答:
Well, minimally,  your solution doesn't account for the fact that there are twice as many executives as there are workers.

Knowing that fact ($2W = E), and knowing that there are 60 employees, we know that

$$ W + E = 60 \\
W + 2W = 60 \\
3W = 60 \\
W = 20 $$

, where $W$ is the number of worker bees, so, there are $E = 2W = 40$ lazy tie-wearers.

Okay.  From the first fact, the average salary across 60 employees is \$12,000, so, the total money spent on salaries each month is $\$12{,}000 * 60 = \$720{,}000$.

As you did, we will let the basic salary unit be $x$.  Worker bees get 2 salary units ($2x$), and tie-wearers get 5 ($5x$).

Remembering that there are 40 execs and 20 workers, we can account for all the salary units:

$$ 20 \times 2x + 40 \times 5x = \$720{,}000 \\
40x + 200x = \$720{,}000 \\
240x = \$720{,}000 \\
x = $3{,}000 $$

So, \$3,000 is the basic salary unit.  Each worker makes 2 salary units, or \$6,000, and each exec pulls in \$15,000.  Each month.  \$180,000/yr.

Listen, that's just the way that I saw the problem.  If you saw it another way, and are now aware of your counting mistake, a piece of the puzzle that you didn't account for, great.  Work it out until it makes sense.

It's not that there's one right way to do it, or that you know the "best" way.  It's that you can find a way, and you can check that your answer is correct.

Title: Why can't I use ratio in this ??

Body:
The Question:

Average salary of 60 employees is 12000 per month.

Number of executives is twice the no of non-executives.

If average salary of non-executives be 2/5th of the average salary of
executives then what is the average salary of non-executive employees
??

Answer is 6000.

I want to know why this is incorrect:

salaries in ratio =2:5

then  2x+5x=12000

x=12000/7

Please help me !!
And if anything is wrong let me know,

Thank you .


问:
How to understand the object in a category
Am a real beginner in category theory... I read that a category can be defined as a $4$-tuple:

$$
\textbf{C}=(\text{Ob}(\textbf{C}),\text{Ar}(\textbf{C}),\mu(\textbf{C}),\text{id}(\textbf{C}))
$$

where $\text{Ob}(\textbf{C})$ is any collection of elements, $\text{Ar}(\textbf{C})$ is a family of sets $\textbf{C}(X,Y)$ indexed by pairs of elements of $\text{Ob}(\textbf{C})$, and $\text{id}(\textbf{C})$ are in $\textbf{C}(X,X)$ that satisfy the identity laws ...

But am still trying to understand $\text{Ob}(\textbf{C})$ ... I read that given a concrete group $G$, with the category of $G$ as $G_{cat}$, we have that $\text{Ob}(G_{cat})$ has only one member (which can be indexed as $\{0\}$) ... However, this member does not consist of $G$, but rather is a fictitious mathematical object on which $G$ acts...

The description above sounds very nebulous ... Does this mean that the single object ($\{0\}$) can simply be any set such that the $G$ can act on it ? In this case, can an infinite set $S$ isomorphic to $\mathbb{N}$ be considered also as "a specific case of" $(\{0\})$?

答:
Forget categories for a moment, and think about some more familiar kind of algebraic structures — e.g. groups themselves.

Take, for instance, the cyclic group $C_5$ (just as a group, not as a category).  It has 5 elements.  What are those 5 elements?  Depending on what construction you use, they might be certain rotations of $\newcommand{\R}{\mathbb{R}}\newcommand{\C}{\mathbb{C}}\R^2$, or certain roots of unity in $\C$, or perhaps equivalence classes of integers modulo 5.  But it doesn’t really matter what the elements are individually — what matters is how many of them there are, and what the group operation does on them.  Precisely, I’ve described three different groups, but they’re all canonically isomorphic to each other — so we think of them as three different constructions of essentially the same group $C_5$, and when we say $C_5$, it doesn’t matter which specific construction we have in mind.  This is what’s abstract in abstract algebra: it doesn’t matter what the elements of the group are, what matters is the structure they form.

Now, back to categories.  The key point is: A category itself is an abstract algebraic structure.  It’s not important what the individual elements of $\mathrm{ob}(\mathbf{C})$ are; what matters is the structure they form, up to isomorphism.*

In particular, viewing a group as a one-object category: it doesn’t matter what the “one object” of the category is.  $\mathrm{ob}(G_{\mathrm{cat}})$ can be any one-element set.  If you choose $\mathrm{ob}(G_{\mathrm{cat}}) = \{G\}$, and I choose $\mathrm{ob}(G_{\mathrm{cat}}) = \{ \emptyset \}$, and someone else chooses $\mathrm{ob}(G_{\mathrm{cat}}) = \{ 5i + 8 \}$, it doesn’t matter: the categories we get will all be isomorphic!  (With categories, we are often even a bit looser than “isomorphism” — we consider categories essentially the same as long as they’re equivalent, and we expect/require all categorical constructions to respect equivalence of categories.  But isomorphism is certainly strong enough to justify considering two categories as the same.)

ContentLicense: CC BY-SA 4.0


问:
Complex number quadratic with conjugation
I have the following equation:
$$
3z^2+z^*-1=0
$$
z is a complex number and $z^*$ is its conjugate. I'm stuck solving it. There's what I did so far:
$$
z = a+bi
$$
$$
3a^2+6abi-3b^2+a+bi=1
$$
Assuming $Im(z) = 0$ I'm getting following equations:
$$
\begin{cases}
3a^2-3b^2+a=1\\
6abi-bi=0
\end{cases}
$$
$$
bi(6a-1)=0=>b=0 \lor a=-\frac{1}{6}
$$

Since $b=0$, I'm getting
$$
3a^2+a=1
$$
which will have 2 solutions. I have no idea what should I do next since I will be getting 3 different $a$'s, but wolfram shows me only 2 solutions.
$$
z = -\frac{1}{6} - \frac{\sqrt{13}}{6}
$$
$$
z = \frac{\sqrt{13}}{6} -\frac{1}{6}
$$

答:
The given equation implies

$$ z=1-3(z^*)^2=1-3(1-3z^2)^2=-2+18z^2-27z^4
$$
which factorizes as
$$(3z^2+z-1)(3z+1)(3z-2)=0$$
and yields the solutions $z=-\frac16\pm \frac{\sqrt{13}}6$, while $z=\frac23,-\frac13$ are invalid.


问:
Is it possible to tile dodecahedrons in a 3D grid without any spaces between them?
Is it possible to tile dodecahedrons in a 3D grid without any spaces between them? Feel free to send me web links about that.</p>

答: The dihedral angle of a regular dodecahedron is $2\arctan \phi \approx 116.565^\circ$.  As this does not divide into $360^\circ$ evenly, there will be gaps between them.


问:
What is the inverse z transform of 1/(z-1)^2?
I'd like to know how to calculate the inverse z transform of $\frac{1}{(z-1)^2}$ and the general case $\frac{1}{(z-a)^2}$</p>

答:
Hint: find the Maclauren series expansion of $\frac{1}{(z-a)^2}$ by differentiating a power series term-by-term.

Solution: Recall that for a sequence $x[n], n\in\mathbb{Z}$, the $\mathcal{Z}$ transform is defined as

$$
\mathcal{Z}\{x[n]\}(z)=\sum_{n\in\Bbb{Z}}x[n]z^{-n}
$$If we can find a power series expansion for $(x-a)^{-2}$, the coefficients of our power series will be the sequence $x[n]$.  This will be easier than using definition of the inverse transform.

Now, for any $a\neq 0$, we have

$$
\frac{1}{z-a}=-\frac{1}{a}\frac{1}{1-z/a}=-\frac{1}{a}\sum_{n=0}^\infty\left(\frac{z}{a}\right)^n
$$Hence, inside the radius of convergence, we have

$$
(z-a)^{-2}=-\frac{d}{dz}(z-a)^{-1}=\sum_{n=1}^\infty\frac{n}{a^{n+1}}z^{n-1}
$$

We now have a power series, so the only thing that remains is to define the appropriate $x[n]$ such that $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$.  Note that we have only nonnegative powers of $z$, so if we define

$$
x[n]=\left\{\begin{array}{cc}
\frac{-n+1}{a^{-n+2}} & n\leq 0\\
0 &n>0
\end{array}\right.
$$ we have $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$, as you can check.

ContentLicense: CC BY-SA 3.0


问:
I need help with tackling this proof that set $\mathbb Q$ of rationals is a $F_\sigma$ set
Show the set $\mathbb Q$ of rationals is a $F_\sigma$ set

I am using Topology without Tears by Morris

Thoughts on it

$\mathbb Q$ is neither open nor closed in $\mathbb R$.

I know $\mathbb Q$ is countably infinite.

Since $\mathbb Q$ is countably infinite we can form a countable

union of such sets {$S_a$}{$a\in  \mathbb N$}

Thus $\mathbb Q$ is $F_\sigma$ set

Any help would be appreciated

答:
Hints:

$1).\ $ Take $r\in \mathbb Q$ and an arbitrary open non-degenerate interval $(a,b)\ni r$. What "kind" of points are contained in $(a,b)?$

$2).\ $ Repeat $1).$ for $\mathbb Q^c$, the complement of $\mathbb Q.$

$3).\ $ Note that points are closed in $\mathbb R$ (why?). Now, use the definition of an $F_{\sigma}$ set together with a suitable property of $\mathbb Q.$


答: It’s just because it is a countable set, and each point forms a closed set.</p>

Title: I need help with tackling this proof that set $\mathbb Q$ of rationals is a $F_\sigma$ set

Body:
Show the set $\mathbb Q$ of rationals is a $F_\sigma$ set

I am using Topology without Tears by Morris

Thoughts on it

$\mathbb Q$ is neither open nor closed in $\mathbb R$.

I know $\mathbb Q$ is countably infinite.

Since $\mathbb Q$ is countably infinite we can form a countable

union of such sets {$S_a$}{$a\in  \mathbb N$}

Thus $\mathbb Q$ is $F_\sigma$ set

Any help would be appreciated


问:
Expectation of function of multiple random variable
Let $X_1,X_2,...,X_n$ be i.i.d random variables and $\mathbb{E}[X^{-1}_1]$ is finite. Define $S_n = \sum_{i=1}^{n}X_i$ .
Show that for any $1 \leq a \leq b \leq n , $
$\mathbb{E}[\frac{S_a}{S_b}] = \frac{a}{b} $.

I don't think we can equate $S_n = nX_i$ that would make the problem too trivial. So my progress is turn the expectation into simpler parts like $\mathbb{E}[\frac{S_a}{S_b}] = \sum_{i=1}^{a}\mathbb{E}[\frac{X_i}{S_b}]$. My guess is that $\mathbb{E}[\frac{X_i}{S_b}]=\frac{1}{b}$ but was stuck proving it, the denominator thing is hard for me. Any Ideas or solutions? Thanks.

答:
For $1 \le i \le b$, the random variables $X_i/S_b$ have the same distribution and thus have the same expectation. Since $\sum_{i=1}^b \frac{X_i}{S_b} = 1$, we must have $E[X_i/S_b]=\frac{1}{b}$.

Edit: The above argument overlooks some technical issues (assumes $S_b \ne 0$, assumes $E[X_i / S_b]$ is finite).


问:
Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$
Just out of curiosity.
Does the numbers of the form
\begin{align}
 \sum_{n=0}^{N} 2^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?

These numbers can never be divisible by three.
It is also easy to discover integers with multiple representations (e.g., $4 = 2^2 3^0 = 2^0 3^0 + 2^0 3^1$).
I am however stuck to prove that they cover all integers not divisible by 3.

Second question: The numbers of the form
\begin{align}
 \sum_{n=0}^{N} 5^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, are numbers congruent to $\{1, 4, 5, 8\} \pmod{12}$.
These are the numbers not divisible by 3 and not $\{ 2, 3 \} \pmod{4}$.
Many other such forms may be considered.
What is the name of such numbers (numbers of such a form)?

答: Well as far as the first question every integer$M$ can be written $\sum_{n=0} a_i3^n$ where $a_i =0,1=2^0$, or $2^1=2$ i.e., $M$'s tertiary expansion. So yes to the first question.


问:
Determinant of a specific matrix
Let $$C_n =\begin{vmatrix} 1 & 2 & 3 & 4 &\cdots & n-1 & n\\ -1 & 0 & 2 & 3 &\cdots &n-2 &n-1\\ -1& -2  & 0 & 3 & \cdots &n-2 &n-1\\ \vdots &\vdots & \vdots & \vdots & \cdots & \vdots &\vdots \\ -1 & -2 &-3 &-4 & \cdots &0 &n-1\\ -1 & -2 & -3 & -4 & \cdots &1-n &0  \end{vmatrix}$$ While computing the determinants for $n = 1,2,3$ and $4$, I think I found the pattern to say that the determinant of $C_n$ is $n!$. So for the general case I started first by doing row$2$+row$1$, row$3$+row$1$ etc ... and got this matrix - $$C_n =\begin{vmatrix} 1 & 2 & 3 & 4 &\cdots & n-1 & n\\ 0 & 2 & 5 & 7 &\cdots &2n-3 &2n-1\\ 0& 0 & 3 & 7 & \cdots &2n-3 &2n-1\\ \vdots &\vdots & \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 0 &0 &0 & \cdots &n-1 &2n-1\\ 0 & 0 & 0 & 0 & \cdots &0 &n \end{vmatrix}$$ I can definitely see the $n!$ forming on the diagonal but I have troubles getting there. I was thinking of using Laplace expansions on the diagonal elements repeatedly, i.e writing it such $C_n$=$1(2(3(...n-1(det(n)))$, but I easily got lost in my tracks and now I have no clue how to move on. Could anyone provide me some help clearing it up? Thank you very much in advance.

答:
As noted in the comments under the Q:

This matrix is triangular, so it is trivial to compute the determinant: just repeatedly expand by the first column to get $n!$.

