问:
Integrating a poisson kernel in $n$ dimensional unit sphere without using that it's the solution of a dirichlet problem
Someone has asked exactly this question but the answer was preciselly the thing I cannot use, so I'll ask again:

Let $$K(x,y) = \frac{1}{n \alpha(n)}\frac{1-\|x\|^2}{\|x-y\|^n}$$ be the poisson kernel for the ball in $\mathbb{R}^n$. Here $\alpha(n)$ is the volume of the $n$ dimensional ball of radius $1$.

I'd like to show that
$$\int_{\partial B_1(0)}K(x,y)dH^{n-1}(y) = 1$$

(here $dH^{n-1}$ is the n-1 dimensional Hausdorff measure)

The thing is, I CANNOT USE the fact that

$K[f](x) = \int_{\partial B_1(0)}K(x,y)f(y)dH^{n-1}(y)$
is the solution of the Dirichlet problem
$$\left\{\begin{matrix}\Delta u = 0 \; \mbox{ in } B_1(0)\\u = f\; \mbox{ in } \partial B_1(0)\end{matrix}\right.$$

as we used the fact that the integral is $1$ in the first place to prove that $K[f](x)$ is a solution of the Dirichlet problem.

On Lawrence Evans' book "partial differential equations" it says simply "A direct calculation, the details which we omit, verifies [...]"
So it should be possible to calculate this integral. I just don't know how.

Any tips, or book references that might help me are appreciated. Thanks in advance.

答:
In this post there is a very elegant proof of the statement using the strong maximum principle for harmonic functions. Although it uses the fact that
$$ \Delta K[f] (x) = 0,$$
you don't need to use the fact that $\int_{\partial B(0,1)} K(x,y) dy =1$ in first place to show that the Poisson kernel for the ball is harmonic. I hope it helps.


问:
Given a Group G, of order 21, acting on X of size 13. I was asked to show the action has atleast one fixed point.
I proceeded as follows:

Orbit - Stabilizer theorem forces the size of the orbits to be divisors of $G$.
Therefore, orbit sizes are of $1$,$3$, or $7$.

Then, since the orbits partition the set $X$,
$X$ must be a combination of the distinct orbits
$\Rightarrow$ $13 = t+3u+7v$.

And I would expect to find no combination with $t=0$ to solve this.

However in this occassion this didnt happen.
What is another approach to this sort of question?

答: More explicitly, for $G = (\mathbb{Z}_3, +) \times  (\mathbb{Z}_7,+)$ we can find a transitive action on $\mathbb{Z}_3$: $(g,f)x = g+x$, and on $\mathbb{Z}_7$: $(g,f)x = f+x$. Then we construct an action of $G$ on a disjoint union $\mathbb{Z}_3 \cup \mathbb{Z}_3 \cup \mathbb{Z}_7$. The idea belongs to @freakish.

ContentLicense: CC BY-SA 4.0


问:
Proving that operator is linear if it is unitary and surjective

Let $X$ be unitary space. Show that surjective operator $T: X\to X$ such that $\langle Tx, Ty\rangle = \langle x, y \rangle \forall x, y \in X$ is linear.

The only thing I was able to conclude (trivial stuff) is that
$$
\langle Tx, Ty\rangle = \langle x, y \rangle \implies \langle Tx, Tx\rangle = \langle x, x \rangle \implies \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} \\
\implies \| Tx\| = \|x\|
$$
so $T$ is isometry.

To show that it is linear I want to have:
$$
T(ax) = aT(x) \\
T(x+y) = T(x)+T(y)
$$

答:
\begin{align}
\langle T(x+y),T(z)\rangle&=\langle x+y,z\rangle\\
&=\langle x,z\rangle +\langle y,z\rangle\\
&=\langle T(x),T(z)\rangle+\langle T(y),T(z)\rangle\\
&=\langle T(x)+T(y),T(z)\rangle
\end{align}
Since $T$ is onto, $T(z)$ can be any vector, hence $T(x+y)=T(x)+T(y)$.

Same proof can be modified for $T(\alpha x)=\alpha T(x)$.


问:
How to compute $\mathbb{P}(B(\tau)\leq a)$, $B(t)$ is standard Brownian motion.
How to compute $\mathbb{P}(B(\tau)\leq a)$, where $\tau$ ~ Exp(λ), which is independent of $B(t),t\geq 0.$

here is my computation:
$ \mathbb{P}\left(B\left(\tau\right)\le a\right) =\mathbb{E}\left[\mathbb{E}\left[1_{\left\{ B\left(\tau\right)\le a\right\} }|\tau\right]\right]
 =\int_{0}^{\infty}\lambda e^{-\lambda s}\mathbb{P}\left(B\left(\tau\right)\le a|\tau=s\right)ds
 =\int_{0}^{\infty}\lambda e^{-\lambda s}\mathbb{P}\left(B\left(s\right)\le a\right)ds
 =\int_{0}^{\infty}\lambda e^{-\lambda s}\mathbb{P}\left(\frac{B\left(s\right)}{\sqrt{s}}\le\frac{a}{\sqrt{s}}\right)ds
 =\int_{0}^{\infty}\lambda e^{-\lambda s}\Phi\left(\frac{a}{\sqrt{s}}\right)ds$.

why we have
$=\int_{0}^{\infty}\lambda e^{-\lambda s}\mathbb{P}\left(B\left(\tau\right)\le a|\tau=s\right)ds
 =\int_{0}^{\infty}\lambda e^{-\lambda s}\mathbb{P}\left(B\left(s\right)\le a\right)ds$ ?

it means that $\mathbb{P}\left(B\left(\tau\right)\le a|\tau=s\right)=\mathbb{P}\left(B\left(s\right)\le a\right) \ \ \mathbb{P}_\tau -a.s.$

how to prove $\mathbb{P}\left(B\left(\tau\right)\le a|\tau=s\right)=\mathbb{P}\left(B\left(s\right)\le a\right) \ \ \mathbb{P}_\tau -a.s.$?

答:
It boils down to more general result. Namely:

Theorem Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $\mathcal G \subset \mathcal F$ a sub-$\sigma-$field. Let $X:\Omega \to (E_1,\mathcal E_1)$ and $Y: \Omega \to (E_2,\mathcal E_2)$ be random variables (where $(E_i,\mathcal E_i)$ are some let's say polish spaces). Assume that $X$ is independent of $\mathcal G$ whereas $Y$ is $\mathcal G$ measurable. Moreover let $F:E_1 \times E_2 \to \mathbb R$ be measurable such that $\mathbb E[|F(X,Y)|] < \infty$. Then $\mathbb E[F(X,Y)| \mathcal G] = H(Y)$ almost surely, where $H(t) = \mathbb E[F(X,t)]$

Proof Take any $A \in \mathcal G$. We need to prove $$ \mathbb E[F(X,Y) 1_A] = \mathbb E[H(Y)1_A] $$ To do this, consider random vector $V=(X,Y,1_A)$ and note that due to independence assumptions, $X$ is independent from $(Y,1_A)$, so that letting $\mu_X,\mu_{(Y,1_A)}$ be respectivelly distributions of $X$ and $(Y,1_A)$ we arrive at $$ \mathbb E[F(X,Y)1_A] = \int_{E_1 \times E_2 \times \mathbb R} F(x,y)z d\mu_{(X,Y,1_A)}(x,y,z) = \int_{E_1}\int_{E_2 \times \mathbb R} F(x,y)z d\mu_X(x)d\mu_{(Y,1_A)}(y,z) = $$ $$ = \int_{E_2 \times \mathbb R} z \int_{E_1} F(x,y) d\mu_X(x) d\mu_{(Y,1_A)}(y,z) = \int_{E_2 \times \mathbb R} z \mathbb E[F(X,y)] d\mu_{(Y,1_A)}(y,z)$$
Where we used Fubinii due to integrability assumptions. Now note that under integral we have our function $H$, hence we showed $$ \mathbb E[F(X,Y)1_A] = \int_{E_2 \times \mathbb R} z H(y) d\mu_{(Y,1_A)}(y,z) = \mathbb E[1_A H(Y)] $$ which proves our theorem.

In your question, you want to use it here: $$ \mathbb E[1_{B(\tau) \le a} | \tau] $$ with $F(B,\tau) = 1_{B(\tau) \le a}$ (we treat here $B$ as a function $\Omega \to C([0,\infty))$ and $\mathcal G = \sigma(\tau)$ getting $$ \mathbb P(B(\tau) \le a) = \mathbb E[ \mathbb E[F(B,\tau) | \tau]] = \mathbb E[ H(\tau)] $$ where $H(t) = \mathbb E[F(B,t)] = \mathbb P(B(t) \le a) = \Phi(\frac{a}{\sqrt{t}})$. Hence $$ \mathbb P(B(\tau) \le a) = \mathbb E[ \Phi(\frac{a}{\sqrt{\tau}})] = \int_0^\infty \lambda e^{-\lambda x} \Phi(\frac{a}{\sqrt{x}})dx $$

Edit: Maybe two words why function $F$ is measurable. Note that $F$ is a superposition of $1_{[-\infty,a]}$ which is measurable and function $G:C([0,\infty) \times \mathbb R \to \mathbb R$ given by $G(f,x) = f(x)$. It is not hard to see that $G$ is continuous, hence measurable. Indeed, taking any such $f,x$ and $\varepsilon > 0$ we have for all $g \in C([0,\infty)$ such that $\|f-g\|_{\infty} \le \frac{\varepsilon}{2}$ and for all $y \in \mathbb R$ such that $|x-y| < \delta$ for appriopate $\delta$ giving $|f(x)-f(y)|<\frac{\varepsilon}{2}$ that $$ |G(f,x)-G(g,y)| = |f(x)-g(y)| \le |f(x)-f(y)| + |f(y)-g(y)| \le \frac{\varepsilon}{2} + \|f-g\|_{\infty} \le \varepsilon $$


问:
what is $ \lim_{x\to 0}\frac{cos(x)-1}{8x^3-6x^2} $
I tried to calculate it using LHopital's rule and got $\frac{1}{12}$, however, symbolab says its infinity

答:
The numerator is in essence $-\frac12 x^2 + O(x^4)$ as $\cos x = 1 - \frac12 x^2 + O(x^4)$, the denominator is $x^2(8x-6)$, so
the fraction is (cancelling the $x^2$ in both terms) $$\frac{-\frac12 +O(x^2)}{(8x-6)}$$ So the limit as $x \to 0$ will be clear now...


问:
weighted average
I'm not sure the correct term for my problem is weighted average. But let me explain.

I've conducted a survey where participants answer on a scale betweeen $1$ and $7$.
The questions fall into three categories. In category one & two there are $12$ questions, and in category three there are four questions.

Lets assume there are nine participants and the distribution of the answers are $579$ points in category one. $450$ points in category two and $87$ in category three.

I want to calculate/show the answer distribution of each category (compared to the total points), taking into account that category three is three times smaller.

I hope I'm making sense.

答:
Just multiply the points in third category by 3 to make points in different categories comparable.
BTW There is a separate forum for Statistics
Cross Validated</p>

ContentLicense: CC BY-SA 3.0

Title: weighted average


问:
Prove that if $p > 1$, then $V$ is dense in $l^p$
In the space $l^p$ with $1 ≤ p < ∞$ we consider the following subset:

$V = ${${(x_1,x_2,x_3,...) \in l^p : \sum_{n=1}^{\infty}}x_n = 0$}

I have to prove that if $p > 1$, then $V$ is dense in $l^p$ and if $p = 1$, then $V$ is not dense in $l^p$.

I know that if we let $X$ be a metric space, the subset $V \subset X$ is called dense when the closure of $V$ equals $X$, thus when $\overline{V} = X$.

For the first statement, the textbook states that we can use the following: Let $e_k = (0, . . . , 0, 1, 0, . . .)$, where the $1$ is at the k-th entry, and show that $e_k ∈ V$  $\forall k ∈ \mathbb{N}$.

I am not sure how to use the hint given by the textbook. Any help would be grateful. Thanks in advance.

答: Fix $k$. Define $$v_n=e_k-\frac1n\sum_{j=1}^n e_j.$$ Show that $v_n\in V$ and $\lim_{n\to\infty}||e_k-v_n||_p=0$. Hence $e_k\in\overline V$.


问:
If there is a $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $\langle x,y\rangle=0$ for all $y\in V$, then $\overline V\ne\ell^q$?
This should be quite obvious, but I struggle to find a formal proof: Let $1\le p<q\le\infty$ with $\frac1p+\frac1q=1$. We know that $\ell^q$ is isometrically isomorphic to $(\ell^p)'$ via $$\iota:\ell^q\to(\ell^p)'\;,\;\;\;y\mapsto\langle\;\cdot\;,y\rangle,$$ where $$\langle x,y\rangle:=\sum_{n\in\mathbb N}x_ny_n\;\;\;\text{for }(x,y)\in\ell^p\times\ell^q.$$

Assume $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $$\langle x,y\rangle=0\;\;\;\text{for all }y\in V\tag1.$$ Why can we conclude that $\overline V\ne\ell^q$?

答:
$\langle x,-\rangle\colon\ell^q\to\mathbb{R}$ is a continuous linear functional on $\ell^q$.  $V$ is contained in the proper closed subspace $(\langle x,-\rangle)^{-1}(\{0\})$ of $\ell^q=(\ell^p)'$ so cannot be $\ell^q$ itself: There is a continuous linear functional on $\ell^p$ which doesn't annihilate $x$, namely $y\in(\ell^p)'=\ell^q$ with
$$
y_i=
\begin{cases}
\frac{\lvert x_i\rvert^{p}}{x_i} & x_i\neq 0\\
0 & x_i=0
\end{cases}.
$$


问:
Equivalence between statements on linear independence
Consider the integers $N,T,J$. In what follows, $I_N$ is the $N\times N$ identity matrix. $0_{N\times N}$ is the $N\times N$ zero matrix.

