问:
Technical lemma about state space of a $C^*$-algebra.
Consider the following proof from the book "$C^*$-algebras and finite-dimensional approximations":

Why does this proof work in the non-unital case? (see the last line). Maybe we have
$$\|a + \lambda \| = \sup_{\varphi \in \mathcal{S}} \{|\varphi(a, \lambda)|\}$$
where the states $\varphi: A  \to \mathbb{C}$ have been extended (uniquely) to states on the unitization?

答:
The states of $A$ extend uniquely (as states) to the unitization. The extension is $\tilde\varphi(a+\lambda)=\varphi(a)+\lambda$. It easy to check that such extension is a state.

And you have an extension $\psi$ of $\varphi$ that is a state, then $\psi(1)=1$ (a positive functional always satisfies $\psi(1)=\|\psi\|$). That gives you
$$
\psi(a+\lambda)=\psi(a)+\psi(\lambda)=\varphi(a)+\lambda.
$$

The only state in the unitization $\tilde A$ of $A$ that does not come from a state in $A$ is the state $\alpha(a+\lambda)=\lambda$. As this state is zero in all elements of $A$, it not in the weak-$*$ closed convex hull of the other states, so it doesn't affect the definition of $S$ nor the proof.

I don't see a way to go about the proof in an obvious way as Nate and Taka make it sound, but here is my take. Write $n=\|a\|$. Then, with $\tilde\varphi$ the unique extension of $\varphi$ to the unitization as a state,
\begin{align}
\|a+n\|&=\sup_{\|b\|\leq1}\|ab+nb\|
=\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\varphi(ab+nb)|\\[0.3cm]
&=\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\tilde\varphi((a+n)b)|\\[0.3cm]
&\leq\sup_{\|b\|\leq1}\sup_{\varphi\in S}\tilde\varphi(b^*b)^{1/2}\tilde\varphi((a+n)^2)^{1/2}\\[0.3cm]
&\leq\sup_{\varphi\in S}\tilde\varphi(\|a+n\|\,(a+n))^{1/2}\\[0.3cm]
&=\|a+n\|^{1/2}\,\sup_{\varphi\in S}\tilde\varphi(a+n)^{1/2}.
\end{align}
Cancelling and squaring we obtain
$$
\|a+n\|\leq\sup_{\varphi\in S}\tilde\varphi(a+n),
$$
which is the non-obvious inequality that is needed.


问:
When do $L^p$ and $L^q$ contain the same functions?
From Rudin:

Given that X is a finite measure space, when do $L^p$ and $L^q$ contain the same functions?

What I know is that $L^p$  contains $L^q$ if p<q.

I was thinking

How to make a function which is in $L^p$ but not in $L^q$, assuming it is possible to do so? And what fails when it is not possible.

I appreciate any help.

答:
The usual "measure level" characterization is that with $1 \leq p<q<\infty$:

$L^q \subset L^p$ if and only if the space has no sets of arbitrarily large finite measure, i.e. there exists $M>0$ such that for all measurable $A$, if $\mu(A)<\infty$ then $\mu(A) \leq M$. For most practical purposes this condition is the same as "the space has finite measure", but it is slightly more general. Extending this to $q=\infty$ requires you to actually assume the space has finite measure.

$L^p \subset L^q$ if and only if the space has no sets of arbitrarily small positive measure, i.e. there exists $\epsilon>0$ such that for every measurable $A$, if $\mu(A)>0$ then $\mu(A)>\epsilon$. For most practical purposes this is the same as "the space is discrete". This condition extends to $q=\infty$ with no issue.

Thus to have $L^p=L^q$ you must have both of these conditions holding.

Side remarks:

The intuition for the first condition is that if there are large measure sets, then there are heavy-tailed functions whose integrability can be improved by raising them to larger powers. You can explore the family $x^r \chi_{[1,\infty)}(x)$ (which are integrable for $r<-1$) to get a feel for this.

The intuition for the second condition is if there are small measure sets, then there are functions with singularities whose integrability can be improved by raising them to smaller powers. You can explore the family $x^r \chi_{(0,1)}(x)$ (which are integrable for $r>-1$) to get a feel for this.

A fun exercise on this topic: with $X$ being the Lebesgue measure space on $\mathbb{R}$, and given $p_0 \in [1,\infty)$, construct a function $f$ which is in $L^p$ for only $p=p_0$.

ContentLicense: CC BY-SA 4.0


问:
Graph neighbourhood matrix in which each pair of rows has at least one common column full of ones
I am trying to understand a particular method of moments estimation algorithm. If you are interested in that, see this for more detailed description.

Long story short, I am trying to understand what it means for the neighbourhood matrix of a directed graph to have the following property: no element-wise product of each pair of rows results in a zero row. To put it another way, each pair of rows in the neighbourhood matrix has at least one common column full of ones. So e.g. (not necessarily a neighbourhood matrix) $\begin{pmatrix} 1&0&1\\1&1&1\end{pmatrix}$ has this property while $\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ does not.

So specifically:

1.) How can you represent this kind or property in a more graph theoretical/linear algebra way? I can only talk about it in a "programmatic" way, i.e. for all pairs of rows it holds that their dot product is not equal to zero.

2.) What can you tell about a graph whose neighbourhood matrix satisfies this property?

答: If neighborhood matrix is another name for adjacency matrix, the property means that every pair of nodes has a common out-neighbor.  That is, for all $i,j\in N$ with $i<j$, there exists at least one $k\in N$ such that $(i,k) \in A$ and $(j,k) \in A$.


问:
How to find 'intersection' sequence of two sequences?
For example, suppose I am given the following:

$$ a_n = \frac{\pi}{8} + (-1)^n \frac{n \pi}{4}$$

$$ b_n = \frac{\pi}{4} + (-1)^n n \frac{\pi}{2}$$

Now, from these two sequences, how do I find the sequences which contains the intersection terms of both(i.e: a new sequence in which terms are common to both a_n and b_n?)

答:
Take a common denominator of 8 everywhere and remove the common factor of $\frac{\pi}{8}$.

The first sequence is $1 + (-1)^{n}2n$ and is always odd.

The second sequence is $2 + (-1)^{n}4n$ and is always even.

The two never overlap. Even numbers cannot be odd and vice versa.


问:
If $4$ distinct integers $a, b, c, d$ are randomly selected from $n$ distinct integers, what is the probability that $a \lt b \lt c \lt d$?

a. If $i$ distinct integers $r_1, r_2, \ldots,r_i$, i $\leq n$, are randomly selected (one after other) from n distinct integers, what is the probability that $r_1 < r_2 < \cdots < r_i$?

b. Does the probability you got in part a of the question change if < is replaced with $\leq$?

I know that a question similar to part 'a' of this  question can be solved using the concept of Decision tree that orders the elements of the list, an application of which is to analyse the complexity of comparison-based sorting algorithm.

I came across a solution to part 'a' of the question that says the required probability is $C(n, i)/P(n,i)$, which is nothing but equal to the solution of the Decision tree based approach, which is $1/i!$.

What I don't understand is this - how $C(n,i)$ is equal to favourable number of outcomes and $P(n,i)$ is equal to total number of outcomes?

Figure: A decision tree for sorting three distinct elements.

Image taken from Discreet Mathematics and its Applications by Kenneth Rosen

答:
Approach 1:

If you take letters $a, b, c, d$, there are $4!$ orders possible.

$a \lt b \lt c \lt d$ is one of them.

Hence the desired probability $\displaystyle  = \frac{1}{4!}$

Approach 2: Place ( $ \ $ a $ \ $ b $ \ $ c $ \ $ d $ \ $) first, and now start placing other $(n-4)$ letters in between. The first letter of $(n-4)$ has $5$ places available. Once you place the first letter, you now have $6$ places for the next letter and so on.

Total number of favorable arrangements $ = 5 \times 6 \times 7 ... \times n$. Divide this by $n!$ which is total number of unrestricted arrangements.

Approach 3 (same as approach 2): As the order is fixed, you cannot permute (a b c d) so the number of arrangements are $\frac{n!}{4!}$. Divide this by $n!$


问:
Identity involving complex sigma function
When trying around with the DivisorSigma function of Mathematica, I found this Identity:

$\#\{a\mid\exists b\in\mathbb{Z}[i]: ab=n\}=\underbrace{\#\{a\mid\exists b\in\mathbb{N}: ab=n\}^2}_{\sigma(0,n)^2}\Leftrightarrow \forall p|n,\;p\text{ prime}: p\equiv 1\mod 4$

(In words: the number of gauss integer divisors of n is equal to the square of the number of integer divisors of n iff n is in A004613)

I could verify it for values up to 10 million. However, I have been unable to find any Identities that could allow me to prove them. I could take several results from exactly that OEIS page, however none of those helped me out. I am missing some result that connects the gaussian integer divisors with the real ones. I would be very happy with any help on this problem.

答:
Thanks to Gerry Myersons comment, I was able to find the general outline of the proof:
Let us denote $\tilde{\sigma}$ as the divisor counting function in $\mathbb{Z}[i]$ ($\sigma$ is the divisor counting function in $\mathbb{Z}$).
We use the identity
$$a=\prod\limits_{k=1}^n p_k^{e_k}$$
$$\sigma(a)=\prod\limits_{k=1}^n (e_k+1)$$
and that (according to Gerry Myerson)
$$p\equiv 1\mod4\Leftrightarrow \exists!\pi,\pi':\;p=\pi\pi'$$
Therefore (we set $p_k=\pi_{2k}\pi_{2k+1}$),
$$a=\prod\limits_{k=1}^n (\pi_{2k}\pi_{2k+1})^{e_k}$$
and
$$\tilde{\sigma}(a)=\prod\limits_{k=1}^{n} (e_k+1)(e_k+1)=\sigma(a)^2$$

All that is left to do is fill in the gaps and discuss, why the other direction works aswell.

ContentLicense: CC BY-SA 3.0


问:
Proving $g:P\to\Bbb R, g(x)=3x+2$ is discontinuous

Define $g:P\to\Bbb R,g(x)=3x+2$ where $P$ is the Cantor set. Show that $g$ can't be continuous under Euclidean metric in $\Bbb R$.

No information is given about the metric on $P$. I assumed it is the Euclidean metric. Then $g^{-1}((a,b))=\left(\frac{a-2}3,\frac{b-2}3\right)\cap P$ which is open in $P$. Then this map is continuous because the inverse image of every open set is open. In fact it seems $g$ will be continuous on any subspace of $\Bbb R$ since it is continuous on $\Bbb R$. Am I doing something wrong?

答:
You are right and the question is wrong.

If we want to spell it out:

Let $a \in P$.  Let $\epsilon > 0$.  Let $\delta = \frac {\epsilon}3$.

For all $x \in P$ so that $|x-a| < \delta$ then

$|(3x + 2) - (3a+2)|= 3|x-a| < 3\delta = \epsilon$ and so $f$ is continuous at all $a \in P$.

Can't get much more plain than that.


问:
Shuffle a poker deck between 4 players, with least required entropy
We are shuffling a standard poker deck of 52 cards between 4 players, each getting 13 cards. The order of cards for a particular player does not matter.

A naive algorithm is to first shuffle the whole deck, then split it into 4 parts. However, this requires $log_2(52!) \approx 226$ bits of entropy.

Combinatorics says that such a shuffle should only require $log_2(52! / 13! / 13! / 13! / 13!) \approx 96$ bits of entropy, more than half less.

What is the algorithm that would produce such an efficient shuffle, using a random generator that produces integers in a range?

I was thinking of this algorithm:

Start with a sorted deck, for each card:

generate a number between 1 and 4, to determine which player the card goes to.

once one of the players gets 13 cards, only generate numbers between 1 and 3 (and so on)

Stop when only one player remains, give them all undealt cards

Would such algorithm produce an unbiased deal, and if not, what is the correct one?

答:
Your approach won't work because:
Let's suppose that the descks have $13-a$, $13-b$, $13-c$, $13-d$ cards respectively. You can assign the rest of the cards in $\frac{(a+b+c+d)!}{a!b!c!d!}$ ways. If you assign the card to the first deck, you can assign the rest in $\frac{(a+b+c+d-1)!}{(a-1)!b!c!d!}$ ways. For the second deck, it is $\frac{(a+b+c+d-1)!}{a!(b-1)!c!d!}$. $\frac{(a+b+c+d-1)!}{a!b!(c-1)!d!}$ for third deck and $\frac{(a+b+c+d-1)!}{a!b!c!(d-1)!}$ for fourth deck. Therefore, you can't assign the next card with equal probabilities. You can easily figure out that the ratio of the probabilities has to be $a:b:c:d$. The easiest way to choose one deck with these probabilities is choosing a random integer from $1$ to $a+b+c+d$ and then look in which deck it falls. The upper bound for your random generator starts at 52 and drops by 1 each time you assign a card to a deck. Now the bits of entropy is: $\sum_{i=1}^{52}log_2(i) = log_2(52!)$. So if you fix your algorithm this way, you are back at the naive generator.