Title: Determinant of a specific matrix


问:
Assumption of solutions for differential equations
When solving a differential equation (ODE or PDE), we often make an assumption about the solution in advance to facilitate (or enable in the first place) the solution process. Take e. g. an ODE of type $$ay(x)''+by(x)'+cy(x)=0$$ which is (in case the characteristic polynomial has non-identical real roots $k_1$ and $k_2$) solved by a function of type $$y=C_1e^{k_1x}+C_2e^{k_2x}$$ which has been proven to be the "one and only" solution. My question: How was that solution (and the generic solutions of other differential equation types) determined "the first time around"? Is there a way to systematically determine them starting from the original ODE/PDE or did they "reverse engineer" them? (Like hmmm let's try an exponential function. Heureka!)

答:
In the case of linear, constant-coefficient, homogeneous:

$$ay''+by'+cy = 0,$$

one might look at the equation and say, "Hmmm...I need a function that is a multiple of its own derivatives...." and decide $y=e^{rx}$ might be worth a try.  For a homogeneous Euler equation:

$$ax^2y''+bxy'+cy = 0,$$

one might say, "Hmmmm, I need a function that every time I take a derivative, and then multiply by $x$, I get multiple of that function.  So when I take the derivative, I reduce the power of $x$ by one...."  and decide that $y=x^r$ might be a good choice.

In the non-homogeneous linear case, one usually find the homogeneous solution $y_h$ first.  Then he might say, "The homogeneous solution almost solves the equation, and any constant multiple $cy_h$ solves the equation, so maybe if I make the constant a variable, I could get a solution to the non-homogeneous part." And decide that $y=u(x)y_h(x)$ might be worth a try.

So it's not just random guessing, but looking at the equation and deducing properties of the solution.  A little bit if guessing, but a lot of thinking.


问:
How is the derivative truly, literally the "best linear approximation" near a point?
I've read many times that the derivative of a function $f(x)$ for a certain $x$ is the best linear approximation of the function for values near $x$.

I always thought it was meant in a hand-waving approximate way, but I've recently read that:

"Some people call the derivative the “best linear approximator” because of how accurate this approximation is for $x$ near $0$ (as seen in the picture below). In fact, the derivative actually is the “best” in this sense – you can’t do better." (from http://davidlowryduda.com/?p=1520, where $0$ is a special case in the context of Taylor Series).

This seems to make it clear that the idea of "best linear approximation" is meant in a literal, mathematically rigorous way.

I'm confused because I believe that for a differentiable function, no matter how small you make the interval $\epsilon$ around $x$, there will always be for any $a$ near $x$ in that interval a line going through $x$ that is either as good an approximation of $f(a)$ as the one given by $f'(x)$ (in case the function is actually linear over that interval), or a better approximation (the case in which the line going through $(x, f(x))$ also goes through (a, f(a)) and any line between this line and the tangent at $x$).

What am I missing?

答:
Let $y=f(x)$ be a (differentiable) function that we are trying to approximate around the point $P(a,f(a))$. The simplest way to approximate its behaviour around that point is by fitting a linear function $g(x)=mx+c$ to it. We can then define 'the best linear approximation' to $f$ as the function $g$ that has the following property:
$$
\lim_{x \to a}\frac{f(x)-g(x)}{x-a}=0 \, .
$$
What this criterion tries to capture is the 'relative error' of $g$. If $x$ is very close to $a$, then $g(x)$ should be closer still to $f(x)$. Simple algebraic manipulation shows that the only function $g$ that satisfies this property is the tangent to $P$. Let
$$
h(x)=\frac{f(x)-g(x)}{x-a} \, .
$$
Then, $f(x)-g(x)=h(x)(x-a)$. This means that
$$
\lim_{x \to a}f(x)-g(x) = \lim_{x \to a}h(x) \cdot \lim_{x \to a}(x-a)=0 \, .
$$
Therefore,
$$
\lim_{x \to a}f(x)=\lim_{x \to a}g(x) \, ,
$$
which implies $f(a)=g(a)$ since both $f$ and $g$ are differentiable and hence continuous. Unsurprisingly, the 'best linear approximation' of a function around the point $x=a$ should be exactly equal to the function at the point $x=a$. Using the point-slope form of the equation of a line, we find that
$$
g(x) = m(x-a) + g(a) = m(x-a) + f(a) \, .
$$
We are now tasked with proving that $m=f'(a)$. Luckily, this is not too difficult:
\begin{align}
& \lim_{x \to a}\frac{f(x)-g(x)}{x-a}=0 \\[4pt]
\implies & \lim_{x \to a}\frac{f(x)-g(x)}{x-a}=0 \\[4pt]
\implies & \lim_{x \to a}\frac{f(x)-f(a)+f(a)-g(x)}{x-a}=0 \\[4pt]
\implies & \lim_{x \to a}\frac{f(x)-f(a)}{x-a} + \lim_{x \to a}\frac{f(a)-g(x)}{x-a}=0 \\[4pt]
\implies & \lim_{x \to a}\frac{f(x)-f(a)}{x-a} + \lim_{x \to a}\frac{g(a)-g(x)}{x-a}=0 \\[4pt]
\implies & f'(a) - g'(a) = 0 \\[4pt]
\implies & f'(a) = g'(a)
\end{align}
Since $g'(a)=m$, we find that $g$ must have the equation
$$
g(x) = f'(a)(x-a) + f(a) \, .
$$
But this is the equation of the tangent to $P$, and so, in this sense, the derivative gives the best linear approximation of $f(x)$ around a certain point.


问:
Prove or disprove, Span
Let $V$ be a vector space, $M, N \subset V $sets.
Prove or disprove:

i) span($M$) $\cap$ span($N$) = Span(M$\cap$N)

ii) let $u,v$ be $\in$ V. Then: Span$(u,v) = $span$(u,u+v)$

I'm sorry, but for i) I can't give any steps, so please point me in the right way.

ii) is ture I think: two cases. First u,v can be linear dependent. Then span$(u)=$span$(v)=$span$(u+v)=$span$(u,u+v). $Second u,v can be linear independent. Then $u,u+v $are still linear independent as per definition of linear dependency, and span$(u,v)=$span$(u,u+v)$

I'm not sure about this either, though... Any help and pointers are gladly welcome!!

Thanks

答: A counterexample for the first question is $V=\mathbf R$, $M= \{1\}$ and $N=\{2\}$, that verify $span(M)=span(N)=V$ but $span(M\cap N)= \{0\}$.

Title: Prove or disprove, Span


问:
What Percentage of Formulas in Propositional Logic is Satisfiable?
Let $P_{n}$ be the set of all formulas in propositional logic with $n$ variables. Let then $Q_{n} \subset P_{n}$ be the maximal set of all non-equivalent formulas in propositional logic of length $n$.

More formally, $Q_{n}:=\hbox{argmax}_{|p_{n}|} \{p_{n} \subset P_{n} \land \forall i, j \in p_{n}: i \not \equiv j\}$

I haven't found any information about the percentage of formulas in $Q_{n}$ that is satisfiable, i.e. $\frac{|\{q_{n} | q_{n} \in Q_{n} \land \hbox{SAT}(q_{n})\}|}{|Q_{n}|}$.

Is it known what that number would be?

答:
Since you propose to consider only classes of inequivalent formulas, the answer is simple.  Exactly one formula is unsatisfiable, because any two unsatisfiable formulas are logically equivalent!

There are $2^{2^n}$ inequivalent formulas with $n$ variables.  (Consider the truth tables.  A truth table with $n$ variables has $2^n$ rows, and the rows may be filled in $2^{2^n}$ ways.)   So the answer is simply $$1-2^{-{2^n}}.$$

(Note that for your example of $n=1$, there are $2^{2^1} = 4$ formulas, and all but one $(\it{False})$ is satisfiable.)


问:
One way functions and P = NP?
How can I show that no one way functions exist under assumption of P = NP?</p>

答:
If one way function exist, you can not prove your proposition!

It is better that ask:

Can we reduce the existance on OWF to P vs NP problem?

Andrej Bogdanov give a negative result. He show that non-adaptive reduction is not sufficient.

This problem in fact is a reduction from average case complexity to worst case complexity. You can see the monograph by Bogdanov and Trevisan.

Title: One way functions and P = NP?


问:
Queueing Theory: Difference between Time in Queue and Waiting Time?
I'm looking at the formulas here: http://www.cs.auckland.ac.nz/courses/compsci742s2c/lectures/p-q.pdf

It has one for r-th percentile of waiting time ($r$% of customers wait less than said time), and another one for r-th percentile of time in queue. But if the customers are in the queue, then aren't they waiting?

The formulas:

$\pi_w(r) = E(w)\log_e\frac{100}{100 - r}$

$\pi_q(r) = E(w)\log_e\frac{100\rho}{100 - r}$

答:
The definitions earlier in the document are:

time in queue $q$

time in system (queue+server) $w = q + s$

so $\pi_w(r) > \pi_q(r)$, as suggested by the formulae if the traffic intensity $\rho <1$, necessary if the queue is expected to clear.


问:
Irreducible Polynomial in $\mathbb F_{256}$.
Let $\mathbb F_{256}$ be the finite field with $2^8 = 256$ elements. Consider the polynomial over this field
$$
  x^2 + x + 1.
$$
I wanted to know if it is irreducible, so I calculated it for all values $0, \ldots, 256$ and considered the remainder modulo 256, I found that it doesn't evaluates to $0$, so I guess it is irreducible, but is this a valid method?</p>

答:
The method you cite should work, because the polynomial is of degree $2$ and hence any factors will correspond to solutions of the equation $x^2+x+1=0$. But you have to do it right and work over the field $\mathbb F_{256}$, which means you must begin by constructing that field etc. Working over the integers modulo 256 (i.e. $\mathbb Z/256\mathbb Z$) will not do.

Here is my suggestion to avoid this work: the polynomial is irreducible over $\mathbb F_2$ but it becomes reducible over $\mathbb F_4$ and extensions thereof. Since $256 = 2^8$ and $8$ is even, it contains $\mathbb F_4$ as a subfield and hence the polynomial is reducible.


问:
Finding area of paraboloid $2z = x^2 - y^2$ inside $x^2 + y^2 \leq 1$ using double integrals
The original problem was to find the surface area of hyperbolic paraboloid
$$2z = x^2 - y^2$$

that lies inside the cylinder $x^2 + y^2 \leq 1$.

My attempt was to parametrize $(x, y, z) = (\cos u, \sin u, v)$ and get $2v = \cos2u$  from the first equation. Then I found $|[r_u, r_v]| = 1$. After this I made assumption (don't know whether it was correct or not) that area of the whole figure is $4$ times area of the part of paraboloid that lies in $y > 0, z > 0$. Now we can calculate
$$\mu S = 4\iint_S dudv = 4\int_{-\pi/4}^{\pi/4}du\int_0^{\frac{1}{2}\cos2u} dv = 2$$

However I've found similar problem in my textbook (but the problem there was to find area for $z > 0$) and the answer for that was $\frac{\pi}{3}(2\sqrt2 - 1)$ which is far from what I've got.

My question is: how to get the correct answer with the parametrization that I made? I've read that the correct parametrization should be simple polar coordinates after setting $|[r_u, r_v]| = \sqrt{1 + z_x^2 + z_y^2}$, but is there a method to compute the area with the parametrization that was made above?

答:
For the surface area of $z>0$, it can be integrated via

$$A=\int_{x^2+y^2\le1} \sqrt{1+z_x^2 + z_y^2} \>dxdy
= 4\int_0^{\frac\pi4}\int_0^1 \sqrt{1+r^2}rdr d\theta= \frac\pi3(2\sqrt2-1)
$$


问:
Why isn't every finite locally free morphism etale?
I have been reading the notes here. There, a finite locally free morphism of schemes is defined as as a morphism of schemes $f: X \rightarrow Y$ which is finite and for which the sheaf $f_{*} \mathcal{O}_{X}$ is locally free as an $\mathcal{O}_{Y}$-module. A finite etale morphism is then defined as a morphism of schemes $f: X \rightarrow Y$ which is finite locally free, and for which the fiber over any point $q \in Y$ is an etale $\kappa(q)$-algebra.

I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this.

Take $f: X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $\operatorname{spec}A$ around $q$ small enough so that $f_{*} \mathcal{O}_{X}$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $\operatorname{spec}(A^{\oplus d}) \rightarrow \operatorname{spec}A$. Then the fiber over $q$ is just $\operatorname{spec}(\kappa(q)^{\oplus d})$. This seems to trivially be an etale $\kappa(q)$-algebra, since tensoring with some algebraic closure $\Omega$ would give $d$ copies of $\Omega$.

Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?

Further to this, the same notes say that for a finite etale morphism, the degree of the morphism at a point $q \in Y$ is the same as the cardinality of the fiber over $q$. But again the above reasoning seems to show this for any finite locally free morphism. Where does etale come into it at all?

答:
It is my impression that this is a student site, where PhD students in algebraic geometry and other fields can ask questions on details. It is difficult to find books giving "all" details, and this forum is a good place to discuss this during a "pandemic".

Question:"Take $f:X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $Spec(A)$ around $q$ small enough so that $f_∗\mathcal{O}_X$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $Spec(A^{\oplus d})→Spec(A)$. Then the fiber over $q$ is just $Spec(κ(q)^{\oplus d})$. This seems to trivially be an etale $κ(q)$-algebra, since tensoring with some algebraic closure $Ω$ would give $d$ copies of $Ω$. Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?"