Let $A$ be an $NT\times (N+J)$ matrix structured as follows
$$
A\equiv \begin{pmatrix}
I_N & A_1\\
I_N & A_2\\
\vdots & \vdots\\
I_N & A_T
\end{pmatrix}
$$
where each of $A_1,...,A_T$ are $N\times J$ matrices.

Could you help to show that (or give me an informal intuition on why)

(1) The columns of $A$ are linearly independent if and only if the columns of
$$
B\equiv \begin{pmatrix}
I_N & A_1\\
0_{N\times N} & A_2-A_1\\
\vdots & \vdots\\
0_{N\times N} & A_T-A_1
\end{pmatrix}
$$
are linearly independent

(2) The columns of $A$ are linearly independent if and only if the columns of
$$
C\equiv \begin{pmatrix}
 A_2-A_1\\
\vdots \\
A_T-A_1
\end{pmatrix}
$$
are linearly independent

答:
You are simply doing row operations (in blocks), subtracting $[I_{N} \ A_{1}]$ from each block below it. Row operations don't effect the linear dependence/independence relations of the columns.

Once you do this, you can clearly see that there is a pivot position in each of the first $N$ columns. Any pivot position in the columns below that would correspond to a pivot position in
$$\begin{bmatrix} A_{2} - A_{1} \\ \vdots \\ A_{T} - A_{1}\end{bmatrix}.$$


问:
Is homotopy of smooth maps $M \to \mathbb{S}^n$ equivalent to smooth homotopy?
I'm trying to solve the following exercise from Milnor's Topology from the Differentiable Viewpoint

Let $M$ be a compact smooth manifold, show that every continuous map $M \to \mathbb{S}^n$ can be uniformly approximated by a smooth map. If  two smooth maps are continuously homotopic, show that they are smoothly homotopic.

I solved the first part approximating every coordinate thanks to Stone-Weierstrass' Theorem and normalizing. I think it should work. Then I tried to use the first part to solve the second one (about smooth homotopy), but I'm not able to understand how I should use approximation to change a continuous map to a smooth one. How could I solve this problem?

答:
The intuitive idea is to just apply a suitable approximation theorem a second time, namely to the homotopy $h \colon M \times I \to \mathbb{S}^n$. But you have to be a bit more careful because you want $h_0$ and $h_1$ to not change. A reference for the fact that you can do this is Theorem 6.21 of Lee's Intro to Smooth Manifolds. I'll more or less recite it.

Theorem. Let $M$ be a smooth manifold with boundary and $F \colon M \to \mathbb{R}^k$ be a continuous function such that $F|_{\partial M}$ is smooth. Then $F$ can be uniformly approximated by a smooth function $\widetilde{F} \colon M \to \mathbb{R}^k$ in such a way that $\widetilde{F}|_{\partial M} = F|_{\partial M}$.

The rough idea is as follows. First, extend the smooth function $F|_{\partial M}$ to a smooth function $G_1 \colon M \to \mathbb{R}^k$. Now let $U = \{x \in M : |G(x) - F(x)| < \varepsilon\}$, which is an open subset containing $\partial M$. Now uniformly approximate $F$ on the interior to some $G_2$ using e.g. Stone--Weierstrass, and then use a partition-of-unity argument to combine $G_1$ on $U$ with $G_2$ on $M \setminus U$ so as to form a global function $\widetilde{F}$ which both approximates $F$ everywhere and still stays fixed at the boundary.


问:
Taylor Series, Newtons Method and Newton Raphson
I am currently learning about different optimization methods. That is, I am interested in finding the maxima/minima of functions. The last method I have studied is Newton Raphson.

I understand Newton Raphson. As far as I understand it, Newtons method is simply the first order case of the Newton Raphson method. Is this correct?

Now, how do both of these methods relate to the concept of Taylor Series? It might be a trivial question but I am a bit confused, because these concepts seem to overlap a bit.

答:
To my knowledge there is no distinction between Newton's method and the Newton-Raphson method, they are the same thing.

As for how Newton's method relates to power series. In the optimization context, Newton's method (iteratively) optimizes the best quadratic approximation to a function $f$ at a point $x_0$, which is given by truncating its power series at the second order term,

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}\approx f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2.$$

Taking the derivative of the quadratic approximation and setting it equal to zero gives us,

$$x=x_0-\frac{f'(x_0)}{f''(x_0)}.$$

Thus optimizing the best quadratic approximation to $f$ gives rise to the familiar root finding algorithm on $f'$, where each iteration in the root finding algorithm is equivalent to reforming the best quadratic approximation at each step of the optimization algorithm.


答: An appropriate choice of $y\in \ell^q\setminus\{0\}$ among the ones that realise the equality case of Hölder's inequality, namely $y_n=\alpha_n\cdot\lvert x_n\rvert^{p/q}$ where $\alpha_n\cdot x_n=\lvert x_n\rvert$, will be in $\ell^q\setminus\overline V$, because for said $y$ $$\langle x,y\rangle=\lVert xy\rVert_1=\lVert x\rVert_p\lVert y\rVert_q\ne0$$

Title: If there is a $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $\langle x,y\rangle=0$ for all $y\in V$, then $\overline V\ne\ell^q$?

Body:
This should be quite obvious, but I struggle to find a formal proof: Let $1\le p<q\le\infty$ with $\frac1p+\frac1q=1$. We know that $\ell^q$ is isometrically isomorphic to $(\ell^p)'$ via $$\iota:\ell^q\to(\ell^p)'\;,\;\;\;y\mapsto\langle\;\cdot\;,y\rangle,$$ where $$\langle x,y\rangle:=\sum_{n\in\mathbb N}x_ny_n\;\;\;\text{for }(x,y)\in\ell^p\times\ell^q.$$

Assume $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $$\langle x,y\rangle=0\;\;\;\text{for all }y\in V\tag1.$$ Why can we conclude that $\overline V\ne\ell^q$?


问:
$L^p$ and $L^q$ spaces proof
I am studying for my final exam and I have found such a task:

For every $p,q \in (1, \infty)$ we know that $\frac{1}{p}+\frac{1}{q}=1$. Show that if $ f \in L^p(R)$ fulfills  $\int_{R}{f(x)g(x)}dx=0$ for every $g \in L^q(R)$ then $f=0$ almost everywhere.

I have no idea how to approach this problem. Any help would be useful.

答:
Hint :

Apply the hypothesis to
$$g=\mathrm{sign}(f) \times |f|^{\frac{p}{q}}$$

Title: $L^p$ and $L^q$ spaces proof


问:
Family of elliptic curves with trivial torsion
I'm wondering, if it is true that the torsion subgroup of $y^2=x^3+p$ (for $p$ some prime, greater than 2), is always trivial?.

I was trying to prove this using Lutz-Nagell, but I can't quite get it.

Thank you

答:
Nagell-Lutz says,as you know, that if a rational point $P=(x,y)$ is torsion, both x and y are integers and either:

$y=0$

y divides the discriminant of your curve (and hence, so must $y^2$).

In your case, the discriminant is $-27p^2$. So if $y^2 | -27p^2$ we see that y can either be $\pm 3$, $\pm p$, $\pm 3p$ or $\pm 1$. Suppose first that $p \neq 5$. Then one can show that, since 5 is a place of good reduction and the ramification degree is 1, $E(\mathbb{Q})_{tors}| E(\mathbb{F}_5)$ (here this is the order of the torsion groups of E over the respective field).It is then clear that $E(\mathbb{Q}_{tors})$ must be either 1,2,3 or 6, since  $E_(\mathbb{F}_5)=6$. It is clear that there can be no point of order 2.

For points of order 3, use the m-division polynomials to find that if P is a 3-torsion point, $\phi_3(x,y) = 3x(x^3+4p)$. Now, $x=0$ is a solution of this equation, and inserting this into your equation you get $y^2=p$, but since p is prime, there can be no points satisfying this equations on your curve. $x^3+4p= 0$ has no rational solutions, so no 3-torsion points exist.

For points of order 4, consider the 4th divisopn polynomial $\phi_4(x,y)/\phi_2(x,y)=2x^6+80px^3-8p^2$. It has no rational roots, so no points of order 4 exist. One knows that point of order 6 exists if points of order 2 and 3 exist. No points of order 2 or 3 exists, so you are done.

So, for the case that $p=5$, you can finish it off easily:
y^3=x^2+5, Nagell-Lutz gives you that $y= \pm 3$, $y= \pm 5$, $y= \pm 1$, $y= \pm 15$. Checking these cases, you are done.


问:
Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$
I saw this question in an old trig text book:

Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$.

I ask:

What solution can people here give ?

Is there a solution that does not (like mine) require the construction of $\sqrt[4]{5}$ ? In other words, a solution without assuming the circle intersection property (See Hartshore: Euclid and Beyond pg. 144)

答:
Outline. Let $A=(0,2)$ and $B=(1,0)$. Consider the circle around $A$ with radius $AB=\sqrt{5}$, which intersects the $y$-Axis at $C=(0, 2+\sqrt{5})$. Let now $\Gamma$ be the circle centered at $O=(0,0)$ through $C$ which intersects the line $l$ through $D=(0,3)$ and parallel to the $x$-Axis at $E$. Then

$$\sin\angle DEO=\frac{OD}{OE}=\frac3{2+\sqrt{5}}$$


问:
Determinant of $n\times n$ matrix $|i-j|$
For integer $n$ greater than or equal to 2, let $A_n$ be $n^{th}$ order square matrix whose $(i,j)$ component is $|i-j|$. That is,
\begin{align*}
    A_n =\begin{pmatrix}
    0& 1& 2& ... & n-1\\
    1& 0& 1& ... & n-2\\
    2& 1& 0& ... & n-3\\
    .& .& .& ... & . \\
    .& .& .& ... & . \\
    .& .& .& ... & . \\
    n-1& n-2& n-3& ... &0
    \end{pmatrix}
\end{align*}
Find the determinant of $A_n$.

We have two general tools for the determinant of $n\times n$ matrix, they are co-factor formula and the Leibniz formula but I tried to build an expression for the determinant of this matrix but failed, Any hints would be appreciated.

Leibniz Formular : $\det A = \sum_{\sigma \in S_n}\bigg(\text{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}\bigg) $ I can figure this out for $3\times 3$ matrices but it is hard to think for $n\times n$ matrices with these permutations $\sigma$.

And how do we find these kind of problems on the internet? Any recommendations would be appreciated

答:
Answering for closure. Let me write $v_1,\ldots,v_n$ for the columns of $A$. Now consider the matrix with columns $v_1,v_2 - v_1,\ldots,v_n - v_{n-1}$. It looks like
$$\begin{pmatrix}
0 & 1 & 1 & 1 & \cdots &1& 1 \\
1 & -1 & 1 & 1 & \cdots & 1&1 \\
2 & -1 & -1 & 1 & \cdots &1& 1 \\
3 & -1 & -1 & -1 & \cdots & 1& 1 \\
\vdots & \vdots & \vdots & \vdots & \ddots &\vdots& \vdots \\
n - 2 & -1 & -1 & -1 & \cdots & -1 & 1\\
n - 1 & -1 & -1 & -1 & \cdots & -1 & -1
\end{pmatrix}$$
No apply Laplace expansion on the bottom row to obtain a recurrence relation between $\det A_n$ and $\det A_{n-1}$. Use this to prove that, in general,
$$\det A_n = (-1)^{n-1} (n-1)2^{n-2}.$$


问:
Is this relationship on subcartesian products of quotient groups over the free group correct?
I am trying to check if the following relationship is correct. Let $F$ be a free group over some set $X$. Let $\alpha_1, \alpha_2, ..., \alpha_n$ be epimorphisms from $F$ to groups $H_1, H_2, ..., H_n$ respectively. Let $K_1, K_2, ..., K_n$ be the kernels of $\alpha_1, \alpha_2, ..., \alpha_n$ respectively. It follows that $K_1, K_2, ..., K_n$ are all normal.

Now, let $K= \bigcap K_i$ for $i=1,...,n$ (i.e. the normal group formed from the intersection of the kernels). We can now form the quotient $F/K$.

Is it true that $F/K$ is isomorphic to a subcartesian product of $F/K_1,F/K_2,...,F/K_n$ ? How could I show it ?

I want to check if this is correct since this would then mean that $F/K$ is also isomorphic to a subcartesian product of $H_1,H_2,...,H_n$ ...

(For context over this question, am trying to understand a proof for Birkhoff's variety theorem which uses the above)

答:
Edit: as pointed out in the comments, "subcartesian product" has a formal meaning and is not the same as a cartesian product of subgroups.

Some thoughts which I found too long for a comment:

The maps $\alpha_i$ define a map

$$
f \colon t \in F \mapsto (\alpha_1(t),\ldots, \alpha_n(t)) \in H_1 \times \cdots \times H_n
$$

Its kernel consists of the $t \in F$ for which $\alpha_i(t) = 1$ for all $i$ simultaneously, i.e. $K = \cap_i K_i$. Hence

$$
F/K \simeq \mathbf{im} f \leq H_1 \times \cdots \times H_n.
$$

We have not used that each $\alpha_i$ is epi nor the fact that $F$ is free. In general, the image of $f$ need not be a cartesian product of subgroups, even with these hypotheses (e.g. if $F = \Bbb Z$, $n = 2$, and $\alpha_1 = \alpha_2 = id$, then $f$ is the diagonal map $\Delta \colon \Bbb Z \to \Bbb Z^2$).

