问:
Continuity with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Is there $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$?
Let $f: \mathbb{R} \to \mathbb{R}$ be continous with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Show there exists $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$.

My question is, do we even need $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$ to show this, or is is sufficient, that f is bounded? What happens if $f$ is monotonous?

I posted a sketch for the proof of the original/task problem below.

答:
This is how I would approach the proof for the original problem/task:

First consider $f:[0,1]\to \mathbb{R}$. Since $f$ is continuos, we can apply the extreme value theorem. Thus, there exists $a:= \min_{x\in [0,1]}$ and $b:=\max_{x\in [0,1]}$.

Now we can use $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. As a result there cannot exists $\xi \in \mathbb{R}$ with $f(\xi)<f(a)$ or $f(\xi)>f(a)$. Has anyone an idea how this can be shown rigorously? As a consequence $a$ is a global minimum and $b$ is a global maximum.

Let $g: \mathbb{R} \to \mathbb{R},\; g(x)=f(x+\sqrt{2})-f(x) $.

Then we have $g(a)=f(a+\sqrt{2})-f(a)\geq 0$, and $g(b)=f(b+\sqrt{2})-f(b)\leq 0$. Because $a$ is a global minimum and $b$ is a global maximum of $f$.

If $g(a)=0$ or $g(b)=0$ we are done. Else, by the intermediate value theorem, there exists $x \in (a, b)$, such that $g(x)=0$.

Why are we done? Because  $g(x)=0 \iff f(x+\sqrt{2})- f(x)=0 \iff f(x)=f(x+\sqrt{2})$.


问:
Linear transformation invertible or not?

Let $ T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear transformation defined such that the inner product of $\langle T(v), v \rangle = 0$ for all $v$ in $\mathbb{R}^2$. Is $T$ invertible or not?

Attempt:
If $T$ has eigenvalues (i.e., it's characteristic polynomial splits), then $0$ is an eigenvalue. This is because $T(v) = kv$ ($k$ is an eigenvalue). Therefore $\langle Tv, v\rangle  = \langle kv, v\rangle  = k\langle v, v\rangle  = 0$. Since $v$ is anything in $\mathbb{R}^2$, it is not necessarily $0$, which implies $k = 0$. So if a transformation has a $0$ eigenvalue, it is not invertible.

If $T$ doesn't have eigenvalues, then it's not injective and thus not invertible.

Help! Am I close?

答:
With the given information, we can determine the form of $T$ quite specifically, and see that it will almost always be invertible. We don't need to consider eignevalues at all.

Suppose $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ is linear and $\forall v \in \mathbb{R}^2, \langle T(v),v\rangle=0$. Then $T$ is a $\frac{\pi}{2}$ rotation, followed by a scaling.

Hint: If we take $e_1,e_2$ to be the standard basis of $\mathbb{R}^2$, then $\langle T(e_1),e_1\rangle = 0 = \langle T(e_2),e_2\rangle$. You can use this to say something about the values of $T(e_i)$. But we also have that $\langle T(e_1+e_2),e_1+e_2\rangle=0$, which tells us something else. (Using the fact that $T$ is linear so $T(v+w)=T(v)+T(w)$.)

ContentLicense: CC BY-SA 3.0


问:
If I know the average value over a radial cross section, can I compute the average value over a disk (assuming radial symmetry)?
Let's say that I am trying to calculate the average velocity of a fluid through a pipe and I only know the average velocity of a cross-section through the center, can I find the average velocity in the pipe as a whole if we know it is radially symmetric? How? I have made two attempts and failed both times. But I'll show my work below. Assume the pipe has radius R:

We are given the average velocity in the two dimensional slice, so we know $\bar{v}_{2D} = \frac{1}{R}\int_0^R v(r)\,dr $ **NOTE: we do not know v(r), just what this result is.

If we knew the function for the velocity profile (we don't) then the average velocity in the pipe would just be $ \frac{1}{\pi R^2}\int_0^{2\pi}\int_0^Rv(r)\,rdrd\theta$. At this point I am tempted to use the first relation to say that:

$\bar{v}_{2D} = \frac{1} {R}\int_0^R v(r)\,dr \longrightarrow \int_0^R v(r)\,dr=\bar{v}_{2D}  R$ and substitute into the above equation but we can't because it is $rdr$ and not just $dr$.

My second line of reasoning is that this must be simpler than I am making. It has to be fundamentally tied with the relationship between area, $\pi$, and radius. I have a feeling that perhaps the ratio between the area of a right triangle and the area of the cone generated by rotating that triangle around its 90 degree junction in a full circle would be equal to the ratio of this average velocity to the average velocity over the pipe as a whole. Perhaps the ratio of every radially symmetric volume to a radial slice is the same? However this is just a guess and I have no Idea how to really verify it. Also, if this guess is wrong, then I have no idea how to proceed.

EDIT possible answer:

Using Peter's advice below I considered the following thought experiment to show this is not enough information to solve the problem: The ratio of the area of a rectangle to the volume of the cylinder it generates is different than the ratio of the area of right triangle to the volume of the cone it generates. However both are proportional to πr
which is kind of cool!So the answer would probably be proportional to $\bar{v}_{2D} \pi r$ but we can't know how...

答:
It is not possible in general. It will suffice to show that there exist two profiles $v(r)$ with the same $v_{2d}$ but different average velocity over the disk. For example

$$
v_1(r)=\frac{r}{R} \\
v_2(r)=1-\frac{r}{R}
$$

Satisfy

$$
v_{2d}=\frac{1}{R}\int\limits_0^Rdr \ v_1(r)=\frac{1}{R}\int\limits_0^Rdr \ v_2(r)=\frac{1}{2}
$$

However, the average velocity is different

$$
\bar{v}_1=\frac{2\pi}{\pi R^2}\int\limits_0^R dr \ rv_1(r)=\frac{2}{3} \\
\bar{v}_2=\frac{2\pi}{\pi R^2}\int\limits_0^R dr \ rv_2(r)=\frac{1}{3} \\
\bar{v}_1 \neq \bar{v}_2
$$

We conclude that in general the average velocity is not uniquely determined by $v_{2d}$.

ContentLicense: CC BY-SA 4.0


答: The function $\arctan x$ is bounded and continuous, but the $x$ the problem asks for does not exist, so the periodicity assumption is indeed important.

Title: Continuity with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Is there $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$?

Body:
Let $f: \mathbb{R} \to \mathbb{R}$ be continous with $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$. Show there exists $x \in \mathbb{R}$ with $f(x)=f(x+\sqrt{2})$.

My question is, do we even need $f(x)=f(x+1)$ $\forall x\in \mathbb{R}$ to show this, or is is sufficient, that f is bounded? What happens if $f$ is monotonous?

I posted a sketch for the proof of the original/task problem below.


问:
A question about the use of successions in the quantifier clause of the definition of satisfaction
In first order logic, why do we say that:
$\exists$ $xi$ $\phi$ is satisfied by succession si iff there is a different succession from s that satisfied $\phi$?

For instance, maybe the only succession that satisfies $\phi$ is actually si, and there is no other succession different from si that satisfies $\phi$. Why there must be a different succession?

答:
The basic fact is to define what does it mean for a structure (or: interpretation) $\mathfrak A$ to satisfy formula $\varphi$ with variable assignment function $s$. In symbols:

$\mathfrak A,s \vDash \varphi$ (alternatively: $\mathfrak A \vDash \varphi[s]$).

Intuitively, the definition means that the "translation" of $\varphi$ determined by $\mathfrak A$ is true, where the variable $x$ is translated as $s(x)$ wherever it occurs free in the formula.

The "variable assignment function" trick is needed in order to manage free occurrences of variables in formulas, because a variable has no fixed meaning in an interpretation.

Consider the simple example of formula $(x_1=0)$ where $\mathfrak A$ is $\mathbb N$.

What is its truth-value? It depends...

For assignment $s_1$ such that $s_1(x_1)=0$ we have that $(x_1=0)$ is true (i.e. $\mathbb N,s_1 \vDash (x_1=0)$), while for $s_2$ such that $s_2(x_1)=1$ we have that $(x_1=0)$ is false.

If so, what does it mean that the formula $\exists x_1 (x_1=0)$ is true in $\mathbb N$?  It means that there is at least one element of $\mathbb N$ - call it $d$ - such that, when variable $x_1$ is "translated" as $d$, the formula $(x_1=0)[x_1 \leftarrow d]$ is true. Obviously, in $\mathbb N$ the element called $d$ will be the number $\text {zero}$.

Consider the formal defintion:

$\mathfrak A,s \vDash \exists x_1 \varphi[s]$ iff for some $d \in \mathfrak A$, we have that $\mathfrak A,s \vDash \varphi [s(x_1 \mid d)]$,

where $s(x \mid d)$ is the variable assignment function which is exactly like $s$ except for the fact that at the variable $x_1$ it assumes the value $d$.

Now, going back to your example: the formula $(x_1=0)$ is satisfied by $s = (0,\ldots)$, because $s(x_1)=0$ and $s(x_1 \mid 0)=s$.

But the formula is satisfied also by $s_1=(1,\ldots)$, because $s_1(x_1 \mid 0)=(0,\ldots)$ and (assuming for simplicity that $x_1$ is the only variable occurring free in $\varphi$) the only relevant fact is that $s(x_1)=0$.


答: You can note that if you have $\xi\in\mathbb{R}$, you have $f(\xi)=f(\xi-\lfloor \xi\rfloor) $ and you have $\xi-\lfloor\xi\rfloor \in [0,1]$, so you get that the maximum and the minimum are global


问:
Are focus and directrix for a conic section curve unique?
Question. Can a conic section curve have two distinct pairs of focus and directrix?

Attempt. I cannot think of a rigorous and logical way to convince myself of the uniqueness. But for me it is like two degrees of freedom (focus and directrix) define a single degree of freedom (a curve based on the constant eccentricity); intuitively for me, it might not be unique?

Comment. I do not need the answer to be rigorous; just some framework or hint would satisfy. Thank you!

答:
We can always find the unique pair of Dandelin spheres from the conic section without ambiguity.

$\hspace{5cm}$

Each Dandelin sphere touches both the right circular cone and the conic section on the corresponding focus.

See also detail discussion with animation in the link here and another post of mine about focal conics and confocal quadrics here.

For a general conic, namely

$$0=
\begin{pmatrix}
  x & y & 1 \\
\end{pmatrix}
\begin{pmatrix}
  a & h & g \\
  h & b & f \\
  g & f & c
\end{pmatrix}
\begin{pmatrix}
  x \\ y \\ 1
\end{pmatrix}$$

Its foci $z=x+yi$ are given by the complex quadratic, namely

$$Cz^2-2(G+Fi)z+(A-B)+2Hi=0$$

where capital letters $A,B, \ldots$ represent co-factors of the corresponding entries of small letters $a,b, \ldots$ of the matrix respectively.  That is

$$
\begin{pmatrix}
  A & H & G \\
  H & B & F \\
  G & F & C
\end{pmatrix}
=
\begin{pmatrix}
  bc-f^2 & fg-ch   & fh-bg \\
  fg-ch  & ac-g^2 & gh-af \\
  fh-bg  & gh-af    & ab-h^2
\end{pmatrix}$$

See another post for finding the principal axes, etc.


问:
Number of ways to distribute 15 indistinguishable balls to 7 kids such that each kid gets a maximum of 3 toys?
I want to find the number of ways to distribute 15 indistinguishable balls to 7 kids such that each kid gets a maximum of 3 toys.

I tried splitting to different cases: first determining how many kids get 3 balls, then the number or kids that get 2, than the kids that get 1. But I got lost and it seemed to be too long anyway (this question was taken from an exam).

The answer is 728.

Any help would be appreciated.

答:
One approach is to use generating functions:

Assuming the kids are distinguishable, this is the coefficient of $x^{15}$ in the expansion of $(1+x+x^2+x^3)^7$ and is indeed $728$.