Answer: If $B:=A[t]/(P(t))$ where $P(t):=t^d+a_1t^{d-1}+\cdots + a_d \in A[t]$, let $S:=Spec(B), T:=Spec(A)$ and $\pi:S\rightarrow T$ be the canonical map. Since there is an isomorphism

$B\cong A\{1,\overline{t},\cdots , \overline{t}^{d-1}\}$ as $A$-modules, it follows $B$ is a free $A$-module of rank $d$ hence the map $\pi$ is "finite locally free" as map of schemes. If $x\in T$ is any point with residue field $\kappa(x)$ we may pass to the fiber at $x$:

$\pi^{-1}(x):=Spec(\kappa(x)[t]/\overline{P(t)})$

where $P_x(t):=\overline{P(t)}\subseteq \kappa(x)[t]$ is the polynomial we get when we "reduce the coefficients modulo $x$". It follows the polynomial $P_x(t)$ may be written as s product

$P_x(t)=\prod_{i=1}^n P_{i,x}(t)^{l_i}$
where $P_{i,x}(t)$ is an irreducible polynomial for all $i$. If $l_i\geq 2$ for some $i$ or if $P_{i,x}(t)$ has repeated roots in the algebraic closure $\overline{\kappa(x)}$ it follows the map $\pi$ is not etale. The fiber ring at $x$ is the direct sum

F1. $\kappa(x)[t]/(P_{1,x}(t)^{l_1}) \oplus \cdots \oplus \kappa(x)[t]/(P_{n,x}(t)^{l_n})$

Your claim: "Then the fiber over $q$ is just $Spec(\kappa(q)^{\oplus d})$.

Answer: No, formula F1 gives a description of the fiber ring in this case, and it is not a direct sum of $d$ copies of $\kappa(x)$ as you claim.

Rings of integers in number fields: At this link

Tensoring is thought as both restricting and extending?

you will find a discussion of the calculation of the fiber of the canonical map

$\pi: C:=Spec(\mathcal{O}_L)\rightarrow S:=Spec(\mathcal{O}_K)$

for a finite extension of number fields $K\subseteq L$.

Example. If $K=\mathbb{Q}$ is the rational numbers, it follows $\mathcal{O}_L$ is a free $\mathbb{Z}$-module of finite rank (this result is proved in an introductory book on algebraic number theory). Hence the map $\pi$ is finite and locally free. The fiber of $\pi$ at a non-zero prime $(p)$ in $S$ is "ramified" in general. By definition it follows

$\pi^{-1}(p):=\mathbb{Z}/(p)\mathbb{Z} \otimes_{\mathbb{Z}} \mathcal{O}_L
\cong $

F1. $\mathcal{O}_L/(p)\mathcal{O}_L \cong  \mathcal{O}_L/\mathfrak{p_1}^{l_1}\cdots \mathfrak{p_n}^{l-n} \cong \mathcal{O}_L/\mathfrak{p_1}^{l_1}\oplus \cdots \oplus \mathcal{O}_L/\mathfrak{p_n}^{l_n}:=R_1\oplus \cdots \oplus R_n$.

The decomposition in F1 is proved in any book on algebraic number theory. The rings $R_i$ are Artinian local rings for all $i=1,..,n$ and the schematic fiber $\pi^{-1}((p))$ is the disjoint union

F2. $\pi^{-1}((p))\cong Spec(R_1)\cup \cdots \cup Spec(R_n)$

The integers $l_i\geq 2$ in general and the field extension $\mathbb{Z}/(p)\mathbb{Z} \subseteq \mathcal{O}_L/\mathfrak{p}_i$ is a non trivial finite extension in general.

Hence the fiber is not a finite direct sum of copies of $\mathbb{Z}/(p)\mathbb{Z}$.

We could define the map $\pi$ to be "etale at a point $x\in C$" iff the module of relative Kahler differentials $\Omega(\pi)_{x}$ is zero at $x$.

Note: There is a problem in positive characteristic. If $char(\kappa(x))=p>0$ and $f(x_1,..,x_n)\in \kappa(x)[x_1^p,..,x_n^p]$ it follows the differential

$df:=\sum_i \frac{\partial f}{\partial x_i}dx_i=0$. let $A:=\kappa(x)[x_1,..,x_n]/(f)$. It follows

$\Omega^1_{A/\kappa(x)}\cong A\{dx_1,..,dx_n\}$ is a free $A$-module of rank $n$. Intuitively $Spec(A)$ is a hypersurface in $\mathbb{A}^n_{\kappa(x)}$
and should have dimension $n-1$. Hence to use the module of Kahler differentials to define "etale" in characteristic $p$ can lead to problems.

In Hartshorne, Exercise II.3.5 you prove that any finite morphism $f:X\rightarrow Y$ of schemes is closed, hence the image $f(Z)$ of any closed subscheme $Z\subseteq X$ is closed in $Y$. If the relative cotangent sheaf $\Omega(f):=\Omega^1_{X/Y}$ is zero on an open subscheme $U\subseteq X$, it follows the complement $Z(f):=X-U$ is closed and hence $D(f):=f(Z(f))\subseteq Y$ is a closed subscheme of $Y$. Hence in the case when $f$ is finite and locally free we may in many cases use the relative cotangent sheaf $\Omega(f)$ to construct an open subscheme $V:=Y-D(f)$ with the property that the induced morphism $f_V: \pi^{-1}(V)\rightarrow V$ is etale. Hence intuitively your locally trivial finite morphism $f$ should be etale over an open subscheme $V\subseteq Y$. In Hartshorne III.10.7 there is a "generic smoothness" theorem valid for a morphism $f$ of algebraic varieties over an algebraically closed field of characteristic zero. In this situation "etale" means "smooth of relative dimension $0$".

Your question: "Or is it just the case that every finite locally free morphism is etale?"

Answer: In many cases there should be an open subscheme $V\subseteq Y$ where the induced morphism $f_V:f^{-1}(V) \rightarrow V$ is etale. You must look for a relative version of HH. Corr III.10.7, the "Stacks Project" is a good place to start.

Given any morphism $\pi:X\rightarrow S$ of schemes, one uses the ideal sheaf of the diagonal to define  the notion differentially smooth. The morphism $\pi$ is "differentially smooth" iff it is flat, $\Omega^1_{X/S}$ is locally free and the canonical map

$Sym^*_{\mathcal{O}_X}(\Omega_{X/S})\rightarrow  \oplus_{i\geq 0}  \mathcal{I}^i/\mathcal{I}^{i+1}$

is an isomorphism of $\mathcal{O}_X$-modules. Here $\mathcal{I} \subseteq \mathcal{O}_{X\times_S X}$ is the ideal of the diagonal. It could be this is a useful notion of smoothness in characteristic $p>0$. This notion is mentioned in [EGA, IV4 no. 32]. You find a copy at numdam.org:

http://www.numdam.org/item/PMIHES_1967__32__5_0/

Example. Note that for any monic polynomial $P(t)\in A[t]$ with $B:=A[t]/(P(t))$ it follows for any prime ideal $\mathfrak{p}\subseteq A$ there is an isomorphism $C:=\kappa(\mathfrak{p})\otimes_A B$ \cong $\kappa(\mathfrak{p})[t]/(P_p(t))$, where $P_p(t)$ is the polynomial we get when we reduce the coefficients of $P(t)$ "modulo $\mathfrak{p}$". It follows
from Matsumuras book that

$C\otimes_B \Omega^1_{B/A} \cong \Omega^1_{C/\kappa(\mathfrak{p})}\cong C\{dt\} /C\{P_p'(t)dt\}$.

The module $\Omega(\mathfrak{p}):=C\otimes \Omega^1_{B/A}$ is the restriction of $\Omega^1_{B/A}$ to the fiber ring $C$, and if $P_p'(t)=0$ in $\kappa(\mathfrak{p})[t]$ it follows $\Omega(\mathfrak{p})$ does not give information on the polynomial $P_p(t)$. Hence in positive characteristic we must use another definition to define "smooth" and "etale".

Question: "I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this."

Answer: for any finite extension  $K \subseteq L$ of number fields, it follows the morphism $\pi: S:=Spec(\mathcal{O}_L) \rightarrow T:=Spec(\mathcal{O}_K)$ is always integral, locally free (and ramified when $K=\mathbb{Q}$). Hence it is not etale in general. There is an open subscheme $U \subseteq T$ with the property that $\pi_U:\pi^{-1}(U) \rightarrow U$ is etale. The fiber  $\pi^{-1}((p))$ in formula F1 above is unramified iff $l_i=1$ for all $i$.
The module of Kahler differentials $\Omega:=\Omega^1_{\mathcal{O}_L/\mathcal{O}_K}$ is zero on the open subscheme $\pi^{-1}(U)$. Note moreover that the field extension

$\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$

is always separable since it is a finite extension of finite fields (here the ideals $\mathfrak{p}, \mathfrak{q}$ are maximal ideals). The ramified prime ideals in $\mathcal{O}_K$ are related to the discriminant of the extension $L/K$ (see in "Algebraic number theory", page 49, by Neukirch). In Neukirch he defines the discriminant using a basis for $L$ over $K$, but it can also be defined using the module of Kahler differentials $\Omega$.

Question: "I noticed that sometimes people are leaving two separate answers for the same question.Sometimes when I answer and leave up to three different methods as the answer, should I be splitting those up into three different answers? Is this being done to pad upvotes, or is it frowned upon or should it be avoided?"

Answer: "It is a matter of understandability. In most cases, I try to combine multiple approaches into one answer. However, if the approaches are long and involved, I will split them into separate answers to reduce confusion and improve readability. I strive to give one answer per question, but on occasion, that is not the best approach."

I agree to this principle, but some users have computers that are slow and where it is difficult to write longer posts (posts with more than 5000 characters) - I believe this forum should allow such users to write multiple posts if this is needed.


问:
partial fractions when the fraction cannot be decomposed
I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can I then go about writing this fraction as the sum $\frac {1}{2}\left [\frac {1}{x^2+1}-\frac{x^2-1}{(x^2+1)^2}\right]$? Is it just manipulation and trial and error?

答:
Just to clarify:

In cases of irreducible denomiantors over the real numbers such as your case, we treat it like a regular fraction with a repeated factor in the denominator but instead of letting the numerators of each fraction produced by the partial fraction decomposition be constant, instead we let the numerator be a polynomial of degree one lower than the factor in question.

That was rather a mouthful! To clarify, in your case:

$$\frac{1}{(x^2+1)^2}\equiv\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$$

Finding those coefficients, however, yields $D=1$ and this is the only non zero coefficient. This is why your fraction is already in  partial fraction form.


问:
Quotient group $Q$: 1) is it normal, 2) does $G=Q\times N$?
I know that if we form the direct product group

$$G=G_1\times G_2$$

then both $G_1$ and $G_2$ are normal subgroups of $G$. I also know that if a group $G$ has a normal subgroup $N\triangleleft G$, then we define a quotient group

$$Q=G/N$$

where the elements are cosets.

My questions:

Is $Q$ also a normal subgroup?

Can be recover $G$ by some product of $Q$ and $N$, for example $Q\times N$?

答:
$Q$ is not necessarily a normal subgroup, take the example of $SO(3)=SE(3)/T(3)$, i.e., rotation around the origin is the quotient of special Euclidean transform up to a translation. However rotating around the origin is not a normal subgroup. $t\ r \ t^{-1}$ is not a rotation around the origin in general, but a rotation around $t$.

Sometimes it can be a direct or semidirect product, but some splitting conditions need to be satisfied.

OwnerDisplayName: Travis Lee

OwnerDisplayName: Y Tong


答:
To put my comments in the form of a formal answer:

The quotient group isn't even a subgroup because the elements of $Q$ are cosets ($gN,\ \forall g\in G$) which are clearly not elements of your group $G$. For a more mathy answer, please see this. Although one can sometimes find isomorphic subgroups to the quotient group, this is not always guaranteed. Let me elucidate a bit more on this. Consider $D_{3}=\{e,c,c^{2},b,bc,bc^{2}\}$ and $H=\{e,c,c^{2}\}$ which is a normal subgroup. Now, the quotient group $Q$ consists of the cosets $gH, \ \forall g\in D_{3}$ which reads
$$Q=D_{3}/H = \{eH=H,bH\}\cong C_{2},$$

i.e. it is isomorphic to $C_{2}$. For something a bit more non-trivial, consider the dicyclic group of order $3$, i.e. $Dic_{3} = \{e,x,a,a^{2},a^{3},a^{4},a^{5},x,ax,a^{2}x,a^{3}x,a^{4}x,a^{5}x\}$ and it is easy to show that it's center $Z = \{e,a^{3}\}$, and which is by definition is a normal subgroup of $Dic_{3}$. Quotienting yields

$$Q = Dic_{3}/Z = \{\{a^{3},e\},\{a,a^{4}\},\{a^{2},a^{5}\},\{x,a^{3}x\},\{ax,a^{4}x\},\{a^{2}x,a^{5}x\}\}\cong D_{3},$$

i.e. it is isomorphic to $D_{3}$, which is a subgroup of $Dic_{3}$ but not a normal subgroup of it. Extending this you can show that $Q = Dic_{n}/Z\cong D_{n}$. This finding of something that is isomorphic to your quotient is not useless since it aids us in completing the character table for higher dimensional representations via the process of lifting. with reference to the latter example, I have provided, you can define a map $\psi: Dic_{3}/Z\to D_{3}$ and convince yourself that $\psi$ is indeed an isomorphism since the inverse to $\psi$ exists (since the image in $D_{3}$ has a unique preimage and $\psi^{-1}$ preserves the group property). We can now identify that there exists a $1:1$ relation between the irreps of $Dic_{3}$ with $Z\leq Ker(\rho)$, where $\rho: Dic_{3}\to GL(n,\mathbb{C})$ and the irreps of $Dic_{3}/Z$, or put in a more succinct way,

$$D^{Dic_{3}}(g) = Dic_{3}^{\left(Dic_{3}/Z\right)}(gDic_{3}).$$

With this identification, you can simply "lift" the corresponding values of the characters in the table of $D_{3}$ to the table of $Dic_{3}$ (for example, $\chi_{Dic_{3}}(a) = \chi_{Dic_{3}}(\psi(\{a^{2},a^{5}\})) = \chi_{D_{3}}(c) = \chi_{D_{3}}(c^{2}) = -1$, and so on).

Title: Quotient group $Q$: 1) is it normal, 2) does $G=Q\times N$?

Body:
I know that if we form the direct product group

$$G=G_1\times G_2$$

then both $G_1$ and $G_2$ are normal subgroups of $G$. I also know that if a group $G$ has a normal subgroup $N\triangleleft G$, then we define a quotient group

$$Q=G/N$$

where the elements are cosets.