It may be that $F/K$ is still isomorphic to some cartesian product through some other morphism.


问:
finding a matched subset, that meets meets a condition
Given a finite set $A$ of random $n$-dimensional vectors and a $n$-dimensional vector $\vec v$, how do I find a subset $B \subset A$ whose vectors average deviate least from $\vec v$.

$$B = \left\{ \vec d_1, \vec d_2 \cdots  \vec d_n \right\}, \vec d_a = \begin{pmatrix} e_{a1} \\ e_{a2} \\ \cdots  \\ e_{am} \end{pmatrix}$$

so that

$$ \lvert \sum_{a=1}^n \frac{d_a}{n} - \vec v \rvert
$$

is minimal.

答:
Because of the fact that each vector has to appear with a weight either 0 or $\frac 1 n$, this makes it hard. Even in dimension 1, this problem is NP-hard : see https://en.wikipedia.org/wiki/Subset_sum_problem.

Of course, if instead of searching for an isobarycenter, you can take any barycentric combination of the vectors (for example $\frac 1 3 \vec d_1 + \frac 1 6 \vec d_2 + \frac 1 2 \vec d_3$), then this just becomes geometry, and solving this becomes an easy problem (though not always fun to code).


问:
How do I solve a differential equation on this form?
I have a differential equation I am a bit confused about. Basically, I want to express $x_1(t)$ in terms of $x_2(t)$.

This is how they are related (I will let $\dot{x} = \frac{d}{dt} x$):

\begin{equation}
\dot{x_2}(t)^2 = -\dot{x_1}(t)^2 e^{2c x_2(t)}
\end{equation}
where c is a constant.

so basically I have two differential equations that are equal to each other. I don't know how to solve this at all.

Could I rewrite it as $$\dot{x_1} = - \dot{x_2} e^{-c x_2}$$
And just solve for $x_2(t)$ treating the L.H.S as a constant? I am not sure that is allowed.

Body:
$$\dot{x_2}(t)^2 = -\dot{x_1}(t)^2 e^{2c x_2(t)}$$
$$(\dot{x_2}(t)^2 -i^2\dot{x_1}(t)^2 e^{2c x_2(t)})=0$$
$$(\dot{x_2}(t) -i\dot{x_1}(t) e^{c x_2(t)})(\dot{x_2}(t) +i\dot{x_1}(t) e^{c x_2(t)})=0$$
Then you have the equations:
$$(\dot{x_2}(t) -i\dot{x_1}(t) e^{c x_2(t)})=0$$
$$(\dot{x_2}(t) +i\dot{x_1}(t) e^{c x_2(t)})=0$$
Then you can integrate both equations.
$$\int e^{-c x_2}dx_2=\pm i\int  dx_1$$


答:
Here is an interpretation of the answers so far given, breaking down the argument, and highlighting an
important intermediate result.

First fact. Let $T$ be a surjective isometry.  Then $T$ admits an adjoint operator.  In other words,
for every $y$ in $X$,  there exists a (necessarily unique) $z$ such that
$$
  \langle T(x), y\rangle  =   \langle x, z\rangle , \quad\forall x\in  X.
  $$

Proof (@Udalricus.S)  Given $y$,  find $z$ such that $T(z)=y$.  Then
$$
  \langle T(x), y\rangle  =   \langle T(x), T(z)\rangle  = \langle x, z\rangle .
  \tag*{$\square$}
  $$

Incidentally,  the vector $z$ referred to above is often denoted by $T^*(y)$,  so the highlited expression
in the statement  becomes the familiar one:
$$
  \langle T(x), y\rangle  =   \langle x, T^*(y)\rangle .
  $$

Second  fact. Let $T:X\to X$ be a function (not necessarily linear,  unitary or surjective) and suppose that there exists another
function $T^*:X\to X$,  such that
$$
  \langle T(x), y\rangle  =   \langle x, T^*(y)\rangle \quad\forall x, y\in  X.
  $$
Then $T$ is linear.

Proof.  (@Chrystomath) Given $x_1$ and $x_2$ in $X$,  we have for all $y$ that
$$
  \langle T(x_1+x_2), y\rangle  =
  \langle x_1+x_2, T^*(y)\rangle  = $$$$ =
  \langle x_1, T^*(y)\rangle  +  \langle x_2, T^*(y)\rangle  =
  \langle T(x_1), y\rangle  +   \langle T(x_2), y\rangle  = $$$$ =
  \langle T(x_1) +   T(x_2), y\rangle .
  $$
Since $y$ is arbitrary, we deduce that $T(x_1+x_2)=T(x_1) +   T(x_2)$.  Similarly one proves that
$T(\lambda x)=\lambda T(x)$.  QED

As an added bonus, should $X$ be a Hilbert space, one can show by means of the closed graph Theorem that
the map $T$ of the second fact above is moreover bounded!

Title: Proving that operator is linear if it is unitary and surjective

Body:
Let $X$ be unitary space. Show that surjective operator $T: X\to X$ such that $\langle Tx, Ty\rangle = \langle x, y \rangle \forall x, y \in X$ is linear.

The only thing I was able to conclude (trivial stuff) is that
$$
\langle Tx, Ty\rangle = \langle x, y \rangle \implies \langle Tx, Tx\rangle = \langle x, x \rangle \implies \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} \\
\implies \| Tx\| = \|x\|
$$
so $T$ is isometry.

To show that it is linear I want to have:
$$
T(ax) = aT(x) \\
T(x+y) = T(x)+T(y)
$$


问:
System of Equation Solver? Online
What site do people use to solve systems of equations with algebraic coefficients. I.E I know wolfram solves

$$
x+y=2,~~y-3x=1
$$
but are there sites that let you put unknown constants?
$$
x+c_1y=2,~~y-c_2x=c_3.
$$

EDIT : (all helpful answers and I would up vote if I could!) Note I tried this in wolfram and it didn't work (this is the beast I wanted solved for me ):

solve v_1=A+B- (c_5 a / c_1(c_8+ c_3c_5/c_1 )),   v_2=Ae^{r_1 h }+Be^{r_2 h}-  (c_5 a / c_1(c_8+ c_3c_5/c_1 )),  x_1=b-(1/c_1)(A(c_2+c_3/r_1)+B(c_2+c_3/r_2)), x_2=b+ah-(1/c_1)(Ae^{r_1 h}(c_2+c_3/r_1)+Be^{r_2 h}(c_2+c_3/r_2)),   for a, b, A, B

答: Yes. Graphically you can do this on Desmos and get a sliding bar to manipulate the variables. Here are your equations.</p>

Title: System of Equation Solver? Online


问:
4 intersecton points of 2 conic circles are always concyclic?

Given 2 conic section of type

$$ax^2+2hxy+by^2+2gx+2fy+c=0$$
have 4 intersection points. Are the points for sure concyclic?

If yes, does the proof come from the fact that I can create an infite amount of new conic section using the linear combination of them and there is for sure a value of lambda that will give me a circle?

答:
No. In geogebra varying coefficients you can easily find counterexamples.

This has solutions

$(1,0),(\frac1{\sqrt{3}},\frac1{\sqrt{3}}),(-\frac1{\sqrt{3}},-\frac1{\sqrt{3}}),(\frac37,-\frac87).$

Which you can check are not concyclic: $(x-1/6)^2+(y+1/6)^2=13/18$ does not contain $(\frac37,-\frac87).$


问:
Determine a limit of a product
Suppose $f(x)\to 0$ as $x\to\infty$.

I would like to show that
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=0
$$

Is this trivial because
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=\lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}((1+f(x))^{2x})
$$
and
$$
\lim_{x\to\infty}((1+f(x))^{2x})=\underbrace{\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}\cdot\ldots\cdot\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}}_{2x-times}=1
$$

I think this is not valid since as $x\to\infty$, the number of factors goes to infinity.

答:
There are two problems in your interpretation of $(1+f(x))^{2x}$.

When $x$ is a real number, the expression $a^x$ ($a>0, a\ne 1$) is not interpreted as the multiplication of "$x$ many $a$'s". It does not make sense to say multiply $\pi$ copies of $a$. Instead, one of its definitions is
$$
a^x=e^{x\ln a}
$$

Even you restrict $x$ to be positive integers,

$$
\lim_{n\to\infty}(a_n)^{n}\ne \lim_{n\to\infty}(\lim_{n\to\infty}a_n)^{n}.
$$

The statement you want to show:

if $\lim_{x\to\infty}f(x)=0$, then $\lim_{x\to\infty}f(x)(1+f(x))^{2x}=0$

is in general not true. The problem is that one of the factor is of the indeterminant form: $1^\infty$. One counterexample has been given in a comment and Yves's answer.

As a convenient sufficient condition, if the function $f$ is in addition such that
$$
\lim_{x\to\infty}x\ln((1+f(x))=0
$$
then by continuity, you have
$$
\lim_{x\to\infty}\exp\big(2x\ln((1+f(x))\big)=1
$$