You can do it this way:

generate a random number $k$ between $1$ and $\frac{52!}{13!13!13!13!}$.

set $a,b,c,d:=13$, $u:=\frac{52!}{13!13!13!13!}$

if all cards are assigned to a deck, end.

set $g:=\frac{u}{a+b+c+d}$

if $k \leq ga$ put card to first deck, set $a:=a-1$, $u:=ga$, return to step 3

else if $k \leq g(a+b)$ put card to second deck, set $b:=b-1$, $u:=gb$, $k:=k-ga$, return to step 3

else if $k \leq g(a+b+c)$ put card to third deck, set $c:=c-1$, $u:=gc$, $k:=k-g(a+b)$, return to step 3

else put card to fourth deck, set $d:=d-1$, $u:=gd$, $k:=k-g(a+b+c)$, return to step 3

It is basically the same with the exception that you don't generate a new number in each step, but you decide according to a number which you generate at the beginning. It is basically an unrank function for permutations with repetitions.


问:
Interpreting divergence of velocity field
The wikipedia article on divergence describes one interpretation of divergence: "The velocity of the air at each point defines a vector field. While air is heated in a region, it expands in all directions, and thus the velocity field points outward from that region."

If we have a vector field which represents a force, I interpret the divergence as representing the strength of the field at whatever point it's taken at. However I'm confused on how to interpret the divergence of a velocity field. Clearly if the divergence is positive gas is expanding outward and if it's negative it's contracting, however what quantity is actually represented?

If I have a velocity field with $m/s$ units, then the div presumably has $m/s^2$ units. Is the quantity we get the actual acceleration of gas away from that point?

答:
The interpretation of the divergence of the velocity field as frequency can lead to confusion and not intuitive understanding of the concept. When analyzing the mass conservation equation, you can verify mathematically the role of the divergent velocity in the (Navier-Stokes) conservation equations, but for an intuitive understanding, imagine an expanding solid body.

$\displaystyle \vec{v} =\frac{d\vec{u}}{dt}$ (where $\displaystyle \vec{v}$ is the velocity and $\displaystyle \vec{u}$ is the displacement)

$\displaystyle \nabla \cdotp \vec{v} =\nabla \cdotp \frac{\partial \vec{u}}{\partial t} =\frac{\partial }{\partial t}( \nabla \cdotp \vec{u})$

And here $\displaystyle \nabla \cdot \vec{u}$ plays the role of the volumetric deformation $\displaystyle \varepsilon_{v}$, and therefore

$\displaystyle \nabla \cdot \vec{v} = \frac{\partial \varepsilon_{v}}{\partial t}$

$\displaystyle \varepsilon_{v}$ has no unit, but represents the volume deformation $(\displaystyle \frac{\Delta V}{V_{0}})$, so we see that although the $\displaystyle \nabla \cdotp \vec{v}$  unit is 1/s, its meaning is volumetric deformation per unit time


问:
question about Morse theory in Hilbert space
This is sade to be the Morse theory in Hilbert space ,and i want to know the definition (or where i can find it ) of :

The qth singular relative homology groupe

The qth critical group

Please;

答:
A definition of relative homology is given for instance on page 115 of Hatcher's book on algebraic topology, which is freely available on his website.

The $q$-th critical group seems to be defined in your text itself as $H_q(\Phi^c\cap U,\Phi^c\cap U\backslash\{u\}:\mathbb{F})$.


问:
M-reducable set
I have found the following task:
Show that K (the halting set) is m-reducable to the zero functions $$ K \leq_M\{x | f_x = 0\}$$

We need to find a total-computable function h, such that:
$$ x \in K \iff h(x) \in \{x | f_x = 0\} $$
The i have found this solution:

h(x,y) = t (if f_x (x)↓}

h(x,y) = ↑ (else)

But shouldn't h be a total-computable function? And h is not-total ? And why does it have two arguments? Any tips? Am I understanding things wrong?

page 5:
http://www.cs.sjtu.edu.cn/~gao-xf/computability/Document/10-Reducibility.pdf

答:
Your understanding is correct; the slides are somewhat confusingly written.

The function $f_{\bf 0}$ is not the desired $m$-reduction. Indeed it can't possibly be since the arity is wrong: an $m$-reduction is unary, but $f_{\bf 0}$ is binary. Instead, $f_{\bf 0}$ is an oddly-packaged description of the outputs of the intended $m$-reduction of $K$ to $\{x: \phi_x\simeq{\bf 0}\}$.

The point is that for each $x$, the function $$\eta_x: y\mapsto f_{\bf 0}(x,y)$$ is a partial computable function with the property that $\eta_x\simeq {\bf 0}$ iff $x\in K$. (I'm writing "$\simeq$," rather than "$=$," for equality of possibly-partial functions.)

The idea is then that our $h$ should be a function sending $x$ to an index for the partial computable function $\eta_x$. That is:

Suppose we have a total computable function $h$ such that for each $x$ we have $$\phi_{h(x)}\simeq \eta_x.$$ Then $h$ is an $m$-reduction of $K$ to $\{x: \phi_x\simeq {\bf 0}\}$.

The existence of such an $h$ is then justified by the s-m-n theorem, applied to the partial computable binary function $f_{\bf 0}$. So when the author more-or-less says "$f_{\bf 0}(x,y)$ demonstrates $K\le_m\{x: \phi_x\simeq {\bf 0}\}$," what they mean is that we can extract an appropriate reduction from $f_{\bf 0}$, not that $f_{\bf 0}$ is itself such a reduction.

Title: M-reducable set


问:
Finding maximum of this function in complex analysis
This particular question is from my assignment in complex analysis and I need to verify my answer and way of solving.

Let $f$ be analytic for $|z| \leq 3$ such that $|f(z)| \leq 1$ for $|z| \leq 3$ and has $n$ roots at $w_{k}=e^{2 k \pi i / 3}(k=0,1,2, \ldots, n-1),$ the $n$ th roots of unity. What is the maximum value of $|f(0)| ?$ Which functions attain a maximum?

Attempt: Instead of using maximum modulus principle, I wrote the function as f(z) = $(z-w_1)... (z-w_k) $ and then putting z=0. $ |f(0)|= e^{ \frac{2k(2k-1) πi} {3}} $ . But here maximum will depend upon value of k. So, how to find maximum?

Also, I don't understand what it means by which functions attain it's maximum.

If there is another method by which it can be solved more elegantly, please tell it.

答:
If we let $g(z)=f(3z), |z| \le 1$ we get a holomorphic function that sends the unit disc to itself and has $n$ zeroes at $z_{k}=e^{2 k \pi i / 3}/3$, where $k=0,1,2, \ldots, n-1$ and $f(0)=g(0)$ so we can use $g$

If there were no zeroes then, of course, $|g(0)| \le 1$ with equality if it is a unit modulus constant, so we need to find which analytic functions preserve the unit disc (including being $1$ in absolute value on the boundary) and have zeroes only at the specified (finitely many) points inside (also under some simple conditions one can actually extend to infinitely many zeroes inside).

But for one zero $|z_0|<1$ we can take the standard automorphism $\phi_{z_0}(z)=\frac{z-z_0}{1-\bar z_0z}$ for which of course $|\phi_{z_0}(0)|=|z_0|, |\phi_{z_0}(e^{it}))|=1$.

This clearly means that the product of such automorphisms (also called a finite Blaschke product), $B(z)=\phi_{z_0}(z)\phi_{z_1}(z)..\phi_{z_{n-1}}(z)$ satisfies that $B$ has zeroes only at the given $n$ points, $B$ preserves the unit disc and $|B|=1$ on the unit circle; hence $g/B$ is a holomorphic function from the unit disc to itself (continuous on the boundary) so by maximum modulus $|g/B| \le 1$ and since $g/B$ has no zeroes, we can actually attain maximum precisely when $g=\alpha B$ for some arbitrary $|\alpha|=1$,
hence $|g(0)| \le |B(0)|$ with equality possible.

Particularizing to our case and the given $z_k$ leads to $|f(0)|=|g(0)| \le |B(0)|=\frac{1}{3^n}$ with equality precisely when $f(z)=\alpha B(z/3), |\alpha|=1, |z| \le 3$


问:
Evaluating definite integral with functions as limits
This was a physics example regarding electric field, but I need some help in the calculation part.

Positive charge $Q$ is uniformly distributed along the $y-axis$ between $y=-a$ and $y=+a$. Find the electric field $P$ on the $x-axis$ at a distance $x$ from the origin

So the electric field is generated by a rod, which is broke into infinitesimal segments.

Rod below

| <== $dy$

|

| -------------$x$------------ $P$

|

|

Each segment is called $dy$

$x$ is the distance from $P$ to the rod

$2a$ is the length of rod

$Q$ is total charge, $Q/2a$ is the linear charge density

After a series of steps, the textbook reached the following step, for the $x$ component of the electric field:
$$ E_{x} = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{x{\,}dy}{(x^2+y^2)^{3/2}} $$

the detail of evaluation was omitted in the textbook, but it's evaluated to  $$ E_{x} = \frac{Q}{4\pi \epsilon_{0}} \frac{1}{x \sqrt {x^2+a^2}} $$

I spent a good deal of time trying to evaluate this integral myself using substitution but all met dead ends. I need some guidance on this.

答:
What you need is essentially the followig indefinite integral:
\begin{align}
\int \frac{1}{(x^2+y^2)^{3/2}}\,dy
&=\int \frac{\cos u}{x^2}du& (y=x\tan(u),\ dy=x\sec^2(u)du)\\
&=\frac{\sin(u)}{x^2}+C\\
&=\frac{\sin(\tan^{-1}\frac{y}{x})}{x^2}+C\\
&=\frac{y}{x^2\sqrt{x^2+y^2}}+C
\end{align}

It follows that
\begin{align}
\int_{-a}^a\frac{x}{(x^2+y^2)^{3/2}}dy
&=2x\int_{0}^a\frac{1}{(x^2+y^2)^{3/2}}dy\\
&=2x\cdot \frac{y}{x^2\sqrt{x^2+y^2}}\big|_{y=0}^{y=a}\\
&=\frac{2a}{x\sqrt{x^2+a^2}}
\end{align}


问:
How to find what power of a number is of another number
In the equation, $a=b^x$, Is there an easy way to find the value of $x$ using a calculator (Other than trying multiple $x$ values)?

答: yes, it is $\log_b(a)$. if your calculator doesn't have an option to do different bases, you can use change of base and do $\frac{\log(a)}{\log(b)}$.


问:
How to define a p.d.f when the c.d.f is discontinuous at infinitely many points.
Let $X$ be a random variable with c.d.f $F_X$ and probability law $\mathbb{P}^X$ on $\mathbb{R}$. Let $D$ be the countable set of discontinuities of $F$ and suppose that $D$ is non-empty. This implies that for some points $x$ we have $$\mathbb{P}^X(\{x\})=\mathbb{P}[X=x]>0.$$
What is a correct way of defining a density function $f_X$ for $X$? Ie we need a function $f_X$ such that
$$\int_{-\infty}^\infty f_X(x)d\mathbb{P}^X(x)=1,$$
and that resembles both a p.d.f and a p.m.f.

I think when $D$ is finite or has all isolated points we can define this function easily. However, what if $D$ contains an accumulation point? Or even worse, what if $D$ is dense?

答:
The simple answer is you don't, or rather you can't. Not all random variables have a PDF. In fact a real valued random variables has a density if and only if its CDF is absolutely continuous (this is a stronger condition than continuity).

Distributions are a generalisation of functions, and you can consider every measure as a distribution. A distribution $T$ is a linear function $C_{\infty}(\mathbb R) \to \mathbb R$ (which satisfies some continuity conditions, but that's not important for this discussion). For example the Dirac delta distribution can be defined by $\delta_0(f) = f(0)$.

All probability measures can be considered as a distribution, if $\mu$ is a probability measure then we can turn it into a distribution by defining
$$\mu (f) = \int_{\mathbb R} f(x) \mu(\text{d} x). $$

CDFs are always just normal functions, the function $F(x) = \mu((-\infty, x])$ is always well defined after all. What we can't guarantee is that the CDF will have a corresponding density


答:
First, notice that $x$ is a constant with respect to the variable of integration, and so

\begin{align}
E_{x} &= \frac{1}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{x{\,}dy}{(x^2+y^2)^{3/2}}\\
& = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}}.
\end{align}

Now, let
\begin{align}
y &= x\tan(u)\\
dy &= x\sec^{2}(u)\,du.
\end{align}

Then, using the identity $\tan^2(u) + 1 = \sec^2(u)$, we obtain
\begin{align}
\int \frac{dy}{(x^2+y^2)^{3/2}} &=\int\frac{x\sec^{2}(u)\,du}{(x^{2} + x^{2}\tan^{2}(u))^{3/2}}\\
&=\int\frac{x\sec^{2}(u)\,du}{(x^{2}\sec^2(u))^{3/2}}\\
&=\frac{1}{x^{2}}\int \cos(u)\,du\\
&=\frac{1}{x^{2}}\sin(u) + C.
\end{align}

Now, $$\tan(u) = \frac{y}{x} \implies \sin(u) = \frac{y}{\sqrt{x^2 + y^2}},$$ and so we have $$\int \frac{dy}{(x^2+y^2)^{3/2}} = \frac{y}{x^{2}\sqrt{x^{2} + y^{2}}} + C.$$  Using this to evaluate the definite integral:
$$\int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}} = \frac{y}{x^{2}\sqrt{x^{2} + y^{2}}}\Bigg|_{-a}^{a} = \frac{2a}{x^{2}\sqrt{x^2 + a^2}}.$$  Putting everything back together, then, we obtain
$$E_{x} = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{dy}{(x^2+y^2)^{3/2}} = \frac{x}{4\pi \epsilon_{0}} \frac{Q}{2a}\frac{2a}{x^{2}\sqrt{x^2 + a^2}}=\frac{Q}{4\pi \epsilon_{0}} \frac{1}{x \sqrt {x^2+a^2}}.$$

Title: Evaluating definite integral with functions as limits

Body:
This was a physics example regarding electric field, but I need some help in the calculation part.