My questions:

Is $Q$ also a normal subgroup?

Can be recover $G$ by some product of $Q$ and $N$, for example $Q\times N$?

OwnerDisplayName: Spoilt Milk


问:
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?

I figured out this bacteria $=500(3)^{96/24}$

but then my mate told me it's bacteria $=500(1+3/1)^{96/24 }$

what do I do what is the next step.

答: Your mate is simply wrong. I see no ambiguity here at all.</p>


问:
Constructive Counting
How many 3-digit numbers have a property that the first digit is at least the twice the second digit?
I looked at the solutions, but is says second digit 0 and the cases of it's
first digit is 1,2,3,4,5,6,7,8,9 but I don't get it since 0 times something is always 0 anyways</p>

答:
First, let's turn this into something more Mathematical:

the first digit is at least the twice the second digit

This can be rephrased into "The second digit is $k$ and the first digit is $\geq 2k$

If the first digit is $0$ then we need any number greater than $2(0)=0$,

We have $1\geq0$ so this works for any number starting $1$

Title: Constructive Counting


问:
Proving multivariable chain rule
Let $f:G\to G'$, $G\subset \Bbb{R^n}$, $G'\subset \Bbb{R^m}$ be differentiable at $x_0 \in G$ and let $g:G' \to \Bbb{R^p}$ be differentiable at $y_0 = f(x_0)\in G'$. Then the composite mapping $g\circ f:G\to \Bbb{R^p}$ is differentiable at $x_0$ and $D_{g\circ f}(x_0)=D_g(f(x_0)\cdot D_f(x_0)$

I'm going over the proof. Let $f$ differentiable at $x_0$ and $g$ differentiable at $y_0 =f(x_0)$. I begin with
$$g(f(x_0 +u))=g(f(x_0)+D_f(x_0)u+\vert u\vert\epsilon_1(u))$$

$$=g(y_0) + D_g(y_0)(D_f(x_0)u+\vert u\vert\epsilon_1(u))+\vert h(u)\vert D_g(y_0)(\epsilon_2(h(u)))$$

How do we arrive at the second equality? We apply $g$ to each term of the sum, and somehow $g((D_f(x_0)u)=D_g(y_0)(D_f(x_0)u+\vert u\vert\epsilon_1(u)).$

答:
From your post, I assume you are ok with the first equality, so we are at

$$ g(f(x_0+u)) = g \Big( f(x_0) + D_f(x_0)u + |u| \varepsilon_1(u) \Big) $$

Then you say:

We apply $g$ to each term in the sum, and somehow ...

We do not apply $g$ to each term in the sum. The next step is actually the same thing that happened in the first step: in the first step, you used the linear approximation of $f$ to say that

$$ f(x_0+u) = f(x_0) + D_f(x_0) \cdot u + |u| \varepsilon_1(u) $$
The next step that you are asking about is doing the same thing, but with $g$. In place of $x_0$ we now have $y_0=f(x_0)$, and in place of $u$, we now have
$$ v = D_f(x_0) \cdot u + |u|\varepsilon_1(u) $$ So we get
$$ g(y_0 + v) = g(y_0) + D_g(y_0) \cdot v + |v|\varepsilon_2(v) $$
Now just plug in the expression for $v$ and you get what you have in your post. Although on your last line you call it $h(u)$ where I call it $v$.

Title: Proving multivariable chain rule


问:
Does $\lim_{x \to a}\frac{f(x)}{g(x)}=0 \text{ and } \lim_{x \to a}g(x)=0 \implies \lim_{x \to a}f(x) = 0$?
The Problem

Say
$$
\lim_{x \to a}\frac{f(x)}{g(x)}=0 \text{ and } \lim_{x \to a}g(x)=0 \, .
$$
Does this imply that $$\lim_{x \to a} f(x) = 0 \, ?$$
I understand that $f(x)$ must become very small relative to $g(x)$ as $x$ approaches $a$, but I'm worried that my intuitions might be failing me here.

What I have tried so far

For some neighbourhood of $a$, it seems evident that $|f(x)|\leq|g(x)|$. (Otherwise, the limit wouldn't equal $0$.) Consider the case where both $f(x)$ and $g(x)$ are nonnegative. Then, we are left with $0 \leq f(x) \leq g(x)$. Since
$$
\lim_{x \to a}g(x)=\lim_{x \to a}0=0 \, ,
$$
the squeeze theorem seems applicable here. However, I'm not sure how to generalise this result to when $|f(x)|\leq|g(x)|$, nor am I certain how to show that $|f(x)|\leq|g(x)|$ in the first place.

答: You can rearrange $~f(x)~$as $~f(x)=\frac{f(x)}{g(x)}\cdot g(x)~$, then $$\lim_{x \to a} f(x) = \lim_{x \to a} \dfrac{f(x)}{g(x)}\cdot g(x) =\lim_{x \to a} \dfrac{f(x)}{g(x)}\cdot \lim_{x \to a}  g(x) = 0 \cdot 0=0~.$$


答: The polynomial you write is the third cyclotomic polynomial. It is reducible in any finite field $\Bbb F_q$ where $3$ divides the order $q-1$ of the multiplicative group. In the case $q=256$ we find that $3$ divides $255$, so the polynomial splits in $\Bbb F_{256}[x]$.</p>

Title: Irreducible Polynomial in $\mathbb F_{256}$.

Body:
Let $\mathbb F_{256}$ be the finite field with $2^8 = 256$ elements. Consider the polynomial over this field
$$
  x^2 + x + 1.
$$
I wanted to know if it is irreducible, so I calculated it for all values $0, \ldots, 256$ and considered the remainder modulo 256, I found that it doesn't evaluates to $0$, so I guess it is irreducible, but is this a valid method?</p>


问:
Solve $\frac{7}{x+\sqrt{x+5}}+\frac{7}{x-\sqrt{x+5}}=8$
Solve the equation: $$\dfrac{7}{x+\sqrt{x+5}}+\dfrac{7}{x-\sqrt{x+5}}=8.$$ I am not sure how to approach the problem. Should we first determine the domain? I think we can also check for every value we get for $x$ if the expression is defined, or not. The answers are $x=-\dfrac{5}{4}$ and $x=4.$ The numerators are equal and we also have resemblance in the denomitators. I am not sure how to use that. Thank you in advance! Happy holidays!

Body:
Note that:

$$\frac{7}{x+\sqrt{x+5}}+\frac{7}{x-\sqrt{x+5}}=\frac{14x}{x^2-x-5}$$


答:
\begin{align*}
\frac{7}{x + \sqrt{x+5}} + \frac{7}{x - \sqrt{x + 5}} = 8 & \Rightarrow \frac{14x}{x^{2} - x - 5} = 8\\\\
& \Rightarrow 8x^{2} - 22x - 40 = 0\\\\
& \Rightarrow 4x^{2} - 11x - 20 = 0\\\\
& \Rightarrow \left(x = - \frac{5}{4}\right)\vee(x = 4)
\end{align*}

Substituting in the original equation, we conclude that they are solutions indeed.

Title: Solve $\frac{7}{x+\sqrt{x+5}}+\frac{7}{x-\sqrt{x+5}}=8$

Body: Solve the equation: $$\dfrac{7}{x+\sqrt{x+5}}+\dfrac{7}{x-\sqrt{x+5}}=8.$$ I am not sure how to approach the problem. Should we first determine the domain? I think we can also check for every value we get for $x$ if the expression is defined, or not. The answers are $x=-\dfrac{5}{4}$ and $x=4.$ The numerators are equal and we also have resemblance in the denomitators. I am not sure how to use that. Thank you in advance! Happy holidays!


Body: $$\dfrac{7}{x+\sqrt{x+5}}+\dfrac{7}{x-\sqrt{x+5}} = \frac{7(x-\sqrt{x+5})}{x^2-x-5} +\frac{7(x+\sqrt{x+5})}{x^2-x-5} \\ = \frac{14x}{x^2-x-5}$$ So the equation is equivalent to $$\frac{14x}{x^2-x-5} =8 \\ 14x =8x^2-8x-40 \\ 8x^2-22x-40=0$$


问:
Looking for a proof of two combinatorial summation identities
While working on a problem I came across the following similar combinatorial identities valid by numerical evidence $(n\ge m)$:
$$\begin{align}
A(n,m)&\equiv\sum_{k\ge1}\binom km\binom{n-1}{k-1}=2^{n-m-1}\binom nm\frac{n+m}{n},\\
B(n,m)&\equiv\sum_{k\ge1}\left[\binom{\left\lfloor\frac k2\right\rfloor}m
+\binom{\left\lceil\frac k2\right\rceil}m\right]\binom{n-1}{k-1}=2^{n-2m}\binom {n-m}{m-1}\frac{n+1}{m}.
\end{align}$$

I am looking for an algebraic proof of these identities.

Is there a smart way to prove:
$$
B(n,m)=A(n-m+1,m)
$$
without explicit evaluation of the sums?

Body:
For the second identity we get for the LHS

$$\sum_{k\ge 1}
\left[ {\lfloor \frac{k}{2} \rfloor \choose m} +
{\lceil \frac{k}{2} \rceil \choose m} \right]
{n-1\choose k-1}
\\ = 2\sum_{k\ge 0} {k\choose m} {n-1\choose 2k-1}
+ \sum_{k\ge 0} {k\choose m} {n-1\choose 2k}
+ \sum_{k\ge 0} {k+1\choose m} {n-1\choose 2k}$$

which we seek to prove is equal to

$$2^{n-2m} {n-m\choose m-1} \frac{n+1}{m}.$$

where we will take $m\ge 1.$ We get for the first term

$$2\sum_{k\ge 0} {k\choose m} {n-1\choose n-2k}
= 2[z^n] (1+z)^{n-1} \sum_{k\ge 0} {k\choose m} z^{2k}
\\ = 2[z^n] (1+z)^{n-1} \sum_{k\ge m} {k\choose m} z^{2k}
= 2[z^{n-2m}] (1+z)^{n-1} \sum_{k\ge 0} {k+m\choose m} z^{2k}
\\ = 2[z^{n-2m}] (1+z)^{n-1} \frac{1}{(1-z^2)^{m+1}}.$$

The second term is

$$\sum_{k\ge 0} {k\choose m} {n-1\choose n-1-2k}
= [z^{n-1}] (1+z)^{n-1} \sum_{k\ge 0} {k\choose m} z^{2k}
\\ = [z^{n-2m-1}] (1+z)^{n-1} \frac{1}{(1-z^2)^{m+1}}.$$

The third term is

$$\sum_{k\ge 0} {k+1\choose m} {n-1\choose n-1-2k}
= [z^{n-1}] (1+z)^{n-1} \sum_{k\ge 0} {k+1\choose m} z^{2k}
\\ = [z^{n-1}] (1+z)^{n-1} \sum_{k\ge m-1} {k+1\choose m} z^{2k}
= [z^{n-2m+1}] (1+z)^{n-1} \sum_{k\ge 0} {k+m\choose m} z^{2k}
\\ = [z^{n-2m+1}] (1+z)^{n-1} \frac{1}{(1-z^2)^{m+1}}.$$

Adding these together we get

$$[z^{n-2m+1}] (1+z^2+2z) (1+z)^{n-1} \frac{1}{(1-z^2)^{m+1}}
= [z^{n-2m+1}] (1+z)^{n+1} \frac{1}{(1-z^2)^{m+1}}
\\ = [z^{n-2m+1}] (1+z)^{n-m} \frac{1}{(1-z)^{m+1}}.$$

The coefficient extractor now yields

$$\sum_{q=0}^{n+1-2m} {n-m\choose q} {n+1-2m-q+m\choose m}
= \sum_{q=0}^{n+1-2m} {n-m\choose q} {n+1-m-q\choose m}
\\ = \sum_{q=0}^{n+1-2m}
 {n-m\choose q} \frac{n+1-m-q}{m} {n-m-q\choose m-1}.$$

Now

$${n-m\choose q} {n-m-q\choose m-1}
= \frac{(n-m)!}{q! \times (m-1)! \times (n+1-2m-q)!}
\\ = {n-m\choose m-1} {n+1-2m\choose q}.$$

We get for the sum

$$\frac{1}{m} {n-m\choose m-1}
\sum_{q=0}^{n+1-2m} (n+1-m-q) {n+1-2m\choose q}
\\ = \frac{1}{m} {n-m\choose m-1}
\sum_{q=0}^{n+1-2m} (n+1-2m-q) {n+1-2m\choose q}
\\ + {n-m\choose m-1}
\sum_{q=0}^{n+1-2m} {n+1-2m\choose q}
\\ = \frac{1}{m} {n-m\choose m-1}
\sum_{q=0}^{n+1-2m} q {n+1-2m\choose q}
+ {n-m\choose m-1} 2^{n+1-2m}
\\ = \frac{n+1-2m}{m} {n-m\choose m-1}
\sum_{q=1}^{n+1-2m} {n-2m\choose q-1}
+ {n-m\choose m-1} 2^{n+1-2m}
\\ = \frac{n+1-2m}{m} {n-m\choose m-1}
2^{n-2m} + {n-m\choose m-1} 2^{n+1-2m}.$$

This simplifies to

$$\bbox[5px,border:2px solid #00A000]{
\frac{n+1}{m} {n-m\choose m-1} 2^{n-2m}.}$$

Addendum. Following the hint by OP we can simplify the
three terms first. We get

$$2\sum_{k\ge m} {k\choose m} {n-1\choose 2k-1}
+ \sum_{k\ge m} {k\choose m} {n-1\choose 2k}
+ \sum_{k\ge m-1} {k+1\choose m} {n-1\choose 2k}
\\ = 2\sum_{k\ge m} {k\choose m} {n-1\choose 2k-1}
+ \sum_{k\ge m} {k\choose m} {n-1\choose 2k}
+ \sum_{k\ge m} {k\choose m} {n-1\choose 2k-2}
\\ = \sum_{k\ge m} {k\choose m} {n\choose 2k}
+ \sum_{k\ge m} {k\choose m} {n\choose 2k-1}
= \sum_{k\ge m} {k\choose m} {n+1\choose 2k}.$$

We then find

$$\sum_{k\ge m} {k\choose m} {n+1\choose n+1-2k}
= [z^{n+1}] (1+z)^{n+1} \sum_{k\ge m} {k\choose m} z^{2k}
\\ = [z^{n+1-2m}] (1+z)^{n+1} \sum_{k\ge 0} {k+m\choose m} z^{2k}
= [z^{n+1-2m}] (1+z)^{n+1} \frac{1}{(1-z^2)^{m+1}}.$$

From this point on the computation continues as before.


