问:
For every integer $y$, there is an integer $x$ such that $x + y < 0$

Determine the truth value of the following statement and explain your answer.

 Statement: For every integer $y$, there is an integer $x$ such that $x + y < 0$.

I think the truth value is true but I'm not sure how to explain why since we're not allowed to use examples in our explanation. I tried converting the statement into the following formula

$$\forall y \exists x \left( x + y < 0 \right)$$

but don't quite know where to go from there.

Can someone please confirm my answer and help me explain it? I will gladly appreciate it!

答:
The expression to hold is
$$x+y<0$$

Now let $x=-y-k$
expression becomes
$$(-y-k)+y<0 \Leftrightarrow$$
$$/\text{ Associative rule} : -y-k=-k-y/$$
$$-k-y+y<0\Leftrightarrow$$
$$-k<0\Leftrightarrow$$
$$0<k$$

Now we can select any $k>0$ and the proposition will hold.


问:
Prime number represented by spiral
There is this image on 9gag, with description: "Comparison between 5,000 and 50,000 prime numbers plotted in polar coordinates"

I thought it might have something to do with Ulam spiral, but something looks wrong in picture (but I'm not sure, of course).

I think this white spiral you can track on left image is too "proper", and then filled in second image?

And what would it mean to represent prime in polar coordinates, and how to do it?

答:
As 3Blue1Brown explains, all you need to do is to make sure that the x and y coordinates in the polar plane (also represented by r (radius) and theta (angle between the line joining the point to the origin and the X-Axis) respectively).

In the second image, the Primes occur at a 'periodic' frequency which forms those straight arms, while the non-Primes are also somewhat periodic and hence form the white arms.

The non-Primes do occur a lot but because the scale is so huge, they are not seen properly.

ContentLicense: CC BY-SA 4.0

Title: Prime number represented by spiral


问:
Integrate an exponential distribution
One of my textbooks on phylogenetics keeps going further than I can follow with my 'college calculus'...a probability function is stated as

$$\int_0^\infty \frac{e^{-\theta\tau}(\theta\tau)^d}{d!}e^{-\tau} d\tau $$

...and this is evaluated as

$$ = \left(\frac{1}{1+\theta}\right)\left(\frac{\theta}{1+\theta}\right)^d   $$

Could anyone help me with the intervening steps? I feel sure the answer must have something to do with the definition of e as the limit of 1/n!

答:
The integral is defined, not undefined

$$\int_0^{\infty}\frac{e^{-\theta\tau}(\theta\tau)^d}{d!}e^{-\tau} d\tau$$

and can be rewritten as follows

$$\frac{\theta^d}{(\theta+1)^{d+1}d!}\underbrace{\int_0^{\infty}[(\theta+1)\tau]^d e^{-\tau(\theta+1)}d[\tau(\theta+1)]}_{=\Gamma(d+1)=d!}=\frac{\theta^d}{(\theta+1)^{d+1}}=\frac{1}{\theta+1}\Bigg(\frac{\theta}{\theta+1}\Bigg)^d$$

EDIT: assuming $d \in N$

Title: Integrate an exponential distribution


问:
How many students in the school?

If they put the rows of 13, there are 8 students left; if they put the rows of 15, there are 3 students left and if they put the rows of 17, there are 9 students left. How many students are there given that the total students are < 5000?

The following congruences are: $$x\equiv 8 \pmod {13} \\ x\equiv 3 \pmod {15} \\ x\equiv 9 \pmod {17}$$

I'm still new to this, how can I apply Chinese Remainder Theorem to these congruences to find the total students?

答:
It's easy since here since the moduli are in A.P. (Arithmetic Progression) $\,13,15,17\,$ so we can apply a simple inversion-free version of CRT that only needs to invert the A.P. step-size (here $2$). First we transform our residues $\,8,3,9\,$ to $\rm\color{#0a0}{con}\color{#90f}{gruent}$ ones in A.P (using the Remark below), so

$$\begin{align}
&x\equiv 8\equiv\ \ \:\!  \color{#90f}{99}\!\!\!\pmod{\!13}\\
&x\equiv 3\equiv\ \ \  \ \:\! 3\!\!\!\pmod{\!15}\\
&x\equiv 9\equiv\! \color{#0a0}{-93}\!\!\!\pmod{\!17}_{\phantom{|_{|_|}}}\end{align}\qquad\qquad$$

thus $\!\bmod \color{#c00}{13\!+\!2n}\!:\ x\equiv 99\!-\!96n\!$ $\iff\! 2x\equiv 198\!-\!96\smash[t]{\overbrace{(\color{#c00}{-13})}^{\large \color{#c00}{2n}}}\equiv 1446\,\smash{\overset{\large \div\,2_{\phantom|}\!}\iff}\, x\equiv\,\bbox[5px,border:1px solid #c00]{723}$

Remark $ $ To get residues in A.P. we solve
$\,3\!-\!(8\!+\!13j) = (9\!-\!17k)\!-\!3 \!\iff\! 17k\!-\!13j\equiv 11$ $ \!\iff\!\!\bmod 13\!:\, 4k^{\phantom{|^|}}\!\!\!\equiv\! 11\!\equiv\! 24\!\iff\! k\equiv 6\,$ so $\,j = 7,\,$ so $\,8\!+\!13j=\color{#90f}{99},\,$ $9\!-\!17k=\color{#0a0}{-93}$

Title: How many students in the school?


问:
Symmetry matrix Q or rank(Q-I)=1
An orthogonal matrix $Q$ that satisfies rank$(Q-I)=1$ is called symmetry.

Based on my research, orthogonal matrices map a vector to another vector where "length" is preserved. It also preserves the angle between two vectors.

What I don't or can't imagine is the significance and geometric interpretation of symmetries. May I ask for help? Thank you so much.

答:
A symmetry, as defined in this manner, is a mirroring of $\Bbb R^n$ across a subspace of dimension $n-1$. For instance,
$$
Q=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}
$$
is a mirroring of $\Bbb R^3$ across the $xy$-plane

Title: Symmetry matrix Q or rank(Q-I)=1


问:
Binomial Coefficients Combinatorics
For a positive integers n, prove that

$$\displaystyle\sum\limits_{v=0}^n \frac{(2n)!}{(v!)^2 ((n-v)!)^2} = \binom{2n}{n}^2.$$

If somebody could please help me with this question, I would greatly appreciated it.

答:
$$ \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\frac{(2n)!}{n!n!}\cdot \left(\frac{n!}{v!(n-v)!}\right)^2=\binom{2n}n \cdot \left(\frac{n!}{v!(n-v)!}\right)^2$$

$$\sum\limits_{v=0}^n= \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\binom{2n}n \cdot\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)^2$$

Now, equate the coefficients of $x^n$ in the following identity $$(1+x)^n(x+1)^n=(1+x)^{2n}$$
 to find  $$\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)\cdot \left(\frac{n!}{v!(n-v)!}\right)=\binom{2n}n$$

ContentLicense: CC BY-SA 3.0

Title: Binomial Coefficients Combinatorics


问:
How to prove $P(T_n\le t< T_{n+1}) = \int_0^t P(T_n = s) P(\xi_{n+1} > t-s) ds$?
Here is a proof from Rick Durrett's Probability, Poisson process, and I am confused with it.

Here $\xi_1,\xi_2\cdots$ are independent random varibales with $P(\xi_i > t) = e^{-\lambda t}$.
$T_n = \sum_{k=1}^n \xi_k$, $N_t=\sup\{n: T_n \le t\}$. Now we have $P(N_t = 0) = P(T_1>t)=e^{-\lambda t}$ and for $n\ge 1$,

$P(N_t=n)=P(T_n\le t< T_{n+1}) = \int_0^t P(T_n = s) P(\xi_{n+1} > t-s) ds$ ...

I am confused about the second equality, how to get it? How to get a integral form from the probability?

答:
Note the following, you are looking for the probability of the event $\{N_t = n\} = \{T_n \leq t < T_{n+1}\}$. Hence, for any $s \leq t$, if $T_n = s$, and $\xi_{n+1}>t-s$, this means that $T_{n+1} = T_n + \xi_{n+1} > s+(t-s)=t$.

Therefore, I hope it is now clear that $\{T_n = s, \xi_{n+1} > t-s\} \subset \{T_n \leq t < T_n+1\}$. And
$$\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\}  = \{T_n \leq t < T_n+1\}$$

Since the events are the same, we know that
$$P(\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} ) = P(\{T_n \leq t < T_n+1\})$$

Finally, since $\xi_{n+1}$ is independent of $T_n$, we have:
$$
P(\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} )= \int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds
$$

To prove the identity above, we assume that $P\circ T_n^{-1}$ is absolutely continuous with respect to the Lebesgue measure. Hence,
$$
\int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds =
\int_{T_n^{-1}([0,t])}P(\xi_{n+1}>t-s)d(P\circ T_n^{-1})(s)
$$

Now, we can partition $[0,t]$ in $k \in \mathbb N$ subdivisions (i.e. $[0,t/k)\cup...\cup[t(k-1)/k,t]$).

As $k \to \infty$, the set
$$A_k:=
\bigcup_{i\in \{1,...,k\}}
\{T_n \in \big[t\cdot (i-1)/k,t\cdot i/k \big ),
\quad \xi_{n+1} > t-(i\cdot t/k)\}
\\
A_k
\to
\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} = A
$$

By the continuity of probability, then
$P(A_k) \to P(A)$.

The only thing left to do is to prove that
$$P(A_k) \to
\int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds
$$

Let's call $I_i = [t\cdot (i-1)/k, t \cdot i/k)$, and define a simple
function
$$f_k = \sum_{i=1}^k P(\xi_{n+1}> t-t\cdot i/k) \mathbb I_{I_i}$$
Hence, we have:
$$
P(A_k) = \sum^k_{i=1}P(T_n \in I_i)P(\xi_{n+1}> t-t\cdot i/k)=
\int_{T_n^{-1}([0,t])}f_kd(P\circ T_n^{-1})(s)
$$

Finally, since $f_k(s) \to f(s) = P(\xi_{n+1}>t-s)$. We can use the Dominated Convergence Theorem to conclude that:
$$
P(A_k) = \int_{T_n^{-1}([0,t])}f_kd(P\circ T_n^{-1})(s)
\to
\int_{T_n^{-1}([0,t])}P(\xi_{n+1}>t-s)d(P\circ T_n^{-1})(s)
$$


问:
Sequences, whose difference converge to 0
We still know that: if a sequence of real numbers converges, it is a bounded sequence and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.

I need an example, which is a bounded sequence of real numbers $\left(a_n \right)_{n\in\mathbb N^*}$ that do not converge and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.

答:
Take any unbounded sequence $a_n$ with differences that go to $0$ (surely you know some such sequences). Consider the sequence $b_n=a_{n+1}-a_n$. This gives $a_n+b_n=a_{n+1}$. Now instead consider the sequence $c_n$ given by $c_0=a_0$ and
$$
c_{n+1}=c_n\pm b_n
$$
where the sign is strategically chosen so that the sequence doesn't converge, but is bounded. Say, for instance, you choose signs so it just grows past $1$, then decreases until it just decreases past $-1$, then up again to $1$, and so on.

You need to show that $c_n$ will keep going up and down indefinitely, and the non-convergence of $a_n$ is crucial here. You also need to show that it is bounded, and $b_n\to 0$, or rather the boundedness of $b_n$, is crucial here.


问:
Chain rule when taking non-dimensionalising an ODE
I have a silly question. So let's say we have:

$$\frac{d^{2}x}{dt^{2}} = kx$$

Now let's say we pick $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$. What I don't understand is, if we plug in $Xx_{c}$ for x, and $Tt_{c}$ for t, how come we get:

$$\frac{x_{c}d^{2}X}{t_{c}^{2}dT^{2}}$$

How come the $t_{c}$ is squared? Can someone do the math and explain it to me?  Our professo r just said. it was. chain rule but I am not sure how

答:
$$\frac{d^{2}x}{dt^{2}}  =\frac{d\left(dx/dt\right)}{dt}$$

$$ =\frac{d\left(d[x_c X]/d[t_cT]\right)}{d[tcT]}$$
Since $x_c,t_c$are constant multipliers pull them out of parentheses. And so it becomes

$$ = \dfrac{x_c}{t_c^2}\cdot\frac{d^2X}{dT^2}$$

Stimmt?


问:
How to find defined integral $ \int_{-\pi}^{\pi} e^{x^{2}} \sin x d x $
$$
\int_{-\pi}^{\pi} e^{x^{2}} \sin x d x
$$
I found out, that it can't be integrated via elementary functions properties. How should I dound it?

答: Observe that $f(x)=e^{x^2}\sin x$ is an odd function ($f(-x)=-f(x)$), so $$\int_{-\pi}^\pi f(x)dx=0.$$


问:
Translated convex sets proof
Let A ⊂ R2 be a fixed convex set and let X1, . . . , Xn ⊂ R2 be any convex sets
such that every three of them intersect a translation of A. Then there exists a translation of A that intersects all sets Xi.

I understood the first two part of the question but I am not able to derive the last part or rather how to prove it. Can someone please help me on how to solve this?

答:
Denote by cm(W) the centre of mass of the set W. For every i, define Yi
in such a way that Xi ∩ A' is not equal to φ if and only if cm(A')∈ Yi, for every translation A' of A. Now using Helly’s Theorem we have a point z ∈ Y1 ∩ ... ∩ Yn, we can find a translation A" such that
cm(A") = z. By definition of the sets Yi, the set A" intersects all sets Xi</p>

Title: Translated convex sets proof


问:
Finding Lagrangian of a system (Oscillations in a parabolic trough)
A particle of mass m slides down on the curve $z=1+\frac{x^2}{2}$ in the $xz$-plane without friction under the action of constant gravity.Suppose $z$-axis points vertically upward.Find the Lagrangian of given system.

