问:
Smoothness of quotient of Holder continuous functions, provided the decay
Let $I=(-1,1)$ and $u \in \text{Lips}(I)$, the space of Lipschitz continuous functions in I. Suppose that
$$
|u(x)|  \le C |x|^\alpha
$$
for $x \in I$, for some $\alpha \in (1,2)$. I would like to estimate the Holder continuity of
$$
v(x)=
\begin{cases}
|x|^{-\beta}\cdot u(x) &\text{ for } x \neq 0
\\
0& \text{ for } x =0
\end{cases}
$$
for some $1<\beta<\alpha$ (so that $|x|^{\beta}$ is also in $\text{Lips}(I)$) in the whole interval $I$.

It is clear that $v$ is in $\text{Lips}(I \setminus (-\varepsilon,\varepsilon))$ (and therefore $C^{\alpha-\beta}(I\setminus (-\varepsilon,\varepsilon))$) for any $\varepsilon \in (0,1)$. and We also know that $v$ is "Holder continuous at $0$", that is, $|v(x)-v(0)| \le C |x|^{\alpha-\beta}$. Is it true that $v$ is Holder of order $\alpha - \beta$? The problem, of course, is to check if the Holder condition holds over sequences $\{x_n\}_n$ and $\{y_n\}_n$ such that $|x_n-y_n|\ll |y_n| \longrightarrow 0$.

I managed to prove that $v$ is in $C^{1-\beta/\alpha}(I)$. Without further hypothesis, is this the best Holder smoothness $v$ can have in $I$?

The choices of $\alpha$ and $\beta$ seemed to be a convenient regime to start, but the question could be done in a more general setting.

EDIT Let me put my proof of the Holder regularity $1- \beta/\alpha$.
Assum w.l.o.g that $|x|<|y|$. We have from hypothesis that
$$
|u(x)-u(y)| \le L |x-y|
$$
and for some constant $L>0$ and that
$$
|u(x)-u(y)| \le C (|x|^\alpha+|y|^\alpha) \le 2C |y|^\alpha.
$$
thefore, writing $|u(x)-u(y)|=|u(x)-u(y)|^{\delta}|u(x)-u(y)|^{1-\delta}$,
we have
$$
\tag{1}
|u(x)-u(y)| \le C^\prime |x-y|^{(1-\delta)}|y|^{\delta\alpha}.
$$

Now, we write
$$
\left | \frac{u(x)}{|x|^\beta}-\frac{u(y)}{|y|^\beta} \right|
=
\left | \frac{u(x)}{|x|^\beta} \frac{u(x)-u(y)}{|y|^\beta}-\frac{u(x)-u(y)}{|y|^\beta} \right|.
$$

Using the triangular inequality and $(1)$ and the fact that $|x|^\alpha$ is Lipchitz, we have
\begin{align*}
\left | \frac{u(x)}{|x|^\beta}-\frac{u(y)}{|y|^\beta} \right|
&
\le C^{\prime \prime} \left [
|x|^{\alpha-\beta} |x-y|
+|y|^{\delta\alpha-\beta}|x-y|^{1-\delta}
\right]
\end{align*}
We then choose $\delta = \beta/\alpha$, and we get that the RHS is bounded by $C^{\prime \prime}|x-y|^{1-\beta/\alpha}$.

答:
No it's not, in fact you found the sharp exponents! The example showcasing sharpness is a function which is 1-Lipschitz, has slope almost always -1 or 1, and oscillates between $|x|^{\alpha}$ and $-|x|^{\alpha}$. I don't think writing it explicitly will help much, so here's a picture instead:

Notice the dimensions in purple (which you can do at any crossing point $x_0$, of which there is infinitely many. Let's see what happens to them when we multiply by $|x|^{-\beta}$. Again, I think a picture does a better job:

That means that for points $x_0$ arbitrarily close to zero, we can find a point $x'$ (the point where $u$ meets with $|x|^{\alpha}$) such that $|x'-x_0|\sim x_0^\alpha$ and $|u(x')-u(x_0)|\sim x_0^{\alpha-\beta}$. From the first of those equalities we get that $x_0\sim |x'-x_0|^{1/\alpha}$. Plugging that into the second equation, we obtain the bound that shows sharpness:

$$|u(x')-u(x_0)|\sim |x'-x_0|^{1-\beta/\alpha}$$

and in particular, $u(x)$ is not $C^\lambda$ for any $\lambda>1-\beta/\alpha$.


问:
Find the matrix representation relative to the standard bases
Let $T : P_{2}(C) \to  C^3$ be a linear transformation such that
$$T(a + bx + cx^2) = (a, a + b, a + b + c)$$

Find the matrix representation $[T]$ relative to the standard bases.

If standard basis for $R^3$,

$T(1,0,0)=(1,0,0)$

$T(0,1,0)=(1,1,0)$

$T(0,0,1)=(1,1,1)$

Therefore,
$[T]$ = \begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}

But it is the same with $C^3$ ?

答:
First, we need to analyze two case:

$\mathbb{C}_{\mathbb{R}}$: Complex space over $\mathbb{R}$.

$\mathbb{C}_{\mathbb{C}}$: Complex space over $\mathbb{C}$.

Then, the answer dependes if in your problem $T:P_{2}(\mathbb{C}_{\mathbb{C}})\to \mathbb{C}_{\mathbb{C}}^{3}$ or $T:P_{2}(\mathbb{C}_{\mathbb{R}})\to \mathbb{C}_{\mathbb{R}}^{3}$.

Suppose for example the second case, also is well-know that $\mathbb{C}_{\mathbb{R}}\cong \mathbb{R}^{2}$ and $\mathbb{C}_{\mathbb{R}}^{3}\cong  \mathbb{R}^{6}$. (In general $\mathbb{C}_{\mathbb{R}}^{n} \cong \mathbb{R}^{2n}$). So, we have that $[T]_{\beta_{1}\to \beta_{2}}$ be a matrix of order $\dim(\mathbb{C}_{\mathbb{R}}^{3})\times \dim(P_{2}(\mathbb{C}_{\mathbb{R}}))$.

Now, in the first case:  $T:P_{2}(\mathbb{C}_{\mathbb{C}})\to \mathbb{C}_{\mathbb{C}}^{3}$, we have that your approach is correct. Indeed $$[T]=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{pmatrix}$$

ContentLicense: CC BY-SA 4.0


问:
Does the following identity involving limits and limsup/liminf hold?
Suppose I have three sequences $(a_n)_{n\ge1}$, $(b_n)_{n\ge1}$ and $(c_n)_{n\ge1}$ on the real line which satisfy:

$a_n = b_n + c_n$

The sequence $a_1,a_2,\ldots$ converges (in R), and all elements of the sequence $(c_n)_{n\ge1}$ are in a bounded subset of the real line (say [0,1]). Then perhaps it holds in general that:

$\lim a_n = \liminf b_n + \limsup c_n$

I haven't been able to track down this or similar results in my textbooks. I have a bit of a messy proof/intuition why this should hold, but I believe it cannot be that complicated. Is there a simple proof of the identity? Is there a counter-example?

答: Let $B_n=b_n-a_n$, $C_n=c_n$. Then $B_n=-C_n$, so $$\liminf B_n=-\limsup C_n.$$On the other hand, since $a_n$ is convergent it's clear that $$\liminf B_n=\liminf b_n-\lim a_n.$$


问:
Given that matrix $A$ is diagonalizable and has eigenvalues of $0$ or $1$, show that $A^2 = A$
Suppose that $A$ is a diagonalizable $n \times n$ matrix such that the characteristic polynomial of $A$ is $p(λ)=λ^k(1−λ)^{n−k}$, where $k$ is a positive integer such that $0≤k≤n$.

I want to prove $A^2=A$. I know that for this, I'll need to prove that for any eigenvector $v$, $A^2 v=Av$. I tried setting up my proof like this:

Since $A$ is diagonalizable, there exists an invertible matrix $P$ such that $A=P^{-1}DP$, where $D$ has all the eigenvalues of $A$ on its diagonal. But since every eigenvalue is either $0$ or $1$, $D= \lambda I$. From there, we get:
$$A = P^{-1}DP = P^{-1}(\lambda I_n)P = (\lambda I_n)(P^{-1}P) = \lambda I_n$$
$$ \implies A \times A = (\lambda I_n) \times (\lambda I_n) = \lambda ^2$$

All I've succeeded in doing here is showing $A^2 = \lambda ^2$. How can I show also that $A^2 = A$? Any guidance is greatly appreciated!

答: Since $A$ is a diagonalizable $n \times n$ matrix and has only eigenvalues $0$ and $1$ it Jordan matrix is like: $$J =\pmatrix{O_k&0\\0&I_{n-k}}\implies J^2 = J$$ where $O_k$ is $k\times k$ zero matrix and $I_{n-k}$ is $n-k\times n-k$ identity matrix. Thus your matrix has the same property.


问:
Calculate $\int \frac{1}{x-\sqrt{x}} \, dx$
I need to calculate $\displaystyle \int \frac{1}{x-\sqrt{x}} \, dx$. Any ideas? Integration by parts?</p>

答: Try the substitution $ x = u^2 $ to get $ \displaystyle \int \frac{2u}{u^2 - u} \ du $, which is easy to integrate. </p>

ContentLicense: CC BY-SA 3.0


问:
Showing results regarding Cayley Transform

Let $\Xi$ be the set of all real $n\times n$ which do not have $1$ as an eigenvalue. Consider the mapping

\begin{equation}\phi:\Xi\rightarrow\mathbb{R}^{n\times n},\quad \phi(A)=(A-I)^{-1}(A+I).\end{equation}

Prove that $\phi(A)\in\Xi$ and that $A$ is skew-symmetric if and only if $\phi(A)$ is orthogonal.

I have managed to show the later part of the question as follows:

$(\implies)$Let $A$ be skew-symmetric: \begin{align}\phi(A)\phi(A)^T&=(A-I)^{-1}(A+I)(-A+I)(-A-I)^{-1}\\
&=(A+I)(A-I)^{-1}(-A+I)(-A-I)^{-1}\\
&=-(A+I)(A-I)^{-1}(A-I)(-A-I)^{-1}\\
&=-(A+I)(-A-I)^{-1}\\
&=(-A-I)(-A-I)^{-1}=I\\\end{align}
Hence $\phi(A)$ is orthogonal.

$(\Longleftarrow)$ Let $\phi(A)$ be orthogonal:
\begin{align}\phi(A)\phi(A)^T&=I\\
(A-I)^{-1}(A+I)[(A-I)&^{-1}(A+I)]^T=I\\
(A-I)^{-1}(A+I)(A+I)&^{T}((A-I)^{-1})^T=I\\
(A-I)^{-1}(A+I)(A^T+&I)(A^T-I)^{-1}=I\\
(A+I)(A^T+I)(A^T&-I)^{-1}=(A-I)\\
(A+I)(A^T+I)=&(A-I)(A^T-I)\end{align}
We know that we require this last line to commute, and we can see that to have these commute we require that $A^T=-A,$ i.e. we require $A$ skew-symmetric:
\begin{align}(A+I)(-A+I)&=(A-I)(-A-I)\\
-(A+I)(A-I)&=-(A-I)(A+I)\\
(A+I)(A-I)&=(A-I)(A+I).\end{align}
Therefore $A$ is skew-symmetric $\iff$ $\phi(A)$ is orthogonal.

I am pretty confident that my $\implies$ is correct, but would like a sanity check on the proof for $\Longleftarrow$.

As for showing that $\phi(A)\in\Xi,$ I am unsure how to tackle this.

答:
We compute:
$$
\begin{aligned}
\phi(A)-I
&=(A-I)^{-1}(A+I)-I\\
&=(A-I)^{-1}[\ (A+I)-(A-I)]\\
&=2(A-I)^{-1}\ ,
\\[3mm]
&\qquad\text{ so}
\\
\det(\phi(A)-I)
&=2^n\det(\ (A-I)^{-1}\ )
\\
&\ne 0\ ,
\end{aligned}
$$
(in characteristic $\ne 2$, yes, we are working over reals,) so $1$ is not an eigenvalue.

The first implication is ok.

For the second implication ($\Leftarrow$), just open the parenteses in $(A+I)(A^T+I)=(A-I)(A^T-I)$ when reaching this point to get $2(A+A^T)=0$.


问:
To show the existence of $f,g\in\mathcal{L}(V,\mathbb{F})$ such that $f(x)\neq g(x) \forall x\in S\subseteq V$

Let $V$ be a finite-dimensional vector space over $\mathbb{F}$. Show that, given any $\emptyset = S ⊆ V \backslash\{0\}$, there exist nonzero linear functionals $f$ and $g$ on $V$ such that, $∀x ∈ S$, $f(x) \neq g(x)$.

I've tried several things here:

First, I came up with a vague argument that requires some gap-filling. Consider two non-zero functionals $f,g\in\mathcal{L}(V,\mathbb{F})$ such that $f(x)=g(x)\neq 0$, for all $x\in S'\subseteq S$. We keep $f$ as it is, and construct a new functional $\alpha g$ from $g$ by multiplying by a suitable scalar $\alpha$. We do this as follows:
Let $y\in S'$, so $f(y) =  g(y) \neq 0$. Then $f(y) \neq \alpha g(y)$ for some choice of scalar $\alpha$. We change $g$ iteratively so that eventually we "cover" all points in $S'$, and end up with two functionals $f$ and $\alpha_0 g$, so that $f(x) \neq \alpha_0g(x)$ $\forall x\in S$.

I have assumed that $f(x) = g(x)\neq 0$ in several places, but it may be too much to ask that none of the functionals go to zero on any point in $S$. I believe this is what's to be shown anyway - that we can find such functionals! How do I do that?

Also, I tried an inductive argument since $V$ is finite-dimensional (on the dimension of $V$), but that didn't go anywhere.

I'd appreciate any hints, thanks!