It would be possible to do the expansion manually in an exam.  A slight short-cut is to notice that by symmetry it is also the coefficient of $x^{3\times7-15}=x^6$ in the expansion of $(1+x+x^2+x^3)^7$ and that can be found from the table below.  Without the shortcut, this table would need nine more columns.

```<code>            0   1   2   3   4   5   6   ...
            _   _   _   _   _   _   _
0   :       1   0   0   0   0   0   0
1   :       1   1   1   1   0   0   0
2   :       1   2   3   4   3   2   1
3   :       1   3   6   10  12  12  10
4   :       1   4   10  20  31  40  44
5   :       1   5   15  35  65  101 135
6   :       1   6   21  56  120 216 336
7   :       1   7   28  84  203 413 728
</code>```

where each term is the sum of the four values above to the left: for example $728=56+120+216+336$

Body:
One approach is to use generating functions:

Assuming the kids are distinguishable, this is the coefficient of $x^{15}$ in the expansion of $(1+x+x^2+x^3)^7$ and is indeed $728$.

It would be possible to do the expansion manually in an exam.  A slight short-cut is to notice that by symmetry it is also the coefficient of $x^{3\times7-15}=x^6$ in the expansion of $(1+x+x^2+x^3)^7$ and that can be found from the table below.  Without the shortcut, this table would need nine more columns.

            0   1   2   3   4   5   6   ...
            _   _   _   _   _   _   _
0   :       1   0   0   0   0   0   0
1   :       1   1   1   1   0   0   0
2   :       1   2   3   4   3   2   1
3   :       1   3   6   10  12  12  10
4   :       1   4   10  20  31  40  44
5   :       1   5   15  35  65  101 135
6   :       1   6   21  56  120 216 336
7   :       1   7   28  84  203 413 728

where each term is the sum of the four values above to the left: for example $728=56+120+216+336$


问:
Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$

Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$

I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.

Background:
Here however is a very similar problem

If $a,b,c>0$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$

There is a beautiful proof of  this by Michael Rozenberg (arqady)

WLOG $a\ge b\ge c$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but  $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$
$$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$

However this method doesnt work...

I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)

I am not planning to disclose how I proved the inequality as it may spoil the fun!

答:
Suppose $c = \min\{a,b,c\},$ then
$$\sum \frac{a^3+abc}{b+c}-(a^2+b^2+c^2) = \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}$$
$$\geqslant \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a} = \frac{(a-b)^2(a^2+ab+b^2-c^2)}{(b+c)(c+a)} \geqslant 0.$$
Therefore
$$\sum \frac{a^3+abc}{b+c} \geqslant a^2+b^2+c^2. \qquad (1)$$
Now, we write the inequality as
$$\sum \frac{a^3+abc}{b+c} + abc \sum \frac{1}{b+c} \geqslant \frac{(a+b+c)^2}{2}.$$
By the Cauchy-Schwarz inequality, we have
$$\sum \frac{1}{b+c} \geqslant \frac{9}{2(a+b+c)}. \qquad (2)$$
From $(1)$ and $(2),$ we need to prove
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)} \geqslant \frac{(a+b+c)^2}{2},$$
equivalent to
$$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
This is Schur inequality. The proof is completed.


问:
Uniform distribution joint $\to$ marginal
Let vector $(X,Y)$ have a uniform distribution on the set $N = \{ (x,y): x<1,y<1,1<x+y\}$. Determine distribution $X-Y$.

So far I've thought of this:
\begin{align}
P[X | Y=y] &\sim U(0,1-x) \space \forall y \in (0,1)\\
P[Y|X=x] &\sim U(0,1-y) \space \forall x \in (0,1)\\
\end{align}
but honestly I don't know how to go about it.

Any hints?

答:
So, random variables $X$ and $Y$ have joint pdf $f(x,y)$:

(source: tri.org.au)

which appears thusly:

(source: tri.org.au)

Let $Z=X-Y$. Then, the cdf of $Z$ is $P(Z<z)$ = $P(X-Y<z)$:

(source: tri.org.au)

where Prob is a mathStatica function.

Simply differentiate to obtain the pdf ...


问:
$Pa\rightarrow \exists y Qy \vdash \exists y (Pa\rightarrow Qy)$ using natural deduction
$Pa\rightarrow \exists y Qy \vdash \exists y (Pa\rightarrow Qy)$

My friend asked me to prove this using natural deduction. He knows I studied logic but I know little about natural deduction since I studied logic only by hilbert style.

Using inference rules he presented I thought a lot, but couldn't find any proof easier than these two. First, introduce $Pa\rightarrow \exists y Qy$ as a premise.

1-1. Assume $\neg\exists y(Pa\rightarrow Qy)$ and derive $\forall y\neg(Pa\rightarrow Qy)$

1-2. Derive $\neg(Pa\rightarrow Qb)$ and find subproof of $(Pa\wedge \neg Qb)$

1-3. Derive $\forall y\neg Qy$ from $\neg Qb$ and $\exists y Qy$ from $Pa$

1-4. Derive $\neg\exists yQy$ from $\forall y \neg Qy$ and conclude there's a contradiction

Here's another.

2-1. Drive $\neg Pa \vee\exists yQy$

2-2. Derive $Pa\rightarrow Qb$ each from $\neg Pa$  and $\exists y Qy$ and conclude $\exists y(Pa\rightarrow Qy)$

Since I haven't tried this in a full content, I don't know how long proof will be but I guess it will be.. I think there should be easier proof than these but I don't know how to find it.

答:
$\def\fitch#1#2{~~~~~\begin{array}{|l}#1\\\hline #2\end{array}}$

It might be worth noting that any proof will necessarily involve something like excluded middle or double negation elimination - the statement is not valid in intuitionistic first-order logic.

Here's Mauro ALLEGRANZA's LEM proof, reformatted.

$$\fitch{~~1.~~Pa\to\exists y~Qy\hspace{21ex}\textsf{Premise}}{~~2.~~Pa\lor\lnot Pa\hspace{23ex}\textsf{Tautology (Law of Excluded Middle)}\\\fitch{~~3.~~\lnot Pa\hspace{25ex}\textsf{Assumption}}{\fitch{~~4.~~Pa\hspace{23ex}\textsf{Assumption}}{~~5.~~\bot\hspace{24ex}\textsf{Negation Elimination 5, 4}\\~~6.~~Qa\hspace{23ex}\textsf{Explosion 5}}\\~~7.~~Pa\to Qa\hspace{20ex}\textsf{Conditional Introduction 4-6}\\~~8.~~\exists y~(Pa\to Qy)\hspace{15ex}\textsf{Existential Introduction 7}}\\\fitch{~~9.~~Pa\hspace{27ex}\textsf{Assumption}}{10.~~\exists y~Qy\hspace{24ex}\textsf{Conditional Elimination 1, 9}\\\fitch{\boxed b 11.~~Qb\hspace{21ex}\textsf{Assumption}}{\fitch{12.~~Pa\hspace{19ex}\textsf{Assumption}}{13.~~Qb\hspace{19ex}\textsf{Reiteration 11}}\\14.~~Pa\to Qb\hspace{16ex}\textsf{Conditional Introduction 12-13}\\15.~~\exists y~(Pa\to Qy)\hspace{11ex}\textsf{Existential Introduction 15}}\\16.~~\exists y~(Pa\to Qy)\hspace{15ex}\textsf{Existential Elimination 10, 11-15}}\\20.~~\exists y~(Pa\to Qy)\hspace{19ex}\textsf{Disjunction Elimination 2, 11-19, 4-10}}$$

Here's the DNE version.

$$\fitch{~~1.~~Pa\to\exists y~Qy\hspace{21ex}\textsf{Premise}}{\fitch{~~2.~~\lnot\exists y~(Pa\to Qy)\hspace{13.5ex}\textsf{Assumption}}{\fitch{~~3.~~Pa\hspace{22.5ex}\textsf{Assumption}}{~~4.~~\exists y~Qy\hspace{19.5ex}\textsf{Conditional Elimination 1, 3}\\\fitch{\boxed b~~5.~~Qb\hspace{16.5ex}\textsf{Assumption}}{\fitch{~~6.~~Pa\hspace{15ex}\textsf{Assumption}}{~~7.~~Qb\hspace{15ex}\textsf{Reiteration 5}}\\~~8.~~Pa\to Qb\hspace{12ex}\textsf{Conditional Introduction 6-7}\\~~9.~~\exists y~(Pa\to Qy)\hspace{7ex}\textsf{Existential Introduction 8}\\10.~~\bot\hspace{20ex}\textsf{Negation Elimination 9, 2}\\11.~~Qa\hspace{19ex}\textsf{Explosion 10}}\\12.~~Qa\hspace{23ex}\textsf{Existential Elimination 5-11}}\\13.~~Pa\to Qa\hspace{20ex}\textsf{Conditional Introduction 3-12}\\14.~~\exists y~(Pa\to Qy)\hspace{15.5ex}\textsf{Existential Introduction 13}\\15.~~\bot\hspace{28ex}\textsf{Negation Elimination 14, 2}}\\16.~~\lnot\lnot\exists y~(Pa\to Qy)\hspace{16ex}\textsf{Negation Introduction 2-15}\\17.~~\exists y~(Pa\to Qy)\hspace{19ex}\textsf{Double Negation Elimination 16}}$$


问:
Orthogonal decomposition of derivative of a curve
Let $\alpha: I \subset \mathbb{R} \rightarrow \mathbb{R^3}$ be a differentiable curve. I'm in the process of proving that $$||\alpha(b) - \alpha(a)|| \leq L(\alpha)_a^b$$ being $L(\alpha)_a^b$ the length of the curve from a to be.

It says in my text book that, if $\alpha(b) \neq 0$ we can separate  $\alpha'$ in $(t, b]$ as follows:

$$
\alpha'= \langle \alpha' , \frac{\alpha}{||\alpha||} \rangle \frac{\alpha}{||\alpha||} + \alpha^\perp
$$

At first i thought of Gram–Schmidt process of orthogonalization but
I don't recognize this decomposition and dont know why this equality holds.

答: If you have two vectors $v$ and $w$, with $w\ne0$, then$$v=\left\langle v,\frac w{\|w\|}\right\rangle\frac w{\|w\|}+\left(v-\left\langle v,\frac w{\|w\|}\right\rangle\frac w{\|w\|}\right).$$This equality expresses $v$ as the sum of a vector which is a multiple of $w$ with a vector which is orthogonal to $w$.


答:
As there are $15$ balls and $7$ children at least one child has to get $3$ balls. Let's count in how many ways we can distribute the balls depending on how many children get $3$ balls.

$1$ child gets $3$ balls: then you have to distribute $12$ balls among $6$ children and each of them has to get at most (and actually exactly) $2$ balls. You can do this in $1$ way (the balls are indistinguishable);

$2$ children get $3$ balls: we have to distribute $9$ balls among $5$ children and each of them has to get at most $2$ balls. You can do this in $5$ ways (you have to choose the one that gets $1$ ball);

$3$ children get $3$ balls: we have to distribute $6$ balls among $4$ children and each of them has to get at most $2$ balls. You can do this in $4+6=10$ ways (either there is one child without any balls, or everyone has at least $1$ ball).

$4$ children get $3$ balls: we have to distribute $3$ balls among $3$ children and each of them has to get at most $2$ balls. You can do this in $1+6 = 7$ ways (either there is a child without any balls, one with $1$ ball, and one with $2$, or everyone has $1$ ball).

$5$ children get $3$ balls: there are no other balls to distribute.

Now let us sum the numbers taking into account the number of choices we have for which children get $3$ balls:

$$ 7 \cdot 1 + 21 \cdot 5 + 35 \cdot 10 + 35 \cdot 7 + 21 = 728.$$

Title: Number of ways to distribute 15 indistinguishable balls to 7 kids such that each kid gets a maximum of 3 toys?

Body:
I want to find the number of ways to distribute 15 indistinguishable balls to 7 kids such that each kid gets a maximum of 3 toys.

I tried splitting to different cases: first determining how many kids get 3 balls, then the number or kids that get 2, than the kids that get 1. But I got lost and it seemed to be too long anyway (this question was taken from an exam).

The answer is 728.

Any help would be appreciated.


问:
Can the inverse to a scalar multiplication be found efficiently in elliptic curves?
Let $P$ be a point over an elliptic curve $F_p$, say Curve25519, and let $k$ be a scalar. Assume I want to find an inverse $k^{-1}$ such that:
$$k \cdot k^{-1} \cdot P = P$$
Does such an inverse always exists and can it be calculated efficiently?

答:
Well, it amounts in a way to the discrete logarithm problem for EC.

Given two points $P,Q$ on an EC curve where $Q$ is a multiple of $P$.
Find $k$ such that $Q=kP$.

This problem is intractable and assures the security of EC cryptography.


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
I think this problem can only be solved either approximately (asympotically exact) or numerically. Karl's approximation is the most natural answer, and I'd go for it. If you instead want an exact result, here's an efficient numerical procedure (computer assisted, though).

Letting $g_A(n)$ be the probability that player $A$ wins given that the boss has power $n$, we have the recursion

$$ g_A(n) = \sum_{j=n}^\infty p_A(j) + \sum_{j=1}^{n-1} p_{ABC}(j)g_A(n-j)$$

where $ p_A(j)$ is the probability mass function of the damage inflicted by player $A$ and $p_{ABC}(j)$ is the same for the three players summed.

Solved numerically in Octave (or Matlab) this gives

$$g_A(1000) =  0.28347$$
pretty near Karl's approximation ( $0.28302$)

The graph shows that that approximation is quite good after $N\approx 900$

Code:

```<code>pa = [1:100];
pa = pa >= 25 & pa <= 50;
pa = pa/sum(pa);
pb = [1:100];
pb = pb >= 30 & pb <= 70;
pb = pb/sum(pb);
pc = [1:100];
pc = pc >= 10 & pc <= 80;
pc = pc/sum(pc);
pab = [0, conv(pa,pb)];
pab = [0, conv(pa,pb)];
pac = [0, conv(pa,pc)];
pbc = [0, conv(pb,pc)];
pabc = [0, conv(pab,pc)];
N=1000;
pabc(N)=0;
ga = zeros(1,N);
ga(1)=1;
for n=2:N
  ga(n) = sum(pa(n:100)) +  flip(ga(1:n-1)) * pabc(1:n-1)';
endfor
ga(N)
</code>```