The position vector of the particle is given by $\overline r=(x,0,1+\frac{x^2}{2})$.
Then the velocity vector is given by
$ \dot {\overline r}=(\dot x,0,x\dot x)$.
So the kinetic energy is given by
$T=\frac {m}{2}((\dot x)^2+x^2(\dot x)^2)$. Now I am confused in potential energy. The potential energy is given by $V=mgz$ but what will be the sign of it? What does it mean that $z$-axis points vertically upward?

答:
The movement is restricted to the manifold $f(x,z) = 1+\frac 12 x^2-z=0$. The potential energy is taken having a referential plane as for instance, the plane $XY$ so the potential energy is given by $m g z$. The Lagrangian is given by

$$
L = T - V +\lambda f = \frac 12 m(\dot x^2+\dot z^2) - m g z + \lambda\left(1+\frac 12x^2-z\right)
$$

giving the movement equations

$$
\cases{
\ddot x = -\frac{x(g+\dot x^2)}{1+x^2}\\
\ddot z = \frac{\dot x^2-g x^2}{1+x^2}\\
\lambda = -\frac{m(g+\dot x^2)}{1+x^2}
}\ \ \ \ \ \ \ \ (*)
$$

Here $\lambda$ represents the reaction force along $f(x,z)=0$

NOTE

According to Euler-Lagrange formalism, the movement equations are

$$
\cases{
m\ddot x -\lambda x=0\\
m\ddot z+\lambda+mg=0\\
1+\frac 12 x^2=z}
$$

now deriving twice the $f(x,z)=0$ manifold we obtain the third equation substitution as

$$
\ddot z= x \ddot x +\dot x^2
$$

now solving

$$
\cases{
m\ddot x -\lambda x=0\\
m\ddot z+\lambda+mg=0\\
\ddot z= x \ddot x +\dot x^2}
$$

for $\ddot x,\ddot z,\lambda$ we obtain $(*)$


问:
Is this equivalence of the compactness theorem true?
In my mathematical logic class the teacher asked us the following question: ¿is the compactness theorem equivalent to: let $\Gamma$ be a set of propositional formulas, and let $\sigma$ be a formula. If $\Gamma \models \sigma$  then there is finite ${ \Gamma  }_{ 0 }\subseteq \Gamma $ such that ${ \Gamma  }_{ 0 }\models \sigma $?, I have tried several times but can't find the correct way to prove it, I appreciate any help.

The statement my teacher gave was: $\Gamma$ is finitely satisfiable if and only if $\Gamma$ is satisfiable.

答:
The version of Compactness you're given is: Γ is finitely satisfiable if and only if Γ is satisfiable. Then:

=> Let Γ be a set of propositional formulas, and let σ be any formula s.t. $Γ \models σ$. Consider the set $Σ = Γ \cup \{¬σ\}$. Clearly this is unsatisfiable, as any model of Γ must satisfy σ, so Γ and ¬σ cannot be satisfied simultaneously. It follows from our assumption that Σ is not finitely satisfiable, so $\exists Σ_0 \subseteq Σ$ finite which is unsatisfiable. Let $Γ_0 = Σ_0 \setminus \{¬σ\}$. We can then see that $Γ_0 \models σ$.

<= Clearly if Γ is satisfiable then it is finitely satisfiable. Suppose that Γ is finitely satisfiable. If Γ is empty then we're done. Otherwise, assume for a contradiction that Γ is not satisfiable, and let $σ \in Γ$. Since Γ is unsatisfiable, we see that $ Γ \setminus \{σ\} \models ¬σ$. By our assumption, we have that $\exists Γ_0 \subseteq Γ \setminus \{σ\}$ finite, such that $Γ_0 \models ¬σ$. We then have that $Γ_0 \cup \{σ\}$ is a finite subset of Γ, which is unsatisfiable. This contradicts our hypothesis, so Γ must be satisfiable.


问:
An inequality with respect to sum of power of prime factors
Let $L > 1 $ be an integer such that $$ L= p_{1}p_{2}...p_{n} = q_{1}q_2 q_3 ... q_m$$
where $p_i$ are primes which are not necessarily distinct and $q_i$ are arbitrary integers.

If I take $L=100$, then $p_1=p_2=2$ and $p_3 = p_4 = 5$ and let $q_1 = q_2 = 2$ and $ q_3 = 25$, then we observe that
$$ \sum_{1}^{4}p_i^x \leq \sum_{1}^{3}q_i^x $$
for any $x \geq 1$

Also if I take many $L$ and test for different $x$ the inequality
$$ \sum_{1}^{n}p_i^x \leq \sum_{1}^{m}q_i^x $$
for any $x \geq 1$
seems to be obvious and very intuitive but proving it seems to be monstrous task.

Is the above inequality true for any $x$ greater than or equal to 1 or does it holds when $x$ is between $0$ and $1$ or does there exist a $L$ such that the above condition is not satisfied?

答:
The above inequality is true for any $x \ge 1$. We only need the following lemma:

$$\forall a,b \in \mathbb N\setminus\{1\}: a^x+b^x\le (ab)^x$$

Which is proven by observing:

$$\frac 1{a^x}+\frac1{b^x}\le \frac1a+\frac1b\le\frac12+\frac12=1$$

and multiplying the result by $(ab)^x$ on both sides. This proves the inequality after strong induction on $n$, i.e. the number of prime divisors (including multiplicities), as follows:

The case $n=1$ is trivial and the case $n=2$ follows from the lemma.

Suppose the result holds for $n=k$. Consider $L = \prod_{i=1}^{k+1} p_i$.

-Suppose $q_i \ne L\  \forall i$, and choose $q_j \ne 1$.

Let $S$ be a subset of $\{1, \dots, n\}$ such that $\prod_{i \in S} p_i = q_j$.

By induction hypothesis, $\sum_{i \in S}p_i^x \le q_j^x$ and $\sum_{i \notin S} p_i^x \le \sum_{i\ne j} q_i^x$.

Hence $\sum_{i=1}^n p_i^x \le q_j^x+\sum_{i\ne j} q_i^x =\sum_{i=1}^m q_i^x$.

-The case $q_j=L$ follows easily from induction as well:

$$\sum_{i=1}^m q_i^x \ge L^x \ge p_1^x + \left(\frac L {p_1}\right)^x \ge p_1^x + p_2^x + \left(\frac L {p_1p_2}\right)^x \ge \dots \ge \sum_{i=1}^n p_i^x$$

This concludes the induction step.

The inequality is not necessarily true for $0<x<1$. For $n>m$, when $x \to 0$:

$$\sum_{i=1}^n p_i^x\to n, \sum_{i=1}^m q_i^x\to m$$

hence eventually the inequality fails. For example, we have $2^{1/2} + 2^{1/2} > 4^{1/2}$.


问:
A property in Hausdorff compact spaces
Let $X$ be a Hausdorff compact space and $A$ and $B$ be two disjoint closed subsets of $X$.

Since $X$ is Hausdorff and compact, it is normal, hence there exists $A_1$ and $B_1$ disjoint neighborhoods of $A$ and $B$. Since $X$ is Hausdorff, $A_1$ and $B_1$ can be chosen s.t. $\overline{A_1}\cap\overline{B_1} =\emptyset.$

In this way we can construct family of neighborhoods $\{ A_n \}$ and $\{ B_n\}$ for $A$ and $B$ with the properties $\overline{A_n} \subseteq A_{n+1},\; \overline{B_n} \subseteq B_{n+1}$ and $\overline{A_n}\cap \overline{B_n}=\emptyset,\; \forall n\in\mathbb{N}.$

My question: Is there a way to construct $\{ A_n \}$ and $\{ B_n\}$ s.t. $X=\bigcup\limits_{n\in\mathbb{N}} A_n \cup \bigcup\limits_{n\in\mathbb{N}} B_n$?

答:
This is not true. Let $X=[0,1], A=\{0\}, B=\{1\}$, $A_n-[0,\frac 1 3 -\frac 1n), B_n=(\frac 2  3-\frac 1 n,1]$. Then $\cup A _n \cup \cup B_n \neq X$.

Answer for the revised verion of the question: This cannot be done if $X$ is connected. If $X=\cup A _n \cup \cup B_n$ then there is a finite subcover for this open cover. In view of monotonicity of the sets this gives $X=A_N \cup B_N$ for some $N$. But this gives a separation of $X$.

Title: A property in Hausdorff compact spaces


问:
Efficient directional nearest-neighbor search among rectangles
I have a large set of rectangles with random orientation and size. Say up to $10000$ rectangles, which are non-overlapping.

For every rectangle, I need to find the nearest compatible neighbor. Two rectangles are said to be compatible when the angle they both form with the line that joins their center is below a fixed tolerance. The distance is computed between the rectangle centers. Note that every rectangle has two orthogonal directions. (In a more advanced version of the problem, the tolerance can also depend on the sizes.)

I can obviously solve this by exhaustive search, trying all pairs of rectangles, but this is an $O(N^2)$ process. I am looking for algorithmic ways to accelerate (parallelization is not an option). I know about efficient solutions for the all-nearest-neighbors search based on the Voronoi diagram, kD-trees and similar techniques, but in my case the compatibility condition makes them unsuitable.

Any suggestion ? A speedup factor of $10$ would already be very welcome.

答:
(If I understand the question correctly,) here is one approach.

Sort for angle of rotation first. Keep offset in the list of +90 degree modulos. Use your favorite sorting algorithm here. Several of the classical ones shall perform $\mathcal O(n\log(n))$.

Now all compatible ones will lie locally close to each other with exception of the edges of the list. This is nice both for cache hits and for ease of writing algorithm.

We can build a double loop with the inner loop breaking as soon as we go outside of threshold. This shall greatly reduce the number of comparisons required if the threshold is small. No unnecessary comparisons will need to be done.


问:
Why does $\frac{1}{z}$ not have an antiderivative but $\frac{1}{z^2}$ does?
On the complex plane $\mathbb C-\{0\}$, it is stated that $\frac{1}{z}$ does not have an antiderivative. But apparently $\frac{1}{z^2}$ does and I find this confusing because both functions have a pole at 0 so that means any closed curve around the origin is not 0 right?

答:
So first off, poles do not inherently mean that integrals around the pole vanish. The important relationship is that all closed integrals vanish if and only if there is an antiderivative, and if the domain is simply connected, then an antiderivative exists. So on simply connected domains, closed integrals vanish. But just because all closed integrals vanish, the domain doesn't have to be simply connected. It's just a one-way implication.

Now to the functions at hand: there are a few different ways to see why one has an antiderivative, while the other does not. For one, we already know that all functions of the form $z^n$ with $n\in\mathbb Z\backslash\{-1\}$ have an antiderivative given by $\frac{1}{n+1}z^{n+1}$. And we can already see from this formula that at $n=-1$ something special should happen, since this formula would force us to divide by $n+1=-1+1=0$, which we can't. So this formula cannot possibly apply. We have to think of something different. Interestingly, we can observe that an antiderivative of $z^{-1}$ must be a holomorphic logarithm, up to an additive constant. If $F$ were such an antiderivative, then the derivative of $F\circ\exp(z)$ would be $\exp(z)F'\circ\exp(z)=\frac{\exp(z)}{\exp(z)}=1$. So $F\circ\exp$ is an antiderivative of $1$, so $x+c$ for some $c\in\mathbb C$. So $F$ is a logarithm, up to the constant $c$. But we know that there is no holomorphic logarithm on the punctured complex plane, so that's impossible.

Or we could directly calculate the integrals for a nice contour, like the unit circle. Then we have
$$\int_{\vert z\vert=1}z^n\mathrm dz=\int_0^{2\pi} \mathrm i\mathrm e^{\mathrm it}(\mathrm e^{\mathrm it})^n\mathrm dt=\int_0^{2\pi}\mathrm e^{\mathrm i(n+1)t}\mathrm dt.$$
Again, we see that for $n=-1$, something special happens. For all other $n$, the integrand is periodic in a way such that the integral will evaluate to $0$. But for $n=-1$, it is constant, so it won't evaluate to $0$. So the integral doesn't vanish for $n=-1$, but it does vanish for all other $n$.


问:
Show that $\int_{\mathbb{S}^{n-1}} Du(x)[v]\,dv=0$
Show that on a riemannian manifold $(M,g)$, if we integrate over the sphere $\mathbb{S}^{n-1}\subset T_x M$, we have that $$\int_{\mathbb{S}^{n-1}} \langle \mbox{ grad}u(x), v\rangle_x \,dv=0.$$

My approach: If we consider $dv$ as the measure of the unit sphere $\mathbb{S}^{n-1}$, then   $$\int_{\mathbb{S}^{n-1}} \langle \mbox{ grad}u(x), v\rangle_x \,dv=\int_{\mathbb{S}^{n-1}} Du(x)[v]\,dv$$

Now, if we define $v'=-v$, then $$\int_{\mathbb{S}^{n-1}} Du(x)[v]\,dv=-\int_{\mathbb{S}^{n-1}} Du(x)[v']\,dv'$$

So, the integral is $0$. But I don't know if the change of variable is correct because we are on the unit sphere $\mathbb{S}^{n-1}$ and is this not the lebesgue measure.

答:
The idea in the OP is correct.

Polar coordinates, and the induced product measure is definable even in the context of Lebesgue measure. (See for example Exercise 6, of Chapter 8 of Rudin Real & Complex Analysis.)

In fact, more generally if $U$ is an orthogonal matrix $n\times n$, then
$$
\int_{S^{n-1}}f(Uw)\,dw =
\int_{S^{n-1}}f(w)\,dw
$$
This holds for $f\in L^1(S^{n-1})$ and in order to prove it we start with characteristic functions, using the facts

Fact A. $m_n(A)=m_n(UA)$, for $A\subset \mathbb R^n$, measurable, where $m_n$ is  the Lebesgue measure.