Positive charge $Q$ is uniformly distributed along the $y-axis$ between $y=-a$ and $y=+a$. Find the electric field $P$ on the $x-axis$ at a distance $x$ from the origin

So the electric field is generated by a rod, which is broke into infinitesimal segments.

Rod below

| <== $dy$

|

| -------------$x$------------ $P$

|

|

Each segment is called $dy$

$x$ is the distance from $P$ to the rod

$2a$ is the length of rod

$Q$ is total charge, $Q/2a$ is the linear charge density

After a series of steps, the textbook reached the following step, for the $x$ component of the electric field:
$$ E_{x} = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{2a} \int_{-a}^{a} \frac{x{\,}dy}{(x^2+y^2)^{3/2}} $$

the detail of evaluation was omitted in the textbook, but it's evaluated to  $$ E_{x} = \frac{Q}{4\pi \epsilon_{0}} \frac{1}{x \sqrt {x^2+a^2}} $$

I spent a good deal of time trying to evaluate this integral myself using substitution but all met dead ends. I need some guidance on this.


问:
Relating the curvature of a plane curve to the curvature of a stretched version
Let $\theta : I \to \mathbb{R}^2$ be a regular plane curve with curvature $ |k_{\theta}|\leq1$ everywhere.

We now define a curve $\theta_{d}$ by stretching $\theta$ in one direction, i.e.,

$\theta = ( \theta_{1}(t), \theta_{2}(t))$,

$\theta_{d} = (2 \theta_{1}(t), \theta_{2}(t))$

How the curvature of  $\theta_{d} , k_{d} $ relates to $\theta$ ?

答:
Using the standard formula for curvature of a plane curve $\alpha(t)$,
$$k = \frac{\|\alpha'(t)\times\alpha''(t)\|}{\|\alpha'(t)\|^3}\,,$$
we find that
$$k_{\theta_d} = \frac{2k_\theta}{(4\theta_1'^2 + \theta_2'^2)^{3/2}}\,.$$
Not so informative, but since the speed of $\theta_d$ is between $1$ and $2$ and you gave the curvature bound on $\theta$, we have
$$\frac{k_\theta}4 \le k_{\theta_d} \le 2{k_\theta}\le 2\,.$$</p>


问:
Why does the intersection of two parametric curves have different t results for each curve?
I'm slightly confused after reading the problem posed here: Points of intersection of two parametric curves

Why is it that the "t" result at the intersection point of the two curves is not the same? I.e. the solution involves the answer "t1" for the first curve and "t2" for the second curve, but if the two curves are being made equal, why is t1 != t2?

Also, is there any way we can "manufacture" two different curves (say two parametric straight lines) that, at a specific t value for BOTH curves, yield the SAME point in space? How would you go about doing that? (given, for example, 3 points in space)

答:
Let's think of the two curves $c_1, c_2$ as the routes two different cars take. The parameter $t$ then represents time, and $c_1(t), c_2(t)$ describe the position of the cars at time $t$.

If both cars stop at the same gas station, does it follow that they have to do it at the same time? Of course not!


问:
How many coefficients of the polynomial $P_n(x_1,...,x_n) = \prod_{1\leq i < j \leq n}(x_i+x_j)$ are odd?
How many coefficients of the polynomial $P_n(x_1,...,x_n) = \prod_{1\leq i < j \leq n}(x_i+x_j)$ are odd?

I computed a couple small cases first. for $n = 1$ the problem doens't make sense. For $n=2$ there are 2 odd coefficients, and for $n = 3$ there are 6. I'm pretty sure I need to plug something in for $x_i$ and do some modular arithmetic but I'm not sure what I need to do.

Some hints would be greatly appreciated.

答:
The answer is $n!$. The cleanest strategy, to my mind, is contained in the comments to saulspatz's answer: working $\bmod 2$ we can equivalently count the number of nonzero monomials in $\prod_{i < j} (x_i - x_j)$ over $\mathbb{F}_2$. This is the Vandermonde determinant. The Leibniz formula expresses the Vandermonde determinant is a sum of $n!$ terms, each of which is a monomial with coefficient $\pm 1$ (so $1 \bmod 2$) and each of which is distinct, so there's no additional cancellation.

(By the way, the product does still make sense when $n = 1$, it's just empty, so it's equal to $1$ and has $1! = 1$ odd coefficients. This is consistent with the $1 \times 1$ Vandermonde determinant $\det(1) = 1$ also.)


问:
Find all roots of a quintic polynomial given 1?
Let's say I have solved for one of the roots $a$ of a quintic equation using Newton's method. I read somewhere that you can "simply" divide the equation by $x - a$ to get a quartic and solve from there.

How would I get this quartic equation from my quintic euqation (using a formula), and after doing this, how would I go about solving it numerically?

答:
You can use polynomial long division to find the quotient.

An example of numerical package that does this is Python's numpy package. The following example shows that the quotient of $x^5-1$ by $x-1$ is $x^4 + x^3 + x^2 + x + 1$ and that the remainder is $0$

```<code>>>> import numpy as np
>>> q, r = np.polydiv([1, 0, 0, 0, 0, -1], [1, -1])
>>> q
array([1., 1., 1., 1., 1.])
>>> r
array([0.])
</code>```

The source code is available in the numpy source tree, see the lines after def polydiv.

You don't need to divide the polynomials to find the roots. Efficient algorithms exist that find simultaneously all the complex roots of the polynomial. With numpy you can simply write this to get all the complex roots

```<code>>>> np.poly1d([1, 0, 0, 0, 0, -1]).r
array([-0.80901699+0.58778525j, -0.80901699-0.58778525j,
    0.30901699+0.95105652j,  0.30901699-0.95105652j,
    1.        +0.j        ])
</code>```

Body:
You can use polynomial long division to find the quotient.

An example of numerical package that does this is Python's numpy package. The following example shows that the quotient of $x^5-1$ by $x-1$ is $x^4 + x^3 + x^2 + x + 1$ and that the remainder is $0$

>>> import numpy as np
>>> q, r = np.polydiv([1, 0, 0, 0, 0, -1], [1, -1])
>>> q
array([1., 1., 1., 1., 1.])
>>> r
array([0.])

The source code is available in the numpy source tree, see the lines after def polydiv.

You don't need to divide the polynomials to find the roots. Efficient algorithms exist that find simultaneously all the complex roots of the polynomial. With numpy you can simply write this to get all the complex roots

>>> np.poly1d([1, 0, 0, 0, 0, -1]).r
array([-0.80901699+0.58778525j, -0.80901699-0.58778525j,
    0.30901699+0.95105652j,  0.30901699-0.95105652j,
    1.        +0.j        ])


问:
Is restriction of scalars well-defined on subspaces?
Let $K/k$ be an extension of fields and let $v_1,\ldots,v_r,u_1,\ldots,u_r\in k^n$.  If the span of the $v$'s over $K$ equals the span of the $u$'s over $K$, must the two spans also be equal over $k$?

$$_K\langle v_1,\ldots,v_r\rangle=_K\langle u_1,\ldots,u_r\rangle\qquad\overset{?}\Longrightarrow\qquad_k\langle v_1,\ldots,v_r\rangle=_k\langle u_1,\ldots,u_r\rangle$$

Here is another way to look at this question: Given a subspace $U\subset K^n$ with some basis in $k^n$, is the procedure of restricting scalars to $k$ well-defined, or does it depend on the choice of basis of $U$?

答:
Step 1: A subset $S \subset k^n$ is $k$-linearly independent iff it is $K$-linearly independent.  This can be shown for instance by expressing linear independence in terms of the reduced row echelon form (rref) of the matrix with columns the elements of $S$ and appealing to the uniqueness of rref.

(Alternate proof: if $S$ is $k$-linearly independent, extend it to a $k$-basis $B$ of $k^n$.  The matrix $M$ with columns the elements of $B$ is invertible, i.e., there is a matrix $N$ with $MN = NM = 1$.   This matrix equation holds over $K$ as well and shows that the columns of $M$ are $K$-linearly independent.)

Step 2: To ask whether

$\langle u_1,\ldots,u_r \rangle_K \subset \langle v_1,\ldots,v_s \rangle_K$

we may assume without loss of generality that $v_1,\ldots,v_s$ are $k$-linearly independent, because by throwing away some of the $v_j$'s we can reduce to this case without changing the span (either over $k$ or over $K$).  Then the above containment holds iff for all $i$, the set $v_1,\ldots,v_s,u_i$ is $K$-linearly dependent.  As above, this holds over $k$ iff it holds over $K$, so the given containment implies

$\langle u_1,\ldots,u_r \rangle_k \subset \langle v_1,\ldots,v_s \rangle_k$.

Step 3: The affirmative answer to your question follows immediately from Step 2.


问:
Checking an introductory Linear Algebra Exercise regarding linear applications.
My final is few days away and I'm doing the list of exercises recommended by the teacher. Not all exercises have solutions however, and the teacher isn't answering my questions, so I just want to check whether I did the following exercise correctly:

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^3 : (2x_1-x_2+x_3,x_2+x_3,x_1+x_2+2x_3)$.

Prove $f$ is a linear application.

Find the matrix $A$ that corresponds with the linear applicaiton $f$.

Find a basis for $\text{Im}(f)$ and $\text{Ker}(f)$.

Proving that $f(c\textbf{v})=cf(\textbf{v})$ and that $f(\textbf{v}+\textbf{u})=f(\textbf{v})+f(\textbf{u})$ is easy enough.

$$A=\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$

This is the bit I'm more concerned about. Vectors of $\text{Im}(f)$ are precisely those of the form

$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\ x_3\end{pmatrix}$$

we can apply elementary column operations since these don't affect the column space ($\text{Im}(f)$),

$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 0 \end{pmatrix}$$

Since the first two columns are clearly linearly independent, and vectors of $\text{Im}(f)$ are preciesly those of the form

$$x_1\begin{pmatrix} 0 \\2 \\ 3\end{pmatrix}+x_2\begin{pmatrix} -1 \\1 \\ 1\end{pmatrix}$$

for $x_1,x_2\in \mathbb{R}$, we must have that $\{ (0, 2, 3),(-1, 1, 1)\}$ is a basis of $\text{Im}(f)$. Solutions of the equation $$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\textbf{v}=\textbf{0}$$

are not changed by multiplying on the left both sides by elementary matrices.

$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -3 & -3 \end{pmatrix}\sim\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$

Thus $\text{Ker}(f)$ is precisely the set of vectors $v$ such that $x_1=x_2=-x_3$, or $$v=t\begin{pmatrix} 1 \\ 1\\ -1\end{pmatrix}$$ for $t\in \mathbb{R}$. Therefore $\{ (1,1,-1)\}$ is a basis of $\text{Ker}(f)$.

答:
That all looks fine to me (except that for item 1, you'd get no credit unless you actually wrote out the proof, of course).

An alternative to your solution for 1 is to do item 2 first; then you have

$$
f(v) = Av
$$
for all $v \in \Bbb R^3$. So
$$
f(v+w) = A(v + w) = Av + Aw = f(v)+ f(w)
$$
by the distributive law for matrix multiplication, and a similar thing works for scalar multiples.