For completeness, here are the other probabilities

```<code>gb = zeros(1,N); % prob of B winning, if B has next turn
gb(1)=1;
 for n=2:N
 gb(n) = sum(pb(n:100)) +  flip(gb(1:n-1)) * pabc(1:n-1)';
endfor
gbb=zeros(1,N);  % prob of B winning, if A has next turn
pa(N)=0;  % extend pa size
for n=2:N
 gbb(n)= flip(gb(1:n-1)) * pa(1:n-1)';
endfor
gcc = 1 - (ga+gbb);
</code>```

Notice that the period of the quasioscillations is around the mean value of the whole turn : $(25+50+30+70+10+80)/2 = 132.50$

Body:
I think this problem can only be solved either approximately (asympotically exact) or numerically. Karl's approximation is the most natural answer, and I'd go for it. If you instead want an exact result, here's an efficient numerical procedure (computer assisted, though).

Letting $g_A(n)$ be the probability that player $A$ wins given that the boss has power $n$, we have the recursion

$$ g_A(n) = \sum_{j=n}^\infty p_A(j) + \sum_{j=1}^{n-1} p_{ABC}(j)g_A(n-j)$$

where $ p_A(j)$ is the probability mass function of the damage inflicted by player $A$ and $p_{ABC}(j)$ is the same for the three players summed.

Solved numerically in Octave (or Matlab) this gives

$$g_A(1000) =  0.28347$$
pretty near Karl's approximation ( $0.28302$)

The graph shows that that approximation is quite good after $N\approx 900$

Code:

pa = [1:100];
pa = pa >= 25 & pa <= 50;
pa = pa/sum(pa);
pb = [1:100];
pb = pb >= 30 & pb <= 70;
pb = pb/sum(pb);
pc = [1:100];
pc = pc >= 10 & pc <= 80;
pc = pc/sum(pc);
pab = [0, conv(pa,pb)];
pab = [0, conv(pa,pb)];
pac = [0, conv(pa,pc)];
pbc = [0, conv(pb,pc)];
pabc = [0, conv(pab,pc)];
N=1000;
pabc(N)=0;
ga = zeros(1,N);
ga(1)=1;
for n=2:N
  ga(n) = sum(pa(n:100)) +  flip(ga(1:n-1)) * pabc(1:n-1)';
endfor
ga(N)

For completeness, here are the other probabilities

gb = zeros(1,N); % prob of B winning, if B has next turn
gb(1)=1;
 for n=2:N
 gb(n) = sum(pb(n:100)) +  flip(gb(1:n-1)) * pabc(1:n-1)';
endfor
gbb=zeros(1,N);  % prob of B winning, if A has next turn
pa(N)=0;  % extend pa size
for n=2:N
 gbb(n)= flip(gb(1:n-1)) * pa(1:n-1)';
endfor
gcc = 1 - (ga+gbb);

Notice that the period of the quasioscillations is around the mean value of the whole turn : $(25+50+30+70+10+80)/2 = 132.50$


答:
Representation of the original problem -

$\displaystyle \sum\limits_{i=1}^{7} x_i = 15, 0 \leq x_i \leq 3$

To make the working easier, we now define  a variable $y_i = 3 - x_i$ and we derive $0 \leq y_i \leq 3$ from the original constraint. Please note that as the new variable $y_i$ varies from $0$ to $3$ for $i \in (1,7)$, the solutions directly map to permissible solutions of the original problem.

So we rewrite the problem as,

$\displaystyle \sum\limits_{i=1}^{7} (3-y_i) = 15, 0 \leq y_i \leq 3 \ $ or, $\displaystyle \sum\limits_{i=1}^{7} y_i = 6, 0 \leq y_i \leq 3$

Without restriction, we have ${{7 + 6 - 1} \choose {7-1}} = 924$ solutions.

Now for exclusions,

There are $7$ combinations if one of the numbers is $6$.

There are $7 \cdot 6 = 42$ combinations if one of the numbers is $5$ (one of the other numbers is $1$)

If one of the numbers is $4$, we either have other number as $2$ which is $7 \cdot 6 = 42$ combinations or we have two other numbers that are $1$ each. That is $7 \cdot {6 \choose 2} = 105$ combinations.

So total number of valid solutions $ = 924 - (7 + 42 + 42 + 105) = \fbox{728}$.


问:
Frattini subgroup is set of nongenerators
The Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$. (If there are none, then $\Phi(G)=G$.) We say that an element $g\in G$ is a nongenerator if whenever $\langle X\cup \{g\}\rangle = G$, we have $\langle X\rangle = G$ for subsets $X\subseteq G$.

If $G$ is finite, show that $\Phi(G)$ is the set of nongenerators of $G$.

So I want to start by assuming that $g\in G$ is in all maximal subgroups of $G$, and $\langle X\cup\{g\}\rangle = G$, and to prove that $\langle X\rangle = G$. At first I thought if $\langle X\cup\{g\}\rangle = G$, then $\langle X\rangle$ might be a maximal subgroup itself, but it seems that this is not necessarily the case. (For example $G$ is a cyclic group of order $4$ generated by $g$, and $X$ is the trivial subgroup.)

答:
Suppose that $\langle X\cup\{g\}\rangle=G$, but $\langle X\rangle\neq G$. Because $\langle X\rangle$ is not all of $G$, it is contained in some maximal subgroup $M\subset G$...

but so is $\{g\}$, so you'd have to have $\langle X\cup\{g\}\rangle\subseteq M$, which is a contradiction.


问:
Can the quotient of a division of a polynomial act as the residual in the remainder theorem?
The problem is as follows:

The polynomial $p(x)$ is divisible separately by $(x-2)$, $(x-3)$,
$(x-4)$, and by $(x-5)$. If $p(x)$ is divided by $(x-9)$ leaves a
remainder equal to $5040$, what is the independent term of such
polynomial?

The alternatives given in my book are as follows:

$$\begin{array}{cc}
1.&\textrm{600}\\
2.&\textrm{360}\\
3.&\textrm{740}\\
4.&\textrm{720}\\
\end{array}$$

The official solution for this problem according to my workbook in precalculus states the following argument:

$p(x)$ is divisible separately by $(x-2)$, $(x-3)$, $(x-4)$ and $(x-5)$ $\Leftrightarrow$ $p(x)$ is divisible by $(x-2)$, $(x-3)$, $(x-4)$ and $(x-5)$.

Using the division algorithm we have:

$$p(x)= (x-2)(x-3)(x-4)(x-5)\cdot q(x)$$

Therefore $\operatorname{grad}[q(x)]=0$ (comment: here the notation grad indicates the degree of the polynomial)

Hence $q(x)=M$

Using the remainder theorem it is obtained:

$$\frac{p(x)}{x-9}$$

$$p(9)=(9-2)(9-3)(9-4)(9-5)q(x)=5040$$

Then:

$$(7)(6)(5)(5)(4)M=5040$$

$$M=6$$

Then to get the independent term we do evaluate the polynomial in zero as: $p(0)$

$$p(0)=(0-2)(0-3)(0-4)(0-5)(6)=6\cdot 20 \cdot 6 =720$$

which corresponds with the option number $5$.

That's where it ends the official solution.

Now my problem is that I'm confused with the way how it was found the degree of the quotient and why it was earlier declared as such and later used as a remainder.

First and foremost, how do we know the degree of the quotient is zero? I'm only assuming that since this polynomial is divisible by those factors it implies an exact division, hence the quotient can only be a constant term, otherwise it will increase an additional factor for which the polynomial is not divisible with. But is this assumption correct?. Can someone explain this?

The second part for which I'm confused is a weird usage of the polynomial remainder theorem.

If my memory serves me right, the remainder theorem goes as this:

When a polynomial is divided by $(x-a)$ then

$$p(x)=d(x)\cdot q(x)+r$$

$$p(x)=(x-a)q(x)+r$$

When we do evaluate the polynomial in a thus:

$$p(a)=r$$

Hence evaluating the polynomial with that $a$ will result in the remainder of the division which it can be another polynomial, right? But when it is applied to a linear quotient then the remainder is a constant term.

But in this case, the way how it was made such interpretation is some confusing because it uses $M$ and $M$ was earlier declared as the quotient.

Shouldn't it be as:

$$p(x)=(x-9)((x-2)(x-3)(x-4)(x-5)\cdot q(x))+r=5040$$

Because of these inconsistencies. I'd appreciate someone could help me to understand what might the author had intended to do, because I'm lost. Perhaps does it exist a better method of argumentation here?. I appreciate if someone could explain with words the right interpretation of the polynomial remainder theorem because this is the part which I'm struggling the most. How is it being used here?.

The part which it confuses me a lot is why is it treating $M$ as if it were $r(x)$ if it was earlier declared as quotient?. Isn't this a contradiction when evaluates the polynomial on $p(9)$?. So all and all, can someone help me to address these doubts please?

答:
For the first problem, as others pointed out, you are right that nothing guarantees $\mathrm{deg}(q)=0$ if we take the exercise as it is written. But perhaps what the author(s) meant is ``$p(x)$ is divisible by [...], and ONLY by those'', which would then imply that $\mathrm{deg}(p)=4$, and so necessarily $\mathrm{deg}(q)=0$.

Now for the second part. The reminder theorem says that for polynomials $p(x), s(x)$, there always exist polynomials $q(x), r(x)$ such that
$$p(x) = s(x)\cdot q(x)+r(x),$$
where (!) $\mathrm{deg}(r)<\mathrm{deg}(s)$. Meaning that we can divide $p$ by $s$ ``as much as possible'', in the sense that whatever remains has degree strictly lower than the polynomial we are dividing by. In the exercise, we have two instances of division, and get two formulas:
$$(1)\ \ p(x)=(x-2)(x-3)(x-4)(x-5)q(x),$$
$$(2)\ \ p(x)=(x-9)\cdot t(x)+r(x).$$
Formula $(1)$ just says that $p$ is divisible by all those linear factors, and $q$ is the quotient. Formula $(2)$ is the division of $p$ by $(x-9)$, where we get quotient $t(x)$ and some reminder $r(x)$, which actually has to be just a number, because $\mathrm{deg}(r)<\mathrm{deg}(x-9)=1$. Moreover, we are actually told that it is just a number, and even which number it is: $r=5040$.

Now we substitute $x\rightarrow 9$ into $(2)$, and get
$$p(9)=(9-9)\cdot t(9)+r=r=5040.$$
But using $(1)$, this means that
$$5040=p(9)=(9-2)(9-3)(9-4)(9-5)q(9)=7\cdot6\cdot5\cdot4\cdot q(9)=840q(9)$$
$$q(9)=5040/840=6$$

Now assuming that $\mathrm{deg}(q)=0$, we know that actually $q=6$ (the authors write this as $M=6$ just to stress out that $q$ is constant), and so
$$p(x)=6\cdot(x-2)(x-3)(x-4)(x-5),$$
and it's easy to obtain the independent term. You can see it directly from the formula, that all you need to do is multiply $6\cdot(-2)\cdot(-3)\cdot(-4)\cdot(-5)=720$, but the authors use the general fact that $p(0)$ is the independent term for every polynomial, and so they evaluate $p(0)$, which is the same multiplication leading to $720$.


问:
Expectation of transition semigroup of process with invariant measure
Let $(Z_t)_{t\geq0}$ be a $d$-dimensional stochastic process with invariant measure $P$ and $Z\sim P$. Let $(P_t)_{t\geq0}$ denote the transition semigroup of the process, i.e. in particular for $h:\mathbb{R}^d \to \mathbb{R}, x\in \mathbb{R}^d$
$$
P_th(x) = \mathbb{E}_P[h(Z_{t,x})],
$$
where $Z_{t,x}$ denotes the state of the process at time $t$ when starting at $x$.
I've come across the equation
$$
\mathbb{E}_P[h(Z)] = \mathbb{E}_P[(P_th)(Z)].
$$
Unfortunately, that is not obvious to me. Can someone explain where that comes from? Thanks a lot!

Edit: Is this obvious, because $P$ is the invariant measure of the process, so $Z_{t,x} \sim P$ regardless of the starting point $x$? Therefore $Z_{t, Z}$ and $Z$ follow the same distribution and
$$
\mathbb{E}_P[(P_th)(Z)] = \mathbb{E}_P[\mathbb{E}_P[h(Z_{t, Z})]] = \mathbb{E}_P[\mathbb{E}_P[h(Z)]] = \mathbb{E}_P[h(Z)]?
$$

答:
This seems to be indeed obvious, and my problem was that I did not find a clear definition of the term "invariant measure". I found that now so I'm going to answer myself. (Maybe someone can confirm that this is correct.)

Apparently, an invariant measure is a distribution that does not change under the flow of the stochastic process, i.e. if $Z_0\sim P$, then $Z_t\sim P$ for all $t\geq0$.
Then by definition, $Z_{t, Z}$ and $Z$ follow the same distribution and the equation in the Edit section should be correct.


问:
Show $R$ is a field
Is this okay or does it require a different approach? Let $R$ be a commutative ring with unity. Suppose that the only ideals of $R$ are $\{0\}$ and $R$, show that $R$ is a field.

Attempt:Since $\{0\}$ and $R$ are the only ideals, $\{0\}$ is maximal, thus $R/\{0\} \cong R$ so $R$ is a field since $R/\{0\}$ is a field.

答:
A general and easy to prove fact:

Fact: Let $R$ be a ring. Then $(r)=R$ if and only if $r\in R^\times$ (i.e. $r$ is a unit in $R$).

Proof. If $r$ is a unit then there is some $s\in R$ such that $s\cdot r=1$. Hence, $1\in(r)$ and so $a=a\cdot1\in(r)$ for all $a\in R$. So $R\subseteq(r)$ and consequently $(r)=R$. Conversely assume that $(r)=R$. Then in particular $1\in(r)$. But then by definition $s\cdot r=1$ for some $s\in R$, that is, $r$ is a unit.$~\square$

Now suppose the only ideals of your ring are $\{0\}$ and $R$. If $r\ne0$, then $(r)\ne\{0\}$ as $r\in(r)$. Thus $(r)=R$ and by the aforegoing fact it follows that $r\in R^\times$. As $r$ was an arbitrary non-zero element of our ring we conclude that every such element is invertible and hence $R^\times=R\setminus\{0\}$. Given that $R$ is commutative and unital we see that then $R$ is in fact a field.

Title: Show $R$ is a field


问:
Irrationality of an "Euler-like" number
Let $(a_n)_{n=0}^{\infty}$ be a sequence of zeroes and ones such that $a_n=1$ for infinitely many $n$. Let $\displaystyle x:=\sum_{n=0}^{\infty} \frac{a_n}{n!} .$

Is $x$ irrational? I believe it is, but I don't know how to prove it. I'll appreciate any help.

答:
As was pointed out by leoli1, this problem can be easily solved by slightly modifying Fourier's proof of irrationality of Euler's number.

Case closed.

Title: Irrationality of an "Euler-like" number


问:
Improving system identification results?
I'm performing system identification of the lateral closed-loop dynamics of a quadrotor.
My model receives a setpoint and should return position and acceleration.
I've proposed a second order model of the type, see equations, and solved it with MATLAB's greyest():

$\begin{bmatrix} v_y \\ a_y \end{bmatrix} = \begin{bmatrix}0&1 \\-\omega^2&-2\xi\omega \end{bmatrix} \begin{bmatrix}y \\ v_y \end{bmatrix} + \begin{bmatrix} 0 \\ \omega^2 \end{bmatrix} y_0$

$\begin{bmatrix} y \\ a_y \end{bmatrix} = \begin{bmatrix}1&0 \\-\omega^2&-2\xi\omega \end{bmatrix} \begin{bmatrix}y \\ v_y \end{bmatrix} + \begin{bmatrix} 0 \\ \omega^2 \end{bmatrix} y_0$

The model performs very well for the position, 97% on a cross-validation dataset, but the performance is very low for acceleration 43%.

Note the acceleration dataset comes out from an accelerometer so it's noisy, I've already tried low pass filtering the noise to improve performance with no positive results.

Any ideas on how to improve this?

Thanks!

答:
What is $\xi$?
Maybe the following (related to dead-reckoning) is way different than what you are actually asking? but...

Results tend to be greatly improved if your sensor-fusion system/algorithm includes magnetometer data: "sensor fusion using magnetometers can reduce the average error in position obtained by the system after 60 seconds from over 150 m to around 5 m." (Woodman 2007)

"errors in orientation propagating through the algorithm are by far the main source of error in the system." (Woodman 2007)

Kalman filtering tends to be quite popular: "There are many techniques for performing sensor fusion, the most popular of which are Kalman and particle filters." (Woodman 2007 page 33)

References

Woodman 2007

Gade 2009


问:
Given $f$ continuous, $T>0$ and supposed $\int_x^{x+T} f(t)dt=\int_0^T f(t)dt$, show that $f$ is periodic with period $T$
Given $T>0$ and supposed $f$ continuous for all $x\in\mathbb{R}$, show that if $\int_{x}^{x+T} f(t)dt=\int_0^T f(t)dt$ then $f$ is periodic for all $x\in\mathbb{R}$ with period $T>0$.

My approach is the following: since by hypothesis $f$ is continuous, then by the fundamental theorem of calculus its integral function is differentiable; so, since by hypothesis $\int_x^{x+T}f(t)dt=\int_0^T f(t)dt$, taking derivatives both sides it is
$$\frac{d}{dx}\int_x^{x+T}f(t)dt=\frac{d}{dx}\int_0^T f(t)dt \implies f(x+T)-f(x)=0 \implies f(x+T)=f(x)$$
That is, $f$ is periodic with period $T>0$.

Is my approach correct? I am unsure because I know that a period is defined as the minimum $T>0$ such that $f(x+T)=f(x)$, so how do I know that the equality $f(x+T)=f(x)$ I've found is satisfied for the minimum $T$? Or in this case the problem is asking just a period and not the minimum one?

I believe that this result holds for non continuous function as well (just integrable ones), so for those functions the proof has to change because I can't use the fundamental theorem of calculus anymore, so maybe I'm using too much regularity to prove this.

I know that if $f$ is continuous for all $x\in\mathbb{R}$ and periodic with period $T>0$, then $\int_x^{x+T}f(x)dx=\int_0^T f(x)dx$, combining this with the result in this problem is correct to affirm that "Given $f$ continuous and periodic for all $x\in\mathbb{R}$, $f$ is periodic if and only if $\int_x^{x+T} f(t) dt=\int_0^T f(t) dt$"? If yes, can this be taken as an equivalent definition of periodic functions?

My textbook writes $\int_x^{x+T} f(t)dt=\int_0^x f(t)dt+\int_x^T f(t)dt+\int_T^{x+T} f(t)dt$ and proceeds in another way. Why he assumes that $x$ is "after" $0$ in the integration interval and $x$ is lesser than $T$? Can he do this WLOG because they are generic values and if it is not like this the proof works the same just by inverting them or is there another reason?