Fact B. If $D\subset S^{n-1}$ measurable, then
$$
\sigma_{n-1}(D)=n\, m_n\big(\{rw: w\in D,\,\,r\in [0,1]\}\big)
$$
where $\sigma_{n-1}$ the surface measure on $S^{n-1}$ and $m_n$ the Lebesgue measure in $\mathbb R^n$.


问:
Krein-Smulian Theorem
I am going through the book Vector Measure by Diestel and Uhl and I got stuck in the proof of Krein-Smulian theorem given on page no.51.

I understood that any sequence $(x_n)$ in $W$ is inside the closed linear span of $\{x_n:n\in \mathbb{N}\}$, that is a separable subspace of $X$. But how $W$ itself can be considered to be separable using Hahn-Banach theorem? Any help will be appreciated.

答: Pick a sequence in the closed convex hull of $W$. Let $Y$ be the closed subspace generated by the sequence $(x_n)$ we have chosen. Then $W_0=W \cap Y$ is a weakly compact set in the Banach space $Y$. If we prove that result in the case of separable Banach spaces it would follows that $(x_n)$ has  weakly convergent subsequence in $Y$. But the subsequence also converges weakly in $X$ since any continuous linear functional becomes a continuous linear functional on $Y$ when restricted to $Y$. That finishes the proof.

Title: Krein-Smulian Theorem


问:
Find minimum value of $|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$ where $z$ is a complex number

Find minimum value of $$|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$$ where $z$ is a complex number.

we notice that we have to minimze the sum of distances from the points in the argand plane  $(0,e^{ix},5e^{ix},7e^{ix},32e^{ix})$.We also see these points are on a straight line which makes an angle $x$ with real axix. .We now rotate the plane by $x$ radians anticlockwise so that the points lie on the real line.Let the coordinates of $z$  after rotation be $(h,k)$.Intuitively this does not change the minimum.So now we have to minimise: $$\sqrt{h^2+k^2}+\sqrt{{(h-1)}^2+k^2}+\sqrt{{(h-5)}^2+k^2}+\sqrt{{h-7)}^2+k^2}+\sqrt{{(h-32)}^2+k^2}$$  which happens when
$k=0$.We minimize  $$f(h)=|h|+|h-1|+|h-5|+|h-7|+|h-32|$$
But $f(h)$ becomes  a lenear function for each of the intervals $h\in [-\infty,0],[0,1],[1,5],[5,7],[7,32],[32,\infty]$.But a lenear function achieves its maximum and minimum at the endpoints of interval.

We seek:$$\min(f(0),f(1),f(5),f(7),f(32))=f(5)=38$$

Is it good?
I am looking for alternative proofs

答:
This is not a formal approach but a trick I devised by myself. I shall be grateful if someone can provide it formally. Here's the approach.

Given, $$\left|z-r_{1}\right|+\left|z-r_{2}\right|+\left|z-r_{3}\right|+\left|z-r_{4}\right|+\left|z-r_{5}\right|+ \dots +\left|z-r_n\right|$$

Open the modulus such that the coefficient of $z$ is either $1$ or $0$. If it is zero, you straight away get the minimum, if it is $1$ just put $z$ such that it is the higher $r_i$ in the interval of expansion.

Example: (solution to the given question)

I will assume $z$ between $r_2,r_3$. (One might do it for $r_3,r_4$ too). We have, $$z+\left(z-e^{ix}\right)+\left(5e^{ix}-z\right)+\left(7e^{ix}-z\right)+\left(32e^{ix}-z\right) = \left(-1+5+7+32\right)e^{ix}-z$$

Consequently $z=5e^{ix}$ (since it is limited between $[e^{ix},5e^{ix}]$) and hence the minimum $38e^{ix}$


问:
Why $\mathbb Z(\sqrt2)$ is not a lattice?
The set $\mathbb Z(\sqrt2) = \{a + b\sqrt2 : a, b \in \mathbb Z\}$ is not a lattice, according to the book of Robeldo =

because when you replace
$a, b \in \mathbb Z$ by $a, b \in  \mathbb R$ we do not obtain all of $C$

but only a $1$-dimensional real space (in this case just $\mathbb R$).

In other words, there are no two points $w_1, w_2$ in $\mathbb Z$ whose coordinates are linearly independent in $R^2$.

This is hard to get for me.

How can I know that if I replace $a, b \in \mathbb Z$ by $a, b \in  \mathbb R$ we do not obtain all of $C$?

答:
Because $\sqrt{2} \in {\mathbb R}$, ${\mathbb R}[\sqrt 2] = \{ a + b \sqrt2 \mid a, b \in {\mathbb R}\}$ is equal to $\mathbb R$ and not to $\mathbb C$.

(I'm not sure what is exactly meant by "In other words, there are no two points $w_1, w_2$ in $\mathbb Z$ whose coordinates are linearly independent in $R^2$.")


问:
Limit of $\frac{x\sin(\pi\|x\|)}{\|x\|}$ as x approaches 0
I have read the following statement. Let $x \in \mathbb{R}^n$, then

$$\lim\limits_{x \to 0} \frac{x \text{sin}(\pi\|x\|)}{\|x\|}=0$$

I have to admit that I haven't worked with limits for a while, which is why I can't formally prove this (although I suspect the proof might be very elementary). The intuition is quite clear and I am only looking for a way to formally prove the equality above, so if anyone knows how to prove the statement formally, I would be very grateful. I know that

$$\lim\limits_{t \to 0}\frac{\sin(t)}{t}=1$$

if this is needed. Thank you in advance for any help.

答:
The limit is $0_{\mathbb{R}^n}$.

As $\pi ||x|| \underset{x \to 0}{\longrightarrow} 0$ and $\frac{\sin(s)}{s} \underset{s \to 0}{\longrightarrow} 1$, one has by composition of limits:

$$
\frac{\sin(\pi ||x||)}{||x||} x =\pi \cdot \frac{\sin(\pi ||x||)}{\pi ||x||} x \underset{x \to 0}{\longrightarrow} \pi \cdot 1 \cdot 0_{\mathbb{R}^n} = 0_{\mathbb{R}^n}
$$


答: Now you have edited it $$\lim_{x\to0}\frac{x\sin(\pi||x||)}{||x||}\\=\lim_{x\to0}\frac{\sin(\pi||x||)\cdot x\pi}{\pi||x||}\\=\lim_{x\to0}\frac{\sin(\pi||x||)}{\pi||x||}\lim_{x\to0}x\pi=\pi\cdot0=0$$

Title: Limit of $\frac{x\sin(\pi\|x\|)}{\|x\|}$ as x approaches 0

Body:
I have read the following statement. Let $x \in \mathbb{R}^n$, then

$$\lim\limits_{x \to 0} \frac{x \text{sin}(\pi\|x\|)}{\|x\|}=0$$

I have to admit that I haven't worked with limits for a while, which is why I can't formally prove this (although I suspect the proof might be very elementary). The intuition is quite clear and I am only looking for a way to formally prove the equality above, so if anyone knows how to prove the statement formally, I would be very grateful. I know that

$$\lim\limits_{t \to 0}\frac{\sin(t)}{t}=1$$

if this is needed. Thank you in advance for any help.


问:
Discrete Maths - Graph Theory Problem
I am trying to do Mathematics for CS course( 6.042) from MIT opencourseware. Could anyone please help me with this problem( from problem set 6. Problem 6).

Let G be a graph. In this problem we show every vertex of odd
degree is connected to at least one other vertex of odd degree in G.

(a) [6 pts] Let v be an odd degree node. Consider the longest walk starting at v that does
not repeat any edges (though it may omit some). Let w be the final node of that walk. Show
that w is not equal to v.

(b) [4 pts] Show that w must also have odd degree.

答:
Question 1: Proof by contradiction

Suppose the w = v. That means in a walk there is no edge left going out of v such that we can walk on that edge. So, we walked on all the edges going out of v. So, for every time we went out of v, we were able to reach back to v(because w = v). That means v must be having even degree. We reached contradiction because v is having odd degree. So, w is not equal to v.

Question 2: Proof by contradiction

Suppose w is having even degree. So, the walk ending on the node with even degree. From w we cannot move further because all the edge have been visited from w. But every time walk enters w it must leave w because it has even degree. So, by that sense the last edge of the walk it enters w and does not leave. So, before the last edge whenever w was reached it was left, that shows even degree of w has been traversed but as the last time it was entered w it didn't left. So, it shows w has odd degree. Hence we reached contradiction

Title: Discrete Maths - Graph Theory Problem


答:
Yes, the two statements below are equivalent. Let us prove it. Let $\Gamma$ be a set of formulas.

For any  formula $\sigma$,  $\Gamma \models \sigma$  if and only if there is a finite $\Gamma_0 \subseteq \Gamma $ such that $\Gamma_0 \models \sigma $;

$\Gamma$ is finitely satisfiable if and only if $\Gamma$ is satisfiable.

Proof that 1. implies 2. We suppose that 1. holds and we prove that the equivalence in 2. holds

$\Rightarrow$: Assume that $\Gamma$ is finitely satisfiable. Then, for every finite $\Gamma_0 \subseteq \Gamma$, there is a structure that satisfies $\Gamma_0$. Let $\sigma$ be a contradiction formula (i.e. a formula that is not satisfied by any structure, for instance $\bot$ or $\alpha \land \lnot \alpha$). So, for every finite $\Gamma_0 \subseteq \Gamma$, there is a structure that satisfies $\Gamma_0$ and not $\sigma$, i.e. $\Gamma_0 \not\models \sigma$.
By 1. (direction left-to-right), $\Gamma \not\models \sigma$ and hence there exists a structure that satisfies $\Gamma$ but not $ \sigma$.
In particular, $\Gamma$ is satisfiable.

$\Leftarrow$:  Assume that $\Gamma$ is satisfiable. Then, there exists a structure that satisfies all the formulas in $\Gamma$. In particular, it satisfies every finite $\Gamma_0 \subseteq \Gamma$. Therefore, $\Gamma$ is finitely satisfiable. (Note that here we do not use the hypothesis that 1. holds)

Proof that 2. implies 1. We suppose that 2. holds and we prove that the equivalence in 1. holds. Let $\sigma$ be a formula.

$\Rightarrow$: Assume that $\Gamma \models \sigma$. Then, every structure that satisfies $\Gamma$ does not satisfies $\lnot \sigma$, i.e. $\Gamma \cup \{\lnot \sigma\}$ is unsatisfiable. By 2. (direction left-to-right), $\Gamma \cup \{\lnot \sigma\}$ is not finitely satisfiable, hence there is a finite $\Gamma_0 \cup \{\lnot \sigma\} \subseteq \Gamma \cup \{\lnot \sigma\}$ that is unsatisfiable. Therefore, for some  finite $\Gamma_0 \cup \{\lnot \sigma\} \subseteq \Gamma \cup \{\lnot \sigma\}$, every structure that satisfies $\Gamma_0$ satisfies $\sigma$ too, i.e. $\Gamma_0 \models \sigma$.

$\Leftarrow$: Assume that $\Gamma_0 \models \sigma$ for some finite $\Gamma_0 \subseteq \Gamma$. Then,  every structure that satisfies $\Gamma$ in particular satisfies $\Gamma_0$ and hence $\sigma$, since $\Gamma_0 \models \sigma$. Therefore, $\Gamma \models \sigma$. (Note that here we do not use the hypothesis that 2. holds)

In both statements 1. and 2., the interesting part is the left-to right direction. Indeed, in statement 1., the right-to-left direction is trivial and holds independently from the hypothesis that 2. holds.  And similarly, in statement 2., the right-to-left direction is trivial and holds independently from the hypothesis that 1. holds.

The left-to-right direction of the compactness theorem (in both formulations 1. and 2.) has the following logical form, at the meta-level:
$$\tag{*}\forall \exists \implies \exists \forall$$
What do I mean? Take the left-to-right direction of statement 2. By making explicit the meaning of "satisfiable" and "finitely satisfiable", it says: if for all finite subsets of $\Gamma$ there exists a structure that satisfies it, then there is a structure that satisfies all formulas in $\Gamma$. This is surprinsing, because $\Gamma$ can be an infinite set of formulas and so compactness theorem says that we can somehow transfer a property (satisfiability) from finite to infinite.

Usually, implications of the form $(*)$ does not hold. Think of natural numbers: the fact that for every natural number $n$ there is a natural number  greater than  $n$, does not imply that there is a natural number  greater all other natural numbers.
Compactness theorem says that when we talk about satisfiability of sets of formulas, the implication $(*)$ holds, even when the set $\Gamma$ of formulas is infinite.

The other direction (the right-to-left one) is of the form
$$\exists \forall \implies \forall \exists$$
which is obvious and indeed it corresponds to the less interesting part of statements 1. and 2.

Title: Is this equivalence of the compactness theorem true?

Body:
In my mathematical logic class the teacher asked us the following question: ¿is the compactness theorem equivalent to: let $\Gamma$ be a set of propositional formulas, and let $\sigma$ be a formula. If $\Gamma \models \sigma$  then there is finite ${ \Gamma  }_{ 0 }\subseteq \Gamma $ such that ${ \Gamma  }_{ 0 }\models \sigma $?, I have tried several times but can't find the correct way to prove it, I appreciate any help.

The statement my teacher gave was: $\Gamma$ is finitely satisfiable if and only if $\Gamma$ is satisfiable.