问:
Circles with numbers

This is a diagram Venn,what choice it discribes (AB)'UC

The choices are :

```<code>a)2,4,5
b)1,2,5
c)something else
d)1,3,4,5
e)1,2,3,4,5
</code>```

答:
Your image should look more like this. The area $A$ and $B$ have in common is a shared area; not a separate circle.

Now $A \cap B$ are the things that $A = \{1,2\}$ and $B = \{2,3\}$ have in common.  They have $2$ in common so $A\cap B = \{2\}$

$(A\cap B)'$ is everything else that isnt in $A\cap B =\{2\}$. That is $1,3,4,5;$ everything except $2$.  So $(A\cap B)' = \{1,3,4,5\}$.

And $C = \{4\}$.  And $(A\cap B)' \cup C$ is every that $(A\cap B)' = \{1,3,4,5\}$ has combined with everything $C = \{4\}$ has.  That is $\{1,3,4,5\}$ combined with $\{4\}$ so that is $\{1,3,4,5\}$.

So $(A\cap B)' \cup C = \{1,3,4,5\}$.

Title: Circles with numbers


问:
Cardinality of a quotient set $\mathbb{N}^\mathbb{N}/\sim$
I'm having trouble with the following task:

On the set of all functions $\mathbb{N}^\mathbb{N}$ an equivalence relation $\sim$ is defined in the following way:

$f \sim g \iff \forall n \in \mathbb{N}. f(f(n)) = g(g(n))$.

Prove that the quotient set $\mathbb{N}^\mathbb{N}/\sim$ is uncountable.

I've tried to find an injective function from $\mathbb{N}^\mathbb{N}$ to $\mathbb{N}^\mathbb{N}/\sim$ and I've tried proving that every equivalence class has only one element, but I haven't succeeded in either of the attempts. I can't even figure out what would be the main idea needed for the proof.

答:
For each $A\subseteq\Bbb N\setminus\{0,1\}$ let

$$f_A:\Bbb N\to\Bbb N:n\mapsto\begin{cases}
0,&\text{if }n\in A\cup\{0\}\\
1,&\text{otherwise;}
\end{cases}$$

then $f_A\circ f_A=f_A$, so the function

$$\varphi:\wp(\Bbb N\setminus\{0,1\})\to{^{\Bbb N}\Bbb N}/\!\sim\;:A\mapsto[f_A]_\sim$$

is an injection.


问:
Show that $Z_t = \sqrt{\sqrt{2t}e^{-\sqrt{2t}}} B_{e^{\sqrt{2t}}-1}$ solves $dZ_t = f(t)Z_tdt +dM_t$ and find $f(t)$.
Let $B=(B_t)_{t\geq0}$ be a standard Brownian motion started at $0$ and define $M=(M_t)_{t\geq0}$ by

$$M_t = \int_0^{e^{\sqrt{2t}}-1} \sqrt{\frac{\log{(1+s)}}{1+s}} dB_s$$

so that M is also a standard Brownian motion. Show that the process $Z=(Z_t)_{t\geq0}$ defined by

$$Z_t = \sqrt{\sqrt{2t}e^{-\sqrt{2t}}} B_{e^{\sqrt{2t}}-1}$$

solves the following stochastic differential equation

$$dZ_t = f(t)Z_tdt +dM_t$$

with $Z_0 = 0$ and determine the function $f$ explicitly.

I'm not really sure how to go about this, any help would be appreciated, thanks.

答:
Yes first see the first differential from the integral form you have.

$$ dM_t= \sqrt{\dfrac{\log(1+s)}{1+s}}dB_s$$ where $s=e^{\sqrt{2t}-1}$

Then from definition of $Z_t$

$$dZ_t=d(\sqrt{\sqrt{2t}e^{-\sqrt{2t}})}B_{e^{\sqrt{2t}-1}}+dB_{e^{\sqrt{2t}-1}}\sqrt{\sqrt{2t}e^{-\sqrt{2t}}}$$

And from here I think you can end :)


问:
How to calculate the hessian of a matrix function $G(\theta)^T Z W Z^T G(\theta)$
I am estimating a model minimizing the following objective function,

$ M(\theta) = (Z'G(\theta))'W(Z'G(\theta)) \equiv G(\theta)^T Z W Z^T G(\theta)$

$Z$ is an $N \times L$ matrix of data, and $W$ is an $L\times L$ weight matrix, neither of which depends on $\theta$. $G(\theta)$ is a function which takes the $K \times 1$ vector of parameters I am estimating into an $N \times 1$ vector of residuals.

I am trying to calculate the Gradient and the Hessian of M to feed to into a Matlab solver ($G(\theta)$ is highly nonlinear). I believe I have the gradient correctly calculated as follows. Let $J = dG/d\theta$ be an $N \times K$ Jacobian matrix, where $J_{i,k}$ is the derivative of element $i$ of $G$ with respect to parameter $k$. Then the gradient vector of M is,

$ \nabla M = 2 (Z'J(\theta))'W (Z'G(\theta))' $

However, I cannot figure out how to take the derivative of this gradient to get the Hessian. I can calculate the vectors of derivatives of each element of $J(\theta)$, but I'm not sure what the order of that derivative matrix should be. I see that I basically need to divide this function up into two parts and use a the product rule, but cannot figure out what the derivative of each part should look like.

Any help would be greatly appreciated. Thank you.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$For ease of typing, use $x$ instead of $\theta$ for the independent variable. Let's also name variables using calligraphic letters for tensors, uppercase letters for matrices, lowercase letters for vectors, and greeks letters for scalars. To that end define
$$\eqalign{
g &= g(x) \\
J &= \p{g}{x} \\
A &= ZWZ^T = A^T \\
\mu &= g^TZWZ^Tg \qquad&\big({\rm objective\,function}\big) \\
 &= A:gg^T \qquad&\big({\rm reformulated}\big) \\
}$$
Calculate the differential and the gradient of the objective function.
$$\eqalign{
d\mu
 &= A:\big(dg\,g^T + g\,dg^T\big) \\
 &= \big(A+A^T\big):dg\,g^T \\
 &= 2Ag:J\,dx \\
 &= 2J^TAg:dx \\
\p{\mu}{x} &= 2J^TAg \;\doteq\; p \qquad\big({\rm gradient}\big) \\
}$$
The Hessian is just the gradient of the gradient, so
$$\eqalign{
dp &= 2J^TA\,dg + 2\,dJ^TAg \\
  &= 2J^TAJ\,dx + 2\big(g^TA^T{\cal K}\,dx\big) \\
  &= 2\left(J^TAJ + g^TA{\cal K}\right)dx \\
\p{p}{x} &= 2\left(J^TAJ + g^TA{\cal K}\right)
   \;\doteq\; H\qquad\big({\rm Hessian}\big) \\
}$$
where ${\cal K}$ is the gradient of $J,\;$ a third-order tensor with components
$$\eqalign{
{\cal K}_{ijk} &= \frac{\partial J_{ij}}{\partial x_k} \\
}$$
I can't say anything more about ${\cal K}$,
since you didn't tell us anything about the function $g(x)$.

In some of the steps above, a colon is used as a convenient product notation for the trace, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) = B:A \\
}$$


问:
Fallacy in solution attempt to Hartshorne exercise chapter I, 2.11
I am somewhat of a beginner in algebraic geometry and have been working on the following problem from Hartshorne:

Let $Y \subset \mathbb P^n$ be a projective linear variety, that is its ideal $I(Y)$ can be generated by homogeneous polynomials of degree $1$. Then for any $r \in [n]$, $\dim Y = r$ if and only if $I(Y)$ can be minimally generated by $n-r$ homogeneous linear polynomials.

This is how I started: since $Y$ is a projective variety, $I(Y)$ is a prime ideal in the integral domain $S:= k[X_0, \cdots , X_n]$ whence by proposition 1.8A(b) (section I.1), we have
$$\text{ht} I(Y) = \dim S - \dim S(Y) = n+1-r$$
and Krull's Principal Ideal Theorem (in conjunction with the Noetherianity of $S$ via the Hilbert Basis Theorem) shows that the minimal number of generators $\mu(I(Y))$ of the ideal $I(Y)$ satisfies the inequality
$$\mu (I(Y)) \geq \text{ht} I(Y) = n-r+1.$$

On the other hand, what I have to show is that $\mu(I(Y))=n-r$. I do not know where the fallacy in the above argument is and would really appreciate some help regarding the same. It would also be nice to know if this argument shows promise. Thank you.

答: Note that we actually have $\dim(S(Y))=\dim(Y)+1=r+1$.


问:
$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$
Let $$f(x) = x\cos\frac{1}{x}$$ for x in $[1,\infty)$

Now I need to prove or disprove the difference $$f(x+2) - f(x) > 2$$ for all $x$ in the domain.

I tried a lot but I don't seem to be getting anywhere. My approach was to try and use graphs, but somehow that doesn't seem to work out that well.

Any ideas or suggestions please?

答: Hint: When $x\ge 1$, then $0 \lt \frac 1x \le 1$.  How does $\cos y$ behave when $0\lt y\le 1$?  </p>


问:
Formula for the 3 hands to coincide in a clock, no matter wether the clock is a conventional one or not
I am trying to find out a formula for the clock's hands to coincide no matter it's a conventional one or not. Here is the progress i made, but I'm stuck. enter image description here</p>

答:
In the time the seconds hand made a full rotation, the minute hand made $\frac{1}{60}$ of a full rotation and the hour hand made $\frac{1}{720}$ (so $\frac{1}{12}$ of the progress of the minute hand) of a full rotation. So if the seconds hand made t rotations, the minute hand made $\frac{t}{60}$ rotations and the hours hand made $\frac{t}{720}$ rotations in that same time.

The first time (after midnight) the hour and minute hand meet occurs when the minute hand made a full rotation and catches up to the hour hand, at which point the amount the minute hand rotated is exactly one full rotation more than the rotation the hour hand made, so $t = \frac{t}{12} + 1$. Solving this equation gives you $t = \frac{12}{11}$, so 1 hour, 5 minutes and about 27 seconds after midnight. Every $\frac{12}{11}$ hours the hour and minute hand meet, that's 22 times per day - but you can check that the only point where the seconds hand is also in the same place is at 12am and 12pm.

Since you tried it using trigonometric functions, here's an alternative way:

Let's suppose the seconds hand moves $\alpha$ degrees. In the amount of time that the seconds hade rotated $\alpha$ degrees, the minute hand rotates $\frac{\alpha}{60}$ of the whole clock, while the hours hand rotates $\frac{\alpha}{720}$ of the whole clock.

We want to know $\alpha$ for which all three hands are in the same place, meaning

$\cos(\alpha) = \cos(\frac{\alpha}{60}) = \cos(\frac{\alpha}{720})$ and

$\sin(\alpha) = \sin(\frac{\alpha}{60}) = \sin(\frac{\alpha}{720})$

This is the case if $\alpha = 1440\pi \cdot n$ with $n\in \mathbb{Z}$, or in other words every 259,200°

Since the seconds hand makes $60\cdot 24=1,440$ full rotations over the whole day, we search for $n\geq 0$ for which $\alpha = 259,200° \cdot n$ is less than $1440\cdot 360°=518,400°$. This is the case only for $n \in \{0,1\}$. So there's 2 times when all hands are in the same place:

After the seconds hand turned $\alpha = 259,200 \cdot 0 = 0$ degrees, so at midnight.

After the seconds hand turned $\alpha = 259,200 \cdot 1 = 259,200$ degrees, that is after 720 full rotations or 12 hours, so at noon.


问:
Russian roullette no reshuffling probability
What are the odds of playing Russian roulette alone surviving to $5$ consecutive trial without reshuffling?

If the event were independent it would be $(5/6)\cdot (4/5)\dotsb (1/2) = 1/6$.
But are the event independent? You must survive previously to be able to play the next move?

I  am not sure how to compute from there.

The Intersection of dependents events is $P(E∩F) = P(E) * P(F|E)$.
$F$ and $E$ being surviving first and second trial respectfully.
Is that correct?
How computing  $P(F|E)$? Is it $F*E$ ?

Am I correct? So the result would be $P(G∩H∩I∩E∩F)$?
(with $G$ $H$ $I$ the fifth forth and third trial.)
I just don't know how to compute that.

I would really appreciate more than the number the method.

Any helps or tips appreciated!

答:
We do know that the total probability is 1/6, as if the cartridge looks like B-O-O-O-O-O,
You'll only survive if you start at the second slot and there are 6 slots.
As for the big question, if the events  were truly independent (i. e. reshuffling), the odds of each are 5/6 so it would be calculated as (5/6)^5. The previous move is factored in as doing (5/6)(4/5)(4/3)...*(1/2). The fact you have to survive the last move doesn't matter; think of it like this: if you have a friend shoot the gun 5 times instead of you, meaning all 5 slots get fired, it wouldn't matter as long as you keep the same cartridge position.


问:
How to transform a non-homogeneous Neumann boundary condition into a homogeneous for the wave equation?
I have a initial/ boundary value problem for standard wave equation
$$
\frac{\partial^2u}{\partial t^2}=c\frac{\partial^2u}{\partial x^2},
$$
where one of the boundary conditions is non-homogeneous: precisely $$
\frac{\partial u(0,t)}{\partial x}=e^t.
$$
Question: is it possible to transform the above problem into a one with homogeneous boundary conditions?

In the Q&A "Solve wave equation and inhomogeneous Neumann Condition with eigenfunction expansion (Fourier Series Solution)" I saw a similar reduction, but I am unable to apply the method proposed there to my problem.

答: You should find a function $v(x, t) = A(t) x^2 + B(t) x + C(t)$ that satisfies boundary conditions (in your case $\partial_x v (0,t) = e^t$). And you can find solution of your problem as $u = v + w$. For function $w$ you get a problem for wave equation with homogenous boundary conditions.