答:
Yes, your approach is fine. But you cannot prove that $T$ is the least positve number with $f(x+T)=f(x)$. Very often Mathematicians  say $f$ is periodic with period $T$ just to mean that $f(x+T)=f(x)$.

There is a theorem of Lebsgue which extends FTC to locally integrable functions. So, as long as the integrals in the question exist we can prove that $f(x+T)=f(x)$ holds for almost all $x$.

The given property is indeed equivalent to the condition $f(x+T)=f(x)$ for continuous functions.

$\int_a^{b}f(x)dx$ is defined as $-\int_b^{a}f(x)d$ when $ a>b$, so your book is not assuming that $0 <x<T$.


问:
$e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$
Why is $e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$?

I know $e^{5 \frac{2 \pi i}{6}} = e^{-\frac{\pi i}{3}}=\overline{e^{\frac{\pi i}{3}}}=\overline{e^{\frac{2 \pi i}{6}}}$, but I still don't see why the equation is true.

答: Because $(\mathbb{Z} / 6 \mathbb{Z},+) \to (E_6,\cdot)$ with $E_6 = \{ e^{\frac{2 \pi i k}{5}} : k = 0,1,\dots,5 \} $ is a group isomorphism, the statement follows because $5+1=6=0$ in $\mathbb{Z} / 6 \mathbb{Z}$.


答:
Some comments;

The question intends you to show $T$ is one possible period, but does not imply $T$ is the least such.  Indeed, the property is true when applied to constant $f(x)$ for all $T$ or $f(x)=\sin x$ for $T=2\pi, 4\pi, 6\pi, $ etc.

It is also true when $f$ is merely integrable over finite intervals of length $T$.  Taken in the sense of the Lebesgue integral, the conclusion $f(x)=f(x+T)$ is true for almost all $x$ (i.e. the set on which it is not true has zero measure).  That follows from the fact that if $f(x)$ is integrable over $[a,b]$, $\int_a^x f(t) dt $ is differentiable and its derivative is equal to $f(x)$ for almost all $x, a \leqslant x\leqslant b$ (see for example, Real Analysis, Stein and Shakarchi, Princeton University Press to see definitions and proof).  Therefore the same argument as in (1) can then be applied.

Your statement is correct but could be better phrased as, "Given $f$ is continuous [integrable .. ] on every finite interval of length $T$ then $f$ is periodic with a period $T$ if and only if $\int_x^{x+T} f(t)dt = \int_0^T f(t) dt$ for every $x$.

The equality in you text book does not appear to be correct regardless of the sign of $x$ unless $x$ happens to be zero.

Title: Given $f$ continuous, $T>0$ and supposed $\int_x^{x+T} f(t)dt=\int_0^T f(t)dt$, show that $f$ is periodic with period $T$

Body:
Given $T>0$ and supposed $f$ continuous for all $x\in\mathbb{R}$, show that if $\int_{x}^{x+T} f(t)dt=\int_0^T f(t)dt$ then $f$ is periodic for all $x\in\mathbb{R}$ with period $T>0$.

My approach is the following: since by hypothesis $f$ is continuous, then by the fundamental theorem of calculus its integral function is differentiable; so, since by hypothesis $\int_x^{x+T}f(t)dt=\int_0^T f(t)dt$, taking derivatives both sides it is
$$\frac{d}{dx}\int_x^{x+T}f(t)dt=\frac{d}{dx}\int_0^T f(t)dt \implies f(x+T)-f(x)=0 \implies f(x+T)=f(x)$$
That is, $f$ is periodic with period $T>0$.

Is my approach correct? I am unsure because I know that a period is defined as the minimum $T>0$ such that $f(x+T)=f(x)$, so how do I know that the equality $f(x+T)=f(x)$ I've found is satisfied for the minimum $T$? Or in this case the problem is asking just a period and not the minimum one?

I believe that this result holds for non continuous function as well (just integrable ones), so for those functions the proof has to change because I can't use the fundamental theorem of calculus anymore, so maybe I'm using too much regularity to prove this.

I know that if $f$ is continuous for all $x\in\mathbb{R}$ and periodic with period $T>0$, then $\int_x^{x+T}f(x)dx=\int_0^T f(x)dx$, combining this with the result in this problem is correct to affirm that "Given $f$ continuous and periodic for all $x\in\mathbb{R}$, $f$ is periodic if and only if $\int_x^{x+T} f(t) dt=\int_0^T f(t) dt$"? If yes, can this be taken as an equivalent definition of periodic functions?

My textbook writes $\int_x^{x+T} f(t)dt=\int_0^x f(t)dt+\int_x^T f(t)dt+\int_T^{x+T} f(t)dt$ and proceeds in another way. Why he assumes that $x$ is "after" $0$ in the integration interval and $x$ is lesser than $T$? Can he do this WLOG because they are generic values and if it is not like this the proof works the same just by inverting them or is there another reason?


问:
Banach space adjoint as a matrix

Question

Let $1< p,q < \infty$ be holder conjugates.

Let $A \colon \ell_n^p(\mathbb{C}) \to \ell_n^p(\mathbb{C})$ be a
$n\times n$ matrix.

Let $A^* \colon \ell_n^q(\mathbb{C}) \to \ell_n^q(\mathbb{C}) $ be
the Banach space adjoint of $A$.

Note $\ell_n^p(\mathbb{C})^* = \ell_n^q(\mathbb{C})$.

Show $A^*$ is the transpose of $A$.

My Understanding

Let
$$x= (x_1. \cdots, x_n), \ y = (x_1, \cdots, y_n) \in \ell_n^p(\mathbb{C})$$

Let
$$f_x(y) = x \cdot y = \sum_{k=1}^n x_k y_k, \text{where } f_x = \sum_{k=1}^n x_k e_k$$

Then we want to show that
$$A^*(f_x)(y)= A^T(f_x)(y)$$

Note that
$$x \cdot y = x^T y = f_x(y)$$

Also, by definition of adjoint we have
$$A^*(f_x)= f_x \circ A$$

We have

$$A^*(f_x)(y)= f_x(Ay)= \sum_{k=1}^n x_k e_k=x^T \cdot (Ay) = (Ay)\cdot x^T = (Ay)^Tx^T= y^TA^Tx^T$$

which became a mess. But what I really want is to have the last equalities be
$$=A^T(x^T  y)= A^T(f_x)(y)$$

Any help is appreciated!

答:
By the Banach space definition of an adjoint, $$(A^*f_x)(y)=(f_x\circ A)y=x\cdot(Ay)=x^\top Ay=(A^\top x)^\top y=(A^\top x)\cdot y$$ Since this is true for all $y$, it follows that $A^*f_x=f_{A^\top x}$.

This is commonly written as $A^*x=A^\top x$ if one abuses the notation and writes $x$ for $f_x$.

Title: Banach space adjoint as a matrix


问:
Linear least squares of factorizable 2D polynomial
I would like to fit 2D data (a set of $x_i$, $y_i$, $z_i = f(x_i, y_i)$) where my function is a product of two 1D polynomials of $n$th order ($n>1$), so it is factorizable.

$$f(x,y) = (a_0 + a_1x + a_2x^2 + ... + a_nx^n)(1 + b_1y + b_2y^2 + ... + b_nx^n)$$

I know that I can solve this using any optimization routine that uses non-linear least squares. Is there a way to use a linear least square method to fit such a factorizable polynomial?

One can expand the polynomial and formulate the problem as

$$
\begin{pmatrix}
1 & x_1 & y_1 & x_1y_1 & x_1^2 & y_1^2 & x_1^2y_1 & x_1y_1^2 & \cdots
\\
&&&& \vdots &&&&
\end{pmatrix}
\mathbf{X}
=
\begin{pmatrix}
f(x_1, y_1) \\ \vdots
\end{pmatrix}
$$
and solve for $\mathbf{X}= (a_0, a_1, a_0b_1, a_1b_1, \cdots)^T$.

Then I have the problem, that the system of equations that define $a_i$ and $b_i$ is overdetermined ($\mathbf{X}$ has length $(n+1)^2$, but I have only $2n+1$ independent variables).

Constrained least squares (using e.g. $\mathbf{L}\mathbf{X} = \mathbf{d}$) also does not seem to work, since the constraints are non-linear.

Is there any (tricky?) way to use linear least squares for this type of problem?

Example of such data that I try to fit to ($n=4$):
$x = [  0.       ,   0.       ,   0.       ,   0.       ,   0.       ,
         0.       ,   0.       , -13.5      , -10.125    ,  -6.75     ,
        -3.375    ,   0.       ,   3.375    ,   6.75     ,  10.125    ,
        13.5      ,  13.510486 ,  13.50262  ,  13.50262  ,  13.510486 ,
         6.755392 ,   6.7513475,   6.7513475,   6.755392 ,  -6.755392 ,
        -6.7513475,  -6.7513475,  -6.755392 , -13.510486 , -13.50262  ,
       -13.50262  , -13.510486 ]$
$y = [ 0.  ,  0.  ,  2.3 ,  1.15,  0.  , -1.15, -2.3 ,  0.  ,  0.  ,
        0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  0.  ,  2.3 ,  1.15,
       -1.15, -2.3 , -2.3 , -1.15,  1.15,  2.3 ,  2.3 ,  1.15, -1.15,
       -2.3 , -2.3 , -1.15,  1.15,  2.3 ]$
$f(x, y) = [1.009518  , 0.99437314, 1.0034443 , 1.0105053 , 1.0142992 ,
       1.0154775 , 0.9043408 , 0.9617672 , 1.000238  , 1.0179688 ,
       1.0113076 , 0.98180985, 0.9331824 , 0.8696933 , 0.79686165,
       0.7879401 , 0.7935547 , 0.7991753 , 0.79890484, 0.93638456,
       0.935677  , 0.92781633, 0.9205955 , 0.9865708 , 0.99512374,
       1.0054537 , 1.0065441 , 0.9099518 , 0.90881765, 0.9004598 ,
       0.892026  , 1.013272  ]$

答:
The method below is not verry accurate but might be sufficient. For example the fitting is not bad with your data as shown below.

$$f(x,y) = g(x)h(y)\quad \begin{cases}
g(x)=(a_0 + a_1x + a_2x^2 + ... + a_nx^n) \\
h(y)=(1 + b_1y + b_2y^2 + ... + b_ny^n)
\end{cases}$$
We approximate $g(x)$ with the function :
$$g(x)\simeq \frac{1}{(c_0 + c_1x + c_2x^2 + ... + c_mx^m)}$$
Note that $m$ can be different from $n$.
$$f(x,y)(c_0 + c_1x + c_2x^2 + ... + c_mx^m) \simeq (1 + b_1y + b_2y^2 + ... + b_nx^n)$$
$$c_0f + c_1xf + c_2x^2f + ... + c_mx^mf-b_1y - b_2y^2 - ... - b_ny^n\simeq 1$$
This is a linear equation wrt the coefficients. Thus $c_0,c_1,...,c_m,b_1,b_2,b_n$ can be approximately computed thanks to linear regression.

Now knowing the approximate values of $b_1,b_2,...,b_n$ we can compute

$$H_k=b_1y_k + b_2y_k^2 + ... + b_ny_k^n \quad \text{from }k=1 \text{ to } N=\text{number of points.}$$

The original equation becomes :
$$f_k\simeq a_0H_k + a_1H_kx_k + a_2H_kx_k^2 + ... + a_nH_kx_k^n$$
This is a linear equation wrt the coefficients. Thus $a_0,a_1,...,a_n$ can be approximately computed thanks to linear regression.

So, approximate values of $a_0,a_1,...,a_n,b_1,...,b_n$ are obtained.

Numerical example with your data :

The calculus was made with $m=4.$

$$c_0f + c_1xf + c_2x^2f + c_3x^3f + c_4x^4f-b_1y - b_2y^2 - b_3y^3 - b_4y^4\simeq 1$$
The linear regression gives :

$c_0=1.027432 \: ; \: c_1=0.004248 \:;\: c_2=-1.322825 \:10^{-4} \:;\: c_3=9.926459 \:10^{-6} \:;\: c_4=4.165036 \:10^{-6}$

$b_1=-0.005547\:;\:b_2=-0.010356\:;\:b_3=0.002587\:;\:b_4=0.002165$

Then the $H_k=b_1y_k + b_2y_k^2 + ... + b_ny_k^n$ were computed.

The linear regression from $\quad f_k\simeq a_0H_k + a_1H_kx_k + a_2H_kx_k^2 + a_3H_kx_k^3 + a_4H_kx_k^4\quad$ leads to :
$$a_0=0.996376 \:;\:
a_1=-0.006457\:;\:
a_2=-0.001046\:;\:
a_3=1.129896\:10^{-5}\:;\:
a_4=2.196121\:10^{-6}$$

To check the result, one compute
$$fcal_k=(a_0 + a_1x_k + a_2x_k^2 + a_3x_k^3 + a_4x_k^4)(1 + b_1y_k + b_2y_k^2 + b_3y_k^3 + b_4y_k^4)$$
$$RMSE=\sqrt{\frac{1}{N}\sum_{k=1}^N(f_k-fcal_k)^2}=0.0409$$
And visually :

Probably the fitting can be slightly improved with higher $m$, for example 5 or 6 , ... , but I didn't try it.

Of course non-linear regression gives a better accuracy of fitting. But this is not easy with such high number of parameters. The softwares for non-linear fitting require to give guessed initial values or guessed ranges for the parameters. In case of 9 parameters this is a big difficulty in practice. Often the calculus goes out of range or takes too long time.

This difficulty can be overcome thanks to the above method. The above method gives appoximate values of the parameters which can be used as good initial values for starting the iterative calculus of non-linear regression.


答:
Can you prove that $\lVert A - B\rVert \leq \lVert A\rVert \lVert B\rVert\lVert A^{-1} -  B^{-1}\rVert$? Then the first inequality is easy. Do the same with $\lVert A^{-1} -  B^{-1}\rVert \leq \lVert A^{-1}\rVert\lVert B^{-1}\rVert\lVert A - B\rVert.$ for the other inequality.

Here we use the submultiplicativity property of matrix norms, i.e.
$$
\lVert AB\rVert \leq \lVert A\rVert \lVert B\rVert,
$$
which is not a property possessed by all norms (only induced and the Frobenius norm as far as I am aware).

Title: Prove that $\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} \le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}.$

Body:
Let $A$ and $B$ be invertible matrices and $K(C)$ the condition number of an invertible matrix $C$. Prove that $$\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} \le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}.$$

I tried
\begin{align}
\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} &\le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{1}{K(B)}\frac{\|A\|-\|B\|}{\|A\|} &\le \frac{\|A^{-1}\|-\|B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B\|-\|A\|}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|A\|\|B\|\|B^{-1}\|} &\le \frac{\|A^{-1}\|}{\|B^{-1}\|}\le \frac{\|A\|\|A^{-1}\|(\|B\|-\|A\|)}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|A\|\|B\|} &\le \|A^{-1}\|\le \frac{\|A^{-1}\|(\|B\|-\|A\|)}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|B\|} &\le K(A)\le \|A^{-1}\|(\|B\|-\|A\|) &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|B\|} &\le K(A)\le \|A^{-1}\|\|B\|-K(A)  &&\Leftrightarrow \\[5pt] \frac{\|A-B\|}{\|B\|} &\le 0 \le \|A^{-1}\|\|B\|-2K(A) &&\Leftrightarrow \\[5pt] \|A-B\| &\le 0 \le \|A^{-1}\|\|B\|-2K(A) &&\Leftrightarrow \\[5pt]
\|A-B\| &\le 0 \le \|B- 2A\|
\end{align}

I don't know how to continue. It doesn't look correct.


答:
Let $g\in\Phi(G)$ and suppose that $g$ is not a  non-generator for the group. Then there exists a subset $X$ of $G$ such that $G=\langle X,g\rangle$ and $G\neq \langle X\rangle$. Then $g\notin \langle X\rangle$, otherwise $\langle X,g\rangle=\langle X\rangle=G$, contradicting our assumptions. Let $\mathscr{S}$ be the set of all subgroups of $G$ containing $\langle X\rangle$ and not containing $g$. First of all, $\mathscr{S}$ is not the empty set, because $\langle X\rangle$ belongs to it. Moreover $(\mathscr{S},\subseteq)$ is an ordered set. If we take a chain of elements of $\mathscr{S}$, and do their insiemistic union, this union also belongs to  $\mathscr{S}$ and evidently is an upper bound for the elements in the chain. By Zorn's Lemma we get that $\mathscr{S}$ has a maximal element, call it $M$. Suppose $M<H\leq G$. Then $g\in H$ and $H=G$ (since $G=\langle X,g\rangle\leq \langle H,g\rangle=\langle H\rangle =H$). It follows that $M$ is a maximal subgroup of $G$, not containing  $g$ , and this is absurd since we've taken $g$ in $\Phi(G)$. So $g\in \Phi(G)\leq M$, thus $G=\langle g,X\rangle=M$, absurd. We conclude that $g$ must be a non-generator of the group.

Conversely, suppose that $g$ is a  non-generator and that $g$ is not an element of Frattini subgroup of $G$. Then there exists a maximal subgroup $M$ of $G$ that does not contain the element $g$. It follows that $M\neq \langle g,M\rangle$ and so, by maximality of $M$ that $G=\langle g,M\rangle$. But $g$ is a non-generator, so that $G=M$, contradicting hypothesis saying that $M$ is a maximal (hence proper) subgroup of $G$

Title: Frattini subgroup is set of nongenerators

Body:
The Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$. (If there are none, then $\Phi(G)=G$.) We say that an element $g\in G$ is a nongenerator if whenever $\langle X\cup \{g\}\rangle = G$, we have $\langle X\rangle = G$ for subsets $X\subseteq G$.

If $G$ is finite, show that $\Phi(G)$ is the set of nongenerators of $G$.

So I want to start by assuming that $g\in G$ is in all maximal subgroups of $G$, and $\langle X\cup\{g\}\rangle = G$, and to prove that $\langle X\rangle = G$. At first I thought if $\langle X\cup\{g\}\rangle = G$, then $\langle X\rangle$ might be a maximal subgroup itself, but it seems that this is not necessarily the case. (For example $G$ is a cyclic group of order $4$ generated by $g$, and $X$ is the trivial subgroup.)


问:
Homotopy push-out squares and exact triangles are colimits
I read somewhere that homotopy push-out squares and exact triangles in a triangulated category can both be interpreted as special cases of higher categorical colimits. Why is this true?

Please note that I am indeed a beginner in category theory, so a rough intuition of the analogy would be more than enough for me. I am fully aware this would sacrifice precise formalism, but I am looking for the general idea here.

答:
Oh, boy, this is a very long story to tell!

I will indeed sacrifice formalism to convey intuition: roughly speaking, a triangulated category arises as a certain kind of homotopy category, i.e. a category $C$ where you have some notion of weak equivalence, i.e. a class of maps you want to treat as if they were invertible.

Now, you can think of homotopy categories as arising as "1-dimensional shadows" of higher categorical concepts: intuitively, a 1-category is objects+morphisms; a higher category has higher-dimensional "cells" "filling" "commutative" "diagrams" (each of these words is better left undefined).

$$
\begin{array}{c}
\text{higher categories} \ni \mathcal K\\
\downarrow \\
\text{1-categories} \ni K
\end{array}
$$ where the correspondence "flattens" all the higher-dimensional information, forgetting all the stuff that lies beyond dimension 1.

If you're familiar with simplicial complexes, the intuition behind truncating a simplicial complex carries over here, again with some magic spell muttered here and there.