答:
Induction on $\dim V.$ If $\dim V=1$ then $V=\{av:a\in \mathbb F\}.$ Set $f(v)=1$ and $g(v)=1+1.$

Suppose the claim is true for all $(V,\mathbb F)$ such that $\dim V=n.$

Suppose now that $\dim V=n+1$ and consider $\mathscr S=\operatorname{span}\ S.$ If $\dim \mathscr S<n+1$ then there are functionals $f,g:\mathscr S\to \mathbb F$ such that $f\neq g$ on $S$. Extend these to $\tilde f,\tilde g$ on $V$ by defining them to be $0$ on $\mathscr S^{\perp}.$

If $\dim \mathscr S=n+1$ then choose a basis $\{v_1,\cdots, v_{n+1}\}$ for $\mathscr S.$ Each $s\in S$ can be written $s=a_1v_1+\cdots +a_{n+1}v_{n+1}.$ There are $f,g: \mathscr S'\to \mathbb F$ where  $\mathscr S'=\{s\in \mathscr S: a_{n+1}=0 \}$, that satisfy $f\neq g$ on $\mathscr S'\cap S.$ Extend $f$ by zero to all of $\mathscr S$ and $g$ to $\mathscr S$ by letting $g(v_{n+1})=1.$

The induction is complete and the claim follows.


问:
Predicate logic difficulty
If $∀x∃y P(x, y)$ is true, can the following be true?

$∀x∀yP(x, y)$
or

$∃x∀yP(x, y)$
or

$∃x∃yP(x, y) $

I understand the order of $X$ and $Y$ matters when the quantifiers are different but can the quantifier rather than the variable change, and can a 'there exists' be upgraded to a 'for all'  or vice versa as in the first and third question?

答:
None of these are implied by the given statement.

$\forall x\exists y P(x,y)\not\to\forall x\forall y P(x,y)$: Let $x,y\in\mathbb N$, and let $P(x,y)$ be the equality relation.

$\forall x\exists y P(x,y)\not\to\exists x\forall y P(x,y)$: Same example as $1$.

$\forall x\exists y P(x,y)\not\to\exists x\exists y P(x,y)$: Let $x,y\in\varnothing$, and $P(x,y)$ be the false relation. $\forall x\exists y P(x,y)$ is true since the statement is true for all $x$ (since there are no $x$). However, there does not exist an $x,y$ such that $P(x,y)$, so that is false.

Title: Predicate logic difficulty


答:
Let $x = t^{2}$

Then $$I=\int \dfrac{dx}{x-\sqrt{x}}=\int \frac{2tdt}{t^{2}-t}=2\int\frac{dt}{t-1}=2ln|t-1|$$

Title: Calculate $\int \frac{1}{x-\sqrt{x}} \, dx$

Body: I need to calculate $\displaystyle \int \frac{1}{x-\sqrt{x}} \, dx$. Any ideas? Integration by parts?</p>


问:
What's the maximum speed that an object can reach given starting speed, distance and acceleration?
I'm new here and using this for my coding project, so please give answers in a simple form if possible!

I have a train. Let's say its moving at 50kph ($s$). The next speed board/marker is for 80kph ($v$) and it is 100 meters ($d$) away. My train accelerates/decelerates at a constant speed of 0.5m/s ($a$). If it starts accelerating now, what's the maximum speed ($m$) it will reach at the speed board?

Any help is greatly appreciated :)
Thanks!

答: Use the equation $v^{2}=u^{2}+2as$ where $v$ is the final (maximum in this case) velocity, $u$ is the initial velocity, and $a$ is the accelaration. This gives  $$v^{2}=13.8^{2}+2(0.5)(100)=292.89$$ (after converting km/h to m/s) and therefore $$v = (292.89)^{1/2} = 17.11~\text{m/s} = 61.6~\text{km/h}$$


答:
$(\mathbb Z/17\mathbb Z)^\times$ is a cyclic group with $16$ elements.

Once you know that $3$ is a generator, the generators are $3^n$ with $\gcd(n,16)=1$,

i.e., the odd powers of $3$.

(You already calculated all the powers of $3$, so you know what they are.)

Title: Generators of multiplicative group of integers modulo a prime number $(\mathbb{Z}/p\mathbb{Z})^\times$

Body:
The problem is

Find all generators of $(\mathbb{Z}/17\mathbb{Z})^\times$.

My attempt:

Since $\varphi(17)=16$, we can check whether powers of integers from $2$ to $16$ modulo $17$ are cyclic of order $16$.

(mod $17$ are omitted)

$2^0=1, 2^1=2, 2^3=8, 2^4=16, 2^5=15, 2^6=13, 2^7=9, 2^8=1$,

so the order is not $16$, $2$ is not a generator. Since $2$ generates $4, 8, 16$, they are not generators either.

$3^0=1, 3^1=3, 3^2=9, 3^3=10, 3^4=13, 3^5=5, 3^6=15, 3^7=11, 3^8=16, 3^9=14, 3^{10}=8, 3^{11}=7, 3^{12}=4, 3^{13}=12, 3^{14}=2, 3^{15}=6, 3^{16}=1,$

so the order is $16$, $3$ is a generator.

$5^0=1, 5^1=5, 5^2=8, 5^3=6, 5^4=13, 5^5=14, 5^6=2, 5^7=10, 5^8=16, 5^9=12, 5^10=9, 5^{11}=11, 5^{12}=4, 5^{13}=20, 5^{14}=15, 5^{15}=7, 5^{16}=1,$

so $5$ is a generator.

$6^0=1, 6^1=6, 6^2=2, 6^3=12, 6^4=4, 6^5=7, 6^6=8, 6^7=14, 6^8=16, 6^9=11, 6^{10}=15, 6^{11}=5, 6^{12}=13, 6^{13}=10, 6^{14}=9, 6^{15}=3, 6^{16}=1,$

so $6$ is a generator.

…

Question:

Is there an easier way to do this problem?


答:
Let $x = t^2$ and $dx = 2tdt$

$\displaystyle \int\frac{1}{x-\sqrt{x}}dx = \int\frac{2t}{t^2-t}dt$
$ = \displaystyle \int\frac{2t-2+2}{t^2-t}dt = 2\int\frac{1}{t}dt+2\int\frac{1}{t.(t-1)}dt$

$ = \displaystyle 2\ln \left|t\right|+2\int\frac{1}{t-1}dt-2\int\frac{1}{t}dt$

$ = \displaystyle 2\ln \left|t\right|+2\ln \left|t-1\right|-2\ln \mid t \mid+C$

$ = \displaystyle 2\ln \mid \sqrt{x}-1\mid+C$


问:
Attracting or parabolic cycles other than fixed points
I am studying the following complex polynomial
$$P(z) = \frac{2z^{4}-2z^{3}+2z^2-z}{2z^{3}-2z^{2}+3z-2}$$
and I would like to know, if there are attracting or parabolic cycles for $P(z)$ different from the attracting fixed points $0,1$ and $\infty$. By numerical computations one can see that the critical points can be associated with those fixed points. Does that mean that there are no such cycles? Or how would I find them?

Also, how one could determine the qualitative local nature of the fixed point at infinity? For example with a local phase diagram. How can I draw such thing at infinity?

答:
The direct answer to your first question, as originally answered in the comments is no - any attracting or parabolic cycle must contain a critical point in its immediate basin. This is stated as theorems 9.3.1 and 9.3.2 of Beardon's Iteration of Rational Functions.

Your function has six critical points, which you find by solving $P'(z)=0$. You can determine the forward orbit of these easily enough by direct computation:

$ 1 \to 1$ (the super-attractive fixed point)

$0.477 \to 0$ (the attractive fixed point with multiplier $1/2$)

$ 0.154441 \pm 0.590415 i \to 0$   (the attractive fixed point with multiplier $1/2$ again)

$ 0.106856 \pm 1.67351 i \to \infty$  (the neutral fixed point with multiplier $1$)

The dynamical picture, with the fixed points marked red and critical points marked green looks like so:

Your second question, originally asked in the comments, is "how do we determine the qualitative local nature of the fixed point at infinity? how can I draw such thing at infinity?" The trick there is to conjugate the function $P$ by the reciprocal function $\varphi(z) = 1/z$. That is, we investigate
$$p(z) = 1/P(1/z)  = \frac{2 z^4-3 z^3+2 z^2-2 z}{z^3-2 z^2+2 z-2}.$$
Note that  $\varphi$  swaps zero  and  infinity. Thus, where $\infty$ is a fixed point of $P$, we have that zero is a fixed   point  of  $p$. Furthermore, the zero is  a neutral fixed point  of $p$ since $p'(0)=1$. Thus, we say that  infinity is a neutral fixed  point of $P$.  We can also draw the Julia set of $p$ to have something like a picture of the Julia set of $P$ near infinity:

Better yet, we could draw the image on the Riemann sphere:

Note that there is a higher quality  interactive animation with the  implementation in this Observable notebook.


问:
Show that there is a $ x \in [0,1]$ that, $f(x) = g(x)$
Let $f, g: [0, 1] → [0, 1]$ be continuous functions such that $f(0)\leqslant g(0)$ and $f(1) \geqslant g(1)$. Show that there is $x ∈ [0, 1]$ such that $f(x) = g(x)$.

Is it enought to draw a grap? Or i need to show something more?

答:
If you are familiar with it, you should use the Intermediate value theorem :

Define the function:

$$h:[0,1]\to\mathbb{R}$$

as:
$$h(x)=g(x)-f(x)$$

Now, note that $h(0)=g(x)-f(0)\geq 0$ and $h(1)=g(1)-f(1)\leq 0$. Now the theorem states that there exists some point $x_0\in[0,1]$ that satisfies $h(x_0)=0$. By the definition of $h$ we got that:

$$g(x_0)-f(x_0)=h(x_0)=0 \\ \Downarrow \\ g(x_0)=f(x_0)$$

And that's what we wanted to show.


答:
Since $f(0) \le g(0), \ \ f(1) \ge g(1)$, we define
$$h(x) = f(x) - g(x) $$
Then, we have $h(0) \le 0$ and $h(1) \ge 0$. Since $h$ is a continuous function, there exists a zero in $[0,1]$; that is, $h(c) = f(c) - g(c) = 0$ for some $c\in [0,1]$ as desired

Title: Show that there is a $ x \in [0,1]$ that, $f(x) = g(x)$

Body:
Let $f, g: [0, 1] → [0, 1]$ be continuous functions such that $f(0)\leqslant g(0)$ and $f(1) \geqslant g(1)$. Show that there is $x ∈ [0, 1]$ such that $f(x) = g(x)$.

Is it enought to draw a grap? Or i need to show something more?


答: By the assumptions, $h(x)\equiv f(x)-g(x)$ is continuous, $h(0)\leqslant0$ and $h(1)\geqslant0$. By the intermediate value theorem there exist some $x_0$ for which $h(x_0)=0$, which implies $f(x_0)=g(x_0)$.


问:
If the line at infinity is a secant line of a conic then the conic is a hyperbola?
I'm learning perspective geometry and I have a question about the line at infinity. Is it true that if the conic has the line at infinity as its secant then the conic represents a hyperbola in Euclidean geometry?

If it is, how can I imagine the shape hyperbola using the line at infinity? I think I can lift up the line to infinity but it seems like a parabola than a hyperbola.

I draw a conic going through $A, B, C, D, E$, and I choose AB as the line at infinity. Then I lift it up and it seems like a parabola

P.S: I'm quite new to this field so I may use the wrong jargon. Thank you

答:
Imagine three dimensional Euclidean space with the origin at $\,O\,$ and a
plane $\,P\,$ not containing $\,O.\,$ The projective plane associated with
the space is all of the lines that pass through $\,O.\,$ Each of these lines
intersects the plane $\,P\,$ except those that lie in the plane $\,Q\,$ that
passes through the origin $\,O\,$ and is parallel to $\,P.\,$ Thus, the points
in $\,P\,$ represent ordinary projective points. Pick a circle $\,L\,$ with
center $\,O\,$ that lies in $\,Q.\,$ Pairs of antipodal points in $\,L\,$ are
the points in the line at infinity.

Now pick a double cone $\,D\,$ with center at $\,O.\,$ The intersection of
$\,D\,$ with plane $\,P\,$ is a conic section $\,C.\,$ The number of points
at infinity of $\,D\,$ is at most two. The case with two points at infinity
are those where $\,C\,$ is a hyperbola because two lines of $\,D\,$ pass
through $\,O\,$ and also intersect $\,L\,$ in two points at infinity. Note
that a Euclidean hyperbola has two connected open real components because
they omit the two points at infinity that connect them. Similarly a Euclidean
parabola has one open real component and is missing one point at infinity.


问:
Basis of fundamental cycles
In graph theory, for a connected, undirected finite graph one can easily find a set of fundamental cycles - I understand that part. This forms a basis of some linear space of all cycles of that graph, meaning any such cycle is a linear combination of these fundamental cycles. What I don't understand is how one defines such linear space, and how to prove last statement. Can anyone direct me to a book or article which would explain this - if there is a simple way of spelling it out here, that would be even better.</p>

答:
To define the cycle space, we first define the edge space of a graph $G$. This can be done in two equivalent ways:

As the set of all spanning subgraphs of $G$ (including all vertices, though possibly with degree $0$). Two subgraphs $H_1, H_2$ can be "added" by taking the symmetric difference of their edge sets: the subgraph $H_1 \mathbin{\Delta} H_2$ has an edge iff exactly one of $H_1$ or $H_2$ had it.

As the set of all functions from $E(G)$ to $\mathbb F_2$ (the field with two elements). This is a vector space over $\mathbb F_2$; scalar multiplication isn't very interesting over this field, and vector addition is just adding two functions pointwise.

Each cycle (together with all vertices not part of it, if necessary) gives us a spanning subgraph of $G$, so it's an element of the edge space. Thinking of it in the second way, given a cycle $C$, we can define its indicator function $f : E(G) \to \mathbb F_2$ by setting $f(e) = 1$ if $e$ is an edge of $C$, and $f(e)=0$ otherwise.

The set of all elements of the edge space which can be obtained by adding cycles is the cycle space. This has an alternate characterization: it is the set of all spanning subgraphs in which every vertex has even degree.