Title: Determine a limit of a product


答:
If you ask Wolfram Alpha

```<code>solve x+c_1y=2, y-c_2x=c_3 for x,y
</code>```

you get

$x = \dfrac{2 - c_1 c_3}{c_1 c_2 + 1}$ and $y = \dfrac{2 c_2 + c_3}{c_1 c_2 + 1}$ and $c_1 c_2 + 1\not=0$

$y = -\dfrac {c_3}{x - 2}$ and $2 c_2 + c_3 = 0$ and $c_3 \not=0$ and $c_1 = \dfrac2{c_3}$

I suspect you may looking for the first of these, though the second deals with the $c_1 c_2+1=0$ situation.

For more complicated situations, you may want to invest in a computer algebra system

Title: System of Equation Solver? Online

Body:
What site do people use to solve systems of equations with algebraic coefficients. I.E I know wolfram solves

$$
x+y=2,~~y-3x=1
$$
but are there sites that let you put unknown constants?
$$
x+c_1y=2,~~y-c_2x=c_3.
$$

EDIT : (all helpful answers and I would up vote if I could!) Note I tried this in wolfram and it didn't work (this is the beast I wanted solved for me ):

solve v_1=A+B- (c_5 a / c_1(c_8+ c_3c_5/c_1 )),   v_2=Ae^{r_1 h }+Be^{r_2 h}-  (c_5 a / c_1(c_8+ c_3c_5/c_1 )),  x_1=b-(1/c_1)(A(c_2+c_3/r_1)+B(c_2+c_3/r_2)), x_2=b+ah-(1/c_1)(Ae^{r_1 h}(c_2+c_3/r_1)+Be^{r_2 h}(c_2+c_3/r_2)),   for a, b, A, B

Body:
If you ask Wolfram Alpha

solve x+c_1y=2, y-c_2x=c_3 for x,y

you get

$x = \dfrac{2 - c_1 c_3}{c_1 c_2 + 1}$ and $y = \dfrac{2 c_2 + c_3}{c_1 c_2 + 1}$ and $c_1 c_2 + 1\not=0$

$y = -\dfrac {c_3}{x - 2}$ and $2 c_2 + c_3 = 0$ and $c_3 \not=0$ and $c_1 = \dfrac2{c_3}$

I suspect you may looking for the first of these, though the second deals with the $c_1 c_2+1=0$ situation.

For more complicated situations, you may want to invest in a computer algebra system


问:
Indefinite integral $\int \cos^k{x} dx$?
From WolframAlpha, we can see
$$\int \cos^k x \;dx = - \frac{\sqrt{\sin^2x}\csc x \cos^{k+1}(x) _2F_1\left(\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};\cos^2{x}\right)}{k+1} $$
for any $k\in\mathbb{R}$ and $_2F_1(a,b;c;z)$ is hypergeometric series.
I want to know its proof.

答:
This integral can be calculated iteratively:

$$\int \cos^kxdx = \frac{1}{k}\cos^{k-1}x \cdot \sin x + \frac{k-1}{k} \int \cos^{k-2}xdx$$


问:
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$?
If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?

I need to solve it using combinatorics.
I've started with calculating the number of possible solutions  without restrictions and I am stuck.

答:
$a + b + c +d = 12 \ $ where $0 \leq a, b, c, d \leq 6$ and $a, b, c, d \ne 3$

Here is how you can solve using Principle of Inclusion-Exclusion, which is straightforward as well but I have taken more space and time to explain the steps.

A) Number of unrestricted solutions for non-negative numbers

$\displaystyle  = {15 \choose 3} = 455$

B) Number of solutions where at least one of the numbers is $3$

$ \displaystyle  = 4 {11 \choose 2} - {4 \choose 2} {7 \choose 1} + {4 \choose 1} - {4 \choose 4} = 181$

Explanation: First term is choosing one of the numbers to be $3$ and rest $3$ numbers sum to $9$ but these will overcount cases where two numbers are $3$ and hence the second term subtracts cases where two numbers are $3$ each and then other two sum to $6$. Then the last term is where three of the numbers are each $3$. Please note that if three numbers are $3$, the fourth will also be $3$. Hence in the first term we count such cases $4$ times and then subtract $6$ times in the second term. As there is only one way for three or four numbers to be $3$ each, we add $3$ so we count it one time. The other way you can write it as ${4 \choose 1} - {4 \choose 4}$ but I wanted to explain the rationale behind it.

C) Number of solutions where at least one of the numbers is greater than $6$

$ \displaystyle  = 4 {8 \choose 3} = 224$

Explanation: Now if one of the numbers is $\gt 6$, none of the other numbers can be greater than $6$. We first set one of the $4$ numbers as $7$ and then count number of ways for $4$ numbers to sum to $5$.

Now B) and C) both count solutions where one of the numbers is greater than $6$ and one of the numbers is $3$.

D) Solutions where one of the numbers is greater than $6$ and one of the numbers is $3$

$\displaystyle = {4 \choose 3} {3 \choose 1} {4 \choose 2} = 72$.

Explanation: There are $4$ ways to choose one of the numbers being $7$ and one of the numbers being $3$ and then we have $3$ numbers sum up to remaining $2$ (including the number $7$ which can also be $8$ or $9$).

So the number of desired solutions is $ = 455 - 181 - 224 + 72 = \fbox {122}$


问:
Is $e^x$ the only morphism for addition and multiplication?
I find it interesting that the only (more or less) function that is equal to its own derivative also happens to be a morphism from the reals under addition to the positive reals under multiplication. This would be even more interesting if it were the only such morphism.

Is this the case?

答:
The exponential function and the trivial morphism are the only group homomorphisms $(ℝ,+,0) → (ℝ_{>0},·,1)$ which are differentiable and equal to their derivatives.

Any differentiable function $f \colon ℝ → ℝ$ such that $f' = f$, observing $\tfrac{f}{\exp}$ is constant by differentiating, can be written as $f = f(0)·\exp$. For $f$ to be a group homomorphism $(ℝ,+,0) → (ℝ_{>0},·,1)$ you need $f(0) = 1$, so then $f = \exp$.

Edit: Okay, I just realized that you wasn’t asking for that and that I misread your question, but here continues the more interesting part for you:

There are non-continuous group homomorphisms $(ℝ,+,0) → (ℝ_{>0},·,1)$, since any $ℚ$-base of $ℝ$ defines a $ℚ$-automorphism of $ℝ$, which also gives an underlying group homomorphism $(ℝ,+,0) → (ℝ,+,0)$. Postcomposing $\exp$ yields many non-differentiable group homomorphisms.


问:
Uniformly selecting a tree from all trees
The following is from a paper by Joel Spencer, called Enumerating Graphs and Brownian Motion.

Let us look at a tree $T$ on vertex set $\{1, \ldots, n\}$ from a computer science
perspective. We'll take $n=7$ and the tree with edges 13,15,16,27,34,37
as an example throughout. Start with vertex 1 and find the tree through a
breadth-first search (BFS), all lists being in numerical order. To do this we
have a queue $Q$, that is initially (1). At each stage we take the vertex $x$ at the
head of the queue, remove $x$ from the queue, and add all "new" neighbors of
$x$ to the queue. We'll call that operation "processing $x$ " and say that $x$ finds $y$ if $y$ is a new vertex adjacent to $x$. In our example, 1 finds $3,5,6,$ then 3
finds $4,7,$ then 5 finds nothing, 6 finds nothing, 4 finds nothing, 7 finds $2,$ and
finally 2 finds nothing. After 2 is processed, the queue becomes empty and
the BFS ends.

Let $x_{t}$ be the number of vertices found by the $t^{\text {th }}$ vertex processed (not
vertex number $t$ ), which, in our example, is 3,2,0,0,0,1,0 . Let $q_{t}$ be the
size of the queue after the $t^{\text {th }}$ vertex is processed. Then $q_{0}=1$ and $q_{t}=
q_{t-1}+x_{t}-1$, which, in our example, is $1,3,4,3,2,1,1,0$ . The necessary and
sufficient conditions on the sequence $q_{t}$ are
$$
q_{n}=0 \quad \text { and } \quad q_{i}>0 \quad \text { for } i<n
$$
Now note that there are precisely $(n-1) ! / \prod x_{i} !$ trees with a given sequence
$x_{t}$. Indeed, any partition of the vertex set excluding source vertex 1 into sets
of sizes $x_{1}, \ldots, x_{n}$ gives a unique such tree $T$.

Random Trees

Let $X_{1}, \ldots, X_{n}$ be independent random variables, each Poisson with mean $1$ .
That is, $\operatorname{Pr}\left[X_{i}=x\right]=e^{-1} / x !$ . Define $Q_{0}=1$ and $Q_{t}=Q_{t-1}+X_{t}-1$ .
Condition on
$$
Q_{n}=0 \quad \text { and } \quad Q_{i}>0 \quad \text { for } i<n
$$
The probability of having values $x_{1}, \ldots, x_{n}$ is then proportional to $1 / \prod x_{i} !$
and the support for the $X_{t}$ is that for trees so that the conditional distribution
is precisely that for the $\left(x_{1}, \ldots, x_{n}\right)$ when a tree $T$ is selected from all trees on $n$ vertices with a uniform distribution.

I do not understand what is meant by "the support for the $X_{t}$ is that for trees so that the conditional distribution
is precisely that for the $\left(x_{1}, \ldots, x_{n}\right)$ when a tree $T$ is selected from all trees on $n$ vertices with a uniform distribution." More specifically the part about support for $X_t$.

Can anyone clarify this for me?

答:
The support of a probability measure is the set where it is non-zero (https://en.wikipedia.org/wiki/Support_(measure_theory)).

What he says is that, once you condition on $Q_n = 0, Q_i > 0$ then your sequence of $(X_1, ..., X_n)$ will look like the sequences he described before. Moreover, he claims that this is the same distribution as if you had taken a uniformly random tree and taken its vertex discovery sequence. I guess he proves these facts just afterwards in the paper.


问:
Why does the Mean Value form of the remainder of the Taylor expansion not work here?
I was solving the following problem:

"Prove the following inequality for all $x \geq 0$: $\arctan(x) \geq x-\frac{x^3}{3}$"

I am aware this is able to be solved using the regular MVT, but when I attempted to solve it using a Taylor expansion at x=0 and the mean value form of the remainder, I found that for all odd powers of the expansion, Taylor's theorem did not hold.
That is, there was no intermediate point c for which the remainder was equal (for example): $\frac{(\arctan(c))^{(4)}}{4!}x^4$ for any $x>0$ (as the remainder is negative for c>1, and checking against desmos not every positive $x$ has an intermediate point such that the Taylor expansion equals $\arctan (x)$.

Even though supposedly all the conditions for the theorem hold ($\arctan(x)$ is infinitely differentiable on $\mathbb{R}$ and in particular around the point $x=0$).

My question is, what makes Taylor's theorem fail for odd powers here?

答: I realised my mistake. I was assuming c had to grow larger as x does, which is not the case, in which case taylor's theorem definitely holds. Thanks anyways!</p>


问:
Is $G\cong \mathbb{C} \setminus \{0\}$ under complex multiplication ?yes/no
Given   $G= \{ z \in \mathbb{C} |  \exists \ n \in \mathbb{Z}^{+} \text{such that} \ z^n=1\} $

Is $G\cong \mathbb{C} \setminus \{0\}$ under complex multiplication? yes/No

My attempt : I thinks No

Here $G$ is cyclics  but  $\mathbb{C} \setminus \{0\}$ is not cyclics

so $G$ not isomorphics to  $\mathbb{C} \setminus \{0\}$

答:
The Group $G$ is a torsion group, i.e. each element has finite order.

On the other hand $\mathbb C \setminus \{0\}$ is not torsion, e.g $2^k \neq 1$ for all $k\in \mathbb Z_{>0}$.

So the two groups cannot be isomorphic.


答:
Let $f(x)=x^{-1/2}$.

Then

$$\lim_{x\to\infty}x^{-1/2}(1+x^{-1/2})^{2x}=\lim_{t\to\infty}t^{-1}(1+t^{-1})^{2t^2}=\lim_{t\to\infty}t^{-1}((1+t^{-1})^t)^{2t}.$$

The last expression is bounded below by $\dfrac{2^{2t}}t$ (for $t\ge1$) and the limit diverges.

Note that with $f(x)=x^{-1}$ you bound above with $\dfrac9x$ and the limit is indeed $0$. It is much more challenging (though IMO possible) to find a function that leads to a finite limit.

Title: Determine a limit of a product

Body:
Suppose $f(x)\to 0$ as $x\to\infty$.

I would like to show that
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=0
$$

Is this trivial because
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=\lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}((1+f(x))^{2x})
$$
and
$$
\lim_{x\to\infty}((1+f(x))^{2x})=\underbrace{\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}\cdot\ldots\cdot\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}}_{2x-times}=1
$$

I think this is not valid since as $x\to\infty$, the number of factors goes to infinity.


问:
Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$

Given $x,y,z$ are  positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$

My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}  \right)$$
$$x^2+1 \geq 2x \Rightarrow \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}\right)\leq \frac{1}{2}\left(x^2y+y^2z+z^2x\right)=\frac{3}{2}$$
$$P \geq x^3y+y^3z+z^3x - \frac{3}{2}$$
I was stuck when I was finding the minium of $x^3y+y^3z+z^3x$

Help me, thank you so much

答:
Too long for a comment.

Considering the minimization problem

$$
\max_{x,y,z} -f(x,y,z)\ \ \text{s. t.}\ \ g(x,y,z) = 0
$$

with

$$
\cases{
f(x,y,z) = \frac{x^5 y}{x^2+1}+\frac{x z^5}{z^2+1}+\frac{y^5 z}{y^2+1}\\
g(x,y,z) = x^2 y + y^2 z + z^2 x - 3
}
$$

and according to the notes, the point $x=y=z=1,\lambda=\frac 56$ is a stationary point on the lagrangian

$$
L=-f+\lambda g
$$

This point can be qualified using the bordered hessian which is

$$
H = \left(
\begin{array}{cccc}
 0 & 2 x y+z^2 & x^2+2 y z & 2 x z+y^2 \\
 2 x y+z^2 & -\frac{2 x^3 \left(3 x^4+9 x^2+10\right) y}{\left(x^2+1\right)^3} & -\frac{x^4 \left(3 x^2+5\right)}{\left(x^2+1\right)^2} & -\frac{z^4 \left(3
   z^2+5\right)}{\left(z^2+1\right)^2} \\
 x^2+2 y z & -\frac{x^4 \left(3 x^2+5\right)}{\left(x^2+1\right)^2} & -\frac{2 y^3 \left(3 y^4+9 y^2+10\right) z}{\left(y^2+1\right)^3} & -\frac{y^4 \left(3
   y^2+5\right)}{\left(y^2+1\right)^2} \\
 2 x z+y^2 & -\frac{z^4 \left(3 z^2+5\right)}{\left(z^2+1\right)^2} & -\frac{y^4 \left(3 y^2+5\right)}{\left(y^2+1\right)^2} & -\frac{2 x z^3 \left(3 z^4+9
   z^2+10\right)}{\left(z^2+1\right)^3} \\
\end{array}
\right)
$$

and at the point of interest we have

$$
H(1,1,1) = \left(
\begin{array}{cccc}
 0 & 3 & 3 & 3 \\
 3 & -\frac{11}{2} & -2 & -2 \\