Now, a fundamental tenet of Platonism, as well as of higher category theory is that very little can be said about an object, judging the shadow it casts: so, it turns out that construction in triangulated categories, like distinguished triangles and homotopy colimits, lack the universal properties they had "upstairs" when regarded as shadows.

In short: the object $\mathcal K$ has a true and very clear universal property, but it is extremely difficult to define and handle the place where it belongs to; otoh, its shadow $K$ is extremely unnatural to define (it's not a real colimit, just a shadow thereof), but the place where it lives is way more ergonomic.


问:
Simplify this expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?
Is it possible to simplify this constant expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Here $\psi(x)$ is digamma function.

Particularly, the constant is real, so it would be good to get rid of the imaginary numbers here.

A similar question about the following:
$$e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$$

答:
Maybe one of these is "simpler"
$$
e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}\\
=e^{2\operatorname{Re}\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\\
=\left(e^{\operatorname{Re}\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\right)^2\\
=\left|e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\right|^2\\
$$


答:
The answer is no, there's no way to avoid using digamma functions here, however it's easy enough to show that this expression is real.

$$\psi(1+z)=-\gamma+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+z} \right)$$

$$\psi(1/2+z)=-\gamma+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n-1/2+z} \right)$$

$$\psi(1/2+z)+\psi(1/2-z)=-2\gamma+\sum_{n=1}^\infty \left(\frac{2}{n}-\frac{1}{n-1/2+z}-\frac{1}{n-1/2-z} \right)=$$

$$=-2\gamma+2\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{n-1/2}{(n-1/2)^2-z^2} \right)$$

So we have:

$$\color{blue}{\psi\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac{1}{2}-\frac{i}{2\sqrt{3}}\right)= \\ =-2\gamma+2\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{n-1/2}{(n-1/2)^2+1/12} \right)}$$

I believe Euler–Maclaurin formula could be useful in calculating this series to high precision, as the integral of the general term w.r.t. $n$ has an elementary form.

$\gamma$ is the Euler-Mascheroni constant

Title: Simplify this expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Body:
Is it possible to simplify this constant expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Here $\psi(x)$ is digamma function.

Particularly, the constant is real, so it would be good to get rid of the imaginary numbers here.

A similar question about the following:
$$e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$$


问:
How do I prove that it takes at most $3$ cuts to divide a $n$ by $m$ chocolate bar up into any ratio? (cocoacoalition)
I'm currently doing a competitive programming problem (cocoacoalition) from the NCPC 2019 contest.

In it, we're asked to find the minimum number of splits to divide an $n$ by $m$ chocolate bar into $a$ and $nm - a$ pieces.

Alice and Bob decide to share a chocolate bar, which is an n by m rectangular grid of chocolate cells. They decide that Alice should get a<n⋅m pieces and that Bob should get b=n⋅m−a pieces. To split the chocolate bar, they repeatedly take a single piece of chocolate and break it either horizontally or vertically, creating two smaller pieces of chocolate. See Figure 1 for an example.

What is the minimum number of splits that Alice and Bob need to perform in order to split the n-by-m chocolate bar into two piles consisting of a and b chocolate cells?

Figure 1: Illustration of a solution to Sample Input 2, showing the original 10-by-10 chocolate bar split three times into pieces of size 10-by-2, 10-by-5, 3-by-3 and 7-by-3. Giving Alice the 10-by-5 and 7-by-3 pieces, she gets a total of 50+21=71 chocolate cells.

The solution to the problem (slide 34) claims you need a maximum of three splits to accomplish this. In the case where $n|a$ or $m|a$, you only need one split. If not, if $a$ can be factored into $xy$ for some $x,y \leq min(n,m)$, you only need two splits.

However, it goes on to claim that $3$ splits are enough for any value of $a$.

How would I go about proving this statement? Trying to phrase the idea of the third split mathematically, I want to prove that $a$ can always be written as any of $xy + kn$, $xy + km$, $xy + k(n-x)$, or $xy + k(m-y)$, but this feels clunky somehow?

Is there a better way of formulating this so that I can prove it better?

答:
Transcription of the slide:

Solution

One split suffices if $a$ is divisible by $n$ or $m$.

Two splits suffice if $a$ can be factored into $a=x \cdot y$ where $x \leq n$ and $y \leq m,$ or if $n \cdot m-a$ can. (Check by trying all $O(n)$ possible values of $x .)$

Three splits are always enough.

I suppose the first point is sufficiently clear. For the second point, you're cutting off a block of size $x\times y$ from one of the corners of the bar.

For the third point, we can now assume that $a$ is not divisible by $n$. Thus
$$ a = kn + r$$ where $r<n$ and $k< m$ (If $k\ge m$, then $a=kn+r\ge nm+r>nm$.)
You can therefore use one cut to make a $k\times n$ block and a $(m-k)\times n$ block. From this second block, you can use two cuts (by step 2) to make a $1\times r$ block.
Now give Alice the $k\times n$ and the $1\times r$ blocks.


问:
Convert Hexadecimal $8D(16)$ to binary in signed magnitude
I'm supposed to covert hexadecimal value, $8D(16)$ into $8$-bit binary if signed magnitude representation is used.

$8D(16)$ $\to$ $1000$ $1101(2)$

For signed magnitude, the left most bit is used to signify whether the value is a $+$ or $-$ value. In this case, how do I represent the value 8 into signed magnitude binary? Isn't $1000$ equals to $-0$. I don't see any possible way to write $8D$ in binary

答: An 8-bit signed integer has a range of -128 to 127.  But hexadecimal 8D is 141.   So what you're trying to accomplish is impossible.</p>


问:
How can this property of definite integrals be true?
In this question, people are saying that the definite integral of $f(x)$ from $0$ to $a$ is equal to the integral of $f(a-x)$ from $0$ to $a$. How can that be true? Simple examples don't work. </p>

答: Put $z=a-x$ then $dz=-dx$. When $x=0, z=a$ and when $x=a,z=0$. Now $$\int_0^af(a-x)dx=\int_a^0f(z)(-dz)=\int_0^af(z)dz$$</p>


问:
Confusion about linear convergence
In my textbook, it says that a sequence $\{P_n\}$ converges to $p$ , if posisitve Constants $\lambda$ and $\alpha$ exists with $\lim _{n \rightarrow \infty} \frac{\left|p_{n+1}-p\right|}{\left|p_{n}-p\right|^{\alpha}}=\lambda$, then if $\alpha=1$ and $\lambda<1$, the sequence is linearly convergent.

However, when I am practicing a problem(linear convergence) as follows:

Show that for any positive integer $k,$ the sequence defined by $p_{n}=1 / n^{k}$ converges linearly to $p=0$.
The solution is :
$$
\frac{\left|p_{n+1}-0\right|}{\left|p_{n}-0\right|}=\frac{1 /(n+1)^{k}}{1 / n^{k}}=\frac{n^{k}}{(n+1)^{k}} \rightarrow 1
$$
as $n \rightarrow \infty$, we have that $p_{n}$ converges linearly to 0 for any integer $k>0$.

My question is, in the above solution, $\lambda=1$, how is this a linear convergence when referring to the definition?

答:
Your textbook's definition is correct, your calculation is correct, but your test problem is poorly stated.

The case of $\alpha = 1$ and $\lambda=1$ is known as sublinear convergence. Linear convergence is slow, but sublinear convergence is fundamentally different and so slow that it is essentially useless in many practical application.

I will now explain the fundamental difference between linear convergence and the sequence given by $p_n = n^{-k}$.

In the case of linear convergence, we have
$$ |p_{n+1}| \approx \lambda |p_n| $$
for $n$ sufficiently large. There is no harm in assuming that
$$ |p_{n+1}| \approx \lambda |p_n| $$
for all $n$. If necessary, we simply discard the initial elements and renumber the rest. In short, linear convergence is characterized by
$$ |p_n| \approx \lambda^n |p_0|.$$
Let us consider the case where $p_n$ measures the error of some relevant computation. Then a relevant goal is to ensure that $$|p_n| \approx 10^{-m}$$ for some suitable value of $m$. Equivalently, we need
$$ n \log \lambda + \log|p_0| \approx -m \log(10) $$
or
$$ n(m) \approx \frac{-m \log(10) - \log|p_0|}{\log \lambda} = m \frac{\log 10}{-\log \lambda} + \frac{\log|p_0|}{- \log \lambda}$$
In particular, the number of iterations needed to decrease the error by a factor of $10$, is independent of $m$, because
$$ n(m+1) - n(m) \approx \frac{\log 10}{-\log \lambda}.$$

In the case of $p_n = \frac{1}{n^k}$ the situation is quite different. We have $$ |p_n| \approx 10^{-m}$$ if and only if $$ - k \log(n) \approx - m \log(10).$$
Equivalently
$$ n(m) \approx \exp \left(\frac{m \log(10)}{k}\right)$$
The number of extra iterations needed to decrease the error by a factor if $10$ is no longer independent of $m$, we have
$$ n(m+1) \approx n(m) \exp\left(\frac{\log(10)}{k}\right) = n(m) 10^{\frac{1}{k}}.$$
Typically, the computational cost (wall time, CPU cycles, energy consumption) is directly proportional to the number of iterations, so in the case of sublinear convergence the situation is dire. The cost of getting a single extra decimal is proportional to all the work we have already done. This explodes rapidly and explains why we really need a different label for the case $\lambda=1$.

Title: Confusion about linear convergence


问:
Prove that $\Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} = 1 + O \Bigl( \frac{1}{\ln x} \Bigr)$
It was my first attempt to use
$\lim\limits_{x \rightarrow \infty} \Bigl( 1 - \frac{1}{x} \Bigr)^{x} = e^{-x}$.
But then I didn't know what to do with $\frac{1}{\ln x}$.
On my second try, I wanted to use
\begin{align*}
\Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} - 1 = e^{-x} \Bigl( e^{x \frac{\ln(x-1)}{\ln(x)}} - e^x \Biggr).
\end{align*}
But the attempt was unsuccessful, because I didn't know how to estimate the term to prove the statement.

答:
Hint

Try to use

$$y=\Big[ 1 - \frac{1}{x} \Big]^{\frac{x}{\log( x)}}\implies \log(y)= {\frac{x}{\log(x)}}\log\Big( 1 - \frac{1}{x} \Big)\sim -\frac{1}{\log(x)}$$
$$y=e^{\log(y)}\implies y\sim\cdots ???$$


问:
find general solution to $\dot{x}=Ax$

Given $A_{n\times n}$ real matrix and $0\neq v,w\in\mathbb{R}^n$ such that for every $t\in\mathbb{R}$ $x(t)=ve^{-t}+wte^{-t}$ is a solution to $\dot{x}=Ax$

For $n=2$ find the general solution.

find a Non-trivial solution to $\dot{x}=A^2x$

I'm not sure what am I suppose to find,because the given $x(t)$ is a general solution or do I need to find general form of $A$?

Edit:
I've tried taking the given equation $x(t)=ve^{-t}+wte^{-t}$ and multiply from the left by $A$ and then $\dot{x}=Ax(t)=Ave^{-t}-wte^{-t}$ and maybe from this equation I can get another solution ?

答: From the form of $x(t)$, it is implied that $A$ has a repeated eigenvalue of $-1$. (Can you see why?) Can you proceed now?

Title: find general solution to $\dot{x}=Ax$


问:
Totally ordered abelian group
Let $\Gamma$ be a totally ordered abelian group (written additively), and let $K$ be a field. A valuation of $K$ with values in $\Gamma$ is a mapping $v:K^* \to \Gamma$ such that

$1)$ $v(xy)=v(x)+v(y)$

$2)$ $v(x+y) \ge \min (v(x),v(y))$,

for all $x,y \in K^*$.

Show that the set $A=\{x\in K^*:v(x) \ge 0\} \cup \{0\}$    is a valuation ring of $K$.

I want to show that $-1 \in A$. From $v(1)=v(1)+v(1)$, $v(1)=0$ and $v(1)=v(-1)+v(-1)$ so that $2v(-1)=0$. Is it possible to conclude that $v(-1)=0$?

In general for any totally ordered abelian group, does $2x=0$ imply $x=0$?

答:
Yes.  If $x$ is positive then so is $nx$ for any $n \in \mathbb{Z}^+$ and if $x$ is negative then $-x$ is positive and $nx = 0$ iff $n(-x) = 0$.

In other words, a totally ordered abelian group is necessarily torsionfree.  More interestingly, the converse also holds: any torsionfree abelian group can be totally ordered (in at least one way). See Section 17.2 of these notes for the proof.

Title: Totally ordered abelian group


问:
How do you arrive at these two probability formulas?
In probability, if you are counting an ordered set with repetitions not allowed, how and why does the formula $N(N - 1)(N - 2)...(N - (n - 1))$ become $\frac{N!}{(N - n)!}$ in its concise form?

I was told that multiplying the formula by $\frac{(N - n)!}{(N - n)!}$ gives the concise form, how do you multiply the first formula by that number?

Second question:
If you are counting an unordered set with repetitions not allowed this time, the formula becomes
$\frac{N}{n} = \frac{N!}{n!(N-n)!}$, where a $n!$ was added in the denominator to remove the overcounting, how and why was the $n!$ added in that specific place?

答:
First question:

Look at $N=5$ and $n=3$.

Then you have

$$5\cdot4\cdot3=5\cdot4\cdot3\cdot\frac{2\cdot1}{2\cdot1}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{5!}{(5-3)!}$$

Second question:

You take $n$ elements out of the set. If you care about the ordering, the first formula holds. But if you dont, how many orderings are there for the $n$ elements? Exactly $n!$.

Thus, if you want to compute the number of different sets, but without ordering you just compute the number of different ordered sets, but divide by $n!$.


答:
For the first part, you have $N$ choices for the first selection, $N-1$ choices for the second selection, and so on up to $N-(n-1)$ choices for the $n$th selection.  So

$$N (N - 1) (N - 2) \cdots (N - (n - 1))
\\=N (N - 1) (N - 2) \cdots (N - (n - 1)) \dfrac{(N - n)(N - n - 1)\cdots 1}{(N - n)(N - n - 1)\cdots 1}
\\= \dfrac{N (N - 1) (N - 2) \cdots (N - (n - 1))(N - n)(N - n - 1)\cdots 1}{(N - n)(N - n - 1)\cdots 1}
\\= \dfrac{N!}{(N-n)!}$$

ways of making the ordered selection of $n$ from $N$.

For the second part, you have $${N \choose n} = \dfrac{N!}{n!(N-n)!}$$ because the $n$ selections can come in any $n!$ different ways which you want to treat as the same selection.

Title: How do you arrive at these two probability formulas?

Body:
In probability, if you are counting an ordered set with repetitions not allowed, how and why does the formula $N(N - 1)(N - 2)...(N - (n - 1))$ become $\frac{N!}{(N - n)!}$ in its concise form?

I was told that multiplying the formula by $\frac{(N - n)!}{(N - n)!}$ gives the concise form, how do you multiply the first formula by that number?

Second question:
If you are counting an unordered set with repetitions not allowed this time, the formula becomes
$\frac{N}{n} = \frac{N!}{n!(N-n)!}$, where a $n!$ was added in the denominator to remove the overcounting, how and why was the $n!$ added in that specific place?


问:
If $-1<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<1$ then there exists the limit $\lim _{n\to \infty }\left(\sum _{k=0}^n\:\:x_k\right)$
Let be $x_n$ a sequence.
Prove that if $$-1<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<1$$ then there exists the limit of the sequence
$$y_n=\sum _{k=1}^nx_k\:$$ and it is finite.

I am not one hundred percent that this is completely true. In any event, I managed to show that if
$$0<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<1$$
then the sequence
$$y_n=\sum _{k=1}^nx_k\:$$
is convergent owing to the fact that it is not only bounded but it is also increasingly from a certain rank $n_0.$

Moreover, I proved that if
$$-1<\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)<0$$
then the sequence
$$y_n=\sum _{k=1}^nx_k\:$$
is convergent as well since it is bounded and decreasingly from a certain rank $n_1$.
However I still have to prove this for the case
$$\lim _{n\to \infty }\left(\frac{x_{n+1}}{x_n}\right)=0$$
Can somebody help me out?

答:
As Kavi Rama Murthy said in the comments, if the limit of $\frac {x_{n+1}}{x_n}$ exists then its absolute value also exists.
You can then apply the ratio test (Dalambert criterion) which requires all elements $a_n \gt 0$ and that $\lim_{n \to \infty} \frac {a_{n+1}}{a_n} = r$.
When $r \lt 1$ then $\sum_{n=1}^{\infty} a_n$ converges.


问:
$A_n$ normal subgroup of $S_n$
If I want to prove that $A_n$ is a normal subgroup of $\:S_n$

I know how to prove that $A_n$ is a subgroup, but if I want to prove that it is a normal subgroup.

Can I prove like that:

Let  $f_1\in S_n$ then we need to provef that: $f_1A_n=A_nf_1$.
$\DeclareMathOperator{\sign}{sign}$
So we know that: $\forall h\in A_n:\:\sign\left(f_1\:\circ \:h\right)=\sign\left(f_1\right)\cdot \sign\left(h\right)=\sign\left(f_1\right)$