问:
Definition of boundary of differential manifold is well defined
Given the following definition of manifold with boundary :

$S \subset \mathbb{R}^{N}, N \in \mathbb{N}$ is a $n-$differential manifold with boudary if : $n \in \mathbb{N}, \forall \hspace{0.1cm} p \in S, \hspace{0.1cm}\exists \hspace{0.1cm} V $open neighborhood of $p$ in $\mathbb{R}^{N}$ and $U$ open in $\mathbb{R}^n$, with $\varphi : U \longmapsto V \hspace{0.1cm} C^\infty$ with differential of rnk $n$ such that $\varphi(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\}) = V \cap S$ is a homeomorphism.
In this setting we say that $\partial S \ni p \iff \varphi(U \cap \{x_n = 0\}) \ni p$

With definition I'd like to prove the well-definess of the boundary of a manifold, without using arguments which involve algebraic topology.

To be more accurate, it's not clear to me how the following follows from the definition :

If $\varphi_1,\varphi_2$ are parametrization with $ p \in \varphi_1(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\}), p \in \varphi_2(U \cap \{x \in \mathbb{R}^n : x_n \geq 0\})$ we have that $$p \in \varphi_1(U_1 \cap \{x_n = 0\}) \iff p \in \varphi_2(U_2 \cap \{x_n = 0\})$$

Any help hint or reference of the proof will be appreciated.

答:
To prove the consistency of the definition for a point $p$ to be in the boundary, it is equivalent to prove the consistency of the converse, i.e the definition for a point to be in the interior of $M$.

Let $p \in M$ and suppose there exists a chart $\varphi : U \to \mathbb{R}^n_+=\{(x_1,\ldots,x_n)\in \mathbb{R}^n~|~x_n \geqslant 0\}$ such that $\varphi(p)_n >0$. Suppose $\psi : V \to \mathbb{R}^n_+$ is another chart with $p \in V$. Let us show that $\psi(p)_n>0$.

By the very definition of charts, $\psi\circ \varphi^{-1} : \varphi(U\cap V) \to \psi (U\cap V)$ is a diffeomorphism. Now, as $\varphi(p)_n>0$, there exists an open subset around $\varphi(p)$ is $\varphi(U\cap V)$ is an open subset $W\subset \varphi(U\cap V)$ such that for all $x \in W$, $x_n>0$ (take an open ball with small enough radius). $W$ is then an open subset of $\mathbb{R}^n_+$ such that every $x$ in it has $x_n >0$: it is an open subset of $\mathbb{R}^n$.

Now, extand the codomain of $\psi\circ \varphi^{-1}$ to be all of $\mathbb{R}^n$. We have a map $\psi \circ \varphi^{-1} : W \to \mathbb{R}^n$ which defined on an open subset of $\mathbb{R}^n$ and such that it is a diffeomorphism onto its image: its image is thus open because of the inverse function theorem. Therefore, it is an open subset of $\mathbb{R}^n$, contained in $\mathbb{R}^n_+$. Consequently, all $x$ in $W$ satisfy $\psi\circ\varphi^{-1}(x)_n >0$. That is, all $q$ in $\varphi^{-1}(W)$ satisfies $\psi(q)_n>0$. Thus, $\psi(p)_n >0$. We have shown that if there exists a chart such that $\varphi(p)_n >0$, then for all chart $\psi$ for which $\psi(p)$ is defined, then $\psi(p)_n>0$.

Conversly, if there exsits a chart for which $\varphi(p)_n = 0$, then for all chart $\psi$ such that $\psi(p)$ is defined, $\psi(p)_n =0$.

Comment: this is an exact reformulation of this but we do not use here the invariance domain theorem because in this context, the fact that a diffeomorphism is open is a straightforward consequence of the inverse function theorem, that is an easier result.

Edit It has been asked several times in comments to explain why transition maps of a smooth manifold are diffeomorphisms. This is directly following from the definitions of:

a diffeomorphism

the transitions maps

Here is the reason. A diffeomorphism $f : U \to V$ between open subset of $\mathbb{R}^n$ is a smooth homeomorphism with smooth inverse.

The transitions maps are $\varphi_{\alpha\beta} = \varphi_{\beta}\circ {\varphi_{\alpha}}^{-1} : \varphi_{\alpha}(U_{\alpha}\cap U_{\beta}) \to \varphi_{\beta}\left(U_{\alpha}\cap U_{\beta} \right)$ and by definition, are smooth homeomorphisms. Now, it is straightforward to show that ${\varphi_{\alpha\beta}}^{-1} = \varphi_{\beta\alpha}$ is also a transition function and thus, is a smooth homeomorphism by definition.

Hence, $\varphi_{\alpha\beta}$ is a smooth homeomorphism with smooth inverse: it is a diffeomorphism.


问:
How to solve this minimization problem in classical mechanics with holonomic constraints?
One should minimize the distance between two points $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$. The holonomic constraint states that $()=^2−2+5$ where $x_1$ is an element of this graph and $x_2$ is an element of the graph of a different function: $()=2−1$
Now the goal is to use the constraints in such a way, to eliminate two of the given coordinates $x_1,x_2,y_1,y_2$. After the elimination one should determine the minimum of the distance between these 2 points. A hint is that this process is reminiscent of the 2. Lagrangian formalism.

I just don't even know where to start off, I would be so grateful if someone could help me out somehow!

答:
Given $p_1 = (x_1,y_1),\ \ p_2 = (x_2, y_2)$ and the restrictions $g_1(x_1,y_1) = y_1 -( x_1^2-2x_1+5)=0,\ \ g_2(x_2,y_2) = y_2 -( 2x_2-1)=0$ the problem can be formulated as the stationary points determination for the lagrangian

$$
L = \|p_1-p_2\|^2+\lambda_1 g_1(x_1,y_1)+\lambda_2 g_2(x_2,y_2)
$$

The stationary points are the solutions for

$$
\nabla L = 0 = \cases{
 \lambda_1(2 x_1-2)+2 (x_1-x_2) \\
 2 \lambda_2-2 (x_1-x_2) \\
 2 (y_1-y_2)-\lambda_1 \\
 \lambda_2+2 (y_1-y_2) \\
 x_1^2-2 x_1-y_1+5 \\
 2 x_2-y_2-1 \\
}
$$


问:
Solve a cubic polynomial (given one root is four times a second root)?
So, I've been stuck on a question for a long time now:

"Solve the equation $10x^3 + 23x^2 + 5x - 2 = 0$ given that one root is four times a second root."

How would you go about solving this?
Any help would be greatly appreciated.

答:
If $a$, $4a$ are the two roots in question and $b$ is the third, then you know by Vieta that $a+4a+b=-\frac{23}{10}$ and $ab+4ab+4a^2=\frac5{10}$ and $4a^2b=\frac2{10}$.

Alternatively (inspired by Taladris' approach, but without the need to know how to solve quadratics! As a penalty, you need to work with a few bigger numbers, though)

With $P(x)=10x^3+23x^2+5x-2$, let $a$ be the root such that $4a$ is also a root. Then $a$ is a root of $P(x)$ and of $P(4x)=640x^3+368x^2+20x-2$.
But $a$ is also a root of any linear combinations of these polynomials (adjusted to eliminate the highest powers of $x$), especially of (essentially we are computing the $\gcd$)
$$ Q(x):=\frac16(P(4x)-64P(x))=-184x^2-50x+21$$
then of
$$R(x):=92P(x)+5xQ(x) = 1866x^2+565x-184$$
and finally of $$933Q(x)+92R(x)=5530x+2665,$$
hence we must have $a=\frac12$. The rest is easy.


问:
Nowhere-Dense-Sets Clarified
If A is a nowhere dense set, it means that "A closure" has an empty interior.
Canwe also say that "A" too has an empty interior?

I believe Yes, we can say so.

Because "A closure" is just the union of A and it's limit points. The set of limit points are always closed(since they are not open). Now when we talk about "A Closure's interior", we refer to A's interior. And so if we say A closure has an empty interior, it means "A Closure" sans the closed subset of A (which is the set of it's limit points) has an empty interior. In other words, A has an empty interior? Am I Right?

答:
Definitely closure of a set having empty interior implies that set also has empty interior.

As far as I understand from your argument, you accept the definition a set is nowhere dense if its closure has empty interior.

You said

A Closure's interior", we refer to A's interior

If you are trying to prove if $A$ has an empty interior then $A$ is a nowhere dense, here is a counter example-
Take $A$ be the set of rational numbers in real metric space with respect to Euclidean metric then the interior of rational numbers is empty but the interior of limit points of $A$ is $\mathbb{R}$.

Title: Nowhere-Dense-Sets Clarified


问:
How to derive discrete system model from satellite continuous model of unscented kalman filter?
How to derive the discrete system model from a continuous model. Below is the satellite's differential equation.

$$\dfrac{dr_x}{dt} = v_x, \dfrac{dr_y}{dt} = v_y, \dfrac{dr_z}{dt} = v_z$$
$$\dfrac{dv_x}{dt} = \dfrac{-\mu r_x}{r^3},\dfrac{dv_y}{dt} = \dfrac{-\mu r_y}{r^3},\dfrac{dv_z}{dt} = \dfrac{-\mu r_z}{r^3}$$

More specifically, the state vector of the unscented Kalman filter is

$$x=[r_x, r_y, r_x, v_x, v_y, v_z]^T$$

The system model has the form of $$x_{k+1}=f(x_k, u_k)$$
Thus, the question is how to derive the discrete model $f$ which has the above form? Or is it even possible?

Any reference or feedback is appreciated.

I come up with the following answer but not sure if it's correct or not. Please point out the error in the derivation if you spot one.

$$v_{x,k+1}=v_{x,k}+\Delta v_x = v_{x,k}+\dfrac{dv_x}{dt}dt = v_{x,k}+\dfrac{-\mu r_{x,k}}{r^3}dt=v_{x,k}+\dfrac{-\mu r_{x,k}}{r^3}\Delta t$$
$\Delta t$ is the time interval.

答:
One approach is to use some kind of integration method for ordinary differential equations. You already described one of the simplest of such methods in your question, what is also called the Euler method. In its entirety this would give

$$
\begin{bmatrix}
p_{k+1} \\ v_{k+1}
\end{bmatrix} =
\begin{bmatrix}
p_{k} \\ v_{k}
\end{bmatrix} +
\begin{bmatrix}
v_k \\ \frac{-\mu\,p_k}{r_k^3}
\end{bmatrix} \Delta t,
$$

with $p_{k} = \begin{bmatrix}r_x & r_y & r_z\end{bmatrix}^\top$ and $v_{k} = \begin{bmatrix}v_x & v_y & v_z\end{bmatrix}^\top$. This can get you close to the true change in position and velocity after a time step $\Delta t$ the smaller the time step gets, but will in general always be a little bit wrong.

There are other higher order integration methods, such as the Runge-Kutta integration methods. Another common integration method often used for systems like this is Verlet integration which conserves energy and can guarantee long term stability of the orbit. However, long term stability of the predicted trajectory is probably not really relevant here since you only require a prediction over the time step $\Delta t$.

It is also (kind of) possible to solve it analytically. Namely, your dynamics can be described as a Kepler orbit whose elements all remain constant except the true anomaly. The true anomaly can be found as a function of time by solving Kepler's equation, which in general can't be solved analytically but one can get arbitrarily close using numerical methods.

It can be noted that the added accuracy of the analytical solution might be overshadowed by the noise on the measurements of the state vector. So in many cases I think that integration methods would yield sufficiently good predictions. Additionally, the integration methods would also allow for external forces, such applied thrust.


问:
Can you use only L'Hôpital's rule in indeterminate forms
On evaluating $$f(x)=\begin{cases} \frac{x\ln(\cos x)}{\ln(1+x^2)} & x\neq0,\\0& x=0\end{cases}$$
all the solutions I've seen used the
L'Hôpital's rule to break it down; that way, you find out that the function is continuous at x=0, becoming 0 there.

I used $\lim_{x^2 \to 0} \frac{\ln(1+x^2)}{x^2}=0$ and logarithmic properties to simplify it to $\lim_{x
\to 0}\ln(\cos x)^{ \frac{1}{x}}$;

Defining h as a small positive value, as x approaches 0 from the right, it becomes h. As cosx approaches 0 from the right, it becomes 1-h. Thus the right-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{1}{h}$. This becomes $ln0=-\infty$

Similarly, as x approaches 0 from the left, it becomes -h. As cosx approaches 0 from the left, it becomes 1-h, the same as before. Thus the left-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{-1}{h}$. This is $\lim_{h \to 0^+}ln\frac{1}{(1-h)^\frac{1}{h}}$, which becomes $ln\infty$, which is $\infty$.

As the limit doesn't seem to exist as per my evaluation, am I supposed to use only the L'Hôpital's Rule in indeterminate forms?

答:
Your idea of using the fact that $\frac {\ln (1+t)} t \to 1$ as $ t \to 0$ is good so the question reduces to that of finding $\lim_{x \to 0} \frac {\ln (\cos x)} x$. But this last limit is $0$ by L'Hopital's Rule. I don't know how you concluded that this limit does not exist.

There are conditions to be satisfied for application of L'Hopital's Rule and a wrong application can lead to a wrong answer.


答:
" Am I supposed to use only the L'Hôpital's rule for indeterminate forms? "

No, not only LH rule for indeterminate forms.

For example, let's see the following limit:
$$\lim_{n\to \infty} \frac{\ln n}{n}$$
(you can write $x$ instead of $n$, as well). Let's try to solve it without LH rule.

Note that
$$0 \le \frac{\ln n}{n} = \frac{2\color{red}{\ln \sqrt n}}{\sqrt n \color{red}{\sqrt n}} \le \frac{2}{\sqrt n}\cdot \color{red}1 \\ 0 \le \frac{\ln n}{n} \le \frac{2}{\sqrt n}$$
Now, by Sandwich Theorem (or Squeeze Theorem), we conclude that $$\lim_{n\to \infty} \frac{\ln n}{n} = 0$$

Title: Can you use only L'Hôpital's rule in indeterminate forms

Body:
On evaluating $$f(x)=\begin{cases} \frac{x\ln(\cos x)}{\ln(1+x^2)} & x\neq0,\\0& x=0\end{cases}$$
all the solutions I've seen used the
L'Hôpital's rule to break it down; that way, you find out that the function is continuous at x=0, becoming 0 there.