问:
A question about using Lagrange multipliers to maximize a function
The function is $f(x,y,z)=13x^4+13y^4+13z^4$, and the constraint is $13x^2+13y^2+13z^2=3$.

I've set $<f_x, f_y, f_z>=\lambda<g_x, g_y, g_z>$ and got:

$$<52x^3, 52y^3, 52z^3>=\lambda<26x, 26y, 26z>$$

Then I equated everything (assuming that $x$, $y$, or $z$ are not equal to 0) to get the three equations:
$$x^2=\frac\lambda2 $$
$$y^2=\frac\lambda2$$
$$z^2=\frac\lambda2$$

After plugging these into the constraint I found
$$3\left(\frac{13\lambda}{2}\right)=3$$
so $\lambda=\frac{2}{13}$

That must mean that $x$, $y$, and $z$ are all equal to $\sqrt\frac{1}{13}$

Plugging those into $f(x,y,z)$ gives $f=\left(\sqrt\frac{1}{13},\sqrt\frac{1}{13},\sqrt\frac{1}{13} \right)=\frac{3}{13}$, which should be the maximum value.
But apparently the maximum value is $\frac{9}{13}$ and $x$, $y$, and $z$ all equal $\sqrt\frac{3}{13}$ . Can someone help me figure out where I went wrong?

答:
As you found, a solution (there are eight) is  $x=y=z=\frac{1}{\sqrt{13}}$.

All solutions are $$(x,y,z) \in \left\{ \left(\mp \frac{1}{\sqrt{13}},\mp \frac{1}{\sqrt{13}},\mp \frac{1}{\sqrt{13}} \right)\right\}$$

$$f(x,y,z) = 13(x^4+y^4+z^4) = 13 \cdot 3\cdot \frac{1}{13^2} = \frac{3}{13}.$$


问:
Check whether a given graph is bipartite except for one edge
Given any graph $G=(V,E)$, how to decide if deleting just one of its edges can make it bipartite?

Trivial solution: check for each edge whether deleting it makes the graph bipartite. This is in $O(|E|\cdot(|E|+|V|))$.

Are there more efficient ways?

答:
There is a linear solution, with a Depth-First Search (DFS), using Tarjan's representation of the graph.

In short, the problem's solution is : run Tarjan's bridge-finding algorithm while also bicoloring the graph and removing the edges incompatible with the coloration you're computing. Once you've done this, find one of the bridges that, upon suppression, allows you to add all the edges you removed. This can be found along the exploration. (see the definition of a bridge and the algorithm in  https://en.wikipedia.org/wiki/Bridge_(graph_theory)#Tarjan's_bridge-finding_algorithm)

Tarjan's representation of a DFS

The idea of Tarjan's representation is simple, but very powerful for a lot of graph connectivity problems (and coloration, etc). I haven't found a generic presentation for reference, so here is a short one, along with the vocabulary for this answer.

First, suppose your graph is connex (else, apply to each connected component).
You start a depth-first search, and you classify the edges when you encounter them :

Some edges are connecting the node you're actually in to a node you've never encountered before. Mark them as "down" edges.

Some edges are connecting the node you're actually in to a node you've already encountered. Mark them as "return" edges.

Now, some very interesting properties happen :

the "down" edges are a rooted spanning tree of the graph (rooted in the first explored node), representing the recursive calls of the depth-first search.

if the graph is non-oriented, then the "return" edges are always between a node and one of its ancestors in the spanning tree. Indeed, you have already encountered the node before, but you haven't yet finished enumerating its nodes, so the depth-first search recursive function hasn't returned yet.

General culture fact : if the graph is oriented, that the "return" edges have more variations (they can go to other branches of the tree, or to children in the tree). These are cool properties to study for other algorithms.

The power of Tarjan's representation lies in these properties, that allow us to send information down and up the recursive calls of the DFS. The solution to this problem works like this.

Algorithm ideas

Computing a first coloration and compatibility of edges

Consider the classic DFS algorithm for computing a bipartition :

color a node in a color

recursively color each unseen neighbor in the opposite color

This gives an initial coloration $C_0$. If we look how it acts on the Tarjan representation, we have :

The nodes are colored according to their depth (modulo 2), so the "down" edges are compatible with the coloration $C_0$

To check whether a "return" edge is compatible with the coloration $C_0$, we only have to check whether the difference of depth between its ends is odd. If it is even, then both its endpoints have the same color, so this edge is a conflict (and we have found an odd-cycle length). If it is odd, then this edge does not conflict with the coloration.

So now, in one pass, we can generate a coloration along with all the conflicts it creates.

Which edges could be the solution to the problem ?

Now, we have to choose one edge $e$ to remove, and edit the coloration $C_0$ to have a bipartition with a new coloration $C(e)$.

Either we are lucky, and there is only one (or zero) conflicting edge ; then we remove it and call it a day.

Or, we have at least two conflicting edges. Then, we have to remove one of the "down" edges ; indeed, if we don't remove one of those, then the spanning tree imposes the coloration $C_0$ we already had, and there will be a conflict.

When we remove a "down" edge, this will split the spanning tree in two trees. We have to invert the coloration of exactly one of these two trees (once we invert one of the nodes, all the nodes in its tree will have to be inverted as well ; and if we invert both trees, then we've done nothing). By symmetry, let's say we choose to invert the part that does not contain the root (we won't explicitly compute it, so it has no influence on the actual algorithm). For a down edge $e$, this gives the new coloration $C(e)$. We call $M(e)$ the main part of the initial spanning tree which contains the root, and $B(e)$ the branch which does not contain the root.

Now :

The remaining "down" edges are still compatible with the new coloration $C(e)$

$(*)$ Let $f$ be a "return" edge that has one end in $M(e)$ and one in $B(e)$, then if it was compatible with $C_0$, it isn't compatible with $C(e)$. Which is impossible. But if $f$ was incompatible with $C_0$, then now it is compatible with $C(e)$.

$(**)$ Let $g$ be a "return" edge that has both ends in $M(e)$, or both ends in $B(e)$, then its compatibility with $C_0$ is the same as its compatibility with $C(e)$. So, if $e$ is the right edge to remove, then $g$ was compatible with $C_0$.

Translating this in computable properties

Because of $(*)$, if we restrict the graph $G$ to the edges compatible with $C_0$, then $e$ has to be a bridge of this restricted graph $G_r$.
If an edge $e$ is a bridge, then removing it and inverting the coloration of $B(e)$ won't break the compatibility of already-compatible edges, because none of them ends up in case $(*)$.

Because of $(**)$, all the edges incompatible with $C_0$ have an end above $e$ and an end below $e$.
If an edge $e$ corresponds to these constraints for all the edges that were incompatible with $C_0$, then removing it and inverting the coloration of $B(e)$ will repair the compatibility of all the precedingly incompatible edges, because all of them end up in case (*).

We need both : a $G_r$-bridge, that is part of every path of the incompatible edges (globally, this means that $e$ is part of every odd-length cycle that is elementary along the spanning tree). For the same edge. Let's do this !

Actual algorithm

Ideas

The task is now to find such an edge $e$. All the computation related to this can be done in a single DFS, but to make things clearer we're gonna do several of them, encountering the nodes in the same order.

We run a first DFS algorithm to compute the coloration $C_0$.

Then we remove the edges incompatible with $C_0$, to compute the restricted graph $G_r$.

Then we run a bridge-finding algorithm on $G_r$, to extract all the bridges.

Then, on the Tarjan decomposition, each incompatible return edge $f = v_{bottom} \to v_{top}$ has an end $v_{top}$ which lies in $M(e)$ and an end $v_{bottom}$ which lies in $B(e)$ ; so $v_{top}$ imposes that $e$ is a descendant of $v_{top}$, and $v_{bottom}$ imposes that $e$ is an ascendant of $v_{bottom}$.

This kind of constraints (be a bridge, be lower than all $v_{top}$, be higher than all $v_{bottom}$, be only one edge) are the kind of constraints that are easy to enforce in a DFS.

Implementation

The exact implementation might be tricky, but conceptually you would be asking to each of your descendants in the DFS "what are the constraints you have for me and my ascendants ?", and expecting an answer among :

I have no constraint

I need a bridge with a depth at least equal to {some node}, but I have found one in my branch that is OK for me

I need a bridge with a depth at least equal to {some node} and haven't found one yet

I have found my bridge in my branch and you can't make me change it

The constraints are too hard to be solved with only one edge removal

Given such an answer for each of your descendants, and all the return edges of which you are the $v_{bottom}$ (and whether they're compatible or not), and whether the edge you came through is a bridge, you can compute the answer for the node you're actually in. A lot of particular cases to think of, but it'll work.

And of course, don't forget that you can be lucky and maybe the first coloration you found already has one or zero incompatible edges. You have to check that independently.

And run that on each connected component.


问:
Do the generators of the subfields of a field extension $K$ "generate" $K$ in a linear sense?
This is kind of a two-parter, I am reading this proof that has some steps that suggest that this true:

Let $K=\mathbb{Q}(\theta)$ with $\mathbb{Q} \subset K$ Galois, and $\mathbb{Q(\theta_1)}, \mathbb{Q(\theta_2)},...... \mathbb{Q(\theta_n)}$ be all the subfields of $K$.

Is $K$ $\mathbb{Q}$-linearly generated by $\theta_1,..\theta_n$? (i.e for every $x \in K$, $x$ can be expressed as $x=a_1\theta_1+...+a_n\theta_n$ for $a_i \in \mathbb{Q}$)

Is there a relation between the linear expression above given and $\theta$'s conjugates? (i.e if $x=a_1\theta_1+...+a_n\theta_n$, can I find $x'$ ($x$ first conjugate) using that equation?)

Now I am guessing the first can be explained by absurd reasoning and generating another subfield.
The second seems logical, since each $\mathbb{Q}(\theta_i)$ is directly associated to a conjugate of $\theta$ by Galois. But I can't seem to figure it out.

答: If $K/\mathbb{Q}$ is cyclic, with Galois group $\mathbb{Z}/n\mathbb{Z}$, then there are as many subfields of $K$ as there are divisors of $n$. Of course unless $n\leqslant 2$ this number is strictly less than $n$, so generators of subfields cannot form a linear basis of $K$, since the dimension of $K$ is $n$.


问:
Make three random cuts to a circle, what is the expected number of areas produced?
For two random cuts, I have 1/3 chance to have 4 and 2/3 chance to have 3. So expectation is 10/3. What about 3 cuts or even n cuts? Are there any general results?</p>

答: In the crude diagram below, we can see $4$, $5$, $6$, and $7$ areas resulting from $3$ cuts. There is no easy way to tell which is more likely.


问:
The differential complex of smooth forms
I'm  learning my self  about differential forms from the book by Loring Tu, and I've come across this sentence which seems important by I didn't understand it: " The differential complex of smooth forms on a manifold can be pulled back under a smooth map, making the complex into a contravariant functor called the de Rham complex of the manifold"

I was thinking that the de Rham complex is the same as the complex of differential forms !

Could you please enlight the difference between de Rham complex and the complex of differential forms and what it is meant by "contravariant functor" and say more about the above sentence .

Thanks a lot!

答:
Terminology aside, the idea is that we can define a contravariant functor $$\Omega^*:(\text{Smooth Manifolds})\to (\text{Chain Complexes of Vector Spaces})$$
by assigning to a manifold $M$ its de Rham Complex
$$
\Omega^0(M)\xrightarrow{d}\Omega^1(M)\xrightarrow{d}\Omega^2(M)\xrightarrow{d}\cdots\xrightarrow{d} \Omega^n(M).