A cycle basis is a minimal set of cycles which generates the cycle space. The fundamental cycle basis is obtained from a spanning tree $T$ as follows: for each edge $e \notin E(T)$, we take the cycle $C_e$ by finding the unique path in $T$ between the endpoints of $e$, and then combining that with $e$ itself to create a cycle.

Your question is, as I understand it: why is the fundamental cycle basis a cycle basis?

Since it consists of cycles, it generates a subspace of the cycle space. It's linearly independent because each $C_e$ is the only element of the fundamental cycle space that includes edge $e$. (Linear independence here means that there's no nontrivial sum that adds to the zero function: no nontrivial symmetric difference that gives the empty graph. That's true here because the symmetric difference of distinct fundamental cycles $C_e \mathbin{\Delta} C_{e'} \mathbin{\Delta} C_{e''} \mathbin{\Delta} \cdots$ will include the edges $e, e', e'', \dots$: it won't be empty.)

The last thing we need from a basis is that it generates every element of the cycle space. So let $H$ be an element of the cycle space.

We begin by taking
$$H \mathbin{\Delta} C_{e_1} \mathbin{\Delta} C_{e_2} \mathbin{\Delta} \cdots \mathbin{\Delta} C_{e_k}$$ where $e_1, e_2, \dots, e_k$ are all the edges of $H$ that are not in $T$. The resulting graph is still in the cycle space, because everything in this expression was. It is also a subgraph of $T$, because every edge $e_i \notin E(T)$ was deleted by taking the symmetric difference with $C_{e_i}$.

Any nontrivial element of the cycle space contains a cycle (start at any vertex with positive degree, and keep walking around until you get back to a vertex you've seen before - you can't get stuck because every vertex you visit has degree at least $2$). But this element of the cycle space can't contain a cycle: it's a subgraph of a tree! So it must be trivial, and therefore $$H = C_{e_1} \mathbin{\Delta} C_{e_2} \mathbin{\Delta} \cdots \mathbin{\Delta} C_{e_k}.$$ We've shown that any element of the cycle space can be expressed in terms of the fundamental cycle basis!

Title: Basis of fundamental cycles


问:
Last three digits of tetration
Find the last three digits of the number:

$7^{7^{7^7...}}$

where there are 1001 sevens.

I know how to do it for when there are 4 and 5 sevens. I get an answer of 343. But how do I find it for such a large number of sevens?

答:
Render $7^4=2401=2400+1$.  Raise to the fifth power with the Binomial Theorem, thus:

$7^{20}=(2400+1)^{5}=(\text{ a multiple of }2400^2)+(5×2400)+1$

Both the "multiple of $2400^2$" and $5×2400=12000$ are obviously multiples of $1000$, so $7^{20}\equiv1\bmod 1000$ forcing $7^n\equiv343$ whenever $n$ is three more than a multiple of $20$.

Title: Last three digits of tetration


问:
Show that $\text{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ (morphism construction)
I'm curious how we can construct a morphism $\text{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \rightarrow S_3$ and show then that it is an isomorphism.

I know that $(12),(23)$ generate $S_3$ and that $([a]_2,[b]_2)$ with $a,b \in \mathbb{Z}/2\mathbb{Z}$ generate $\mathbb{Z}/2\mathbb{Z}$. But, I don't really see how we can construct it.

If you could give me some ''hints'' I would appreciate it.

Thank you in advance

答:
Automorphisms are bijections (of a group into itself) which preserve the order of the elements. Now $\Bbb Z_2\times \Bbb Z_2\cong V_4=\{e,a,b,ab\}$, where all the elements have order $2$. Therefore, an automorphism of $V_4$ maps:

$e$ to $e$;

$a$ to any among $a,b,ab$;

$b$ to any between the two elements left available to by previous choice;

$ab$ to the only element left available by the previous two choices.

Therefore, the elements of $\operatorname{Aut}(V_4)$ actually permutes $3$ objects, just like the elements of $S_3$ do (by definition), and you can map any of the former to the "corresponding" element of $S_3$. Formally, define a bijection $f\colon \{a,b,ab\} \to \{1,2,3\}$. For example, set: $f(a)=2, f(b)=1, f(ab)=3$; then, take $\alpha\in \operatorname{Aut}(V_4)$ defined e.g. by (see the bullet points bove): $\alpha(a)=ab, \alpha(b)=b, \alpha(ab)=a$. Now, let the bijection $f$ to come into play:

\begin{alignat}{4}
&\alpha(a)=ab &&\iff f(\alpha(a))=f(ab)=3 &&&\iff f\alpha(f^{-1}(2))=3 &&&&\iff (f\alpha f^{-1})(2)=3 \\
&\alpha(b)=b &&\iff f(\alpha(b))=f(b)=1 &&&\iff f\alpha(f^{-1}(1))=1 &&&&\iff (f\alpha f^{-1})(1)=1 \\
&\alpha(ab)=a &&\iff f(\alpha(ab))=f(a)=2 &&&\iff f\alpha(f^{-1}(3))=2 &&&&\iff (f\alpha f^{-1})(3)=2 \\
\end{alignat}

Therefore, within this framework, the sought morphism maps $\alpha\mapsto (23)$, and similarly for the others elements of $\operatorname{Aut}(V_4)$.