I used $\lim_{x^2 \to 0} \frac{\ln(1+x^2)}{x^2}=0$ and logarithmic properties to simplify it to $\lim_{x
\to 0}\ln(\cos x)^{ \frac{1}{x}}$;

Defining h as a small positive value, as x approaches 0 from the right, it becomes h. As cosx approaches 0 from the right, it becomes 1-h. Thus the right-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{1}{h}$. This becomes $ln0=-\infty$

Similarly, as x approaches 0 from the left, it becomes -h. As cosx approaches 0 from the left, it becomes 1-h, the same as before. Thus the left-hand limit becomes $\lim_{h \to 0^+}ln(1-h)^\frac{-1}{h}$. This is $\lim_{h \to 0^+}ln\frac{1}{(1-h)^\frac{1}{h}}$, which becomes $ln\infty$, which is $\infty$.

As the limit doesn't seem to exist as per my evaluation, am I supposed to use only the L'Hôpital's Rule in indeterminate forms?


问:
Sequences, whose difference converge to 0
We still know that: if a sequence of real numbers converges, it is a bounded sequence and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.

I need an example, which is a bounded sequence of real numbers $\left(a_n \right)_{n\in\mathbb N^*}$ that do not converge and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.

答: Take, for instance, the sequence$$0,1,\frac12,0,\frac13,\frac23,1,\frac34,\frac24,\frac14,0,\frac15,\ldots\tag1$$For any $N\in\Bbb N$, you have $|a_{n+1}-a_n|<\frac1N$ of $n$ is large enough, which means that $\lim_{n\to\infty}(a_{n+1}-a_n)=0$. However, the sequence doesn't converge. Actually, every element of $[0,1]$ is the limit of some subsequence of $(1)$.


问:
The ordering of the field $\mathbb{C}$ such that $\mathbb{R}$ has the same order
The field consisting of the set $\mathbb{C}$ and the usual +,$\times$ cannot be ordered. But we can order the set $\mathbb{C}$ by mapping it to the set $\mathbb{R}$ and defining different addition and multiplication operations satisfying the field axioms. My question is can we make an ordered field on the set $\mathbb{C}$ with some addition and multiplication operations such that the order of $\mathbb{R}$ is kept the same as the usual order? If such operations don't exist prove that they don't exist.

答:
The ordering on $\Bbb R$ can be extended to an ordering on $\Bbb R(x)$ by postulating that $x>\alpha$ for all $\alpha\in\Bbb R$ and that for any two non-zero polynomials $p,q$ the rational function $\frac{p(x)}{q(x)}$ has the same sign as the ratio of their leading coefficients. Then you can consider a bijection $\Phi:\Bbb C\setminus\Bbb R\to \Bbb R(x)\setminus\Bbb R$ and the map \begin{align}&\Psi:\Bbb C\to\Bbb R(x)\\&\Psi(u)=\begin{cases}u&\text{if }u\in\Bbb R\\ \Phi(u)&\text{if }u\in\Bbb C\setminus\Bbb R\end{cases}\end{align}

The operations \begin{align}a+_\Psi b&=\Psi^{-1}(\Psi(a)+\Psi(b))\\ a\times_\Psi b&=\Psi^{-1}(\Psi(a)\Psi(b))\\ a\le_\Psi b&\Leftrightarrow \Psi(a)\le\Psi(b)\end{align}

accomplish what you want and $(\Bbb R,+_\Psi,\times_\Psi,\le_{\Psi})$ is an ordered subfield which is isomorphic to the usual $\Bbb R$ via the tautological map.


答:
Hint:

We actually don't need The Intuition behind l'Hopitals Rule Or Series Expansion as we can write

$$2f(x)=-x\cdot\dfrac{\ln(1-\sin^2x)}{-\sin^2x}\cdot\dfrac{x^2}{\ln(1+x^2)}\cdot\left(\dfrac{\sin x}x\right)^2$$


问:
Converting $3$-variable truth table to 3CNF
I have a truth table for length $3$ binary strings, say $110$ and $001$ map to $1$ and everything else to $0$. Is there an an algorithm to represent this table as a 3CNF which is satisfied only by $110$ and $001$?

答:
Yes, rewrite rules can be used:
$$(A \land B \land  C) \rightarrow (A \land B) \land C$$
$$ A \lor (B \land C) \rightarrow (A \lor B) \land (A \lor C)$$
$$(A \land B) \lor C \rightarrow C \lor (A \land B)$$

The input is
$$(X \land Y \land \neg Z) \lor (\neg X \land \neg Y \land Z)$$ where $X$, $Y$ and $Z$ are Boolean variables.


问:
Triple integral with absolute values in the domain
I want to calculate the following integral:

$$ I = \int_T x z dx dy dz $$
With:
$$ T = \left\{ (x,y,z)\ {\Large\mid}\ \vert x - z \vert + \vert x - 2y \vert \leq \vert y \vert \leq 1 \right\} $$
I am struggling on picturing the set T, so the first I did was defining the following transformation:
$$ \phi(x,y,z) = (x-z,x-2y,y) = (u,v,w) $$
In this way, I have:
$$ A = \phi(T) = \{ (u,v,w) \vert \vert u \vert + \vert v \vert \leq \vert w \vert \leq 1\}$$
and then I need to calculate:
$$ I = \int_{A} (v+2w)*(v+2w-u)*\frac{1}{d_J}dudvdw $$
where $d_J$ is the determinant of the jacobian of the transformation $\phi$.
Still at this point I struggle to see a fast way to solve the integral after the transformation.
Does someone have a tip?

答:
$T = \{ (x,y,z) \vert \vert x - z \vert + \vert x - 2y \vert \leq \vert y \vert \leq 1 \}$

Here is how I am visualizing the region -

It is defined by set of $6$ planes,

i) $y = 1$

ii) $y = -1$

For $x \geq z, x \geq 2y, \,$ or $x \leq z, x \leq 2y, \,$ we have bounds as

iii) $2x-3y-z = 0$

iv) $2x-y-z = 0$

For $x \leq z, x \geq 2y, \,$ or $x \geq z, x \leq 2y, \,$ we have bounds as

v) $3y - z = 0$

vi) $y - z = 0$

Note the last $4$ planes pass through the origin at different angles and they are bound along y-axis by planes $y = 1, y = -1$.

So the region is split into two pyramids with vertex at $(0, 0, 0)$ and its base as a parallelogram (one pyramid above $XZ$ plane in first octant and one below it in eighth octant).

The parallelogram base is formed by intersection of $(i), (iii), (v); (i), (iii), (vi); (i), (iv), (v)$ and $(i), (iv), (vi)$ (length along $x$ axis is $1$, it goes from $x = 1$ to $2$ for $z = 1$ and from $x = 2$ to $3$ for $z = 3$.

Similar construct with $(ii)$ below $XZ$ plane.


问:
Let ($a_n$) be an Integer sequence convergent to $L$ , Prove that from som place $a_n$ is constant
I got this question , I tried to answer it in a few ways but got stuck in every one of them.

The question :
Let $a_n$ be an integer sequence convergent to $L$.
Prove that for some $n$ , the sequence $a_n$ is constant.

My try:

I tried to assume that the sequence won't be constant for every $n$.

we will take the closest number integer to $L$.

decrease the $ε$ to recieve a smaller area , take the number $a_n+1$ and prove he is not integer.

and by this to prove that the sequence is constant for some $n$.

Thank you!

答:
Being convergent, the sequence will also be Cauchy.
Therefore, there exists $n_0$ such that $|a_n-a_m|<1$ for $m$, $n> n_0$. This implies ( the differences are integral) $a_m= a_n$ for $m, n> n_0$.

$\bf{Added:}$

If you want to avoid Cauchy: let $n_1$ such that $|a_m-L|< \frac{1}{2}$ for $m>n_1$. We get for $m, n> n_1$
$$|a_m- a_n| < |a_m - L| + |L-a_n| < \frac{1}{2}+ \frac{1}{2} = 1$$
Now proceed as before.


答:
You can do it easier, bu I'm going to do in two steps.

STEP 1: $L\in\Bbb z$.

Indeed, If it it not were the case, let $\varepsilon=\frac{1}{2} {\rm dist}(L,\Bbb Z)>0$, so since $a_n\to L$, there is some $N\in\Bbb N$ such that $|a_n-L|<\varepsilon$ for $n\ge N$. So, since $a_n\in\Bbb Z$ $${\rm dist}(L,\Bbb Z)\le {\rm dist}(a_n,L)=|a_n-L|<\frac{1}{2} {\rm dist}(L,\Bbb Z)$$ but this is a contradiction. So $L\in \Bbb Z$.

STEP 2: $a_n=L$ for $n$ large enough.

Since $L\in\Bbb Z$, take $\varepsilon=\frac{1}{3}$, so $a_n\to L$ wold imply that there is some $N\in\Bbb N$ such that $|a_n-L|<\frac{1}{3}$ for $n\ge N$, that is, $L-1/3<a_n<L+1/3$. But since $a_n\in\Bbb Z$ and the ONLY integer in $(L-1/3,L+1/3)$ is $L$, one concludes that $a_n=L$ for $n\ge N$.

Title: Let ($a_n$) be an Integer sequence convergent to $L$ , Prove that from som place $a_n$ is constant

Body:
I got this question , I tried to answer it in a few ways but got stuck in every one of them.

The question :
Let $a_n$ be an integer sequence convergent to $L$.
Prove that for some $n$ , the sequence $a_n$ is constant.

My try:

I tried to assume that the sequence won't be constant for every $n$.

we will take the closest number integer to $L$.

decrease the $ε$ to recieve a smaller area , take the number $a_n+1$ and prove he is not integer.

and by this to prove that the sequence is constant for some $n$.

Thank you!


问:
What is the expected value of distinct numbers in an array of integer numbers of size n?
We have an array of size $n$ and the elements of the array can be from $1$ to $N$, in a way that the number $1 \leq i \leq N$ has probability $P_i$ to be chosen for any place in array. Define $X$ a random variable for each array that shows how many distinct numbers exist in the array.

What is the mean(expected value) of $X$?

To demonstrate an example, if $N = 5$ and $n = 7,$ an array can be $[3, 5, 5, 1, 3, 1, 5],$ thus $X = 3$ because the set of distinct numbers is $\{1, 3, 5\}$

答:
Hints:

given a particular $i$, what is the probability it does not get selected in the first position?

given a particular $i$, what is the probability it does not get selected in any position?

what is the expected number of $i$s not selected in any position?

what is the expected number of $i$s selected in at least one position?


问:
The matrix $B = \left(\begin{smallmatrix}A& 0\\A & A\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ is diagonalizable.
Let $A\in\mathcal {M}_{n}(\mathbb {C}), $ I am trying to prove that:

The matrix $B = \left(\begin{smallmatrix}A& 0\\A & A\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ is diagonalizable.

My effrot :

$(\Rightarrow) $ we suppose that $B $ is diagonalizable then $m_B$ is a product of distinct linear polynomials.  Let $m_A(x)$, $m_B(x)$,  be the minimal polynomials of $A$, and $B$, respectively.

We have
$$
0 = m_B(B) = \pmatrix{m_B(A) & 0\\(*)&m_B(A)}
$$
so, $m_B(A) = 0$.  It follows that $m_A$ divides $m_B$, thus $m_A $  is a product of distinct linear polynomials. Then $A $ is diagonalisable.

For $(\Leftarrow )$ I have no idea

An idea please.

答:
This is not true: $\pmatrix{ 1 & 0 \\ 1& 1}$.

If $A$ is diagonalizable, then $B$ is similar to a matrix of the form $\pmatrix{D & 0 \\ D& D}$ where $D$ is diagonal.
By permuting, rows and columns, we get a block-diagonal matrix  with diagonal blocks of type $\pmatrix{d_i& 0 \\ d_i & d_i}$. These blocks are diagonalizable if and only if $d_i=0$. Then $B$ is diagonalizable if and only if $D=0$ which is equivalent to $A=0$.


问:
Derivative of a integral with limits as functions of other variable
I'm being asked to calculate, explicitly the following derivative:

$$
\frac{d}{d x}\left[\int_{e^{-x}}^{e^{x}} \sqrt{1+(\ln t)^{2}} d t\right]
$$

How can I do that?

I thought about using the fundamental theorem of calculus and chain rule but I'm not sure about this approach. I think that the integral itself can not be done analyticalally, so I can not integrate it first to do the derivative latter.

答:
Let $f(x)$ be a continuous function (where it is needed).

Then $F'(x)=f(x)$, where $F(x)=\int_a^xf(t)dt$ (for some appropiate $a$). This is FTC.

Now, $$G(x)=\int_{a(x)}^{b(x)}f(t)dt=\int_a^{b(x)} f(t)dt- \int_a^{a(x)}f(t)dt =F(b(x))-F(a(x))$$ (or simply, use Barrow).
So, applying Chain rule, and taking into account that $F'=f$:
$$G'(x)=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x).$$
Now, apply THAT formula to your situation.


Title: Derivative of a integral with limits as functions of other variable

Body:
I'm being asked to calculate, explicitly the following derivative:

$$
\frac{d}{d x}\left[\int_{e^{-x}}^{e^{x}} \sqrt{1+(\ln t)^{2}} d t\right]
$$

How can I do that?

I thought about using the fundamental theorem of calculus and chain rule but I'm not sure about this approach. I think that the integral itself can not be done analyticalally, so I can not integrate it first to do the derivative latter.