$$
You can find the definition of a functor between categories formally defined all over (like on wikipedia) but I'll sketch the idea here. Being a (contravariant) functor means that $\Omega^*$ assigns to each manifold a chain complex like above, and to each smooth map of manifolds $f:M\to N$ an associated morphism of chain complexes
$$\Omega^*(f):\Omega^*(N)\to \Omega^*(M).$$
The contravariance indicates the above flipping of directionality of the maps, i.e. $\Omega^*(f)$ goes from $\Omega^*(N)$ to $\Omega^*(M)$ rather than the other way around. More has to be true: $\Omega^*$ should respect compositions in that $\Omega^*(f\circ g)=\Omega^*(g)\circ \Omega^*(f)$ and have $\Omega^*(\Bbb{1}_M)=\Bbb{1}_{\Omega^*(M)}$. Anyway, so much for the formalities. The picture here is quite simple: we have specified $\Omega^*(M)$ explicitly already, and associated to $f:M\to N$ we define $\Omega^*(f)$ to be the morphism of chain complexes $f^*$ as
$\require{AMScd}$
\begin{CD}
    \Omega^0(N) @>d>> \Omega^1(N)@>d>>\cdots@>d>>\Omega^n(N)\\
    @V f^* V V @V f^*V V @VV V @V f^* VV\\
    \Omega^0(M) @>>d> \Omega^1(M)@>>d>\cdots@>>d>\Omega^n(M).
\end{CD}
Note that the diagram commutes by $f^*\circ d=d\circ f^*$. You can check the functoriality properties listed above.


问:
Geodesic equation for the motion of a particle
i know i can write the geodesic equation for a massive particle as:

\begin{equation}
\dot{x}^{\nu}\nabla_\nu \dot{x}^{\mu}=0
\end{equation}

and then we can express this using the 4 momentum, $ p^\mu = mu^\mu=m\dot{x}^{\mu}$,

\begin{equation}
p^{\nu}\nabla_\nu p^{\mu}=0
\end{equation}

I want to show that this can be written as

\begin{equation}
m\frac{dp_\mu}{d\tau}=\frac{1}{2}p^\sigma
p^\rho\partial_\mu g_{\rho \sigma}
\end{equation}

i expanded the covariant derivative out and lowered some of the indices using the metric to obtain that

\begin{equation}
g^{\mu\rho}p^\nu[\partial_\nu p_\rho-\Gamma^\alpha_{\nu \rho}p_\alpha]=0
\end{equation}

the first term in that expression can be written as $$g^{\mu \rho}m\frac{\partial p_\rho}{\partial \tau}$$
and i know i can get a 1/2 out of the christofell symbol but everything i try ends in a mess of metrics derivatives and indices.

I feel like im so close. Any help is much appreciated.

答:
Consider the term

$$ p^\nu \Gamma^\alpha_{\nu \rho}p_\alpha = p^\nu p_\alpha \frac{g^{\alpha \beta}}{2}
\left(  \partial_\nu g_{\rho \beta} + \partial_\rho g_{\nu\beta} -\partial_\beta g_{\nu\rho}\right)$$

$$ =\frac{1}{2}\left[ p^\nu p^\beta \partial_\nu g_{\rho \beta} + p^\nu p^\beta \partial_\rho g_{\nu\beta} - p^\nu p^\beta \partial_\beta g_{\nu\rho} \right]$$

Relabel the indices $ \nu \leftrightarrow \beta$ in the last term

$$ =\frac{1}{2}\left[ p^\nu p^\beta \partial_\nu g_{\rho \beta} + p^\nu p^\beta \partial_\rho g_{\nu\beta} - p^\nu p^\beta \partial_\nu g_{\beta\rho} \right]$$

The first and last terms cancel, leaving the term you want

$$ =\frac{1}{2}p^\nu p^\beta \partial_\rho g_{\nu\beta}
$$


问:
List of small groups
I was checking this website https://en.wikipedia.org/wiki/List_of_small_groups and wondering why don't we have $(\mathbb Z/5 \mathbb Z)^\times$ in the list of groups with 4 elements for example ?

$(\mathbb Z/5 \mathbb Z)^\times$ has 4 elements and is not in the list of groups with 4 elements and I do not know why ? Does it have to be isomorphic to $\mathbb Z/4 \mathbb Z$ ?
Same thing for $(\mathbb Z/7 \mathbb Z)^\times$, $(\mathbb Z/9 \mathbb Z)^\times$ and $(\mathbb Z/14 \mathbb Z)^\times$ in the groups of 6 elements.

答:
They list a representative up to isomorphism, and the group you are thinking of is already isomorphic to one of the ones on the list just represented differently. The cyclic group of order 4 in that particular case.

I will put the mapping to help:

$$1 -> 0$$
$$2 -> 1$$
$$4 -> 2$$
$$3 -> 3$$

Title: List of small groups


问:
Can continuous functions on Tychonoff spaces tell missing limit points apart?
Let $X$ be a Tychonoff space. Let $\{x_i\}$ and $\{y_i\}$ be nets that don’t necessarily converge in $X$, but for every continuous function $f:X\rightarrow [0,1]$, $\{f(x_i)\}\rightarrow a_f\in \mathbb{R}$ and  $\{f(y_i)\}\rightarrow b_f\in \mathbb{R}$. If the two nets are eventually ‘outside’ each other, is the existence of an f such that $a_f\neq b_f$ guaranteed?

答:
Let $\mu$ and $\nu$ be the nets in question; I find it a bit easier to work with filters than with nets, so let $\mathscr{F}$ and $\mathscr{G}$ be the corresponding filters on $X$. $\mathscr{F}$ and $\mathscr{G}$ are filter bases on $\beta X$, so they can be extended to ultrafilters $\mathscr{U}$ and $\mathscr{V}$, respectively, on $\beta X$. Your condition separating the nets presumably implies that there are $F\in\mathscr{F}$ and $G\in\mathscr{G}$ such that $F\cap G=\varnothing$, so $\mathscr{U}\ne\mathscr{V}$. $\beta X$ is compact, so $\mathscr{U}$ and $\mathscr{V}$ converge to distinct points $p$ and $q$, respectively, in $\beta X$.

If $f:X\to[0,1]$ is continuous, and $\beta f$ is its unique continuous extension to $\beta X$, then $a_f=(\beta f)(p)$ and $b_f=(\beta f)(q)$. $\beta X$ is Tikhonov, so there is a continuous $g:\beta X\to[0,1]$ such that $g(p)=0$ and $g(q)=1$, and if we set $f=g\upharpoonright X$, then $a_f\ne b_f$.


问:
how can i graph this function analytically? i can't properly use first derivative and second derivative helps me a bit but not enough
$y={\frac{x^3-4}{(x-1)^2}}$, so I have to graph this equation analytically, but when I use the first derivative to find the maximum, or minimum, I get the very bad cubic equation. so I can't properly find the intervals, where the function is increasing or decreasing etc.  but with second derivative I just find the point of inflection. so is there any good method to graph this equation.

答:
Hint.

What you have is a rational function. The first step to sketch such functions is to find the vertical asymptotes and horizontal asymptotes, if any.

The general procedure is outlined here: https://tutorial.math.lamar.edu/classes/alg/graphrationalfcns.aspx

This function can be written as
$$
f(x)=g(x)+\frac{3x-6}{(x-1)^2}=g(x)+\frac{3}{x-1}-\frac{3}{(x-1)^2}
$$
where $g(x)=x+2$.

Observe that
$$
\lim_{x\to +\infty}f(x)-g(x)=0,\quad \lim_{x\to -\infty}f(x)-g(x)=0
$$
and that
$$
\lim_{x\to 1-}f(x)=\lim_{x\to 1+}f(x)=-\infty.
$$


问:
example of a map in a prime ring
Let $R$ be a prime ring, and $f: R\mapsto R~$ an additive map such that
$$f(xy)=f(x)f(y)+f(x)y+xf(y)\quad \forall~x,y\in R $$
It is clear that if we take any endomorphism $g$ of R then $g-id_R$ verify the above formula.

My question is : can we provide another example ?

答: Hint: To show that given any such map $f$, it "comes from" such a ring endomorphism $g$, why not set $g := id_R + f$ and try to show  that $g$ a ring endomorphism of $R$?

Title: example of a map in a prime ring


问:
Disjoint union?
Consider $n$ sets $A_1,...,A_n$. Futhermore consider the set of indices $\{i_1,..,i_k\}\subset \{1,..,n\}$ and its complement  $\{j_1,..,j_m\}$with $m=n-k$. Thus
$$\bigcap_{l=1}^k A_{i_l} = \bigcup_{k_1,...,k_m\in \{1,c\}}\left(\bigcap_{l=1}^k A_{i_l} \cap \bigcap_{s=1}^m A^{k_j}_{j_s}\right) $$

with  $A^1=A$. $A^c$ refers to the complement of $A$.

Why should this be a disjoint union?

答:
For brevity let $B=\bigcap_{\ell=1}^kA_{i_\ell}$. Let $k_1,\ldots,k_m,k_1',\ldots,k_m'\in\{1,c\}$, and assume that there is some $t\in\{1,\ldots,m\}$ such that $k_t=1$ and $k_t'=c$; we want to show that

$$\left(B\cap\bigcap_{s=1}^mA_{j_s}^{k_j}\right)\cap\left(B\cap\bigcap_{s=1}^mA_{j_s}^{k_j'}\right)=\varnothing\,.\tag{1}$$

And

$$\begin{align*}
\left(\bigcap_{s=1}^mA_{j_s}^{k_{j_s}}\right)\cap\left(\bigcap_{s=1}^mA_{j_s}^{k_{j_s}'}\right)&=\bigcap_{s=1}^m\left(A_{j_s}^{k_{j_s}}\cap A_{j_s}^{k_{j_s}'}\right)\\
&\subseteq A_t^{k_{j_t}}\cap A_t^{k_{j_t}'}\\
&=A_t\cap A_t^c\\
&=\varnothing\,,
\end{align*}$$

from which $(1)$ follows immediately.

Title: Disjoint union?


问:
Prove or disprove that if $A\setminus B = A\setminus C$ then $A \cap B = A \cap C$
I've been having trouble solving this problem and I have no clue where to go at this point. If anyone could help me out and explain along the way I'd appreciate it greatly. </p>

<p>Let $A,B,C$ be a sets.  Supppose that $A\setminus B = A\setminus C$, then $A \cap B = A \cap C$. Prove or disprove with a counterexample. </p>

答:
Expanding $A \setminus B = A \setminus C$ using the definition $$x \in A \setminus B \;\equiv\; x \in A \land \lnot(x \in B)$$ and then simplifying results in
$$
\begin{align}
& A \setminus B = A \setminus C \\
\equiv & \;\;\;\;\;\text{"extensionality"} \\
& \langle \forall x :: x \in A \setminus B \equiv x \in A \setminus C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\setminus$, twice"} \\
& \langle \forall x :: x \in A \land \lnot(x \in B) \equiv x \in A \land \lnot(x \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: move common conjunct out of $\equiv$"} \\
& \langle \forall x :: x \in A \Rightarrow (\lnot(x \in B) \equiv \lnot(x \in C)) \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify by negating both sides of $\equiv$"} \\
& \langle \forall x :: x \in A \Rightarrow (x \in B \equiv x \in C) \rangle \\
\end{align}
$$

Now, in the same way expand $A \cap B = A \cap C$ using the definition $$x \in A \cap B \;\equiv\; x \in A \land x \in B$$ and compare the results.


问:
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$
I came across this integral

$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$

According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.

A promising way might be the following

\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
 &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3}   \right )^2 + \frac{8}{3}}} \\
 &= \cdots
\end{align*}

A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )

So, any ideas how to evaluate it?

P.S: Are there techniques available for these type of problems?

答:
This is an elliptic integral of the first kind, so there is no analytical way to evaluate it.

More on Elliptical Integrals: https://en.wikipedia.org/wiki/Elliptic_integral

In any case, we might try to give your integral a numerical result because it's bounded between $0$ and $1$.

We can expand the whole integrand in Taylor-MacLaurin series. Notice that the very first term of the expansion would just be $f(0) = \frac{1}{\sqrt{3}}$, so:

Order zero

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}}$$

Hence

$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} \approx 0.57735(...)$$

Let's go on with another term.

Second Order

Going on with the expansion we find:

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} + O(x^4)$$

Whence

$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\int_0^1 x^2\ \text{d}x = \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\cdot \frac{1}{3}  = \frac{8}{9 \sqrt{3}} \approx 0.5132(...)$$

We can go on with the terms.

Numerical Exact Result

Via the help of W. Mathematica, the numerical inteegration gives the result

$$0.500539(...)$$

Which is rather near to our last result, as you can see.

Let's do better.

Fourth Order

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} - \frac{x^4}{3 \sqrt{3}} + O\left(x^6\right)$$

So again

$$\int_0^1 f(x)\ \text{d}x \approx \frac{8}{9 \sqrt{3}} - \int_0^1 \frac{x^4}{3 \sqrt{3}} = \frac{37}{45 \sqrt{3}} \approx 0.47471(...) $$

It starts to oscillate around $0.500$. You can go on with the terms, what you will have to integrate are just polynomials.

A Big Order

Say we go on to order ten:

$$\frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}+O\left(x^{12}\right)$$

Whence

$$\int_0^1 f(x)\ \text{d}x \approx \int_0^1 \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}\ \text{d}x = \frac{238984}{280665 \sqrt{3}} \approx 0.491609(...)$$

Got it?


问:
$(X,d)$ is metric space. $(X,d)$ is compact if and only if any continuous function on $X$ has a maximum.
$(X,d)$ is metric space. $(X,d)$ is compact if and only if any continuous function on $X$ has a maximum.

I dont know whether these functions real valued or not but only real valued functions may make sense, I think.

