2[i];
            }
        }
        return ret;
    }

    private static int[] swap(int[] ints, int i, int j) {
        int[] ints1 = new int[ints.length];
        System.arraycopy(ints, 0, ints1, 0, ints.length);
        ints1[i] = ints[j];
        ints1[j] = ints[i];
        return ints1;
    }

    private static String permutationToString(int[] ints) {
        char[] chars = {'8', 'L', 'H'};
        final StringBuilder sb = new StringBuilder();
        int i = 0;
        for(int j=1; j <= 5; j++) {
            for(int k=0; k < (5-j); k++) {
                sb.append(' ');
            }
            for(int k=0; k < j; k++) {
                sb.append(chars[ints[i]]);
                sb.append(' ');
                i++;
            }
            sb.append('\n');
        }
        return sb.toString();
    }

    private static String diffToString(int[] diff, int[] permutation) {
        char[] chars = {'8', 'L', 'H'};
        final StringBuilder sb = new StringBuilder();
        int i = 0;
        for(int j=1; j <= 5; j++) {
            for(int k=0; k < (5-j); k++) {
                sb.append(' ');
            }
            for(int k=0; k < j; k++) {
                sb.append(diff[i] != 0 ? chars[permutation[i]] : '.');
                sb.append(' ');
                i++;
            }
            sb.append('\n');
        }
        return sb.toString();
    }
}