Body:
$$\frac{d}{dx}\int_{e^{-x}}^{e^x} \sqrt{1+(\ln t)^2} dt= e^x \sqrt{1+x^2}+e^{-x} \sqrt{1+x^2}=\sqrt{1+x^2}(e^x+e^{-x})=2\cosh x \sqrt{1+x^2}$$

Note that:
$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt= a'(x) f(a(x))-b'(x) f(b(x))$$


问:
Binomial Coefficients Combinatorics
For a positive integers n, prove that

$$\displaystyle\sum\limits_{v=0}^n \frac{(2n)!}{(v!)^2 ((n-v)!)^2} = \binom{2n}{n}^2.$$

If somebody could please help me with this question, I would greatly appreciated it.

答:
A combinatorial proof. The right-hand side corresponds to choosing two (possibly overlapping) sets of size $n$ from $2n$ elements.

Now we observe that:

$$\dfrac{2n!}{v!^2(n-v)!^2} = \binom{2n}{n} \binom n v \binom n{n-v}$$

If we divide the set of $2n$ elements in two fixed sets of $n$ elements, then picking $n$ from the original set amounts to picking $v$ from the first set, and $n-v$ from the second, for $v =0 \ldots n$.

Thus we find that

$$\sum_{v=0}^n \binom nv \binom n{n-v} = \binom{2n}n$$

which finally leads to the desired

$$\sum_{v=0}^n \frac{2n!}{v!^2(n-v)!^2} = \sum_{v=0}^n\binom{2n}{n} \binom n v \binom n{n-v} = \binom{2n}n \sum_{v=0}^n \binom nv\binom n{n-v} = \binom{2n}n^2$$

Title: Binomial Coefficients Combinatorics


问:
How to get ILT of $e^{-\alpha \sqrt{s}}/s^{3/2}$ by residue theorem?
I posted some examples before such as

$$\mathscr{L}^{-1}\left[\frac{e^{-\alpha \sqrt{s}}}{\sqrt{s}(s-A)}\right]=\frac{e^{-\alpha \sqrt{A}+At}}{\sqrt{A}} - \frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{1/2}(r+A)}dr$$ and they have been solved by friends here (How to get the inverse Laplace Transform?).

I have been trying $$\frac{e^{-\alpha \sqrt{s}}}{s^{3/2}}$$ by the same way but failed, because the result is $$-\frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{3/2}}dr$$ and it is divergent.

Moreover, $$\lim_{A \to 0}\left[\frac{e^{-\alpha \sqrt{A}+At}}{\sqrt{A}} - \frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{1/2}(r+A)}dr]\right] \nrightarrow -\frac{1}{\pi}\int_{0}^{\infty}\frac{\cos{(\alpha \sqrt{r})}e^{-rt}}{r^{3/2}}dr$$

I think the branch cut along the minus axis might be wrong.

Would you like to help me?

答:
Mathematica can fulfill this inverse Laplace transform:
$$\mathscr{L}^{-1}\left[\frac{e^{-\alpha \sqrt{s}}}{s^{3/2}}\right] = -\alpha + \frac{2\sqrt{t}}{\sqrt{\pi}}e^{-\frac{\alpha^2}{4t}} +\alpha erf\left(\frac{\alpha}{2\sqrt{t}}\right)$$.

But I really want to know this fulfillment by residue theorem.


答:
You can prove inductively that $$\forall m\in\Bbb{N},\quad B^m=\begin{pmatrix}A^m&0\\mA^m&A^m\end{pmatrix}$$ so that for every polynomial $P$ it holds that $$P(B)=\begin{pmatrix}P(A)&0\\AP'(A)&P(A)\end{pmatrix}.$$ Therefore, letting $\mu_A$ be $A$'s minimal polynomial and $P\in\Bbb{C}[X]$ arbitrary we get that $P$ annihiliates $B$ if and only if $\mu_A|P$ and $\mu_A|XP'$. This implies that if we let $$\mu_A=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i}$$ with the product running over the nonzero eigenvalues of $A$ (and $\nu\in\Bbb{N}$ possibly $=0$) we get that $\mu_B=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i+1}$.

The only way $\mu_B$ has simple roots is to have $\mu_A=X$, i.e. $A=0$. In other words, if $B$ is diagonalizable then $A$ is the zero matrix.

EDIT. Let $P=X^\nu\prod_{i=1}^r(X-\lambda_i)^{m_i+1}$. We prove that $P=\mu_B$ in two steps:

$P$ divides $\mu_B$: this we prove by showing that $0$ has multiplicity $\geq\nu$ as a root in $\mu_B$ and if $\lambda$ has multiplicity $m\geq 1$ as a root in $\mu_A$ then it has multiplicity $\geq m+1$ in $\mu_B$

$P(B)=0$ and so $\mu_B | P$

This implies $P=\mu_B$.

Suppose $0$ is an eigenvalue of $A$ i.e. $\nu\geq 1$. Then $\mu_A|\mu_B$ implies that $0$ has multiplicity $\geq \nu$ in $\mu_B$ and the extra $X$ in $X\mu_B'$ implies that $X^\nu$ automatically satisfies $X^\nu | X\mu_B'$ so $0$ will have the multiplicity $\geq \nu$ as a root of $\mu_B$.

Suppose $\lambda$ is a nonzero eigenvalue of $A$ (if there is one) with multiplicity $m\geq 1$ as a root of $\mu_A$. Then $(X-\lambda)^m|\mu_B$ and $(X-\lambda)^m|\mu_B'$ since $(X-\lambda)^m$ is coprime with $X$. Thus $\lambda$ is a root of multiplicity $\geq m+1$ of $\mu_B$.

Combining the two results above yields $P|\mu_B$. Conversely, $P(A)=0$ and $AP'(A)=0$ since $\mu_A$ divides $P$ by construction and $\mu_A$ divides $XP'$ because of multiplicities. Thus $\mu_B|P$ and so $\mu_B=P$.

Body:
Let $A\in\mathcal {M}_{n}(\mathbb {C}), $ I am trying to prove that:

The matrix $B = \left(\begin{smallmatrix}A& 0\\A & A\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ is diagonalizable.

My effrot :

$(\Rightarrow) $ we suppose that $B $ is diagonalizable then $m_B$ is a product of distinct linear polynomials.  Let $m_A(x)$, $m_B(x)$,  be the minimal polynomials of $A$, and $B$, respectively.

We have
$$
0 = m_B(B) = \pmatrix{m_B(A) & 0\\(*)&m_B(A)}
$$
so, $m_B(A) = 0$.  It follows that $m_A$ divides $m_B$, thus $m_A $  is a product of distinct linear polynomials. Then $A $ is diagonalisable.

For $(\Leftarrow )$ I have no idea

An idea please.


问:
Proving A Propositional Logic Statement Using Only Equivalence Laws
I'm trying to figure out how to prove

$$ \neg (p \vee q) \vee (\neg p \wedge q) \quad \Leftrightarrow \quad \neg p \tag{*} $$

using purely equivalence laws in propositional logic, the equivalence laws being here: link.

What I've done so far is to use de Morgan's law on one of the brackets, then use the distributive law. However, I wasn't able to get $\neg p$ from doing so. Could someone explain how to show $\text{(*)}$ using purely equivalence laws?

Thanks in advance.

答:
We want to prove that
$$ ¬(p ∨ q) ∨ (¬p ∧ q) ⇔ ¬p$$

To do this, we'll have to prove two statements:

$$\big(¬(p ∨ q) ∨ (¬p ∧ q)\big) \to ¬p$$

$$¬p \to \big(¬(p ∨ q) ∨ (¬p ∧ q)\big)$$

The second statement is the converse of the first statement.

Part 1. Prove that
$$¬(p ∨ q) ∨ (¬p ∧ q) \to ¬p$$

If this statement is true, it means that if we assume that ¬(p ∨ q) ∨ (¬p ∧ q) is true, then we can deduce ¬p.

From the identity:
$$ (p \to r) \land (q \to r) = (p \lor q) \to r$$
We get:
$$ (¬(p ∨ q) \to ¬p) \land ((¬p ∧ q) \to ¬p) $$
We need to prove that $¬(p ∨ q) \to ¬p$ and that $(¬p ∧ q) \to ¬p $.

From the fact that ($p \land q) \to p$.
$$¬(p \lor q) = (¬p \land ¬q) \to ¬p$$.

Similarly
$$(¬p \land q) \to ¬p$$.

And we are done. We prove the converse statement by following our steps bottom to top.


问:
Transform derivatives from 2D Cartesian to axisymmetric cylindrical coordinates
Consider the 1st and 2nd derivatives (differential operations) of a function $z=f(x)$ with respect to horizontal coordinate $x$ in 2D Cartesian coordinates $(x,z)$,

$$\frac{df}{dx} \quad \text{and} \quad \frac{d^2f}{dx^2}.$$

If $f(x)$ is axisymmetric about the vertical $z$ axis and we want to transform the derivatives into cylindrical coordinates $(r,z)$, can they be written straightforward as follows:

$$\frac{1}{r}\frac{d(rf)}{dr} \quad \text{and} \quad \frac{1}{r}\frac{d}{dr}(r\frac{df}{dr}), $$
where $z=f(r)$ and $r$ is the radial coordinate.

Update:

For the 1st derivative, I believe the answer of @David K is correct. However, I don't understand the 2nd derivative. Actually, I got the transformation of the 2nd derivative by comparing the Laplace operators  in Cartesian coordinates ($z=f(x)$) and in axisymmetric cylindrical coordinates ($z=f(r)$):

In Cartesian coordinates:

$$\nabla^2 f=\frac{d^2f}{dx^2}$$

In cylinderical coordiantes:

$$\nabla^2 f=\frac{1}{r}\frac{d}{d r}(r\frac{d f}{d r})$$

答:
You have no angular component in your "cylindrical" coordinates,
so all you have actually is Cartesian coordinates in which one coordinate is named $r$ instead of $x.$
The transformation of the derivative is indeed extremely straightforward:

\begin{align}
\frac{\mathrm df}{\mathrm dx} &\to \frac{\mathrm df}{\mathrm dr}, \\
\frac{\mathrm d^2f}{\mathrm dx^2} &\to \frac{\mathrm d^2f}{\mathrm dr^2}.
\end{align}

Why would it be anything else?

You could extend your function $f$ from a function on the $x$ axis to a function $\bar f$ over a plane $\pi_0$ through the origin, perpendicular to the $z$ axis, by setting up cylindrical coordinates around the $z$ axis, thereby assigning polar coordinates $r,\theta$ to the plane $\pi_0$, and defining the function

$$ \bar f(r, \theta) = f(r). $$

You can then take partial derivatives with respect to $r$ and $\theta$:

\begin{align}
\frac{\partial \bar f}{\partial r} &= \frac{\mathrm df}{\mathrm dr}, \\
\frac{\partial \bar f}{\partial \theta} &= 0. \\
\end{align}

The gradient of $\bar f$ is then

$$ \nabla \bar f
 = \hat{\mathbf r} \frac{\partial \bar f}{\partial r}
 + \hat{\mathbf \theta} \frac 1r \frac{\partial \bar f}{\partial \theta}.
$$

(See How to obtain the gradient in polar coordinates and its answers.)
Note that there is a factor of $\frac1r$ outside one of the partial derivatives, but it's on the partial derivative with respect to $\theta,$ not $r,$
and there is no factor $r$ inside any partial derivative.


问:
Show $\int f\phi = \int g \phi$ for all test functions $\phi$ implies $f=g$ for continuous $f$ and $g$

Suppose that $f$ and $g$ are continuous in $\mathbb{R}^n$. Then $$ \int_{\mathbb{R}^n} f(x)\phi(x)dx=\int_{\mathbb{R}^n} g(x)\phi(x)dx, \forall\phi\in \mathcal{D}$$ implies that $f(x)=g(x),\forall x\in\mathbb{R}^n$.

All kinds of variants of this question have been asked on this site before, f.ex. If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$. and If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e. . But what about simply continuous functions $f$ and $g$? Every continuous function is locally integrable, so by the proof provided in the second link (generalized to $\mathbb{R}^n$), we would have that $f-g=0$ a.e. in $L^1$. How do I actually show that $f=g$ everywhere in $\mathbb{R}^n$?

Is there another way to prove this, without using local integrability?

Thanks.

答:
Since $f$ and $g$ are continuous, $f-g$ is also continuous. By linearity, this means that you should verify whether $\int_{\mathbb{R}^n} f(x)\phi(x)dx=0$ for all continuous compactly supported functions implies $f\equiv0$.

Assume $f(x_0)> 0$, then by continuity there exists a ball $B_\delta(x_0)$ such that $f(x)>\frac{1}{2}f(x_0)$ for all $x$  in said ball. There is also a ball $B_{\delta'}(x_0)\supsetneq B_{\delta}(x_0)$ for which $f(x)>0$. Take a continuous function $\phi$ such that $\phi_{B_\delta(x_0)}=1$ and $\phi_{\big( B_{\delta'}(x_0) \big)^c}=0$. Check the integral $\int_{\mathbb{R}^n}f(x)\phi(x)dx$.


问:
Integral of test function is a test function? (Application in tensor theory of distributions)
Let $f:\mathbb{R}^m\to\mathbb{C}$ and $g:\mathbb{R}^n\to\mathbb{C}$, then we can define the tensor product $f\otimes g$ of $f$ and $g$ as the function $\mathbb{R}^{m+n}\to\mathbb{C}$ for which $f\otimes g(x,y)=f(x)g(y)$.