问:
How can I estimate the number of ads a snack bar owner needs to use along a road to achieve 99/100 of driver's attention
I snack bar owner performed a research and noticed that with one ad on the road he achieved 1/2 of driver's attention (the ad was seen). He was unsatisfied he wanted a value above 99/100  how many more ads he must use at minimum?

```<code>a) 99
b) 51
c) 50
d) 6
e) 1
</code>```

I don't even now how to solve it. I promese I tried but probability has no middleground or you know how to approach it or you just stand still. How can I approach a problem like this?

I corrected the question it is not how many but how many more ads he must use.

答: After placing $n$ ads on the road, $\frac1{2^n}$ of drivers have not seen at least one ad, assuming that drivers see each ad with equal probability. This falls below the target value of $\frac1{100}$ at seven ads, so you should put $6$ as the answer.


问:
Is my proof to this number theory question valid?
I had the following question:

Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?

$P(x)$ has no rational root;

For every positive integer $n$, there exist an integer $m$ such that $n$ divides $P(m)$.

I created a proof showing that there was no polynomial satisfying both of these conditions:

Suppose that we have a nonzero polynomial with integer coefficients $P(x)=\sum_i c_i x^i$ without a rational root, and for all positive integer $n$, we have an integer $m_n$ such that $n|P(m_n)$. This would imply $P(m_n)\equiv0\pmod n\Rightarrow \sum_i c_i m_n^i\equiv0$. By Freshman's Dream we have $P(m_n+an)=\sum_i c_i(m_n+an)^i\equiv\sum_i c_im_n^i+c_ia^in^i\equiv\sum_ic_im_n^i=P(m_n)\equiv0\pmod n$ for some integer $a$. Therefore if $b\equiv m_n\pmod n$ then $P(b)\equiv P(m_n)\equiv0\pmod n$.

Now the above conditions and findings imply for all prime $p$, we have a number $m_p$ such that $p|P(m_p)$, and that if $b\equiv m_p\pmod p$ then $P(b)\equiv 0\pmod p$. Consider the set of the smallest $n$ primes $\{p_1,p_2,p_3,\cdots,p_n\}$. By Chinese Reamainder Theorem there exists an integer $b$ such that $b\equiv m_{p_1}\pmod{p_1},b\equiv m_{p_2}\pmod{p_2},b\equiv m_{p_3}\pmod{p_3},\cdots,b\equiv m_{p_n}\pmod{p_n}$ by Chinese Remainder Theorem. Then $p_1,p_2,p_3\cdots,p_n|P(b)\Rightarrow p_1p_2p_3\cdots p_n|P(b)$. As $n$ approaches infinity (as there are infinitely many primes), $p_1,p_2,p_3\cdots,p_n$ approaches infinity. Therefore either $P(b)=\infty$ or $P(b)=0$. Since for finite $b$ and integer coefficients $P(b)$ must be finite, then $P(b)=0$. However as $a$ is an integer, this implies $P$ has a rational root, a contradiction.

I'm not sure if my proof is correct, and my main concern is that I am incorrectly using Chinese Remainder Theorem since I am not sure if I can apply it to infinitely many divisors.

Is this proof correct, and if not, how do I solve this question?

EDIT: It appears not only is my proof incorrect (as$b$ does not converge) as Paul Sinclair has shown, but that according to Jaap Scherphuis there are examples of polynomials that satisfy the conditions. Therefore, my question now is how one can prove the existence of these polynomials while using elementary methods (as this is an IMO selection test problem).

答:
Here is an elementary way to construct such a polynomial explicitely. I will need the fact that if $p\nmid a$, then $a$ is invertible modulo $p$, which can be proven using Bézout theorem.

The proof of Fact 1 is extremely long, because i did not want to use group theory but only congruences. It can be shorten into a 5 lines proof if you allow quotient group and the group $(\mathbb{Z}/p\mathbb{Z})^\times$.

Fact 1. Let $p$ be an odd prime number. Then the number of nonzero squares modulo $p$ is $\dfrac{p-1}{2}$. In particular, if $a,b$ are integrs both prime to $p$, then one of the integer $a,b$ or $ab$ is a square mod $p$.

Proof. Set $m=\dfrac{p-1}{2}$. An integer coprime to $p$ is congruent to some integer $\pm k$, where $1\leq k\leq m$. Since $(-k)^2\equiv k ^2 \mod p$, there is at most $m$ nonzero squaremod $p$.

Now if $1\leq k_1,k_2\leq m$ satisfy $k_1^2\equiv k_2^2 \mod p$, then $p\mid (k_1-k_2)(k_1+k_2)$. Note that $2\leq k_1+k_2\leq 2m=p-1$, so $p\nmid k_1+k_2$. Therefore $p\mid k_1-k_2,$ meaning $k_1=k_2$ since $-p<-m\leq k_1-k_2\leq m<p$. Consequently, the integrs  $k^2,1\leq \leq m$ are all distinct modulo $p$, and there is exactly $m$ non zero square mod $p$.

Now let $a,b$ be two integers coprime to $p$. If $a $or $b$ is a square modulo $p$, we are done.  Assume $a,b$ are not squares modulo $p$. Since $a$  is coprime to $p$; it is invertible modulo $p$, so the elements of  $A=\{a k^2, 1\leq k\leq m\}$ are pairwise distinct modulo $p$. Now the elements $B=\{k^2, 1\leq k\leq m\}$ are all distinct modulo $p$.

Note that it is not possible to have $ak_1^2\equiv k_2^2 \mod p$ for some $1\leq k_1,k_2\leq m$, since otherwise $a$ would be a square modulo $p$, as we can see by multiply by the square of an inverse mod $p$ of $k_1$.

A counting argument then shows that an integer coprime to $p$ can be represented by a unique element of $A\cup B$ modulo $p$ .

Thus, if $b$ is not a square mod $p$, then $b$ is congruent to an element of $B.$ Consequently $b\equiv a k^2\mod p$, and $ab\equiv (ak)^2\mod p$ is a square mod $p$.

Fact 2. Let $p$ be a prime number, and  let $P\in\mathbb{Z}[X]$. Assume that there is $x_0\in \mathbb{Z}$  such that $P(x_0)\equiv 0 \mod p $ and $P'(x_0)\not\equiv 0 \mod p$. Then for all $m\geq 0$, there exist $x_m\in\mathbb{Z}$ such that $P(x_m)\equiv 0 \mod p^{m+1}$ and $x_{m}\equiv x_0 \mod p.$

Prof. By induction on $m$, the case $m=0$ being part of the assumption. Assume that such $x_m$ exists for some $m\geq 0$ and let us show the existence of some $x_{m+1}$.
By assumption, $P(x_m)=\mu p^{m+1}$. Let $0\leq \lambda\leq p-1$ such that $\lambda$ is an inverse of $P'(x_0)$ modulo $p$, and set $x_{m+1}=x_m-\mu \lambda p^{m+1}$.

Note that for all $x,y\in\mathbb{Z}$, we have $P(x)=P(y)+(x-y)P'(y)+(x-y)^2 Q(y)$ for some $Q\in \mathbb{Z}[X]$. Applying this to $x=x_{m+1}$ and $y=x_m$, we get  $P(x_{m+1})=\mu p^{m+1}-\mu\lambda p^{m +1}P'(x_m) \ mod p^{m+2}$.

Since $x_m\equiv x_0 \mod p$, we have $P'(x_m)\equiv P'(x_0) \mod p$ and thus $\lambda P'(x_m)\equiv \lambda P'(x_0)\equiv 1 \mod p$. Consequently $\lambda p^{m+1}P'(x_m)\equiv p^{m+1} \mod p^{m+2}$  and $P((x_{m+1})\equiv 0 \mod p^{m+2}$. Finalement note that by construction, $x_{m+1}\equiv x_m \mod p^{m+1}$, so $x_{m+1}\equiv x_m \equiv x_0 \mod p$, which shows this induction step.

Thm. Let $P=(X^2+X+2)(X^2-17)(X^2-19)(X^2-17\times 19)$. Then $P$ has no rational roots, but has a root modulo $n$ for every integer $n\geq 2$.

Proof. Clearly, $P$ has no rational roots. By CRT, it is enough to prove that $P$ has a root modulo $p^m$ for all prime $p$ and all $m\geq 1$.

Note that $1$ is a root of $X^2+X+1$ mod 2. The derivative at $1$ of this polynomial is $3$, which is $\not\equiv 0 \mod 2$. Hence $X^2+X+1$. has a root mod $2^m$ for all $m\geq 1$ by Fact 2.

Let $p$ be an odd prime number, $p\neq 17,19$. By Fact $2$, one of the integers $17,19,17\times 19$ is a nonzero square mod $p$. Let $a$ be this integer. Notice now that the derivative of $X^2-a$ at an integer $x_0$ which is nonzero $p$ is $2x_0$, which is also nonzero modulo $p$. By Fact 2, $X^2-a$ has a root mod $p^m$ for all $m\geq 1$.

Assume that $p=17$. Then $6^2\equiv 19 \mod 17$, so $X^2-19$ has a root mod $p$, hence mod $p^m$ for all $m\geq 1$ as previously.

Assume that $p=19$. Then $6^2\equiv 17 \mod 19$, so $X^2-17$ has a root mod $p$, hence mod $p^m$ for all $m\geq 1$ as previously.

All in all $P$ has a root modulo $p^m$ for all prime $p$ and all $m\geq 1$.


问:
Fourier transform of $xf''(x)$
I am very unsure of how to approach this Fourier transform and have not been able to find anything which has cleared up my confusion. I know that the transform of $xf(x)=ig'(\alpha)$, but I fail to see how I can extend this to $xf''(x)$.

Title: Fourier transform of $xf''(x)$

Body:
We know that Fourier transform, $$\mathcal{F}\{f(x):u\}=F(u)=\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(t)e^{-iut}dt,$$
$$\mathcal{F}\{f'(x):u\}=\mathcal{F}\left\{\frac{d}{dx}f(x):u\right\}=iu~F(u)$$
$$\text{and}~~~~~~~~\mathcal{F}\left\{\frac{d^n}{dx^n}f(x):u\right\}=(iu)^nF(u)$$

So  $$\mathcal{F}\left\{\frac{d^2}{dx^2}f(x):u\right\}=(iu)^2F(u)$$
Now$$\mathcal{F}\{xf''(x)\}=\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}xf''(x)e^{-iux}dx=i\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f''(x)\dfrac{\partial}{\partial u}\left(e^{-iux}\right)dx=i\dfrac{\partial}{\partial u}\left(\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f''(x)e^{-iux}~dx\right)=i\dfrac{d}{d u}\left[(iu)^2F(u)\right]=-i\left\{2uF(u)+u^2F'(u)\right\}$$


问:
If p (mod 4) = 3 and p is a Gaussian Prime. How to show that Z[i]/(p) is equal to GF(p^2)/(x^2+1)?
I currently work with Gaussian Integer. I try to use prime Gaussian Integer field for Elliptic Curve instead of prime field.
.
We know that every finite field isomorphic to polynomial field with modulus irreducible polynomial and all finite fields of the same size are isomorphic.
.
When I counting the order of the elliptic curve group, I found that Z[i]/(p) and GF(p^2)/(x^2+1) give the same output (same order).
But if I change the modulus with another irreducible polynomial, the output is not same.
.
Why? Is any special thing that I missed?
.
Thankyou, and sorry for my bad english.

答:
(This is a late answer, just found this question while looking for something specific related to elliptic curves.)

We have the sequence of isomorphisms of rings / fields. (I will use $=$ instead of $\cong$ to mark isomorphic objects, thus working the corresponding category.)
I will also use the notation $\Bbb F_q$ for "the" field with $q$ elements instead of $GF(q)$.
$$
\begin{aligned}
\Bbb Z[i]/p
&=(\Bbb Z[x]/(x^2+1))\ /\ (p)\\
&=\Bbb Z[x]\ /\ (x^2+1,\ p)\\
&=(\Bbb Z[x]/(p))\ /\ (x^2+1)\\
&=\Bbb F_p[x]\ /\ (x^2+1)\\
&=\Bbb F_{p^2}\ ,
\end{aligned}
$$
where at the last step we use $p=-1$ modulo four, so that $-1$ is not a quadratic residue modulo $p$, i.e. $(x^2+1)$ has no roots in $\Bbb F_p$, i.e. it is an irreducible = prime polynomial.

Let $i$ be the class of $x$ in the field $\Bbb F_{q}$, $q=p^2$, obtained above, so $i^2=-1$.

Then note that $GF(p^2)[y]/(y^2+1)$ is not a field, since the polynomial $y^2+1=(y+i)(y-i)$ splits in two factors over $GF(p^2)=GF(q)=\Bbb F_q$. This is related to the question of Ihf in the comments. The field with $q=p^2$ elements is the one mentioned in the above chain at forelast place, $\Bbb F_p[x]\ /\ (x^2+1)$.


问:
no. of real roots of exponential equation in three questions
How Can i calculate no. of real roots of exponential equation in three questions

(1) $2^x = 1+x^2$

(2) $2^x+3^x+4^x = x^2$

(3) $3^x+4^x+5^x = 1+x^2$

My Try::

(1) Let $f(x) = 1+x^2-2^x$

now Diff. both side w.r. to $x$

$f^{'}(x) = 2x-2^x.\ln(2)$

and $f^{''}(x) = 2-2^x.\ln^2(2)$

Now for Max. or Min. $f^{'}(x) = 0$

means $2x = 2^x.\ln(2)$

Now How can I calculate after that

答:
Don't get caught up in differentiating more than you need to. With a transcendental equation like yours, it's probably not going to give you an exact algorithmic solution.

For the first one, note that they are equal for $ x = 0 $. If $ x < 0 $, then $ 2^x $ has gotten smaller while $ 1 + x^2 $ has gotten bigger, so they certainly aren't equal for $ x < 0 $.

Now, note that they are also equal at $ x = 1 $. For $ x > 0 $, at first $ 2^x $ is increasing faster. However, at some point in $ (0, 1) $, $ x^2 $ begins increasing faster than $ 2^x $ and "catches up" at $ x = 1 $. Because $ x^2 $ is increasing faster, $ x^2 > 2^x $ for $ x $ near $ 1 $. However, it's clear that eventually $ 2^x $ is larger eventually so at some point they will intersect again.

Here's a little bit more math. $ \frac{d}{dx}[1 + x^2] = 2x $ and $ \frac{d}{dx} 2^x = \ln 2 \cdot 2^x $. Hence, $ 1 + x^2 $ is hardly increasing at all near $ x = 0 $. However, at $ x = 1 $, $ 1 + x^2 $ is increasing at a rate of $ 2 $ while $ 2^x $ is increasing at a rate of $ 2 \ln 2 < 2 $. So at some point, $ x^2 + 1 $ began increasing faster. However, at $ x = 4 $, $ 2^x $ is increasing at a rate of $ 16 \ln 2 $ which is much larger than $ 8 $. There aren't going to be any more switches about which one is increasing faster. This is because a line can only intersect an exponential function at most twice, the line being $ y = 2x $ and the exponential function being $ y = \ln 2 \cdot 2^x $. We already know that there were two intersections, so there definitely aren't any more.


问:
A possible general formula for this planting flowers problem?

In a flower garden, one can grow a variety of roses, and each variety can have different degrees of growth and height. Therefore, people will adhere to the following principles of growing flowers in the garden:

All flowers will be planted on the left and upper sides, but higher flowers will be more prioritized than the shorter one so that the flowers can get the best light in the morning.

Plant flowers in rows from top to bottom, ensuring that the number of plants in the upper rows is no less than the lower rows.

For example, let say there are $2$ types of flowers with $2$ different heights are $1m$ and $2m$. Each type has $2$ flowers. So there will be a total of $8$ ways to plant according to the rules:

Let say there are N types of flower, each type has a different height $h_i$  and has $k_i \,\, (1 \leq i \leq N)$ flower(s) of that types. Is there a general formula to find all possible ways to plant flowers with respect to the rules?

For the case $N = 1$, I found the following recurrence relation:

$$x_k = x_{k-1} + 2^{\left\lfloor{\frac{k-2}{2}}\right\rfloor}$$

where $k$ is the number of flowers.

Edit $1$: As @MikeEarnest mentioned in the comment, the recurrence I found was wrong for $k = 8$. The number of arrangements for the case $N = 1$ is actually the partition function $p(k)$

答:
We have $m$ different species of flowers with heights $h_1 >h_2 > \cdots > h_m$,  and we have quantities $n_1, n_2, \cdots, n_m$ of each category,
for a total of $N$.

Now, if I properly understood  the problem, you want to plant them on a rectangular grid

in such a way that the higher flowers precede the lower either top-down and left-right.

That implies that the quantity of an intermediate species in a row might be higher than in the preceding row (see e.g. $q_{\,2, \, 3}$ in the sketch).

However the total number of flowers in each row shall be not higher than that in the upper row.

An algorithm to tackle the problem could be as follows.

Let's denote with $s_{\,j,\,k}$ the cumulative number of flowers of species $1$ to $k$ in row $j$, i.e. :
$$
s_{\,j,\,k}  = \sum\limits_{i = 1}^k {q_{\,j,\,i} }
$$

Start by choosing a partition of $N$ into any allowed number of parts, call it $\pi _{\,m}$, and assign the resulting (non-increasing) parts
as the values of $s_{\,j, \,m}\; | \, 1 \le j \le \pi_{\,m}$.

That is, such a partition will define the rightmost bound of the rows (the perimeter in blu in the 2nd sketch).

Then choose a partition of $N-n_{\,m}$, with a number of parts $\pi _{\,m-1} \le \pi _{\,m}$.

Assign the result to $s_{\,j, \,m-1}$, thereby defining the perimeter of the species till $m-1$.

Proceed with the partition of $N-n_{\,m}- n_{\, m-1}$ ... ,

and so on,  till arriving to the partition of $n_1$ into a number of parts not greater than the preceding.

The total number of planting schemes will correspond to the total number of such "nested" partitions, wihich can be computerized but not expressed by a closed formula.


问:
How to prove $\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$?
I need help in solving the following inequality---
$$\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$$

The inequality is used in proving Margolus Levitin theorem as can be seen here.

I tried using the Cauchy Mean Value theorem, in similar ways as mentioned in Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$, and rearranging the inequality to a more manageable form, but couldn't proceed further.

Essentially the same question has been asked on Physics SE here
but the currently accepted answer is a bit heuristical because it assumes a specific form of expression for $\cos(x)$ with adjustable parameters and then finds the parameters by several arguments.

I am looking for a more formal and direct proof probably using some well-known inequalities (like Cauchy-Schwarz inequality) or theorems (like Mean Value Theorem).

EDIT: The domain should be $x>0$ because clearly the inequality doesn't hold for $x<0$ as pointed out in a comment by Greg Martin.
This is physically justifiable while using in the Margolus Levitin theorem because there $x\propto t$ and time $t>0$.

答:
This is equivalent to
$$
\frac{1-\cos x}{x+\sin x}\leq \frac{2}{\pi}
$$
for $x>0$. Take derivative of left we get $\frac{x\sin x}{(x+\sin x)^2}$, this is $\geq 0$ until $x=\pi$, when the left equals to $\frac{2}{\pi}$; after that the left decreases to $0$ until $x=2\pi$. After $2\pi$ the denominator is even bigger, e.g. the left $\leq \frac{1+1}{2\pi-1}<\frac{2}{\pi}$. Overall the maximum of left side over$[0, \infty)$ is $\frac{2}{\pi}$, reached at $x=\pi$.


问:
Does the zero product property hold for orthogonal projections on Hilbert spaces?
I have two orthogonal projections $P_{1}$ and $P_{2}$ on a Hilbert space H, where $P_{1} \neq 0$, $P_{2} \neq 0$, $P_{1} \neq P_{2}$, and I define $P = \lambda_{1}P_{1} + \lambda_{2}P_{2}$. I am currently trying to prove that if ran$P_{1}$ $\perp$ ran$P_{2}$, then $PP^{*} = P^{*}P$. I end up with $P_{1}P_{2} = 0$ and $P_{2}P_{1} = 0$. If this is correct, then the equality follows. Am I violating the assumption that $P_{1} \neq 0$, $P_{2} \neq 0$ ? Any responses would be greatly appreciated.

答:
Indeed, if ${\rm ran}\,P_1\perp{\rm ran}\,P_2$ then $P_1P_2v=0$ for all $v$ since $P_2v\in{\rm ran}\,P_2\perp{\rm ran}\,P_1$ and $P_1$ is an orthogonal projection, so $({\rm ran}\,P_1)^\perp\subseteq \ker P_1$.

Similarly $P_2P_1=0$ also follows.


问:
Height of a spherical cap from volume and radius
$\hskip 2.2 in$

I'm trying to calculate the height of a spherical cap on a sphere where you know the volume of the cap and the radius of the sphere.

I was able to find the equation for the volume of the cap with radius $r$ and height $h$ which is

$$V=\frac{\pi h^2}{3}(3r-h)$$

I feel really dumb but for the life of me I can't seem to rewrite the equation in such a way to find $h$ when given a known volume and radius.

答:
By solving the volume formula $[V=\frac{\pi h^2}{3}(3r-h)]$ for h to obtain the cubic equation $[\frac{\pi}{3}h^3-\pi r h^2+V=0]$, converting to the depressed cubic form of the equation $[t^3+pt+q]$, using the trigonometric solution $\Bigg[\Bigg(t_k=2\sqrt{-\frac{p}{3}}\cos\bigg[\frac{1}{3}\cos^{-1}\bigg(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\bigg)-\frac{2\pi k}{3}\bigg]\Bigg)$ for $k=0,1,2\Bigg]$ to solve for the root where k=1, converting back to the original form of the cubic equation, and simplifying, the following solution can be obtained for h in terms of V and r.

$$
h=r[\sqrt{3}\sin(\frac{1}{3}cos^{-1}A)-\cos(\frac{1}{3}\cos^{-1}A)+1]
$$
$$
A=1-\frac{3V}{2\pi r^3}=1-\frac{2V}{V_{sphere}}
$$


问:
Avoiding Russell's paradox with ZF axioms
I recently learned about the axioms of Zermelo–Fraenkel and I am trying to understand how those axioms avoid paradoxes such Russell's paradox. Recall that the "set" from Russell's paradox is:

$$R=\{x:x\notin x\}$$
Which led to the obvious contradiction $R\in R\iff R\notin R$.

I think that I understand it, my interpretation is:

Under $\mathsf{ZF}$ axiomatic system, the only way we can construct such a 'set' is from existing sets. So, using the axiom of separation we must have some set $B$ such that:

$$R=\{x\in B:x\notin x\}$$

Moreover, by the axiom of regularity it's not hard to prove that every set $A$ is not an element of itself, That is, $A\notin A$. Therefore, the formula $\varphi(x):=[x\notin x]$ is always true. Therefore:

$$R=\{x\in B:\overbrace{x\notin x}^{\text{always true}}\}=B$$

So no new set was created and no contradiction has been occurred.

Is my interpretation valid? Is that how $\mathsf{ZF}$ axioms 'avoid' such axioms?

Feedbacks will be appreciated!

答:
Not quite; you must still consider why the original contradiction $R \in R \iff R \not \in R$ no longer applies.
That is because once you limit to $B$, you can no longer conclude $R \in R \iff R \not \in R$ anymore; you can only conclude that under the additional assumption that $R \in B$.  So the conclusion is simply that $R \not \in B$, and there is no contradiction.

Note that this holds even if you remove the axiom of regularity.


问:
How to integrate $ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $?
The problem is to make the following integral stationary:
$$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$
to simplify the Euler equation, I tried to change the independent variable:
$$ \int_{y_1}^{y_2} \frac{\sqrt{1+x'^2}}{y^2}dy, \: \: \: \: \: \: \: \: y=y\left( x \right),\: y'=\frac{dy}{dx} $$
with the correspondent Euler equation:
$$ \frac{d}{dy}\frac{\partial F}{\partial x'}-\frac{\partial F}{\partial x}=0 $$