<span class="math-container">```
```

Body:
Well I turns out the answer is 4.

Since the maximum number of balls that can be in the wrong place is 8.

I brute forced solution by generating all 51480 permutations and compared each against the 12 solutions.

It would be trivial to find out distributions of distances to a solution, if anyone was interested.

Below is an example of a worst case scenario:

Nearest solution:
    L
   H L
  L 8 H
 H L H L
L H L H H

Permutation:
    H
   L H
  8 L L
 H L H L
H L L H H

Diff:
    H
   L H
  8 L L
 . . . .
H L . . .

Below javacode (takes about 10 mins to run on a decent laptop):

import java.util.ArrayList;
import java.util.List;

public class Main {
    private static final int B = 0;
    private static final int L = 1;
    private static final int H = 2;
    public static void main(String[] args) {
        int[][] solutions = {
                {
                         L,
                        H,L,
                       L,B,H,
                      H,L,H,L,
                     L,H,L,H,H
                },
                {L, L, H, H, B, L, L, H, L, H, H, H, L, H, L},
                {H, L, H, H, B, L, L, H, L, H, H, L, H, L, L},
                {H, H, L, L, B, H, H, L, H, L, L, L, H, L, H},
                {L, H, H, L, L, L, H, H, B, H, H, L, H, L, L},
                {H, H, L, L, H, H, H, L, B, L, L, H, L, H, L},
                {H, L, L, H, H, H, L, L, B, L, L, H, L, H, H},
                {L, L, H, H, L, L, L, H, B, H, H, L, H, L, H},
                {H, L, H, H, H, L, L, B, L, H, L, H, L, H, L},
                {L, H, H, L, L, L, H, B, H, H, L, L, H, L, H},
                {L, H, L, L, L, H, H, B, H, L, H, L, H, L, H},
                {H, L, L, H, H, H, L, B, L, L, H, H, L, H, L}
        };

        final List<int[]> permutations = permutations(0, solutions[0]);
        int[][][] diffs = new int[permutations.size()][][];
        for (int i = 0; i < diffs.length; i++) {
            diffs[i] = diffs(permutations.get(i), solutions);
        }

        int[][] diffCounts = new int[diffs.length][];
        for (int i = 0; i < diffs.length; i++) {
            diffCounts[i] = diffCount(diffs[i]);
        }

        int[] minIndexes = new int[diffCounts.length];
        for (int i = 0; i < diffCounts.length; i++) {
            minIndexes[i] = indexOfMin(diffCounts[i]);
        }

        int[] mins = new int[minIndexes.length];
        for (int i = 0; i < mins.length; i++) {
            mins[i] = diffCounts[i][minIndexes[i]];
        }

        int indexOfMax = indexOfMax(mins);
        int max = diffCounts[indexOfMax][minIndexes[indexOfMax]];

        System.out.println("number of permutations:" + permutations.size());
        System.out.println("max of min diffs:" + max);
        System.out.println(permutationToString(solutions[minIndexes[indexOfMax]]));
        System.out.println(permutationToString(permutations.get(indexOfMax)));
        System.out.println(diffToString(diffs[indexOfMax][minIndexes[indexOfMax]], permutations.get(indexOfMax)));
    }

    private static int indexOfMax(int[] mins) {
        int maxIndex = 0;
        for (int i = 0; i < mins.length; i++) {
            if(mins[i] > mins[maxIndex]) {
                maxIndex = i;
            }
        }
        return maxIndex;
    }

    private static int[] diffCount(int[][] diffs) {
        int[] diffCount = new int[diffs.length];
        for (int j = 0; j < diffs.length; j++) {
            diffCount[j] = diffCount(diffs[j]);
        }
        return diffCount;
    }

    private static int[][] diffs(int[] test, int[][] solutions) {
        int[][] diffs = new int[solutions.length][];
        for (int i = 0; i < solutions.length; i++) {
            diffs[i] = sub(solutions[i], test);
        }
        return diffs;
    }

    private static int indexOfMin(int[] diffCounts) {
        int indexOfMin = 0;
        for (int j = 1; j < diffCounts.length; j++) {
            if(diffCounts[j] < diffCounts[indexOfMin]) {
                indexOfMin = j;
            }
        }
        return indexOfMin;
    }

    private static String toStringRaw(int[] test) {
        final StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < test.length; i++) {
            stringBuilder.append(i == 0 ? '{' : ',');
            stringBuilder.append(String.format("% d", test[i]));
        }
        stringBuilder.append('}');
        return stringBuilder.toString();
    }

    private static int diffCount(int[] ints) {
        int ret = 0;
        for (int i = 0; i < ints.length; i++) {
            if(ints[i] != 0) {
                ret++;
            }
        }
        return ret;
    }

    private static int[] sub(int[] ai0, int[] ai1) {
        int[] ret = new int[15];
        for (int i = 0; i < 15; i++) {
            ret[i] = ai0[i] - ai1[i];
        }
        return ret;
    }

    private static List<int[]> permutations(int offset, int[] ints) {
        final List<int[]> ret = new ArrayList<>();
        if(offset == ints.length - 1) {
            ret.add(ints);
        } else {
            List<int[]> p = permutations(offset + 1, ints);
            for (final int[] ints1 : p) {
                for (int i = offset; i < ints.length; i++) {
                    final int[] ints2 = swap(ints1, offset, i);
                    if(!contains(ret, ints2)) {
                        ret.add(ints2);
                    }
                }
            }
        }
        return ret;
    }

    private static boolean contains(List<int[]> list, int[] ints2) {
        boolean ret = false;
        for (int[] ints : list) {
            ret |= equals(ints, ints2);
        }
        return ret;
    }

    private static boolean equals(int[] ints, int[] ints2) {
        boolean ret;
        if(ints.length != ints2.length) {
            ret = false;
        } else {
            ret = true;
            for (int i = 0; (i <ints.length) && ret; i++) {
                ret &= ints[i] == ints2[i];
            }
        }
        return ret;
    }

    private static int[] swap(int[] ints, int i, int j) {
        int[] ints1 = new int[ints.length];
        System.arraycopy(ints, 0, ints1, 0, ints.length);
        ints1[i] = ints[j];
        ints1[j] = ints[i];
        return ints1;
    }

    private static String permutationToString(int[] ints) {
        char[] chars = {'8', 'L', 'H'};
        final StringBuilder sb = new StringBuilder();
        int i = 0;
        for(int j=1; j <= 5; j++) {
            for(int k=0; k < (5-j); k++) {
                sb.append(' ');
            }
            for(int k=0; k < j; k++) {
                sb.append(chars[ints[i]]);
                sb.append(' ');
                i++;
            }
            sb.append('\n');
        }
        return sb.toString();
    }

    private static String diffToString(int[] diff, int[] permutation) {
        char[] chars = {'8', 'L', 'H'};
        final StringBuilder sb = new StringBuilder();
        int i = 0;
        for(int j=1; j <= 5; j++) {
            for(int k=0; k < (5-j); k++) {
                sb.append(' ');
            }
            for(int k=0; k < j; k++) {
                sb.append(diff[i] != 0 ? chars[permutation[i]] : '.');
                sb.append(' ');
                i++;
            }
            sb.append('\n');
        }
        return sb.toString();
    }
}

```


问:
Example of zero diameter on a metric space
I want to find an example of an open ball in a metric with $diam(A)=0$ ,$A=B(x_{0},r)$

my example , which i am not sure if it is correct:

$(\mathbb{R},\left | \cdot  \right |) $  and $A=B(\frac{1}{2},\frac{1}{n})$

then with a triangle inequality we can easily see that $diam(A)\leq \frac{2}{n}$ if $n \to \infty  \Rightarrow  diam(A)\leq 0  \Rightarrow  diam(A)=0 $

does this example do the job ?, is there a more interesting one?