If $f$ and $g$ are locally integrable on resp. $\mathbb{R}^m$ and $\mathbb{R}^n$, then so is $f\otimes g$ on $\mathbb{R}^{m+n}$. We can therefore consider the regular distribution $T_f$,$T_g$ and $T_{f\otimes g}$ associated with each of these functions. The latter one is defined as $$ \langle T_{f\otimes g},\phi(x,y)\rangle = \int_{\mathbb{R}^{m+n}} f\otimes g(x,y)\phi(x,y)dxdy = \int_{\mathbb{R}^m} f(x) \left(\int_{\mathbb{R}^n} g(y)\phi(x,y)dy\right) dx, \text{ for } \phi\in \mathcal{D}(\mathbb{R}^{m+n}).$$ For this last integral to be well-defined, we would need that $\psi(x) := \int_{\mathbb{R}^n} g(y)\phi(x,y)dy$ is a test function over $\mathbb{R}^m$, i.e. $\psi \in \mathcal{D}(\mathbb{R}^m)$. I don't see how I could show this rigorously. Does continuity of $\psi$ follow directly? Can I be sure that the integral will be continuous? As for the support, is taking the intersection of $[\phi]$ with a compact $K_1\subset \mathbb{R}^m$ for which $[\phi]\subset K_1\times K_2$ (and $K_2\subset \mathbb{R}^n$) enough?

Thanks in advance.

答:
First we should note that you really don't need this to show that $f\otimes g$ defines a distribution. You can just note that $f\otimes g$ is locally integrable, so $\langle T_{f\otimes g},\phi(x,y)\rangle = \int_{\mathbb{R}^{m+n}} f\otimes g(x,y)\phi(x,y)dxdy$ defines a distribution with no  problem.

But in fact yes, $\psi$ is a test function. You can use Leibniz' rule for differentiating under the integral sign to show $\psi$ is smooth. Choose $R>0$ so $\phi(x,y)=0$ for all $y$  if $|x|>R$. Now it's clear that $\psi(x)=0$ for $|x|>R$.


问:
Prove the following holds for all Boolean algebras?
For all x,y: (x ∨ y) ∧ x = x

This is my attempt, I am just getting back to the question:

(x ∨ y) ∧ x

= x ∧ (x ∨ y)

= (x ∧ x) ∨ (x ∧ y)

= x ∨ (x ∧ y)

= (x ∨ x) ∧ (x ∨ y)

= x ∧ (x ∨ y)

So where am I going wrong here?

答:
You are not "going wrong"; $x \land (x \lor y)$ follows. Its truth table has the same result as $x$.From line 4 above, we have
$$\neg(\neg x) \lor (x \land y)$$

which is $\neg x \implies (x \land y)$.
Applying the absorption rule as in the comment gives
$$\neg x \implies (\neg x \land x \land y) $$

This is $x \lor F$ or $x$, as required.


问:
Proper implies "universally closed"
Let $f:X\to Y$ and $g:Z\to Y$ be continuous maps of locally compact Hausdorff second-countable topological spaces.  Assume that    $f$ is proper (the preimage of compact set is  compact).  Is it necessarily true that $X\times_Y Z\to Z$ is a closed map (here $X\times_Y Z$ is the fiber product i.e. the pullback of f and g in TOP)? I know it is a standard result that $f$ is "universally closed" in this case  (i.e. $X\times_{pt} Z\to Y\times_{pt} Z$ is always closed),  but here I am working with a slightly different notion of "universally closed" and I'm not sure if the two are equivalent.

答:
You have closed immersions $i: X \times_Y Z \rightarrow X\times_{pt} Z$, $j: Z \rightarrow Z \times_{pt} Y$, and you know that $f’: X \times_{pt} Z \rightarrow Y \times_{pt} Z$ is closed.

Let $b: X \times_Y Z \rightarrow Z$, then $j \circ b=f’ \circ i$.

Let $F \subset X \times_Y Z$ be a closed subset. Then $f’(i(F))$ is a closed subset of $Z \times Y$, and it is $j(b(F))$. As $j$ is a closed immersion, $b(F)$ is closed in $Z$. QED.

Title: Proper implies "universally closed"


问:
If in the pullback p is shrinkable, the p' is shrinkable.
A map $p:E \rightarrow B$ is said shrinkable if $p$ is homotopy equivalent to $id_B$ in $TOP_B$. Show that if $p$ is shrinkable in the
pullback

then $p'$ is shrinkable too.

答:
Choose a section $s:B\rightarrow E$ of $p$ and a fibre homotopy $G:sp\simeq_B id_E$. Then the pair of the identity $id_{B'}$ and the composite $sf:B'\rightarrow E$ induces a map $s':B'\rightarrow E'$ with $p's'=id_{B'}$. Notice that if $\widetilde p:E\times I\rightarrow B$ is the map $\widetilde p(e,t)=p(e)$, then $E'\times I$ is the pullback of the cotriad $B'\xrightarrow{f}B\xleftarrow{\widetilde p}E\times I$. The map $\widetilde p':E'\times I\rightarrow B'$, $(e',t)\mapsto p'(e')$ together with the composite $E'\times I\xrightarrow{f'\times 1} E\times I\xrightarrow{G}E$ now induce a homotopy $G':E'\times I\rightarrow E'$. It is a fibre homotopy over $B'$ by defintion, and including in at the end points we see that it verifies $s'p'\simeq id_{E'}$.

If you would prefer to work with point sets, then you may find it easier to identify $E'=\{(b',e)\mid f(b')=p(e)\}$ before running the details.


答: The complement of a measure zero set in $\mathbb{R}$ is of course dense. So if $f-g$ is continuous and equal to $0$ on a dense set, it must be zero everywhere.

Title: Show $\int f\phi = \int g \phi$ for all test functions $\phi$ implies $f=g$ for continuous $f$ and $g$

Body:
Suppose that $f$ and $g$ are continuous in $\mathbb{R}^n$. Then $$ \int_{\mathbb{R}^n} f(x)\phi(x)dx=\int_{\mathbb{R}^n} g(x)\phi(x)dx, \forall\phi\in \mathcal{D}$$ implies that $f(x)=g(x),\forall x\in\mathbb{R}^n$.

All kinds of variants of this question have been asked on this site before, f.ex. If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$. and If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e. . But what about simply continuous functions $f$ and $g$? Every continuous function is locally integrable, so by the proof provided in the second link (generalized to $\mathbb{R}^n$), we would have that $f-g=0$ a.e. in $L^1$. How do I actually show that $f=g$ everywhere in $\mathbb{R}^n$?

Is there another way to prove this, without using local integrability?

Thanks.


问:
Alternative ways to say "if and only if"?
There are some scenarios about which I would like to get some confirmation:

when defining a concept A,

We call A, if ... [definition of
  concept A]

Does "if" here mean equivalence
instead of just sufficiency? Is it
incorrect to replace "if" with "if
and only if"?

For precisely what condition is
   B satisfied?

Does "precisely" mean asking for
necessary and sufficient condition
for B?

Some other ways to say "if and only
if" / "necessary and sufficient"?

Some references that summarize some standard terminologies in Mathematics  such as this one?

Thanks and regards!

答:
The demand for economy of expression means that our shared understanding must be exploited to the fullest, and this sometimes involves importing some presumptive structures from ordinary language into the writing of mathematics materials. An important instance of this is the introductory “if”, which does not mean “if”, but rather “given that” (which is only one third as many syllables, and much less than one third of text). We see this in textbook exercises, and on tests, all the time, e.g., “If x = 3, evaluate 5x + 2.” “So,” muses the mischievous student, “if x is NOT 3, then I don’t have to bother with doing the evaluation, right?” But, in reply to the mischievous student, this really means, “Given that x = 3, evaluate 5x + 2.”

Regarding definitions, in support of conciseness, there is also operative the default of presuming maximality. That is, the stated condition is presumed to be maximal, and therefore necessary. In ordinary language, uttering non-maximal statements, intentionally or unintentionally, is highly misleading, and stomped on when detected, as in this classic exchange.:

A. “90% of Science Fiction is trash.”

B. “Of course, 90% of ANYTHING is trash.”

As far as theorems are concerned, I would prefer using something (short) that DOES distinguish aurally between “if” and “if, and only if,”. I would like to see the adoption of “fif” for this purpose, e.g., “A set M is compact fif it is closed and bounded.”

Regards,
Mike Jones


问:
conic sections, ellipse
A particle is travelling clockwise on the elliptical orbit given by
$$\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$$
The particle
leaves the orbit at the point $(-8, 3)$ and travels in a straight line tangent to the ellipse.
At which point will the particle cross the $y$-axis?</p>

答:
HINT:

From Article $262$ of this, the equation of the tangent at $P(h,k)$ of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{   is      }\frac{x\cdot h}{a^2}+\frac{y\cdot k}{b^2}=1$$

So, the equation of the tangent here will be $$\frac{x\cdot (-8)}{100}+\frac{y\cdot 3}{25}=1$$

Now to cross the $y$ axis, $x=0$

Title: conic sections, ellipse


问:
Helping me understand the formal definition of ODE's
Currently I am writing a project at Uni, where we define Ordinary differential equations (ODE's). However, I seem to have some trouble comprehending the definition. It's taken from Lecture notes on Ordinary Differential Equations. Annual Foundation School, IIT Kanpur, Dec.3-28, 2007. By S. Sivaji Ganesh.

It basically states a hyphothesis first that says:

Let $\Omega \subseteq \mathbb{R}^{n+1}$ be a be a domain and $\mathbb{I} \subseteq \mathbb{R}$ be an interval. Let $F : \mathbb{I} × \Omega → \mathbb{R}$ be a function
defined by $(x, z, z_1, . . . z_n) \mapsto F(x, z, z_1, . . . z_n)$ such that $F$ is not a constant function in the variable $z_n$.

Then we have the definition:

Assume the above hypothesis. An ordinary differential equation of order
$n$ is defined by the relation
$$F(x,y,y',y'',...,y^{(n)}) = 0$$
where $y^{(n)}$ stands for $n$'th derivative of unknown function $x \mapsto y(x)$ with respect to the independent variable $x$.

Now I understand the difference between $\mapsto$ and $\to$. And I understand, that $\subseteq$ means subset and what that is.

What I don't understand is why $\Omega \subseteq \mathbb{R}^{n+1}$, since, as far as I know, $\mathbb{R}^{n+1}$ is a vector space of $n+1$ dimensions. Since I don't understand that, our cartesian product doesn't really make sense for me aswell.

Lastly, I don't understand why $F$ can't be a constant function in the variable $z_n$, my guess would be, it's because we can't solve / there is no solution to:
$$F(x,y,y',y'',...,y^{(n)}) = k$$
for any given constant $k$.

Sorry about multiple questions in one question, I hope that someone can help me along to understanding it.

答:
The reason we set to RHS to zero is because we can always set it to be such. Consider for example the ODE
$$F(x,y,y')=x^3-yy''+y'=6$$
Of course however we can just move the $6$ over:
$$G(x,y,y')=x^3-yy''+y'-6=0$$
Since we can always do this it makes sense to only consider examples where $F=0$ for simplicity.

The reason $F$ can't be a constant function in $z_n$ (which is actually $y^{(n)}$) is because then it wouldn't be an $n$th order ODE, which contradicts our starting point. Consider the "third order" ODE
$$y''\sqrt{y'+y}+0\cdot y'''=0$$
Of course this is not actually third order.


问:
Proving $f^{(n)} (0)$ s unbounded where $f(x)= \tan^{-1} x$
The following question was asked in real analysis quiz and I am looking for an more elegant solution.

Question :Let $f(x)= \tan^{-1}x $ for all $x\in \mathbb{R}$. Prove that sequence $\{f^{(n)}(0)\}$ is unbounded.

There is a solution given by differentiating $n$ times the expression $(1+x^2) f(x)$ and then using the recurrence relation $(1+x^2) f^{(n+1)} (x)+ 2nx f^{(n)} (x) + n(n-1) f^{(n-1)} (x)=0$ and $f'(0)=1$. But is there a more elegant solution using any other concepts of other branches of pure mathematics or analysis (real, complex, functional).

I thought I should ask here as I don't wanted to use that method although that is clear to me!

Body:
$$f'(x)=\frac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$
Use $$D^n(x+a)^{-1}=(-1)^n n! (x+a)^{-n-1}$$
Then $$f^{n}(x)=(2i)^{-1} (-1)^{n-1} (n-1)! \left(\frac{1}{(x+i)^{n}}-\frac{1}{(x-i)^{n}}\right).$$
$$f^{n}(0)=\frac{(-1)^n-1}{2i^{n+1}} (n-1)!$$


问:
How to show that $a_{n+2}=\frac{a_{n+1} ^2 -1}{a_n}$ . is bounded?
Let $a_{n+2}=\frac{a_{n+1} ^2  -1}{a_n}$  be a sequence of real numbers where $a_n>0$ for all $n \in \mathbb{Z}_{+}$. It is given that $a_1=1 , a_2=b>0$. It is given that $1<b<2$ .Is it possible to show that $a_n$ is bounded from this given information?

答:
It seems that for all integers $n\geq 1$, $a_n$ will be a rational fraction in
$b$ of degree $n-1$, with leading coefficient $1$ (by that I mean that for all fixed $n\geq 1$, you can write that $a_n\sim b^{n-1}$ as $b$ goes to infinity.

So my intuition is that if $b>1$, then $a_n$ will behave like a diverging geometric sequence.


问:
Uniform convergence on compact sets and limits at infinity
Suppose that we have a sequence of continuous functions $(f_n)_{n \in \mathbb{N}}$,

$f_n: \mathbb{R} \rightarrow \mathbb{R}$,

which converges uniformly on compact sets to a function $f$, and that we have moreover that all that limits $a_n:=\lim_{x \rightarrow \infty} f_n(x)$, as well as $a:= \lim_{x \rightarrow \infty} f$ exist. Can we necessarily conclude that $\lim_{n \rightarrow \infty} a_n=a$?