 3 & -2 & -\frac{11}{2} & -2 \\
 3 & -2 & -2 & -\frac{11}{2} \\
\end{array}
\right)
$$

now according to the notes, $n = 3, m = 1$ so we need to consider the two minors $H_3$ and $H_4$. Following we have $\det(H_3) = 63 > 0$ and $\det(H_4) = -\frac{1323}{4} < 0$. The signs are $\text{sign}(\det(H_3)) = (-1)^{m+1} = 1$ and $\text{sign}(\det(H_4))$ should alternate as negative.

This characterizes the point $(x,y,z)=(1,1,1)$ as a local minimum for $f$ constrained.

NOTE

Taking from the restriction $x$ as

$$
x = \phi(y,z) = \frac{\sqrt{z^4-4 y^3 z+12 y}-z^2}{2 y}
$$

and substituting into $f(x,y,z)$ we obtain a surface

$$
u = f(\phi(y,z),y,z)
$$

which can be depicted from the following plot. In blue the relative minimum point with coordinates $u=\frac 32, y=1,z=1$


答:
If we look at

\begin{equation}
\dot{x_2}(t)^2 = -\dot{x_1}(t)^2 e^{2c x_2(t)},
\end{equation}

then we have $\dot{x_2}(t)^2 \ge 0$ and $e^{2c x_2(t)}>0.$

Hence $-\dot{x_1}(t)^2  \ge 0$, thus $\dot{x_1}(t)^2 \le 0$, which gives $\dot{x_1}(t)=0,$

Then

\begin{equation}
\dot{x_2}(t)^2 = -\dot{x_1}(t)^2 e^{2c x_2(t)},
\end{equation}
becomes

$$\dot{x_2}(t)=0$$

Consequence: each pair of constant function satiesfies the differential equation.

Title: How do I solve a differential equation on this form?

Body:
I have a differential equation I am a bit confused about. Basically, I want to express $x_1(t)$ in terms of $x_2(t)$.

This is how they are related (I will let $\dot{x} = \frac{d}{dt} x$):

\begin{equation}
\dot{x_2}(t)^2 = -\dot{x_1}(t)^2 e^{2c x_2(t)}
\end{equation}
where c is a constant.

so basically I have two differential equations that are equal to each other. I don't know how to solve this at all.

Could I rewrite it as $$\dot{x_1} = - \dot{x_2} e^{-c x_2}$$
And just solve for $x_2(t)$ treating the L.H.S as a constant? I am not sure that is allowed.


问:
Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.
The equation of a curve is $y=x^2e^{-x}$.

Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.

Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.

This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.

答:
The stationary point of a curve is simply the point where the derivative vanishes. You can see the behavior of these points from this link.
https://mathworld.wolfram.com/StationaryPoint.html

The first question asks for the $x$-coordinates of the stationary points. You can start by taking the derivative of the curve. Then, you should equate it to zero and solve the equation to find the roots.

$$\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=e^{-x}x(2-x)=0\space \Rightarrow \space x=0 \space\text{or}\space x=2$$

At $x=0$ and $x=2$, the derivative is zero which means they are the stationary points.

We should verify the equation for the normal line of the curve at $x=1$. At this point $y$-coordinates of the normal line and the original curve is the same. Since tangent and normal lines to a curve at some particular point are perpendecular to each other, multiplication of their slopes yields $-1$. At $x=1$, the slope of the tangent line is $\frac{1}{e}$. This means the slope of the normal line is $-e$. Below you may see the normal line equation.  $$ax+b=y$$ $$x_0=1,\space \space a=-e, \space \space y_0=\frac{1}{e}$$ If we substitute these values into the line equation, we find $b$ to be $e+\frac{1}{e}$. $$-ex+e+\frac{1}{e}=y \space \rightarrow \space e^2+1=e^2x+ey$$


问:
On the functional square root of $x^2+1$
There are some math quizzes like:

find a function $\phi:\mathbb{R}\rightarrow\mathbb{R}$
such that $\phi(\phi(x)) = f(x) \equiv x^2 + 1.$

If such $\phi$ exists (it does in this example), $\phi$ can be viewed as a "square root" of $f$ in the sense of function composition because $\phi\circ\phi = f$.  Is there a general theory on the mathematical properties of this kind of square roots? (For instance, for what $f$ will a real analytic $\phi$ exist?)

答:
(Update:  Oh sorry; I'm new here. Didn't notice the much more substantial link to the mathoverflow. But perhaps it is interesting for some newbie anyway...)

If the function is a polynomial (or powerseries) with constant term $\ne0$ this is difficult, but sometimes a meaningful solution can be given.

If the function is as above, but the constant term is zero, then it is just the question of relatively simple manipulation of the formal powerseries/polynomial to construct a "compositional" (or "iterative") root.

There is a very good deal in L.Comtet "Advanced Combinatorics" at pages around 130..150 (don't have it around).

Also the keywords "Bell-matrix", "Carleman-matrix" are helpful: these matrices transform the problem of funtion composition/iteration to matrix-products/matrix-powers. (The matrix-notation just implies the required formal powerseries-manipulations) . This is well established for functions $f(x)= ax + bx^2 + cx^3 +\cdots$.

If the constant term does not vanish, $f(x) = K + ax + bx^2 + cx^3 +\cdots$, then things are usually much more difficult. But there is one -I think: again well established- workaround:

Rewrite $f(x)$ as some function $g(x+x_0) - x_0$ such that now $g(x)$ has no constant term and then apply the above mentioned procedures (solving for iterative square root and the like) to $g(x)$.

Example: denote the iterative root of a some function $f(x)$ by $f^{[1/2]}(x)$ then solve

$$f^{[1/2]}(x) = g^{[1/2]}(x-x_0)+x_0$$

where you first must find $x_0$.

[update 2] I've tried to generate that power series for $g^{[1/2]}(x)$, which has then complex coefficients. I got
$$ g^{[1/2]}(x) \approx  \sqrt{2 x_0} x + (0.20412415 - 0.22379688 i) x^2 + (0.050024298 + 0.048042380 i) x^3 + (-0.022112213 + 0.028383580 i) x^4 + (-0.023623808 - 0.010393981 i) x^5 + (0.00074679924 - 0.021136421 i) x^6 + O(x^7)$$
where $f^{[1/2]}(x) = g^{[1/2]}(x-x_0) + x_0 $ and the fixpoint is $x_0 = \exp(i \pi /3 ) \approx 1/2 + 0.86602540 i $ and $\sqrt{2 x_0}=(1.2247449 + 0.70710678 i) $ . The range of convergence is small; however using Euler-summation (of complex orders) I could manage to reproduce $f(0)$ to $f(1)$ by applying two times the half-iterative to about 6 to 8 correct digits. [end update 2]

For a very basic introduction you may try my small essay at

go.helms-net.de/math/tetdocs

and look for the article "continuous functional iteration".


问:
Finding an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$
For an odd integer $n$, find an explicit isomorphism between $\mathbb{Z}^{\times}_n$ and $\mathbb{Z}^{\times}_{2n}$.

How do I do this? I don't really know where to start. I can easily find bijections but am yet to find a structure preserving one.

答: HINT $\: $ It's very easy. The natural quotient map $\rm\ a+2\:n\ \mathbb Z\ \mapsto\ a+n\ \mathbb Z\ $ is surjective (so injective) on units, $\:$  since $\rm\:n\:$ odd $\:\Rightarrow\: $ one of $\rm\:a,\ a+n\:$ is odd, $\:$ so is coprime to $\rm\:2\:n\iff$  coprime to $\rm\:n\:.$


答:
Once you know that $\def\R{\mathbb{R}}(\R_{>0},\cdot)$ is isomorphic to $(\R,+)$ via $\exp$, what you're asking is classifying the additive isomorphisms of $(\R,+)$ onto itself.

If the isomorphism is supposed to be continuous, then it is of the form $x\mapsto ax$, for some $a\in\R$, $a\ne0$. But there are many others, because $\R$ is a vector space over $\def\Q{\mathbb{Q}}\Q$ and vector space isomorphisms are in particular additive isomorphisms of $(\R,+)$ onto itself.

Since the cardinality of a basis of $\R$ over $\Q$ is the same as $|\R|$, there are at least $|\R|$ permutations of the basis, which give as many isomorphisms.

Title: Is $e^x$ the only morphism for addition and multiplication?

Body:
I find it interesting that the only (more or less) function that is equal to its own derivative also happens to be a morphism from the reals under addition to the positive reals under multiplication. This would be even more interesting if it were the only such morphism.

Is this the case?


问:
Is this a statistic?
If I have $n$ random variables, $X_1,X_2,...,X_n$.

Is this a statistic:

$$\Theta = 1/2(X_1+X_2)$$

I know the definition of statistic is the random variable represented by a function on my $n$ random variables, since I cannot think of a function on $n$ R.Vs that would yield this surely it is not a statistic?

答:
Or course $\frac{X_1+X_2}{2}$ is a Statistic!

to understand and detect immediately a Statistic: It is a function of the Data (of the data only...)

Example

$X_1,\dots X_5$ is a random sample drawn from a population $X\sim U(0;\theta)$

These are statistics:

$T_1=\Sigma_i X_i$

$T_2=\frac{1}{5}\Sigma_i X_i$

$T_3=\max(X_i)$

$T_4=\min(X_i)$

$T_5=\{X_1,X_2\}$

On the contrary,

$T_6=\frac{\max(X_i)}{\theta}$

is not a Statistic because is depends both from the data and from the parameter. Anyway $T_6$ has a very important Statistic meaning. What kind of quantity is it?

Title: Is this a statistic?


问:
Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$
I saw this question in an old trig text book:

Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$.

I ask:

What solution can people here give ?

Is there a solution that does not (like mine) require the construction of $\sqrt[4]{5}$ ? In other words, a solution without assuming the circle intersection property (See Hartshore: Euclid and Beyond pg. 144)

Body: $$|OB'|= \sqrt{5} \quad\to\quad |C'D|=3(\sqrt{5}-2) \quad\to\quad \sin A'' = \frac{3(\sqrt{5}-2)}{1}=\frac{3}{\sqrt{5}+2}$$


问:
Least number of terms one would have to take in alternating series
What would be the least number of terms so that the absolute error of the alternating series $\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{k-1}}{k^{10}}$ is less than $0.000287$

答:
The error bound for an alternating series states that the error in the truncation to $n$ terms is at most the absolute value of term $n + 1$. Then, we wish to find $k$ such that:

$$\vert a_{k + 1}\vert\leq 0.000287$$

$$\bigg\vert\frac{(-1)^{k}}{(k + 1)^{10}}\bigg\vert\leq 0.000287$$

$$\frac{1}{(k + 1)^{10}}\leq 0.000287$$

$$(k + 1)^{10}\geq \frac{1}{0.000287}$$

$$k \geq (0.000287)^{-\frac{1}{10}}-1\approx 1.2605$$

$$k\geq 2$$

Thus, you can truncate the series to $\boxed{2\text{ terms.}}$


问:
A continuous map $M \to \mathbb{S}^n$ is homotopic to a constant if $dim(M) < n$?
I'm trying to solve the following exercise from Milnor's Topology from the Differentiable Viewpoint, but I'm not sure I understood the statement.

Let $M$ be a smooth manifold with $dim(M) = m < n$, show that every map $f : M \to \mathbb{S}^n$ is homotopic to a constant.

I was able to proove this under the assumption that $f$ is a smooth map. In this case, since $m < n$, $df_x$ is not surjective $\forall x\in M$. By Sard's Theorem exists $p_0\in \mathbb{S}^n$ regular value for $f$, so $f^{-1}(p_0)$ must be empty and the map is not surjective and I can build and homotopy between $f$ and the constant $-p_0$ (they are always not opposite).

If $f$ is not smooth, can I still say that it's not surjective? I'm not able to prove that, but I'm not even sure that this is the right way to proceed if $f$ is just continuous. Sometimes in the book the author assumes that maps are smooth, so does this statement still hold for continuous maps?

答:
The reason  is that each continuous $f :  M \to S^n$ can be approximated arbitrarily closely by smooth maps. That is, for  each $\epsilon > 0$ there exists a smooth map $g : M \to S^n$ such that $\lVert g(x) - f(x) \rVert < \epsilon$ for all $x \in M$. Note that for $\epsilon \le 2$ the maps $g, f$ are homotopic via
$$H(x,t) = \frac{tg(x) - (1-t)f(x)}{\lVert tg(x) - (1-t)f(x) \rVert}.$$
Milnor does not prove this, but addresses it in the exercises, problem 4 (at least fort compact $M$). See also his proof of the Brouwer fixed point theorem. Quotation:

The procedure employed here can frequently be applied in more
general situations: to prove a proposition about continuous mappings,
we first establish the result for smooth mappings and then try to use an
approximation theorem to pass to the continuous case. (Compare 58,
Problem 4.)


问:
Can you draw a curve that captures light?
Suppose we had a differentiable curve $C$ in $\mathbb{R}^2$ that serves as our "light container". Light is shining in from all directions, so the space of incoming light-beams is $\mathbb{R} \times S^1$ where the $S^1$ parametrises the angle that the light is coming in from, and the $\mathbb{R}$ parametrises the offset from the origin. When the light hits $C$ it bounces as you would expect it to off a mirror. Some of the incoming light beams will be trapped inside $C$, and keep bouncing forever without escaping, however most will just bounce off.

Is there such a finite curve $C$ such that it actually captures a non-zero amount of light? (take the natural measure on $\mathbb{R} \times S^1$, is there a curve $C$ that captures a non-null subset of the light?) My intuition seems to be telling me that there isn't (because I can't seem to write down any convincing examples), but I can't really think of a reason why not.

If so, what's the curve $C$ that captures the most light? (for a given length $L$. Or, I guess any collection of curves $C_i$ whose total length is $L$).

Sorry about the confusion, I mean that the light is shining in from infinity. Here for example, the black curve C is our jar, and the blue lines represent incoming light. Most just reflect off, some may get trapped forever. We can draw such C that captures a discrete amount of light, but can we draw C that captures a non-trivial amount of light?