问:
Does this integral have simple result expression?
I am evaluating the integral

$$\int_0^\infty\frac{1}{(1+x)^a(1+bx)^2}\,\mathrm{d}x,$$

where $a>0$ and $b>1$.

By expressing $(1+x)^{-a}$ and $(1+bx)^2$ in terms of Meijer-G function, I can obtain the result for the above integral, but the result contains the Meijer-G function.

I wonder if the above integral can have a simple expression result, i.e., without containing any special functions.

Thanks!

答:
After Claude Leibovici's answer, let me calculate
$$J_n(B) = \int_0^\infty \frac{1}{(1+x)^n (B+x)^2} dx$$
when $n$ is an integer $\ge 2$. (Note that $I_n$ and $J_n(B)$ are different; we have
$$I_n = \int_0^\infty \frac{1}{(1+x)^n (1+bx)^2} dx = \frac{1}{b^2}\int_0^\infty \frac{1}{(1+x)^n (b^{-1}+x)^2} dx = b^{-2}J_n(b^{-1})$$
and $J_n$ is more convenient to deal with.)

One can assert $$ \frac{1}{(1+x)^n(B+x)^2} = \frac{q_n(x)}{(1+x)^n} + \frac{1}{(B+x)^2}(\alpha + \beta (B+x))$$
where $q_n$ is some polynomial of degree $\le n-1$ and $\alpha,\beta$ are constant. Multiply $(1+x)^n (B+x)^2$ to the both side, put $x = -B$ to have $\alpha = (1-B)^{-n}$, differentiate with respect to $x$ and put $x= - B$ to have $\beta = -n (1-B)^{-n-1}$; thus we have

$$ \frac{(1-B)^{n+1}}{(1+x)^n(B+x)^2} = \frac{\tilde{q}_n(x)}{(1+x)^n} + \frac{1}{(B+x)^2}(1-B -n (B+x))$$
where $\tilde{q}$ is some polynomial of degree $\le n-1$.

Now

\begin{align*}
(B+x)^2\tilde{q}_n(x)&=  (1-B)^{n+1} - \left(1-B -n (B+x))\right)(1+x)^n\\
& =(1-B)^{n+1} +\left(nx -1+B +nB)\right)(1+x)^n\\
x^2 \tilde{q}_n(x-B) & =(1-B)^{n+1} +\left(nx -1+B)\right)(x+1 - B)^n
\end{align*}
one can apply the binomial theorem and do bunch of calculations to get
\begin{align*}
\tilde{q}_n(x) &= \sum_{k=0}^{n-1} (k+1) \binom{n+1}{k+2} (1-B)^{n-k-1} (x+B)^k\\
& = \sum_{j=0}^{n-1}a_{n, j}(1-B)^{n-j-1}(1+x)^j
\end{align*}
where $$a_{n, j}= \sum_{k=0}^{n-j-1} (k+1) \binom{n+1}{k+j+2} \binom{k+j+1}{j}(-1)^k.$$
One can check that $a_{n, j}=j+1$.

Thus,
\begin{align*}
(1-B)^{n+1} J_n(B)&= (1-B)\int_0^\infty\frac{1}{(x+B)^2} dx - n \int_0^\infty\frac{1}{x+B}dx+\\
&\phantom{=+}+\sum_{j=0}^{n-2}a_{n, j}(1-B)^{n-j-1}\int_0^\infty \frac{1}{(1+x)^{n-j}}dx + n \int_{0}^\infty \frac{1}{x+1}dx\\
& = \frac{1-B}{B} + n \log B+(1-B)\sum_{j=0}^{n-2}(1-B)^{n-2-j}\frac{j+1}{n-1-j} \\
J_n(B)&= \frac{1-B +nB \log B}{B(1-B)^{n+1}} + \frac{1}{(1-B)^n}\sum_{j=0}^{n-2}(1-B)^{n-2-j}\frac{j+1}{n-1-j}
\end{align*}

Note that the coefficient in the last summation term $\frac{j+1}{n-1-j} $ is symmetric, in the sense of $\frac{j+1}{n-1-j}  \cdot \frac{(n-2-j)+1}{n-1-(n-2-j)} =1$.


问:
What can we represent any integer $x$ by $x = p^k \cdot y (\mod p)$?
This is a problem I encountered in real analysis.
Let $\mathbb{Z}$ be the set of integers and $p$ be a fixed prime number. Represent any $x ∈ Z$ as

$x = p^k · y$, where $y\not\equiv 0$ (mod $p$) ($k ∈ N∪{0},y ∈ Z$)

The problem then defines d(x,y) and asks us to show that d is a metric.
But I don't know why the above equation works. Like if $x=3$ and $p=2$ then what are k and y?

答:
We can't represent $x = 0$.

But if $x \ne 0$ it can be done.  Take the prime factorization of $x$.  Take the highest factor of $p$ that divides into $x$. Call that $k$. (If $p\not \mid x$ then $k=0$).  Let $y = \frac {x}{p^k}$.  As we have divided out all powers of $p$, we have $p \not \mid y$ and so $y \not \equiv 0 \pmod p$.

So you your example of $x=3$ and $p =2$ then $k = 0$ and $y = 3$ and $3 = 2^0\cdot 3$.

=====

The problem then defines d(x,y) and asks us to show that d is a metric.

Just out of curiousity HOW does the problem define $d(x,y)$?


答:
First: looking for roots to deduce a polynomial over some field is irreducilbe works as long as the polynomials degree $\,\le 3\,$ .

Second: the polynomial $\,x^2+x+1\,$ is defined over $\,\Bbb F_2\,$ and it's irreducible over this field. Since $\,2\mid 8\;$, we have that $\,\Bbb F_{2^2}\le\Bbb F_{2^8}\,$ , and this means that all the roots of $\,x^2+x+1\,$ are already in $\,\Bbb F_{2^8}\,$ so either you didn't check correctly or else, as a comment above hints, you confused $\,\Bbb F_{256}\,$ with $\,\Bbb Z_{256}\,$ .


问:
Frogs spinning in a pot
In the classic jumping frogs game, we have 15 frogs on a circular dish, each in a pot. The dish is spinning and every few seconds it stops abruptly, so the frogs jump and land again either in their original pot or in one of the two neighboring.

In how many ways can they be rearranged after they fall? All frogs are different.

I have calculated it to be $610+(377+1)*2$ but it doesn't seem correct to me.

Any help is much appreciated.

答:
In the linear case we get Fibonacci numbers because each of the $n$ frogs they either stay put or exchange position with one of their neighbors. One can think of this as tiling some $1\times n$ space with $1\times 1$ dominos (corresponding to the frogs that stay put) and $1\times 2$ dominos (corresponding to the pair of frogs that exchange positions). Denote $g_n$ to be this number of linear arrangements for $n$ frogs. Then $g_1 = 1$ and $g_2 = 2$, and for each $n\ge 3$, we have $g_n = g_{n-1} + g_{n-2}$, because if the first frog stays put, then we have $g_{n-1}$ many ways to arrange the rest, and if the first frog exchanges with the second frog, then we have $g_{n-2}$ many ways. Hence the recurrence, which is just the Fibonacci numbers (shifted). Here we have $g_{13} = 377$, $g_{14} = 610$.

For the circular case, there are several interpretations of what you may want.

In the simplest situation where the pots are all fixed in place (so we will ignore circular symmetry for now), and assuming the frogs will either stay in place or exchange with one of the neighbors. Denote $a_n$ to be the number of outcomes for $n$ many frogs. Then looking at frog number 1, it can either

(1) stay in place, so the rest of the $n-1$ frogs are in a line, giving $g_{n-1}$ many ways;

(2) exchange with frog 2, giving $g_{n-2}$ many ways;

(3) exchange with frog $n$, giving $g_{n-2}$ many ways.

So this gives $a_n = g_{n-1} + 2g_{n-2}$.

Now, there is also the situation where all the frogs jump to the left, or all the frogs jump to the right. Then in this case we have $a_n = g_{n-1} + 2g_{n-2} +2 $. (However if you take rotational symmetry into consideration, these two extra configurations are just rotationally the same no frog moved.) With $n = 15$, it seems to agree with what you got.

However, as the circular dish is spinning, it seems to me that we should take rotational symmetry into consideration, then we will get $a_n = g_{n-1} + 2g_{n-2}$.

If there are any other missing ways of how the frogs can move, then we will need to account for those. Otherwise we have our result.

Title: Frogs spinning in a pot


问:
Prove or disprove $f^{-1}(A - B) = f^{-1}(A)-f^{-1}(B)$
I'm getting that we can neither prove nor disprove it, but it seems there's a definitive answer for this. How to prove or disprove this?</p>

答:
This is true since preimage commutes with set difference.

Theorem: Let $A$ and $B$ be sets, and consider a function $f: A \rightarrow B .$ Let $B_{1} \subseteq B$ and $B_{2} \subseteq B$ be any subsets. Then we have
$$
f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right)
$$
Proof: For any given $x \in A$, we have
$$
\begin{aligned}
x \in f^{-1}\left(B_{1} \backslash B_{2}\right) & \Longleftrightarrow f(x) \in B_{1} \backslash B_{2} \\
& \Longleftrightarrow f(x) \in B_{1} \wedge f(x) \notin B_{2} \\
& \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \wedge x \notin f^{-1}\left(B_{2}\right) \\
& \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right)
\end{aligned}
$$
and therefore, since they comprise the same elements of $A$,
$$
f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right)
$$


答:
To see why this works, consider the prime factorisation of $x$. If it is of the form
$$ x = p^\alpha \cdot z$$ where $z$ does not divide $p$ (i.e. we put all the $p$ fators in $p^k$), then your $y=k$. The condition $y\not\equiv 0\ (mod\ p)$ is same as this divisibility condition mentioned above. If $p$ does not divide $x$, then you can set $k=0$ and $y=x$.

Title: What can we represent any integer $x$ by $x = p^k \cdot y (\mod p)$?

Body:
This is a problem I encountered in real analysis.
Let $\mathbb{Z}$ be the set of integers and $p$ be a fixed prime number. Represent any $x ∈ Z$ as

$x = p^k · y$, where $y\not\equiv 0$ (mod $p$) ($k ∈ N∪{0},y ∈ Z$)

The problem then defines d(x,y) and asks us to show that d is a metric.
But I don't know why the above equation works. Like if $x=3$ and $p=2$ then what are k and y?


问:
Question about completeness of the space of bounded linear operators.

Let $(X,\|\cdot\|_X)$ be a Banach space and $(Y,\|\cdot\|_Y)$ be a normed linear space. Can we say that $(B(X,Y),\|\cdot\|_{\text {op}})$ is Banach?

I know that if $Y$ is Banach then regardless of what the space $X$ is, the space $B(X,Y)$ is necessarily Banach. One of my friends claimed that if $Y$ is not Banach then $B(X,Y)$ cannot be Banach. What he claimed is even much stronger than that. He claimed that $Y$ and $B(X,Y)$ are isometrically isomorphic regardless of the space $X.$ I don't think so because then what is the necessity of $X.$ Can anybody please shed some light on it?

Thanks for your time.

答:
$\textbf{Proposition}$. If $B(X,Y) $ is a Banach space (with the operator norm), then $Y$ is also complete.

Proof.  Let $ (y_n)_n \subset Y $ be a Cauchy sequence and let  $ f \in X^* \setminus \{ 0 \}$. For all  $ n \in \mathbb N $, define  $  T_n \colon X \to Y$, by  $ T_n(x) = f(x) y_n $. Then,  $ T_n $ is linear and bounded, for all $  n \in \mathbb N $. To see that it is bounded, notice that
$ || T_n(x)||  = | {f(x)}| \ ||{y_n}|| \leq  ||{f}|| \ ||{y_n}|| \ ||{x}||$, for all $ x \in X $. Moreover, $ || {T_n-T_m} ||= \sup_{x \in B_X} |{f(x)}| \ ||{y_n - y_m}|| \xrightarrow{n.m \to \infty} 0$. Hence,   $ (T_n)_n \subset  B(X,Y) $ is a  Cauchy sequence. By completeness, we can find  $ T \in B(X,Y) $ such that

$ T_n \to  T$ in the operator norm. Pick any  $ x \in X \setminus \ker f $, then  $ T_n(x) \to T(x) $ and thus  $ y_n \to T(x) / |{f(x)}| \in Y $.

$\textbf{EDIT}$: It is true that $B(X,Y)$ is isometric to $Y$. Let $f \in X^*$ with $||f||=1$. Consider the map
$$ Y \ni y \mapsto T_y \ \colon X \to Y ~~~\text{ where } ~~~T_y(x)=f(x) y. $$
This map is linear and for all $y \in Y$, $||T_y|| = \sup_{x \in B_X}|f(x)| \ ||y||= ||y|| \ ||f||=||y||$. In other words, it's a linear isometry.


问:
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)

My approach:

Solving the quadratic, I get:

$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$

Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $c = 0$.

$a + bp^\frac{1}{3} = 0$

$-\dfrac{a}{b} = p^\frac{1}{3}$

This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?

答:
There's a typo in the first step.  @mjw caught it.  Note that you're applying the quadratic formula on $p^{1/3}$, so $$p^{1/3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}, \text{ if }c \ne 0.$$
Another missing link in your proof is your lack of argument about the claimed irrationality of $± \sqrt{b^2 - 4ac} - 2cp^{2/3}$.