On another hand: $\forall h\in A_n:\:\sign\left(h\:\circ f_1\right)=\sign\left(h\right)\cdot \sign\left(f_1\right)=\sign\left(f_1\right)$

Because we know that $\sign(h)=1$

So we can say that: $f_1A_n=A_nf1$

Is this correct?

答: This is correct. However, you could use a more general result. $\mathfrak A_n$ is the kernel of the signature and the kernel of a group homomorphism is always a normal subgroup.

Title: $A_n$ normal subgroup of $S_n$


问:
Why don't I understand "What is mathematics?" by Richard Courant
I am 14 years old and can do A level maths happily. However, I want to further my knowledge of mathematics to pave my way for more discrete maths and harder analysis etc. So I decided to pickup "What is mathematics?" by Richard Courant because I heard it was very useful and it would be good for someone of my knowledge.

I am having a lot of trouble understanding it though, and I'm not sure what it is that is stopping me. All I know is that it takes me a long time and some SE questions to understand even a page. It could be because of the way it is written which I am not used to?

If anyone has ever encountered this before then could you please offer me some advice and maybe a new way to approach it.

答:
Questions like this are very subtle, but I think I can probably highlight a few points.

Firstly, this book is rather elderly. It's been around for a while, and I've seen it described as old-fashioned and outdated. The outdated bit isn't such an issue, but the language used is perhaps a little stiff for an introduction to mathematics (though fairly true to what textbooks often look like). Looking quickly through a preview,

Not everybody likes the same style of mathematics. There are several aspects to this. Firstly, you might not like the writing style/level of detail the author provides. Some people prefer terse books with the core concepts, so that you have to fill in the details; some people prefer very explanatory books which walk you through at least the first few calculations. Secondly, some people lean more towards applied mathematics, and away from pure mathematics. You might find yourself much more interested in how mathematical objects relate to concrete problems than in how the mathematical objects relate to each other. In this case, I still recommend learning more 'pure' mathematics as far as you can, but be aware that this is a very different ball game to A-level mathematics (which is more like the simple bits at the start of an engineering degree than a mathematics degree). You should try looking through books before buying them as much as you can in order to see whether or not they're the right kind of book for you.

This is the first time you've tried reading something with any actual rigorous mathematics in it. This is always hard at first. It's a very different way of thinking to normal thought for most people, and that's fine. Work at it, and if you can, get people to critique your arguments. (One thing you might like to have a go at to supplement your learning is programming. Learning a little bit of how to solve simple problems on computers might help you with the mix of creative and practical logic you need for maths. Even just staring at some problems on Project Euler might help you think more clearly about what you're doing. That's just a random thought.)

Don't be afraid to ask questions. It's perfectly normal to have lots of questions when you're first getting into gear with a new set of ideas. Particularly so when they're presented without clear motivation or a consistent theme - and "What Is Mathematics?" looks to me like it jumps around a lot! Talking new ideas through with fellow students is one of the best things about universities, in fact. You shouldn't always expect a pure mathematics definition, for instance, to be so obvious that you don't read it twice. Most of the time, you will find yourself reading a definition 3 or 4 times before you have any real sense of what's going on behind the scenes. It might take the best part of a year to get to grips with more abstract definitions, depending on how you think, and how much help you get.

In closing, you might want to try a book I've never read but which many people I know have: How to Prove It. This is an introduction to problem solving which you might find a lot more useful in figuring out how mathematicians think. As I said above, though, have a look through it before thinking about buying it.

(Have you ever read any Murderous Maths books, by the way? When I was younger I thought they were great fun, and probably helped to clear my thinking up a lot. The material is much more basic than A-level but I once loved them!)


问:
Prove that the sum of derivatives of a non-negative function is non-negative
Let $f(x)$ be a polynomial of degree $n$ with real coefficients such that $f(x)$ is non-negative for all real $x$. Let $g(x)=f(x)+f'(x)+f''(x)+\dots$ be the sum of $f(x)$ and the first $n$ derivatives of $f(x)$. Show that $g(x)$ is non-negative for all real $x$.

Thanks for any comments/answers in advance!

答: Note that $f(x) $ must have even degree and a positive leading coefficient (unless it is zero, in which case it's trivial) and thus so must $g(x)$. Thus, $g$ has a global minimum, say at $y$. From the definition of $g$, $g'(y)=0$ then implies that $g(y) = f(y) \ge 0$.


问:
What two functions can i graph to produce a circle?
So I'm of the understanding that a function is defined as only ploting one point per input. I'm being taught implicit differentiation so am now thinking about circles, ovals and rotated quadratics. I feel the route to my answer is probably through the rotated quadratic seens as a circle is fundementaly a closed ... ("system"?).

What two functions can i graph to produce an enclosed circular shape?
Edit
I'm looking for two equations in terms of y

答:
We could use
$$ y = \sqrt{r^2-x^2}$$
$$ y = -\sqrt{r^2-x^2}$$

For a circle of radius $r$ centered at $O$, with the equations representing the top and bottom semi-circles respectively.


问:
A property of uniform spaces
Is it true that $E\circ E\subseteq E$ for every entourage $E$ of every uniform space?</p>

答:
As noted in the comments, it clearly is not true in general.

If an entourage $E$ has the property that $E\circ E\subseteq E$, then $E$ is a transitive, reflexive relation on $X$, and the entourage $E\cap E^{-1}$ is an equivalence relation on $X$. A uniform space whose uniformity has a base of equivalence relations is non-Archimedean, the uniform analogue of a non-Archimedean metric space; such a space is zero-dimensional, since it clearly has a base of clopen sets. Conversely, it’s known that every zero-dimensional topology is induced by a non-Archimedean uniformity. (I’ve seen this last result cited to B. Banaschewski, Über nulldimensionale Räume, Math. Nachr. $\bf13$ ($1955$), $129$-$140$, and to A.F. Monna, Remarques sur les métriques non-archimédiennes, Indag. Math. $\bf12$ ($1950$), $122$-$133$; ibid. $\bf12$ ($1950$), $179$-$191$, but I’ve not seen these papers.)

Title: A property of uniform spaces


问:
How fast does the sequence converge?
We have the sequence $$a_1=\frac{9}{4} \quad \qquad a_{n+1}=2\frac{a_n^3-a_n^2+1}{3a_n^2-4a_n+1}$$ that converges to $\frac{9}{4}$.

I want to check how fast the convergence is.

Do we have to calculate $\lim_{n\rightarrow \infty}\frac{|a_{n+1}-9/4|}{|a_n-9/4|}$ ?

答: You want to find how many terms it takes in order for $|a_n-\frac{9}{4}|<\epsilon$, where $\epsilon$ is some chosen positive constant. For example, how many terms would it take for $a_n-\frac{9}{4}<10^{-3}$?

Title: How fast does the sequence converge?


问:
Aces distribution between 4 card players
I recently was playing a card game with a few people and we got to talking about probabilities. I intuitively claimed that it's much more likely that (a) 1 of 4 players gets all 4 aces than (b) all players get 1 ace each. But after some discussion it became clear that for most people (b) seems more likely than (a).

We assume that a full card deck is distributed between the players.

I'm not very good with probability, but from elementary considerations we have the following:

$$p_a=\frac{1}{4^4} \cdot 4=\frac{1}{4^3}$$

This is because each player drawing each ace is an independent event, while the players themselves are indistinguishable, so there's $4$ possible cases satisfying the condition.

$$p_b=\frac{1}{4^4}$$

And here there's only one possible case, so the probability is much smaller.

Is my reasoning correct?

And, going beyond mathematics, do most non-mathematicians expect a "fair" distribution to be more probable than an "unfair" one? Is it a common enough mistake?

答:
You calculated $p_a$ correctly, but your result of $p_b$ is much smaller than its actual value.

Let the four players be $A, B, C, D$.  Distributing the four aces independently to one of the four players, we obtain the equiprobable sample space
$$
\Omega = \{A, B, C, D\}^4
$$
where, say, $(A, B, C, C)$ means that the four aces went to $A, B, C,$ and $C$, respectively.

Therefore,
\begin{align*}
p_b &= \frac{\text{number of permutations of $(A, B, C, D)$}}{\text{total number of possibilities}} \\
&= \frac{4!}{4^4} = \frac{3}{32} = 9.375 \%,
\end{align*}
whereas $p_a = 1/64 = 1.5625 \%$.


答:
For the “each player gets one Ace each” case, you undercounted by a factor of $4!$

This is because player 1 can get any of the 4 aces; player 2 can get any of the 3 remaining aces, and so on.

Title: Aces distribution between 4 card players

Body:
I recently was playing a card game with a few people and we got to talking about probabilities. I intuitively claimed that it's much more likely that (a) 1 of 4 players gets all 4 aces than (b) all players get 1 ace each. But after some discussion it became clear that for most people (b) seems more likely than (a).

We assume that a full card deck is distributed between the players.

I'm not very good with probability, but from elementary considerations we have the following:

$$p_a=\frac{1}{4^4} \cdot 4=\frac{1}{4^3}$$

This is because each player drawing each ace is an independent event, while the players themselves are indistinguishable, so there's $4$ possible cases satisfying the condition.

$$p_b=\frac{1}{4^4}$$

And here there's only one possible case, so the probability is much smaller.

Is my reasoning correct?

And, going beyond mathematics, do most non-mathematicians expect a "fair" distribution to be more probable than an "unfair" one? Is it a common enough mistake?


问:
If $y=\frac{1}{3u}\frac{du}{dx}$, find $\frac{dy}{dx}.$
If $y=\frac{1}{3u}\frac{du}{dx}$, find $\frac{dy}{dx}.$

My thoughts:

Differentiating with respect to x gives:

$\frac{dy}{dx}=\frac{1}{3u}\frac{d^2u}{dx^2}-\frac{1}{3u^2}\frac{du}{dx}$

However, it should be:

$\frac{dy}{dx}=\frac{1}{3u}\frac{d^2u}{dx^2}-\frac{1}{3u^2}(\frac{du}{dx})^2$

Could someone explain this?

答:
$$y=\frac{1}{3u}\frac{du}{dx}$$
D.w.t. $x$
$$\frac{dy}{dx}=\frac{-1}{3u^2}\left(\frac{du}{dx}\right)^2+\frac{1}{3u}\frac{d^2u}{dx^2}$$
When you differentiate $u$, $du/dx$ will multiply.


问:
What does Skolem's paradox have to do with "unrelativized quantification"?
Timothy Bays' excellent essay on Skolem's paradox includes the following claim:

... Skolem's Paradox doesn't introduce contradictions into various forms of axiomatized set theory, even when these axiomatizations are themselves understood formalistically or model-theoretically. From a proof-theoretic standpoint, for example, there is a difference between unrelativized quantification and quantification which has been explicitly relativized to some formula in our language (where this formula is one that, from an intuitive perspective, serves to “pick out” the domain of countable model of ZFC). So, there's no a priori reason to think that a sentence with unrelativized quantifiers will conflict with that sentence's fully-relativised counterpart.

What does he mean by "relativized" or "unrelativized quantification" from a "proof-theoretic standpoint"? Later, in the same paragraph, he speaks about domain-restricted quantification, so it cannot be that.

答:
Suppose that $V$ is our universe, and $A$ is a subset of that universe. We can relativise the formulas to $A$ by recursion:

If $\varphi(\vec x)$ is an atomic formula, then $\varphi^A(\vec x)$ is simply $\varphi(\vec x)\land\vec x\in A$. (Here $\vec x\in A$ is understood as the conjunction of $x_i\in A$ for the variables appearing in $\vec x$).

If $\varphi=\varphi_0*\varphi_1$ (or $\lnot\varphi_0$), then $\varphi^A$ is $\varphi_0^A*\varphi_1^A$ (or $\lnot\varphi_0^A$), where $*$ is any connective.

If $\varphi=\exists x\varphi_0(x)$, then $\varphi^A=\exists x(x\in A\land\varphi_0(x))$.

Of course, if $A$ is not definable, we need to add it as a predicate. We can deduce from this that $(\forall x\varphi(x))^A=\forall x(x\in A\to\varphi^A(x))$, assuming the logic we're working with is close enough to first-order logic.

The point is that since ZFC proves that $\Bbb R$ is uncountable, it proves "for all $x$, if $x$ is a function from $\Bbb N$ to $\Bbb R$, then $x$ is not surjective". When we take a countable model of ZFC, the quantifiers (and the interpretation of $\Bbb N$ and $\Bbb R$) all relativise to that model. That means that we simply have:

For all $x$ in the model, if $x$ is a function from $\Bbb N$ of the model to $\Bbb R$ of the model, then $x$ is not surjective (in the model).

Note that we can have a reverse Skolem paradox: if ZFC is consistent, then it has a model $M$ such that $\Bbb N^M$ is uncountable (externally!).


问:
Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$
Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc  \ge 3
$$

Equality holds for $a=b=c=\frac13$.  One could try homogenizing:

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4  -  \sum_{cyc} a^3(b+c) - 2  \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$

but I didn't manage to advance further.

答:
From the Schur inequality, we have
$$\sum a^2(a-b)(a-c) \geqslant 0,$$
equivalent to
$$a^4+b^4+c^4+abc(a+b+c) \geqslant \sum a^3(b+c). \qquad (1)$$
After expand, we write your inequality as
$$2\left[\sum a^4 + a b c(a+b+c)\right] \geqslant \sum a^3(b+c) + 2 \sum a^2 b^2.$$
Now, from $(1)$ we need to prove
$$\sum a^3(b+c) \geqslant 2 \sum a^2 b^2.$$
Which is true because by the AM-GM inequality, we have
$$\sum a^3(b+c) = \sum ab(a^2+b^2) \geqslant \sum ab \cdot 2ab = 2 \sum a^2 b^2.$$
Done.


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
This problem can be modelled as a Markov chain and we can use known results about hitting probabilities to calculate the exact answer. I will first explain how to model it as a Markov chain and then give the exact answer as implementation in matlab.

Let $S$ denote the set of states: For each player $P$ and each possible HP value $x$ of the boss (between $0$ and $1000$) we add the state ($P$, $x$) to $S$ (note that $0$ is representative of any HP value less or equal to $0$ here). The random variable $X_i$ is the state of the system after $i$ rounds. The state transition probabilities (i.e. the probability $P[X_{i + 1} = (P, x) | X_i = (P', x')]$) are now given by the problem description.

As an example, assume we know already that $X_i = (A, 1000)$ and we want to compute the (conditional) probability that $X_{i + 1} = (B, x)$. Since $A$ can do damage between $25$ and $50$ we have that $P[X_{i + 1} = (B, x) \, | \, X_i = (A, 1000)] = \frac{1}{26}$ for $x \in \{950, \dots, 975 \}$ and it is $0$ for other values of $x$. Of course, the probability of reaching any state with player $C$ next is $0$.

The hitting probability of a state $(P, x) \in S$ is the probability that we ever reach $(P, x)$ in this random process. Thus the hitting probability of the state $(A, 0)$ is exactly the probability that $C$ delivered the killing blow. Similarly, the hitting probability of $(B, 0)$ and $(C, 0)$ are the probabilities of $A$ and $B$ respectively delivering the killing blow.

These probabilities can be computed as the solution of a linear system of equations. I have coded this up in matlab, see below. States are represented as numbers from $1$ to $3003$. If a state $s$ satisfies $s =_3 0$ it is a state where $A$ attacks next. Similarly, $B$ and $C$ are associated with states in residue classes $1$ and $2$ respectively.

The HP of the boss in state $s$ is encoded as $ \lceil s / 3 \rceil - 1$. For example, state $3003$ is a state where $A$ attacks next and where the boss has $1000$ hitpoints.

Matlab has a built-in function for finding hitting probabilities which I use here. If you run the code below, it will tell you that we get probabilities $0.2835, 0.3494, 0.3671$ for players $A, B$ and $C$ respectively.

Let me know if you have any questions.