答:
The isosceles right triangle with an opening in one of the edges close to a vertex away from the right angle has this property.

First let us consider a case where only a null subset of the light.

Consider $ABC$ an isosceles triangle with right angle at $C$. Let's $\mathcal{C}= (AB] \cup [BC] \cup [CA)$. Then according to Lemma 3.1 of Tokarsky (1995), light beams entering through $A$, i.e. in $\{0\} \times (0,\frac{\pi}{4})$, never exit.

G. W. Tokarsky, Polygonal Rooms Not Illuminable from Every Point, The American Mathematical Monthly, Vol. 102, No. 10 (Dec., 1995), pp. 867-879. https://doi.org/10.2307/2975263

His proof is quite intuitive, thanks to the fact that you can easily represent multiple reflections of the triangle by its faces as a lattice of $\mathbb{R}^2$ :

Then, it is easy to see that to exit a ray has to cross some reflection of $A$, and that before doing so it will always encounter first the reflection of another vertex. This corresponds to absorption with infinite reflections.

From there, I believe you can extend this to any curve $\mathcal{C}'$, $[BC] \varsubsetneq \mathcal{C}' \subset \mathcal{C}$. Indeed, what you want is to find elements of $S^1$ for offsets corresponding to points in $\mathcal{C}'\setminus\mathcal{C}$ such that reflections of $B$ or $C$ are encountered at shorter distances than $A$.

Let us take such a curve $\mathcal{C}' = (AB] \cup [BC] \cup [CD]$, with an "opening" $AD = \varepsilon > 0$.

Let us consider any ray passing through $(s,0)$ with $0 < s < \varepsilon$, then all rays with slope $(1+2j)/(2k+2j-s)$ with $(j,k) \in \mathbb{N}\times\mathbb{N}^*$ cross reflections of point $B$ and are thus trapped. The set of corresponding angles is dense in $(0,\pi/4)$, so we have a non-zero amount of light trapped.


问:
confused about the limit of a trigonometric function
I am trying to calculate the limit of the following function for general $a$:

$$\lim_{x\to a}[\cos(2 \pi x)-\sin(2 \pi x) \cot(\frac{\pi x}{a})]$$

I was believing this is infinite. But Mathematica calculates for $a=3$ for instance the limit is $5$

And for $a$ in general Mathematica returns an infinit term that says:

DirectedInfinity[a Sin[2 a [Pi]]]

I can not understand the role of $a\sin(2a \pi)$

May someone here ehlp me out of confusion?

答:
HINT:

$$F=\cos 2\pi x-\sin 2\pi x\cot \frac{\pi x}a$$

$$=\frac{\sin \frac{\pi x}a\cos 2\pi x-\sin 2\pi x\cos \frac{\pi x}a}{\sin \frac{\pi x}a}$$

$$=\frac{\sin(\frac{\pi x}a-2\pi x)}{\sin \frac{\pi x}a}$$

$$=\frac{\sin(\frac{\pi x}a(1-2a))}{\sin \frac{\pi x}a}$$

If $1-2a$ is integer, $\lim_{x\to a}F=\frac00,$ we can apply L' Hospital Rule

Else the  denominator goes to zero, but the numerator doesn't, hence, the limit goes to $\infty$


问:
Is every set a subset of the empty set?
So I am really a beginner to set theory and I read that the empty set is a subset of every set vacously.
however is it true that every set is a subset of the empty set for the same reason?</p>

答:
No, normally if $A \subset B$, then $B \subset A$ is not true.
For example: The set $\{1\}$ is not a subset of $\{\}$.

Title: Is every set a subset of the empty set?


问:
$\ln x \leq x-1$ for all $x\ge 1$

Let $x \geq 1$. Show that $\ln x \leq x-1$.

I used the mean value theorem to show that $\ln x \leq x-1$.
But then they ask us to deduce that $e^t \geq x+1$. I know that $\ln x$ and $e^x$ are inverse but I am not able to solve it.

答:
If you know that $\log x \le x-1$ for all $x \ge1$ then

$$e^{\log(x)} \le e^{x-1}  \qquad \Rightarrow \qquad e^x\ge ex \ge x+1$$

Title: $\ln x \leq x-1$ for all $x\ge 1$


问:
Examples of Categories that are not Complete
I have been studying Category Theory and I am now trying to become more familiar with the definition of complete categories. I understand the definition well, but I was hoping to develop some sort of intuition to predict when a category may be complete or not. I think looking at counter-examples may be helpful, but I have not been able to find many.

Taking the definition: A category is complete if it has all small limits. I know that many familiar categories are complete, such as Ab, Set and Top. Are there examples of familiar categories that are not complete? If there aren't examples of "familiar" categories that are not complete, what are some examples of "unfamiliar" ones?

答: The category of fields fails awfully in this regard: if $K$ and $L$ are fields of different characteristics, there are no maps $K \to L$. Hence we cannot form a (co)product between $K$ and $L$ in $\mathsf{Field}$.


答:
There is a whole class of examples obtained by deliberately destroying an otherwise perfectly complete category. For instance, take the complete category $\mathbf {Set}$ of sets and functions. Now, remove from it all sets with precisely $4$ distinct elements. You still have a category but now there is no product of any two doubleton sets.

Other examples are infinite lattices that are not complete lattices. Any lattice is a special category (one where hom sets are sub-singletons). In a lattice you have all finite joins and meets, and these are, respectively, coproducts and products. All other finite colimits and limits exist automatically. However, not all lattices are complete, namely they need not have all meets. In other words, any lattice that is not complete is a category that if finitely complete but not complete. Of course, there are many many posets that do not even have finite products.

Then there are examples coming from algebra. For instance, every field has a characteristic and the characteristic is preserved by the morphisms (this is a very strong property that is quite unique to fields). Consequently, the product of two fields of different characteristics cannot exist.

And finally you can form non-complete categories by forcing them not to be sufficiently large, in a sense. For example, the category of all finite sets has all finite limits but not all infinite products.

Title: Examples of Categories that are not Complete

Body:
I have been studying Category Theory and I am now trying to become more familiar with the definition of complete categories. I understand the definition well, but I was hoping to develop some sort of intuition to predict when a category may be complete or not. I think looking at counter-examples may be helpful, but I have not been able to find many.

Taking the definition: A category is complete if it has all small limits. I know that many familiar categories are complete, such as Ab, Set and Top. Are there examples of familiar categories that are not complete? If there aren't examples of "familiar" categories that are not complete, what are some examples of "unfamiliar" ones?


问:
Need an explanation of this paragraph "Measure Theory"
I will just quote a part of one proof in "On uniformly regular topological measure spaces by Babiker: page 781" vol43 No4 Duke Math. J. 1976.

Let $I$ be the unit interval endowed with Lebesgue measure $m$. If $\mu$ is nonatomic probability (Radon) measure on (compact) space $X$, then there is $E\subset X$, $D\subset I$ with $\mu(E)=m(D)=0$ and a map $\psi\colon X\rightarrow I$ such that $\psi$ is a homeomorphism from $X\setminus E$ onto $I\setminus D$ and $m=\psi(\mu)$.

The author refer to this reference: N. Bourbaki, Elements De Mathematique Integration I, chaps 1-5, Paris, 1965: in chapter V page 85.

I do not know French to look at it. Could someone give me some steps proving the above result.

答:
This is what I found in my Russian edition of this book.

Lemma 1. Let $\mu$ be positive Borel measure on locally compact topological space $X$ and $A\subset X$ be $\mu$-measurable set with $\mu(A)>0$. Assume for all $\mu$-measurable $B\subset A$ either $\mu(B)=0$ or $\mu(B)=A$, then there exist $a\in A$ such that $\mu(A)=\mu(\{a\})$.

Idea. Consider intersection $P$ of all compact sets $K\subset A$ so that $\mu(K)=\mu(A)$. It will be the desired set.

Lemma 2. Let $\mu$ and $\nu$ be two positive Borel measures on $\mathbb{R}$. Assume that $\mu([a,b))=\nu([a,b))$ for all $a<b$. Then $\mu=\nu$.

Lemma 3. Let $X$ be compact topological space and $\mu$ be positive non-atomic probability measure on $X$. Then there continuous map $\pi:X\to I$ to the unit interval $I$. Such that image of $\mu$ under $\pi$ is Lebesgue measure $\lambda$ on $I$.

Sketch. Following idea of the proof of the Uryson's lemma we construct a family of open sets $U(t)$ where $t\in I$ such that $U(0)=\varnothing$, $U(1)=X$, $\overline{U(t)}\subset U(t')$ for $t'<t$ and finally $\mu(U(t))=t$ for all $t\in I$. Using lemma 1 we can show that conditions $V$, $W$ two open $\mu$-measurable with $\overline{V}\subset W$ and $\mu(V)<\mu(W)$ implies existence of $\mu$-meausrable open set $U$ such that $\overline{V}\subset U\subset\overline{U}\subset W$ such that $1/3\mu(W\setminus V)\leq\mu(U\setminus V)\leq 2/3\mu(W\setminus V)$. Now we apply lemma 2.

Theorem 1. Assume all assumptions of previous lemma are satisfied plus $X$ is metrizable, then there exist $\mu$-measurable set $N\subset X$ of measure $0$ and $\lambda$-measurable subset $M\subset I$ of measure zero such that we have a homeomorphism $\pi:I\setminus M \to X\setminus N$ with the following property once $\pi$ is continuously (in fact arbitrarily) extended up to $\varphi:I\to X$ and $\pi^{-1}$ similarly extended to $\psi:X\to I$, then $\varphi\circ\lambda=\mu$ and $\psi\circ\mu=\lambda$

Sketch. For each $n\in\mathbb{N}$ we can find finite partition of $X$ consisting of set of $\mu$-measure $0$ and open $\mu$-measurable sets of diameter $\leq1/n$ and measure $\leq 1/n$. (Don't ask why $-$ this is a long story) Then using induction and taking the limit we get existence of continuous map $f:I\setminus S\to X$ (where $S$ is countable) such that $f\circ\lambda=\mu$. Moreover $f$ can be chosen to be homeomorphism from $I\setminus S$ onto the subset of $X$ of full measure.


答:
Let $y\geq 0$. Then $e^y\geq 1$, and from what you already proved,  it follows
$$y= \ln (e^y)\leq e^y-1 \quad \Longrightarrow y+1 \leq e^y\,.$$

Title: $\ln x \leq x-1$ for all $x\ge 1$

Body:
Let $x \geq 1$. Show that $\ln x \leq x-1$.

I used the mean value theorem to show that $\ln x \leq x-1$.
But then they ask us to deduce that $e^t \geq x+1$. I know that $\ln x$ and $e^x$ are inverse but I am not able to solve it.


问:
How many combinations of size $n+1$ at least in a group of $2n$ elements?
How many combinations of size $n+1$ at least in a group of $2n$ elements ?

It seems that total amount of combinations is equal to $2^{2n}$ in this case, but what if I only want the number of combinations strictly bigger than n elements ?

答: You would have to subtract $\binom{2n}{n}$ from $2^{2n}$ and then divide by $2$.


答:
As pointed out in the comments, lets try the following: Let $\{1\} \subseteq \varnothing$. This would mean that $1 \in \{1\}$ and $1 \in \varnothing$. However, by definition of the empty set, we have $\forall x : x \notin \varnothing$. Therefore, having $1 \in \varnothing$ would mean that $\varnothing \neq \varnothing$, which isn't the case. Therefore, $\{1\} \nsubseteq \varnothing$.

You can now generalize this to any set $A$, by using $a \in A$.

Hope that helps.

Title: Is every set a subset of the empty set?

Body:
So I am really a beginner to set theory and I read that the empty set is a subset of every set vacously.
however is it true that every set is a subset of the empty set for the same reason?</p>


问:
How does the filled square retract to three of its boundary lines?
In Example 0.15 of Allen Hatcher's Algebraic Topology, he uses the fact that, for $I = [0,1]$, $I \times I$ retracts to $(I \times \left\{ 0 \right\}) \cup (\partial I \times I)$. At first this seemed obvious, but I'm having a hard time constructing an actual retraction. For example $f(s,t) = \begin{cases} (0,t), s\leq 1/2 \\ (1,t), s> 1/2 \end{cases}$ if $t>0$ wouldn't work because it isn't continuous. Nor would fixing a point in the interior of $I\times I$ and doing something similar to the retraction of the punctured space $\mathbb{R}^n$ to $S^{n-1}$, because we wouldn't know where to send that point. So what would be a retraction?

答:
The easiest way is to do it in two stages.

First stage: get to the one-point union of two triangles, one with vertices $(0,0),(0,1),(\frac12,0)$ and the other $(1,0),(1,1),(\frac12,0)$.  This is just "pulling down" along $y$-direction for the half we want to vacate.

Second stage: retract for each triangle to the relevant edges.  This is just pulling in, e.g., direction $(\pm m,-1)$ where $m>\frac12$.


问:
Understanding the definition of the complexity class $LOGSPACE$.
In the area of complexity classes, the complexity class $LOGSPACE$ or $L$ is defined as:

A language $A$ is in $LOGSPACE$ iff. there exists a deterministic Turing Machine $M$ so that $A=T(M)$ and $M$ modifies at max logarithmic many memory cells.

What does it mean by modifying "logarithmic" many cells? How is that possible? I am a little confused by the wording.

答:
First, consider time complexity once again. The basic idea is simple. When considering a Turing machine $T_M$, we simply "count" the steps $T_M$ needs to halt for any given input, whereby we usually consider a "step" being a transition made by $T_M$. We can consider time complexity as a function $t_c:\Sigma^* \to \mathbb{N}$ of some algorithm.