As you handled the degenerate case "$c = 0$ and $b \ne 0$" well, we'll keep the assumption "$c \ne 0$ or $b = 0$" for the rest of the proof.  Also we assume that $\sqrt{b^2-4ac}$ is irrational.  Multiply $2c$ on both side of the above equality, then cube it.

\begin{align}
8c^3p =& -b^3 \pm 3b^2 \sqrt{b^2-4ac} - 3b(b^2-4ac) \pm (b^2-4ac)^{3/2} \\
=& -4b^3+12abc \pm 4(b^2-ac) \sqrt{b^2-4ac} \\
2c^3p =& -b^3 + 3abc \pm (b^2-ac) \sqrt{b^2-4ac}
\end{align}

Make $\sqrt{b^2-4ac}$ the subject of the above equality.  If $b^2-ac \ne 0$,

$$\sqrt{b^2-4ac} = \pm \frac{2c^3p + b^3 - 3abc}{b^2-ac} \in \mathbb{Q},$$
contradicting our assumption on the irrationality of $\sqrt{b^2-4ac}$ if $b^2 - ac \ne 0$.

Assume that $b^2 - ac = 0$.  Then $b^2 - 4ac = b^2 = -3b^2$, and $$p^{1/3} = \frac{(-b\pm\sqrt3|b|i)}{2c} = \begin{cases} \frac{b}{c} e^{2\pi i/3} \text{ or } \frac{b}{c} e^{4\pi i/3} \text{ if } b > 0 \\ \frac{b}{c} e^{\pi i/3} \text{ or } \frac{b}{c} e^{5\pi i/3} \text{ if } b < 0 \end{cases}.$$

In this case, if $b \ne 0$, we don't have the desired conclusion, since $p = \left(\dfrac{b}{c}\right)^3$ might not be an integer, so it's not a perfect cube.

If $b = 0$, we use the assumption $b^2 = ac$, we've $a = 0$ or $c = 0$.

If $a = 0$, only the term $cp^{2/3} = 0$ is left in the original equation, but $p \ne 0$ as $p$ can't be a perfect cube, so $c = 0$.

If $c = 0$, the original equation becomes $a = 0$. Done.


问:
How can I show that 0 is not an eigenvalue of the Hamiltonian matrix?
I have the following Hamiltonian matrix

$$
\begin{align}
H = \begin{pmatrix}A&-BR^{-1}B^{\intercal}\\-C^{\intercal}C&-A^{\intercal}\end{pmatrix}
\end{align}
$$

that admits eigenvalues in pairs, i.e. if $\lambda$ is an eigenvalue, then so is $-\lambda$. Now, $H$ is clearly a block-matrix and has has the dimension $H\in\mathbb{R}^{2n\times2n}$ (each of the blocks is in $\mathbb{R}^{n\times n}$). For this case we have that $R$ is positive definite and the special relations that

$$
\begin{align}
\Gamma = \begin{bmatrix}B&AB&A^{2}B&A^{3}B&\dots&A^{n-1}B\end{bmatrix}\quad \text{and}\quad \Omega =\begin{bmatrix}C\\CA\\CA^{2}\\CA^{3}\\\vdots\\CA^{n-1}\end{bmatrix}
\end{align}
$$

are of full rank. In control theory, or systems theory, we say that $(A,B)$ is reachable and $(C,A)$ is observable when $\Gamma$ and $\Omega$ are of full rank, respectively (just a side note). Since my knowledge of matrix algebra and linear algebra is weak, I have yet to come up with any fruitful approach that would solve this problem. I have tried assuming, for contradiction, that 0 is an eigenvalue, which implies that

$$
\begin{align}
Hx = 0
\end{align}
$$

where $x$ is the corresponding eigenvector, but that is about as far as that takes me. Can I use, somehow, that the eigenvalues come in pairs? Is it not the case that $H^{\intercal}H$ shares eigenvalues with $H$, although they are called singular values?

Edit:

Proof that eigenvalues come in pair:

$$
\begin{align*}
    \lambda\begin{bmatrix}v\\w\end{bmatrix} &= \begin{bmatrix}A&-BR^{-1}B^{\intercal}\\-C^{\intercal}C&-A^{\intercal}\end{bmatrix}\begin{bmatrix}v\\w\end{bmatrix} = \begin{bmatrix}Av-BR^{-1}B^{\intercal}w\\-C^{\intercal}Cv-A^{\intercal}w\end{bmatrix}
\\ & \implies \left\{\begin{matrix}Av-BR^{-1}B^{\intercal}w=\lambda v\\-C^{\intercal}Cv-A^{\intercal}w=\lambda w \end{matrix}\right.
\end{align*}
$$

We investigate what happens when we right-multiply $H^{\intercal}$ by the vector $\begin{bmatrix}w&-v\end{bmatrix}^{\intercal}$:

$$
\begin{align*}
    H^{\intercal}\begin{bmatrix}w\\-v\end{bmatrix}&=
\begin{bmatrix}A^{\intercal}&-C^{\intercal}C\\-BR^{-1}B^{\intercal}&-A\end{bmatrix}\begin{bmatrix}w\\-v\end{bmatrix} = \begin{bmatrix}A^{\intercal}w + C^{\intercal}Cv\\-BR^{-1}B^{\intercal}w +Av\end{bmatrix}
\\ & = \begin{bmatrix}-\lambda w\\\lambda v\end{bmatrix}= -\lambda \begin{bmatrix} w\\- v\end{bmatrix}
\end{align*}
$$

Since $H$ and $H^{\intercal}$ share eigenvalues (same characteristic polynomial) we have showed what we wanted.

Edit:

The Hamiltonian matrix is associated with the algebraic Riccati equation:

$$
\begin{align*}
    A^{\intercal}P+  PA - PBR^{-1}B^{\intercal}P + C^{\intercal}C = 0.
\end{align*}
$$

Here $P\in\mathbb{R}^{n\times n}$ is a symmetric positive definite solution that can be shown to exist and be unique when $\Gamma$ and $\Omega$ are of full rank. Note that $C^{\intercal}C \geq 0$ (symmetric positive semidefinite).

答:
Suppose that $P$ is positive definite and solves the CARE. It is known that the feedback control law $u = -R^{-1}B^TP$ minimizes the cost function
$$
\int_0^\infty (x^TC^TCx + u^TRu)\,dt,
$$
and that this minimal cost is finite. Define $K = A - BR^{-1}B^TP$. The feedback input/state system is governed by
$$
\frac {dx}{dt}  = Kx.
$$
Because the cost-function integral converges, it must be the case that $\lim_{t \to \infty} x(t) = 0$ for any initial state $x(0)$. This can only be the case if $K$ is stable, which is to say that the eigenvalues of $K$ have negative real part.

On the other hand, we find that
$$
PK = -PBR^{-1}B^TP + PA = -C^TC - A^TP \implies\\
H \pmatrix{I\\P} = \pmatrix{A & -BR^{-1}B^T\\-C^TC & -A^T} \pmatrix{I \\ P} = \pmatrix{K\\PK} = \pmatrix{I\\P} K.
$$
Let $D$ denote the Jordan form of $K$, and let $X$ be such that $K = XDX^{-1}$. We have
$$
H \pmatrix{I\\P} = \pmatrix{I\\P} XJX^{-1} \implies
H \pmatrix{X\\PX} = \pmatrix{X\\PX} D.
$$
In other words, each eigenvalue of $K$ is also an eigenvalue of $H$, and the associated eigenvectors are the columns of $\pmatrix{X\\PX}$.

Thus, $H$ has $n$ eigenvalues with negative real part. Because the eigenvalues of $H$ come in $\pm $ pairs, it must hold that $H$ has $n$ more eigenvalues with positive real part. Thus, all eigenvalues of $H$ have either negative or positive real part, which means that none of them can have zero real part.

In particular, we conclude that none of the eigenvalues of $H$ can be $0$.


问:
Irreducible Polynomial in $\mathbb F_{256}$.
Let $\mathbb F_{256}$ be the finite field with $2^8 = 256$ elements. Consider the polynomial over this field
$$
  x^2 + x + 1.
$$
I wanted to know if it is irreducible, so I calculated it for all values $0, \ldots, 256$ and considered the remainder modulo 256, I found that it doesn't evaluates to $0$, so I guess it is irreducible, but is this a valid method?</p>

答:
As "Myself" observed, you are apparently misconstruing $\mathbb F_{2^8}$ as being isomorphic to $\mathbb Z/2^8$, which it is not. For example, the latter has many zero-divisors, and is far from being a field.

Also, in any case, for human-executable work checking 256 cases is usually a terrible method. Instead, look for some structure or meaning. Here, the polynomial $x^2+x+1$ is the third cyclotomic polynomial $(x^3-1)/(x-1)$, and has the property that a root of it (in any field) has multiplicative order $3$. Thus, if there is a root, by Lagrange's theorem $3$ divides the order of the multiplicative group. Conversely, by Cauchy's theorem, if $3$ divides that order then the equation has a root. Thus, it has a root in the field with $2^8$ elements if and only if the multiplicative group has order divisible by $3$. Since $2^8-1=255=5\cdot 51=5\cdot 3\cdot 17$, there is a factor of $3$, and the polynomial has a root, and is reducible.


问:
Problem based on Rigid Body Mechanics (Newtonian-Mechanics)
I got this problem in a tutorial sheet given to me by my Professor.
I am confused as to how to apply the torque about point C, which is hinged. Please be kind to ignore any wrong format of asking the question.
Without any further ado, Here's the question:

The rigid triangular element shown is hinged at C and supported at pin B (BC is horizontal). Assuming all contacts to be frictionless, given F=84.2 N in the direction shown, calculate

(a) the magnitude of reaction force at pin B, and

(b) the magnitude of the net reaction force at hinge C.

答:
Calling

$$
\cases{
A = (x_a,y_a)\\
B = (x_b, y_b)\\
C = (x_c, y_c)\\
F_A = f_a(-\cos(\alpha_1),\sin(\alpha_1))\\
F_B = f_b(\cos(\alpha_2),\sin(\alpha_2))\\
F_C = (f_{c_x},f_{c_y})
}
$$

with

$$
\cases{
\alpha_1 = \frac{\pi}{12}\\
\alpha_2 = \frac{\pi}{6}\\
f_a = 84.2N\\
x_a =-0.075m\\
y_a =-0.075m\\
x_b =0m\\
y_b=0m\\
x_c=0.150m\\
y_c=0m
}
$$

The equilibrium equations are

$$
\cases{
F_A+F_B+F_C = 0\\
F_A\times(A-B)+F_C\times (C-B) = 0
}
$$

or equivalently

$$
\left\{
\begin{array}{l}
 f_a \sin (\alpha_1) (x_a-x_c)+f_a \cos (\alpha_1) (y_a-y_c)+f_b \sin (\alpha_2)
   (x_b-x_c)+f_b \cos (\alpha_2) (y_c-y_b)=0 \\
 f_b \cos (\alpha_2)+2 f_{c_x}=0 \\
 f_b \sin (\alpha_2)+2 f_{c_y}=0 \\
\end{array}
\right.
$$

with solution

$$
\left\{
\begin{array}{l}
f_b= \frac{f_a (\sin (\alpha_1) (x_c-x_a)+\cos (\alpha_1) (y_c-y_a))}{\sin (\alpha_2) (x_b-x_c)+\cos (\alpha_2)(y_c-y_b)} = -146.71N\\
f_{c_x}= \frac{f_a \cos (\alpha_2) (\sin (\alpha_1) (x_a-x_c)+\cos (\alpha_1) (y_a-y_c))}{2 \sin (\alpha_2)(x_b-x_c)+2 \cos (\alpha_2) (y_c-y_b)}= 63.53N \\
f_{c_y} =\frac{f_a \sin (\alpha_1) (x_a-x_c)+f_a \cos (\alpha_1) (y_a-y_c)}{2 (\cot (\alpha_2)(y_c-y_b)+x_b-x_c)}=36.68N \\
\end{array}
\right.
$$

or

$$
|f_b| = 146.71N,\ \ |f_c| = 73.35N
$$

NOTE

$F_C$ is normal to the groove surface.


问:
Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$
Solve for $x$ in

$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$

I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?

The equation I got was:

$$x^4-3x^3-22x^2-48x-32=0$$

答:
(Fill in the gaps as needed. If you're stuck, show what you've tried.)

Expanding everything, the equation is

$$ x^4 - 3x^3 - 22x^2 - 48x - 32 = 0. $$

Factorize this.

Apply the quadratic formula to find the roots.


问:
Product of elements of a finite abelian group

Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.
  Prove that if there is more than one element of order $2$ then $x=e$.

What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)

$G$ has an even number of elements with order $>2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd.

If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$.

If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)

I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?

Thanks in advance. Please excuse my English.

edit: Not a canonical answer but a simple one (if such even exists).

答:
Note that set of all elements of order 2 and the identity element form a subgroup.

It’s enough to prove the theorem for such a subgroup/group.

Consider two distinct elements $a$ and $b$ of order 2. The subgroup generated by these two elements is,
$$H_1 = \{ a, b, ab, e\}$$

Product of all elements of $H_1$ is $e$.

Proof 1:

Let’s try to generate the whole group from scratch.  We can keep extending this subgroup to generate bigger subgroup by adding  new elements.

At any stage new subgroup can be formed from $H_{n}$ by extending from new element $x$ of order $2$ as shown below,

$$H_{n+1} = x H_{n} \cup H_{n}$$

Consider the product of all elements of $H_{n+1}$ denoted by $P_{n+1}$

$$P_{n+1} = x^{2k}P_n^2,$$
Where $k = 2^n$.

Hence proved (by induction).

Proof 2:

Consider the distinct cosets of $H = H_1$ in $G$,
$$G = \cup _ {x \in G} xH.$$

The product of all the elements in each such coset is equal to $e$. Hence, the product of all elements in the group $G$ is $e$.


问:
Weak and $*$-weak convergence of functionals in $C[0,1]$
For a sequence of functionals $F_n \in C[0,1]^*$,
where
$$
F_n(f) = \int_0^1 f(x^n)dx
$$
I want to know if this sequence converges weakly.