thus
$$
\begin{aligned}
\frac{\partial F}{\partial x}=0 \Rightarrow \frac{\partial F}{\partial x'} &= k\\
                                               \frac{x'}{y^2 \sqrt{1+x^2}} &= k\\
                                                        \frac{dx}{dy} = x' &= \frac{ky^2}{\sqrt{1-k^2y^4}}
\end{aligned}
$$
and I get:
$$ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $$
Now, can I change $ ky^2 $ to a new single arbitrary variabel to simplify the integrand? Or are there a more effective method?

答:
Do it, substitute $u=ky^2,\frac{du}{dy}=2\sqrt{ku}$ (assuming $k>0$):
$$x=\frac1{2\sqrt k}\int\sqrt{\frac u{1-u^2}}\,du$$
This is an elliptic integral. Say we're integrating from $0$ to $u$, then Byrd and Friedman 235.06 gives
$$x=\sqrt{\frac2k}\left(E(\varphi,m)-F(\varphi,m)/2-\sqrt{\frac{u(1-u)}{2(1+u)}}\right)+C$$
where $\sin^2\varphi=\frac{2u}{1+u}$ and $m=1/2$. You will have to use numerical methods to get $y$ in terms of $x$.


问:
Why aren't the lines in taxicab geometry the geodesics of the $L^1$ metric?
This page claims that the proper lines for taxicab geometry should be the same lines as for Euclidean geometry (with the $L^2$ norm on $\mathbb{R}^2$ instead of the $L^1$ norm for taxicab geometry).

Lines in the Taxicab Plane look like lines in the Euclidean plane...

The same claim also appears to be implicit in the Wikipedia page for taxicab geometry, on this webpage, on this one, and also in the book by Millman and Parker, Geometry: A Metric Approach with Models. All of the sources claim as a result that taxicab satisfies all of the same axioms as Euclidean geometry except for the SAS postulate.

Question: Isn't this a very bad, artificial, and unnatural definition of lines for taxicab geometry?

$\mathbb{R}^2$ with the metric induced by the $L^1$ norm is a length space, according to Exercise 11 here, so the lengths of its geodesics do correspond to the distance between points -- so why shouldn't the "lines" in taxicab geometry be the geodesics with respect to the metric induced by the $L^1$ norm?

If we want to study taxicab geometry, and not Euclidean geometry, shouldn't we only study structures which arise naturally from the geometry's definition, rather than definitions of angle, line, and triangle which are unnaturally "imported" from Euclidean geometry unchanged?

Motivation: The answers to this question will allow me to answer my previous question.

In particular, because of the choice of lines for taxicab geometry, even though it is a length space, this is not a case where $d(A,B)+d(B,C) = d(A,C)$ implies that $A,B,C$ are collinear (although it would be if we chose the lines to be the geodesics of the length space). Thus, in order to define a notion of betweenness, we need to say that $B$ is between $A$ and $C$ if and only if (1) $A,B,C$ are collinear, and (2) $d(A,B) + d(B,C) = d(A,C)$. With a more natural definition of line, like that in Euclidean space, the first condition would be redundant, and the notion of betweenness more intimately and naturally related with our choice of metric.

答:
If one considers general $L^p$ norms where the distance between two points, $\mathbf{x}$ and $\mathbf{y}$, is defined by
$$ d(\mathbf{x},\mathbf{y}) = \left(\sum_{i}\left| x_i - y_i\right|^p\right)^{1/p}, $$
one can show that for all $p>1$, the geodesics are straight, i.e. Euclidean, lines.

The taxicab metric is the $L^p$ metric with $p=1$. Since all the $L^p$ metrics with $p>1$ have straight lines as geodesics, and, although there are no unique geodesics in taxicab geometry, straight lines do give the shortest distance in the taxicab metric, one could make the argument that straight lines are the natural geodesics for the taxicab geometry that follow from its metric. At least if taxicab geometry is seen as a limiting case of case of $L^p$ geometries with $p>1$.

In order to show that the geodesics for $L^p$ metrics are straight lines, we need to use the calculus of variations. Let $\mathbf{x}(t)=(x_1(t),..., x_n(t))$ be a parametrization of a curve in $\mathbb{R}^n$ . First we need to calculate the distance for two points infinitesimally close on the path $\mathbf{x}(t)$

$$d(\mathbf{x}(t+dt),\mathbf{x}(t)) = \left(\sum_{i}\left| x_i(t+dt) - x_i(t)\right|^p\right)^{1/p}
= \left(\sum_{i}\left| x_i+\frac{dx}{dt} dt- x_i\right|^p\right)^{1/p}
=\left(\sum_{i}\left| \frac{dx}{dt}\right|^p \right)^{1/p}dt.$$

Now let's define the function

$$f(\mathbf{x},\dot{\mathbf{x}}) \equiv \left(\sum_{i}\left|\dot{x}_i\right|^p \right)^{1/p}$$

where $\dot{x_i}=\frac{dx_i(t)}{dt}$. We now need to minimize the integral

$$\int f\left(\mathbf{x}(t),\dot{\mathbf{x}}(t)\right)dt = \int \left(\sum_{i}\left|\dot{x}_i\right|^p \right)^{1/p}dt.$$

This can be done via the Euler–Lagrange equations:

$$\frac{\partial f}{\partial x_i} - \frac{d}{dt}\left(\frac{\partial f}{\partial \dot{x_i}}\right)=0.$$

For our problem, the Euler–Lagrange equations give

$$0-\frac{d}{dt}\left(\dot{x}_i |\dot{x}_i|^{p-2}\left(\sum_{j}\left| \dot{x}_j\right|\right)^{1/p-1}\right)=0.$$

Integrating these equations with respect to $t$, and dividing pairs of them gives

$$ \frac{a_i}{a_j}=\frac{\dot{x}_i}{\dot{x}_j} \left|\frac{\dot{x}_i}{\dot{x}_j} \right|^{p-2}$$
where the $a_i$ are integration constants. There are two cases to check, depending on the sign of $\frac{\dot{x}_i}{\dot{x}_j}$, but it is pretty easy to see that in both cases the equations reduce to
$$\dot{x}_j=\left(\frac{a_i}{a_j}\right)^\frac{1}{p-1}\dot{x}_i$$

(maybe up to a minus sign). These equations can be immediately integrated, and the results are lines. The only issue is when $p=1$, where the equations becomes singular.


问:
Are there any two triangular numbers that add up to a perfect cube?
The triangular numbers are all of these numbers:

$$\sum_{n=1}^x n$$ For $x\gt 0$ and that $x$ are only integers.

Here's a list: $$1,3,6,10,15\cdots$$

Are there any $2$ triangular numbers in which the sum of those is a cube?

EDIT: Are there an infinite number of these solutions?

答:
Use Determine if a number is the sum of two triangular numbers.. Then $n^3$ is a sum of two triangular numbers if and only if the $3 \pmod 4$ prime factors of $4n^3+1$ have even exponent.

If $n = m^2$ for some $m$, then $4n^3+1 = 4m^6+1 = (2m^3)^2+1$, so no prime factor of this is $3$ mod $4$. Hence squares $n$ work.


答:
Defining $R:=\{x\in B|x\notin x\}$, $R\in R\iff R\in B\land R\notin R$, which reduces by the tatutology $(p\iff q\land\neg p)\equiv(\neg p\land\neg q)$ to $R\notin R\land R\notin B$, rather than a contradiction. Regularity tells us further that $R=B$.

One more point: in theories such as ZF, we can do more than just prove certain sets exist. We can also prove a certain set doesn't exist by deriving a contradiction, including of course a hypothetical set whose elements are precisely those $x$ with $x\notin x$, by the usual argument. So we don't only fail to prove such a set does exist; we can also prove it doesn't. (A related concept is to say that the class of all sets $x$ with $x\notin x$ is not a set, but rather a proper class.)

Title: Avoiding Russell's paradox with ZF axioms

Body:
I recently learned about the axioms of Zermelo–Fraenkel and I am trying to understand how those axioms avoid paradoxes such Russell's paradox. Recall that the "set" from Russell's paradox is:

$$R=\{x:x\notin x\}$$
Which led to the obvious contradiction $R\in R\iff R\notin R$.

I think that I understand it, my interpretation is:

Under $\mathsf{ZF}$ axiomatic system, the only way we can construct such a 'set' is from existing sets. So, using the axiom of separation we must have some set $B$ such that:

$$R=\{x\in B:x\notin x\}$$

Moreover, by the axiom of regularity it's not hard to prove that every set $A$ is not an element of itself, That is, $A\notin A$. Therefore, the formula $\varphi(x):=[x\notin x]$ is always true. Therefore:

$$R=\{x\in B:\overbrace{x\notin x}^{\text{always true}}\}=B$$

So no new set was created and no contradiction has been occurred.

Is my interpretation valid? Is that how $\mathsf{ZF}$ axioms 'avoid' such axioms?

Feedbacks will be appreciated!


问:
If the line at infinity is a secant line of a conic then the conic is a hyperbola?
I'm learning perspective geometry and I have a question about the line at infinity. Is it true that if the conic has the line at infinity as its secant then the conic represents a hyperbola in Euclidean geometry?

If it is, how can I imagine the shape hyperbola using the line at infinity? I think I can lift up the line to infinity but it seems like a parabola than a hyperbola.

I draw a conic going through $A, B, C, D, E$, and I choose AB as the line at infinity. Then I lift it up and it seems like a parabola

P.S: I'm quite new to this field so I may use the wrong jargon. Thank you

答:
I assume you are experimenting in Geogebra to get an intuition for this concept, and I'll answer from that point of view.

What you're currently doing is making $A,B$ approach the same point at infinity.  As a result you are constructing a conic that meets the line at infinity at a single point, and thus has the line at infinity as a tangent, i.e. the conic is a parabola.

Instead, draw two lines $a,b$ intersecting at the origin, and then create a third  line $c$. Let points $A,B$ be the intersections of $c$ with $a,b$.  Then create a conic that goes through the points $A,B,C,D,E$ and move $c$ towards infinity. You'll tend towards a hyperbola. (It's more interesting if you choose $C,D,E$ so that your conic starts off as an ellipse.).  Using this construction you'll ensure that $A,B$ approach different points at infinity, and you'll get a conic for which the line at infinity is a true secant.


问:
How to prove two polynomials have no zeroes in common?
The question asked:
Divide the polynomial $P(x) = x^3 + 5x^2 - 22x - 6$ by  $G(x) = x^2 - 3x + 2$. I did, and got the answer: $(x+8)(x^2-3x+2)-22$.

However, it now asks to: "Show that $P(x)$ and $G(x)$ have no zeros in common."

How do I prove this?

Thanks.

答: The roots of $G(x)$ are $1$ and $2$.However $P(1)$ and $P(2)=-22$.Hence they share no common roots.</p>


问:
Let $A_1,A_2\in L(V,V)$, $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Is $\dim\ker (A_2A_1)=n_1+n_2$ even if $\dim V=\infty$?
Let $V$ be a vector space that is not necessarly finite dimensional. Consider two surjective linear operators $A_1,A_2\in L(V,V)$ with finite-dimensional kernel: $A_i(V)=V$ and $\dim\ker A_i=n_i\in\mathbf{N}$. Can it be proven that $\dim\ker (A_2A_1)=n_1+n_2$?

In words: Is the dimension of the kernel of the composition of surjective linear operators with finite dimensional kernels the sum of the dimensions of the kernels?

Looking at this formula, one might think that the formula $\dim\ker A_2A_1=n_1+_2$ is valid, but I have looked at the answers and they are based on the rank nullity theorem, which I can't use, since $V$ is not finite dimensional.

Motivation: I want to prove that the solution space of homogeneous linear differential equations of order $n$ is $n$-dimensional by writing the $n$-th order differential operator as composition of first-order differential order with one-dimensional kernel. Actually, the answer to this question completes my proof.

答:
Here's a proof using quotient spaces. We note that the map $[A_1] : V/\ker(A_1) \to V$ induced by $A_1$ is a vector space isomorphism. Thus, $A_2\circ [A_1]:V/\ker(A_1) \to V$ has a kernel with dimension equal to that of $A_2$.

On the other hand, $A_2 \circ [A_1] = [A_2 \circ A_1]$. We note that
$$
\ker([A_2 \circ A_1]) = \ker(A_2 \circ A_1)/\ker(A_1) \implies\\
\dim \ker A_2 = \dim \ker([A_2 \circ A_1])
= \dim \ker(A_2 \circ A_1) - \dim \ker(A_1) \implies\\
\dim \ker(A_2 A_1) = \dim \ker A_1 + \dim \ker A_2.
$$
To put it another way, we have proved the result with the following commutative diagram:

For those familiar with short exact sequences, this amounts to considering the sequence

$$
0 \to \ker A_1 \overset{\iota}{\to} \ker A_2A_1 \overset{A_1}{\to} \ker A_2
\overset{A_2}{\to} 0,
$$

where $\iota$ denotes the inclusion map.


答: You have that $P(x) = (x+8)G(x) - 22$. What happens when $G(x)$ is zero?</p>

Title: How to prove two polynomials have no zeroes in common?

Body:
The question asked:
Divide the polynomial $P(x) = x^3 + 5x^2 - 22x - 6$ by  $G(x) = x^2 - 3x + 2$. I did, and got the answer: $(x+8)(x^2-3x+2)-22$.

However, it now asks to: "Show that $P(x)$ and $G(x)$ have no zeros in common."

How do I prove this?

Thanks.


问:
Prove $f(\xi)$ behaves as $e^{\xi^2}$ if $\lambda<0$ in $\frac{d^2f}{d\xi^2}-2\xi\frac{df}{d\xi}+2\lambda f(\xi)=0$

How does one show that $f(\xi)$ will behave as $e^{\xi^2}$ if $\lambda<0$ in the Hermite differential equation $\frac{d^2f}{d\xi^2}-2\xi\frac{df}{d\xi}+2\lambda f(\xi)=0$

This problem comes in solving the linearly independent Schrodinger equation correpsonds to a 1D harmonic oscillator.

It says since $f(\xi)$ behaves as $e^{\xi^2}$ if $\lambda<0$, $f(\xi)\to +\infty$ as $x\to \pm \infty$, the corresponding solutions are physically not desirable, thus $\lambda \geq0$.

how do I prove that $f(\xi)$ behaves as $e^{\xi^2}$ if $\lambda<0$ mathematically ?

Note: I understand how to solve the Hermite differential equation when $\lambda\geq 0$ to obtain Hermite functions of different order using the Power series method.

Reference: Page 74 of ONE DIMENSIONAL HARMONIC OSCILLATOR

Possible Way

As I have seen in some reference,

Making use of the Frobenius method, taking $f(\xi)=\xi^s\sum_{r=0}^\infty a_r\xi^r$

$$
\frac{a_{r+2}}{a_r}=\frac{2(r+s-\lambda)}{(r+s+2)(r+s+1)}=\frac{2(1+\frac{s}{r}-\frac{\lambda}{r})}{r(1+\frac{s+2}{r})(1+\frac{s+1}{r})}\bigg|_{r\to\infty}\to \frac{2}{r}
$$

Consider the series $e^{\xi^2}=1+\xi^2+\frac{\xi^4}{2!}+\cdots$. The ratio of the adjacent terms is
$$
\frac{b_{r+2}}{b_r}=\frac{(r/2)!}{(\frac{r+2}{2})!}=\frac{\bigg(\dfrac{r}{2}\bigg)!}{\bigg(\dfrac{r+2}{2}\bigg)!}=\frac{1}{\frac{r}{2}+1}\bigg|_{r\to\infty}\to \frac{2}{r}
$$
asymptotically, $f(\xi)$ is behaving like $e^{\xi^2}$, thus $\psi=e^{-\xi^2/2}f(\xi)$ goes like $e^{+\xi^2/2}$.

Is it a well defined explanation ?

答:
The long story short is that the exponential approximation is only valid if $\boldsymbol{|\lambda|}$ is close to $\boldsymbol1$. The explanation is below.

Putting $\lambda\to -\lambda$ into the equation we have
$$y''-2xy'-2\lambda y=0$$
The solution to this equation can be obtained via a power series approach. We can write
$$y(x;\lambda)=c_1\cdot M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)+ c_2\cdot 2^{-\lambda}U\left(\frac{-\lambda}{2};\frac{1}{2}~\bigg|~x^{2}\right) $$
These are the confluent hypergeometric functions of the first and second kind. The strange delimiters in the arguments is a stylistic choice due to their relation to the generalized hypergeometric function. IF $y$ satisfies the convenient conditions $y(0)=1$, $y'(0)=0$ then we can discard the confluent hypergeometric of the second kind and write
$$y(x;\lambda)= M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)$$
$M$ is defined as
$$M(a;b~|~z):=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{a+i}{b+i}\frac{x^k}{k!}\right]$$
Which would mean that
$$M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{\lambda/2+i}{1/2+i}\frac{x^{2k}}{k!}\right]=\sum_{k=0}^\infty \left[\prod_{i=0}^{k-1}\frac{1+2i/\lambda}{1+2i}\frac{(\sqrt{\lambda}~x)^{2k}}{k!}\right]$$
Assuming $\lambda$ is close to $1$ the product vanishes and we have
$$y(x;\lambda)=M\left(\frac{\lambda}{2};\frac{1}{2}~\bigg|~x^2\right)\approx\sum_{k=0}^\infty \frac{(\sqrt{\lambda}~x)^{2k}}{k!}=\exp\left(\lambda x^2\right).$$

Making use of the Frobenius method, taking $f(\xi)=\xi^s\sum_{r=0}^\infty a_r\xi^r$

$$
\frac{a_{r+2}}{a_r}=\frac{2(r+s-\lambda)}{(r+s+2)(r+s+1)}=\frac{2(1+\frac{s}{r}-\frac{\lambda}{r})}{r(1+\frac{s+2}{r})(1+\frac{s+1}{r})}\bigg|_{r\to\infty}\to \frac{2}{r}
$$
Consider the series $e^{\xi^2}=1+\xi^2+\frac{\xi^4}{2!}+\cdots$. The ratio of the adjacent terms is
$$
\frac{b_{r+2}}{b_r}=\frac{(r/2)!}{(\frac{r+2}{2})!}=\frac{\bigg(\dfrac{r}{2}\bigg)!}{\bigg(\dfrac{r+2}{2}\bigg)!}=\frac{1}{\frac{r}{2}+1}\bigg|_{r\to\infty}\to \frac{2}{r}
$$
asymptotically, $f(\xi)$ is behaving like $e^{\xi^2}$, thus $\psi=e^{-\xi^2/2}f(\xi)$ goes like $e^{-\xi^2/2}$.

This is OK, but we can think of pathological examples that don't follow this reasoning. Consider the two functions $e^{z}=\sum_{n=0}^\infty a_n z^n$ and $\sin(z)=\sum_{n=0}^\infty b_nz^n$ where $a_n=\frac{1}{n!}$ and
$$b_{n} =\begin{cases}
0 & n=2k\\
\frac{1}{n!} & n=4k+1\\
\frac{-1}{n!} & n=4k+3
\end{cases}$$
It is true that for odd $n$,
$$\frac{a_{n+4}}{a_n}=\frac{b_{n+4}}{b_n}=\frac{1}{(n+1)(n+2)(n+3)(n+4)}$$
But I think it would be foolish to say $\sin(x)\to e^x$ for $x\to\infty$.


问:
Can someone tell me why this is wrong
$L[t*\sqrt{t}]
= L[t*t^{1/2}]
= L[t^{3/2}]$

$L[t^{n}] =\frac{\Gamma n}{s^{n+1}}$    so   $L[t^{3/2}]= \frac{\Gamma \frac {3}{2}}{s^\frac {5}{2}}$

$ = \frac{\frac {1}{2} \Gamma \frac {1}{2}}{s^\frac {5}{2}}$

$ = \frac {\sqrt{\pi}} {2 s^\frac {5}{2}}$

but my texbook and this answer says that the correct answer is $\frac{1}{s^{5/2}} \cdot \frac{3\sqrt{\pi}}{4}$

where did I make a mistake?

Title: Can someone tell me why this is wrong

Body:
Actually, in Laplace transformation, $$ F(p)={\mathcal{L}}\left\{ x^n:p \right\}=\dfrac{\Gamma(n+1)}{p^{n+1}}~~~~~~~~~~~~\text{for}~~~p >-1$$
and $$ F(p)={\mathcal{L}}\left\{ x^n:p \right\}=\dfrac{n!}{p^{n+1}}~~~~~~~~~~~~\text{for}~~~p =0,~1,~2,~\cdots$$
For your case $~n=3/2~,$ so $$L[t^{3/2}]= \frac{\Gamma \left(\frac {3}{2}+1\right)}{s^\frac {5}{2}}=\frac{\frac {3}{2}\cdot\Gamma \left(\frac {3}{2}\right)}{s^\frac {5}{2}}=\frac{\frac {3}{2}\cdot\frac {1}{2}\cdot\Gamma \left(\frac {1}{2}\right)}{s^\frac {5}{2}}=\frac{\frac {3}{4}\cdot\sqrt \pi}{s^\frac {5}{2}}=\dfrac{3\sqrt \pi}{4}s^{-5/2}~.$$
as $~ Γ(n+1)=nΓ(n)~$ and $~Γ(1/2)=\sqrt \pi~.$


问:
Calculus Made Easy Exercise 10 Question 9
(9) The efficiency u of an electric generator at different values
of output x is expressed by the general equation:

$$u = \frac{x}{a + bx + cx^2}$$

where $a$ is a constant depending chiefly on the energy losses in
the iron and $c$ a constant depending chiefly on the resistance of
the copper parts. Find an expression for that value of the output
at which the efficiency will be a maximum.

I am trying to solve this question but I am not sure how to go about approaching question.
If you need I can show my steps so far. Please do not give away the answer but guide me in the right direction. That way I can learn more.

Many thanks and stay safe!

答:
Hint:

Write $$u = \frac{1}{{a\over x} + b + cx}$$ so you have to find a minimum for  ${a\over x} + cx$. This should be easy now, even without calculus...


问:
If $g$ is a continuous and increasing function of $x$, prove that $g(X)$ is a random variable.
Exercise 2.3.12 from Grimmet Stirzaker's Probability and Random processes asks the following. I'd like, if you guys can help verify my solution.

Let $X$ be a random variable and $g:\mathbb{R} \to \mathbb{R}$ be continuous and strictly increasing. Show that $Y = g(X)$ is a random variable.

My Solution.

As $g$ is a monotonically increasing function, it is injective(one-to-one). That is, if $x_1 < x_2$, then $g(x_1) < g(x_2)$. Therefore, $x_1 \ne x_2 \implies g(x_1) \ne g(x_2)$.

I am not sure how to deduce, that $g$ is surjective (onto).

If $g$ is bijective, the inverse function $g^{-1}$ exists and is well-defined.

Hence, the set

\begin{align*}
&\{ \omega : g(X(\omega)) \le x \}\\
=&\{ \omega : (X(\omega) \le g^{-1}(x) \} \in \mathcal{F}
\end{align*}

since $X$ is a random variable. Consequently, $g(X)$ is a random variable.

答:
The continuity and the strict monotonicity of $g$ are irrelevant. What is required is that $g$ is a Borel function. Note that either condition "$g$ is continuous", "$g$ is monotonic increasing" implies that $g$ is a Borel function.

Suppose that $g$ is a Borel function. Let $A\in\mathcal{B}(\mathbb{R})$. Observe that $g(X)^{-1}(A) = X^{-1}(g^{-1}(A))\in\mathcal{F}$ because $g^{-1}(A)$ is a Borel set. Hence $g(X)$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable, i.e., a random variable.


问:
Proving a floor function is injective/surjective
Is the function $\lfloor x/2\rfloor$ injective or surjective? If so why? The domain is $\mathbb R$ and the co-domain is $\mathbb Z$. I think it is not injective as if we take x to be $20$ and $y$ to be $21$, we end up with $f(x)=10=f(y)$, but $x$ is not equal to $y$. I know that to check for surjection, we have to solve in terms of $x$, but since this is a floor function, I'm not sure what to do with the floor symbol while solving for $x$.

Thanks in advance.

答:
Your argument about injection is correct.

Hint: For surjection you should figure out whether you can get every integer as a value somehow. No need to think formally about "solving for $x$".


问:
How can I estimate the number of ads a snack bar owner needs to use along a road to achieve 99/100 of driver's attention
I snack bar owner performed a research and noticed that with one ad on the road he achieved 1/2 of driver's attention (the ad was seen). He was unsatisfied he wanted a value above 99/100  how many more ads he must use at minimum?