Now, space complexity is different. Let us be very naive for a second, then space complexity would be very simple. We just consider the maximum number of cells that $T_M$ ever reads on any input of size $n$, which trivially yields to $n+1$ ($+1$ since $T_M$ needs to evaluate the blank symbol $\epsilon$).

But the general idea is different. Assume that $T_M$ is a three tape Turing machine, whereby one tape is the input tape, one is the output tape and one is the so called work tape. We then define the maximum number of cells used on the work tape as the space used by some algorithm during computation and call it the space complexity. We usually say that the work tape is the only tape which is writable (modifiable) (which makes it equivalent to what we consider as "memory"). We often call the work tape, the "memory" tape of $T_M$.  Hence, for any input $x$ of size $n$ we could use substantially less cells then we do on the input tape, e.g. we could have an algorithm using only $log(n)$ cells on the memory tape. And we call the set of problems solvable by using only $O(log(n))$ space $LOGSPACE$ problems.

To summarize: A language $L$ is in $LOGSPACE$ if it can be decided by a deterministic Turing machine using $O(log(n))$ space on the memory tape.


问:
Why $G/C_G(a) \leq G/\zeta(G)$ in this lemma?

My question:

Could you please tell me why $G/C_G (a)\leq G/\zeta (G)$ in the last sentence? Here $\zeta (G)$ is the center of $G$. Thanks in advance.

答:
I am sorry, but the answer of Chris Custer is not correct. You have $\zeta(G) \subseteq C_G(a) \subseteq G$, and this only implies that $G/C_G(a) \cong (G/\zeta(G))/(C_G(a)/\zeta(G))$. Probably you need divisibility and locally finiteness along the line to show that $G/C_G(a)$ is isomorphic to a direct summand of $G/\zeta(G)$.

Note added later Aha! Now I see the full proof. So indeed the statement you are questioning is totally wrong. As I pointed out above, $G/C_G(a)$ is a quotient of a divisible group hence divisible. That is what is needed. And of course that a bounded divisible group is trivial. So thanks for adding the full proof and good question you asked! I already +1-ed your post.


答:
Slightly poorly phrased question, but a pretty simple answer. We know
$$
{a\choose b} = {a\choose a-b}
$$

and so we have that
$$
{2n\choose x} = {2n\choose 2n-x}
$$

This gives us a nice symmetry in your 2n elements between the combinations of size less than $n$ and greater than $n$. Thus if we take $2^{2n} - {2n\choose n}$ we have exactly twice the number of combinations of size greater than $n$. This gives us our final answer as
$$
\frac{2^{2n} - {2n\choose n}}{2}
$$

Title: How many combinations of size $n+1$ at least in a group of $2n$ elements?

Body:
How many combinations of size $n+1$ at least in a group of $2n$ elements ?

It seems that total amount of combinations is equal to $2^{2n}$ in this case, but what if I only want the number of combinations strictly bigger than n elements ?


问:
Approximation for $L_{\text{loc}}^{\infty}(U)$ is this proof correct?
Let $U\subset\mathbb{R}^n$ be open and bounded. I am trying to extend Evans' proof (in his PDE book) for approximating functions in $L_{\text{loc}}^{p}(U)$ for the case that $p=\infty$ using mollifiers. In particular I would like to show that
\begin{equation}%\label{eq: wts}
f\in L_{\text{loc}}^{\infty}(U)\Rightarrow f^{\varepsilon}\rightarrow f\text{ in } L_{\text{loc}}^{\infty}(U).
\end{equation}
I modified his proof as follows. Is it ok?

Let $V, W$ be open subsets of $U$ such that $V\subset\subset W\subset\subset U$. Now if $f\in L_{\text{loc}}^{\infty}(U)\Rightarrow f\in L_{\text{loc}}^{p}(U)$ for $1\leq p<\infty$. As in Evans (page 631) we have for sufficiently small $\varepsilon>0$
\begin{equation*}
\|f^{\varepsilon}\|_{L^{p}(V)}\leq\|f\|_{L^{p}(W)}.
\end{equation*}
We know that the Lebesgue measure is a Radon measure, $W$ is locally compact and Hausdorff and therefore $\overline{C_c(W)}=L^{p}(W)$ for $1\leq p<\infty$. So given any $\delta>0$ there is a $g\in C_c(W)$ such that:
\begin{equation*}
\|g-f\|_{L^{p}(W)}<\delta.
\end{equation*}
Then
\begin{align*}
\|f^{\varepsilon}-f\|_{L^{p}(V)}&\leq \|f^{\varepsilon}-g^{\varepsilon}\|_{L^{p}(V)}+\|g^{\varepsilon}-g\|_{L^{p}(V)}+\|g-f\|_{L^{p}(V)}\\
&\leq 2\|g-f\|_{L^{p}(W)}+\|g^{\varepsilon}-g\|_{L^{p}(V)}\\
&\leq 2\delta+\|g^{\varepsilon}-g\|_{L^{p}(V)}.
\end{align*}
Since $\mathcal{L}^{n}(V)<\infty$,
\begin{equation*}
\|f^{\varepsilon}-f\|_{L^{\infty}(V)}\leq 2\delta+\|g^{\varepsilon}-g\|_{L^{\infty}(V)}
\end{equation*}
as $p\rightarrow \infty$.
Since $g^{\varepsilon}\rightarrow g$ uniformly on $V$, we have
\begin{equation*}
\limsup_{\varepsilon\rightarrow 0}\|f^{\varepsilon}-f\|_{L^{\infty}(V)}\leq 2\delta.
\end{equation*}
This is true for all $\delta>0$ so we conclude
\begin{equation*}
f^{\varepsilon}\rightarrow f\text{ in } L_{\text{loc}}^{\infty}(U).
\end{equation*}

答:
Each $f^\epsilon$ is continuous. If $f^\epsilon$ converges locally in $L^\infty$ then $f$ must be equal to a continuous function almost everywhere. There must be something wrong in your proof.

Consider this example: $U=(-1,1)\subset\mathbb{R}$, $f(x)=1$ if $0\le x\le1$, $f(x)=0$ if $-1\le x<0$ and $\phi(x)=(1-|x|)_+$. Then $\|f^\epsilon-f\|_{L^\infty(V)}=1/2$ for any open set $V\subset U$ containing $0$ (see the graph below.)


问:
Questions about group extensions, and finding inverse elements
Let $I=(x^4+x+1)\vartriangleleft\mathbb F_2 [x]$ and $R = \mathbb F_2[x] / I$. I am to:

List irreducible polynomials of $\mathbb F_2$ with degree smaller than 5.

Show that $R$ is a field and find the number of elements in $R$.

Find the inverse of $x^3+x^2+x+1$ in this field.

Show that $R^*$ is a cyclic group by explicitly naming a generator.

I approached these questions in the following way. Since a polynomial in $\mathbb K[X]$ of degree $\leq 3$ is invertible iff it has no roots, first I began by listing the irreducible polynomials of different degrees.

Degree one we have $X, X+1$ and of degree two we only have one: $X^2+X+1$.  I found this by letting $f(X)=X^2+aX+c \in \mathbb F_2[X]$ be irreducible, $\implies \not \exists$ a root in $\mathbb F_2$ hence $f(0) = c = 1$ and $f(1)=1+a+c=1+a+1=a=1$.

With the same logic, the only two irreducible polynomials of degree 3 are: $X^3+X+1, X^3+X^2+1$.

Finally, the irreducible polynomials are: $$X^4+X+1, X^4+X^3+1, X^4+X^3+X^2+X+1$$ which gives us the polynomial we need to find an inverse of.

Now, since $X^4+X+1$ is irreducible in $\mathbb F_2 [x]$, $(X^4+X+1)$ is a maximal ideal, and thus R is indeed a field. Now, $[R : \mathbb F_2] = \deg(X^4+X+1)=4$, and thus $|R|=2^4 = 16$.

I also know that the multiplicative group $R^*$ has order $|R|-1=15$.I know that it is cyclic because $(X+1)^3 \not = 1 \not = (X+1)^5$.

Now my question is how do I find the inverse element of the polynomial? And how do I find the explicit generator?

答:
''Now my question is how do I find the inverse element of the polynomial? And how do I find the explicit generator?''

Hints:

Use the extended Euclidean algorithm to divide $x^3+x^2+x+1$ into $x^4+x+1$. This gives you the inverse.

An explicit generator is the residue class of $x$ in $F = \Bbb F_2[x]/\langle x^4+x+1\rangle$, since the polynomial $x^4+x+1$ is primitive.

Note that the residue class of $x$ in $F = \Bbb F_2[x]/\langle x^4+x^3+x^2x+1\rangle$ is not a generator, since the polynomial $x^4+x+1$ is not primitive as it divides $x^5+1$.


问:
Is there any solution for converting any irregular shape to any constructible regular polygon without changing area through pure construction?
Is there any solution for converting any irregular shape to any constructible regular polygon without changing area through pure construction?</p>

答:
Starting from a polygon having $n$ sides, we can construct an equivalen polygon having $n-1$ sides.

Draw diagonal $DF$ and parallel line to $DF$ passing through $E$. This parallel intersects $CD$ in $H$. The triangles $FDE$ and $FDH$ are equivalent because they have the same base and the same height. So the polygon $ABCHFG$ is equivalent to $ABCDEFG$ and has one side less.

This construction can ideally be done until we get a triangle equivalent to the starting polygon.

A rectangle  equivalent to the triangle can be easily constructed.

See the second picture below. $D$ is the midpoint of $AB$, $FE\parallel AB$ and lines $AF,DE$ are perpendicular to $AB$.

A square equivalent to the rectangle $ABCD$ can be constructed in the following way (see third picture below).

With center in $B$ draw an arc having radius $BC$ which intersects the line $AB$ in $F$.

Let $M$ be the midpoint of $AF$. With center in $M$ and radius $AM$ draw a semicircle that intersects the line $CB$ in $H$. The $AHF$ is a right triangle and $BF\cdot AB=BH^2$.

So the red square is equivalent to the blue rectangle.


问:
Is there a software that can graph parametric equations where x and y are defined implicitly?
For example, I am working on an exercise from a Calculus book that asks to find the slope of $x=f(t), y=g(t)$ at $t=2$, where

$$
\begin{align}
x^2-2tx+2t^2&=4\\
2y^3-3t^2&=4
\end{align}
$$

I have already worked out the solution and found that $dy/dx|_{t=2}=0$, but I was interested in plotting the graph just for kicks and giggles.  I don't think my TI Nspire can do it, or at least I don't know how. Could I use Octave or WolframAlpha somehow?  Thank you.

答:
Geogebra is your friend in this case.

Create a slider for values of $t$.

Create a list of the possible values of $x$: listofx=Solutions(x^(2)-2 t x+2 t^(2)=4)

Create a list of the possible values of $y$: listofy=Solutions(2 y^(3)-3 t^(2)=4)

Geogebra will give you as many values of $x$ and $y$ as the equations will yield and it's happy with implicit equations.

Then turn those coordinates into points using the "Element" command.

A=(Element(listofx,1),Element(listofy,1))

B=(Element(listofx,2),Element(listofy,1))

Create a trace of the points A and B as you adjust $t$

Teaching a lesson now, but will post image later.


问:
Minimizing the $L^2$ norm of the derivative (Schilder's theorem)
We define the following functional on $C^0([0,1],\mathbb{R})$:
$$
I(f)=\begin{cases}
\frac{1}{2}\int_0^1(f')^2=:\frac{1}{2}\|f'\|_2^2&\text{ if $f$ is absolutely continuous with square-integrable derivative}\\
\infty&\text{ otherwise}
\end{cases}
$$
and we would like to minimise this over the set $C=\{f:\int_0^1f^2\geq 1\}$, i.e. find the quantity
$$
\inf_{f:\|f\|_2\geq 1}I(f)=\inf_{f:\|f\|_2\geq 1}\tfrac{1}{2}\|f'\|_2^2.
$$
This problem actually comes from p.17 in "Large Deviations" by Varadhan, where we're trying to apply Schilder's theorem for large deviations for Brownian motion to the set $C$ above. Varadhan just says this is an eigenvalue problem without elaborating, but I'd welcome some help on understanding how to solve this!

答:
This is mainly a consequence of Lagrange multipliers. First note that if $||f||_2 > 1$, you can rescale your energy with $f \mapsto \frac{1}{||f||_2}f$ and decrese $I(f)$ that way. So we can simplify the constraint to $||f||_2^2=1$.

For $I=(0,1)$ and $\langle \cdot, \cdot \rangle$ the standard $L^2(I)$ scalar product, the Euler-Lagrange associated to $I(f)$ is given by
$$
\langle f',\phi' \rangle=0  \; \; \forall \phi \in C_c^{\infty}(I)
$$
and for your constraint it is given by
$$
\langle f, \phi \rangle =0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
Using the methods of Lagrange multipliers, you know that a solution of your variational problem satisfies
$$
\langle f',\phi' \rangle + \lambda \langle f,\phi \rangle=0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
for some $\lambda \in \mathbb{R}$. Now you can apply regularity theory to deduce that $f \in C^2(I)$ and then integrate by parts to arrive at
$$
\langle f''-\lambda f, \phi \rangle=0 \; \; \forall \phi \in C_c^{\infty}(I)
$$
from which follows
$$
f''=\lambda f
$$
This is an eigenvalue problem for the linear operator defined by $Tf=f''$.

Solving this problem can be done via abstract existence theory for eigenvalues or the direct method of the calculus of variations - there could also be a more elementary approach to this problem which I can not recall as of now. There are many routes to take, just pick one that you are familiar with.


问:
Names for specific rings

Is there a name for a ring $A$ (commutative with $1 \neq 0$) all of whose ideals are radical?

Is there a name for a ring $A$ (commutative with $1 \neq 0$) such that for each $x \in A$, there is $n \geq 2$ such that $x^n = x$?

I am currently calling $A$ in 1 a radical ring and $A$ in 2 a semi-Boolean ring.

答:
(I'm pretty sure that "radical" here means that $\sqrt{I}=I$)

Under the conditions of #1, all ideals of the ring are semiprime. Taking any element $a$, $(aR)^2$ is a semiprime ideal, so it must contain $aR$, and we have proven that $aR=(aR)^2$. Since this says that $a=ara$ for some $r\in R$, we have also shown that $R$ is von Neumann regular. An observation that we'll need shortly is that $ar$ is idempotent, and $aR=araR\subseteq arR\subseteq aR$, so $aR=eR$ for an idempotent $e=ar$.

Conversely, if $R$ is von Neumann regular, we can show that all its ideals are semiprime, since they are all idempotent.

As observed before, for any element $a\in R$, there exists an idempotent $e$ such that $eR=aR$. Suppose $I\lhd R$ and $I^2\neq I$. Clearly $aR\subseteq I$, and $(aR)^2\subseteq I^2$. Pick $a\in I\setminus I^2$. Then $(aR)^2=(eR)^2=e^2R=eR=aR$, and that implies that $a\in (aR)^2\subseteq I^2$, a contradiction. Hence $I^2=I$ for all ideals, and $I$ is semiprime (=radical).