答:
No. Consider a sequence of function $f_n$ defined as follows: $f_n$ is $1$ on the interval $[-n,n]$, it is zero outside $[-n-1,n+1]$ and continuous on $[-n-1,-n]\cup [n,n+1]$.

For example $$f_n(x)=\left\{\begin{array}{ll} 1 & \text{ for } |x| < n\\
0 & \text{ for } |x|>n+1\\
-x+n+1 &\text{ for } n\leq x\leq n+1\\
x+n+1 &\text{ for } -n-1<x<-n  \end{array}\right.$$

The functions $f_n$ converge to the constant function $1$ on compact sets. Buth with your terminology we have $a_n=0$ forall $n$ and $a=1$.


问:
Evaluate $6\log_8(4)$ without using a calculator
I am to evaluate $6\log_8(4)$ without a calculator. The answer is provided as 4 but I cannot see how to arrive at 4.

Ignoring the 6 at the beginning of the expression, $log_8(4)$ can be written as $8^x=4$.

Without the use of a calculator I cannot simplify that part any further. I know that the root or 3rd root of 8 is not 4.

I'm not sure if the leading 6 helps or if I am to multiply the end result by 6 or some part of the expression by 6 while doing the working out part?

Completely stuck here. How can I arrive at 4? Granular, baby steps appreciated.

[EDIT]
I'd like to add that this is the textbook chapter I am working of. It's the absolute beginning of learning about logarithms. Looking at the comments and answer so far, there's reference to dividing logs which has not been covered in this book so far. Given the content of this chapter, I wonder if it's expected of me that I know how to solve this question?

答:
Using $\log_ac=\dfrac{\log_bc}{\log_ba}$ when all the logarithms remain defined,

$$\log_84=\dfrac{\log_24}{\log_28}=\dfrac{\log_2(2^2)}{\log_2(2^3)}=?$$


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

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

Thank you very much.

答:
I might have misunderstood anything, but if by "action" in the title you mean the following:

\begin{alignat*}{2}
  \mathcal{A}:(\operatorname{Aut}(A)\times\operatorname{Aut}(G))\times T & \longrightarrow & T \\
  ((\sigma,\rho),\alpha) & \longmapsto & (\sigma,\rho)\cdot\alpha: G & \longrightarrow \operatorname{Aut}(A) \\
  && g &\longmapsto \sigma^{-1}\alpha(\rho(g))\sigma \\
\tag 0
\end{alignat*}

then seemingly it fails to fulfil the "Action#2" property for actions (see $(1)$ and $(2)$ hereunder). In that case, group actions wouldn't assist the matter.

Good definition: for every $\sigma\in \operatorname{Aut}(A)$, $\rho \in\operatorname{Aut}(G)$ and $g,h\in G$, indeed $\sigma^{-1}\alpha(\rho(g))\sigma\in \operatorname{Aut}(A)$ and:
\begin{alignat}{1}
((\sigma,\rho)\cdot\alpha)(gh) &= \sigma^{-1}\alpha(\rho(gh))\sigma \\
&= \sigma^{-1}\alpha(\rho(g)\rho(h))\sigma \\
&= \sigma^{-1}\alpha(\rho(g))\alpha(\rho(h))\sigma \\
&= \sigma^{-1}\alpha(\rho(g))\sigma\sigma^{-1}\alpha(\rho(h))\sigma \\
&= (\sigma^{-1}\alpha(\rho(g))\sigma)(\sigma^{-1}\alpha(\rho(h))\sigma) \\
&= ((\sigma,\rho)\cdot\alpha)(g)((\sigma,\rho)\cdot\alpha)(h) \\
\end{alignat}
whence indeed $(\sigma,\rho)\cdot\alpha\in T$

Action#1: $\forall \alpha\in T,\forall g\in G$,
\begin{alignat}{1}
((\iota_A,\iota_G)\cdot\alpha)(g) &= \iota_A^{-1}\alpha(\iota_G(g))\iota_A \\
&= \alpha(g) \\
\end{alignat}
whence $\forall\alpha\in T$:
$$(\iota_A,\iota_G)\cdot\alpha=\alpha$$

Action#2: $\forall \sigma,\sigma'\in\operatorname{Aut}(A),\forall \rho,\rho'\in\operatorname{Aut}(G), \forall \alpha\in T,\forall g\in G$,
\begin{alignat}{1}
(((\sigma,\rho)(\sigma',\rho'))\cdot\alpha)(g) &= ((\sigma\sigma',\rho\rho')\cdot\alpha)(g) \\
&= (\sigma\sigma')^{-1}\alpha((\rho\rho')(g))\sigma\sigma' \\
&= \sigma'^{-1}\sigma^{-1}\alpha(\rho(\rho'(g)))\sigma\sigma' \\
\tag 1
\end{alignat}
while
\begin{alignat}{1}
((\sigma,\rho)\cdot((\sigma',\rho')\cdot\alpha))(g) &= \sigma^{-1}(((\sigma',\rho')\cdot\alpha)(\rho(g))\sigma \\
&= \sigma^{-1}\sigma'^{-1}\alpha(\rho'(\rho(g))\sigma'\sigma \\
\tag 2
\end{alignat}
Therefore, $\mathcal{A}$ defined in $(0)$ is not an action.

However, the map $\mathcal{A}$ becomes indeed an action if you narrow the group which acts on $T$ to $Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))$, as in this case $(1)$ and $(2)$ match. In that case:

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

So, for instance, if you'd know that the action is transitive (so it had one orbit, only), then you could conclude that (Burnside's Lemma):

$$\sum_{(\sigma,\rho)\in Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))}|T^{(\sigma,\rho)}|=|Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))|$$

(I realized just now that I didn't take into accout the fact that $A$ and $G$ are finite abelian groups, by assumption, which should simplify this all...)


答: Or you can see that $\log(a/b)=\log a -\log b$. So $\log_8 4=\log_8{(8/2})=\log_88-\log_8 2=1-\frac 1 3=\frac 2 3$

Title: Evaluate $6\log_8(4)$ without using a calculator

Body:
I am to evaluate $6\log_8(4)$ without a calculator. The answer is provided as 4 but I cannot see how to arrive at 4.

Ignoring the 6 at the beginning of the expression, $log_8(4)$ can be written as $8^x=4$.

Without the use of a calculator I cannot simplify that part any further. I know that the root or 3rd root of 8 is not 4.

I'm not sure if the leading 6 helps or if I am to multiply the end result by 6 or some part of the expression by 6 while doing the working out part?

Completely stuck here. How can I arrive at 4? Granular, baby steps appreciated.

[EDIT]
I'd like to add that this is the textbook chapter I am working of. It's the absolute beginning of learning about logarithms. Looking at the comments and answer so far, there's reference to dividing logs which has not been covered in this book so far. Given the content of this chapter, I wonder if it's expected of me that I know how to solve this question?


Body: $6\log_8(4)=\log_8(4^6)=\log_8((2^2)^6)=\log_8(2^{12})=\log_8((2^3)^4)=\log_8(8^4)=4$


Body: $$6\log_84=6\log_{2^3}{2^2}=6\times\frac{2}{3}\times\log_22=4\times 1=4$$


答:
$\newcommand{\R}{\mathbb{R}}$
Since k.defaoite already gave an answer I will focus on your question regarding $\Omega \subseteq \R^{n+1}$. It is not entirely clear to me what you are having trouble with, so I'll just guess what might help you.

Indeed $\R^{n+1}$ is an $n+1$-dimensional vector space. So you can think of it as the plane (this would be $n=1$) and then $\Omega$ is a domain (usually this means an open connected subset). So it might be the disc with radius 1 (without the boundary).
Or it might be something more funky. This is basically where you want the function to be defined. We often want to consider functions that are not defined everywhere, and by restricting ourselves to some domain, we gain a lot of flexibility.

Then we 'add' the interval (note that the connected open subsets of $\R$ are exactly the open intervals) to 'make space' for the variable $x$.
Every object of $\mathbb{I}\times \Omega$ can be written as $(x,z_1,\ldots, z_n)$, BUT NOT VICE VERSA. If $(z_1,\ldots, z_n)$ is not in $\Omega$, then $(x,z_1,\ldots, z_n)$ will of course not be in $\mathbb{I}\times \Omega$.

In general, there will be no 'good' description of $\Omega$, but in every case you are likely to come across it will be something like a disc or so. I would suggest doing some exercises, then you might see how the different parts of the definition fit together.

Title: Helping me understand the formal definition of ODE's

Body:
Currently I am writing a project at Uni, where we define Ordinary differential equations (ODE's). However, I seem to have some trouble comprehending the definition. It's taken from Lecture notes on Ordinary Differential Equations. Annual Foundation School, IIT Kanpur, Dec.3-28, 2007. By S. Sivaji Ganesh.

It basically states a hyphothesis first that says:

Let $\Omega \subseteq \mathbb{R}^{n+1}$ be a be a domain and $\mathbb{I} \subseteq \mathbb{R}$ be an interval. Let $F : \mathbb{I} × \Omega → \mathbb{R}$ be a function
defined by $(x, z, z_1, . . . z_n) \mapsto F(x, z, z_1, . . . z_n)$ such that $F$ is not a constant function in the variable $z_n$.

Then we have the definition:

Assume the above hypothesis. An ordinary differential equation of order
$n$ is defined by the relation
$$F(x,y,y',y'',...,y^{(n)}) = 0$$
where $y^{(n)}$ stands for $n$'th derivative of unknown function $x \mapsto y(x)$ with respect to the independent variable $x$.

Now I understand the difference between $\mapsto$ and $\to$. And I understand, that $\subseteq$ means subset and what that is.

What I don't understand is why $\Omega \subseteq \mathbb{R}^{n+1}$, since, as far as I know, $\mathbb{R}^{n+1}$ is a vector space of $n+1$ dimensions. Since I don't understand that, our cartesian product doesn't really make sense for me aswell.

Lastly, I don't understand why $F$ can't be a constant function in the variable $z_n$, my guess would be, it's because we can't solve / there is no solution to:
$$F(x,y,y',y'',...,y^{(n)}) = k$$
for any given constant $k$.

Sorry about multiple questions in one question, I hope that someone can help me along to understanding it.


Body:
To deal with $8^x=4$, note $8=2^3$ and $4=2^2$.

So that $$ 8^x = 4 $$

$$ \Rightarrow (2^3)^x = 2^2 $$

$$ \Rightarrow 2^{3x} = 2^2 $$

$$ \Rightarrow 3x = 2 $$

$$ \Rightarrow x = \dfrac{2}{3} $$


问:
Integral of $\int^1_0 \frac{dx}{1+e^{2x}}$
I am trying to solve this integral and I need your suggestions.

I think about taking $1+e^{2x}$ and setting it as $t$, but I don't know how to continue now.

$$\int^1_0 \frac{dx}{1+e^{2x}}$$
Thanks!

答:
With the change of variable $u=e^{x}$, you get
$$
\int_{[0,1]}\frac{dx}{1+e^{2x}} = \int_{[1,e]}\frac{1}{u(1+u^2)}du
$$</p>


问:
Problem based on sylow theorem
Let G be a group of order 3179. (a) State the orders of the Sylow p-subgroups of G. (b) Use the Sylow Theorems to determine the possible answers to the question: "How many Sylow p-subgroups can G have?'</p>

答: Hint: $3179=11 \cdot 17^2$ and $11 \not\equiv 1$ mod $17$ and $17 \not\equiv 1$ mod $11$ and $17^2 \not\equiv 1$ mod $11$. What can you conclude for $\#Syl_{11}(G)$ and $\#Syl_{17}(G)$?

Title: Problem based on sylow theorem


问:
In how many ways can we distribute $m$ balls to $n$ baskets when $n > m$

Given $m$ balls and $n$ baskets, in how many ways can we distribute the $m$ balls to the $n$ baskets when given that $n > m$ (the number of baskets is greater than the number of balls)?

So I said, that it does not differ if $n>m$ the solution is still ${n+m-1 \choose n-1}$, is that correct? I merely considered it as the number of solutions $x_1 + x_2 + ... + x_n = m$ when $0 \leq x_i$. Is that correct?

Also, as an addition:

In addition to the question, what if there can be only a maximum of one ball in each basket?

So I said, it is still the same $x_1 + x_2 + ... + x_n = m$ but this time $0 \leq x_i \leq 1$ and it is still the same solution. Is that correct?

答:
since there are not any restriction of putting any number of balls in a basket.
so after putting every ball every basket will be available.

so for 1st ball there are n ways for 2nd ball n ways.....and for mth ball there are still n ways.
so answer will be: $$n\times n\times n\times n\times ........m \,times$$
$$n^m$$


问:
What is the number of partitions of a set of size $16$ into $4$ subsets of size $4$?
What is the number of partitions of a set of size $16$ into $4$ subsets of size  $4$ ?

What I have done:

I have tried ${16 \choose 4} \cdot \frac{{12 \choose 4}}{4!}$ however this gives an answer of $37537.5$ which has a decimal. Am I taking the right approach to this question?

Body: $$\frac{\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{{4!}}$$


问:
Prove that there is no rational root of $x^3+x+1=0$ by contradiction
I have found that this equation only has one root between $-1$ and $0$. I try to let this root=$p/q$ where $gcd(p,q)=1$ and get $p^3+pq^2+q^3=0$. I want to know how to prove it by the method of contradiction.

答:
You did a great job. Only a couple of steps left.

$0=p^3+pq^2+q^3=p(p^2+q^2)+q^3 \implies p|q^3$, but $gcd(p,q)=1$, which means that $p=\pm 1$. Similarly $0=p^3+pq^2+q^3=q(pq+q^2)+p^3 \implies q|p^3$, but $gcd(p,q)=1$, which means that $q=\pm 1$, therefore $x=\pm 1$. But $f(1)=3 \neq 0$ and $f(-1)=-1 \neq 0$, so there is no rational root.