```<code>a) 99
b) 51
c) 50
d) 6
e) 1
</code>```

I don't even now how to solve it. I promese I tried but probability has no middleground or you know how to approach it or you just stand still. How can I approach a problem like this?

I corrected the question it is not how many but how many more ads he must use.

答:
What's given can be rephrased as "Each ad on the road gains $1/2$ of the drives' attention" now it might be tempting to say you need only $2$ ads to gain all the driver attention but that would be wrong, imagine you have $100$ drivers, if the first ad gains half the attention then you have the attention of $50$ drivers and there is another $50$ remaining, add another ad and you will have the attention of half of those remaining so $25$, in total you will have the attention of $75$ drivers with $25$ drivers remaining, etc...

So basically you have to find $k$ such that $\sum_{n=1}^{k} (\frac{1}{2})^{n} >0.99$

Since you are given options then the easiest thing to do would be to plug them in for $k$, we have $\sum_{n=1}^{6} (\frac{1}{2})^{n}=0.9843...$ which is still below our target.

We have $\sum_{n=1}^{50} (\frac{1}{2})^{n}=1$ which is above our target, so the answer is $50$.

Note that to find the sum you have two options, either put the series in a calculator or since it is a geometric series you can do the following:

$S_n=a_1(\frac{(1−r^n)}{1−r}), r\neq1$

Where $n$ is the number of terms, $a_1$ is the first term and $r$ is the common ratio

so for $\sum_{n=1}^{6} (\frac{1}{2})^{n}$ we have $n=6, r=\frac{1}{2}, a_1=\frac{1}{2}$ so our sum would be

$\frac{\frac{1}{2}(1-(\frac{1}{2})^6)}{1-\frac{1}{2}}=\frac{\frac{1}{2}(\frac{63}{64})}{\frac{1}{2}}=\frac{63}{64}=0.9843...$

NOTE: this is the minimum without taking into account there is an ad already placed, so if we have no ads at all, BETWEEN THESE OPTIONS we would need to choose $50$, however, if the question actually asks for how many ads other than the one already placed we need then we would go for option of $6$ ads as $7$ ads are enough to get our desired number, we have: $\sum_{n=1}^{7} (\frac{1}{2})^{n}=0.9921875$


答:
Optimization problems generally follow the same procedure. Namely,

Determine your constraint ($\textit{i.e.} x < a, a \leq x \leq b, etc$)

Find the critical numbers

Determine whether each critical number is a max, min, or neither,

Evaluate your function at the relative extrema and endpoint (if any)

Choose the largest (or lowest) value

Title: Calculus Made Easy Exercise 10 Question 9

Body:
(9) The efficiency u of an electric generator at different values
of output x is expressed by the general equation:

$$u = \frac{x}{a + bx + cx^2}$$

where $a$ is a constant depending chiefly on the energy losses in
the iron and $c$ a constant depending chiefly on the resistance of
the copper parts. Find an expression for that value of the output
at which the efficiency will be a maximum.