答: Consider $\mathbb{N}$ with the usual metric defined by $d(n,m)=|n-m|$. Then the open ball $B(1,1)$ of center $1$ and of radius $1$ is equal to $B(1,1)=\lbrace 1 \rbrace$, so it has diameter $0$.


问:
Ambient isotopy on closed unit ball
Consider a point $a\in\mathbb{R}^n$. $F(x,t)=x-ta$ is an ambient isotopy from $\{a\}$ to $\{0\}$. By the homeomorphism $f:\mathbb{R}^n\rightarrow B^n:f(x)=\frac{x}{1+\parallel x\parallel}$ (where $B^n$ is the unit open ball), we can extend the $F$ to an ambient isotopy $G:D^n\times [0,1]\rightarrow D^n$ taking $\{f(a)\}$ to $0$, relative to $S^{n-1}$, where $D^n$ is the closed unit ball and $S^{n-1}$ is the sphere.

I define $G$ as $G(x,t)=f(f^{-1}(x)-ta)$ $\forall x\in B^n,t\in[0,1]$, and $G(x,t)=x$ $\forall x\in S^{n-1}$. But now I have the problem proving this $G$ is continuous.

Any hint please?

答:
Let us first observe that $f^{-1} : B^n \to \mathbb R^n$ is given by $f^{-1}(x)  = \frac{x}{1-\lVert x \rVert}$.

$G$ is continuous on the open set $B^n \times I$ and on the closed set $S^{n-1} \times I$. Thus it remains to show that if $(x_n,t_n)$ is a sequence in $B^n \times I$ which converges to $(x,t) \in S^{n-1} \times I$ (i.e. $x_n \to x$ and $t_n \to t$), then $G(x_n,t_n) \to G(x,t) = x$. W.l.o.g. we may assume that all $x_n \ne 0$.

Let $r_n = \lVert x_n \rVert$. Thus $r_n \to 1$ and $x_n - (1-r_n)t_na \to x$. We have
$$f^{-1}(x_n) - t_na = \frac{x_n}{1-r_n} - t_na = \frac{x_n - (1-r_n)t_na}{1-r_n} ,$$
$$G(x_n,t_n) = \frac{x_n - (1-r_n)t_na}{1-r_n} / \left(1 + \left\lVert \frac{x_n - (1- r_n)t_na}{1-r_n} \right\rVert \right) = \frac{x_n - (1-r_n)t_na}{1- r_n + \lVert x_n - (1-r_n)t_na \rVert} \to \frac{x}{0 + \lVert x \rVert} = x $$

Title: Ambient isotopy on closed unit ball


问:
Laplace transform $f(t) = f(t + \frac{2\pi }{a}) = \frac{\sin at}{|\sin at|}$
I had done the laplace transform with the period fucntion (T): $F(s) = \frac{1}{1 - e^{ - sT}}\int_o^T {e^{ - st}}f(t) \, dt $
Howerver, in this fuction:

$$f(t) = f \left(t + \frac{2\pi}{a}\right) = \frac{\sin at}{|\sin at|}$$

What is the period? And You can give me detail of solution

Body:
I did by let u = at, so the period T = 2pi

$\begin{array}{l}
 f(t) = f(t + \frac{{2\pi }}{a}) = \frac{{\sin at}}{{|\sin at|}} \\
 Let:f(u) = (u + 2\pi ) = \frac{{\sin u}}{{|\sin u|}} \\
 F(s) = L{\rm{[f(u)]}} = \frac{1}{{1 - {e^{ - 2\pi s}}}}\int_o^{2\pi } {{e^{ - st}}\frac{{\sin {a^2}u}}{{|\sin {a^2}u|}}du}  \\
  = \frac{1}{{1 - {e^{ - 2\pi s}}}}{\rm{\{ }}\int_o^\pi  {{e^{ - st}}du}  - \int_\pi ^{2\pi } {{e^{ - st}}du} {\rm{\} }} \\
  = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\
 L{\rm{[f(u)]}} = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\
  =  > L{\rm{[f(at)] = }}\frac{1}{a}F(\frac{s}{a}) = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{a\frac{s}{a}({e^{\frac{{\pi s}}{a}}} + 1)}} = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{s({e^{\frac{{\pi s}}{a}}} + 1)}} =  Yours\\
 \end{array}$

So did i work corecttly??