It turns out that for commutative rings, being VNR is exactly the same as saying that every ideal is an intersection of maximal ideals. As a result, every proper quotient ring of $R$ has zero Jacobson radical. Since Jacobson used "radical ring" to describe a ring in which the Jacobson radical is the entire ring, this is an unfortunate conflict with your choice of term. Alternatives appear in the list below.

In the noncommutative case, things separate a bit, and I can tell you some more about terminology.

If all ideals are semiprime, the ring is sometimes called fully semiprime, and sometimes fully idempotent, since all ideals being semiprime is equivalent to having all ideals idempotent. VNR rings are fully idempotent (same proof as above) but clearly not all fully idempotent rings are VNR: there exist non-VNR simple rings which are trivially fully idempotent.

If all ideals are the intersection of maximal right ideals, the ring is called a right V-ring (where V stands for Villemajor). There are non-VNR V-rings and non-V VNR rings. This is less relevant to your situation than the above point, but interesting enough to mention :)

I don't immediately recall a name for condition 2, but I'm sure there is one. Of course, condition 2 implies condition 1. You might be interested to know that you don't have to assume commutativity: a theorem by Jacobson proves that if for every $x$ there is $n>1$ such that $x=x^n$, then the ring is commutative.

Title: Names for specific rings


问:
Why is this quotient space the zero space?
The text says “The quotient space $V/V$ is the zero space as $x+V=V$ for all $x \in V$.
I can only seem to think of $V/V = \{V\}$, since the coset $x+V$ (with $x \in V$) always represents the set $V$ itself.

答:
You are right: $V/V$ consists of a single element, which is $V$. And the only vector space with a single element is the zero space.

Besides, note that, in $V/V$,\begin{align}V+V&=(0+V)+(0+V)\\&=(0+0)+V\\&=0+V\\&=V.\end{align}So, $V$ is the null vector of $V/V$ (as it would be expected).


问:
Where is the equivalence of the relation used in the construction of the numbers?
Given $\mathbb N=\{0,1,2,...\}$, the set of integers is defined by
$$\mathbb Z=\Big\{\{(x,y)\mid (x,y)R(a,b)\}\mid (a,b)\in\mathbb N\times \mathbb N\Big\},$$
where $R$ is the binary relation defined by
$$(a,b)R(c,d)\quad\Leftrightarrow\quad a+d=b+c.$$

When this construction is presented, usually the first step is to prove (or, at least to mention) that $R$ is an equivalence relation. The same is true about the construction of $\mathbb Q$ and $\mathbb R$.

Questions:

Why do we have to know that the relations used to construct the numbers are equivalence relations?

Are the reflexivity, symmetry, and transitivity really needed in the construction? Or are these properties just a coincidence?

答:
Equivalence relations are a generalisation of the identity. Their properties are reflecting exactly that.

Now, why are we even using them? Well, we have a structure given to us, e.g. $\Bbb N$, and we want to define $\Bbb Z$ as the closure of $\Bbb N$ under additive inverses. We observe that we can think about an element of $\Bbb Z$ as the "difference between two natural numbers". Only that there is no difference operation, we only have $+$. So we want to understand how would the difference behave, given $+$.

We are smart, and we know what we're looking for, we want to identify differences, i.e. $a-b=c-d$. But we understand that $x-y=z$ should be the same as $x=z+y$, and so we get $a+d=c+b$. And this defines the equivalence relation on $\Bbb{N\times N}$ that we're looking for.

Now, because we were smart and used $+$ to define $-$, we can show that the operations of $+$ and $\cdot$, and the order $\leq$, all of which are defined in terms of $+$,  will in fact extend to $\Bbb{N\times N}/{\sim}$, where $\sim$ is our equivalence relation, under the obvious embedding $n\mapsto[(n,0)]_{\sim}$.

And this understanding extends to $\Bbb{Q,R,C}$, and other type of constructions of this sort.


问:
Does $\sum^\infty_{n=3}\frac {\ln^2(n)}{n^{3/2}}$ converge or diverge?
I am trying to find a way to decide if this sum converges or diverges, I'm trying to use the integral test and bound the integrand or use the limit comparison test, but I am not finding a good function to compare my integrand with at infinity.
 I would appreciate any help or tips in how to find a function that behaves like $\frac {\ln^2(x)}{x^{3/2}}$ at $x\to \infty$ that I can use it for limit comparison test.

Thanks in advance.

Body: Hint: $\displaystyle\lim_{n\to\infty}\frac{\frac{\log^2(n)}{n^{3/2}}}{\frac1{n^{5/4}}}=0$.


问:
Not understanding a rule of inference.
I am reading about a axiom system like this:

Connectives:$\lnot,\rightarrow$

Axioms:

(A1) $A_1 \rightarrow (A_2 \rightarrow A_1)$

(A2) $(A_1 \rightarrow (A_2 \rightarrow A_3)) \rightarrow ((A_1 \rightarrow A_2) \rightarrow (A_1 \rightarrow A_3))$

(A3) $(\lnot A_2 \rightarrow \lnot A_1) \rightarrow ((\lnot A_2 \rightarrow A_1 ) \rightarrow A_2)$

Rule of inference:
(i)MP (which I know)
(ii)Substitution rule (which I don't know)

The description of substitution rule was like this:-

In addition to modus ponens, we have a substitution rule: we may substitute any wf for all occurrences of statement letter in a given wf.

But I still didn't understand this inference.Can someone explain this to me with a concrete example?

答:
You have to consider the difference between [page 3] the symbols $A_1,A_2,\ldots$, called statement letters and the symbols $\mathscr A, \mathscr B, \ldots$ that are statement forms.

The fist one are the symbols of the alphabet of the formal language; we use them with connectives to produce formulas (wf): $A_1 \to \lnot A_2$.

The statement forms [commonly called schemata (singular: schema)] are not part of the formal language but are used in the meta-language as names for formulas.

As you see [page 28 and 29] the three logical axioms as well as Modus Ponens rule are formulated with statement forms.

Consider e.g. logical axiom (A1): $\mathscr B \to (\mathscr C \to \mathscr B)$.

Formula $A_1 \to (A_2 \to A_1)$ as well as $(A_1 \to A_1) \to (A_2 \to (A_1 \to A_1))$ and many more are instances of the shema.

The alternative approach is to state axioms not as schemata but directly as formula of the language [page 39]: $A_1 \to (A_2 \to A_1)$.

In this case, without schematic axioms, we have lost the "rule" to produce instances of axiom.

To restore this possibility, we need a supplementary rule:

In addition to modus ponens, we have a substitution rule: we may substitute any wf for all occurrences of a statement letter in a given wf.

See also John Corcoran, Schemata: The Concept of Schema in the History of Logic (2006).

Title: Not understanding a rule of inference.


问:
confused about the limit of a trigonometric function
I am trying to calculate the limit of the following function for general $a$:

$$\lim_{x\to a}[\cos(2 \pi x)-\sin(2 \pi x) \cot(\frac{\pi x}{a})]$$

I was believing this is infinite. But Mathematica calculates for $a=3$ for instance the limit is $5$

And for $a$ in general Mathematica returns an infinit term that says:

DirectedInfinity[a Sin[2 a [Pi]]]

I can not understand the role of $a\sin(2a \pi)$

May someone here ehlp me out of confusion?

答:
DirectedInfinity is, I believe, a term that Mathematica uses to indicate a limit of infinity in the complex plane. If $a_n$ is an increasing sequence of positive numbers and $k$ is some complex number, then the limit of $ka_n$ will be expressed as DirectedInfinity[k]. (And the same for sequences that approach this type of sequence.) In the case where $a$ is real, it just means that the expression tends to either $\infty$ or $-\infty$, and the sign of $a \sin(2a\pi)$ will tell you which one.

As for the first part of your question, I think that Mathematica is substituing $x=3$ and rearranging your expression to get something like
$$ \cos(\pi) [ \frac{\cos(6\pi)}{\cos(\pi)} - \frac{\sin(6\pi)}{\sin(\pi)} ]$$.

Since $\cos(6\pi)$ can be expressed as a sixth-degree polynomial in $\cos(\pi)$ and the same for $\sin$, the part in brackets probably ends up evaluating to $-5 + 0$ or similar.

This might not be what you want, but it could be what someone else wants when writing this expression. Try the limit for $a$ not an integer, and this simplification shouldn't happen.


答:
I'm not sure whether there's a cleaner way to do it, but taking a detour via a semicircle provides one way to see that a retraction exists.

Let $\phi$ be a homeomorphism from the filled square to a filled semicircle, which maps the three boundary lines which we wish to retract onto to the curved edge of the semicircle, and the final edge to the diameter of the semicircle (hopefully it's clear that this exists, but I can expand a bit here if it's not obvious)

Then to retract from the semicircle to its boundary we can project every point in a direction perpendicular to the diameter until they hit the boundary of the semicircle

Explicitly, if our semicircle was $\{(x,y) ∈ \mathbb{R}^2 | x^2 + y^2 = 1, y \geq 0 \}$ then $(x,y)\mapsto (x, \sqrt{1-x^2})$ suffices as a retraction

If we call this semicircle retraction $r'$, then $\phi^{-1}\circ r'\circ\phi$ should be a valid retraction for the square case

Title: How does the filled square retract to three of its boundary lines?

Body: In Example 0.15 of Allen Hatcher's Algebraic Topology, he uses the fact that, for $I = [0,1]$, $I \times I$ retracts to $(I \times \left\{ 0 \right\}) \cup (\partial I \times I)$. At first this seemed obvious, but I'm having a hard time constructing an actual retraction. For example $f(s,t) = \begin{cases} (0,t), s\leq 1/2 \\ (1,t), s> 1/2 \end{cases}$ if $t>0$ wouldn't work because it isn't continuous. Nor would fixing a point in the interior of $I\times I$ and doing something similar to the retraction of the punctured space $\mathbb{R}^n$ to $S^{n-1}$, because we wouldn't know where to send that point. So what would be a retraction?


问:
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution?
The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:

$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)

答:
Substitute $x=\frac{t}{2}-\frac2t $ instead to integrate
$$\int \frac1{\sqrt{4+x^2}}dx = \int \frac1tdt=\ln t+C
$$


问:
How do i prove f(0) = 0 (Calculus problem)
I need some help to get started with this question.

$$f(x) + \ln(1+f(x)) = \frac{\sin x}{1+x}$$

Show $f(0) = 0$ and use that to find $f'(0)$.

Where do i start to show $f(0) = 0$? $f(x)$ inside $\ln(1+f(x))$ is confusing me a a lot.

答:
For $x=0$, you get
$$f(0)+\ln(1+f(0))=0$$

so $f(0)$ is solution of the equation $y+\ln(1+y)=0$. Now, the function $g : y \mapsto y +\ln(1+y)$ is strictly increasing on its domain, so it is injective. Moreover, $g(0)=0$, so $0$ is the only solution to $g(0)=0$. So
$$f(0)=0$$


问:
Global invertibility of a given function
Say we have $f:\Bbb{R}\rightarrow \Bbb{R}$ s.t $f\in C^\infty$ and for every $x\in \Bbb{R}$, $|f'(x)|<1$. I have to prove that the function $F(x,y)=(x+f(y), y+f(x))$ is locally invertible for every $x_0\in\Bbb{R}$. Moreover, if also $|f'(x)|<1-\varepsilon$, then  $F$ is globally invertible.

First part was quite easy, the Jacobian matrix at $(x_0,y_0)$ is:\begin{pmatrix}
1 & f'(y_0) \\
f'(x_0) & 1
\end{pmatrix}

Which is clearly invertible because $det(J)=1-f'(y_0)f'(x_0)>0$. However, I don't see how to do the second part.

Any help would be apperciated.

答:
For the second part: Let $L := 1 - \varepsilon \in (0, 1)$. Note that for $u, v \in \mathbb{R}$ we have:
$$
\lvert f(u) - f(v) \rvert = \left \lvert \int^u_v f'(s)~\mathrm{d}s \right \rvert \leq \left \lvert \int^u_v \lvert f'(s) \rvert~\mathrm{d}s \right \rvert \leq L \left \lvert \int^u_v 1~\mathrm{d}s \right \rvert = L \lvert u - v\rvert
$$
Now let $G(x, y) := (f(y), g(x))$. Then we have for $x, y, u, v \in \mathbb{R}$ by the above:
$$
\lVert G(x, y) - G(u, v) \rVert _2 \leq \sqrt{L^2 \lvert y - v\rvert^2 + L^2 \lvert x - u\rvert^2} = L \lVert (x, y) - (u, v)\rVert_2
$$

Now we want to prove injectivity of $F$: Let $x, y, u, v \in \mathbb{R}$ such that $F(x, y) = F(u, v)$. Then we have:
$$
\lVert (x, y) - (u, v) \rVert_2 = \lVert F(x, y) - G(x, y) - F(u, v) + G(u, v) \rVert_2 = \lVert G(u, v) - G(x, y) \rVert_2 \leq L \lVert (x, y) - (u, v) \rVert_2
$$
Since $L < 1$, this inequality is only fulfilled, if and only if
$$
\lVert (x, y) - (u, v) \rVert_2 = 0,
$$
i.e. $(x, y) = (u, v)$. This means that $F$ is injective.

Now we need to prove surjectivity: Let $\bar x, \bar y \in \mathbb{R}$. We want to find $(c, d) \in \mathbb{R}$ such that $F(c, d) = (\bar x , \bar y)$. We define $\bar G(x, y) := G(\bar x-x, \bar y-y)$. Then, we know that for all $x, y, u, v \in \mathbb{R}$ we have:
$$
\lVert \bar G(x, y) - \bar G(u, v) \rVert_2 \leq L (\bar x - x, \bar y - y) - (\bar x - u, \bar y - v) \rVert_2 = L\lVert (x, y) - (u, v) \rVert_2
$$
So $\bar G$ is a contraction. According to Banach's fixed point theorem there exist $a, b \in \mathbb{R}$ such that
$$
G(\bar x - a,  \bar y - b) = (a, b).
$$
Let $(c, d) := (\bar x - a,  \bar y - b)$. Then:
$$
G(c, d) = (\bar x - c, \bar y - d) ~ \iff ~ G(c, d) + (c, d) = (\bar x, \bar y) ~ \iff ~ F(c, d) = (\bar x, \bar y)
$$
This means that $F$ is surjective. But then $F$ is bijective, hence invertible.