My attempt.
I can show that there is a weak-star convergence for this sequence. This sequence weakly-star converges to $F\in C[0,1]^*$, where
$$
F(f) = f(0), \ f\in C[0,1]
$$

But I can not prove or disprove weak convergence. Using the fact that $C[0,1]^*\cong BV_0[0,1]$, where $BV_0[0,1]$ is a space of bounded variation functions $g$ such that $g(0) = 0$ and  $g(x-0) = g(x)$. There is a correspondence between functionals  $F_n\in C[0,1]^*$ and functions $g_n = x^{\frac{1}{n}} \in BV_0[0,1]$. And a function $g(x)$ defined by
$$
g(x) = \begin{cases}
0, & x = 0\\
1 & x\in (0, 1]
\end{cases}
$$
corresponds to $F$.
So we need to investigate the convergence $G(g_n)$ to $G(g)$ for every functional from $BV_0[0,1]^*$. I have tried many functionals but all of them converged but I can not prove this in general case. Here I stuck.

答:
Consider the functional $G$ defined on $BV_0[0,1]$ by
$$
  G(g) = \lim_{x\to 0_+}g(x).
  $$
Then
$
  G(g_n) = 0,
  $
but $G(g) = 1$.


问:
Decomposition of a prime in a tower of extension
Let $ \mathbb{Q} \subseteq \mathbb{E} \subseteq \mathbb{F}$ be tower of extensions such that $[\mathbb{F}: \mathbb{E}]=[\mathbb{E}:\mathbb{Q}]=2$. Let $p$ be a prime such that $p \nmid \text{disc}_{\mathbb{E}/\mathbb{Q}}$ but $p | \text{disc}_{\mathbb{F}/\mathbb{Q}}$. Hence $p$ is unramified in $\mathbb{E}$ i.e. either $p\mathcal{O}_\mathbb{E}$ is a prime ideal or $p\mathcal{O}_\mathbb{E}=\mathfrak{q}_1\mathfrak{q}_2$ for some prime ideals $\mathfrak{q}_1,\mathfrak{q}_2 \subseteq \mathcal{O}_\mathbb{E}$ but $p$ ramifies in $\mathbb{F}$.

I would like to know how would the decomposition of $p{\mathcal{O}_\mathbb{F}}$ look like as a product of prime ideals of $\mathcal{O}_\mathbb{F}$.

Since $p$ ramifies in ${\mathcal{O}_\mathbb{F}}$, $p{\mathcal{O}_\mathbb{F}}= \prod \mathfrak{p}_i^{e_i}$ with at least one $e_i>1$. Can we be more specific about the decomposition of $p\mathcal{O}_\mathbb{F}$ considering the decomposition of $p\mathcal{O}_\mathbb{E}$? For example, can we say that $p\mathcal{O}_\mathbb{F}$ is either $\mathfrak{p}_1^2$ or $\mathfrak{p}_1^4$ or $\mathfrak{p}_1^2\mathfrak{p}_2^2$ for some prime ideals $\mathfrak{p}_1,\mathfrak{p}_2 \subseteq \mathcal{O}_\mathbb{F}$?

Any suggestions will be appreciated.

Thanks in advance.

答:
If $\newcommand{\OO}{\mathcal{O}}\newcommand{\EE}{\mathbb{E}}\newcommand{\FF}{\mathbb{F}}\newcommand{\QQ}{\mathbb{Q}}\newcommand{\pri}{\mathfrak{p}}\newcommand{\qri}{\mathfrak{q}}p\OO_\EE$ is prime, then $p\OO_\FF = \pri^2$, where $\pri$ is a prime ideal of $\OO_\FF$. This is because this is the only way for a prime to ramify in the quadratic extension $\FF/\EE$.

If $p\OO_\EE = \qri_1 \qri_2$, where $\qri_i$ are distinct prime ideals of $\OO_\EE$, then by assumption, one of these two prime ideals has to ramify in $\FF/\EE$, but the other one could be split, inert, or also ramified. So the factorization of $p \OO_\FF$ has one of the following three forms (where $\pri_1, \pri_2, \pri_3$ are prime ideals of $\OO_\FF$): $\pri_1^2 \pri_2 \pri_3$, $\pri_1^2 \pri_2$, or $\pri_1^2 \pri_2^2$. If $\FF/\QQ$ is a Galois extension, then only the last possibility (both $\qri_1$ and $\qri_2$ are ramified) can occur, since there's a $\QQ$-automorphism of $\FF$ interchanging $\qri_1$ and $\qri_2$.

Here's an example of how one of the other two possibilities can occur. Let $\EE = \QQ(i)$, where $i^2 = -1$. In $\mathbb{Z}[i]$, we have $5 = (2 + i)(2 - i)$. Let $\FF = \EE(\sqrt{2+i})$. (This is a $D_4$ extension of $\QQ$.) A quick computation in Sage (or you can do it by hand) shows that
$$(2 - i) \OO_\FF = (i\sqrt{2+i} - i + 1)(i \sqrt{2+i} + i - 1),$$
and these are distinct prime ideals of $\OO_\FF$. So
$$5 \OO_\FF = (\sqrt{2+i})^2 (i\sqrt{2+i} - i + 1)(i \sqrt{2+i} + i - 1).$$

I'll leave it to you to find an example where one prime ramifies and the other is inert, rather than split.


问:
Is there an isomorphism of $F[x]/\langle x^2+a\rangle$ to $F[x]/\langle x^2+x+b\rangle$ just as fields?
Question

If $F$ is a field of characteristic $2$, can there be irreducible polynomials $x^2+a$ and $x^2+x+b$ over $F$ such that $F[x]/\langle x^2+a\rangle$ is isomorphic to $F[x]/\langle x^2+x+b\rangle$ just as fields?

My Attempt

I know they cannot be isomorphic over $F$ because, if they were so, let $u$ be the representative of $x$ in $F[x]/\langle x^2+a\rangle$, so that $u^2=a$, and let $v$ be the representative of $x$ in $F[x]/\langle x^2+x+b\rangle$, so that $v^2=v+b$, then we can suppose, without loss of generality, that $F[u]=F[v]$ is the same field with $F$ as common subfield, and there would exist $\alpha,\beta\in F$ such  that $v=\alpha+\beta u$, so:
$$v+b=v^2=(\alpha+\beta u)^2=\alpha^2+2\alpha\beta u+\beta^2a=\alpha^2+\beta^2a,$$
so that $v\in F$, a contradiction.

答:
It depends very much on the field $F$. For instance, every field homomorphism must preserve the prime subfield, so if $F=\mathbb{F}_2$ then the answer is no, by your argument above. (Edit: as Daniel Schepler says below, it can be argued much more straightforwardly in that case, since there are of course no irreducible polynomials of the form $x^2+a$ in $\mathbb{F}_2[x]$. So, in particular, there are vacuously no isomorphic fields of the form you want. The same argument can be used to show that the answer is "no" whenever $F$ is a finite field extension of $\mathbb{F}_2$.)

On the other hand, let $F=\mathbb{F}_2(t)$, the field of rational functions in $t$ with coefficients from $\mathbb{F}_2$, and let $a=b=t$. For convenience denote $$F_1:=F[x]\big/\langle x^2+a\rangle=\mathbb{F}_2(t)[x]\big/\langle x^2+t\rangle$$ and $$F_2:=F[y]\big/\langle y^2+y+b\rangle=\mathbb{F}_2(t)[y]\big/\langle y^2+y+t\rangle.$$ Now, we claim that $\overline{y+t}$ is transcendental over $\mathbb{F}_2\subseteq F_2$. To see this, first note that, since $\mathbb{F}_2(t)$ is a field, the canonical map $\mathbb{F}_2(t)\to F_2$ taking $t\mapsto\overline{t}$ is an injection. In particular, since $t$ is transcendental over $\mathbb{F}_2\subseteq\mathbb{F}_2(t)$, this means $\overline{t}$ is transcendental over $\mathbb{F}_2\subseteq F_2$ as well. Now, suppose for contradiction that $\overline{y+t}$ is algebraic over $\mathbb{F}_2$. Then, since $\overline{y}$ is algebraic over $\mathbb{F}_2(\overline{y+t})$ and algebraic extensions of algebraic extensions are algebraic, $\overline{y}$ would also be algebraic over $\mathbb{F}_2$. But this would mean $\overline{t}=\overline{y+t}+\overline{y}$ is algebraic over $\mathbb{F}_2$ too, a contradiction. Thus $\overline{y+t}$ is transcendental over $\mathbb{F}_2$, and so we may define a map $\mathbb{F}_2(t)\to F_2$ by sending $t\mapsto \overline{y+t}$.

By the universal property of polynomial algebras, we may extend this to a map $f:\mathbb{F}_2(t)[x]\to F_2$ that sends $x\mapsto\overline{y}$.  We claim $\ker f=\langle x^2+t\rangle$, whence $f$ will descend to an isomorphism $F_1\cong F_2$. First note that $f(x^2+t)=f(x)^2+f(t)=\overline{y}^2+\overline{y+t}=\overline{0}$, so indeed $\ker f\supseteq \langle x^2+t\rangle$. On the other hand, $\langle x^2+t\rangle$  is a maximal ideal of $\mathbb{F}_2(t)[x]$, and $f$ is a non-zero map, so we have $\ker f=\langle x^2+t\rangle$, as desired. Thus this gives an example of two isomorphic fields of the form you are looking for.


问:
How many five digit numbers can be formed that are both odd and less than $40,000$ with the set $\{2,3,4,5,6\}$
I have to find how many possible five digit numbers which are both less than 40,000 and odd can be made with the set $\{2,3,4,5,6\}$ by using each number once only .

So far I have noted that there are only two possible choices for the first digit; $2,3$

And only two possible choices for the last digit; $3,5$

Hence, by using dashes to represent the possible number of choices for each corresponding digit:

$$\underline{2} \space \underline{}\space \underline{}\space \underline{}\space \underline{2}$$

Filling in the remaining three middle digits:

$$\underline{2} \space \underline{3}\space \underline{2}\space \underline{1}\space \underline{2}$$
$$2*3*2*1*2=24$$
However I know this is incorrect, I am a bit confused on how to deal with the fact that both the first digit and the last digit have an overlapping choice of the number $3$, I have tried to split it into two cases with one case where the first digit is $2$ and the other case where the first digit is $3$ but it

How would I go about breaking down this problem logically and solving it?

答:
Assuming you cannot repeat digits...

Case 1: the first digit is a 2.

There are 2, candidates for the last digit, and 6 ways to fill the remaining digits.

12 arrangements under this condition.

Case 2: the first digit is a 3.

The last digit is a 5. There are still 6 ways to fill the remaining digits.

$6+12 = 18$


问:
$f(x) < g(x) \Rightarrow \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ $. Why not in both directions?
Let $f$ and $g$ be continuous functions on the interval $\left[ a,b\right]$ and $a<b$

$ f(x) < g(x)  \Rightarrow \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ $ for $x \in \left[ a,b\right]$

Why doesn't the statement apply in the other direction?

$ \int_{a}^{b} f(x) dx < \int_{a}^{b} g(x)dx \ \not\Rightarrow  f(x) < g(x)$

答:
Intuitive answer: because of width. A function could go to a very large value for a very small interval and the area will be still very small. For example, consider
$$
f(x) = \{x \in [0,0.5]: 0; x \in (0.5,1] : 10\}
$$
(in other words, ${f(x)}$ is $0$ when ${0\leq x \leq 0.5}$, and ${f(x)=10}$ when ${0.5< x \leq 1}$) easily you can see
$$
\int_{0}^{1}f(x)dx = 10\times 0.5 = 5
$$
Now consider
$$
g(x) = 6
$$
then ${\int_{0}^{1}g(x)dx=6}$. So ${\int_{0}^{1}f(x)dx < \int_{0}^{1}g(x)dx}$; however, clearly ${f(x)}$ is not smaller than ${g(x)}$ as ${f(x)}$ takes on the value ${10}$, which is greater than $6$.


问:
Finding the nth integral of $\ln(1+z/k)$
I can find the nth integral of $\ln(z)$ as follows:
\begin{aligned}
\left(\frac d{dz}\right)^{-n}\ln(z)&=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln(t)dt\\
&=\frac1{n!}\left[\int\limits_0^z\frac1t(z-t)^ndt-z^n\ln(0)\right]\\
&=\frac1{n!}\left[\int\limits_0^1\frac{z^nt^n-z^n+z^n}{1-t}dt-z^n\ln(0)\right]\\
&=\frac{\ln(z)-H_n}{n!}z^n,
\end{aligned}
but I can't get very far with $\ln(1+z/k)$. I was able to come up with this but it is only a conjecture:
\begin{aligned}
\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln\left(1+\frac tk\right)dt&=\frac{\ln\left(1+\frac{z}{k}\right)-H_n}{n!}(k+z)^n+\sum ^{n}_{i=0}\frac{H_{i}}{i!(n-i)!}z^{n-i}k^i\\
&=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=0}\frac{H_{n} -H_{i}}{i!( n-i) !} z^{n-i} k^{i}.
\end{aligned}
I'm only using $k,n\in\mathbb{N}$ and $z\in\mathbb{R}$ for now. Any help proving this would be appreciated.

答:
I realized that showing how I came up with my conjecture could probably provide the basis of a proof. I'm not sure if I should post this as a separate answer or as part of the question, but since it kind of answers the question I'm posting as an answer. I will not, however, accept this.