问:
I need help to understand a proof that there is at most one operator $*$ satisfying $\eta\wedge*\zeta=\langle\eta,\zeta\rangle\omega$
This is from some lecture notes I found online (LMU Munich):

If $*_1,*_2$ are two operators satisfying the defining property of the Hodge star operator, then $\eta\wedge(*_1\zeta-*_2\zeta)=0$. If the second factor is non-zero for some $\zeta$,
then one can easily find $\eta$ so that the product is a mulitple of the volume form $\omega$.
This proves uniqueness.

According to the lecture notes, this is also valid for indefinite metrics.

Question: Can someone explain the bold sentence? What $\eta$ is the author talking about?

My thoughts:

As far as I understand, this is a proof by contradiction: Assuming $*_1\zeta\neq *_2\zeta$, we can find $\eta$ such that $\eta\wedge(*_1\zeta-*_2\zeta)\neq 0$, but because of $\eta\wedge*_1\zeta=\langle\omega,\eta\rangle\omega=\eta\wedge*_2\zeta$ we should have $\eta\wedge(*_1\zeta-*_2\zeta)=\eta\wedge*_1\zeta-\eta\wedge*_2\zeta=0$.

My first guess was $\eta=\zeta$, but this doesn't work. My second guess was $\eta=*_1\zeta-*_2\zeta$, but this didn't work either. (Or maybe they do work and I just don't get it.)

答:
Suppose $\alpha$ is a $k$-form and define the application $g_{\alpha} : \Lambda^{\dim M-k}T^*M \to \Lambda^{\dim M} T^*M$ defined by $f_{\alpha}(\beta) = \alpha\wedge \beta$.
Your question can be rewritten as: why $g_{\alpha}$ is not the null-application if $\alpha \neq 0$?

This is a local question: in local coordinates around $p\in M$, $\alpha = \sum_{i_1<\cdots < i_k} \alpha_{i_1,\ldots,i_k} \mathrm{d}x^{i_1}\wedge\cdots\wedge \mathrm{d}x^{i_k}$. If $\alpha$ is non-vanishing, then there exists $i_1<\ldots<i_k$ such that $\alpha_{i_1,\ldots,i_k}\neq 0$ on a neighbourhood of $p$. Without loss of generallity (by re-ordering the coordinates), suppose $(i_1,\ldots,i_k) = (1,\ldots,k)$. Then,
\begin{align}
\alpha\wedge \left(\mathrm{d}x^{k+1}\wedge \cdots \wedge \mathrm{d}x^{\dim M}\right) = \alpha_{1,\ldots,k}\mathrm{d}x^1\wedge\cdots\wedge \mathrm{d}x^{\dim M} \neq 0
\end{align}
We have then found a local $\beta$ such that $\alpha \wedge \beta$ is non vanishing in a neighbourhood of $p$. An argument of partition of unity show that $\beta$ can be defined globally (just extend it with $0$), and there exists $\beta$ such that $\alpha \wedge \beta$ is not the zero $\dim M$-form.

This shows that if $\eta \wedge (*_1\zeta - *_1\zeta) = 0$ for every $k$-form $\eta$, then $*_1\zeta - *_2\zeta$ the zero $(\dim M -k)$ form.

Comment: the fact that $M$ is orientable is not required to show this property, that is the uniqueness of the Hodge operator. But it is required to show its existence!

Edit: here is how to extend globally $\beta$. Suppose $\alpha$ is non zero at $p$. Suppose $U$ is an open subset containing $p$, and on which we have the above coordinates. Let $\psi : M \to \mathbb{R}$ be a smooth function with compact support, and with $\mathrm{supp}\psi \subset U$ and $\psi(p) = 1$. Then, the differential form $\beta$ defined to be $0$ outside of $U$ and $\psi \cdot \mathrm{d}x^{k+1}\wedge \cdots \wedge \mathrm{d}x^{\dim M}$ in $U$ (in the above coordinates) is well-defined, smooth, is non zero, and $\alpha \wedge \beta$ is a non-zero $\dim M$-form, because $\alpha\wedge\beta (p) \neq 0$.