In that case

$\Rightarrow$ is easy.

About $\Leftarrow $

My 1st attempt:

I thought some special functions that I can use e.g. $$\varphi:X\to \mathbb R\\ \varphi(x)=d(x_0,x)$$ for some fix $x_0\in X$

Since this is a continuous function then it attains its maximum on $X$, that implies $X$ is bounded .

With this motivation I can define $$\psi:X\to \mathbb R \\\psi(x)=d(x,X)=\inf\limits_{a\in X}d(x,a)$$

since it attains its maximum so $X=\overline X$

So I have boundness and closedness but these dont really imply compactness in metric spaces.

My 2nd attempt was:

Considering any sequences in $X$ and using $f(x_{k_n})$ is convergent iff $x_{k_n}$ is convergent. By showing every sequence has convergent subsequence and using continuity of any $f:X\to\mathbb R$.

I am stuck is there hint or answer you can give me?

答:
HINT: Prove the contrapositive: assume that $X$ is not compact, and construct a continuous real-valued function that does not attain a maximum. Perhaps the simplest approach is to show that if $X$ is not compact, it has a countably infinite closed discrete subset $D=\{x_n:n\in\Bbb N\}$. Define $f:D\to\Bbb N:x_n\mapsto n$ and apply the Tietze extension theorem.

Added. It’s more work, but you can actually avoid the Tietze extension theorem. Every subset of $D$ is closed, and $X$ is normal, so we can carry out the following construction. There are open sets $U_0$ and $V_0$ such that $x_0\in U_0$, $D\setminus\{x_0\}\subseteq V_0$, and $\operatorname{cl}U_0\cap\operatorname{cl}V_0=\varnothing$. Suppose that $n\in\Bbb N$, and we already have an open set $V_n$ containing every $x_k$ with $k>n$; then there are open subsets $U_{n+1}$ and $V_{n+1}$ of $V_n$ such that $x_{n+1}\in U_{n+1}$, $x_k\in V_{n+1}$ for each $k>n+1$, and $\operatorname{cl}U_{n+1}\cap\operatorname{cl}V_{n+1}=\varnothing$. In this way we recursively construct open sets $U_n$ with pairwise disjoint closures such that $x_n\in U_n$ for each $n\in\Bbb N$.

Let

$$\mathscr{U}=\left\{X\setminus\bigcup_{n\in\Bbb N}U_n\right\}\cup\{U_n:n\in\Bbb N\}\,;$$

$\mathscr{U}$ is an open cover of $X$. $X$ is a metric space, so it is paracompact, and $\mathscr{U}$ therefore has a locally finite open refinement $\mathscr{R}$. For each $R\in\mathscr{R}$ there is a $U_R\in\mathscr{U}$ such that $R\subseteq U_R$. For each $U\in\mathscr{U}$ let $V_U=\bigcup\{R\in\mathscr{R}:U_R=U\}$, and let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. It is not hard to show that $\mathscr{V}$ is locally finite, and it is clearly an open refinement of $\mathscr{U}$, since $V_U\subseteq U$ for each $U\in\mathscr{U}$.

For $n\in\Bbb N$ let $V_n=V_{U_n}$. Note that $U_n$ is the only member of $\mathscr{U}$ containing $x_n$, so $x_n\in V_n$. $X$ is completely regular, so for each $n\in\Bbb N$ there is a continuous function $f_n:X\to[0,1]$ such that $f(x_n)=1$ and $f_n[X\setminus V_n]=\{0\}$. For $x\in X$ let

$$f(x)=\sum_{n\in\Bbb N}nf_n(x)\,;$$

clearly $f$ is a function from $X$ to the non-negative reals, and $f(x_n)=n$ for each $n\in\Bbb N$, so $f$ is unbounded above. It’s also clear that $f\upharpoonright V_n=f_n$ for each $n\in\Bbb N$, so $f$ is continuous at each point of $\bigcup_{n\in\Bbb N}V_n$.

Finally, let $x\in X\setminus\bigcup_{n\in\Bbb N}V_n$; clearly $f(x)=0$. If $x$ has an open nbhd $W$ disjoint from $\bigcup_{n\in\Bbb N}V_n$, then $f[W]=\{0\}$, and $f$ is continuous at $x$. If not, then $x\in\operatorname{cl}\bigcup_{n\in\Bbb N}V_n$. $\mathscr{V}$ is locally finite, so $x$ has an open nbhd $W$ such that $F=\{n\in\Bbb N:W\cap V_n\ne\varnothing\}$ is finite. Then

$$x\in\operatorname{cl}\bigcup_{n\in F}V_n=\bigcup_{n\in F}\operatorname{cl}V_n\,,$$

and the sets $\operatorname{cl}V_n$ are pairwise disjoint, so there is a unique $n(x)\in\Bbb N$ such that $x\in\operatorname{cl}V_{n(x)}$. I’ll leave the very last step to you: use this fact to show that $f$ is continuous at $x$, thereby completing the proof that $f$ is continuous.


问:
Parametric equation of an ellipse in the 3D space
I have found here that an ellipse in the 3D space can be expressed parametrically by

$$\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$$

with $\mathbf c = (c_1,c_2,c_3)$ being the center of the ellipse and the lenghts of the half-axis being the lengths of the vectors $\mathbf u = (u_1,u_2,u_3)$ and $\mathbf v = (v_1,v_2,v_3)$.

How could these three vectors $\mathbf c$, $\mathbf u$ and $\mathbf v$  be related to the directions of the axis of the ellipse? Is there maybe a parametric equation for the ellipse in an arbitrary plane of the space whose elements have a more intuitive meaning?

答:
In the parametric equation $\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$, we have: $\mathbf c$ is the center of the ellipse, $\mathbf u$
is the vector from the center of the ellipse to a point on the ellipse with maximum curvature, and
$\mathbf v$ is the vector from the center of the ellipse to a point with minimum curvature.  I assumed $\|u\| > \|v\|$.


答:
Remember that the most general form of the solution of the 1D wave equation is
$$u(x,t)=f(x-ct)+g(x+ct)$$
So,
$$\partial_x u(x,t)=f'(x-ct)+g'(x+ct)$$
$$\partial_x u(0,t)=f'(-ct)+g'(ct)=e^t$$
Since we still have two unknown functions $f,g$ your condition is not yet enough to determine a unique solution.

Title: How to transform a non-homogeneous Neumann boundary condition into a homogeneous for the wave equation?

Body:
I have a initial/ boundary value problem for standard wave equation
$$
\frac{\partial^2u}{\partial t^2}=c\frac{\partial^2u}{\partial x^2},
$$
where one of the boundary conditions is non-homogeneous: precisely $$
\frac{\partial u(0,t)}{\partial x}=e^t.
$$
Question: is it possible to transform the above problem into a one with homogeneous boundary conditions?

In the Q&A "Solve wave equation and inhomogeneous Neumann Condition with eigenfunction expansion (Fourier Series Solution)" I saw a similar reduction, but I am unable to apply the method proposed there to my problem.


问:
Characteristic polynomial of a A agrees with its minimal polynomial if and only if all matrices that commutes with A is a polynomial of A

Let $A \in \mathcal{M}_n (\mathbb{C})$. Over $\mathbb{C}$, show that the following two statements are equivalent:

Characteristic polynomial of a $A$ agrees with its minimal polynomial;

All matrices that commutes with $A$ is a polynomial of $A$.

For 1 > 2, from 1 I can only get that $A$ has a Jordan canonical form that each eigenvalue has only one Jordan block (i.e. maximum size), and I have no idea afterwards.

2 > 1 is a complete no-go for me.

Any hints or solutions are appreciated.

答:
$(2)\to (1)$ Suppose for the sake of contradiction that (1) does not hold. Let $J=T^{-1}AT$ be a Jordan canonical form of $A$ and $T$ be a non-singular matrix. By Claim from Ben Grossmann’s answer, $J$ has distinct Jordan cells $J_1$ and $J_2$ corresponding to the same eigenvalue $\lambda$. Let $R_k$ be the set of rows occupied by $J_k$ for $k=1,2$. Let $D=\|d_{ij}\|$ be a diagonal matrix such that $d_{ii}$ equals $1$, if $i\in R_1$, and equals $0$, otherwise. Let $B=\|b_{ij}\|$ be any polynomial of $J$. Then $b_{ii}=b_{jj}$ for any $i\in R_1$ and $j\in R_2$, so $D$ is not a polynomial of $J$. On the other hand, it is easy to check that $DJ=JD$. Then a matrix $TDT^{-1}$ commutes with $A$, but $TDT^{-1}$ is not a polynomial of $A$, a contradiction.

$(1)\to (2)$ Gerry Myerson proved here that it suffices to show that there exists a vector $v$ such that vectors
$A^0v,A^1v,\dots,A^{n-1}v$ are linearly independent. Let $$\chi(x) = (x - \lambda_1)^{d_1} \cdots (x - \lambda_k)^{d_k}$$  be the
characteristic polynomial of $A$ and $\lambda_i$’s are distinct. For each $i=1,\dots, k$ put $\chi_i(x)=\chi(x)/(x-\lambda_i)$ and
$\bar\chi_i(x)=\chi(x)/(x-\lambda_i)^{d_i}$. Since $\chi_i(x)$ is not a minimal polynomial of $A$, there exists a vector $w_i$ such that $\chi_i(A)w_i\ne 0$. Let $\varphi_i(x)$ be a polynomial of minimal degree such that $\varphi_i(A)w_i=0$.  The minimality of degree of $\varphi_i(x)$ easily implies that $\varphi_i(x)|\chi(x)$. Put $v_i=\bar\chi_i(A)w_i$ and $v=v_1+\dots+v_k$. Suppose for the sake of contradiction that there exists a polynomial $\psi(x)$ with minimal degree $\deg \psi(x) <\deg \chi(x)$  such that $\psi(A)v=0$. The minimality of degree of $\psi(x)$ easily implies that $\psi(x)|\chi(x)$. Since $\deg\psi(x)<\deg\chi(x)$, there exists $i=1,\dots, k$ such that $\psi(x)|\chi_i(x)$. Then $$0=\chi_i(A)v=\chi_i(A)v_i=\chi_i(A)\bar\chi_i(A)w_i.$$ The minimality of degree of $\varphi_i(x)$ easily implies that $\varphi_i(x)| \chi_i(x)\bar\chi_i(x)$. Since $\varphi_i(x)|\chi(x)$, we have that
$\varphi_i(x)|(\chi_i(x)\bar\chi_i(x),\chi(x))=\chi_i(x)$. But then $\chi_i(A)w_i=\varphi_i(A)w_i=0$, a contradiction.


问:
Is $n = 0$ in proofs involving mathematical induction a rigorous expression?
I have had this confusion from high school. In proofs involving mathematical induction, we always say for $n = 0$, blablabla, so that a certain condition holds for $n = 0$. After I learned logic, I found "$n = 0$" may not be a rigorous way of expression. It is in fact a replacement of $n$ by $0$ in $Q\left(n\right)$, where $Q$ is an arbitrary predicate, and we want to prove $\left(\forall n \in \mathbb{N}\right)Q\left(n\right)$. Replacement does not indicate any equality at all. And assume that you would like to finish the proof by asserting $Q\left(n+1\right)$ from $Q\left(n\right)$, then everyone would agree that "for $n = n+1$" is a ridiculous expression. I am wondering why people are still insisting on using the type of expression "$n = 0$" in proofs without explanation in a book at an introductory level. Especially in classes, wouldn't it cause confusions to students who do not understand logic?

答:
I don't find anything wrong with the expression $n=0$.

Let's say you want to prove that for all nonnegative integer $n$, the statement $Q(n)$ is true.

In the base step you first show that when $n=0$, $Q(n)$ is true, i.e., $Q(0)$ is true.

In the induction step, you then show that if $Q(n)$ is true, then $Q(n+1)$ is also true.

"for $n=n+1$" is nonsense does not imply that "$n=0$" is evil.


问:
Proving $g:P\to\Bbb R, g(x)=3x+2$ is discontinuous

Define $g:P\to\Bbb R,g(x)=3x+2$ where $P$ is the Cantor set. Show that $g$ can't be continuous under Euclidean metric in $\Bbb R$.

No information is given about the metric on $P$. I assumed it is the Euclidean metric. Then $g^{-1}((a,b))=\left(\frac{a-2}3,\frac{b-2}3\right)\cap P$ which is open in $P$. Then this map is continuous because the inverse image of every open set is open. In fact it seems $g$ will be continuous on any subspace of $\Bbb R$ since it is continuous on $\Bbb R$. Am I doing something wrong?

答:
Theorem If $A\subset X$ and $f\colon X\to Y$ is continuous function, then $f|A\colon A\to Y$ is continuous function.

Now, $g\colon \Bbb R\to \Bbb R$  and $g(x)=3x+2$ is definitely continuous function. Put, $$f=g|P \ \ \ \text{for all } \  \ \ x\in P$$ So, $f\colon P\to \Bbb R$ where $f(x)=3x+2$ is continuous function by Theorem.


问:
Prove that $\Phi$ is surjective. $\Phi: \ ^C(^B A) \to\ ^{C \times B}A $ where $\Phi (f)(c,b) = \big(f(c)\big)(b)$
I would like to prove that the following function...

$\Phi: \ ^C(^B A) \to\   ^{C \times B}A $ where $\Phi (f)(c,b) = \big(f(c)\big)(b)$

...is surjective. I don't believe this is useful for the task at hand, but I have proven that $\Phi$ is injective.

Any confusion with notation can be addressed here: Notation used in the proof of $^C(^B A) \approx \ ^{C \times B}A$: $\ \ \ \Phi (f)(c,b)$ and $\big(f(c)\big)(b)$

Here is what I have so far, but I am concerned that there may be a leap in my logic...or at least an unacceptable degree of informality:

Claim to Prove - $\forall g \in \  ^{C \times B}A \Big [ \exists f \in \  ^C(^B A) \ \big [\forall c \in C, b \in B \big (g(c,b)=f(c)(b) \big) \big] \Big ]$

Consider an arbitray $g' \in \  ^{C \times B}A$.

$g' \in \  ^{C \times B}A \rightarrow \forall c \in C, b \in B \Big(
\exists ! a \in A \big(g'(c,b)=a\big) \Big)$

Consider $c' \in C$ and $b' \in B$. Then we know that $g'(c',b')=a'$.

There are two families of functions relevant for this next part:

i) functions of the form... $\quad f: C \to \ ^B A$ where $\quad f \in \ ^C(^B A)$

ii) functions of the form... $\quad h: B \to \ A$ where $\quad h \in \  ^B A$

Note that for $c\in C$, $f(c)$ necessarily maps to a function of the "ii" form.

By definition of $^B A$, there exists at least one $h \in \  ^B A$ that maps $b'$ to $a'$. In other words, $\exists h  \in \  ^B A \Big (h(b')=a' \Big)$. Call one such function $h'$.

By definition of $^C(^B A)$, there exists at least one function in $^C(^B A)$ that maps $c'$ to $h'$. In other words, $\exists f  \in \  ^C(^B A) \Big (f(c')=h' \Big)$. Call one such function $f'$.

Therefore, we can see that the following is true: $f'(c')(b')=h'(b')=a'=g'(c',b')$.

I will refer to this block as Procedure 1

Now, the issue that I believe I am running into is that I have only demonstrated that a function $f$ exists that can "handle" a single arbitrary pair of elements from $C$ and $B$. However, what I need to show is that there exists an $f$ that can handle all pairs of elements from $C$ and $B$.

Essentially, I believe I proved:

$\forall g \in \  ^{C \times B}A \Big [  \forall c \in C, b \in B \big [\exists f \in \  ^C(^B A) \Big(g(c,b)=f(c)(b) \Big ) \ \big] \Big]$

Where I am stuck is whether or not the immediately above statement implies the surjective claim because of the definition of $ ^C(^B A)$. If that implication is true, I am unsure of how to formally express it.

Added thoughts:

It seems like for a given $g'$, I could repeat Procedure 1 as many times as necessary using all of the different combinations of elements from $C$ and $B$. Necessarily, this would create subsets of functions from $ ^C(^B A)$ that satisfy the criteria used in Procedure 1.

For example, I would have the following sets:

$\{\text{functions in } ^C(^B A) \text{ that satisfy } c_1, b_1\}, \{\text{functions in } ^C(^B A) \text{ that satisfy } c_1, b_2\},...,\{\text{functions in } ^C(^B A) \text{ that satisfy } c_2, b_1\},...$

It seems like I could take the intersection of this family of sets and if I could prove that the resulting set is non-empty, then I would demonstrate that there is a function that satisfies the necessary criteria of $\exists f \in \ ^C(^B A) \Big [ \forall c \in C, b \in B \big (g'(c,b)=f(c)(b) \big) \Big ]$

答:
The idea is correct, but I think your proof fails on the moment you just say $\exists h\in ^B\! A$ such that $h(b')=a'$, somehow you are missing a lot information.
Take $g\in ^{C\times B}\!\!A$ and define $f\in ^C\!\!(^B\! A)$ as $f(c):=g(c,-)\in ^B \!\!A$ for all $c\in C$ (in other words, for each $c\in C$, $f(c)$ is the function in $^BA$ that maps every $b\in B$ to $g(c,b)$). Then, by definition, $\Phi(f)(c,b)=f(c)(b)=g(c,b)$ for all $(c,b)\in C\times B$.

Two functions are equal if they map same elements to same elements, and they have same range and domain, so $ \Phi(h)=g$. Then $\Phi$ is surjective.

Note that your question is almost trivial and by definition if you don't overthink it, otherwise can seem very difficult to proof. Also note that you must start your reasoning on the "highest set" (in this case $^C(^B A)$), otherwise you won't be able to puzzle all your pieces from down to up.


问:
Why is the set of well formed formulas are defined as the smallest set of strings
Consider this definition of Well formed formulas in mathematical logic -
The set of all well formed formula is the smallest set of strings , WFF that satisfies

All Boolean variables are in WFF, and so are the symbols Τ and F. We call
such formulae atomic.

If A and Β are any strings in WFF, then so are the strings (not A), (A and B),
(A V β), (A —> β), (A = B)

Wouldn't there be exactly one such set of strings satisfying these properties of WFF ?

I don't understand the need for using the word smallest.

答:
Consider first a simpler example:

The set of even numbers $E$ is the smallest $X\subseteq \mathbb{N}$ with the following two properties:

$0\in X$, and

if $n\in X$ then $n+2\in X$.

The two bulletpoints alone do not pin down $E$! For example, both $\mathbb{N}$ itself and $E\cup\{n\in\mathbb{N}: n\ge 17\}$ satisfy them. Basically, the minimality clause is required to make sure that no "unintended" elements enter the set we're defining.

Turning back to the example in the question, note that the set of all finite strings of symbols satisfies points $1$ and $2$ of your definition; it's only the minimality clause that tells us that that's not what we have in mind. See also here.


问:
Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$
I'm practicing for my maths term test mainly on binomial coefficients. I can't seem to find out how to prove the following identity. Any advice?

$$ \sum\limits_{k=1}^n (-1)^{k+1} k{{n}\choose k} = 0 $$

Thanks in advance.

答:
Hint: Note that for $x\ge 1$, $x\dbinom{n}{x}=n\dbinom{n-1}{x-1}$. This can be verified by using the representation of the binomial coefficients in terms of factorials.

We can bring the $n$ to the front, and end up with something that is perhaps familiar. If it is not familiar, expand $(1-1)^m$ using the Binomial Theorem.


问:
Trace and Jordan normal form of Block matrix
Given the Matrix $A \in \mathbb{R}^{2\times2}$ with $Spec(A) = \{300,333\}$ I have the Block matrix

$M = \begin{bmatrix}A&A\\A&A\end{bmatrix} \in \mathbb{R}^{4\times4}$ for which I have to determine the Trace and Jordan normal form.

My attempt (not formally written but roughly the plan) :

Considering that $M$ consist of four blocks the same Matrix $A$ and therefore of equally partitioned blocks we could easily deduce that $tr(M) = tr(A) + tr(A)$. Because a matrix's trace is the sum of its eigenvalues we have that $tr(A) = 300 + 333 = 633 \Rightarrow tr(M) = 633 + 633 = 1266$

We clearly have at least one row identical to another row which means the $det(M) = 0$ and because the determinant is the product of its eigenvalues we could deduce that at least one of the eigenvalues is $0$.

If I conpute the characteristic polynomial mechanically I get:$\mathcal{p}_M=-\frac{λ^2\left(-λ^3d-λ^2bc+2λ^2d^2+3λ^2ad+2λabc-6λad^2-2λa^2d+6λbcd+4b^2c^2+4a^2d^2-8abcd\right)}{bc-d\left(a-λ\right)}$ for $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ which is very ineffective but I can at least deduce that the algebraic multiplicity of the eigenvalue $0$ is at least $2$ which tells me that the sum of the sizes of all Jordan blocks corresponding to the eigenvalue $0$ is at least $2$.

I am unsure how to proceed past this point. Any reminder about any theorems I'm clearly not remembering would be greatly appreciated!

答:
We already know that $A$ has zero as an eigenvalue. Moreover, because the rank of $M$ is $2$, the geometric multiplicty of the zero-eigenvalue is $4 - 2 = 2$. We just need to find the other eigenvalues.

One approach is to try to go directly for the eigenvectors of $M$. Suppose that $v,w \in \Bbb R^2$, and let $(v,w)$ denote the corresponding "block vector" in $\Bbb R^4$. In order for $v,w$ to be an eigenvector of $M$ associated with $\lambda \neq 0$, we need to have
$$
\pmatrix{A & A\\ A & A} \pmatrix{v\\w} = \lambda \pmatrix{v\\w} \implies \begin{cases}Av + Aw = \lambda v\\ Av + Aw = \lambda w\end{cases}.
$$
We see that $\lambda v = Av + Aw = \lambda w$, and because $\lambda \neq 0$ it must be the case that $v = w$. With that substitution, both equations simply become
$$
Av + Av = \lambda v \implies Av = \frac{\lambda}{2}v.
$$
In other words, there is an eigenvalue $\mu$ (non-zero) of $A$ for which $\mu = \frac{\lambda}{2} \implies \lambda = 2 \mu$.

We know that $A$ has the non-zero eigenvalues $300,333$. It follows that $M$ has the eigenvalues $2 \cdot 300 = 600$ and $2 \cdot 333 = 666$.

Now, $M$ has a two dimensional eigenspace associated with zero and two other eigenvalues, which means that the dimensions of the eigenspaces adds up to $4$. It follows that $M$ is diagonalizable.


问:
Least-Squares Solution that Minimizes Two Expressions
I am trying to solve for $B_k$ in the following equation, as the function of two separate least-squares expressions:

$$
B_k = \arg \min ||X_k - B_kD_kA^T||^2 + \mu_k||B_k - P_kB^*||^2
$$

This is part of a larger algorithm for non-negative PARAFAC2 that I am trying to implement, and is applicable for 3rd order tensors. However, for the purposes of this expression we are only considering one slice of the tensor.

I understand that to solve for $B_k$, we would set $X_k = B_kD_kA^T$ and take the right inverse of $D_kA^T$ to isolate $B_k$. However, I am unsure of how to set up the expression such that $B_k$ is the least-squares solution to both of the expressions. It must be solved  using non-negative least squares, as per the article.

Link to the paper

答:
After some more research, I think that I have come up with a solution to the problem. First, take the derivative with respect to $B_k$, and set it to 0.

$$
\frac{\partial}{\partial B_{k}}(||X_k - B_kD_kA^T||^2 + \mu_k||B_k-P_kB^*||^2) = 0
$$

Understanding that the sum of squared residuals can also be described as the square of the Frobenius Norm of the difference between $X_k$ and $B_kD_kA^T$, or $B_k$ and $P_kB^*$ via:

$$
||A||_F = \sqrt{\sum_{i=1}^m\sum_{j=1}^n|a_{ij}|^2} = \sqrt{tr(A^TA)} = \sqrt{tr(AA^T)}
$$

We can rearrange the original equation, adding an arbitrary constant, $\frac{1}{2}$, to aid with simplification later on.

$$
\frac{\partial}{\partial B_{k}}\frac{1}{2}tr((X_k - B_kD_kA^T)(X_k - B_kD_kA^T)^T) + \frac{\partial}{\partial B_{k}}\frac{1}{2}\mu_k(tr((B_k - P_kB^*)(B_k-P_kB^*)^T) = 0
$$

Expanding the equations, where $(X_k - B_kD_kA^T)^T = X_k^T - AD_kB_k^T$ and $(B_k - P_kB^*)^T = B_k^T - B^{*T}P_k^T$:

$$
\frac{\partial}{\partial B_k}\frac{1}{2}(tr(X_kX_k^T) -tr(X_kAD_kB_k^T) - tr(B_kD_kA^TX_k^T) + tr(B_kD_kA^TAD_kB_k^T)) +\frac{\partial}{\partial B_k}\frac{1}{2}\mu_k(tr(B_kB_k^T)-tr(B_kB^{*T}P_k^T) - tr(P_kB^*B_k^T) + tr(P_kB^*B^{*T}P_k^T)) = 0
$$

This equation can be simplified by using some convenient identities from the Matrix Cookbook:
$$
\frac{\partial}{\partial{B_k}}tr(X_kX_k^T) = 0
$$

$$
\frac{\partial}{\partial{B_k}}tr(X_kAD_kB_k^T) = X_kAD_k
$$

$$
\frac{\partial}{\partial{B_k}}tr(B_kD_kA^T) = X_kAD_k
$$

$$
\frac{\partial}{\partial{B_k}}tr(B_kD_kA^T) = X_kAD_k
$$

$$
\frac{\partial}{\partial{B_k}}tr(B_kD_kA^TAD_kB_k^T) = 2B_kD_kA^TAD_k
$$

$$
\frac{\partial}{\partial{B_k}}tr(B_kB^{*T}P_k^T) = PkB^*
$$

$$
\frac{\partial}{\partial{B_k}}tr(P_kB^*B_k^T) = P_kB^*
$$

$$
\frac{\partial}{\partial{B_k}}tr(P_kB^*B^{*T}P_k^T) = 0
$$

$$
\frac{\partial}{\partial{B_k}}tr(B_kB_k^T) = 2B_k
$$

Substituting into the previous equation yields:

$$
-X_kAD_k + B_kD_kA^TAD_k + \mu_kB_k - \mu_kP_kB^* = 0
$$

Multiplying by the inverse of $D_kA^TAD_k$:

$$
-X_kAD_k(D_kA^TAD_k)^{-1} + B_k + \mu_kB_k(D_kA^TAD_k)^{-1} - \mu_kP_kB^*(D_kA^TAD_k)^{-1} = 0
$$

Solving for $B_k$:

$$
-X_kAD_k + \mu_kB_k - \mu_kP_kB^* = -B_k(D_kA^TAD_k)
$$

$$
-X_kAD_k - \mu_kP_kB^* = -B_k(D_kA^TAD_k) - \mu_kB_k
$$

Yields the final form of the equation:
$$
B_k = \frac{X_kAD_k + \mu_kP_kB^*}{D_kA^TAD_k + \mu_kI_R}
$$

Although the notation used differs from the notation used in the paper (i.e. $B_k$ is described as belonging to the first mode and not the second as was described in the paper), the solution is identical to the solution found in the published code.


问:
Algorithm to find the exact roots of solvable high-order polynomials?
It is not generally possible to determine the roots of a polynomial whose grade is bigger than 4 in terms of roots and basic operations. But I heard that it is possible to give a criteria whether a polynomial has such a solution.

For instance Wikipedia tells that the roots of a polynomial of degree $5$ are expressible in terms of roots and basic operations, if it is representable in the form

$$x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0,$$

where $\mu$ and $\nu$ are rational numbers.

Given the case that the polynomials roots are expressible in such a form, is it possible to give an algorithm that computes the solution in this form?

I am not an expert on that topic, just a student who is interested in maths.

答: Yes, it's called Galois Theory.</p>

Body: Yes, it's called Galois Theory.