I am trying to solve this question but I am not sure how to go about approaching question.
If you need I can show my steps so far. Please do not give away the answer but guide me in the right direction. That way I can learn more.

Many thanks and stay safe!


问:
$T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation of $\mathbb{R}^n$. Then which of the following statements are true?
$T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation of $\mathbb{R}^n$, where $n \ge 3$ and $ \lambda_1, .. ,\lambda_n \in \mathbb{C}$ be eigen values of T. Then which of the following statements are true?

(A) If $\lambda_i = 0$ for some $i$, then T is not surjective.

(B) If T is injective, then $\lambda_i = 1$ for some $i$.

(C) If there is a 3-dimensional subspace $U$ of $V$ s.t. $T(U)=U$ then $\lambda_i \in \mathbb{R}$ for some $i$.

My attempt :

(A) $\lambda_i = 0$ then $det(T)=0$, which means $T$ is singular and $dim(kerT) \gt 0$. Therefore not surjective.

(B) Let, $\lambda_i$'s are distinct non-zero eigen values of $T$(all of them are real) such that none of them is equal to $1$.

Then, $det(T) \ne 0$. Thus T is injective. Therefore the statement is false.

Bit about C I couldn't proceed much further. I thought if the restriction of $T$ to $U$, namely $T_U$.

Since $T_U(U) = U$. Then $T_U$ is non-singular. But does that say that it must have at least one real eigen value? And also, is it true that eigen values of $T_U$ is also eigen values of $T$? I don't know surely.

答: When $T(U)=U$ for a $3$-dimensional subspace $U\subseteq\mathbb R^n$, the restriction $T|_U\colon U\to U$ has $3$ of the eigenvalues of $T$ (counting algebraic multiplicity). The non-real complex eigenvalues of a real linear transform always come in pairs: When $\lambda\in\mathbb C$ is a complex eigenvalue, then so is the complex conjugate $\overline \lambda$. Hence, the number of non-real complex eigenvalues of any real linear transform is even. Since $3$ is odd, $T|_U$ has one or three real eigenvalues.


答: Noting $f(x)=\lfloor(x/2)\rfloor$ we have, for example $f(3)=f(3.7)$ then $f$ is not injective. Besides $f(\mathbb R)=\mathbb Z$ then none real non integer can have un inverse image by $f$ so $f$ is not surjective.

Title: Proving a floor function is injective/surjective

Body:
Is the function $\lfloor x/2\rfloor$ injective or surjective? If so why? The domain is $\mathbb R$ and the co-domain is $\mathbb Z$. I think it is not injective as if we take x to be $20$ and $y$ to be $21$, we end up with $f(x)=10=f(y)$, but $x$ is not equal to $y$. I know that to check for surjection, we have to solve in terms of $x$, but since this is a floor function, I'm not sure what to do with the floor symbol while solving for $x$.

Thanks in advance.


问:
Range of stability for iterative map
Using linear stability analysis, I would like to compute the range of stability of the fixed points and the $2$-cycles of the following iterative map: $x_n = x_{n-1}^{2} - 3\mu$.

Setting $x = x^{2} - 3\mu$, I calculated the fixed points: $x_1 = \dfrac{\sqrt{1+12\mu}}{2}$ and $x_2 = \dfrac{\sqrt{1-12\mu}}{2}$.

And the two cycles are the fixed points and two other additional points: $x_3 = \dfrac{-1-\sqrt{3(-1+4\mu)}}{2}$ and $x_4 = \dfrac{-1+\sqrt{3(-1+4\mu)}}{2}$.

Now I am struggling to find the range of stability of the fixed points and $2$-cycles.

What I know is we need to compute the slope of the map at the points. So if we differentiate $x^{2} - 3\mu$ and calculate the derivative at the fixed point $x_1 = \dfrac{\sqrt{1+12\mu}}{2}$, we get $2(x_1) = 1+\sqrt{1+12\mu}$. And similarly for $x_2$,  $2(x_2) = 1-\sqrt{1+12\mu}$. Then we set the absolute values < 1. So we get $|1+\sqrt{1+12\mu}| < 1$ and $|1-\sqrt{1+12\mu}| < 1$. Now here is the problem, I can only find the range of stability for $x_2$ but not for $x_1$. Here is my (hands) work:

$$\begin{align} |1-\sqrt{1+12\mu}| < 1\\ \implies-1<1-\sqrt{1+12\mu} < 1 \\\implies-2<-\sqrt{1+12\mu} < 0\\ \implies-\frac{1}{3} < \mu < \frac{3}{13}\end{align}$$

But now I try to find the range of stability of $x_1$:

$$\begin{align}|1+\sqrt{1+12\mu}| < 1 \\ \implies -1<1+\sqrt{1+12\mu} < 1 \\ \implies -2<\sqrt{1+12\mu} < 0 \end{align}$$ and there is no solution for this inequality.

My questions are:

1. How can we compute the range of stability for $x_1$?

2. What can we conclude for the range of stability for $x_1$? Am I correct to say that its range of stability is zero?

3. Are there any other methods to compute range of stability of fixed points and n-cycles?

Many thanks in advance.

答:
It is always slightly surprising when the graph of the function $u$ defining a dynamical system $x_{n+1}=u(x_n)$ one is interested in, is not even mentioned.

In the present case, the graph $y=u(x)$ of the function $u:x\mapsto x^2-3\mu$ meets the diagonal $y=x$ at $x=x^*$ with $x^*=\frac12(1+\sqrt{1+12\mu})$ and at another point $x^{**}$ which will not interest us. (Your choice of the notations $x_1$ and $x_2$ for these fixed points is rather unfortunate since $x_1$ and $x_2$ already denote members of the sequence.)

Since the derivative at $x^*$ is $u'(x^*)\gt1$, $x^*$ is repulsive in the sense that, for every $x$ close enough to $x^*$, $x\gt x^*$ implies $u(x)\gt x$, while $x\lt x^*$ implies $u(x)\lt x$. Furthermore, $u'(x)\gt1$ uniformly on $x\geqslant x^*$ hence for every starting point $x_0\gt x^*$, the sequence $(x_n)$ is increasing and $x_n\to+\infty$.

The situation for $x_0\lt x^*$, even when $x_0$ is close to $x^*$, is more complicated. Truly enough, the first few terms of the sequence $(x_n)$ are decreasing but  what happens afterwards is not determined by the local behaviour of $u$ at $x^*$.

Thus, if "range of stability" (a phrase I never met before, except with an altogether different meaning) means "set of initial conditions $x_0$ such that the sequence $(x_n)$ converges", I am afraid that an awful more lot of work is needed to determine it. Note for example that, even when $x_0$ is close to $x^*$ with $x_0\lt x^*$, it may well happen that, for some well chosen values of $x_0$, $x_n=x^*$ for every $n$ large enough (or even that $x_n=x^{**}$ for every $n$ large enough).

Title: Range of stability for iterative map


问:
Application of Lawvere Theories
This is supposed to be more of a reference request than an actual question, but anyway...

A Lawvere theory is a category $\mathcal{T}$ with finite products, st. every object is a power of some distinguished object $X$. The category of models is the full subcategory $[\![\mathcal{T},\mathsf{Set}]\!]_\times$ of product preserving functors. Most algebraic categories (like groups, modules, rings etc.) are in fact realized as a category of models of a Lawvere theory. So far so good.

When reading about Lawvere theories it is frequently mentioned that they are special in the sense that they allow us to do algebra in any category $\mathcal{C}$ with finite products, giving a common framework for say topological groups, Lie groups and group schemes etc. Yet I am unable to find any resources on such applications of Lawvere theories beyond the fact that the concept of a $\mathcal{C}$-valued Lawvere theory applies.

I would expect that questions like „under which conditions on $\mathcal{C}$ do colimits exists in the category of models in $\mathcal{C}$? How do we compute them?“ or „how do abelian groups in a good category $\mathcal{C}$ compare to ordinary abelian groups? Under which conditions do we have a calculus of short exact sequences? What diagram lemmas hold true in $\mathcal{C}$?“ are already studied in this context, but I could not find a reference.

PS: I guess that by Kelly‘s transfinite constructions internal algebraic constructions on locally presentable categories are monadic, as we can write down a presentation of the monad. Hence in particular the categories of models are monadic and we know how to compute colimits. Yet my question arose from trying to prove things about topological groups and compact Hausdorff groups and both $\mathsf{Top}$ and $\mathsf{CHaus}$ are far from being locally presentable.

答:
The best result I'm aware of is the following:

let $T$ be an algebraic theory such that the category of $\sf Set$-models is protomodular; then the category of models of $T$ in any finitely complete category $\cal E$ is protomodular.

If you call "pointed protomodular" an algebraic theory such that its category of $\sf Set$-models is pointed and protomodular, then the category of topological $T$-algebras is homological, complete and cocomplete. It is, however, almost never a Barr exact category.

Title: Application of Lawvere Theories


答:
"I'm not sure what to do with the floor symbol while solving for x."

Ah! Subtle point.  As $f$ is not injective you CAN'T solve for a unique $x$.  You must solve for a set of possible values.  And it's easy to do once you "wrap your head around it".

=======

To show serject you let $k$ be an arbitrary elemet of $\mathbb Z$ and see if $[\frac x2] = k$ is always possible.

It's possible if $k \le \frac x2 < k+1$ is possible.

And that is possible if $2k \le x < 2k + 2$ is possible.

So for any $k$ is it always the case that there exist real numbers that between $2k$ (inclusive) and $2k+2$ (exclusive)?  The answer is, of course.  $2k$ is a  possible value so is $2k + 1$ so is $2k + .75$ and $2k + 1.415273860$ etc.

tl;dr

For every $k\in \mathbb Z$ then for $x =2k$ then $f(x) = [\frac {2k}2] = [k] = k$ so yes it is surjective.

To prove (non)injective you try to see if $f(x) = y$ always has a unique solution.

If $f(x) = [\frac x2] = y$ then

$y\le \frac x2 < y+1$ and $2y \le x < 2y+2$.  But that's as far as we can go. Any $x' \in [2y, 2y+2)$ could give us that.  So for example:

$f(x) = 10\implies 20 \le x < 22$ and $x_1 = 20$ and $x_2 = 21$ we have $f(x_1) = f(x_2)$ but $x_1\ne x_2$ so not injective.


问:
Let $c(f)$ be a content of a polynomial over $\mathbb{Z}$ and $a \in \mathbb{Z}$. Show that $c(a \cdot f)= a \cdot c(f)$
My problem:

Let $c(f)$ be a content of a polynomial over $\mathbb{Z}$ and $a \in \mathbb{Z}$. Show that $c(a \cdot f)= a \cdot c(f)$.

The content of a polynomial over $\mathbb{Z}$ is the gcd (greatest common divisor) of its coefficients.

My attempt:

Let $f(x)=a_0+a_1x+a_2x+\ldots +a_nx^n$. We have:

$$\begin{align*}a \cdot f(x) &= a(a_0+a_1x+a_2x+\ldots +a_nx^n)\\
&=a\cdot [c(f) \cdot \tilde{f}],
\end{align*}$$

where $\tilde{f}$ is a primitive part of $f$.

I don't know how to continue from this. I will be grateful for any help.

答:
Let $\lbrace c_0, ..., c_n \rbrace$ be the coefficients of $f$. Then the coefficients of $af$ are $\lbrace ac_0, ..., ac_n \rbrace$.

Let's prove that
$$\gcd(ac_0, ..., ac_n) = a \gcd(c_0, ..., c_n)$$

First, for each $k \in \lbrace 0, ..., n \rbrace$, $\gcd(c_0, ..., c_n)|c_k$, and $a|a$, so $a\gcd(c_0, ..., c_n)|ac_k$, and because this is true for all $k$, you deduce that $$a\gcd(c_0, ..., c_n)|\gcd(ac_0, ..., ac_n) \quad \quad\quad \quad (1)$$

Conversely, for each $k \in \lbrace 0, ..., n \rbrace$, $\gcd(ac_0, ..., ac_n)|ac_k$, so $\frac{1}{a}\gcd(ac_0, ..., ac_n)|c_k$, and because this is true for all $k$, you deduce that $$\frac{1}{a}\gcd(ac_0, ..., ac_n)|\gcd(c_0, ..., c_n), \quad \text{i.e.} \quad \gcd(ac_0, ..., ac_n)|a\gcd(c_0, ..., c_n)\quad \quad\quad \quad (2)$$

So you get with $(1)$ and $(2)$ that $$\gcd(ac_0, ..., ac_n)=a\gcd(c_0, ..., c_n)$$

which is exactly the same as saying that $$c(af)=ac(f)$$


答:
You do not need, in this case, to invert the floor function explicitly to determine if it is or is not surjective. Just ask yourself: "can I find something in its codomain that the floor function cannot produce as an output?"

And the answer is right there in the definition:

the floor function is that function, from reals to reals, which produces from its single input argument the integer which is no greater than that input.

So, given that, what sorts of numbers can it not produce? Are any of these in the given codomain (the reals)? You might also note that the answer here depends on what we choose the codomain to be. What if we choose it to be the integers, instead? How does the answer change, if it does? Why?


问:
Is $\pi$ more transcendent than $e$?
I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?

答: This question at MathOverflow has several answers discussing several senses in which one can say that one real number is more irrational than another, including irrationality measures and other hierarchies of complexity for real numbers.</p>

Title: Is $\pi$ more transcendent than $e$?


问:
Minimal generating set for (R,+)
Let $\mathbb{R}$ be a group under the common addition of real numbers. Does there exist a subset $A\subset \mathbb{R}$ such that $\left=\mathbb{R}$ and for every $\alpha\in A$, $\left\ne \mathbb{R}$?