I already knew the answer would be in the form $$\left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n+P(z,k,n),$$ for some polynomial $P$, thanks in part to the nth integral of $\ln(z)$:
\begin{aligned}
\left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)&=\left(\frac d{d(1+z/k)}\frac{d(1+z/k)}{dz}\right)^{-n}\ln\left(1+\frac zk\right)\\
&\rightsquigarrow\left(\frac d{d(1+z/k)}\right)^{-n}\ln\left(1+\frac zk\right)\left(\frac{d(1+z/k)}{dz}\right)^{-n}\\
&=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}\left(1+\frac zk\right)^n\left(\frac1k\right)^{-n}\\
&=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n.
\end{aligned}
I made a table of $P$ to hopefully find a pattern
\begin{matrix}
n & P(z,k,n)\\
0 & 0\\
1 & k\\
2 & kz+\frac{3k^2}4\\
3 & \frac{kz^2}2+\frac{3k^2z}4+\frac{11k^3}{36}\\
4 & \frac{kz^3}3+\frac{3k^2z^2}8+\frac{11k^3z}{36}+\frac{25k^4}{288}\\
\dots & \dots
\end{matrix}
and indeed several nice patterns appear. The table shows that we can expect $P$ to be in the form $$P(z,k,n)=\sum_{i=0}^n\frac{H_i}{i!(n-i)!}z^{n-i}k^i,$$ which then leads to
\begin{aligned}
\left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)&=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n+\sum_{i=0}^n\frac{H_i}{i!(n-i)!}z^{n-i}k^i\\
&=\frac1{n!}\ln\left(1+\frac zk\right)(k+z)^n-\sum_{i=0}^n\frac{H_n-H_i}{i!(n-i)!}z^{n-i}k^i.
\end{aligned}
Thus, all that really has to be done is prove this form of $P$.


问:
Optimal control problem proving an abs min.
I'm having a trouble with this optimal control problem: $$\int_0^4(\dot{x}^2+x)dt \to extr$$ $$\vert \dot{x} \vert \leqslant 1 $$ $$x(4)=0$$ I found an extreme function $\hat{x}(t)$ (Sorry for wrong notions.I don't know an english terminology for this) $$\hat{x}(t)=\begin{cases} \frac{t^2}{4}-3 , t \in[0;2] \\ t-4 , t\in[2;4] \end{cases}$$ and control $$\hat{u}(t)=\begin{cases} \frac{t}{2} , t \in[0;2] \\ 1 , t\in[2;4] \end{cases}$$ This is considered to be true (in the book it is marked as solution). But I'm stucked in proving that $\hat{x}(t)$ is abs min. What I've already done: $$B_0(\hat{x}(t)+h)-B_0(\hat{x}(t))$$
Here $h$ is perturbation. So in order to $\hat{x}(t)$ be the abs minimum this difference should be greater or equal $0$. So
$$\int_0^4(\dot{h}^2+2\hat{x}'\dot{h}+h)dt$$ here the the hatch is derivative (as $\dot{x}$).
$$\int_0^4hdt = h t \vert_0^4 - \int_0^4(\dot{h}t)dt$$ First term is zero ($h(4)=0$)
So we have $$\int_0^4\dot{h}^2dt+\int_0^4(2\hat{x}'-t)\dot{h}dt$$ On $t\in [0;2]$ $\hat{x}'(t) = \frac{t}{2}$ So the integral over this segment is zero. Finally we have:
$$\int_0^4\dot{h}^2dt+\int_2^4(2-t)\dot{h}dt$$ as $\hat{x}'(t)=1$ on $[2;4]$. And from this part I'm stucked. I don't know how to prove that the last expression is greater or equal $0$.

答: So I have found problem. $\dot{h}(t) \in [-2;0]$ Since $\vert 1+\dot{h}\vert \leqslant 1$ And the second term always greater or equal $0$. The statement about values of $\dot{h}$ always was in front of my eyes, but I thought about $h$ which is greater or equal $0$. It was horrible mistake.


问:
Solve equation $x^2 \cdot |x| = 2 \cdot \bar{x}$ in complex numbers
Well, to solve it I firstly tried to use exponential form. However, it didn't give me expected result. Also, I tried using trigonometric form but I am a bit inexperienced in it, so I couldn't observe anything interesting with it either.

I am new to complex numbers, so I would love to hear some advices on how to think of solution in such equations and maybe some insights on how to solve this one.

答:
The given equation $2 \bar{x} =x^2 |x|$ implies
$$2x=\bar{x}^2|x|=\left(\frac{x^2|x|}2\right)^2|x|=\frac14x^4|x|^3\implies x(x^3|x|^3-8)=0
$$
which leads to $x=0$ and $x|x|=2e^{i \frac{2\pi n}3}$. The second equation leads to $|x|=\sqrt2$, and in turn the corresponding solutions
$$x=\sqrt2 e^{i \frac{2\pi n}3}$$


问:
Prove that relation is partial order, and if it is, find its minimal and maximal elements.
Let $A$ be relation $[((a,b),(c,d)) \in \mathbb{N}^2 \times \mathbb{N}^2; (\exists k \in \mathbb{N}) (c=ka \land d=kb)]$

Prove that it is reflexive, antisymmetric, and transitive.

1.) Rexlexivity: $((a,b),(a,b)) \in A \to a=ka \land b=kb \to k=1 \land k=1$

It is reflexive because we get a tautology.

2.)Antisymmetricity: $((a,b),(c,d)) \in A \land ((c,d),(a,b)) \in A \implies (a,b)=(c,d)$

$(c=ka \land d=kb) \land (a=kc \land b=kd) \to (k=\frac{c}{a} \land k=\frac{d}{b}) \land (k=\frac{a}{c} \land k=\frac{b}{d})\to (\frac{c}{a}=\frac{d}{b})\land (\frac{a}{c}=\frac{b}{d})\to$

$ \to (\frac{a}{c}=\frac{b}{d})=(\frac{a}{c}=\frac{b}{d}) \to \frac{a}{c}=\frac{b}{d}$

This can only be true when $(a,b)=(c,d)$, so $a=c \land b=d$

3.) Transitivity: $((a,b),(c,d)) \in A \land ((c,d),(x,y)) \in A \implies ((a,b),(x,y)) \in A$

$(c=ka \land d=kb) \land (x=kc \land y=kd) \to (k=\frac{c}{a}\land k=\frac{d}{b}) \land (k=\frac{x}{c} \land k=\frac{y}{d}) \to (\frac{c}{a}=\frac{d}{b})\land(\frac{x}{c}=\frac{y}{d})\to$

$\to (\frac{c}{d}=\frac{a}{b})\land(\frac{x}{y}=\frac{c}{d})$ So $\frac{c}{d}=\frac{a}{b}=\frac{x}{y}$ which means that $\frac{a}{b}=\frac{x}{y}$ and from this we know that the relation is transitive.

In conclusion, the relation is reflexive, antisymmetric, and transitive. It doesn't have a maximum so it doesn't have a maximal element either. It does have a minimal element which is and the minimal element is $((0,0)(0,0))$

Is this correct?

Thanks!

答:
For reflexivity, you have the right idea, but I think that you don't phrased it correctly.
It should be something along these lines:

since $k=1$ is a natural number, and $a=1a$ and $b=1b$, then $((a,a),(a,a)) \in A$.

Thus, $A$ is reflexive. (Notice that you inverted the order of the reasoning; I'm not sure if it is wrong or just confusing.)

Now, for anti-symemtry, suppose $(a,b)A(c,d)$ and $(c,d)A(a,b)$.

Then there exist $k,k' \in \mathbb N$ such that $c=ka$, $d=kb$, $a=k'c$ and $b=k'd$.

Thus, $a=k'ka$, whence $k'k=1$, and it follows that $k=k'=1$.

Hence $a=c$ and $b=d$, and therefore $(a,b)=(c,d)$.

We conclude that $A$ is anti-symmetric.

For transitivity, suppose that $(a,b)A(c,d)$ and $(c,d)A(e,f)$.

Then there exist $k,k' \in \mathbb N$ such that $c=ka$ and $d=kb$, and $e=k'c$ and $f=k'd$.

Thus, $e=k'ka$ and $f= k'kb$, yielding $(a,b)A(e,f)$.

Hence, $A$ is transitive.

It follows that $(\mathbb N^2,A)$ is a partially ordered set.

It has a maximum element (thus, the only maximal element), which is $(0,0)$;

indeed, for every $(a,b) \in \mathbb N^2$, we get $0=0a$ and $0=0b$, so $(a,b)A(0,0)$ for all $(a,b) \in \mathbb N^2$.

Concerning minimal elements, it has infinitely many:
for example, $(1,1)$ is a minimal element, and it is below precisely the elements of the form $(a,a)$, with $a \in \mathbb N$.

Otherwise, we get those of the form $(a,b)$ with $\gcd(a,b)=1$;

indeed, if $d|a$ and $d|b$ then $(a/d,b/d)A(a,b)$;

conversely if $(a,b)A(c,d)$, then $c=ka$ and $d=kb$, for some $k \in \mathbb N$;
if $k=1$ this is just reflexivity; otherwise $k|c$ and $k|d$.


问:
What precisely is a vacuous truth?
Is there a proper and precise definition that goes something like this?

Definition. A statement $S$ is a vacuous truth if ... ...

答:
An implication is said to be vacuously true if its antecedent is false.

$$\neg A \implies (A\implies B)$$

This logical principle is a tautology, and can be used to introduce an arbitrary consequent in an implication.

Example 1: If pigs can fly, then the President is a genius.

Proof: The antecedent, "pigs can fly," is false. Therefore the above implication is true, whether or not the consequent, "the President is a genius," is true. Note: We cannot infer from this that the President is not a genius. Nor can we infer the contrary that he is a genius.

Example 2: $\forall x: (x\in \emptyset \implies 0=1)$

Proof: $y\in \emptyset$ is always false. Then the implication $y\in \emptyset \implies 0=1 $ is true (vacuously so). Generalizing, we obtain as required: $$\forall x: (x\in \emptyset \implies 0=1)$$

Title: What precisely is a vacuous truth?


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答: I use Pari/GP. SAGE includes this as a component too, but I really like GP alone, as it is. In fact, GP comes with integer relation finding functions (as you mentioned) and has enough rational/series symbolic power that I have been able to implement Sister Celine's method for finding recurrence relations among hypergeometric sums in GP with ease.</p>

Title: Free online mathematical software


问:
Prove that $x$ is either zero or a zero divisor

Let $x$ be a nilpotent element of a commutative ring $R$.
Prove that $x$ is either zero or a zero divisor.

My attempt : I got the answer here but i didn't understand the answer

My proof  :since  $x$ is nilpotent  , so there exist some index $ m$  with $x^m=0.$

if $m=1 $ then $x=0$

if $m\neq 1$ ,then $x\neq 0$ i,e $xx^{m-1}=0$ where $x^{m-1}=0$ but $x\neq 0$

for example  take $x= \begin{bmatrix} 0&1\\0&0\end{bmatrix} \neq0$ but $x^{m-1}= \begin{bmatrix} 0&0\\0&0\end{bmatrix}$ where $m-1 >0$

Is my proof is correct or not ?

答:
Suppose as you said that $x$ is nilpotent where the index $m$ is the lowest integer for which we have $x^m = 0$.

If $m = 1$ we have as you said $x=0$.

If $m > 1$ then $x \ne 0$ and $x\cdot x^{m-1} = 0$

But at the beginning we supposed $m$ as the lowest integer for which we have $x^m = 0$, so $x^{m-1} \ne 0$.

Then you can see that the case $m > 1$ lead us to the conclusion that $x$ is a zero-divisor.


问:
Prove that $ 2^{6ℓ + 2} ≡ 4\pmod{18}$
Prove that  $2^{6ℓ + 2}≡ 4 \pmod{18}$
for $0\leℓ ∈ ℤ$.

答:
Hint: This follows from the fact that
$$
2^6 \cdot 4 \equiv 4 \pmod{18}.
$$

Title: Prove that $ 2^{6ℓ + 2} ≡ 4\pmod{18}$


问:
$\lim_{y \to 0^{+}} \int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \frac{\pi}{2}$
I have shown using the application of Dominated Convergence Theorem (i.e., interchange of differentiation and integral) that
$$
\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \frac{\pi}{2}-\tan^{-1}y
$$
for $(x,y) \in [0,\infty) \times (0,\infty)$.

Now I want to conclude that
$$
\int_{0}^{\infty} \frac{\sin x}{x} dx =\int_{0}^{\infty} \lim_{y \to 0+}e^{-yx} \frac{\sin x}{x} dx =\lim_{y \to 0+}\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x} dx = \lim_{y \to 0+} \frac{\pi}{2}-\tan^{-1}y =  \frac{\pi}{2}
$$
(The hint which is given with this part is to use DCT.)

To interchange the limit and the integral sign, using DCT, I need to find a non-negative function $g$ which dominates $ |(e^{-yx}\sin x) /x|$ and is integrable, i.e., $g \in L^{1}([0,\infty))$.

The only function I could think of is (obviously) $|\sin x/x|$. But I don't know whether $|\sin x/x|$ is integrable.

答:
The interchange of the limit and integral is in question here. Rather than the dominated convergence theorem, we can apply the following theorem and the Dirichlet test.

If $f:[a,\infty)\times [0,\infty) \to \mathbf{R}$ is continuous and
convergence of the improper integral $F(y) =\int_a^\infty f(x,y) \,
 dx$ is uniform, then $F$ is continuous on $[0,\infty)$ and $$\lim_{y
 \to 0+}F(y) = \int_a^\infty \lim_{y \to 0+}f(x,y) \, dx  =
 \int_a^\infty f(x,0) \, dx$$

With $f(x,y) =  e^{-yx} \frac{\sin x}{x}$ for $x > 0$ we can extend to a continuous function by defining $f(0,y) = 1$. Note that $\int_0^c \sin x \, dx$ is uniformly bounded for all $c > 0$ and $y\geqslant 0$. Furthermore $\left|\frac{e^{-yx}}{x}\right| \leqslant \frac{1}{x}$, and, hence, $\frac{e^{-yx}}{x} \searrow 0$ monotonically and uniformly for all $y \geqslant 0$. By the Dirichlet test the improper integral of $x \mapsto f(x,y)$ over $[,\infty)$ converges  uniformly for all $y \geqslant 0$.

Thus,

$$\lim_{y \to 0+}\int_{0}^{\infty} e^{-yx} \frac{\sin x}{x}\, dx =\int_{0}^{\infty}  \lim_{y \to 0+}e^{-yx} \frac{\sin x}{x}\, dx = \int_{0}^{\infty}  \frac{\sin x}{x}\, dx $$