```<code>% build transition matrix P for N hitpoints
N = 1000;
P = zeros(3*(N + 1), 3*(N + 1));
% states 1, 2, 3 are endstates (i.e. random process loops there)
for i = 1:3
    P(i, i) = 1;
end
% all other transitions
for i = 4:(3*(N + 1))
    % player A
    if mod(i, 3) == 0
        for j = 25:50
            P(i, max(2, i - 1 - (3 * j))) = P(i, max(2, i - 1 - (3 * j))) + 1/26;
        end
    % player B
    elseif mod(i, 3) == 1
        for j = 30:70
            P(i, max(1, i + 2 - (3 * j))) = P(i, max(1, i + 2 - (3 * j))) + 1/41;
        end
    % player C
    else
        for j = 10:80
            P(i, max(3, i - 1 - (3 * j))) = P(i, max(3, i - 1 - (3 * j))) + 1/71;
        end
    end
end
% build markov chain from transition matrix
mc = dtmc(P);
% find hitting probabilities for player A, B, C
C = hitprob(mc, 1);
A = hitprob(mc, 2);
B = hitprob(mc, 3);
% output
A(3*(N+1))
B(3*(N+1))
C(3*(N+1))
</code>```

Body:
This problem can be modelled as a Markov chain and we can use known results about hitting probabilities to calculate the exact answer. I will first explain how to model it as a Markov chain and then give the exact answer as implementation in matlab.