答:
No such set exists. The same argument given for the rationals here works.

Suppose $A$ is a subset of $\mathbb{R}$ such that $\langle A\rangle=\mathbb{R}$. Let $a\in A$. I claim that $A\setminus\{a\}$ still generates $\mathbb{R}$ as an abelian group.

Let $B=\langle A\setminus\{a\}\rangle$. Then $\mathbb{R}/B$ is cyclic (possibly trivial) since it is generated by $a+B$. Therefore $\mathbb{R}/B\cong\mathbb{Z}$ or $\mathbb{R}/B\cong\mathbb{Z}/n\mathbb{Z}$, $n\geq 1$. But the additive group of $\mathbb{R}$ is divisible, hence so is any quotient, but no nontrivial cyclic group is divisible. Therefore $\mathbb{R}/B$ must be trivial, i.e., $B=\mathbb{R}$, so $A\setminus\{a\}$ also generates $\mathbb{R}$.

By induction, if $A$ is a generating set for the additive group of reals, then $A\setminus F$, for any finite subset $F$, is also a generating set; there is no minimal generating set for $\mathbb{R}$.

Title: Minimal generating set for (R,+)


问:
Is there a way of thinking when solving for the unknown variables using the partial derivatives from the Lagrange function?
For example,

$L=5xy+8xz+3yz-λ(2xyz+1920)$

$L(x) = 5y+8z-2λyz=0$

$L(y) = 5x+3z-2λxz=0$

$L(z) = 8x+3y-2λxy=0$

$L(λ) = -2xyz+1920=0$

Where do I go from here?

Body:
$$
\begin{cases}
-2 \lambda  y z+5 y+8 z=0\\
-2 \lambda  x z+5 x+3 z=0\\
-2 \lambda  x y+8 x+3 y=0\\
1920-2 x y z=0\\
\end{cases}
$$
$$
\begin{cases}
\lambda = \frac{5 y+8 z}{2 y z}\\
-2 \frac{5 y+8 z}{2 y z}  x z+5 x+3 z=0\\
-2 \frac{5 y+8 z}{2 y z}  x y+8 x+3 y=0\\
1920-2 x y z=0\\
\end{cases}
$$
$$
\begin{cases}
\lambda = \frac{5 y+8 z}{2 y z}\\
-\frac{x (5 y+8 z)}{y}+5 x+3 z=0\to -z (8 x-3 y)=0 \\
-\frac{x (5 y+8 z)}{z}+8 x+3 y=0\to -y (5 x-3 z)=0\\
1920-2 x y z=0\\
\end{cases}
$$
$$
\begin{cases}
\lambda = \frac{5 y+8 z}{2 y z}\\
y= \frac{8 x}{3}\\
z=\frac{5x}{3}\\
1920-2 x \frac{8 x}{3} \frac{5x}{3}=0\to x^3=216\to x=6\\
\end{cases}
$$
$$
\begin{cases}
\lambda = \frac{1}{2 }\\
y= 16\\
z=10\\
x=6\\
\end{cases}
$$


问:
How to find the value of $\Gamma (0^+)$?
This question was part of my complex analysis assignment and I am not able to solve it.

Find the value of $\Gamma (0^+)$ . (It could be $-\infty $ or $+\infty$.)

I used the formula of gamma function which is $\Gamma (z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt $ and I got by putting $z =0^+ $, $\Gamma (0^+)=\int_{0}^{\infty}(1/x ) e^{-x} dx$ and if I integrate it by parts I get
it equal $-\infty-\int_{0}^{\infty}(1/x^2) e^{-x}dx$ .

If I again use integrating by parts  to $\int_{0}^{\infty}(1/x^2) e^{-x}dx$ and do it sucessively the power of $x$ will become more negative. So, I am not able to solve the integral.

Can you please tell how to do it?

答:
$$\Gamma(0)=\int_0^\infty \frac{e^{-x}}{x}\mathrm{d}x$$
Let $u=\log(x)$. Then
$$\Gamma(0)=\int_{-\infty}^\infty \exp(-e^u)\mathrm{d}u$$
Which certainly diverges to $+\infty$ since the integrand is $\approx 1$ for $u\to-\infty$ and $\approx 0$ for $u\to\infty$.


问:
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$

I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.

Any help is appreciated! Thanks.

Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$

Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$

答: Let $ t_n := \frac{s_n}{3^n} $. Hence, $ t_n = t_{n-1} + 3^{-n}\left(2^{n-2} - 1\right) $. Also, $ t_1 = 0 $. Hence, $$ t_n = \sum_{k = 2}^n \frac{1}{4}\left(\frac{2}{3}\right)^n - 3^{-n} $$ This is a geometric series and easily evaluated to arrive at  $$ t_n = \frac{1}{2}\cdot 3^{-n} \cdot \left(1 - 2^{n}\right) + \frac{1}{6} \implies s_n = \frac{1 + 3^{n-1}- 2^{n}}{2} $$</p>


问:
Bounded solution ODE
I have the system

$$x'(t) = y(t)$$
$$y'(t) = -y(t)^3 - p(x(t))$$

with initial condition $(x(0),y(0)) = (x_0,y_0)$ and $p(x(t))$ is a polynomial of odd degree with positive leading coefficient. I have to show that the solution $(x(t),y(t))$ is bounded for all $t \geq 0$. I am not sure how to start. I am guessing I need to look for a monotone quantity. Any help would be appreciated.

答:
Let
$$p(t)=\sum_{k=0}^{2n+1}a_kt^k\qquad\qquad a_0,\ldots, a_{2n+1}\in\mathbb{R}.$$
Assume that $x$ is not bounded as $t\to+\infty$. The linearized system around $t_0>0$ reads
$$\begin{pmatrix}x_0'(t)\\y_0'(t)\end{pmatrix}=\begin{pmatrix}0&1\\-p'(x(t_0))&-3y(t_0))^2\end{pmatrix}\begin{pmatrix}x_0(t)\\y_0(t)\end{pmatrix}.$$
The matrix has two eigenvalues $\lambda\pm$ which satisfy
$$\lambda_-+\lambda_+=-3y(t_0)^2,\qquad\lambda_-\lambda_+=p'(x(t_0)).$$
We find:
$$\lambda_\pm=-\frac{3y(t_0)^2}{2}\pm\frac{1}{2}\sqrt{9y(t_0))^4-4p'(x(t_0))}.$$
By assumption, the leading term $(2n+1)a_{2n+1}(x(t_0))^{2n}$ of $p'(x(t_0))$ is positive, so $\lambda_\pm$ have negative real part. It follows that the solutions $(x_0,y_0)$ of the linearized system are decaying in time. This implies that the solutions $(x,y)$ are also bounded (as they must be close to $(x_0,y_0)$ when $t$ is close to $t_0$ for any $t_0>0$). This contradicts the assumption on $x$.

Hence $x(t)$ is bounded for $t\in[0,+\infty)$. Now the equation $x'(t)=y(t)$ implies that $y$ is bounded too.

Title: Bounded solution ODE


问:
maximal torus and semisimple elements
I am being stupid here. $G$ is a connected algebraic group and $s$ is a semisimple element. Let $T$ be a maximal torus. Then if $T$ is contained in $C_G(s)$, then $s$ is in $T$. I got stuck here. I know semisimple elements are in maximal tori. But here, could $s$ be in a different torus $T'$ conjugate to $T$? I am trying to show the two tori are in fact one.

This question is from P112 of Gunter Malle's "linear algebraic group and finite group of Lie type".

Proposition 14.1) Let $G$ be connected, $s ∈ G$ semisimple, $T ≤ G$ a maximal torus. Then $s ∈ T$ if and only if $T ≤ C_G(s)^◦$. In particular, $s ∈ C_G(s)^◦$.

Proof: As $T$ is abelian, $s ∈ T$ if and only if $T ≤ C_G(s)$, which is equivalent to $T ≤ C_G(s)^◦$ as $T$ is connected.

答:
I'm not quite sure what you're getting at, but I think it seems like maybe you want the following:

Fact: Let $G$ be a connected reductive group over a field $F$ and
let $s\in G(F)$ be semisimple. Then, for a maximal torus $T$ of $G$
the following are equivalent:

$s\in T(F)$.

$T$ is a maximal torus of $C_G(s)^\circ$.

Proof: If $s\in T(F)$ then since $T$ is abelian we know that $T$ centralizes $s$ and thus $T\subseteq C_G(s)$. Since $T$ is connected we then also have that $T\subseteq C_G(s)^\circ$.

Remark: Note that the above argument also shows, since every semisimple element of $G(F)$ is in $T(F)$ for some maximal torus $T$ of $G$, that $s\in C_G(s)^\circ$ (it's obvious that $s\in C_G(s)$, but less obvious it's in $C_G(s)^\circ$).

Conversely, suppose that $T$ is a maximal torus of $C_G(s)^\circ$. Since $C_G(s)^\circ$ is reductive (e.g. see [Humphreys, Theorem in §2.2]--it's easy to reduce to the semisimple case) we know that $Z(C_G(s))\subseteq T$ (e.g. see [Milne, Proposition 21.7]). But, since $s\in Z(C_G(s))(F)$ this implies that $s\in T(F)$. $\blacksquare$

References:

[Humphreys] Humphreys, J.E., 2011. Conjugacy classes in semisimple algebraic groups (No. 43). American Mathematical Soc..

[Milne] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

Title: maximal torus and semisimple elements


问:
Isomorphism between $\mathbb{Z}_{2} \times \{0\}$ and $\langle (0,2) \rangle$
I'm new to group theory and I'm learning about isomorphisms and quotient groups.

I'm stuck trying to show that that $\mathbb{Z}_{2} \times \{0\} \cong \langle (0,2) \rangle$.

I understand that two groups being isomorphic is defined as there existing a homomorphism between them that is bijective.

So my thought process was to try and define some function that bijectively maps $\{(0,0),(1,0)\}$ to $\{(0,0),(0,2)\}$.

I believe that this is probably a lot more simple than I'm making it out to be, but would like to check to make sure that I'm on the right path.

Also, please tell me if there are any formatting mistakes in this post and I will try to correct them.

答:
Both groups are made of two elements -a trivial one (identity) and another element of order 2. There is only one way to create an isomorpism from one group to the other. Map the identity to the indentity and the element of order 2 to the element of order two of the other group.

Note that those two groups are basically the same - up to changing the 'names' of the elements, they behave in the exact same way. That's exactly the meaning of an isomorphism.


问:
Bounded Variation Chain Rules
Let $\Omega \subseteq \mathbb{R}^d$ (or for simplicity take $\Omega = \mathbb{R}^d$). Let $u\in L^1(\mathbb{R}^d)$ such that $u\in \text{BV}(\Omega)$ i.e its distributional derivative denoted $Du$ is representable as a finite Radon measure. How can I relate $D(u^m)$ to $Du$, for some power $m\in \mathbb{N}$. I guess I'm asking for a chain rule for distributional derivatives.

My main confusion with these things is that at this moment I cannot show $u \in W^{1,1}(\mathbb{R}^d)$, hence don't have access to the usual chain rule.

答:
Setting $f(u) := u^m$ and $v := f\circ u = u^m$, the chain rule in BV says that
$$
\begin{cases}
\widetilde{D} v = f'(u) \nabla u \, \mathcal{L}^d + f'(\widetilde{u}) D^c u,\\
D^j v = [f(u^+) - f(u^-)]\, \mathcal{H}^{d-1}\lfloor J_u.
\end{cases}
$$
Since $f$ is not (globally) Lipschitz continuous, you should know beforehand that $v$ belongs to $BV$.

(For notations and the proof of the result, see Ambrosio-Fusco-Pallara, "Functions of Bounded Variations and Free Discontinuity Problems", Thm. 3.96.)

Title: Bounded Variation Chain Rules


问:
Rational Equations
So, I got this question:
Essentially, a person wants to open an art venue, but the cost is $\$400$ and the foodservice she's working with is charging $\$3.75$ per person for food and drink. The task is to create an equation that evenly splits the cost of the venue between tickets.
My equation was $3.75x + \frac {400}x$

A friend is insisting that it's $\frac {3.75x +400}{x}$
Who is correct? I thought that his equation can't be correct because dividing whatever $x$ is by itself would just be one, and then $3.75$ would be multiplied by $1$.

答:
The equation you are probably looking for is

$$T=3.75+{400\over x}$$

where $T$ represents the price of a ticket and $x$ represents the number of tickets you plan to sell (i.e., the number of people who will attend the event). There are two important points here. First, when formulating a mathematical description of a problem, it's important to explicitly say what the variables you use represent. Second, when writing down an equation for a mathematical description of a problem, it important to actually write down an equation, and not just one side of it.

Where I said "probably," it's because you might actually be looking for the equation in the equivalent form

$$T={3.75x+400\over x}$$

(which uses your friend's expression on its right hand side). An interpretation of the equation in this form is that you are computing the total cost of the event in the numerator on the right hand side and then dividing by the number of tickets to get the ticket price, whereas in the first equation you are thinking of the ticket price as comprised of two parts: the cost of the ticket buyer's food (the $3.75$) and their share of the venue (the $400/x$).

Title: Rational Equations


问:
Spectral norm of banded binary matrix
Let $M\in \mathbb{R}^{n\times n}$ defined as $M_{ij}=1$ if $|i-j|\leq k$, and $0$ otherwise.

Is it true that $\|M\|_2\leq O(k)$?

I have shown that $\|M\|_2\leq O(k^{3/2})$, by decomposing the matrix $M$ into a sum of k block-diagonal matrices, each with spectral norm $O(\sqrt{k})$.

答:
The answer is yes.

Note that because $M$ is symmetric, its spectral norm is equal to its spectral radius (maximal absolute value of an eigenvalue). It follows that
$$
\|M\|_2 \leq \|M\|_\infty \leq 2k + 1 = O(k),
$$
where $\|M\|_\infty$ denotes the induced $\infty$-norm.

Equivalently, we could apply the Gershgorin circle theorem to reach the same conclusion.

Title: Spectral norm of banded binary matrix