Let $S$ denote the set of states: For each player $P$ and each possible HP value $x$ of the boss (between $0$ and $1000$) we add the state ($P$, $x$) to $S$ (note that $0$ is representative of any HP value less or equal to $0$ here). The random variable $X_i$ is the state of the system after $i$ rounds. The state transition probabilities (i.e. the probability $P[X_{i + 1} = (P, x) | X_i = (P', x')]$) are now given by the problem description.

As an example, assume we know already that $X_i = (A, 1000)$ and we want to compute the (conditional) probability that $X_{i + 1} = (B, x)$. Since $A$ can do damage between $25$ and $50$ we have that $P[X_{i + 1} = (B, x) \, | \, X_i = (A, 1000)] = \frac{1}{26}$ for $x \in \{950, \dots, 975 \}$ and it is $0$ for other values of $x$. Of course, the probability of reaching any state with player $C$ next is $0$.

The hitting probability of a state $(P, x) \in S$ is the probability that we ever reach $(P, x)$ in this random process. Thus the hitting probability of the state $(A, 0)$ is exactly the probability that $C$ delivered the killing blow. Similarly, the hitting probability of $(B, 0)$ and $(C, 0)$ are the probabilities of $A$ and $B$ respectively delivering the killing blow.

These probabilities can be computed as the solution of a linear system of equations. I have coded this up in matlab, see below. States are represented as numbers from $1$ to $3003$. If a state $s$ satisfies $s =_3 0$ it is a state where $A$ attacks next. Similarly, $B$ and $C$ are associated with states in residue classes $1$ and $2$ respectively.

The HP of the boss in state $s$ is encoded as $ \lceil s / 3 \rceil - 1$. For example, state $3003$ is a state where $A$ attacks next and where the boss has $1000$ hitpoints.

Matlab has a built-in function for finding hitting probabilities which I use here. If you run the code below, it will tell you that we get probabilities $0.2835, 0.3494, 0.3671$ for players $A, B$ and $C$ respectively.

Let me know if you have any questions.

% build transition matrix P for N hitpoints
N = 1000;
P = zeros(3*(N + 1), 3*(N + 1));
% states 1, 2, 3 are endstates (i.e. random process loops there)
for i = 1:3
    P(i, i) = 1;
end
% all other transitions
for i = 4:(3*(N + 1))
    % player A
    if mod(i, 3) == 0
        for j = 25:50
            P(i, max(2, i - 1 - (3 * j))) = P(i, max(2, i - 1 - (3 * j))) + 1/26;
        end
    % player B
    elseif mod(i, 3) == 1
        for j = 30:70
            P(i, max(1, i + 2 - (3 * j))) = P(i, max(1, i + 2 - (3 * j))) + 1/41;
        end
    % player C
    else
        for j = 10:80
            P(i, max(3, i - 1 - (3 * j))) = P(i, max(3, i - 1 - (3 * j))) + 1/71;
        end
    end
end
% build markov chain from transition matrix
mc = dtmc(P);
% find hitting probabilities for player A, B, C
C = hitprob(mc, 1);
A = hitprob(mc, 2);
B = hitprob(mc, 3);
% output
A(3*(N+1))
B(3*(N+1))
C(3*(N+1))


答:
Method 1let $p=a+b+c,q=ab+bc+ca,r=abc$

Now by schur $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)\ge 0$$ $$\iff p^4-5p^2q+4q^2+6pr\ge 0$$ Now we have to prove $$4 p^4 - 18 p^2 q + 8 q^2 + 30 p r\ge 0$$ or $$4(p^4-5p^2q+4q^2+6pr)+2(p^2q+3pr-4q^2)\ge 0$$ whcih is true because  $$p^2q+3pr-4q^2=a^3(b+c)+b^3(a+c)+c^3(a+b)-2a^2b^2-2b^2c^2-2c^2a^2\ge 0$$

Method 2 I see that i have clashed with an existing answer  so here is a different way

let $3u=a+b+c,3v^2=ab+bc+ca,w^3=abc$ then without even exanding we see that its lenear in $w^3$ Hence by uvw method it suffices to check when

$c=0$

The inequality achieves the equality

$a=b$ also let $t=\frac{c}{a}$ The inequality is $$ 2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but$$4{(t-1)}^2(t+1)\ge 0$$

Method 3 This method is ugly  although it can be got by hand ,i used wolfram alpha to speed up my calculation
WLOG $a\ge b\ge c$ let $$f(a,b,c)=
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4
$$ $$f(a,b,c)-f(\frac{a+b}{2},\frac{a+b}{2},c)=1/2 (a - b)^2 (8 a^2 + 12 a b + 8 b^2 - 5 a c - 5 b c - 6 c^2)\ge 0$$

As inequality is homogenous WLOG   $c=1$  and $t=\frac{a+b}{2}$ we then have to prove $$f(t,t,1)\ge 0$$ $$\iff  2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but $$4{(t-1)}^2(t+1)\ge 0$$

Title: Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$

Body:
Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc  \ge 3
$$

Equality holds for $a=b=c=\frac13$.  One could try homogenizing:

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4  -  \sum_{cyc} a^3(b+c) - 2  \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$

but I didn't manage to advance further.